Category: Inter 1st Year

Andhra Pradesh BIEAP AP Inter 1st Year Zoology Study Material Lesson 5 Locomotion and Reproduction in Protozoa Textbook Questions and Answers.

AP Inter 1st Year Zoology Study Material Lesson 5 Locomotion and Reproduction in Protozoa

Very Short Answer Type Questions

Question 1.
Draw a labeled diagram of the T.S of the flagellum.
Answer:

Question 2.
List any two differences between a flagellum and a cilium.
Answer:

Question 3.
What are dynein arms? What is their significance?
Answer:
‘A’ tube of each peripheral doublet bears paired arms along its length called dynein arms made up of protein dynein.
The dynein arms of the ‘A’ tubule face the tubule ‘B’ of the adjacent doublet.

Question 4.
What is a kinety?
Answer:
In the ciliate protozoans, a longitudinal row of kinetosomes together with kinetodesmata constitute a unit called kinety.

Question 5.
Distinguish between synchronous and metachronous movements.
Answer:
Synchronous movement: Cilia in a transverse row beat simultaneously in one direction. It is called synchronous movement.
Metachronous movement: The sequential movement of cilia, in a longitudinal row, one after the other in one direction is called metachronous movement.

Question 6.
Why do we refer to the offspring, formed by the asexual method of reproduction, as a clone?
Answer:
As a result of the asexual method, the offsprings are, not only identical to one another but also exact copies of their parent. The term ‘clone’ is used to describe such morphologically and genetically similar individuals.

Question 7.
Distinguish between proter and opisthe.
Answer:
During transverse binary fission in Paramecium two daughters, individuals are formed. The anterior one is called proter and the posterior is called opisthe.

Question 8.
How is sexual reproduction advantageous in evolution?
Answer:
Sexual reproduction results the advantageous in evolution in genetic recombination occurs in sexual reproduction.

Question 9.
Distinguish between lobopodium and filopodium. Give an example to each of them.
Answer:
Lobopodium: The blunt and finger-like tubular pseudopodia containing both ectoplasm and endoplasm is called lobopodium.
Ex: Amoeba proteus
Filopodium: The slender filamentous pseudopodia with pointed tips, composed of only ectoplasm are called Filopodium.
Ex: Euglypha

Question 10.
Define conjugation with reference to ciliates. Give two examples.
Answer:
Conjugation is a temporary union between two senile ciliates that belong to two different ‘mating types’ for the exchange of nuclear material and its reorganization. – Wichterman.
Ex: Paramecium and Vorticella.

Short Answer Type Questions

Question 1.
Name the system that controls the fastest swimming movement of protozoans and write its components.
Answer:
Ciliary locomotion is the fastest swimming movement of protozoans, Hence, ciliates are the fastest protozoans.
Cilia are small hair-like structures found in ciliate protozoans like Paramecium.

Infraciliary system: It is located just below the pellicle in the ectoplasm of a ciliate. It includes kinetosomes, kinetodesmal fibrils, and kinetodesmata. The kinetosomes are present at the bases of cilia in transverse and longitudinal rows. The kinetodesmal fibrils are connected to the kinetosomes and run along the right side of each row of kinetosomes as a ‘cord of fibres’ called kinetodesmata. A longitudinal row of kinetosomes together with kinetodesmata constitute a unit called ‘kinety’.

All the kineties together form an infraciliary system, which is connected to a ‘motorium’, located near the cytopharynx. The infraciliary system and motorium form the ‘neuromotor system’ that controls and coordinates the movement of cilia.

Question 2.
Write the mechanism of bending the flagellum and explain effectively recovery strokes.
Answer:
The bending movement of a flagellum is brought about by the sliding of microtubules past each other due to the functioning of ‘dynein arms’ utilizing ATP. A flagellum pushes the fluid medium at right angles to the surface of its attachment, by its bending movement.

Bending movement of flagella and cilia: Dynein arms show a complex cycle of movements using energy provided by ATP (dynein arms are the sites of ATPase activity in the cilia and flagella). The dynein arms of each doublet attach to an adjacent doublet and pull the neighbouring doublet. So the doublets slide past each other in opposite directions. The arms release and reattach a little further on the adjacent doublet and again ‘puli’. As the doublets of a flagellum or cilium are physically held in place by the radial spokes, the doublets cannot slide past much. Instead, they curve and cause the bending of flagellum or cilium. Such bending movements of flagella and cilia play an important role in flagellar and ciliary locomotion.

The flagellar movement of many organisms is a sidewise-lash which consists of two strokes namely the effective or propulsive stroke and the recovery stroke.

• Effective stroke: Flagellum becomes rigid and starts bending to one side beating against the water. This beating against water is at right angles to the body axis and the organism moves forwards.
• Recovery stroke: Flagellum becomes comparatively soft so as to offer the least resistance to water and moves back to its original position. It is called ‘recovery stroke’.

Question 3.
What are lateral appendages? Based on their presence and absence, write the various types of flagella giving atleast one example for each type.
Answer:
Lateral appendages: Some flagella bear one or two or many rows of short, lateral hair-like fibrils called lateral appendages. They are of two types namely ‘mastigonemes’ and ‘flimmers’.
Types of Flagella: Based on the presence or absence and/or the number of rows of lateral appendages, five types of flagella are recognized.
(a) Stichonematic: This flagellum bears one row of lateral appendages on the axoneme.
E.g. Euglena and Astasia.
(b) Pantonematic: This flagellum has two or more rows of lateral appendages on the axoneme.
E.g. Peranema and Monas.
(c) Acronematic: This type of flagellum does not bear lateral appendages and the terminal part of the axoneme is naked without the outer sheath at its tip.
E.g. Chlamydomonas and Polytoma

(d) Pantacronematic: This type of flagellum is provided with two or more rows of lateral appendages and the axoneme ends in a terminal naked filament.
E.g. Urceolus.
(e) Anematic or simple: In this type of flagellum, lateral appendages and terminal filament are absent. Hence, it is called anematic (a-no; nematic-threads)
E.g. Chilomonas and Gryptomonas.

Question 4.
Describe the process of transverse binary fission in paramecium.
Answer:
Transverse binary fission is performed by Paramecium. Binary fission is the most common method of sexual reproduction in protozoans. During favourable conditions, Paramecium stops feeding after attaining its maximum growth.

At first, the micronucleus divides by mitosis and the macronucleus divides into two daughter nuclei by amitosis. The oral groove disappears. After karyokinesis, a transverse constriction appears in the middle of the body, which deepens and divides the parent cell into two daughter individuals, the anterior proper and the posterior opisthe. The proper receives the anterior contractile vacuole, cytopharynx, and cytosome from its parent individual. It develops a posterior contractile vacuole and a new oral groove.

The opisthe receives the posterior contractile vacuole of its parent. It develops a new anterior contractile vacuole, cytopharynx, cytostome and a new oral groove. Binary fission is completed in almost two hours, in favourable conditions and Paramecium can produce four generations of daughter individuals by binary fission in a day. The transverse binary fission is also called homothetogenic fission because the plane of fission is at right angles to the longitudinal axis of the body. As it occurs at right angles to the kineties, it is also called perkinetal fission.

Question 5.
Describe the process of longitudinal binary fission in Euglena.
Answer:
Binary fission is the most common method of asexual reproduction in protozoans. Longitudinal binary fission is performed by Euglena. In this type of binary fission, the body divides into two halves longitudinally, hence called longitudinal binary fission.

During the process of binary fission, the nucleus, basal granules, chromatophores, and cytoplasm undergo division. The nucleus divides by mitosis into two daughter nuclei. Then the kinetosomes and the chromatophores also divide. At first, a longitudinal groove develops in the middle of the anterior end. This groove extends gradually towards the posterior end until the two daughter individuals are separated. One daughter Euglena retains the parental flagella. The other daughter individual develops new flagella from the newly formed basal granules. The stigma, paraflagellar body, and contractile vacuole of the parent disappear. They develop afresh in both the daughter Euglenae. The longitudinal binary fission is known as symmetrogenic division because the two daughters Euglenae resemble each other like mirror images.

Question 6.
Write a short note on multiple fission.
Answer:
Multiple fission: It is the division Of the parent body into many smaller individuals (Multi-many; Fission- splitting). Normally multiple fission occurs during unfavourable conditions. During multiple fission, the nucleus first undergoes repeated mitotic divisions without cytokinesis. This causes the formation of many daughter nuclei. Then the cytoplasm also divides into as many bits as there are nuclei. Each cytoplasmic bit encircles one daughter nucleus. This results in the formation of many smaller individuals from a single-parent organism. There are different types of multiple fissions in protozoans such as Schizogony, malegametogony, sporogony in plasmodium, sporulation in Amoeba, etc.

Question 7.
Give an account of pseudopodia.
Answer:
Locomotion in protozoans is performed by cellular extensions such as pseudopodia found in rhizopods organisms. The pseudopodia are temporary extensions of cytoplasm that develop in the direction of the movement. These temporary structures are useful to move on the substratum as our legs do, hence the name ‘pseudopodia’. There are four kinds of pseudopodia in protozoans.

1. Lobopodia: blunt + finger like Pseudopodia. Ex: Amoeba, Entamoeba.
2. Filopodia: fiber like pseudopodia contain ectoplasm. Ex: Euglypha.
3. Reticulopodia: net like pseudopodia. Ex: Elphidium.
4. Axopodia or heliopodia: Sun ray-like pseudopodia. Ex: Actinophiys.

Question 8.
Give an account of the ultrastructure of an axoneme.
Answer:
Ultrastructure of flagellum: The axial filament or axoneme shows 9+2 organisation under the electron microscope. Two central singlets are enclosed by a fibrous inner sheath. Nine peripheral doublets form a cylinder between the inner sheath and the outer sheath. The “A” microtubule of each doublet is connected to the inner sheath by radial spokes. It also has pairs of arms all along the length and is directed towards the neighbouring doublet.

These arms are made of a protein called dynein. These arms create the sliding force. The peripheral doublets are surrounded by an outer membranous sheath called a protoplasmic sheath, which is the extension of the plasma membrane. Some flagella bear lateral appendages called flimmers or mastigonemes along the length of the axoneme above the level of the pellicle. Each flagellum arises from a basal granule that lies below the cell surface in the ectoplasm.

Question 9.
Draw a neat labelled diagram of Euglena.
Answer:

Question 10.
Draw a neat diagram of paramecium and label its important structures/components.
Answer:

Andhra Pradesh BIEAP AP Inter 1st Year Zoology Study Material Lesson 4 Animal Diversity-II: Phylum Chordata Textbook Questions and Answers.

AP Inter 1st Year Zoology Study Material Lesson 4 Animal Diversity-II: Phylum Chordata

Very Short Answer Type Questions

Question 1.
List out the characters shared by chordates and echinoderms.
Answer:
Chordates and echinoderms are enterococci, deuterostomes, and bilaterally symmetrical.

Question 2.
Write four salient features of cyclostomes.
Answer:

1. Cyclostomes are jawless aquatic forms.
2. The body is scaleless, long, slender, and eel-like in shape.
3. Endoskeleton is cartilaginous.
4. Vertebrae are represented by imperfect neural arches in some.
5. The mouth is circular and suctorial, Hence there are called Cyclostomes. Ex: Petromyzon.

Question 3.
What is the importance of endostyle in lancelets and ascidians?
Answer:
Endostyle is useful for accumulating and moving food particles to the oesophagus in lancelets and ascidians.

Question 4.
Name the type of caudal fin and scales that are present in a Shark and Catla respectively.
Answer:
The caudalfin in shark is heterocercal and scales are placoid. The caudal fin is catla is homocercal and scales are cycloid.

Question 5.
What is the importance of air bladder in fish?
Answer:
Fishes have an ‘air bladder’ acting as a ‘hydrostatic organ’ helping the fish float easily at the desired level without much expenditure of energy.

Question 6.
How do you justify the statement ‘heart in fishes is a branchial heart’?
Answer:
The heart of fish is two-chambered and is described as a branchial heart as it supplies blood only to the gills.

Question 7.
What are claspers? Which group of fishes possesses them?
Answer:
Claspers are formed from the posterior portion of pelvic fins in male cartilaginous fish. They serve as intermittent organs used to channel semen into the female’s cloaca during mating.
Ex: Chondrichthyes fishes possess Claspers.

Question 8.
How does the heart of an amphibian differ from that of a reptile?
Answer:
The heart of an amphibian is three-chambered. The heart of a reptile is incompletely four-chambered.

Question 9.
Name the structures that appeared for the first time in amphibians, in the course of evolution.
Answer:
The two pairs of pentadactyl limbs appeared for the first time in amphibians in the course of evolution.

Question 10.
How do you distinguish a male frog from a female frog?
Answer:
The male frog can be distinguished by the presence of sound amplifying vocal sacs and a copulatory pad on the first digit of each forelimb.

Question 11.
What is a ‘force pump’ in a frog? Why is named so?
Answer:
In frogs, during pulmonary respiration, the buccopharyngeal cavity acts like a ‘force pump’. Due to the elevation of the buccopharyngeal cavity the air forces the glottis to open and enter the lungs.

Question 12.
What are corporabigemina? Mention their chief function.
Answer:
Midbrain is represented by a pair of optic lobes called corpora bigemina. The optic lobes are associated with the sense of sight.

Question 13.
Distinguish between mesorchium and mesovarium.
Answer:
The testes are attached to the kidneys and dorsal body wall by a double fold of the peritoneum called mesorchium.
The ovaries are attached to the kidneys and dorsal body wall by a double fold of the peritoneum called mesovarium.

Question 14.
Distinguish between milt and spawn.
Answer:
During amplexus, the mass of eggs and the mass of sperm released by the female and male are called spawn and milt.

Question 15.
What are the ‘Golden ages’ of the first jawed vertebrates and the first amniotes?
Answer:
The Devonian period is considered the ‘golden age of first jawed vertebrates (Fishes).
The Mesozoic era is considered the golden age of amniotes (Reptiles).

Question 16.
Name two poisonous and non-poisonous snakes found in south India.
Answer:
Poisonous Snakes:

1. Naja naja (Cobra)
2. Bungarus (Krait)
3. Vipera russelli (Chain viper)

Non-poisonous Snakes:

1. Ptyas (rat snake)
2. Tropidonotus (Pond or grass snake).

Question 17.
In which features does the skin of a reptile differ from that of a frog?
Answer:

• The skin of reptiles is rough and dry, covered by horny epidermal scales, and shields.
• The skin of a frog is thin, scaleless, and moist.

Question 18.
Describe a cat and a lizard on the basis of their chief nitrogenous wastes excreted.
Answer:
Based on the nitrogenous wastes excreated by lizards are Uricotelic and cats are ‘Ureotelic’ animals.

Question 19.
Name the four extraembryonic membranes.
Answer:
The extraembryonic membranes namely amnion, allantois, chorion, and yolk sac.

Question 20.
What are Jacobson’s organs? What is their function?
Answer:
Jacobson’s organs are the specialized olfactory structures, that are highly developed in lizards and snakes.

Question 21.
What are pneumatic bones? How do they help birds?
Answer:
The main bones in birds are extensions of air sacs without bone marrow are called pneumatic bones. These are helpful in flying birds.

Question 22.
What is a ‘wishbone? What are the skeletal components that form it?
Answer:
In birds, both the clavicles are fused with the interclavicular to form a ‘V-shaped bone, called fiircula or ‘wish hone’ or ‘Merrythought bone’.

Question 23.
What is continuous oxygenation of the blood? How is it made possible in birds?
Answer:
The lungs of birds are compact, spongy, undistensible lungs associated with air sacs. Air sacs facilitate continuous air supply is called ‘continuous oxygenation of the blood.

Question 24.
Distinguish between the crop and the gizzard in birds.
Answer:

• The Oesophagus of birds is often dilated into a crop for the storage of food.
• The stomach is usually divided into glandular proventriculus and muscular gizzard a grinding mill.

Question 25.
Distinguish between altricial and precocial hatchlings.
Answer:

• Altricial: Young ones of flying birds’ hatchlings are altricial.
• Precocial: Young ones of flightless bird hatchlings are precocial.

Question 26.
In which group of animals do we find three ear ossicles on each side and what are their names from the innermost to the outermost?
Answer:
The middle ear possesses three ear ossicles in the Mammalia group of animals. They are malleus, incus and stapes.

Question 27.
How does a mature RBC of a mammal differ from that of other vertebrates?
Answer:

• In mammals, mature RBC is enucleated and biconcave.
• In other vertebrates RBC is nucleate.

Question 28.
Name the characteristic type of vertebrae found in reptiles, birds, and mammals.
Answer:

• Reptiles’ vertebrae are procoelons.
• Birds’ vertebrae are heterologous.
• Mammalian vertebrae are amphiplatyan.

Question 29.
Name the three meninges. In which group of animals do you find all of them?
Answer:
Mammals have three meninges. They are the outer dura mater, middle arachnoid mater, and inner diameter.

Question 30.
Name the vertebrate groups in which ‘renal portal system1 is absent.
Answer:
The renal portal system is absent in aves (birds) in vertebrate animals.

Short Answer Type Questions

Question 1.
Give three major differences between chordates and non-chordates and draw a sketch of a chordate’s body showing those features.
Answer:
The major differences between chordates and non-chordates.

Question 2.
Name the four ‘hallmarks’ of chordates and explain the principal function of each of them.
Answer:
The hallmarks of chordates: All the chordates exhibit four fundamental characteristics. They are Notochord, Dorsal tubular nerve cord, Pharyngeal slits or clefts, and Post-anal tail.

Notochord: It is a flexible rod-like structure situated along the mid-dorsal line between the gut and the nerve cord. It is derived from the embryonic chorda mesoderm. It is firm but flexible. It is present throughout life in the lancelets and cyclostomes. It is present in the tail of the tadpole larva of an ascidian, It is present in embryonic stages, but is replaced partly or wholly by the vertebral column in the adults of higher chordates. Remnants of notochord occur as nucleipulposi in the intervertebral discs of mammals.

Dorsal tubular nerve cord: A single, hollow tubular, and fluid-filled nerve cord is situated above the notochord and below the dorsal body wall. It is derived from the ectoderm of the embryo. In the higher chordates, it gets enlarged to form a distinct brain at the anterior end the rest of it becomes the spinal cord.

Pharyngeal slits of clefts: These are slit openings present on the Pharyngeal wall and meant for the exit of the water from the pharyngeal cavity. They are present throughout life in the protochordate, fishes, and some amphibians. These are present in larval stages in amphibians. They develop by in-pushing of ectoderm and corresponding out pursing of the endoderm. In land vertebrates, the gills become vestigial and nonfunctional and are restricted to embryonic stages only.

Post-anal tail: Chordates have a tail extending posteriorly to the anus. It is lost in many species during late embryonic development. It contains skeletal elements and muscles, coelom and visceral organs are absent in it.

Question 3.
Describe the features of a tunicate that reveals its chordate identity.
Answer:

1. The body of these animals is covered by cellulose-like covering tunicin, hence called tunicates.
2. These possess Notochord in the tail region during the larval stage, hence called Urochordata.
3. These are sedentary or pelagic marine forms.
4. The notochord is present only in larval tails and degenerated in adults.
5. Open type of blood vascular system with blood pigment vanadium.
6. Indirect development with tadpole larva.
7. Adults (mostly) show degenerate characters.
8. The nervous system is represented in the adult by a single dorsal ganglion.
9. They are bisexual or hermaphrodites.
10. Ex: Ascidia, Salpa, Doliolum, Pyrosoma and Oikopleura.

Question 4.
Compare and contrast sea squirts and lancelets.
Answer:
Sea squirts: These are included in class – Ascadiaceae of subphylum – Urochordata. These are sessile. These are solitary or colonial. The body is enclosed in a permanent test and un-segmented. All these are marine and occur from the surface water to greater depths. Coelom in the absent, pharynx is large and is perforated by numerous gill slits. Branchial aperture in anterior and atrial aperture is dorsal. The digestive tract is ‘complete’. The circulations system is of an open type, the heart is the tubular and the ventral heart. These are bisexual. Development generally includes a free-swimming tadpole larva. Notochordcontinued to the tail hence the name Urochordata.
Ex: Ascidia

Lancelets: Cephalochordates are also called Lancelets. These are marine animals and are small fish without paired fins. These are typical chordates because they possess the notochord, tubular nerve cord, and pharyngeal slits throughout this life. The coelom is enterocoelic respiration mostly across the external body surface. The circulatory system is of a closed type, the heart, blood corpuscles, and respiratory pigments are absent. Excretion by protonephridia fertilization is external and development is indirect.
Ex: Branchiostoma (amphioxus or Lancelet)

Question 5.
List out eight characteristics that help distinguish a fish from the other vertebrates.
Answer:
General characters:

1. Fishes are completely aquatic poikilothermic (cold-blooded) animals.
2. The body of a fish is usually streamlined and differentiated into the head, trunk, and tail.
3. The exoskeleton consists of mesodermal scales or bony plates. A few are scaleless.
4. The endoskeleton may be cartilaginous or bony. Skull is monocondylic. Vertebrae are amphicoelous. Centrum is concave at both anterior and posterior faces.
5. Locomotion is assisted by unpaired (median and caudal) fins along with paired (pectoral and pelvic) fins.
6. The mouth is ventral or terminal. Teeth are usually acrodont, homodont, and polyphyodont.
7. The exchange of respiratory gases is performed by the gills. The heart is ‘two-chambered’.
8. Kidneys are mesonephric. Fishes are mostly ammonotelic and some are ureotelic. (cartilaginous fishes).
9. Cranial nerves are 10 pairs, Meninx Primitiva is the only ‘meninx’ enveloping the central nervous system.
10. The internal ear consists of three semicircular canals. Lateral-line sensory system (to detect movement and vibration in the surrounding water) is well-developed.
11. Eyes are without eyelids and each eyeball is protected by a nictitating membrane.
12. Sexes are separate. Fertilization is internal or external. Development may be direct or indirect.

Question 6.
Compare and contrast cartilaginous and bony fishes.
Answer:

Question 7.
Describe the structure of the heart of the frog.
Answer:
The blood vascular system consists of the heart, blood vessels, and blood. The heart is a muscular organ situated in the upper part of the body cavity. It has two separate atria and a single undivided ventricle. It is covered by a double-layered membrane called the pericardium. A triangular chamber called sinus venosus joins the right atrium on the dorsal side. It receives blood through three vena cavae (caval veins). The ventricle opens into the conus arteriosus on the ventral side. The conus arteriosus bifurcates into two branches and each of them divides into three aortic arches namely carotid, systemic and pulmocutaneous. Blood from the heart is distributed to all parts of the body by the branches of the aortic arches. Three major veins collect blood from the different parts of the body and carry it to the sinus venosus.

Question 8.
Write eight salient features of the class – Amphibia.
Answer:
General characters of Amphibia:

1. They are the first tetrapods and lead a dual mode of life, i.e. on land and in freshwater.
2. The body is divided into distinct ‘head’ and ‘trunk’. The tail may or may not be present.
3. Skin is soft, scale-less (except for the members of Apoda), moist and glandular.
4. The body bears two pairs of equal or unequal pentadactyle limbs (caecilians are limbless).
5. Skull is dicondylic as in mammals. Vertebrae are mostly precocious (centrum is concave at its anterior face only) in the anurans, amphicoelous in the caecilians, and usually opisthocoelous (centrum is concave at its posterior face) in the urodeles. Sternum appeared for the first time in the amphibians.
6. The mouth is large; teeth are acrodont, homodont, and polyphyodont.
7. Respiratory gaseous exchange is mostly cutaneous; pulmonary and buccopharyngeal respiration also occurs. Branchial respiration is performed by larvae and some adult urodeles.
8. The heart is three-chambered with sinus venosus and conus arteriosus. Three pairs of aortic arches and well-developed portal systems are present; erythrocytes are nucleated.
9. Kidneys are mesonephric; ureotelic.
10. Meninges are the inner pia mater and outer dura mater; cranial nerves are 10 pairs.
11. The middle ear consists of a single ear ossicle, the columella Auris which is the modified ‘hyomandibula’ of the fishes.
12. Tympanum, lacrimal and harderian glands appeared for the first time in the amphibians.
13. Sexes are separate and fertilization is mostly external. Development is mostly indirect.
14. e.g. Bufo (toad), Rana (frog), Hyla (tree frog), Salamandra (salamander), Ichthyophis (limbless amphibian), Rhacophorus (flying frog).

Question 9.
Describe the male reproductive system of a frog with the help of a labelled diagram.
Answer:
Male Reproductive System of frog: The male reproductive system consists of a pair of yellowish and ovoid testes, which are attached to the kidneys and dorsal body wall by a double fold of peritoneum called mesorchium. Each testis is composed of innumerable seminiferous tubules which are connected to form 10 to 12 narrow tubules, the vasa efferentia. They enter the kidneys and open into the Bidders canal which is connected to the ureter through transverse canals of the kidney. The urinogenital ducts of both sides open into the cloaca.

Question 10.
Write short notes on organs of special senses in frogs.
Answer:
Special senses: Frog has sense organs such as the organs of touch, taste, smell, sight, and hearing. The well-organized structures among them are eyes, and internal ears, and the rest are ‘cellular aggregations’ around nerve endings. The receptors of touch occur in the skin. Organs of taste are called taste buds that lie on small papillae of the tongue. The organs of smell are a pair of nasal chambers.

The organs of sight are a pair of eyes located in the orbits of the skull. Eyes are protected by eyelids. The upper eyelid is immovable. The lower eyelid is folded into a transparent nictitating membrane, which can be drawn across the surface of the eye. The retina of the eye contains both rods and cones. Cones provide ‘colour vision’ and rods are helpful in ‘dim light vision’.

The ear is useful for hearing and balance. It consists of a middle ear closed externally by a large tympanic membrane (ear drum) and a columella that transmits vibrations to the inner ear. The inner ear consists of a utriculus with three semicircular canals and a small sacculus.

Question 11.
List out the salient features of Exo and endoskeleton in reptiles.
Answer:
The exoskeleton of reptiles occurs in the form of horny epidermal scales, shields, and claws.
Endoskeleton:

1. Skull is monocondylic and many have temporal fossae.
2. Each half of the lower jaws is formed by six bones.
3. Vertebrae are mostly procoelous.
4. The first two cervical vertebrae are specialized into the atlas and axis.
5. The vertebral column is distinguished into cervical, thoracic, lumbar, sacral, and caudal regions.
6. Most living reptiles possess two sacral vertebrae.
7. Interclavicular is associated with the pectoral girdle.
8. Ribs are single-headed except in crocodilians.

Question 12.
List out the extant orders of the Class – Reptilia. Give two examples for each Order.
Answer:

1. Chelonia – Chelone (marine green turtle), Testudo (terrestrial form), Trionyx (freshwater form)
2. Rhynchocephalia – Sphenodon (a ‘living fossil’, endemic to New Zealand)
3. Crocodilia – Crocodylus pulustris (Indian crocodile or mugger), Alligator (alligator), Gavialis gangeticus (Indian gavial or gharial)
4. Squamata
   • Lizards – Hemidactylus (wall lizard), Chameleon, Draco (flying lizard)
   • Snakes
     • Poisonous Snakes: Naja naja (cobra), Ophiophagus hannah (King cobra), Bungarus (krait), Daboia/Vipera russelli (chain viper)
     • Non-Poisonous Snakes: Ptyas (rat snake), Tropidonotus (grass snake or pond snake)

Question 13.
What are the modifications that are observed in birds that help them in flight?
Answer:
So many modifications are observed in birds that help them in flight.

1. Exo and endo skeletons and body structure features might have contributed to their successful aerial mode of life.
2. The exoskeleton consists of epidermal feathers. Feathers are unique to birds. They are useful for flight, particularly the Quill feathers help in flight.
3. The body is boat-shaped and streamlined.
4. Four limbs are modified into wings.
5. Many bones are neumatic with extensions of air sacs.
6. All modern flying birds are provided with powerful breast muscles (flight muscles) chiefly the pectoralis major and pectoralis minor.
7. Lungs are associated with air and seas.

Question 14.
What are the features peculiar to ratite birds? Give two examples of ratite birds.
Answer:
Ratite birds:

1. These are modern flightless running birds.
2. They are ‘discontinuous’ in their distribution like the lungfishes and marsupials.
3. They are characterized by the presence of reduced wings.
4. Feathers are without an interlocking mechanism.
5. Rectrices are absent or irregularly arranged.
6. Prren gland is absent.
7. Pygostyle is rudimentary or absent.
8. The sternum is like without a keel.
9. Clavicles are absent, and syrinx is absent.
10. The male animal has a penis.
11. Young ones are precocial.
12. Ex: Struthio camelus – (African ostrich); Dromaeus (Emu) Kiwi.

Question 15.
Mention the most important features of the nervous system and sense organs in mammals.
Answer:

• The nervous system and sense organs are well developed in mammals.
• Mammals have relatively large brains when compared to that other animals in relation to body size.
• The four optic lobes constitute corpora quadrigemina.
• The two halves of the cerebrum are connected by the corpus callosum.
• The central nervous system is enveloped by three meninges.
• Eyes have movable eyelids with eyelashes.
• The external ear has a large pinna middle ear and possesses three ear ossicles.
• They are malleus, incus, and stapes, Cochlea of the internal ear is spirally coiled and bears the organ of Corti which is the receptor of sound.
• Skin is one of the sense organs.

Question 16.
Write short notes on the following features of the eutherians.

1. Dentition
2. Endoskeleton

Answer:

1. Dentition: The Dental formula of eutherians is i 3/3; c 1/1j pm 4/4, m 3/3; dentition is the codon, heterodont, diphyodont.
2. Endoskeleton: Skull is dycondylic. Most mammals have seven cervical vertebrae; vertebrae are of the amphiplatyan type, sacral vertebrae are two or five, and ribs are double-headed.

Question 17.
Give an example for each of the following.

1. A viviparous fish
2. A fish possessing electric organs
3. A fish possessing poison sting
4. An organ that regulates buoyancy in the body of a fish
5. An oviparous animal with milk-producing glands.

Answer:

1. Scoliodon fish is viviparous fish.
2. Torpedo fish is possessing electric organs.
3. Dasyatis/Trygon fish possess poison sting.
4. The air bladder regulates buoyancy in the body of a fish.
5. Ornithorhynchus anatinus (Duck-billed platypus) is an oviparous animal with milk-producing glands.

Question 18.
Mention two similarities between
(a) Aves and mammals
(b) A frog and a crocodile
(c) A lizard and a snake
Answer:
(a) Aves and mammals:

• Aves and mammals are Triploblaste and bilaterally symmetrical.
• The heart is four-chambered in both.

(b) A frog and a crocodile:

• Erythrocytes are nucleated in both.
• Frogs and crocodiles are uriotelic animals.

(c) A lizard and a snake:

• Lizards and snakes are reptilian animals.
• The heart is incompletely four-chambered.
• Jacobson’s organs, the highly developed specialized olfactory structures are present.

Question 19.
Name the following animals.

1. A limbless amphibian
2. The largest of all living animals
3. An animal possessing dry and cornified skin
4. ‘National animal’ of India.

Answer:

1. Ichthyophis is a limbless amphibian.
2. Balaenoptera musculus (Blue whale) is the largest of all living animals.
3. Crocodylus is an animal possessing dry and cornified skin.
4. Panther Tigris (tiger) is the National animal of India.

Question 20.
Write the generic names of the following.

1. An oviparous mammal
2. Flying fox
3. Blue whale
4. Kangaroo

Answer:

1. An oviparous mammal’s generic name is Ornithorhynchus (Duckbilled platypus).
2. Flying fox’s generic name is Pteropus.
3. The blue whale’s generic name is Balaenoptera musculus
4. Kangaroo generic name is Macropus

Andhra Pradesh BIEAP AP Inter 1st Year Zoology Study Material Lesson 3 Animal Diversity-I: Invertebrate Phyla Textbook Questions and Answers.

AP Inter 1st Year Zoology Study Material Lesson 3 Animal Diversity-I: Invertebrate Phyla

Very Short Answer Type Questions

Question 1.
What physical feature pertaining to the organism and its medium do you notice in a sponge body form in which sponges can be/were identified as animals and not plants? What do you call the region in the sponge body in which you noticed that feature?
Answer:
Sponges are primitive multicellular and sessile animals and have a cellular level of organisation. The body wall is composed of two layers separated by matrix mosohyl, and are heaving canal system for transport of water through Ostia, having a cavity in the body called a spongocoel hence the sponge are animals and are not plants.

Question 2.
What are the different structures that make up the internal skeleton of a sponge? What are the chemicals involved in the formation of these structures?
Answer:
The internal skeleton of a sponge is made up of different types of spicules.
Calcareous spicules made up of CaCO₃.
Ex: Sycon
Siliceous spicules – are made up of Silicon dioxide – glass.
Ex: Euplectella
Spongin fibres.
Ex: Spongilla

Question 3.
What are the functions of the canal system of sponges?
Answer:
The functions of the canal system of a sponge are gathering of food, respiratory exchange of gases, and removal of wastes.

Question 4.
What are the two chief morphological ‘body forms’ of cnidarians? What are their chief functions?
Answer:
The body form of Cnidarians is polyp and medusa. Polyp produces medusae by asexual reproduction. Medusae produce polyps by sexual reproduction.

Question 5.
What is metagenesis? Animals belonging to which phylum exhibit metagenesis?
Answer:
Cnidarians show two basic body forms called polyp and medusa. Cnidarians which exist in both forms exhibit alternation of generations called metagenesis.

Question 6.
What is the cnidarian group with quantitatively/relatively large mesoglea? What is the significance of such a well-developed mesoglea pertaining to the aquatic life of that group?
Answer:
The Scyphozoa of cnidarian animals have large mesoglea, it is the significance of these animals.

Question 7.
What is the chief difference between the hydrozoans and the rest, of the cnidarians regarding the germinal layer (s) in which its ‘defencive structures or cells of defence occur?
Answer:
The defencive structures Cnidocytes or Cnidoblasts occur only in the ectoderm, in the hydrozoans in the rest of the Cnidarians the cnidocytes occur in both ectoderm and endoderm.

Question 8.
What are the excretory cells of flatworms called? What is the other important function of these specialized cells?
Answer:
The excretory cells of flatworms are flame cells. Another important function of these specialized cells is osmoregulation.

Question 9.
Distinguish between amphids and phasmids.
Answer:
Amphids: These are the cuticular depressions present on the lips surrounding the mouth in the nematodes such as Aphasmidia animals and serve as Chemoreceptors.
Phasmids: These are the well-developed sensory organs and they occur in some nematodes such as phasmidia animals.

Question 10.
What is the essential difference between a ‘flat worm’ and a ’round worm’ with reference to the perivisceral area of the ‘bodies’.
Answer:
With the reference to the perivisceral area of the body, the flatworms have dorso-ventrally flattened bodies. The body is not segmented, but some of the animals exhibit pseudometamerism. In the Nematoda the body is circular in cross-section, hence the name roundworms, body is not segmented.

Question 11.
How do you account for the origin of the perivisceral space in the body of a nematode and an annelid?
Answer:
The perivisceral space in the body of a nematode is circular in cross-section. Hence the name ‘wound worms’ body unsegmented. In an annelid, the body is segmented by septa into segments or metameres (annulus – little rings) some of them (Nereis) possess lateral appendages parapodia.

Question 12.
What is metamerism? What is the essential difference between the mode of formation of individual morphological body units of a tapeworm and those of an earthworm?
Answer:
The body is divided into segments like units called metameres. Like the divination in known as metamerism.
Ex: Earthworm.
In tapeworm body segments are pseudometameres.
In earthworms, body segments are true segments or metameres.

Question 13.
How do you distinguish a ‘hirudineaun’ from the rest of the annelids, based on the morphological features pertaining to metamerism? How does the coelom of a leech differ from the coelom of an earthworm with reference to its contents?
Answer:
In hirudinean like Leach, the body is with a definite number of segments. The segments are externally sub-divided into annute, internal segmentation’ is absent.
In Leech coelom is filled with a characteristic tissue called botryoidal tissue. In earthworms, the coelom is filled with coelomic fluid.

Question 14.
What do you call the locomotor structures of Nereis? Why is Nereis called a polychaete?
Answer:
Locomotor structures of Nereies are parapodia. The parapodia bear many setae that help in locomotion hence the name Polychaeta.

Question 15.
What is botryoidal tissue?
Answer:
The coelom of Leech is filled with a characteristic tissue called botryoidal tissue, it is resembling a bunch of grapes. They range from excretion to storage of iron, calcium, and revascularization in areas of injury.

Question 16.
What is the difference between the epidermis of a Nematoda and that of an annelid? How does a nematode differ from an annelid with reference to the musculature of the body wall?
Answer:
The epidermis of Nematoda is syncytial and the epidermis of annelid animals is informed by one cell thick ectodermal epithelial cells.

Question 17.
What do you call the first and second pairs of cephalic appendages of a scorpion?
Answer:
The first and second pairs of cephalic appendages of a Scorpion are Chelicerae and Pedipalpi.

Question 18.
What is the uniqueness of the first two pairs of cephalic appendages of a crustacean compared to those of the other extant arthropods?
Answer:
In crustaceans, cephalic appendages are two pairs of antennae (antennules and antennae). It is the unique feature of Crustaceans compared to those of the other extent arthropods.

Question 19.
What is the sub-phylum to which ‘ticks’ and ‘mites’ belong? How do you distinguish them from insects with reference to their walking legs?
Answer:
Ticks and mites belongs to the sub-phylum Chelicerata and class Arachnida. These have four pairs of walking legs.

Question 20.
What are the respiratory structures of Limulus and Palamnaeus respectively?
Answer:
The respiratory structures of Limulus are book gills, and in palamnaeus are book-lungs.

Question 21.
What ara antennae? What is the arthropod group without antennae?
Answer:
Antennae are the sensory organs, of the animals of sub-phylum-Mandibulata of arthropod bear antennae.

Question 22.
What do you call the perivisceral cavity of an arthropod? Where from is it derived during development?
Answer:
The perivisceral cavity of an Arthropoda is a haemocoel, it is not true coelom, but derived from mostly the embryonic blastocoel.

Question 23.
Which arthropod, you have studied, is called a living fossil? Name its respiratory organs.
Answer:
The arthropod animal Limulus is called a living fossil, it is respiratory organs are book-gills.

Question 24.
How do you identify a Chiton from its external appearance? How many pairs of gills help in the respiration of Chiton?
Answer:
Chiton is bilaterally symmetrical and dorsoventrally flattened. Shell is dorsal and consists of eight transverse plates. Poat is ventral elongated and flat. Gills are 6 to 88 pairs helps in respiration.

Question 25.
What is the function of the radula? Give the name of the group of mollusks that do not possess a radula.
Answer:
The buccal cavity contains a file-like rasping organ called radula for feeding, except for the bivalves and tusk of Molluscs.

Question 26.
What is the other name for the gill of a mollusc? What is the function of osphradium?
Answer:
The other name for the gill of a mollusc is Ctenidia. The main function of Osphradium is to test the purity of water.

Question 27.
What is Aristotle’s lantern 7 Give one example of an animal possessing it?
Answer:
In the mouth of the sea Urchin a complex five Jawed masticatory apparatus called Aristotle’s Lantern.
Ex: Echinus.

Question 28.
What is the essential difference between the Juveniles and adults of echinoderms, symmetry-wise?
Answer:
The adult echinoderms are radially symmetrical (pentamerous radial symmetry), but Juveniles (Larvae) are bilaterally Symmetrical.

Question 29.
What are blood glands in pheretima?
Answer:
Blood glands are present in the 4th, 5th, and 6th segments of pheretima. They produce blood cells and haemoglobin which is dissolved in the plasma.

Question 30.
What are spermathecae on the body of pheretima?
Answer:
In Pheretima there are four pairs of spermothecae are located in the segments 6th to 9th as one pair in each segment. This receives and stores spermatozoa during copulation.

Short Answer Type Questions

Question 1.
Write short notes on the salient features of the anthozoans.
Answer:

1. Anthozoans are commonly referred to as sea anemones.
2. Anthozoa includes sea anemones, corals, and sea pens.
3. All are marine forms. These are solitary or colonial.
4. They are sedentary and only have polypoid information.
5. Coeienteron is divided into several compartments by vertical septa called mesenteries.
6. Mesoglea contains connective tissue.
7. Cnidocytes occur both in the ectoderm and endoderm and are cellular and contain amoebocytes.
8. Germ cells are derived from the endoderm. Ex: Adamsia (sea anemone), Gorgonia (sea fan), Pennatula (sea pen).

Question 2.
What is the class to which the flukes belong? Write short notes on the chief characters of the group.
Answer:
Flukes belong to the class Trematoda of Phylum-Platyhelminthes.

1. Trematoda organs are commonly called flukes.
2. These are parasitic on other animals.
3. The body is covered by a thick cuticle and bears two suckers, an oral and a ventral.
4. The mouth is anterior and the intestine is bifurcated.
5. These are bisexual (monoecious).
6. Life history is complex with many hosts and different types of stages – miracidium, sporocyst, redia, cercaria, etc. Ex: Fasciola (Liver fluke), Schistosoma (blood fluke).

Question 3.
What are the salient features exhibited by Polychaetes?
Answer:

1. These are commonly known as bristle worms.
2. All are marine. Many are burrowing, others are free swimming or crawling or tubicolous.
3. Head is distinct with sensory structures like eyes, antennae, palps, and cirri.
4. Clitellum is absent.
5. Each segment has a pair of lateral appendages called parapodia in which bundles of setae are arranged.
6. Animals are unisexual. Most segments bear glands. Gonoducts are absent.
7. Gametes are shed into the coelom.
8. Fertilization is external.
9. Development includes a trochophore larva. Ex: Nereis

Question 4.
How do the hirudineans differ from the polychaetes and oligochaetes?
Answer:

1. Definite number of body segments are present in hirudinean but many segments are present in polychaetes and earthworms.
2. Locomotion in leeches is by suckers but body setae in oligochaetes and parapodia in polychaetes. Parapodia also help in respiration.
3. Temporary clitellum during the breeding season is present in leeches but clitellum is absent in polychaetes and permanent clitellum is present in oligochaetes.
4. Hirudineans are bisexuals, oligochaetes are bisexual and polychaetes are unisexual animals.
5. Coelom is reduced on leeches, but coelom is spacious in oligochaetes and polychaetes.
6. Development is direct in leeches and earthworms but indirect in polychaetes.
7. Nutrient tissue called botryoidal tissue fills the coelom in hirudinean.
8. Anterior and posterior suckers are present in hirudineans. Such suckers are absent in polychaetes and oligochaetes. Ex: Pheretima, Tubifex

Question 5.
What are the chief characteristics of crustaceans?
Answer:

1. This includes prawns, crabs, lobsters, crayfishes, etc.
2. Mostly marine, a few are fresh water and some are adapted to terrestrial life.
3. In most species, the head and thorax fuse to form a cephalothorax.
4. Cephalic appendages are five pairs – first antennae (antennules) second antennae, mandibles, first maxillae and second maxillae.
5. Thoracic and abdominal appendages are typically biramous
6. Respiration is by gills.
7. Excretory organs are green glands or antennal glands.
8. Sense organs include statocysts, compound eyes, and antennae.
9. Gonopores are paired.
10. Development is direct or indirect involving several larval stages. The basic larva is nauplius. Ex: Palaemon (Prawn); Cancer (Crab).

Question 6.
Mention the general characters of Arachnida.
Answer:

1. This includes scorpions, spiders, ticks, and mites.
2. Primarily they are all terrestrial.
3. Prosoma bears one pair of pre-oral chelicerae, one pair of post-oral pedipalps, and four pairs of walking legs.
4. In spiders each chelicera bears a fang into which the poison gland opens.
5. Abdominal appendages are modified into book lungs, spinnerets, pectines, etc.
6. Telsun is usually absent. It is present as a sting in scorpions.
7. Respiration is by book lungs or tracheae.
8. Excretory organs are coaxial glands and malpighian tubules.
9. Scorpions are viviparous.
10. Development is direct. Ex: Palamnaeus (scorpion); Aranea (spider).

Question 7.
Compare briefly a centipede and a millipede.
Answer:

Question 8.
Cephalopods show several unique or advanced features when compared to the other molluscs. Discuss briefly.
Answer:

1. The class Cephalopoda includes cuttlefishes, squids, octopuses, nautilus, etc.
2. The Head is discrete and bears very conspicuous eyes.
3. Shell is either present (e.g.: Sepia) or absent (e.g.: Octopus). When present it may be multi-charactered and external (e.g.: Nautilus) or internal (e.g.: Loligo).
4. The foot is modified into eight to ten arms (tentacles) present around the mouth and siphons.
5. Some Cephalopods (e.g: Sepia) possess an ink gland as a defensive adaptation.
6. Ctenidia are two or four in number – dibranchiate. e.g.: Sepia and tetrabranchiates. (e.g.: nautilus)
7. The brain is complex and is protected by a cartilaginous cranium.
8. Eyes are superficially similar to those of vertebrates.
9. Development is direct. Ex: Architeuthis (giant squid).

Question 9.
Which class of Mollusca represents the primitive molluscs? What are their chief features?
Answer:
The primitive molluscs are represents the class Aplacophora of Phylum-Mollusca.
These are primitive forms with ‘worm-like’ bodies. These are marine forms without mantle, shell, foot, and nephridia. The Head is poorly developed. A rasping organ radula is present in the buccal cavity. Cuticle contains calcareous spicules. Eyes, statocysts, and tentacles are absent. The heart consists of a single auricle and a ventricle. A pair of gonads are present in some, there is a mid-ventral groove that is homologous to the foot of the other molluscs.
Ex: Neonmia, Chaetoderma.

Question 10.
What are the salient features of the echinoids?
Answer:

1. It includes sea urchins, heart urchins, sand dollars, etc. The body is ovoid or discoidal and covered by movable spines.
2. Arms are absent, tube feet are arranged in five bands, and bear suckers.
3. Ossicles of the body unite to form a rigid test or corona or case.
4. Pedicellaria is “three jawed”.
5. Anus and madreporite are aboral in position.
6. Ambulacral grooves are closed.
7. A complex five-jawed masticatory apparatus called Aristotle’s lantern is present just inside the mouth. It is absent in heart urchins.
8. Life history includes a larval form called echinopluteus.
9. Specialized gills called peristomial gills as present in sea urchins. Eg: Salmacis (Sea urchin), Echino Cardium (Heart urchin), Clypeastoer (Cake Urchin).

Question 11.
Mention the salient features of Holothuroidea.
Answer:
Holothuroidea: This class includes sea cucumbers. Body elongated in the oro-aboral axis. Arms, spines, and pedicellariae are absent skin are soft and leathery (Coriaceous). The dermis contains microscopic, isolated ossicles. The madreporite is internal, suspended in the perivisceral coelom. Tube feet are provided with suckers. The mouth is surrounded by retractile feeding tentacles, which are modified tube feet, chief gas exchange organs are a pair of respiratory trees that arise from the wall of the cloaca and form branched tubes in the perivisceral coelom. The development includes auricularia and doliolaria larvae.
Ex: Cueumaria, synaptic, Thyone.

Question 12.
What is the function of nephridia?
Answer:

• The nephridia of pheretima are ectodermal in origin and are metanephridia.
• Several types of nephridia occur in pheretima but are fundamentally similar in structure.
• Which opens outside through the nephridiopore – The nephridia tolled open nephridia. Ex: Septal nephridia.
• Those who do not have nephridiopore are called closed-type nephridia. Ex: Pharyngeal nephridia.
• Those open at the outer surface are called exonephridia.
• The nephridia play an important role in osmoregulation.
• Earthworms mostly excrete urea as the excretory product and are described as ureotelic animals.

Question 13.
How many types of nephridia occur in pheretima and how do you distinguish them?
Answer:
In Pheretima three types of nephridia are present.

1. Septal nephridia: The septal nephridia are present on the intersegmental septum from 15 & 16 segments onwards to last and are opened into the alimentary canal.
2. Integumentary nephridia: The integumentary nephridia attached to the inner body wall from the 3rd segment to the last. They open to the exterior on the body surface by nephridiopores.
3. Pharyngeal nephridia: The pharyngeal nephridia present three paired tufts in the segments 4tfl, 5th, and 6th. They open into the buccal cavity and pharynx.

Question 14.
Give an account of the hearts in the circulatory system of pheretima.
Answer:
Hearts in Pheretima: The dorsal blood vessel and the ventral blood vessel are connected by a pair of pulsatile hearts, in each of the seventh, ninth, twelfth, and thirteenth segments. Of these four pairs, the anterior two pairs connect only the dorsal blood vessel to the ventral blood vessel. Hence they are ’ called lateral hearts. The posterior two pairs connect both the dorsal blood vessel and the supra-oesophageal blood vessel with the ventral blood vessel. Hence, they are called lateral oesophageal hearts. These two types of hearts also differ in the number and arrangement of their valves. Four pairs of valves are present in each lateral heart, while three pairs of valves are present in each lateral oesophageal heart. Hearts allow the blood to flow into the ventral blood vessel only.

Long Answer Type Questions

Question 1.
Draw a labelled diagram of the reproductive organs of Pheretima.
Answer:
Reproductive organs of pheretima: Pheretima is a hermaphrodite (bisexual). There are two pairs of testes. One pair each present in the 10th and 11th segments. Their vasa deferentia run up to the 18th segment where they join the prostatic ducts. Two pairs of seminal vesicles present in the 11th and 12th segments are sacs in which spermatogonia mature into spermatozoa. The common prostatic and spermatic ducts open to the exterior by a pair of male genital pores on the ventrolateral sides of the 18th segment. Two pairs of accessory glands’ are present one pair each in the 17th and 19th segments. Four pairs of spermathecae are located in the segments 6th to 9th (one pair in each segment). They receive and store spermatozoa (spermatophores) during copulation.

One pair of ovaries is attached to the posterior face of the inter-segmental septum of the 12th and 13th segments. Oviducal funnels are present beneath the ovaries and they continue into oviducts (14th segment). They join together and open to the exterior on the ventral side of the 14th segment by a single median female genital pore.

Question 2.
Describe the digestive system and process of digestion in pheretima.
Answer:
The digestive system in pheretima: The alimentary canal is a straight tube and runs from the first to the last segment of the body. The mouth opens into the buccal cavity (1-3 segments) which leads into the muscular pharynx (4th segment). A small narrow tube, oesophagus (5-7 segments), continues into a muscular gizzard (8th segment). It helps in grinding the small particles of food in the decaying leaves (grinding mill). The stomach extends from segments 9 to 14. The food of earthworms is decaying leaves and other organic matter mixed with the soil. Calciferous glands, present in the stomach, neutralise the humic acid present in the humus of the soil. The intestine starts from the 15th segment and continues till the last segment.

A pair of short and conical intestinal caeca project from the intestine in the 26th segment. An internal median fold of the dorsal wall of the intestine called typhiosole, helping in increasing the area of absorption, is poorly developed in Pheretima (between the 26th and the rectum which occupies the last 23 to 28 segments). The alimentary canal opens to the exterior by a small rounded aperture called the anus. The ingested soil rich in organic matter passes through the digestive tract where digestive enzymes break down complex food into smaller absorbable units. These simpler molecules are absorbed through intestinal membranes and are utilized for various metabolic activities.

Process of digestion: Digestion in earthworms is extracellular. Earthworm obtains their nourishment from the organic debris (detritus) present in the soil. So it is called a detritivore. The pharynx is ejected due to the inside out of the buccal chamber. The pharynx, with the help of its radial-dilator muscles, works as a suction pump in feeding.

The organic food along with the swallowed soil particles is sucked into the pharynx, where it mixes with the salivary secretion. The mucin in the saliva lubricates the gut wall for the easy passage of food and also helps in the formation of the bolus. The proteolytic enzyme in the saliva partly digests the proteins. Then the food reaches the gizzard. Its circular muscle and the thick cuticle grind the food into fine particles. In this state, the food is easily acted upon by the digestive enzymes in the stomach and intestine.

The intestinal juice of an earthworm is comparable to the pancreatic juice of higher animals. All these enzymes like proteases, amylases, and lipases act upon the finely ground food and digest the organic matter in it. Proteases digest proteins into amino acids, amylases digest carbohydrates into glucose and lipases digest lipids into fatty acids and glycerol.

The digested food is absorbed by the intestinal epithelium in the typhlosolar region. The extensive capillary network of blood vessels of the intestine plays a vital role in absorption. The typholosole helps in increasing the area of absorption. The undigested food then passes to the rectum, where water is absorbed from the undigested food. Then the undigested matter is egested out through the anus in the form of worm castings.

Andhra Pradesh BIEAP AP Inter 1st Year Zoology Study Material Lesson 2 Structural Organisation in Animals Textbook Questions and Answers.

AP Inter 1st Year Zoology Study Material Lesson 2 Structural Organisation in Animals

Very Short Answer Type Questions

Question 1.
The body of sponges does not possess a tissue level of organisation, though it is made up of thousands of cells. Comment on it.
Answer:
Even though the sponge’s body is made up of thousands of cells, they exhibit a cellular grade of organization, due to the absence of sensory and nerve cells spongs do not possess a tissue level of organisation.

Question 2.
What is the tissue level of organisation among animals? Which metazoans exhibit this organisation?
Answer:
The tissue level of organisation is the lowest level of organisation among the eumetazoans, exhibited by diploblastic animals like cnidarians.

Question 3.
Animals exhibiting which level of organisation lead the relatively more efficient way of life when compared to those of the other levels of organisation? Why?
Answer:
Animals exhibit an organ-system level of organisation of the animals and are exhibited by the triploblastic animals.
Organ level of organisation is a further advancement over the tissue level in the evolution of levels of organisation.

Question 4.
What is monaxial heteropolar symmetry? Name the group of animals in which it is the principal symmetry.
Answer:
Monaxial heteropolar symmetry: When any plane passing through the central axis of the body divides an organism into two identical parts is called Monaxial heteropolar symmetry or radial symmetry. It is the principal symmetry of diploblastic animals such as Cnidarians and ctenophores.

Question 5.
Radial symmetry is an advantage to sessile or slow-moving organisms. Justify this statement.
Answer:
Animals showing radial symmetry live in water and they can respond equally to stimuli that arrive from all directions. Thus, radial symmetry is an advantage to sessile or slow-moving animals.

Question 6.
What is cephalization? How is it useful to its possessors?
Answer:
Cephalization: Concentration of nerve (Brain) and sensory cells at the anterior end of the body is called Cephalization. As a result of cephalization, these animals can sense the new environment and are more efficient than the other animals in seeking food, locating mates, and avoiding or escaping from predators.

Question 7.
Mention the animals that exhibited a ‘tube-within-a-tube’ organisation for the first time? Name their body cavity.
Answer:
Cnidarians and some flat warms are the first animals to exhibit a ‘tube-within-a-tube’ organisation. The body cavity is pseudocolor.

Question 8.
Why is the true coelom considered a secondary body cavity?
Answer:
During the embryonic development of the coelomates, the blastocoel is replaced by a true coelom derived from the mesoderm. So, the true coelom is also called ‘the secondary body cavity.

Question 9.
What are retroperitoneal organs?
Answer:
Certain organs such as the kidneys of the vertebrates are covered by the parietal peritoneum only on their ventral side. Such a peritoneum is called the ‘retroperitoneum’ and the organs lined by it are called ‘retroperitoneal organs.

Question 10.
If the mesentoblast cell is removed in the early embryonic development of protostomes what would be the fate of such animals?
Answer:
In the protostomes, the mesentoblast cell of the early embryo divides to form mesodermal blacks between the ectoderm and the endoderm and replaces the blastocoel. The split that appears in each mesodermal black leads to the formation of Schizocoelom. If the mesentoblast cell is removed in the early embryonic development of protostomes it will cause no ceolome in their animals.

Question 11.
What is enterocoelom? Name the enterocoelomate phyla in the animal kingdom.
Answer:
Animals in which the body cavity is formed from the mesodermal pouches of archenteron are called ‘enterocoelomates’.
Echinoderms, hemichordates, and chordates are enterocoelomates.

Question 12.
Stratified epithelial cells have a limited role in secretion. Justify their role in our skin.
Answer:
The main function of stratified epithelial cells is to provide protection against chemicals and mechanical stress. Hence the stratified epithelial cells have a limited role in secretion.

Question 13.
Distinguish between exocrine and endocrine glands with examples.
Answer:
Exocrine glands are provided with ducts. Secrete mucus, saliva, earwax, oil, milk, digestive enzymes, and other cell products.
Endocrine glands are ductless and their products are hormones that are not sent out via ducts but are carried to the target organs by blood.
Ex: Pituitary gland.

Question 14.
Distinguish between holocrine and apocrine glands.
Answer:
Apocrine glands in which the apical part of the gland cell in pinched off along with the secretory product.
Ex: Mammary glands
Holocrine glands, in which the entire cell disintegrates to discharge the contents.
Ex: Sebaceous glands

Question 15.
Mention any two substances secreted by mast cells and their functions.
Answer:
Mast cells secrete heparin – an anticoagulant, histamine, bradykinin – vasodilators, and serotonin – vasoconstrictor. Vasodilators cause inflammation in response to injury and infection.

Question 16.
Distinguish between a tendon and a ligament.
Answer:
Tendons are the collagen fibrous tissue of dense regular connective tissue which attaches the skeletal muscles to bones.
Ligaments are also the collagen fibers tissue of dense regular connective tissue which attach bones to other bones.

Question 17.
Distinguish between brown fat and white fat.
Answer:
White fat: It is the predominant type in adults, the adipocyte has a single large lipid droplet. White fat is metabolically not active.
Brown fat: It is found in fetuses and infants. Adipocyte of Brown fat has several small ‘lipid droplets’ and are metabolically active and generates heat to maintain body temperature required by infants.

Question 18.
What is the strongest cartilage? In which regions of the human body, do you find it?
Answer:
The fibrous cartilage is the strongest of all types of cartilage. It occurs in the intervertebral discs and pubic symphysis of the pelvis.

Question 19.
Distinguish between osteoblasts and osteoclasts.
Answer:
Osteoblasts are immature bone cells that secrete the organic components of the matrix and also play an important role in the mineralization of bone and become Osteocytes. Osteoclasts are phagocytic cells involved in the resorption of bone.

Question 20.
Define Osteon.
Answer:
In compact bone structure, a Haversian canal and the surrounding lamellae and lacunae are collectively called a Haversian system or Osteon.

Question 21.
What are Volkmann’s canals? What is their role?
Answer:
In compact bone structure, the Haversian canals communicate with one another, with the periosteum and also with the marrow cavity by transverse or oblique canals called Volkmann’s canals.

Question 22.
What is a Sesamoid bone? Give an example.
Answer:
Sesamoid bones are formed by ossification in tendons.
Eg: Patella (Knee cap) and Pisiform bone of the wrist of a mammal.

Question 23.
What is lymph? How does it differ from plasma?
Answer:
Lymph is a colourless fluid. It lacks RBC, platelets, and large plasma proteins, but has more number of leucocytes. It is chiefly composed of plasma and lymphocytes.

Question 24.
What is the hematocrit value?
Answer:
The percentage of the total volume occupied by RBCs in the blood is called the hematocrit value.

Question 25.
What are intercalated discs? What is their significance?
Answer:
The dark lines across cardiac muscle are called intercalated discs (IDS). These discs are highly characteristic of the cardie muscle.

Question 26.
“Cardiac muscle is highly resistant to fatigue”. Justify.
Answer:
The cardiac muscle is highly resistant to fatigue because it has numerous acrosomes, many molecules of myoglobin (Oxygen storing pigment), and a copious supply of blood which facilitate continuous aerobic respiration. The muscles are immune to fatigue and work tirelessly from the embryonic state until death.

Question 27.
Distinguish between ‘nucleus’ and ‘ganglion’ with respect to the nervous system.
Answer:
A group of cell bodies in the Central Nervous System is called a ‘nucleus’, and in the Peripheral Nervous System, it is called a ‘ganglion’.

Question 28.
Distinguish between tracts and nerves with respect to the nervous system.
Answer:
Groups of axons (nerve fibers) in the central nervous system (CNS) are called tracts’ and in the peripheral nervous system (PNS) they are called ‘nerves’.

Question 29.
Name the glial cells that form the myelin sheath around the axons of the central nervous system and peripheral nervous system respectively.
Answer:
In the central nervous system, the glial cells are called ‘Oligodendrocytes’, in the peripheral nervous system, the glial cells are called Schwann cells.

Question 30.
Distinguish between white matter and Greymatter of ‘CNS’.
Answer:
Myelinated nerve fibers occur in the white matter of the CNS, and in most peripheral nerves, and Non-myelinated axons are commonly found in the grey matter of the CNS and autonomous nervous system.

Question 31.
What are microglia and what is their origin add a note on their function.
Answer:
Microglial cells are the Neurogila (supporting cells) of cells of CNS which are phagocytic cells, of mesodermal origin.

Question 32.
What are pseudounipolar neurons? Where do you find them?
Answer:
In a unipolar neuron, the soma or cyton is found in the dorsal root ganglion they are called pseudounipolar neurons. These are found in spinal nerves.

Short Answer Type Questions

Question 1.
Describe the four different levels of organization in metazoans.
Answer:
The levels of organisation in metazoans are as follows:
1. Cellular level of organisation: It is the lowest level of organisaiton among the metazoans and ‘ is exhibited by the sponges. Different types of cells are functionally isolated due to the absence of sensory and nerve cells. There is no uniformity of labor among the cells and they don’t form tissues.

2. Tissue level of organisation: This is the lowest level of organisation among the eumetazoans, exhibited by diploblastic animals like the cnidarians. In these animals, the cells which perform the same function are arranged into tissues due to the presence of nerve cells and sensory cells.

3. Organ level of organisation: An aggregation of different kinds of tissues that are specialized for a particular function is called an organ. Organ level of organisation is a further advancement over the tissue level in the evolution of levels of organisation. It is the first time appeared in the members of platyhelminths.

4. Organ-system level: It is the highest level of organisation among the animals and is exhibited by triploblastic animals such as the flatworms, nematodes, annelids, arthropods, molluscs, echinodermates, and chordates. High specilized sensory and nerve cells bring about a higher level of coordination and integration among the various organ systems to lead an efficient way of life.

Question 2.
In which group of bilaterians do you find solid bauplan? Why is it called so?
Answer:
The bilaterian’s body plan is the solid bauplan when only one plane that passes through the identical axis divides an organism into two identical parts, it is called bilateral symmetry. It is the principal type of symmetry in triploblastic animals.

Bilaterally symmetrical animals are more efficient than the other animals in seeking food locating mates and in avoiding or escaping from predators, because of cephalization. As a result of cephalization, bilaterally symmetrical animals can sense the new environment into which they enter and respond more efficiently and quickly.

Question 3.
Mention the advantages of coelom over pseudocolor.
Answer:
Advantages of coelom over pseudocoelom:
1. Visceral organs of eucoelomates are muscular (because of their association with mesoderm) and so they can contract and relax freely independent of the muscular movements of the body wall in the coelomic space, e.g.: peristaltic movements of the alimentary canal.
2. Gametes are released into the coelom in some invertebrates (which do not have glnoducts) and in the female vertebrates.
3. Coelomic fluid receives excretory products and stores them temporarily before their elimination.
4. In the eucoelomates, the mesoderm comes into contact with the endoderm of the alimentary canal, and it causes ‘regional specialization of the gut, such as the development of gizzard, stomach, etc., This is referred to as ‘primary induction’. In the case of the pseudocoelomates, due to the absence of such a contract between the gut and the mesoderm, the wall of the gut does not show complex and highly specialized organs.

Question 4.
Describe the formation of schizocoelom and enterocoelom.
Answer:
Schizocoelom: During embryonic development, a specialized cell called 4d blastomere is formed. The cells formed from these cells divide and redivide and develop blocks of mesoderm in blastocoel. The blocks fuse and form the mesodermal band which later on splits to form the Schizocoelom. This type of coelom is present in Annelida, Arthropoda, and Mollusca.

Enterocoelomates: Animals in which the body cavity is formed from the mesodermal pouches of archenteron are called enterocoelomates. Echinoderms, hemichordates, and chordates are enterocoelomates. In these animals, mesodermal ouches that evaginate from the wall of the archenteron into the blastocoel are fused with one another to form the enterocoelom. All the enterocoelomates are deuterostomes and they show radial and indeterminate cleavage.

Question 5.
Describe briefly the three types of intercellular junctions of epithelial tissues.
Answer:
The specialized ‘junctions’ provide both structural and functional links between the individual cells of an epithelium. They show different types of junctions so as to serve the specific needs of that tissue. They are
(A) Tight junctions: These junctions between epithelial cells prevent ‘leakages’ of body fluids. For example, they prevent leakages of water into the surrounding cells in our sweat glands. The plasma membranes of adjacent cells are tightly pressed against each other and are bound together by specific proteins.

(B) Desmosomes: Muscle cells are provided with “desmosomes” (anchoring junctions) which act as ‘rivets’ binding the cells together into strong sheets. Intermediate filaments made of protein ‘keratin’ anchor desmosomes in the cytoplasm.

(C) Gap junctions: They provide continuous cytoplasmic channels between adjacent cells. Various types of ions, sugar molecules amino acids, etc. can pass from a cell to an adjacent cell through ‘gap junctions. They occur in many types of tissues including the ‘cardiac muscles’ where they allow rapid conduction of impulses or depolarization.

Question 6.
Give an account of glandular epithelium.
Answer:
Glandular epithelium: Some of the columnar or cuboidal cells that get specialised for the production of certain secretions, form glandular epithelium. The glands are of two types.
Unicellular glands: Consisting of isolated glandular cells such as goblet cells of the gut.
Multicellular glands: Consisting of clusters of cells such as salivary glands. On the basis of the mode of pouring of their secretions, glands are divided into two types namely exocrine and endocrine glands.
Exocrine glands: These glands provided with ducts secrete mucus, saliva, earwax, oil, milk, digestive enzymes, etc.
Endocrine glands: These glands are ductless and their products are hormones, which are carried to the target organs by blood.
Based on the mode of secretion, Exocrine glands are further divided into three types.
1. Merocrine glands: Which release the secretory granules without the loss of other cellular material.
Ex: Pancreas
2. Apocrineglands: The apical part of the cell is pinched off along with the secretory product.
Ex: Mammary glands
3. Holocrine glands: The entire cell disintegrates to discharge the contents.
Ex: Sebaceous gland

Question 7.
Give a brief account of the cells of areolar tissue.
Answer:
Areolar tissue: It is one of the most widely distributed connective tissue in the body. It forms the packing tissue in almost all organs. Areolar tissue forms the subcutaneous layer of the skin. It has cells and fibres. Cells of the areolar tissue are fibroblasts, mast cells, macrophages, adipocytes, and plasma cells.
1. Fibroblasts are the most common cells which secrete fibres. The inactive cells are called fibrocytes.
2. Mast cells secrete heparin (an anticoagulant), histamine, bradykinin (vasodilators), and serotonin vasoconstrictor). Vasodilators cause inflammation in response to injury and infection.
3. Macrophages are amoeboid cells, Phagocytic in function, and act as internal scavengers. They are derived from the monocytes of blood. Tissue-fixed macrophages’ are called histiocytes and others are ‘wandering macrophages’.
4. Plasma cells are derived from the B-lymphocytes and produce antibodies.
5. Adipocytes are specialized cells for the storage of fats.

Question 8.
Describe the three types of cartilage.
Answer:
Cartilage is a solid, but semi-rigid (flexible) connective tissue. Cartilage is of three types, which differ from each other chiefly in the composition of the matrix.

Cartilage is a solid, but semi-rigid (flexible) connective tissue It resists compression. Matrix is firm but somewhat pliable. It has collagen fibres, elastic fibres (only in the elastic cartilage), and matrix-secreting cells called chondroblasts. These cells are enclosed in fluid-filled spaces called lacunae. Chocolates are the inactive cells of cartilage. Cartilage is surrounded by a fibrous connective tissue sheath called perichondrium. Cartilage is of three types, which differ from each other chiefly in the composition of the matrix.

1. Hyaline cartilage: It is bluish-white, translucent, and glass-like cartilage. Matrix is homogeneous and shows delicate collagen fibres. It is the weakest and the most common type of cartilage. It is found in the walls of the nose, larynx, trachea, and bronchi.

2. Elastic cartilage: It is yellowish due to elastic fibres. Matrix has an abundance of yellow elastic fibres in addition to collagen fibres. It provides strength and elasticity. Perichondrium is present. It is found in the pinnae of the external ears, Eustachian tubes, and epiglottis.

3. Fibrous cartilage: Matrix has bundles of collagen fibres. The perichondrium is absent. It is the strongest of all types of cartilage. It occurs in the Intervertebral discs and pubic symphysis of the pelvis.

Question 9.
Explain the Haversian system.
Answer:
The compact bone consists of several structural units called Osteons or Haversian systems arranged around and parallel to the bone marrow cavities.

The Haversian system consists of a Haversian canal that runs parallel to the marrow cavity. It contains an artery, a vein, and a lymphatic vessel. The Haversian canal is surrounded by concentric lamellae. Small fluid-filled spaces called ‘lacunae’ provided with minute canaliculi lie in between the lamellae. Canaliculi connect the lacunae with one another and with the Haversian canal.

Each lacuna encloses one osteocyte (an inactive form of osteoblast). The cytoplasmic processes of osteocytes extend through canaliculi. A Haversian canal and the surrounding lamellae and lacunae are collectively called a Haversian system or osteon. The Haversian canals communicate with one another, with the perio¬steum and also with the marrow cavity by transverse or oblique canals called Volkmann’s canals. Nutrients and gases diffuse from the vascular supply of Haversian canals.

Question 10.
Write short notes on lymph.
Answer:
Lymph: Lymph is a colourless fluid. It lacks RBC, platelets, and large plasma proteins, but has more number of leucocytes. It is chiefly composed of plasma and lymphocytes. When compared to the tissue fluid, it contains very small amounts of nutrients and oxygen but has abundant CO2 and other metabolites.

The most important site of the formation of lymph is the interstitial space. As blood passes through the blood capillaries, some portion of blood that includes water, solutes, and proteins of low molecular weight passer: through the walls of capillaries, into the interstitial spaces due to hydrostatic pressure at the arteriolar ends. This fluid forms the interstitial fluid (tissue fluid). Most of the interstitial fluid is returned directly to the capillaries due to osmotic pressure at the venular ends.

A little amount of this tissue fluid passes through a system of lymphatic capillaries (lymph capillaries of the intestinal villi are called ‘lacteals’), vessels, and ducts and finally, reaches the blood through the subclavian veins. The extracellular ’tissue fluid’ that passes into the lymph capillaries and lymph vessels is called ‘lymph’. The lymphatic system represents an ‘accessory route’ by which interstitial fluid flows from tissue spaces into blood.

Question 11.
Describe the structure of a skeletal muscle.
Answer:
Skeletal (striped and voluntary) muscle:
It is usually attached to skeletal structures by ‘tendons’. In a typical muscle such as the ‘biceps’ muscle, skeletal muscle fibre is surrounded by a thin connective tissue sheath, the endomysium. A bundle of muscle fibres is called a fascicle. It is surrounded by a connective tissue sheath called perimysium. A group of fascicles forms a ‘muscle’ which is surrounded by an epimysium (outermost connective tissue sheath). These connective tissue layers may extend beyond the muscle to form a chord-like tendon or sheet-like aponeurosis.

A skeletal muscle fibre is a long, cylindrical, and unbranched cell. It is a multinucleated cell with many oval nuclei characteristically in the ‘peripheral’ cytoplasm (a syncytium formed by the fusion of cells). Sarcoplasm has many myofibrils which show alternate dark and light bands. So it is called striped or striated muscle.

Question 12.
Describe the structure of a cardiac muscle.
Answer:
Cardiac (striped and involuntary) muscle: The cardiac muscle is striated like the skeletal muscle (shows sarcomeres). Cardiac muscle is found in the ‘myocardium’ of the heart of vertebrates. The cardiac muscle cells or the myocardial cells’ are short, cylindrical, mononucleate, or binucleate cells whose ends branch and form junctions with other cardiac muscle cells. Each myocardial cell is joined to adjacent myocardial cells by ‘electrical synapses’ or ‘gap junctions. They permit ‘electrical impulses to be conducted along the long axis of the cardiac muscle fibre. The dark lines across cardiac muscle are called intercalated discs (IDs). These discs are highly characteristic of the cardiac muscle.

Question 13.
Give an account of the supporting cells of nervous tissue.
Answer:
Nervous tissue is of two types.
1. Nervous
2. Supporting cell or neuroglia

Neuroglia (supporting cells): These are the supporting and non-conducting cells that provide a microenvironment suitable for neuronal activity. Unlike neurons, they continue to divide throughout life. Neuroglial cells of the CNS include oligodendrocytes (that form myelin sheath as mentioned above); astrocytes (star-shaped cells) that form an interconnected network and bind neurons and capillaries (helping in providing the blood-brain barrier); ependymal cells, which are ciliated cells that line the cavities of the brain and spinal cord to bring movements in the cerebrospinal fluid; microglial cells, which are phagocytic cells, of mesodermal origin. Neuroglial cells of the peripheral nervous system include the satellite cells and Schwann cells. Satellite cells surround the cell bodies in ganglia, and Schwann cells form neurolemma around axons.

Question 14.
Describe the structure of a multipolar neuron.
Answer:
Multipolar neurons have one axon and two or more dendrites. Most neurons in our body are multipolar neurons.
A neuron usually consists of a cell body with one too many dendrites and a single axon.

Neurons: Neurons are the ‘functional units of nervous tissue. These are electrically excitable cells that receive, initiate, and conduct/transmit impulses. When a neuron is stimulated, an electric disturbance (action potential) is generated which swiftly travels along its plasma membrane. A neuron usually consists of a “cell body” with one to many dendrites and a single axon.

Cell body: It is also called perikaryon, cyton, or soma. It contains abundant granular cytoplasm and a large spherical nucleus. The cytoplasm has Nissl bodies (they represent RER, the sites of protein synthesis), neurofibrils, and lipofuscin granules.

Dendrites: Several short, branched processes that arise from the cyton are called dendrites. They also contain Nissl bodies and neurofibrils. They conduct nerve impulses towards the cell body (afferent processes).

Axon: An axon is a single, long, cylindrical process that originates from a region of the cyton called the axon hillock. The Plasmalemma of an axon is called axolemma, and the cytoplasm is called axoplasm, which contains neurofibrils. However, Nissl bodies are absent. An axon may give rise to collateral branches. Distally it branches into many fine filaments called telodendrites, (axon terminals), which end in bulb-like structures called synaptic knobs or terminal boutons. Synaptic knobs possess ‘synaptic vesicles’ containing chemicals called neurotransmitters.

Question 15.
Write short notes on (a) Platelets and (a) Synapse.
Answer:
(a) Blood platelets (Thrombocytes): These are colourless non- nucleated, round or oval biconvex discs. The number of platelets per cubic mm of blood is about 2,50,000-4,50,000. They are formed from giant megakaryocytes produced in the red bone marrow by fragmentation. The average lifespan of blood platelets is about 5 to 9 days. They secrete thromboplastin and play an important role in blood clotting. They adhere to the damaged endothelial lining of capillaries and seal minor vascular openings.

(b) Synapse: It is the place in between the two neurons or inter-neuronal or neuromuscular junctions. Nerve cells do not have direct contact with each other. There is a microscopic gap of about 200-300 A° present between the nerve cells called a synapse. The nerve cell present before synapse is called the presynaptic neuron and the behind is called the post-synaptic neuron. A neurotransmitter substance called acetylcholine is secreted in the synapse by presynaptic neurons’ telodendrites. Acetylcholine helps in the conduction of nerve impulses from one neuron to another neuron.

Long Answer Type Questions

Question 1.
What is coelom? Explain the different types of coelom with suitable examples and neat labelled diagrams.
Answer:
The body cavity, which is lined by mesoderm, is called coelom more elaborately, coelom is a fluid-filled space between the body wall and visceral organs and lined by mesodermal epithelium peritoneum. Based on the body cavity triploblastic animals can be classified into acoelomentes, pseudo-coelo mates, and coelomates.

Acoelomate bilaterians: The bilaterian animals in which the body cavity is absent are called acoelomates, e.g. Platy-helminthes (lowest bilaterians). In these animals, the mesenchyme derived from the thrid germinal layer, called mesoderm, occupies the entire blasto coel, between the ectoderm and the endoderm, so that the adults have neither the primary cavity (blasto-coelom) nor the secondary cavity (coelom). As there is no body cavity, the acoelomates exhibit a solid body plan.

Pseudocoelomate bilaterians: In some animals, the body cavity is not lined by mesodermal epithelia. Such animals are called Pseudocoelomates. They include the members of phylum Aschelminthes (Nematoda, Rotifera, and some minor phyla). During embryonic development mesoderm (mesenchyme) occupies only a part of the blastocoel adjoining the ectoderm. The unoccupied portion of the blastocoel persists as pseudocoelom.

Eucoelomate bilaterians: Coelom or ‘true coelom’ is a fluid-filled cavity, that lies between the body wall and the visceral organs and is lined by mesodermal epithelium, the peritoneum. The portion of the peritoneum that underlines the body wall is the parietal peritoneum or somatic peritoneum. The portion of the peritoneum that covers the visceral organs is the splanchnic peritoneum or visceral peritoneum. In coelomates, the visceral organs are suspended in the coelom by the peritoneum.

During the embryonic development of the eucoelomates, the blastocoel is replaced by the true coelom derived from the mesoderm. So, the true coelom is also called the ‘secondary body cavity. Based on the mode of formation of coelom, the eucoelomates are classified into two types:

I. Schizocoelomates: Animals in which the body cavity is formed by the ‘splitting of mesoderm’ are called schizocoelomates. Annelids, arthropods, and mollusks are schizocoelomates in the animal kingdom. All the schizocoelomates are protostomians and they show ‘holoblastic’, ‘spiral’, and ‘determinate’ cleavage. The 4d blastomere or mesentoblast cell of the early embryo divides to form mesodermal blocks between the ectoderm and the endoderm and replaces the blastocoel. The split that appears in each mesodermal block leads to the formation of Schizocoelom.

II. Enterocoelomates: Animals in which the body cavity is formed from the mesodermal pouches of archenteron are called enterocoelomates. Echinoderms, hemichordates, and chordates are the enterocoelomates. In these animals, mesodermal pouches that evaginate from the wall of the archenteron into the blastocoel are fused with one another to form the enterocoelom. All the enterocoelomates are deuterostomes and they show radial and indeterminate cleavage.

Question 2.
What is symmetry? Describe the different types of symmetry in the animal kingdom with suitable examples.
Answer:
Symmetry: The concept of symmetry is fundamental in understanding the organisation of an animal. Symmetry in animals is the balanced distribution of paired body parts. The body plan of a vast majority of metazoans exhibits some kind of symmetry. However, most of the sponges and snails show asymmetry (lack of symmetry). The symmetry of an animal and its mode of life are correlated.

Asymmetry: The animals, which cannot be cut into two equal parts (antimeres) in any plane passing through the centre of the body is called asymmetrical, e.g.: most sponges and adult gastropods. In asymmetrical animals, the body lacks a definite form.

Symmetry: The regular arrangement of body parts in a geometrical design relative to the axis of the body is called symmetry. In a symmetrical animal, paired body parts are arranged on either side of the plane passing through the principal axis, such that they are equidistant from the plane. The unpaired body parts are located mostly on the plane, passing through the principal axis.
Basically, the symmetry in animals is of two kinds:
(i) Radial symmetry
(ii) Bilateral symmetry

(i) Radial Symmetry or Monaxial heteropolar Symmetry: When any plane passing through the central axis (oro-aboral axis/principal axis) of the body divides an organism into two identical parts, it is called radial symmetry. The animals with radial symmetry are either sessile or planktonic or sluggish forms. It is the principal symmetry of the diploblastic animals such as the cnidarians and ctenophores symmetrical (as it is five-angled, it is also called pentamerous radial symmetry). Radially symmetrical animals have many planes of symmetry, whereas pentamerous radially symmetrical animals have five planes of symmetry.

(ii) Bilateral symmetry: When only one plane (median sagittal plane) that passes through the central axis (anterior-posterior axis) divides an organism into two identical parts, it is called bilateral symmetry. It is the ‘principal type of symmetry’ in triploblastic animals. Among the triploblastic animals, some gastropods become secondarily asymmetrical though they have primarily bilaterally symmetrical larvae.

Bilaterally symmetrical animals are more efficient than other animals in seeking food, locating mates, and in avoiding or escaping from predators, because of cephalization (concentration of nerve and sensory cells at the anterior end). As a result of cephalization, bilaterally symmetrical animals can sense the new environment into which they enter and respond more efficiently and quickly.

Question 3.
Classify and describe the epithelial tissues on the basis of structural modification of cells with examples.
Answer:
Epithelium (epi-upon; thelia – growing) forms the outer covering of the body and the living of internal organs and cavities. There are two types of epithelial tissues namely ‘simple epithelia1 and ‘compound epithelia’ based on the number of layers or strata.

Simple epithelium is composed of a single layer of cells and forms the lining of body cavities, ducts, and vessels. It helps in the diffusion, absorption, filtration, and secretion of substances. On the basis of the shape of the cells, it is further divided into three types:
(i) Simple squamous epithelium (Pavement epithelium): It is composed of a single layer of flat and tile-like cells, each with a centrally located ‘ovoid nucleus’. It is found in the endothelium of blood vessels, mesothelium of body cavities (pleura, peritoneum, and pericardium), wall of Bowman’s capsule of the nephron, lining of alveoli of lungs, etc.

(ii) Simple cuboidal epithelium: It is composed of a single layer of cube-like cells with centrally located spherical nuclei. It is found in germinal epi – thelium, proximal, and distal convoluted tubules of the nephron. Cuboidal epithelium of proximal convoluted tubule of nephron has ‘microvilli’.

(iii) Simple columnar epithelium: It is composed of a single layer of tall and slender cells with oval nuclei located near the base. It has mucus-secreting ‘goblet cells’ in some places. It is of two types:
(a) Ciliated columnar epithelium: Columnar epithelial cells have cilia on their free surface. It is mainly present in the inner surface of hollow organs like fallopian tubes, ventricles of the brain, the central canal of the spinal cord, bronchioles, etc.

(b) Non-ciliated columnar epithelium: Columnar cells are without cilia. It is found in the lining of the stomach and intestine. Microvilli are present in the columnar epithelium of the intestine to increase the surface area of absorption.

(II) Compound epithelium (stratified epithelium): It is made up of more than one layer of cells. Its main function is to provide protection against chemical and mechanical stress. It covers the dry surface of the skin as stratified, keratinized, squamous epithelium. It covers the moist surface of the buccal cavity, pharynx, esophagus, and vagina as stratified non-keratinized squamous epithelium. It forms the inner lining of the larger ducts of salivary glands, sweat glands, and pancreatic ducts as stratified cuboidal epithelium. It forms the wall of the urinary bladder as transitional epithelium.

(III) Glandular epithelium: Some of the columnar or cuboidal cells that get specialised for the production of certain secretions, form glandular epithelium. The glands are of two types – unicellular glands consisting of isolated glandular cells such as goblet cells of the gut, and multicellular glands, consisting of clusters of cells such as salivary glands. On the basis of the mode of pouring of their secretions, glands are divided into two types namely exocrine and endocrine glands. Exocrine glands are provided with ducts; that secrete mucus, saliva, earwax (cerumen), oil, milk, digestive enzymes, and other cell products. In contrast, endocrine glands are ductless and their products are ‘hormones’, which are not sent out via ducts, but are carried to the target organs by blood.

Based on the mode of secretion, exocrine glands are further divided into
(i) merocrine glands (e.g.: pancreas) which release the secretory granules without the loss of other cellular material.

(ii) Apocrine glands (e.g.: mammary glands) in which the apical part of the cell is pinched off along with the secretory product and
(iii) Holocrine glands (e.g.: sebaceous glands), in which the entire cell disintegrates to discharge the contents.

Question 4.
Describe the various types of connective tissue properly with suitable examples.
Answer:
Connective tissue proper is of two types as
(A) Loose connective tissue proper
(B) Dense connective tissue proper

(A) Loose connective tissue: Cells and fibres are loosely arranged in a semi-fluid ground substance there are three types of loose connective tissues-areolar tissue adipose tissue and reticular tissue.
(i) Areolar tissue: It is one of the most widely distributed connective tissues in the body. It forms the packing tissue in almost all organs. Areolar tissue forms the subcutaneous layer of the skin. It has cells and fibres. Cells of the areolar tissue are the fibroblasts, mast cells, macrophages, adipocytes, and plasma cells.
1. Fibroblasts are the most common cells with secreted fibres. The inactive cells are called
fibrocytes.
2. Mast cells secrete heparin (an anticoagulant), histamine, bradykinin (vasodilators), and serotonin (vasoconstrictor), vasodilators cause inflammation in response to injury and infection.
3. Macrophages are amoeboid cells, phagocytic in function, and act as internal scavengers. They are derived from the monocytes of blood. ‘Tissue fixed macrophages’ are called histiocytes and others are ‘wandering macrophages’.
4. Plasma cells are derived from the B-lymphocytes and produce antibodies.
5. Adipocytes are specialized cells for the storage of fats.

(ii) Adipose tissue: It is specialized for fat storage. It consists of a large number of adipocytes and few fibres. The adipose tissue which is found beneath the skin provides. Adipose tissue is of two types: white adipose tissue, and brown adipose tissue. Excess nutrients which are not used immediately are converted into fats and stored in this tissue.
White adipose tissue (WAT): It is the predominant type in adults, and the adipocyte has a single large lipid droplet (monocular). White fat is metabolically not active.
Brown adipose tissue (BAT): It is found in fetuses and infants. Adipocyte of BAT has several small ‘lipid droplets’ (multilocular) and numerous mitochondria. Brown fat is metabolically active and generates ‘heat’ to maintain the body temperature required by infants.

(iii) Reticular tissue: It has specialized fibroblasts called reticular cells. They secrete ‘reticular fibers’ that form an interconnecting network. It forms the ‘supporting framework’ of lymphoid organs such as bone marrow, spleen, and lymph nodes and forms the reticular lamina of the ‘basement membrane’.

(B) Dense connective tissue: This tissue consists of more fibres, but fewer cells. It has very little ground substance. Based on the arrangement of fibres, the dense connective tissue is of three types.
(i) Dense regular connective tissue: In this tissue, collagen fibres are arranged parallel to one another in bundles. Tendons that attach the skeletal muscles to bones and ligaments which attach bones to other bones are examples of this type of connective tissue.

(ii) Dense irregular connective tissue: In this type of connective tissue, bundles of collagen fibres are irregularly arranged. Periosteum, endosteum, pericardium, heart valves, joint capsule, and deeper region of the dermis of skin contain/are made up of this type of connective tissue.

(iii) Elastic connective tissue: It is mainly made of yellow elastic fibres, capable of considerable extension and recoil. This tissue can recoil to its original shape, when the forces of stretch are released. It occurs in the wall of arteries, vocal cords, trachea, bronchi, and ‘elastic ligaments’ present between vertebrae.

In addition to the above-mentioned connective tissues, mucous connective tissue occurs as foetal or embryonic connective tissue. It is present in the umbilical cord as Wharton’s jelly.

Question 5.
What is skeletal tissue? Describe the various types of skeletal tissue.
Answer:
Skeletal tissue (supporting tissue): It forms the endoskeleton of the vertebrates. It supports the body, protects various organs, provides a surface for the attachment of muscles, and helps in locomotion. It is of two types:
(A) Cartilage (Gristle): Cartilage is a solid, but semi-rigid (flexible) connective tissue. It resists compression. Matrix is firm but somewhat pliable. It has collagen fibres, elastic fibres (only in the elastic cartilage), and matrix-secreting cells called chondroblasts. These cells are enclosed in fluid-filled spaces called lacunae. Chondrocytes are the inactive cells of cartilage. Cartilage is surrounded by a fibrous connective tissue sheath called perichondrium.

Cartilage is of three types, which differ from each other chiefly in the composition of the matrix.
1. Hyaline cartilage: It is bluish-white, translucent, and glass-like cartilage. Matrix is homogeneous and shows delicate collagen fibres. It is the weakest and the most common type of cartilage. It is found in the walls of the nose, Larynx, trachea, and bronchi.

2. Elastic cartilage: It is yellowish due to elastic fibres. Matrix has an abundance of yellow elastic fibres in addition to collagen fibres. It provides strength and elasticity. Perichondrium is present. It is found in the pinnae of the external ears, Eustachian tubes, and epiglottis.

3. Fibrous cartilage: Matrix has bundles of collagen fibres. The perichondrium is absent. It is the strongest of all types of cartilage. It occurs in the intervertebral discs and pubic symphysis of the pelvis.

(B) Bone (osseous) tissue: Bone is highly calcified (mineralized), solid, hard and rigid connective tissue. It is the major component of the endoskeleton of most adult vertebrates.

Bone has an outer fibrous connective tissue sheath called periosteum, the inner connective tissue sheath that lines the marrow cavity called the endosteum, a non-living extracellular matrix, living cells, and bone marrow. Bone cells include osteoblasts, osteocytes, and osteoclasts. Osteoblasts (immature bone cells) secrete the organic components (collagen fibres) of the matrix and also play an important role in the ‘mineralization of bone’ and become osteocytes (mature bone cells). Osteocytes are enclosed in fluid-filled lacunae. Osteoclasts are phagocytic cells involved in the resorption of bone.

Types of bones based on the method of formation:
(i) Cartilage bones (replacing bones or endochondral bones) are formed by ossification within the cartilage e.g. bones of limbs, girdles, and vertebrae.
(ii) Investing bones (membrane bones or dermal bones) are formed by the ossification in the embryonic mesenchyme e.g. most of the bones of the cranium.
(iii) Sesamoid bones are formed by ossification in tendons e.g.: patella (knee cap) and pisiform bone of the wrist of a mammal
(iv) are formed by ossification in the soft tissues, e.g.: Oscordis (inside the heart of ruminants), and Ospenis (inside the glans-penis of many mammals such as rodents, bats, and carnivores).

Types of bones based on the structure:
1. Spongy bone (Cancellous bone or trabecular bone): It occurs in the epiphyses and metaphyses
of long bones. It looks spongy and contains columns of bone called ‘trabeculae’ with irregular interspaces filled with red bone marrow.
2. Compact bone: The diaphysis of a long bone is made up of ‘compact bone. It has a dense continuous lamellar matrix between the periosteum and endosteum.

Structure of a compact bone: Diaphysis (shaft) is a part of a long bone that lies in between expanded ends. The diaphysis is covered by a dense connective fibrous tissue called the periosteum. The diaphysis of a long bone has a hollow cavity called marrow cavity which is lined or surrounded by the endosteum. In between periosteum and endosteum, the matrix of the bone is laid down in the form of ‘lamellae’. Outer circumferential lamellae are located immediately beneath the periosteum; inner circumferential lamellae are located around the endosteum.

Between the outer and inner circumferential lamellae, there are many Haversian systems (osteons – units of bone). The spaces between the Haversian systems are filled with interstitial lamellae. Haversian system consists of a Haversian canal that runs parallel to the marrow cavity. It contains an artery, a vein and a lymphatic vessel. Haversian canal is surrounded by concentric lamellae. Small fluid filled spaces called ‘lacunae’ provided with minute canaliculi lie in between the lamellae.

Canaliculi connect the lacunae with one another and with Haversian canal. Each lacuna encloses one osteocyte (inactive form of osteoblast). The cytoplasmic processes of osteocytes extend through canaliculi. A Haversian canal and the surrounding lamellae and lacunae are collectively called a Haversian system or osteon. The Haversian canals communicate with one another, with the periosteum and also with the marrow cavity by transverse or oblique canals called Volkmanns canals. Nutrients and gases diffuse from the vascular supply of Haversian canals.

Question 6.
Give an account of the “formed elements” of Blood.
Answer:
Blood is a red-coloured, opaque, and slightly alkaline fluid. It is composed of blood plasma and formed elements or blood cells – the RBC, WBC, and platelets.
Formed elements of Blood cells: The blood corpuscles (RBC and WBC) constitute 45% of the total blood by volume.

(i) Red blood corpuscles (Erythrocytes): Erythrocytes of mammals are circular (elliptical in camels and Liamas), biconcave and enucleate. The biconcave shape provides a large surface area-to-volume ratio, thus providing more area for the exchange of gases. These are 7.8 pm in diameter. The number of RBCs per cubic millimeter of blood is about 5 million in a man and 4.5 million in a woman.

(ii) White blood corpuscles (Leucocytes): These are nucleate, colourless, complete cells. They are spherical or irregular in shape and are capable of exhibiting amoeboid movement into the extravascular areas by diapedesis. WBC are two main types: 1) Granulocytes and 2) Agranulocytes.

Granulocytes: They possess cytoplasmic granules that may take three different types of stains, neutral or acidic, or basic. The nucleus of the granulocytes is divided into lobes and assumes different shapes, hence, these are also called polymorphonuclear leucocytes. Based on the staining properties these are of three tvoes.

Basophils: They constitute about 0.4% of the total leucocytes. The nucleus is divided into irregular lobes. Cytoplasmic granules are ‘fewer’ and ‘irregular’ in shape. They take basic stains. They produce heparin, histamine, etc. They supplement the function of mast cells when needed.

Eosinophils (acidophils): They constitute about 2.3% of the total leucocytes. The nucleus is distinctly bilobed. The cytoplasm has large granules which stain with acidic dyes such as ‘eosin’.

Neutrophils: They constitute about 62% of the total leucocytes. The nucleus is many lobed (2-5). Specific cytoplasmic granules are small and abundant. They stain with ‘neutral dyes’.

Agranulocytes: Cytoplasmic granules are absent in agranulocytes. divided into lobes. These are of two types:
(a) Lymphocytes: They constitute about 30% of the total leucocytes. They are small, spherical cells with large spherical nuclei and scanty peripheral cytoplasm. There are functionally two types of lymphocytes – ‘B’ lymphocytes, which produce ‘antibodies’ and T lymphocytes which play a key role in the immunological reactions of the body. Some lymphocytes live only a few days while others survive for many years.

(b) Monocytes: They constitute about 5.3% of the leucocytes. The nucleus is kidney-shaped (reniform). These are the largest, motile phagocytes. They engulf bacteria and cellular debris. They differentiate into macrophages when they enter the connective tissues.

(iii) Blood platelets (Thrombocytes): These are colourless non-nucleated, round, or oval biconvex discs. The number of platelets per cubic mm of blood is about 2,50,000-4,50,000. They are formed from giant megakaryocytes produced in the red bone marrow by fragmentation. The average lifespan of blood platelets is about 5 to 9 days. They secrete thromboplastin and play an important role in blood clotting. They adhere to the damaged endothelial lining of capillaries and seal minor vascular openings.

Question 7.
Compare and contrast the three types of muscular tissues.
Answer:
Muscular tissue is mesodermal in origin. Muscles show three essential properties such as excitability, conductivity, and contractility. The study of muscular tissues is known as mycology. Muscular tissue has elongated cells called ‘muscle fibers’ (myocytes) which are surrounded by a connective tissue sheath. The extracellular matrix is absent. The plasma membrane of a muscle fiber is called sarcolemma.

The cytoplasm of a muscle fibre is called Sarco plasm, the endoplasmic reticulum, the sarcoplasmic reticulum, and the mitochondria, the Sarcosomes. The cytoplasm of a muscle fibre has several myofibrils. Each myofibril has thick (myosin) and thin (actin) myofilaments. The regular arrangement of myosin and actin filaments is responsible for the alternate dark and light bands of a ‘striated muscle’. Sarcoplasm also contains ATP, phosphocreatine, glycogen, and myoglobin. Muscles are of three types – skeletal, smooth, and cardiac.

Skeletal (striped and voluntary) muscle: It is usually attached to skeletal structures by ‘tendons’. A typical muscle such as the ‘biceps’ muscle/skeletal muscle fibre is surrounded by a thin connective tissue sheath, the endomysium. A bundle of muscle fibres is called a fascicle.

It is surrounded by a connective tissue sheath called perimysium. A group of fascicles forms a ‘muscle’ that is surrounded by an epimysium (outermost connective tissue sheath). These connective tissue layers may extend beyond the muscle to form a chord-like tendon or sheet¬like aponeurosis.

A skeletal muscle fibre is a long, cylindrical, and unbranched cell. It is a multinucleated cell with many oval nuclei characteristically in the “peripheral” cytoplasm (a syncytium formed by the fusion of cells). Sarcoplasm has many myofibrils which show alternate dark and light bands. So it is called striped or striated muscle.

Smooth (unstriped and involuntary muscle): It is located in the walls of the visceral organs such as blood vessels, trachea, bronchi, stomach, intestine, excretory and genital ducts, and so this is also called ‘visceral muscle’. As cross striations are absent, it is called ‘smooth muscle. It is also found in the iris and ciliary body of the eye and in the skin as ‘arrector pili muscles that are attached to hair follicles.

Usually, smooth muscles are arranged in ‘layers’/’sheets’. A smooth muscle fibre is a spindle-shaped (fusiform), uninucleate cell. Myofibrils do not show alternate dark and light bands due to the irregular arrangement of actin and myosin fibres. They do not work under conscious control, and so they are called involuntary muscles. Smooth muscles exhibit ‘slow’ and ‘prolonged’ contractions. They may remain contracted for long periods without fatigue (show sustained involuntary contractions called ‘spasms’). The contraction of smooth muscles is under the control of the autonomous nervous system.

Cardiac (striped and involuntary) muscle: The cardiac muscle is striated like the skeletal muscle (shows sarcomeres). Cardiac muscle is found in the ‘myocardium’ of the heart of vertebrates. The cardiac muscle cells or the ‘myocardial cells’ are short, cylindrical, mononucleate, or binucleate cells whose ends branch and form junctions with other cardiac muscle cells. Each myocardial cell is joined to adjacent myocardial cells by ‘electrical synapses’ or gap junctions. They permit ‘electrical impulses’ to be conducted along the long axis of the cardiac muscle fibre. The dark lines across cardiac muscle are called intercalated discs (IDs). These discs are highly characteristic of the cardiac muscle.

Andhra Pradesh BIEAP AP Inter 1st Year Zoology Study Material Lesson 1 Diversity of Living World Textbook Questions and Answers.

AP Inter 1st Year Zoology Study Material Lesson 1 Diversity of Living World

Very Short Answer Type Questions

Question 1.
Define the term metabolism. Give any one example.
Answer:
The sum total of all the chemical reactions occurring in the bodies of organisms constitutes metabolism.
Ex: Photosynthesis is one of the metabolic processes in living organisms.

Question 2.
How do you differentiate between growth in a living organism and a non-living object?
Answer:
Growth is one of the fundamental characteristics of living beings growth in living beings is growth from the inside, whereas growth in non-living things is by the accumulation of material on the surface.

Question 3.
What is biogenesis?
Answer:
Life comes only from life is called biogenesis. Living organisms produce young ones of their kind using molecules of heredity.

Question 4.
Define the term histology. What is it otherwise called?
Answer:
Histology is the study of the microscopic structure of different tissues. It is also called Micro anatomy.

Question 5.
Distinguish between embryology and ethology.
Answer:
Embryology: It is the study of events that lead to fertilization, cleavages, early growth, and differentiation of a zygote into an embryo.
Ethology: The study of animal behaviour based on systematic observation, with special attention to physiological, ecological, and evolutionary aspects is called ethology.

Question 6.
In a given area, remains of animals that lived in the remote past are excavated for study. Which branch of science is it called?
Answer:
The branch of science Palaeontology deals with that. In a given area, remains of animals that lived in the remote past (fossilized remains) are excavated for study.

Question 7.
Zoos are tools for ‘classification’ Explain.
Answer:
Zoos are places where wild animals are taken out of their natural habitat and are placed in protected environments under human care. This enables us to learn about the animal’s external features, habits, behaviour, etc. These observations enable us to systematize the organism and position it in the animal world.

Question 8.
Where and how do we preserve skeletons of animals dry specimens etc?
Answer:
The Skeletons and dry specimens are preserved in Museums and are usually stuffed and preserved.

Question 9.
What is trinominal nomenclature? Give an example.
Answer:
The trinominal nomenclature is the extension of the binominal system of nomanclature. It permits the designation of subspecies with a three-worded name called ‘trinomen’.
Ex: Homo Sapiens Sapiens, Corvus splendns spelendns.

Question 10.
What is meant by tautonymy? Give two examples.
Answer:
The practice of naming animals or organisms, in which the generic name and species name are the same, is called Tautonymy.
Ex: Axis axis – spotted dear
Naja naja – The Indian Cobra

Question 11.
Differentiate between Protostomia and Deuterostomia.
Answer:
Protostomia (Gr. mouth first) are the organisms in which blastopore develops into the mouth.
Deuterostomia (Gr. second mouth) are the organisms in which blastopore develops into the anus, the mouth is formed later.

Question 12.
‘Echinoderms are enterocoelomates’. Comment.
Answer:
The animals of phyla Echinodermata have a true coelom, which is an ‘enterocoel’. It is formed from the archenteron.

Question 13.
What does ICZN stand for?
Answer:
ICZN stands for ‘International Code of Zoological Nomenclature which specifies the mandatory rules to be followed for the naming of animals by the International congress (ICZ) in 1898.

Question 14.
Give the names of any four protostomian phyla.
Answer:
The phylum Platyhelminthes, Nematoda, Annelida, Arthropoda, and Mollusca are the protostomian phyla.

Question 15.
Nematoda is a protostomian but not a coelomate justify the statement.
Answer:
Animals of group Nematoda are protostomian but they have no true coelom/secondary body cavity as it is not lined by mesodermal epithelial layers. Pseudocoel is a remnant of the embryonic blastocoel. Hence they are protostonian. Pseudocoelomata, but not coelomates.

Question 16.
What is ecological diversity? Mention the different types of ecological diversities.
Answer:
Diversity at a higher level of organization, i.e., at the ecosystem level is called ‘Ecological diversity.
The other ecological diversities are Alpha, Beta, and Gama diversities.

Question 17.
Define species richness.
Answer:
The more the number of species in an area (unit area) the more species richness.

Question 18.
Mention any two products of medicinal importance obtained from Nature.
Answer:
Anticancer drugs Vinblastin from the plant Vinco rosa and Digitalin from the plant for gloves are obtained from nature.

Question 19.
Invasion of an Alien species leads to the extinction of native species. Justify this with two examples.
Answer:
When alien species are introduced into a habitat, they turn invasive and establish themselves at the cost of the native species.
Ex: Nail perch introduced into lake Victoria, in east Africa led to the extinction of 200 species of Cichlid fish in the lake. The illegal introduction of exotic African catfish for aquaculture purposes in posing a threat to the native catfish.

Question 20.
List out any four sacred groves in India.
Answer:
The following are the Sacred Groves in India.

1. Khasi and Jaintia Hills – Meghalaya
2. Aravalli Hills – Rajasthan and Gujarat
3. Western Ghat region – Karnataka and Maharashtra
4. Sarguja, Bastar – Chhattisgarh
5. Chanda – Madhya Pradesh

Question 21.
Write the full form of IUCN. In which book threatened species are enlisted.
Answer:
IUCN – International Union for the Conservation of Nature and Natural Resources.
All the threatened species are enlisted in the Red Data Book Published by IUCN.

Short Answer Type Questions

Question 1.
Explain the phylogenetic system of biological classification.
Answer:
Phylogenetic classification is an evolutionary classification based on how a common ancestry was shared. Cladistic classification summarizes the ‘genetic distance’ between all species in this ‘Phylogenetic tree’. In Cladistic classification characters such as analogous characters (characters shared by a pair of organisms due to convergent evolution e.g. wings in sparrows and patagia (wing-like structures) in flying squirrels) and homologous characters (characters shared by a pair of organisms, inherited from a common ancestor e.g. wing of sparrows and finches) are taken into consideration. Ernst Haeckel introduced the method of representing Phylogeny by ‘tree’ branching diagrams.

Question 2.
Explain the hierarchy of classification.
Answer:
Human beings are not only interested in knowing more about different kinds of organisms and their diversities, but also the relationships among them. This branch of study is referred to as systematics. Systematics is the branch of science that deals with the vast diversity of life. It also reveals the trends and evolutionary relationships of different groups of organisms. These relationships establish the phylogeny of organisms. A key part of systematics is taxonomy. The taxonomic hierarchy includes seven obligate categories namely kingdom, phylum, class, order, family, genus, and species, and other intermediate categories such as subkingdom, grade, division, subdivision, subphylum, superclass, subclass, superorder, suborder, superfamily, subfamily, subspecies, etc.

Question 3.
What is meant by classification? Explain the need for classification.
Answer:
Classification is defined as the process by which anything is grouped into convenient categories based on some easily observable characteristics. It is impossible to study all living organisms. So, it is necessary to devise some means to make this possible. This process is called classification. The scientific term used for these categories is ‘TAXA’. Taxa can indicate categories at different levels, e.g. Animalia, Chordata, Mammalia, etc. represent taxa at different levels.

Hence based on characteristics, all living organisms can be classified into different taxa: This process of classification is called taxonomy. External and internal structures, along with the structure of cells, developmental processes, and ecological information of organisms are essential and they form the basis of modern taxonomic studies, Hence characterization, identification, nomenclature, and classification are the processes that are basic to taxonomy. To understand the interrelationships among the diversified animal groups, a systematic classification is necessary.

Question 4.
Define species. Explain the various aspects of ‘species’.
Answer:
Species: Species is the ‘basic unit’ of classification. Species is a Latin word meaning ‘kind’ or ‘appearance’. John Ray in his book ‘Historia Generalis Plantarum’ used the term ‘species’ and described it on the basis of common descent (origin from common ancestors) as a group of morphologically similar organisms. Linnaeus considered species, in his book ‘Systema Naturae’, as the basic unit of classification. Buffon, in his book ‘Natural History, proposed the idea of the evolution of species which is the foundation for the biological concept of evolution. This biological concept of species (dynamic nature of species) became more popular with the publication of the book “The Origin of Species” by Charles Darwin.

Buffon’s biological concept of species explains that species is an interbreeding group of similar individuals sharing the common ‘gene pool’ and producing fertile offspring. Species is considered as a group of individuals which are:

1. Reproductively isolated from the individuals of other species – a breeding unit.
2. Sharing the same ecological niche – an ecological unit.
3. Showing similarity in the karyotype – a genetic unit.
4. Having similar structure and functional characteristics – an evolutionary unit.

Question 5.
What is genetic diversity and what are the different types of genetic diversity?
Answer:
Genetic diversity is the diversity of genes within a species. A single species may show high diversity at the genetic levels over its distributional range. For e.g. Rauwolfia vomitoria, a medical plant growing in the Himalayas ranges shows great genetic variation, which might be in terms of potency and concentration of the active chemical (reserpine extracted from it is used in treating high blood pressure) that the plant produces. India has more than 50,000 different strains of rice and 1,000 varieties of mangoes. Genetic diversity increases with environmental variability and is advantageous for its survival.

Question 6.
What are the reasons for greater biodiversity in the tropics?
Answer:
Reasons for greater biodiversity in the tropics:
Reason 1: Tropical latitudes have remained relatively undisturbed for millions of years and thus had a long ‘evolutionary time’. The as long duration available in this region for speciation led to species diversification. (Note: The temperate regions were subjected to frequent glaciations in the past).

Reason 2: Tropical climates are relatively more constant and predictable than that temperate regions. A constant environment promotes niche specialization (how an organism responds, and behaves with the environment and with other organisms of its biotic community) and this leads to greater species diversity.

Reason 3: Solar energy, resources like water, etc., are available in abundance in this region. They contribute to higher productivity in terms of food production, leading to greater diversity.

Question 7.
What is the ‘evil quartet’?
Answer:
The following are the ‘four major causes (The Evil Quartet) for accelerated rates of species extinction in the world.
Habitat loss and Fragmentation: These are the most important reasons for the loss of biodiversity.

• Deforestation leads to species extinction in forests.
  e.g: tropical rainforests once covered 14% of the earth’s land surface now not more than 4%.
• Conversion of forest land to agricultural land.
  e.g: the amazon rainforest, called the lungs of our planet, harbouring innumerable species is cut and cleared to cultivate soybeans or convert to grasslands for raising beef cattle.
• Pollution enhances the degradation of habitats and threatens the survival of many species as pollutants change the quality of the environment.
• Fragmentation of habitat leads to population decline.
  e.g: mammals and birds requiring large territories and certain animals with migratory habits are badly affected.

Question 8.
Explain in brief ‘Biodiversity Hot Spots’.
Answer:
Biodiversity hot spots: A Biodiversity hot spot is a Biogeographic Region that is both a significant reservoir of biodiversity and is threatened with destruction.
The concept of biodiversity originated from Norman Myers. There ate about 34 biodiversity hot spots in the world. As these regions are threatened by destruction habitat loss is accelerated.
e.g.: (I) the Western Ghats and Srilanka
(II) Indo Burma
(III) Himalayas in India.

Ecologically unique and biodiversity-rich regions are legally protected as in

• Biosphere Reserves – 14
• National Parks – 90
• Sanctuaries – 448

Biosphere Reserves: An area that is set aside, minimally disturbed for the conservation of the resources of the biosphere is the ‘Biosphere reserve. The latest biosphere reserve (17th biosphere reserve in India) is Seshachalam hills.

National Parks: A National Park is a natural habitat strictly reserved for the protection of natural life. National Parks, across the country, offer a fascinating diversity of terrain, flora, and fauna. Some important National Parks in India are – Jim Corbett National Park (the first National Park in India located in Uttarakhand), Kaziranga National Park (Assam), Kasu Brahmananda Reddy National Park, MahavirHarinaVanasthali National Park (AP). Keoladeo Ghana National Park (Rajasthan), etc.

Sanctuaries: Specific endangered faunal species are well protected in wildlife sanctuaries which permits eco-tourism (as long as animal life is undisturbed). Some, important Sanctuaries in India (AP) include-Koringa Sanctuary, Eturnagaram Sanctuary, and Papikondalu Sanctuary.

Question 9.
Explain the ‘Rivet Popper’ hypothesis.
Answer:
What if we lose a few species? Will it affect man’s life? Paul Ehrlich experiments Rivet popper, hypothesis, taking an aeroplane as an ecosystem, explains how the removal of one by one ‘rivets’ (species of an ecosystem) of various parts can slowly damage the plane (ecosystem)-shows how important a ‘species’ is in the overall functioning of an ecosystem. Removing a rivet from a seat or some other relatively minor important parts may not damage the plane, but the removal of a rivet from a part supporting the wing can result in a crash. Likewise, the removal of a ‘critical species’ may affect the entire community and thus the entire ecosystem.

Question 10.
Write short notes on In-situ conservation.
Answer:
In-situ conservation (On-site conservation): In-situ conservation is the process of protecting an animal species in its natural habitat. The benefit is that it maintains recovering populations in the surrounding where they have developed their distinctive properties. Conservationists identified certain regions by the name ‘Biodiversity hot spots’ for maximum protection as they are characterized by very high levels of species richness & high degree of endemism. By definition ‘A biodiversity hot spot’ is a ‘Biogeographic Region’ with a significant reservoir of biodiversity that is under threat of extinction from humans. They are Earth’s biologically ‘richest’ and ‘most threatened’ Terrestrial Ecoregions.

Andhra Pradesh BIEAP AP Inter 1st Year Chemistry Study Material 13th Lesson Organic Chemistry-Some Basic Principles and Techniques Textbook Questions and Answers.

AP Inter 1st Year Chemistry Study Material 13th Lesson Organic Chemistry-Some Basic Principles and Techniques

Very Short Answer Questions

Question 1.
Write the reagents required for the conversion of benzene to methyl benzene.
Answer:
Benzene reacts with methyl chloride in presence of anhydrous AlCl₃ to form methyl benzene.

This reaction is called as Friedal craft’s alkylation.

Question 2.
How is nitrobenzene prepared.
Answer:
Benzene undergoes nitration with nitration mixture (cone. HNO₃ + Cone H₂SO₄) at less than 60° C to form nitrobenzene.

Question 3.
Write the conformations of ethane.
Answer:

Question 4.
How do you prepare ethyl chloride from ethylene?
Answer:
Ethylene reacts with hydrogen chloride to form Ethyl chloride in presence of anhydrous AlCl₃ catalyst.

Question 5.
Write the IUPAC names of :
a) CH₃ – CH₂ – CH₂ – CH = CH₃
b)
c)
Answer:

Question 6.
Write the structures of Trichloroethanoic acid, Neopentane, P – nitro benzaldehyde. [Mar. ’13]
Answer:
Trichloro ethanoic acid – CCl₃ – COOH

Question 7.
Discuss Lassaigne’s test.
Answer:
Lassaigne’s Test: –

• A small dry sodium piece is taken in a fusion tube and heated gently till the sodium melts.
• To this molten sodium a small amount of organic compound is added and heated until it changes red hot.
• This red hot tube is plunged in a china dish containing water and this content is boiled, cooled and filtered.
• The obtained filtrate is known as Lassaigne’s extract.
• This test is used to detect the elements Nitrogen, Sulphur, halogens etc.,

Detection of Nitrogen : –
The Lassaigne’s extract is made alkaline by adding few drops of dil. NaOH. to this freshly prepared FeSO₄ solution is added and warmed then few drops of FeCl₃ are added followed by acidification with Cone. HQ (or) H₂SO₄. A bluish green colouration indicates the nitrogen.
Na + C + N → NaCN
2NaCN + FeSO₄ → Na₂SO₄ + Fe (CN)₂
Fe(CN)₂ + 4NaCN → Na₄ [Fe(CN)₆]
3Na₄ [Fe(CN)₆] 4- 4FeCl₃ → Fe₄ [Fe(CN)₆]₃ + 12 NaCl (Prussian blue colouration)

Question 8.
Explain the principle of chromatography.
Answer:
Chromatography has been developed as a method of separation components of a mixture generally between the stationary phase and a mobile phase.
Chromatography involves the three steps given below :
a) Adsorption and retention of a mixture of substances on the stationary phase and separation of adsorbed substances by the mobile phase to different distances on the stationary phase.
b) Recovery of the substances separated by a continuous flow of the mobile phase (known as elution) and
c) Qualitative and quantitative analysis of the eluted substances.

Question 9.
Explain why an organic liquid vaporizes at a temperature below it’s boiling point in it’s steam distillation.
Answer:
In steam distillation, the liquid boils when the sum of vapour pressures due to the organic liquid (P₁) and that due to water (P₂) becomes equal to the atmospheric pressure (P) i.e., P = P₁ + P₂.
∴ P₁ is lower than P the organic liquid vaporises at lower temperature than it’s boiling point.

Question 10.
Explain the following :
a) Crystallisation
b) Distillation
Answer:
a) Crystallisation:

• This technique is used for the purification of solid organic compounds.
• This method is based on the difference in the solubilities of the compound and the impurities in a suitable solvent.
• The impure compound is dissolved in a solvent in which it is partially soluble at room temperature and  appreciably soluble at higher temperature.
• The solution is concentrated to get a nearly saturated solution on cooling the solution pure compound crystallises.
• On repeating this process finally gets the very pure compound.

b) Distillation :
This method is used to separate

1. 1. Volatile liquids from non volatile impurities and
   2. The liquids having enough difference in their boiling points.
2. Liquids having different boiling points vapourises at different temperatures.
3. These vapours are cooled and the liquids formed are collected separately.
   Eg : CHCl₃ (b.pt. 334 K) and C₆H₅NH₂ (b.pt. 457 K) are easily separated by distillation technique.

Short Answer Questions

Question 1.
Complete the following reaction and name the products A, B and C. [T.S. Mar. ’15] [A.P. Mar. 16]

Answer:
CaC₂ + 2H₂O → Ca(OH)₂ + C₂H₂
Acetylene (= A)

A = C₂H₂ (Acetylene)
B = C₆H₆ (Benzene)
C = C₆H₅CH₃ (Methyl benzene)

Question 2.
Name the products A, B and C formed in the following reactions. Give the equations for the reactions.

Answer:

A = 1, 2- dibromo ethane
B = Acetylene
C = 1, 1, 2, 2 – tetra bromo ethane

Question 3.
How does acetylene react with : a) Bromine b) Hydrogen ? Write the balanced equations for the above reactions. Name the products.
Answer:
a) Addition of Br₂: Acetylene reacts with bromine to form finally 1, 1, 2, 2 – tetra bromo ethane.

b) Addition of hydrogen : Acetylene undergoes addition reaction with hydrogen in the presence of Ni catalyst and gives ethylene and ethane.

Question 4.
What is substitution reaction ? Explain any two substitutuin reactions of benzene.
Answer:
Substitution : When an atom or a group in molecule is substituted by another atom or group then the reaction is called electrophillic substitution reaction.
e.g. : 1) Friedal craft’s alkylation : C₆H₆ reacts with chlorine in the presence of AlCl₃ and form chloro benzene.

e.g. : 2) Nitration : C₆H₆ when heated with a mixture of cone. HNO₃ and cone. H₂SO₄, below 60° C, to give nitrobenzene.

Question 5.
What is dehydrohalogenation ? Write the equation for the formation of alkene from alkyl halide.
Answer:
Removal of hydrogen and halogen from the adjacent carbon atoms is called dehydrohalogenation.
Alkene from alkyl halide : When ethyl chloride is heated with alcoholic KOH, ethylene is formed due to dehydrohalogenation

Question 6.
Which type of compounds react with Ozone ? Explain with one example.
Answer:
1) Unsaturated hydrocarbons react with ozone and form addition compound called ozonide. On hydrolysis the ozonide gives carbonyl compound. This is called ozonolysis.

2) Unsaturated hydrocarbons usually react with ozone. Ozonolysis is used for the location of the double bond in unsaturated compounds like alkene, alkyne and benzene.
Example : Ethylene undergoes addition reaction with ozone and form ozonide. If, on hydrolysis, in presence of Zn dust, gives formaldehyde and H₂O₂.

Question 7.
Give two examples each for position and functional isomerism. [A.P. Mar. ’16] [Mar. ’13]
Answer:
Position isomerism : This type of isomerism arises due to the difference in the position of substituent group or in the position of multiple bond.
e.g.:

e.g.: 2. CH₃ – CH₂ – C = CH 1 – butyne
CH₃ – C = C – CH₃ 2 – butyne

Functional group isomerism : This type of isomerism arises in carbon compounds having the same molecular formula but with different functional groups.
e.g.: 1. CH₃ – CH₂ – CH₂ – OH 1 – proparal (C₃H₃O)
CH₃ – CH₂ – O – CH₃ methoxy ethane (C₃H₈O)
e.g.: 2. CH₃ – CH₂ – OH Ethanol
CH₃ – O – CH₃ methoxy methane

Question 8.
Explain the mechanism of halogenations of methane.
Answer:
Methane undergoes halogenation (with Cl₂) to form finally tetrachloro methane. This is a substitution reaction and takes place in presence of light.

Mechanism:
Halogenation process involves the free radical chain mechanism
This mechanism involves 3 steps
i) Initiation
ii) Propagation
iii) Termination

i) Initiation : The reaction initiated by the cleavage of chlorine molecule in presence of light. Cl – Cl bond is weaker than the C – C and C – H bonds

ii) Propagation : Chlorine free radical attacks the CH₄ molecule and generates methyl free radical.

The methyl radical formed attacks the 2^nd molecule of chlorine to form another chlorine free radical.

Propagation takes place in several steps as follows.

iii) Termination : The reaction terminated after sometime due to consumption of reactants. The following are the possible chain terminating steps

Question 9.
How is ethylene prepared from ethyl alcohol ? Write the equation.
Answer:
Ethylalcohol reacts with Conc.H₂SO₄ at 170°C to form Ethylene.

The above reaction is dehydration reaction. (H₂O removal takes place)

Question 10.
Explain the reactions of acetylene with : a) Na in NH₃ b) chromic acid. Write the equations and name the products.
Answer:
a) With Na in NH₃ : Acetylene reacts with Na in liquid NH₃ to give monosodium acetylide and disodium acetylide.


b) Chromic acid : Acetylene is oxidised to acetic acid by chromic acid.

Question 11.
Explain crystallisation and sublimation phenomenon which are used in the purification of organic compounds.
Answer:
Crystallisation: The principle used here is that the impurities are insoluble or soluble in the given solvent at any temperature, but the substance to be purified is sparingly soluble at room temperature but highly soluble at highest temp. (i.e.,) almost near the boiling temp. of the solvent.
This method is useful to purify solid compounds.

Sublimation : Some solid substances when heated pass directly into vapour state without melting. Those vapour on cooling form directly solid with out condensing to liquid. This phenomenon is called sublimation.

Impure compound is taken in a beaker covered with a watch glass and heated on an electric plate. The compound sublimates and solidifies on the lower surface of the watch glass. Impurities settle in the beaker. Pure compounds in separated by scratching the watch glass.
This is also a purification method of solids.

Question 12.
Describe solvent extraction method to purify a compound.
Answer:
Solvent extractiân : Suppose an organic compound ‘A’ is more soluble in an organic solvent that in water but present in aq. solution. The aq. solution is shoken with the organic solvent. ‘A’ goes into the organic solvent which is immiscible with water. The organic layer is separated and distilled to remove the liquid solvent. The compound remains with distillation flask.

Question 13.
Explain the estimation of phosphorus and sulphur in the given organic compounds.
Answer:
Estimation of phosphorus : Known mass of organic compound (‘a’ gm) is heated with fuming HNO₃ in a carius tube. ‘P’ is oxidised to H₃PO₄ acid. This acid is precipitated as ammonium phosphomolybdate (‘b’ gm) by adding ammonia and ammonium molybdate solutions.
% g of phosphorus = \(\frac{100}{a}\) × \(\frac{b}{1877}\) × 31 gm
Molecular mass of (NH₄)₃ PO₄. 12 MO O₃ (ammonium phosphomolybdate) = 1877
Estimation of sulphur : Known mass of organic compound (‘a’ gm) is heated with sodium peroxide in a carius tube. If sulphur is present it is oxidised to sulphuric acid. The acid is precipitated as barium sulphate (big) by adding excess of aq. BaCl₂ solution. The ppt. is filtered, washed, dried and weighed.
% g of sulphur = \(\frac{100}{a}\) × \(\frac{b \times 32}{233}\) g
Molecular mass of BaSO₄ = 233

Question 14.
Explain addition of HBr to propene with the ionic mechanism.
Answer:

Electrophilic addition mechanism : (Ionic Mechanism)
Step (i) ; Formation of carbonium ion

Step (ii) : Stable carbonium ion is attacked by Br^⊖

Question 15.
What is the product formed when sodium propionoate is heated with soda lime, (imp)
Answer:
Sodium propionate is heated with sodalime to form ethane. This reaction is decarboxylation reaction. NaOH + CaO is called as Decarboxylating agent.

Long Answer Questions

Question 1.
Explain the classification of hydrocarbons.
Answer:
Classification of hydrocarbons : Hydrocarbons are classified as aliphatic hydrocarbons and aromatic hydrocarbons. Aliphatic hydrocarbons are again classified as open chain hydrocarbons and closed chain hydrocarbons i.e., cyclohydrocarbons. Both open chain and closed chain hydrocarbons are again classified as hydrocarbons containing C – C single bonds. > C = C < compounds, – C = C – hydrocarbons. Aromatic compounds are called benzenoids as they are related to benzene.

Question 2.
Write IUPAC names of the following compounds :
a)
b) CH₂ = CH – C C – CH₃
c) CH₃ – CH = C (CH₃)₂
d) CH₂ – CH₂ – CH = CH₂
e)

Answer:
a) 1, 3 – Buta diene
b) CH₂ = CH – C ≡ C – CH₃
Pent, 1 -ene 3 – yne
c) CH₃ – CH = C (CH₃)₂
2 – Methyl 2 – Butene
d)
e)

4 – ethyl Dec 1, 5, 8 – Triene

Question 3.
Describe two methods of preparation of ethane. Give any three reactions of ethane.
Answer:
Methods of preparation of ethane:
1. Decarboxylation:
Ethane is prepared by heating sodium propionate with sodalime.

2. Kolbe’s electrolysis:
Ethane is obtained by the electrolysis of potassium acetate solution.

H₂ gas evolves at cathode.
Ethane and CO₂ are formed at anode.

Chemical properties of ethane :
1. Halogenation:
Ethane reacts with chlorine in the presence of sunlight. Hydrogen atoms are substituted by halogen atoms successively. The final product is Hexachloro ethane.

2. Nitration:
Ethane reacts with nitric acid at 400° C and gives nitro ethane.

3. Pyrolysis:
When ethane is heated in the absence of oxygen it decomposes giving ethylene and hydrogen.

Question 4.
Write the structural formulas and IUPAC names for all possible isomers having the number of double (or) triple bond as indicated :
a) C₄H₈ (one double bond)
b) C₅H₈ (one triple bond)
c) C₅H₁₂ (No multiple bonds).
Answer:
a) The possible Isomers of C₄H₈ (one double bond)

b) The possible Isomers of C₅H₈ (one triple bond)

c) The possible Isomers of C₅H₁₂ (No multiple bonds)

Question 5.
Write chemical equations for combustion reaction of the following hydrocarbons.
a) Butane
b) Pentene
c) Hexyne.
Answer:
a) Combustion of Butane :
C₄H₁₀ + \(\frac{13}{2}\) O₂ → 4CO₂ + 5H₂O + Energy

b) Combustion of Pentene:
C₅H₁₀ + \(\frac{15}{2}\) O₂ → 5CO₂ + 5H₂O + Energy

c) Combustion of Hexyne:
C₆H₁₀ + \(\frac{17}{2}\) O₂ → 6CO₂ + 5H₂O + Energy

Question 6.
Addition HBr to propene yields 2 – bromopropane, while in the presence of benzoyl peroxide, the same reaction yields 1 – bromo propane. Explain and give mechanism.
Answer:

• Addition of HBr to propene yields 2 – Bromopropene. Here the reaction mechanism is electrophillic. addition mechanism.
• Addition of HBr to propene in presence of benzoyl peroxide the reaction proceeds through free radical mechanism.
  

Electrophilic addition mechanism : (Ionic Mechanism)
Step (i) : Formation of carbonium ion
HBr → H⁺ + Br^⊖

Question 7.
Describe two methods of preparation of ethylene. Give equation for the reactions of ethylene with the following,
a) Ozone
b) Hypohalous acid
c) Cold and dil. alkKMno₄
d) Heated with O₂ at high pressure
Answer:
Preparation of ethylene:
1. Dehydrohalogenation of alkyl halide: When ethylbromide is heated with alcoholic KOH, ethylene is formed.

2. Dehydration of ethanol: When ethanol is heated with cone. H₂SO₄ at 170°C, ethylene is obtained.

Chemical properties :
a) With O₃: Ethylene undergoes addition reaction with ozone and gives a cyclic compound called ozonide. It undergoes hydrolysis in the presence of zinc dust to give formaldehyde.

b) With HOCl : Ethylene reacts with hypochlorous acid and gives ethylene chlorohydrin.

c) Cold and dil. alk. KMnO₄: Pink coloured cold and dil. alkaline KMn04 solution is Bayer’s reagent. Ethylene decolourises Bayers reagent to give ethylene glycol. This is test for unsaturation.

d) Polymerisation : When Ethylene is heated in the presence of 02 at a temperature of 200° C and high pressure of 1500 – 2000 atmospheres gives polyethene.

Question 8.
How does ethylene react with the following reagents ? Give the chemical equations and names of the products formed in the reactions,
a) Hydrogen halide
b) Hydrogen
c) Bromine
d) Water
e) Oxygen in presence of Ag at 200° C.
Answer:
(a) Reaction with hydrogen halide : Ethylene reacts with Hydrogen halides to give ethyl halides.
H₂C = CH₂ + HX → CH₃ – CH₂ X
e.g.: H₂C – CH₂ + HCl → C₂H₅Cl Ethyl chloride

(b) With Hydrogen : Ethylene reacts with H₂ in presence of Ni, Pt or Pd to give ethane.

(c) With Bromine : Ethylene decolourises red coloured Br₂ in CCl₄ to give 1,2- dibromo ethane.

(d) With water : Ethylene reacts with dil. H₂SO₄ to give ethyl alcohol.

(e)With Sulphur monochloride : Ethylene reacts with S₂Cl₂ to give musturd gas.

(f) With O₂ at presence of Ag: Air oxidises ethylene to ethylene oxide known as epoxide in presence of Ag catalyst at 200 – 400°C.

Question 9.
An alkene ‘A’ on ozonolysis gives a mixture of ethanal and pentan 3-one. Write the reaction, structure of the products and alkene – A. Give the IUPAC name of alkene – A.
Answer:

Question 10.
An alkene ‘A1 contains three C – C, eight C – H bonds and one C = C bond. ‘A’ on ozonolysis gives two moles of an aldehyde of molar mass 44o . Write IUPAC name of ‘A’.
Answer:

IUPAC name of ‘A’ is 2 – Butene

Question 11.
Give two methods of preparation of acetylene. How does it react with water and Ozone ?
Answer:
Preparation of acetylene :
1. Dehydrohalogenation : Acetylene is obtained when dibromo ethane is heated with alcoholic KOH.

2. Acetylene is obtained by heating iodoform with silver powder.

3. Addition of water : Acetylene undergoes addition reaction with water in the presence of mercuric sulphate and sulphuric acid. Vinyl alcohol formed in the reaction undergoes rearrangement and gives acetaldehyde . [Mar. 06, June 04]

4. Reaction with ozone : Acetylene reacts with ozone to form an ozonide which on hydrolysis in presence of Zn forms glyoxal. [June 04]

Question 12.
How does acetylene react with the following reagents ? Give the corresponding equations and name the products formed in the reactions
a) Acetic acid
b) Water
c) Hydrogen
d) Halogens
e) Hydrogen halide
f) Ammonical AgNO₃ and Cl₂Cl₂ [T.S. Mar. 16]
Answer:
(a) Reaction with Acetic acid : Acetylene on treatment with acetic acid gives vinyl acetate in the 1st step and then gives ethylidene acetate. Hg²⁺ ions act as catalyst.

(b) Addition of water: Acetylene undergoes addition reaction with water in the presence of mercuric sulphate and sulphuric acid. Vinyl alcohol formed in the reaction undergoes rearrangement gives acetaldehyde.

(c) Addition of hydrogen : Acetylene undergoes addition reaction with hydrogen in the presence of Ni catalyst and gives ethylene and ethane.

(d) Addition of halogens : Acetylene undergoes addition reaction with halogens to give 1, 1, 2, 2 tetra haloethane. Dihalo alkene is formed in the first step which then adds on another molecule of halogen to give tetra halo ethane.

(e) With Hydrogen halide : Acetylene on addition with HCl gives Vinyl chloride and finally 1, 1 – dichloroethane.

(f) With Ammonical AgNO₃ solution : Acetylene gas is passed through ammonical AgNO₃ solution to form a white precipitate of silver acetylide. .

With Ammonical Cu₂Cl₂ solution : Acetylene gas is passed through Ammonical cuprous chloride solution to give a red precipitate of cuprous acetylide.

Question 13.
Describe any two methods of preparation of benzene with corresponding equations. Benzene does not behave like an alkenes why ? How do we get methyl benzene from benzene ?
Answer:
Preparation of Benzene:
1. When sodium benzoate distilled with sodalime. Benzene is formed.

2. Polymerisation of Acetylene : When Acetylene gas is passed through red hot Cu or Fe tubes, it polymerises and gives Benzene.

Constitution of Benzene : Benzene molecule cannot be represented by a single structure. It has the following Resonance structures.

The resonance energy of Benzene is 150.48 KJ/mole. Thus, it has more stability. Hence does not undergo addition reactions like alkenes. Benzene is an aromatic molecule.
Benzene to toluene (Friedal Craft’s Alkylation) :
Benzene reacts with methyl chloride in presence of AlCl₃ and forms methyl benzene (Toluene).

Question 14.
How do we get benzene from acetylene ? Give the corresponding equation. Explain the halogenations, alkylation, acylation, nitration and Sulphonation of benzene.
Answer:
Preparation of Benzene from acetylene: On passing acetylene gas through red hot iron tubes, it trimerises to give benzene.

Chemical Properties:
(1) Halogenation : Benzene reacts with chlorine in the presence of FeCl₃ or AlCl₃ to give chioro-benzene.

(2) Friedel – Craft’s Alkylation : Benzene reacts with alkyl halides in the presence of AlCl₃ to give alkyl benzene.

(3) Friedel – Craft’s Acylation : Benzene reacts with acetyl chloride in the presence of AlCl₃ to give acetophenone.

(4) Nitration : Benzene when heated with a mixture of cone. H₂SO₄ and cone. HNO₃ below 60°C to give Nitrobenzene.

(5) Sulphonation : With fuming H₂SO₄, benzene, reacts to form benzene sulphonic acid.

Question 15.
Explain the differences between structural isomers and stereo isomers.
Answer:
Structural isomerism : When the isomerism is due to difference in the arrangement of atoms within the molecule, without any reference to space, the phenomenon is known as Structural isomerism. In this type of isomerism, the isomers possess the same molecular formula, but different structural formula.
Structural isomerism is further classified into different types :

Stereoisomerism : The stereoisomers have the same molecular and structural formula, but differ in the arrangement of atoms or groups in space. Thus the phenomenon exhibited by two or more compounds with the same molecular and structural formulae, but different spatial arrangement of atoms or groups. The spatial arrangement of atoms or groups is also referred to as configuration of the molecule. It is further classified into three types.

Question 16.
What is the differences between conformation and configuration in open chain molecules.
Answer:
Configurational Isomerism : (Optical and geometrical isomerism): These are the stereo – isomers which are discrete, stable and isolable substances. They cannot be inter-converted into one another without making and breaking of new bonds. These isomers cannot be superimposed on each other.
These isomers are further classified as enantiomers, diastereomers and geometrical isomers.

Conformational Isomers : These are the stereo – isomers that easily convert into one another by the rotation around “C – C” bonds. These are in dynamic equilibrium with one another and cannot be separated under ordinary conditions.
This type of isomerism is in alkanes such as n – butane.

Question 17.
What do you understand about geometrical isomerism ? Explain the geometrical isomers of 2 – butene.
Answer:
Geometrical isomerism (Cis – trans isomerism): The isomers which possess the same structural formula but differ in spatial arrangement of the groups around the double bond are called geometrical isomers and the phenomenon is known as geometrical isomerism.
When the same groups lie on the same side of double bond, it is Cis – isomer, while when the same groups lie on opposite side of double bond, the isomer is trans.

Question 18.
Explain the method of writing E – Z configurations for geometrical isomers taking CHCl = CFBr as your example.
Answer:
E – Z configurations for geometrical isomers : The following procedure is followed in specifying the configurations of compounds.
i) Arrange the atoms / groups attached to each doubly bonded carbon in the order of their atomic numbers.

ii) Choose the atom/group higher priority on each doubly bonded carbon. If the atoms/groups of higher priority on each carbon are on the same side of the molecule, the letter “Z” is used to denote the configuration of such isomer. When the atoms/groups of higher priority on each carbon are on the opposite sides of the molecule, the letter ‘E’ is prefixed before the name to indicate configuration.
Example : CHCl = CFBr
Among H and Cl, ₁₇Cl gets more priority than ₁H
Among H and Br, ₃₅Br gets more priority than ₉F

Question 19.
If an alkene contains on carbons at double bond Cl. Br and – CH₂ – CH₂ – OH – CH (CH₃)₂, Write the E and Z configurations of it.
Answer:
From the given data, the structural formulae of the molecule is

Question 20.
Write a note on :
a) Distillation
b) Fractional distillation
c) Distillation under reduced pressure
d) Steam distillation.
Answer:
a) Distillation : This process is useful for the purification of liquids contaminated with nonvolatile impurities. The impure liquid is boiled in a distillation flask and the vapours are condensed and collected in a receiver. This method can also be used to separate liquids if only their boiling points differ by above 40°C. However, in the case of liquids that have boiling point difference less than 40°C fractional distillation method is used.

b) Fractional distillation : Fractional distillation for liquids that have B.Pt difference less than 40°C. The technique here is that vapours of liquid mixture. When pass through long fractionating column vapours of the liquid with high B.Pt condense and those with low B.Pt pass over through the condenser get condensed and collected in the receiver.
Here, long tubes having different shapes and designs to fit for particular requirements are used.

These are called fractionating columns. The liquid mixture is taken in a distillation flask i.e., fitted with a fractionating column at its mouth. At the upper end of the column, there is a provision to connect it to the water condenser.

c) Distillation under reduced pressure : This method is useful to purify liquids that have very high boiling points and those which decompose at or below their boiling points. If the external pressure is reduced the liquid boils at lower temperature than its normal boiling point without decomposition.

d) Steam distillation : Here liquids which are immiscrible in water passes high boiling point and steam volatile are purified. In this method steam is passed into the hot liquid that is to be purified. The mixture of steam and vapour of volatile organic compound come out. This is because of the sum of the vapour pressures of steam and the liquid to be purified become equal to the atmospheric pressure. They are passed through the condenser and condensed and finally collected in the receiver. The water layer and the organic liquid layer are separated using a separating funnel.

Question 21.
Write a brief note on chromatography.
Explain the following :
a) Column chromatography
b) Thin layer chromatography
c) Partion chromatography.
Answer:
Chromatography has been developed as a method of separation components of a mixture generally between the stationary phase and a mobile phase.
Chromatography involves the three steps given below :
a) Adsorption and retention of a mixture of substances on the stationary phase and separation of adsorbed substances by the mobile phase to different distances on the stationary phase.

b) Recovery of the substances separated by a continuous flow of the mobile phase (known as elution) and

c) Qualitative and quantitative analysis of the eluted substances.

Classification :
Two general chromatography techniques are discussed below. They are :

1. adsorption chromatography
2. partition chromatography.

Adsorption chromatography is based on the adsorption of different compounds on an adsorbent to different degrees. Generally used adsorbents are silica gel and alumina. A mobile phase is allowed to move over a stationary phase, the adsorbent. The components of the mixture move to different distances over the stationary phase.

Differential adsorption principle is used in
a) column chromatography and
b) thin layer chromatography.

Column chromatography : In the column chromatography the components of a mixture are separated by a column of adsorbent packed in a glass tube. The column is fitted with a stopcock at its lower end. The mixture to be adsorbed on the adsorbent is placed at the top of the stationary phase. A suitable eluant, either a single solvent or a mixture of solvents is allowed to flow down the column slowly. Depending on the degree to which the compounds are adsorbed the components are separated. The most readily absorbed substances are retained near the top and other come down accordingly to various distances.

Thin layer chromatography (TLC): This also involves adsorption differences. Here the adsorbent, say silica gel or alumina is coated over a glass plate of suitable size in thin layer. The plate is called TLC plate or chromoplate. The solution of the mixture to be separated is applied as a small spot at about 2 cm from the bottom of the plate. The plate is then kept in a closed jar containing the eluant. As the eluant rises up the plate, the components of the mixture move up along with the eluant to various distances depending on their degree of adsorption.

The relative adsorption of a component of the mixture is expressed in terms of its RETARDATION FACTOR (R_f) value.
R_f = \(\frac{\text { Distance moved by the subs tance from base line }(\mathrm{X})}{\text { Dis tance moved by the solvent from base line }(\mathrm{Y})}\)
Partition chromatography : This is based on continuous differential partitioning of components of a mixture between the Stationary phase and the mobile phase. In paper chromatography, for example, a special paper called chromatography paper contains water trapped in it which acts as the stationary phase. The chromatography paper spotted with the solution of the mixture at the base is suspended in a suitable solvent or a mixture of solvents.

This solvent acts as the mobile phase. The solvent rises up the paper by capillary action and moves over the spot. The paper selectively retains different components as per their differing partition in mobile and stationary phase. The paper strip so developed is known as chromatogram. The spots of the separated coloured compounds are detected and for colourless compounds other methods like spraying suitable reagent, are used.

Question 22.
Explain the estimation of nitrogen of an organic compound by
a) Duma’s method
b) Kjeldahl’s method.
(or)
Describe Duma’s and Kjeldahl’s method for the estimation of Nitrogen.
Answer:
Estimation of nitrogen : There are two methods to estimate nitrogen in the given organic compound known as
i) Duma’s method and
ii) Kjeldahed’s method.

i) Duma’s method : In this method a known weight of organic compound is heated strongly with coarse cupric oxide. Carbon and hydrogen get oxidised to carbon dioxide and water vapour respectively. Nitrogen if present is converted to nitrogen gas. Even if some nitrogen is converted to its oxides. They are reduced by hot copper gauze to nitrogen. The product gases are collected over a solution of potassium hydroxide. CO₂ is absorbed by KOH solution. Nitrogen is collected over potassium hydroxide solution and its volume is found out.

Suppose that ‘a’ g of organic compound gives V₁ ml of N₂ at room temperature TK. If atmosphere pressure is P and aqueous tension at TK is P, then the pressure of nitrogen gas at TK is (P – p). Let (P – p) = P₁ Reduce the volume of nitrogen to standard temperature 273 K and standard pressure 760 mm.
Volume of nitrogen at STP is v = \(\frac{\mathrm{P}_1 \mathrm{~V}_1}{\mathrm{~T}_1} \times \frac{273}{760}\)
28 g of nitrogen at STP occupies 22400 ml
? g. of nitrogen at STP occupies Vml. of nitrogen.
28 × \(\frac{V}{22400}\)g
‘a’ g. of organic compound \(\frac{28 \times V}{22400}\) g of nitrogen.
100 g organic compound has ? of nitrogen = \(\frac{100}{a} \times \frac{28 \times V}{22400}\) g

ii) Kjeldah’s method : This is another method to estimate nitrogen. In this method, the compound is heated with concentrated sulphuric acid in the presence of small amount of CuSO₄.
Nitrogen is quantitatively converted into ammonium sulphate. The contents of the flask are transferred to another flask and heated with excess of sodium hydroxide solution to liberate ammonia gas. Ammonia gas so liberated is passed and absorbed in a known volume of known concentrated sulphuric acid that is relatively more in amount than that is required to neutralise NH₃ gas. Now, the excess of acid remained after the neutralisation by NH₃ is titrated against a standard solution of alkali. From the above, the amount of H₂SO₄ used to neutralise NH₃ is calculated. From this, the mass of ammonia formed is calculated and from that percentage of nitrogen is calculated.
Organic compound + H₂SO₄ →(NH₄)₂ SO₄
(NH₄)₂SO₄ + 2 NaOH → Na₂SO₄ + 2 H₂O + 2 NH₃
2 NH₃ + H₂SO₄ → (NH₄)₂SO₄
Calculation :
Let the mass of organic compound taken be ‘a’ g.
Let the volume of H₂SO₄ initially taken be ‘V_ml‘ and its molarity M.
After passing the NH₃ gas into the above acid, if the remaining acid is titrated with M molar NaOH and it consumes V₁ ml. of NaOH for complete neutralisation, then from the formula
\(\frac{\mathrm{MV}_1}{\mathrm{n}_1}\) (NaOH) = \(\frac{\mathrm{MV}_2}{\mathrm{n}_2}\) (H₂SO₄)
From the stoichiometric equation
2 NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O
n₁ = number of moles of NaOH = 2; n₂ = number of moles of H₂SO₄ = 1
\(\frac{\mathrm{MV}_1}{2}\) = \(\frac{\mathrm{MV}_2}{1}\) or V₂ = \(\frac{\mathrm{V}_1}{2}\) ml
Therefore, the volume of H₂SO₄ neutralised by NH₃ is [V – \(\frac{\mathrm{V}_1}{2}\)] ml
(or) it is equal to 2 [V – \(\frac{\mathrm{V}_1}{2}\)] ml. of M molar NH₃ solution.
1000 ml. of 1M NH₃ solution contains 17g. of NH₃ or 14g. of N₂
2[V – \(\frac{\mathrm{V}_1}{2}\)] ml ml. of ‘M’ NH₃ solution contains
\(\frac{14 \times M \times 2\left[V-\frac{V_1}{2}\right]}{1000}\) g. of nitrogen.
Percentage of nitrogen = \(\frac{14 \times M \times 2\left[V-\frac{V_1}{2}\right]}{1000} \times \frac{100}{a}\)

Question 23.
Explain inductive effect with a suitable example.
Answer:
Inductive effect: The electron donating or electron with drawing effect of a groups an atom that is transmitted by the polarisation of electrons in σ bonds is called inductive effect.
Illustration : Consider the molecule CH₃ – CH₂ – CH₂ – Cl. There is a ‘σ’ covalent bond between carbon atom and chlorine atom. The electron pair between them is not equally shared. The more electronegative chlorine atom tends to attract the shared pair more towards itself. Due to this, the electron density tends to be greater nearer chlorine atom than carbon atom. It is generally represented
as But, carbon atom bonded to chlorine atom is itself attached to other carbon atoms. Therefore, the effect can be transitted further
“Inductive effect is defined as the polarisation of a bond caused by the polarization of adjacent σ bond”.
(-I) effect i.e., electron with drawing effect is in the order
⁺NH₃ > NO₂ > CN > SO₃H > CHO > CO > COOH > COCl > COOR > CONH₂ > F > G > Br > I > OH > OR > NH₂ > C₆H₅ > H.
+ I effect is – \(\bar{N} R\) > – \(\overline{O}\) ; – Se > S > -O ; – C(CH₃)₃ > – CH(CH₃)₂ > – CH₂CH₃ > – CH₃.

Question 24.
Write a note on mesomeric effect.
Answer:
Mesomeric effect:
“The electron pair displacement caused by an atom or group along a chain by a conjugative mechanism is called the mesomeric effect of that atom or group”.
Salient features of the mesomeric effect:
i) Permanent effect operating in the ground state of the molecule.
ii) Lone pairs and n electrons are involved and operate through conjugative mechanism of electron displacement.
iii) It influences the physical properties, reaction rates etc.

Groups which tend to increase the electron density of the rest of the molecule are said to have (+M) effect. Such groups tend to posses lone pairs of electrons.
e.g. :
Groups that decrease the electron density of the rest of the molecule are said to have (-M) effect. Unsaturated groups having polar character have – M effect.
. Here C = O group decreases the electron density of the remaining molecule. It has – M effect.
+ M effect is – F > – Cl > – Br > – I;
-NR₂ > OR > F;
– NH₂ > – OH > – F;
– OR > – SR > SeR
– O > – OR
– M effect is = O > = NR > = CR₂
= NR₂< =NR ≡ N > ≡ CR

Question 25.
Describe resonance effect with one example.
Answer:
Resonance effect: It is the polarity produced in a molecule by the interactions of two n bonds or between a π bond and a lone pair of electrons present on adjacent atoms. This effect is transmitted through the chain.
If the transfer of electron is away from the atoms or substituent group attached to the conjugated system, then the molecule gets some of its positions high electron density as in aniline and it is given (+ R). If the shift of electrons are towards the atom or substituent group it is (- R) as in nitrobenzene.

Groups showing (+ R) are X, – OH, – OR, – COOR, – NH₂, – NHR, – NR₂, – NHCOR etc.
(- R) effect are – COOH, CHO, > C = O, – CN, – NO₂.

Question 26.
Explain how many types of organic reactions are possible.
Answer:
Organic reactions are mainly classified into four types known as
i) Addition rections.
ii) Substitution reactions
iii) Elimination reactions
iv) Molecular rearrangements.

i) Addition reactions : In these reactions the reagent and the substract combine together to give a single product.
e.g.:
Depending on the reagent added in the slow rate determining step, addition reactions are again a) Electrophilic classified addition reactions, b) Nucleophilic addition reactions c) Free radical addition reactions.

ii) Substitution reactions : In these reactions on atom or a group of the substrate species is replaced by another atom or group. These are again classified as a) Electrophilic substitution b) Nucleophilic substitution and c) Free radical substitution reactions on the basis of the reagent involved in the rate determining step.
e.g.-: \(\overline{\mathrm{O}}\)H_(aq) + R – X → HO – R + X_(aq)

iii) Elimination reactions : In these reactions two or more atoms or groups of an organic substrate are removed to form multiple bonds.

iv) Molecular rearrangements : Here one organic species (generally less stable) rearranges to other species (generally more stable). For example. Fries rearrangement.

Question 27.
Write the possible conformations of ethane and explain which is more stable.
Answer:
Conformation of ethane : The conformational isomers of a given alkane are obtained by rotation about C – C bond and they are represented by Newman projections or line – Wedge or Sawhorse projections. Newman projections and energy diagrams for the various conformations of ethane are given below.

Staggered (S) occurs at Dihedral angle 60°, 180°, 300°
Rotation about the carbon – carbon bond in ethane is though very rapid, not completely free. Two conformations of ethane known as staggered conformation (S) and the eclipsed conformation (E) are very important though infinite number of confomation are possible.

The C – H bonds in the staggered conformation are arranged so that each one bisects the angle defined by two C – H bonds on adjacent carbon. In the eclipsed conformation each C – H bond is aligned with a C – H bond on adjacent carbon.

In the staggered one the distance between the hydrogen nuclei is 2.55 A° but in eclipsed 2.29 A°. The staggered and elipsed conformations are interconvertible by rotation of one carbon with respect to the other around the a bond that connects them. Different conformations of the same molecule are also called conformers or rotamers.
Stability :
Staggered (S) form is more stable than eclipsed. Because in case of eclipsed form electronic repulsions are high and in case of staggered form electronic repulsions are minimum.

Question 28.
Explain aromatic electrophillic substitution reactions of benzene.
Answer:
1) Halogenation : Benzene reacts with bromine or chlorine in the presence of Lewis acids like FeCl₃, AlCl₃ etc., to give corresponding halobenzene.
e.g. :

Similarly with bromine, bromobenzene is formed.

2) Nitration : Benzene undergoes nitration when heated with a mixture of 1 : 1 (by volume) concentrated nitric acid and concentrated sulphuric acid (nitration mixture) at a temperature below 60° C.

3) Benzene reacts with fuming sulphuric acid (oleum) and gives benzene sulphonic acid.

4) Friedel – Craft’s alkylation and acylation : Benzene reacts with alkyl halides and acylh in the presence of Lewis acids (AlCl₃, FeCl₃) and gives alkyl benzenes and acyl benzenes.

Benzene reacts with acetyl chloride in the presence of anhydrous aluminium, chloride and acetophenone.

General Mechanisms : Electrophilic substitution reaction (S_E) proceeds in two steps as
I. Generation of electrophile
II. a) Formation of carbocation intermediate
b) Removal of proton from carbocation intermediate.

I. Generation of electrophile E⁺: In the reactions halogenation, alkylation and acylatio benzene, anhydrous AlCl₃, the Lewis acid produces electrophile X⁺ say Cl⁺, R⁺ and RCO⁺ by reacting the reagent chlorine, alkyl halide and acylhalide respectively.

II. a) Formation of carbocation : Electrophile generated above attacks one of the benz carbons to change it to sp3 hybridised. The carbocation (Arrhenium ion) is stabilised through resonal.

Sigma complex loses aromatic character due to delocalisation of electrons stopping at sp³ carb

b) Losing of proton : To regain aromatic character the \(\stackrel{+}{\mathrm{C}}\) loses one proton to sp³ carbon on attack
of (AlCl₄)^– in case of halogenation, alkylation and acylation and HSO₄^– in case of nitration.

Question 29.
Explain electrophilic addition reactions of ethylene with mechanism.
Answer:
Electrophilic addition reactions of ethylene :
1. Addition of Hydrogen : Ethylene react with hydrozen in the presence of Pt, Pd or Ni catalyst to form ethane.

2. Addition of halogens : Halogens (Cl₂ or Br₂) react with ethylene in the presence of an inert solvent like CCl₄ to form dihalo derivative

Mechanism :

3. Addition of hydrogen halides : Ethylene reacts with HBr to form ethyl bromide

4. Addition of water : In presence of few drops of cone. H₂SO₄, water adds to ethylene to form alcohol.

Question 30.
With the help of mechanism explain free radical halogenations of alkanes.
Answer:
Methane undergoes halogenation (with Cl₂) to form finally tetrachloro methane. This is a substitution reaction and takes place in presence of light.

Mechanism :

• Halogenation process involves the free radical chain mechanism
• This mechanism involves 3 steps
  i) Initiation
  ii) Propagation
  iii) Termination

i) Initiation : The reaction initiated by the cleavage of chlorine molecule in presence of light. Cl – Cl bond is weaker than the C – C and C – H bonds

ii) Propagation : Chlorine free radical attacks the CH₄ molecule and generates methyl free radical.

The methyl radical formed attacks the 2^nd molecule of chlorine to form another chlorine free radical.

iii) Termination : The reaction terminated after sometime due to consumption of reactants.
The following are the possible chain terminating steps

Question 31.
Discuss Markownikov’s rule and Kharash effect.
Answer:
Statement of Markownikoff’s rule : The rule states that when an unsymmetrical reagent adds to a double bond, the positive part of the adding reagent attaches itself to a carbon of the double bond so as to yield the more stable carbocation as an intermediate.

Mechanism : A pair of electrons from the double bond attacks the electrophilic HX to produce an achiral trigonal planar carbocation intermediate. Thehalideion X then adds to either face of the positively charged carbon forming alkyl halide product.

Anti Mark Kownikoffs addition or peroxide effect or Kharasch effect : In the presence of peroxide (R – O – O – R) the addition of HBr to unsymmetrical alkene like propene takes place against Markowni- koff’s rule. As per Anti Markownkoff’s rule, the addition of HBr to an unsymmetrical alkene like propene takes place in such a way that the hydrogen atom becomes attached to the carbon atom with the fewer hydrogen atoms.


The 2° free radical is more stable than 1°. Therefore. 1 – bromopropane is major product.

Question 32.
How would you convert benzene in to following compounds ?
a) Chioro benzene
b) Toluene
c) p – nitrotoluene.
Answer:
a) Benzene reacts with chlorine inpresence of FeCl₃ to form chlorobenzene.

b) Benzene reacts with methyl chloride in presence of AlCl₃ to form toluene (Friedal Craft’s alkylation)

c) From benzene p – nitro Toluene is obtained as follows.

Question 33.
Why is wurtz reaction not preferred for the preparation of alkanes containing odd number of carbon atoms ? Illustrate with one example.
Answer:
Wurtz reaction Alkyl halides reacts with sodium metal in presence of dry ether to form higher alkanes.
R – X + 2Na + R – X R – R + 2NaX

• Wurtz reaction is preferred to prepare even no.of Carbons containing alkanes but not for odd no.of Carbons containing Alkanes.
• For getting odd no.of Carbons containing alkanes we have to consider two different alkyl halides.

The products obtained are in mixture form, So the percentage of the desired hydrocarbon is less, i.e., yield of the desired product is low.
Methane can not be prepared by this reaction.

Question 34.
Write the equations involved in the detection of Nitrogen, Halogens and sulphur in organic compounds.
Answer:
Lassaigne’s Test : –

• A small dry sodium piece is taken in a fusion tube and heated gently till the sodium melts.
• To this molten sodium a small amount of organic compound is added and heated until it changes red hot.
•  This red hot tube is plunged in a china dish containing water and this content is boiled, cooled and filtered.
• The obtained filtrate is known as Lassaigne’s extract.
• This test is used to detect the elements Nitrogen, Sulphur, halogens etc..

Detection of Nitrogen : –
The Lassaigne’s extract is made alkaline by adding few drops of dol. NaOH. to this freshly prepared FeSO₄ solution is added and warmed then few drops of FeCl₃ are added followed by acidification with Cone. HCl (or) H₂SO₄. A bluish green colouration indicates the nitrogen.
Na + C + N →NaCN
2NaCN + FeSO₄ → Na₂SO₄ + Fe (CN)₂
Fe(CN)₂ + 4NaCN → Na₄ [Fe(CN)₆]
3Na₄ [Fe(CN)₆] + 4FeCl₃ → Fe₄ [Fe(CN)₆]₃ + 12 NaCl (Prussian blue colouration)

b) Detection of Halogens : The sodium fusion extract is acidified with HNO₃ and is treated with AgNO₃ solution.
Ag + X → AgX

• White PPt indicates Cl^Θ ion
• Pale yellow PPt indicates Br^Θ ion
• Yellow PPt indicates I^Θ ion

c) Detection of Sulphur : To the sodium extract freshly prepared sodium nitroprusside solution. A deep violet colouration takes place.
S^-2 + [Fe(CN)₅NO]²⁺ → [Fe(CN)₅ NOS]^4- Violet

Question 35.
Explain how carbon and Hydrogen are quantitatively determined in an organic compound.
Answer:
Estimation of carbon and hydrogen : Both the elements are estimated in the same experiment simultaneously. A known weight of the organic substance is taken and completely burnt in excess of air and copper (II) oxide carbon changes to CO₂ and Hydrogen to H₂O.
C_xH_y + (x + \(\frac{y}{4}\))O₂ → x CO₂ + (\(\frac{y}{2}\)) H₂O
CO₂ and H₂O SO obtained are passed through already weighed ‘U‘ – tubes containing anhy, CaCl₂ and caustic potash respectively. The increased weigh is of these two tubes give the weigh of H₂O formed and weight of CO₂ formed.
Suppose ‘a’ g of organic compound on combustion gives ‘b’ grams of water vapour and. ‘C g of CO₂. Then
% g of Carbon = \(\frac{12}{44}\) × \(\frac{100}{a}\) × C g
% g of Hydrogen = \(\frac{2}{18}\) × \(\frac{100}{a}\) × b g

Question 36.
How do you determine sulphur, phosphorous and Oxygen are determined quantitatively in an organic compound ?
Answer:
Estimation of phosphorus : Known mass of organic compound (‘a’ gm) is heated with fuming HNO₃ in a carius tube. ‘P’ is oxidised to H₃PO₄ acid. This acid is precipitated as ammonium phosphomolybdate (‘b’ gm) by adding ammonia and ammonium molybdate solutions.
% g of phosphorus = \(\frac{100}{a}\) × \(\frac{b}{1877}\) × 31 gm
Molecular mass of (NH₄)₃ PO₄. 12 MO O₃ (ammonium phosphomoly bdate) = 1877
Estimation of sulphur : Known mass of organic compound (‘a’ gm) is heated with sodium peroxide in a carius tube. If sulphur is present it is oxidised to sulphuric acid. The acid is precipitated as barium sulphate (big) by adding excess of aq. BaCl₂ solution. The ppt. is filtered, washed, dried and weighed.
%g of sulphur = \(\frac{100}{a}\) × \(\frac{b \times 32}{233}\) × g
Molecular mass of BaSO₄ = 233
Estimation of Oxygen : The percentage of oxygen in an organic compound is usually found by difference between the total percentage composition (100) and the sum of the percentages of all other elements. However, oxygen can also be estimated directly as follows :

A definite mass of an organic compound is decomposed by heating in a stream of nitrogen gas. The mixture of gaseous products containing oxygen is passed over red-hot coke when all the oxygen is converted to carbon monoxide. This mixture is passed through warm iodine pentoxide (I₂O₅) when carbon monoxide is oxidised to carbon dioxide producing iodine.

On making the amount of CO produced in equation (A) equal to the amount of CO used in equation (B) by multiplying the equations (A) and (B) by 5 and 2 respectively; we find that each mole of oxygen liberated from the compound will produce two moles of carbondioxide.
Thus 88 g carbon dioxide is obtained if 32 g oxygen is liberated.
Let the mass of organic compound taken be mg
Mass of carbon dioxide produced be m₁ g
∴ m₁ g carbon dioxide is obtained from \(\frac{32 \times m_1}{88}\) g O₂
∴ Percentage of oxygen = \(\frac{32 \times m_1 \times 100}{88 \times m}\)%

Question 37.
Explain Carius method for the determination of Halogens quantitatively in an organic compound.
Answer:
Estimation of halogens : (Carius method) Known mass of organic compound is heated with framing nitric acid in the presence of AgNO₃ in a hard glass tube called carius tube. Carbon and hydrogen of the compound are oxidised to CO₂ and H₂O. Halogens forms silver halide. So obtained silver halide (Agx) is filtered, washed, dried and weighed.
% g of halogen = \(\frac{100 \times b \times \text { atomic mass of ‘ } x \text { ‘ }}{a \times \text { molecular mass of } \mathrm{Agx}}\)
Where, a = mass of organic compound, b = mass of Agx formed
x = halogen atom.

Question 38.
What is carcinogenicity ? Explain with two examples.
Answer:
Benzene and several polynuclear hydrocarbons like 1, 3 – benzanthracene, 3 – methyl cholanthrene, 1, 2- benzpyrene etc., are toxic and said to carcinogenic (cancer producing).
Most of these are formed due to incomplete combustion of tobacco, coal, petroleum etc.
They undergo various chemical changes in human body and finally damage DNAto cause cancer.

Solved Problems

Question 1.
How many a and it bonds are present in each of the following molecules?
(a) HC ≡ CCH = CHCH₃
(b) CH₂ = C = CHCH₃
Solution:
(a) σ_C-C : 4; σ_C-H : 6; π_C=C: 1 ; π C ≡ C : 2
(b) σ_C-C : 3; σ_C-H : 6; π_C=C : 2.

Question 2.
What is the type of hybridisation of each carbon in the following compounds?
(a) CH₃Cl
(b) (CH₃)₂CO
(c) CH₃CN,
(d) HCONH₂
(e) CH₃CH = CHCN
Solution:
(a) sp³,
(b) sp³, sp²,
(c) sp³, sp,
(d) sp²,
(e) sp³, sp², sp², sp

Question 3.
Write the state of hybridisation of carbon in the following compounds and shapes of each of the molecules.
(a)H₂C = O,
(b) CH₃F,
(c) HC ≡ N.
Solution:
(a) sp² hybridised carbon, trigonal planar;
(b) sp³ hybridised carbon, tetrahedral;
(c) sp hvbridised carbon, linear.

Question 4.
Expand each of the following condensed formulas into their complete structural formulas.
(a)CH³CH²COCH²CH³
(b) CH³CH = CH(CH²)³CH³
Solution:

Question 5.
For each of the following compounds, write a condensed formula and also their bond – line formula.
(a) HOCH₃CH₂CH₂CH(CH₃)CH(CH₃)CH₃
(b)

Solution:
Condensed formula:
(a) HO(CH₂)₃CH(CH₃)CH(CH₃)₂
(b) HOCH(CN)₂
Bond-line formula :

Question 6.
Expand each of the following bond-line formulas to show all the atoms including carbon and hydrogen.

Solution:

Question 7.
Structures and IUPAC names of some hydrocarbons are given below. Explain why the names given in the parentheses are incorrect.

Solution:
(a) Lowest locant number, 2, 5, 6 is lower than 3, 5, 7,
(b) substituents are in equivalent position; lower number is given to the one that comes first in the name according to alphabetical order.

Question 8.
Write the IUPAC names of the compounds i-iv from their given structures.

Solution:

• The functional group present is an alcohol (OH). Hence the suffix is ‘-01’.
• The longest chain containing – OH has eight carbon atoms. Hence the corresponding saturated hydrocarbon is octane.

The – OH is on carbon atom 3. In addition, a methyl group is attached at 6th carbon. Hence, the systematic name of this compound is 6-Methyloctan-3-ol.

i)

Solution:
The functional group present is ketone (>C = 0), hence suffix ‘-one’. Presence of two keto groups is indicated by ‘di’, hence suffix becomes ‘dione’. The two keto groups are at carbons 2 and 4. The longest chain contains 6 carbon atoms, hence, parent hydrocarbon is hexane. Thus, the systematic name is Hexane – 2, 4 – dione.

ii)

Solution:
Here, two functional groups namely ketone and carboxylic acid are present. The principal functional group is the cat ooxyuc acid group; hence the parent chain will be suffixed with ‘oic’ acid. Numbering of the chain starts from carbon of – COOH functional group. The keto group in the chain at carbon 5 is indicated by ‘oxo’. The longest chain including the principal functional group has 6 carbon- atoms; hence the parent hydrocarbon is hexane. The compound is, therefore, named as 5 Oxo- hexanoic acid.

iii)

Solution:
The two C=C functional groups are present at carbon atoms 1 and 3, while the C = C functional group is present at carbon 5. These groups are indicated by suffixes ‘diene’ and ‘yne’ respectively. The longest chain containing the functional groups has 6 carbon atoms; hence the parent hydrocarbon is hexane. The name of compound, therefore, is Hexa-1, 3-dien-5-yne.

Question 9.
Derive the structure of
(i) 2-Chlorohex- ane,
(ii) Pent-4-en-2-ol,
(iii) 3- Nitro-cydohexerte,
(iv) Cyclohex-2-en-l-ol,
(v) 6-Hydroxy-heptanal.
Solution:
(i) ‘hexane’ indicates the presence of 6 carbon atoms in the chain The functional group chloro is present at carbon 2. Hence, the structure of the compound is CH₃CH₂CH₂CH₂CH(Cl)CH₃.

(ii) ‘pent’ indicates that parent hydrocarbon contains 5 carbon atoms in the chain, ‘en’ and ‘ol’ correspond to the functional groups C=C and -OH at carbon atoms 4 and 2 respectively. Thus, the structure is CH₂ = CHCH₂CH (OH)CH₃.

(iii) Six membered ring containing a carbon- carbon double bond is implied by cyclo-hexene, which is numbered as shown in (I). The prefix 3-nitro means that a nitro group is present on C-3. Thus, complete structural formula of the compound is (II). Double bond is suffixed functional group whereas -NO₂ is prefixed functional group therefore double bond gets preference over -NO² group:

(iv) ‘1 -0l’ means that a -OH group is present at C-1. OH is suffixed functional, group and gets preference over C = C bond. Thus the structure is as shown in (II) :

(v) ‘heptanal’ indicates the compound to be an aldehyde containing 7 carbon atoms in the parent chain. The ‘6-hydroxy’ indicates that -OH group is present at carbon 6. Thus, the structural formula of the compound is : CH₃CH(OH)CH₂CH₂CH₂CH₂CHO. Carbon atom of – CHO group is included while numbering the carbon chain.

Question 10.
Write the structural formula of :
(a) o-Ethylanisole
(b)p-Nitroaniline,
(c) 2,3 – Dibromo -1 – phenylpentane,
(d) 4-Ethyl – 1 – fluoro-2-nitrobenzene.
Solution:

Question 11.
Using curved-arrow notation, show the formation of reactive intermediates when the following covalent bonds undergo heterolytic cleavage.
(a) CH₃ – SCH₃,
(b) CH₃ – CN,
(c) CH₃ – Cu
Solution:

Question 12.
Giving justification, categorise the fol-lowing molecules/ions as nucleophile or electrophile :
HS^–, BF₃, C₂H₅O^–, (CH₃)₃ N :,

Solution:
Nucleophiles: HS^–, C₂H₅O^–, (CH₃)₃ N:, H²N^– These species have unshared pair of electrons, which can be donated and shared with an electrophile.
Electrophiles:
Reactive sites have only six valence electrons; can accept electron pair from a nucleophile.

Question 13.
Identify electrophilic centre in the fol-lowing: CH₃CH = O, CH₃CN, CH₃I.
Solution:
Among , the starred carbon atoms are electrophilic centers as they will have partial positive charge due to polarity of the bond.

Question 14.
Which bond is more polar in the following pairs of molecules:
(a) H₃C – H, H₃C – Br
(b) H₃C – NH₂, H₃C – OH
(c) H₃C – OH, H₃C – SH
Solution:
(a) C – Br, since Br is more electronegative than H, (b) C – O, (c) C – O

Question 15.
In which C – C bond of CH₃CH₂CH₂Br, the inductive effect is expected to be the least ?
Solution:
Magnitude of inductive effect diminishes as the number of intervening bonds increases. Hence, the effect is least in the bond between . carbon-3 and hydrogen.

Question 16.
Write resonance structures of CH₃COO^– and show the movement of electrons by curved arrows.
Solution:
First, write the structure and put unshared pairs of valence electrons on appropriate atoms. Then draw the arrows one at a time moving the electrons to get the other structures.

Question 17.
Write resonance structures of CH₂ = CH – CHO. Indicate relative stability of the contributing structures.
Solution:
[I: Most stable, more number of covalent

bonds, each carbon and oxygen atom has an octet and no separation of opposite charge II: negative charge on more electronegative atom and positive charge on more electropositive atom; III: does not contribute as oxygen has positive charge and carbon has negative charge, hence least stable].

Question 18.
Explain why the following two structures, I and 11 cannot be the major contributors to the real structure of CH₃COOCH₃.

Solution:
The two structures are less important contributors as they involve charge separation. Additionally, structure I contains a carbon atom with an incomplete octet.

Question 19.
Explain why is more stable than and is the least stable cation.
Solution:
HyperconjugatiOn interaction in is greater than in as the has nine C – H bonds. In , vacant p orbital is perpendicular to the plane in which C – H bonds lie; hence cannot overlap with it. Thus, laoks hyper conjugative stability.

Question 20.
On complete combustion, 0.246 g of an organic compound gave 0.1989/ of carbon dioxide and 0.10149 of water. Determine the percentage composition of carbon and hydrogen in the compound.
Solution:
Percentage of carbon = \(\frac{12 \times 0.198 \times 100}{44 \times 0.246}\)
= 21 .95%
Percentage of hydrogen = \(\frac{2 \times 0.1014 \times 100}{18 \times 0.246}\)
= 4.58%

Question 21.
In Dumas’ method for estimation of nitrogen, 0.3g of an organic compound gave 5OmL of nitrogen collected at 300K temperature and 715mm pressure. Calculate the percentage composition of nitrogen in the compound. (Aqueous tension at 300K = 15 mm)
Solution:
Volume of nitrogen collected at 300K and 715mm pressure is 50 mL
Actual pressure = 715 – 15 = 700 mm
Volume of nitrogen at STP = \(\frac{273 \times 700 \times 50}{300 \times 760}\)
= 41.9 ML
22,400 ml of N₂ at STP weighs = 28 g
41.9 mL of nitrogen weighs = \(\frac{28 \times 41.9}{22400}\) g
Percentage of nitrogen = \(\frac{28 \times 41.9 \times 100}{22400 \times 0.3}\)
= 17.46%

Question 22.
During estimation of nitrogen present in an organic compound by Kjeldahl’s method, the ammonia evolved from 0.5 g of the compound in Kjeldahl’s estimation of nitrogen, neutralized 10 mL of 1 M H₂SO₄. Find out the percentage of nitrogen in the compound?
Solution:
1 M of 10 mL H₂SO₄ = 1 M of 20 mL NH₃ 1000 mL of 1 M ammonia contains 14 g nitrogen
20 ml of 1 M ammonia contains \(\frac{14 \times 20}{1000}\) g nitrogen
Percentage of nitrogen = \(\frac{14 \times 20 \times 100}{1000 \times 0.5}\)
= 56.0%

Question 23.
In Canus method of estimation of halogen1 0.15 g of an organic compound gave 0.12 g of AgBr. Find out the percentage of bromine in the compound.
Solution:
Molar mass of AgBr = 108 + 80
= 188 g mol^-1
188 g AgBr contains 80 g bromine
0.12 g AgBr contains \(\frac{80 \times 0.12}{188}\) g bromine
Percentage of bromine = \(\frac{80 \times 0.12 \times 100}{188 \times 0.15}\)
= 34.04%

Question 24.
In sulphur estimation, 0.157 g of an organic compound gave 0.4813 g of barium sulphate. What is the percentage of sulphur in the compound ?
Solution:
Molecular mass of BaSO₄ = 137 + 32 + 64
= 233g
233 g BaSO₄ contains 32 g sulphur
0.4813 g BaSO₄ contains \(\frac{32 \times 0.4813}{233}\) g sulphur
Percentage of sulphur = \(\frac{32 \times 0.4813 \times 100}{233 \times 0.157}\)
= 42.10%

Question 25.
Write structures of different chain isomers of alkanes corresponding to the molecular formula C₆H₁₄. Also write their IUPAC names.
Solution:

Question 26.
Write structures of different isomeric alkyl groups corresponding to the molecular formula C₅H₁₁. Write IUPAC names of alcohols obtained by attachment of -OH groups at different carbons of the chain.
Solution:

Question 27.
Write IUPAC names of the following compounds :

1. (CH₃)₃ C CH₂C(CH₃)₃
2. (CH₃)₂ C(C₂H₅)₂
3. tetra – tert-butylmethane

Solution:

1. 2, 2, 4, 4-Tetramethylpentane
2. 3 3-Dimethylpentane
3. 3, 3-Di-tert-butyl -2, 2, 4, 4 – tetramethylpentane

Question 28.
Write structural formulas of the following compounds:
(i) 3, 4, 4, 5-Tetramethylheptane
(ii) 2, 5-Dimethyhexane
Solution:

Question 29.
Write structures for each of the following compounds. Why are the given names incorrect? Write correct IUPAC names.
(i) 2-Ethylpentane
(ii) 5-Ethyl – 3-methylheptane
Solution:

Longest chain is of six carbon atoms and not that of five. Hence, correct name is 3-Methyl- hexane.

Numbering is to be started from the end which gives lower number to ethyl group. Hence, correct name is 3-ethyl-5-methyl-heptane.

Question 30.
Sodium salt of which acid will be needed for the preparation of propane? Write chemical equation for the reaction.
Solution:
Butanoic acid,

Question 31.
Write IUPAC names of the following compounds:

Solution:
(i) 2, 8 – Dimethyl – 3, 6 – decadiene;
(ii) 1, 3, 5, 7 Octatetraene;
(iii) 2 – n – Propylpent – 1 – ene;
(iv) 4 – Ethyl – 2, 6 – dimethyl – dec – 4 – ene;

Question 32.
Calculate number of sigma (σ) and pi (π) bonds in the above structures (i-iv).
Solution:
σ bonds : 33, π bonds : 2
σ bonds: 17, π bonds:4
σ bonds : 23, π bond : 1
σ bonds: 41, π bond : 1

Question 33.
Write structures and IUPAC names of different structural isomers of alkenes corresponding to C₂H₁₀.
Solution:

Question 34.
Draw cis and trans isomers of the following compounds. Also write their IUPAC names:
(i) CHCl = CHCl
(ii) C₂H₅CCH₃ = CCH₃C₂H₅
Solution:

Question 35.
Which of the following compounds will show cis-trans isomerism ?
(i) (CH₃)₂C = CH – C₂H₅
(ii) CH₂ = CBr₂
(iii) C₆H₅CH = CH – CH₃
(iv) CH₃CH = CCl CH₃
Solution:
(iii) and (iv). In structures (i) and (ii), two identical groups are attached to one of the doubly bonded carbon atom.

Question 36.
Write IUPAC names of the products obtained by addition reactions of HBr to hex- 1-ene
(i) in the absence of peroxide and
(ii) in the presence of peroxide.
Solution:

Question 37.
Write structures of different isomers corresponding to the 5^th member of alkyne series. Also write IUPAC names of all the isomers. What type of isomerism is exhibited by different pairs of isomers ?
Solution:
5^th member of alkyne has the molecular formula C₆H₁₀. The possible isomers are :


Position and chain isomerism shown by different pairs.

Question 38.
How will you convert ethanoci acid into benzene?
Solution:

Andhra Pradesh BIEAP AP Inter 1st Year Chemistry Study Material 12th Lesson Environmental Chemistry Textbook Questions and Answers.

AP Inter 1st Year Chemistry Study Material 12th Lesson Environmental Chemistry

Very Short Answer Questions

Question 1.
Define the terms atmosphere and biosphere.
Answer:
Atmosphere: The blanket of gases present around the earth is called the atmosphere. It maintains the heat balance on earth. The atmosphere contains nitrogen and oxygen in large proportions.

Biosphere: Living organisms like plants, animals, and human beings constitute the biosphere.
The biosphere is related to other environment segments.

Question 2.
Explain the terms Lithosphere and Hydrosphere.
Answer:
Lithosphere : The outer mantle of the solid earth consists of minerals present in earth gust and soil. The earth inner surface contains minerals and deeper inner layers contain natural gas and soil. Mountains and hills these are all constitutes lithosphere.

Hydrosphere : All the natural water resources together constitute the hydrosphere. Hydrosphere include oceans, seas, rivers, lakes, streams etc.

Question 3.
Define the term Soil Pollution.
Answer:
The concentration (or) accumulation of natural bodies is called as soil. Soil gets polluted due to industrial wastes, urban wastes, agricultural pollutants, chemical, radio active pollutants etc.

Question 4.
What is Chemical Oxygen demand (COD).     [A.P. Mar. 16] [Mar. 14]
Answer:
The amount of oxygen required to oxidise organic substances present in polluted water is called Chemical Oxygen Demand (COD).
It is an index for amount of organic substances present in water.

Question 5.
What is Bio Chemical Oxygen Demand (BOD) ? [A.P. Mar. 16] [Mar. 14]
Answer:
The amount of oxygen used by the suitable micro organisms present in water during five days at 20° C is called as Bio Chemical Oxygen Demand (BOD).

Question 6.
What are Troposphere and Stratosphere ?
Answer:
Troposphere : The major portion of the atmosphere which contains air is called troposphere.
It is present 0-11 Km from the earth.

Stratosphere : Stratosphere present 11 – 50 Km from the earth and it mainly contains ozone layer. It absorbs the harmful UV radiations coming from sun.

Question 7.
Name the major particulate pollutants present in Troposphere.
Answer:
The major particulate pollutants present in troposphere are dust, mist, fumes, smoke, smog etc.

Question 8.
List out four Gaseous Pollutants present in the polluted air.
Answer:
Oxides of Sulphur,Nitrogen and Carbon, Ozone, Hydrocarbons etc., are gaseous pollutants present in polluted air.

Question 9.
Green house effect is caused by and gases.
Answer:
Green house effect is caused by gases such as CO₂, CH₄, O₃, CFCs (Chloro Fluoro Carbons) and water vapour in the atmosphere.

Question 10.
Which oxides cause acid rain ? and What is its pH value ? [Mar. 13]
Answer

• Oxides of Nitrogen, Sulphur and Carbon dissolved in rain water forms acid rain.
• Acid rain has pH value lessthan 5.6.

Question 11.
Name two adverse effects caused by acid rains. [A.P. Mar. 16] [T.S. Mar. 15]
Answer:
Effects of acid rains

• Acid rains are harmful for agriculture, trees and plants because it dissolves and washes away nutrients needed for their growth.
• Acid rains affects the plants and animal life in aquatic ecosystem.
• Acid rains damages the old buildings and historical monuments like Taj mahal.
• Acid rains corrodes water pipes which lead to decrease the quality of drinking water.

Question 12.
What are smoke and mist ?
Answer:
Smoke : The solid particles (or) mixture of solid and liquid particles formed by the combustion of organic matter are called smoke particulates.
Eg : Cigarette smoke, oil smoke etc.

Mist: The particles produced by the spray liquids and by condensation of vapours in air is called mist.
Eg : H₂SO₄ – mist, herbicides, insecticides etc.
These miss their targets and travel through atmosphere to form mist.

Question 13.
What is classical smog ? and What is its Chemical Character (Oxidizing / reducing) ?
Answer:

• The mixture of smoke, fog and sulphur dioxide is called classical smog. It exists in cool humid climate.
• The chemical character of classical smog is reducing character. Hence it is also called as reducing smog.

Question 14.
Name the common components of Photo Chemical smog.
Answer:
The common components of Photo Chemical smog are O₃, NO, acrolein, formaldehyde and Peroxy Acetyl Nitrate (PAN).

Question 15.
What is PAN ? What effect is caused by it ?
Answer:

• Peroxy Acetyl Nitrate is called as PAN.
• Peroxy Acetyl Nitrate is (PAN) is a powerful eye irritant.

Question 16.
How is Ozone formed in the Stratosphere ?
Answer:

• UV radiations react with dioxygen (O₂) molecules and split into free oxygen (O) atoms. These free oxygen atoms (0) combined to form ozone molecule.
• The following are the reactions that takes place during the formation of ozone in stratosphere.
  

Question 17.
Give the Chemical equations involved in the Ozone depletion by CF₂Cl₂.
Answer:
CF₂Cl₂ which is released in the atmosphere mix with the normal atmospheric gases and reaches the stratosphere.

• In stratosphere CF₂Cl₂ react with powerful UV radiations and liberates chlorine free radicals.
  
• These chlorine radical reacts with ozone present in stratosphere to form chlorine monoxide radicals and molecular dioxygen.
  Cl* + O₃ → ClO* + O₂
• ClO radical react with atomic oxygen and produce more Cl – radicals.
  ClO* + O → Cl* + O₂

Question 18.
What is Ozone hole ? Where was it first observed ?
Answer:
The depletion of ozone layer is commonly known as ozone hole.

• It was first observed in Antarctica over the south pole.
• It was reported by atmospheric scientists working in Antarctica.

Question 19.
What is the value of dissolved Oxygen in pure cooled, water ?
Answer:
The value of dissolved oxygen in pure cooled water is around 10 ppm.

Question 20.
Give the possible BOD values of clean water and the polluted water.
Answer:

• The BOD value of clean water is less than 3 ppm.
• The BOD value of water is greaterthan 4 ppm then it is said to be polluted.
• Highly polluted water has BOD value more than 17 ppm.

Question 21.
Name three industrial Chemicals that pollute water.
Answer:
Detergents, paints, pesticides, dyes and pharmaceuticals etc.

Question 22.
What agrochemicals are responsible for water pollution ?
Answer:
Agrochemicals like chemical fertilisers, chemicals used for killing insects, fungi and weeds in crop etc., are responsible for water pollution.

Short Answer Questions

Question 1.
What are different segments of the earth’s environment ?
Answer:
Environment can be divided into four segments.

1. Atmosphere
2. Hydrosphere
3. Lithosphere
4. Biosphere

1) Atmosphere : The layer of air present around the Earth is called the atmosphere. The atmospheric air contains N₂ and O₂ in large proportions, while the rest of the gases like C02 are present only in smaller proportions. Atmosphere absorbs harmful radiations coming from the Sun. It plays an important role in maintaining the heat balance on Earth. If the proportions of the gases, especially O₂ and N₂ are disturbed by human activity, the equilibrium of the echo system is lost. It leads to disastrous consequences.

2) Hydrosphere : Hydrosphere includes all the surface and ground water resources i.e., oceans, rivers, lakes, polar ice caps etc., 97% of earth’s water is locked up in oceans. 3% is trapped in polar ice caps. Only small percentage of water is available for drinking, agricultural and industrial purpose. 80% of the earth’s surface is covered with water.

3) Lithosphere : One fifth of the total Earth surface is in the form of land. Inner layers of Earth contain minerals. Deeper inner layers of Earth contain Natural gas and oil. All these things, including hills and mountains come under Lithosphere. Plants, animals and human beings are occupied by it.

4) Biosphere : All living organisms like plants, animals and human beings constitute the Biosphere. Biosphere and other segments of the environment are interrelated. Biosphere is dependent on Atmosphere and Hydrosphere. Polluted atmosphere arrest the plant growth and bring health hazards among animals and human beings. Contaminated water causes many diseases and also death of aquatic animals.

Question 2.
Define the terms Sink, COD, BOD and TLV. [T.S. Mar. 16]
Answer:
Sink : The medium which retains and interacts with long lived pollutant is called the sink.
Eg : Oceans are important sinks for atmospheric CO₂.

COD : The amount of oxygen required to oxidise organic substances present in polluted water is called Chemical Oxygen Demand (COD).
It is an index for amount of organic substances present in water.

BOD : The amount of oxygen used by the suitable micro organisms present in water during five days at 20° C is called as Bio Chemical Oxygen Demand (BOD).

TLV : (Threshold Limit Value): The permissiable level of the toxic substances (or) pollutants in the atmosphere which affects a person adversly when he is exposed to this for 7 – 8 hrs. in a day is called TLV.

Question 3.
Name the gaseous pollutants present in the air and explain their formation.
Answer:
Gaseous pollutants present in air are
a) Oxides of sulphur
b) Oxides of nitrogen
c) Oxides of carbon
d) Hydro carbons

a) Oxides of sulphur :

• The oxides of sulphur formed by the fossil fuel containing sulphur are burnt.
  S + O₂ → SO₂
• Sulphur dioxide oxidises to form sulphur trioxide in presence of catalyst.
  2SO₂ + O₂ ⇌ 2SO₃
• Sulphur trioxide can also be formed by the reaction of SO₂ with O₃ (or) H₂O₂.
  SO₂ + O₃ ⇌ SO₃ + O₂
  SO₂ + H₂O₂ → H₂SO₄
• SO₂ is the most common oxide and causes the following adverse effects.
  a) It causes respiratory problems like asthma, bronchitis etc.
  b) It is poisonous to both animals and plants.
  c) It causes irritation to the eyes which results in tears and redness.

b) Oxides of nitrogen :
The major gases present in air are oxygen and nitrogen.

• These do not combine at normal temperature.
• Dinitrogen and dioxygen combined at high altitudes in presence of light to form oxides of nitrogen.
  
• No reacts with oxygen to form NO₂.
  2NO + O₂ → 2NO₂
• NO₂ is quickly formed by the reaction of NO with O₃.
  NO + O₃ → NO₂ + O₂

Adverse Effects

• NO₂ damages the leaves of plants and effects the efficiency of photosynthesis.
• NO₂ causes lung problems.
• NO₂ causes respiratory problems.
• NO₂ affects the textile fibres and metals.

c) Oxides of carbon : –
i) Carbon monoxide (CO)

• CO is one of the most harmful air pollutants.
• CO gas is produced by the incomplete combustion of carbon (coal, firewood, petrol etc.)
  C_(s) + \(\frac{1}{2}\)O_2(g) → C0(g)
• CO is mainly released into atmosphere by automobiles.

Adverse effects :

• It stops the transportation of oxygen to the organs and tissues.
• When it is inhaled by a human being it forms stable complex with haemoglobin of blood. This complex is named as carboxy haemoglobin.
  CO + Hb → CO – Hb
• This reduces the oxygen transportation in body and results into headache, eye problems, nerveous problems and heart problems.

ii) Carbondioxide :

• CO₂ is entered into atmosphere mainly by respiration process.
• CO₂ is also formed by the burning of fossil fuels.
  C + O₂ → CO₂ + heat energy
• CO₂ is formed by the decomposition of lime stone.
  CaCO₃ Cao + CO₂

Effects :

• Due to deforestation and burning of fossil fuels increases CO₂ release and the balance O₂ – CO₂ in atmosphere disturbed.
• The increased level of CO₂ causes global warming which causes several problems like dengue, malaria etc.
• Due to global warming efficiency of photosynthesis decreases.

d) Hydrocarbons:

• These mainly constitutes hydrogen and carbon.
• These are formed by the incomplete combustion of fuel used in auto mobiles.

Adverse Effects:

• These cause cancer.
• These are harmful to plants causing ageing, break down of tissues, flowers and leaves are shedded.

Question 4.
What is green house effect ? and how is it caused ?
Answer:
a) Global Warming : The Earth is heated by sunlight and some of the heat that is absorbed by the Earth is radiated back into space. But some gases like CO₂, CFCs, O₃, NO and water vapour present in the lower atmosphere do not allow the Earth to re-radiate the heat into space. A part of the heat so trapped in these atmospheric gases is re-emitted to the earth’s surface. This phenomenon is called the “Green house effect” (or) Global Warming.
The gases which are responsible for Green house effect are CO₂, CFC’s, O₃, NO, CH₄ water vapour etc. and they are termed as Green house gases. The effect of increase in CO₂ causes global warming.

b) Effects of global warming :

1. If there is a 1° C increase in the temperature, the ice caps of the polar region melt and level of the sea water increases by 90 cm. Due to this, so many coastal countries will be submerged.
2. Due to global warming, the rate of evaporation of water from the seas, rivers, ponds will increase. This leads to ultimately rains, cyclones and hurricanes.
3. Agriculture sector will be badly affected due to the fast evaporation of surface water. There will be a shortage in the supply of water for agricultural purpose.
4. Unseasonal rains.
5. Increase the infectious diseases like dengue, malaria, yellow fever, sleeping sickness etc.

Prevention : To reduce the level of CO₂ on the earth’s atmosphere, one must increase the number of sinks to absorb CO₂. Plants absorb a major portion of the CO₂. Therefore, more plants, trees, forests should be grown. The blue green algae present in the sea also gets extinct, due to water pollution. This should be prevented stopping the production of CFC etc.

Question 5.
Explain, with Chemical equations involved, the formation of acid rain.
Answer:
Acid rains are due to oxides of N, S and C (NO₂, SO₂ and CO₂).

• These oxides dissolve in rain water and formed as acids (HNO₃, H₂SO₄ and H₂CO₃).
• These come down to earth as rain and deposited on the earth’s surface.
• Acid – rain is more in industrial areas.

Chemical equations involved
NO₂ + NO₃ → N₂O₅
N₂O₅ + H₂O + 2HNO₃
CO₂ + H₂O → H₂CO₃
SO₃ + H₂O → H₂SO₄
The pH of acid rain is less than 5.6.

Question 6.
Explain in detail the adverse effects caused by the acid rain.
Answer:
Acid rains are harmful because

1. The life of old buildings will be considerable reduced.
2. The pH of the soil changes affecting its fertility.
3. Ammonium salts formed by acid rains can be seen as atmospheric haze.
4. Ammonium salts in rain drops result in wet decomposition.
5. Acid rains are harmful to agriculture, trees, plants etc.
6. Due to acid rains aquatic life disturbed.
7. Due to acid rains the historical monuments like Tajmahal are damaged.

Question 7.
How is Photochemical Smog formed ? What are its ill effects ?
Answer:

• When unsaturated hydrocarbons and nitrogen oxides produced from auto mobiles, factories reacts with sunlight and forms photo chemical smog.
• This occurs in warm, dry and hot climate.
• This has high concentration of oxidising agents. Hence it is called as photo chemical smog.

Formation :

• Fossil fuels are burnt, many pollutants are entered into troposphere.
• Out of these many pollutants two pollutants are main constituents of photo chemical smog. These are hydrocarbons and nitric oxide.
• These pollutants interacts with sun light and following reaction takes place.
  2NO_(g) + O_2(g) → 2NO_2(g)
  NO_2(g) → NO_(g) + O_(g)
• This oxygen atoms formed in the above reaction combine with O₂ to produce ozone.
  O_(g) + O_2(g) ⇌ O_3(g)
  NO_(g) + O_3(g) → NO_2(g) + O_2(g)
• O₃ is poisonous gas and both NO₂ and O₃ are strong oxidising agents.
• These react with unburnt hydro carbons in the polluted air to produce organic substances like HCHO, PAN etc.
• The common components of photo chemical smog are ozone, NO, acrolein, form aldehyde and PAN.

Question 8.
How is Ozone layer depleted in the atmosphere and what are the harmful effects caused by Ozone layer depletion ? [T.S. Mar. 16] [A.P. Mar. 15]
Answer:

• Ozone layer is present in stratosphere of the atmosphere.
• Ozone in the stratosphere is due to the following reactions.
  
• But due to industrialisation, certain chemical substances enter into stratosphere and destroy the ozone.
• The following are the major substances that causes depletion of ozone layer.
  1. CFCs (Chloro fluoro carbons (or) freons)
  2. NO
  3. Cl₂ (Chlorine)
• CFCs are colourless, odourless, lighter, non flammable, non toxic organic molecules which are used in refrigeratos, air conditioners, in the manufacturing of plastic foam and for cleaning computer parts.
• These CFCs enter in the stratosphere and deplete the ozone as follows.
  
• Chlorine radicals (Cl*) are continuously generated and decompose O₃.
• NO released by super sonic jet planes and that formed by burning fossil fuels enters into stratosphere and decompose ozone.
• Cl₂ decompose into Cl* radicals and these Cl* deplete O₃ by chain reactions.

Effects of depletion of the ozone layer (ozone hole): Due to depletion of ozone more U.V. enters into troposphere. This UV radiation leads to

1. ageing of skin
2. cataracts
3. skin burns
4. skin cancer
5. damage to fish production
6. killing many phytoplanktons
7. effect plant proteins leading to the harmful mutation of cells
8. evaporate surface water through the stomata of the leaves.
9. decrease moisture content of the soil.
10. damage paints, fibres, by fading them faster.
11. effect on photo synthesis etc.

Question 9.
List out the industrial wastes that cause water pollution and what are the international standards fixed for drinking water ?
Answer:
Industrial wastes from food processing plants and paper and pulp mills are oxygen – demanding wastes. So they cause depletion of D.0 from the water.

Salts, trace of elements like Copper, Zinc, Arsenic etc., metals coming out of chromium plating Industry pollute water. They effect the human health and aquatic animals.

In Japan, at Minimata, water was polluted with mercury released from industry. The mercury enter the fish and the people who consumed the fish are adversely effected.

Mining and Nuclear power plant pollute the water with radioactive substances.

International standards for drinking water

Fluoride concentration:

• The drinking water sample is tested for fluoride ion concentration. Its deficiency causes diseases like tooth decay etc.
• The permissible level of concentration of fluoride in water is up to 1 ppm.
• Fluoride concentration above 2 ppm causes brown mottling of teeth.
• Fluoride concentration greater than 10 ppm causes harmful effects to bones and teeth.

Lead :

• The prescribed upper limit of concentration limit of lead in drinking water is about 50 ppm.
• Lead can damage kidney, liver, reproductive system etc.

Sulphate :

• Sulphate concentration greater than 500 ppm in drinking water causes laxative effect.
  Nitrate :
• Nitrate concentration maximum limit in drinking water is about 50 ppm.
• Excess of nitrate in drinking causes disease such as methemoglobinemia.

Metals :

Maximum prescribed concentration of some common metals recommended in drinking water given in the following table.
Maximum prescribed concentration of Some Metals in Drinking Water.

Question 10.
Explain in detail the strategies adopted in Green Chemistry to avoid environment pollution.
Answer:
The ways of using the knowledge and the principles of chemistry and other sciences to develop methods to reduce as far as possible the pollution of the environment are known as green chemistry.
Ex : In the dry cleaning of clothes, earlier (CCl₂ = CCl₂) tetrachloroethane was used. This com-pound contaminates the ground water and is a suspected carcinogen. Therefore, using this compound is replaced by a process in which liquefied carbon dioxide with a suitable detergent is used. This would not pollute ground water much. Now-a-days hydrogen peroxide is used for bleaching clothes in laundries. This gives better results and decreases the consumption of water.
Ex : Synthesis of ethanol is now commercially prepared by one step by the oxidation of ethane in presence of catalyst.

Long Answer Questions

Question 1.
What is environmental pollution ? How many types of pollution are encountered ?
Answer:
Environmental pollution is defined as the addition of any external material like organic, Inorganic biological, radio logical (or) any change in nature which may harm (or) affect badly the living organism directly (or) indirectly, immediately (or) slowly.
The types of pollution encountered are the following :

1. Air pollution
2. Water pollution
3. Soil pollution
4. Oil pollution
5. Noise pollution etc.

Question 2.
Explain the following in detail.
a) Global warming
b) Ozone depletion
c) Acid rain
d) Eutrophication.
Answer:
a) Global Warming : The Earth is heated by sunlight and some of the heat that is absorbed by the Earth is radiated back into space. But some gases like CO₂, CFC’s, O₃, NO and water vapour present in the lower atmosphere do not allow the Earth to re-radiate the heat into space. A part of the heat so trapped in these atmospheric gases is re-emitted to the earth’s surface. This phenomenon is called the “Green house effect” (or) Global warming.
The gases which are responsible for Green house effect are CO₂, CFC’s, O₃, NO, CH₄ water vapour etc. and they are termed as Green house gases. The effect of increase in CO₂ causes global warming.

b) Effects of depletion of the ozone layer (ozone hole) : Due to depletion of ozone more U.V. enters into troposphere. This UV radiation leads to

1. ageing of skin
2. cataracts
3. skin burns
4. skin cancer
5. damage to fish production
6. killing many phytoplanktons
7. effect plant proteins leading to the harmful mutation of cells
8. evaporate surface water through the stomata of the leaves.
9. decrease moisture content of the soil.
10. damage paints, fibres, by fading them faster.
11. effect on photo synthesis etc.

c) Acid rain :

• Acid rains are due to oxides of N, S and C.(NO₂, SO₂ and CO₂)
• These oxides dissolve in rain water and formed as acids. (HNO₃, H₂SO₄ and H₂CO₃)
• These come down to earth as rain and deposited on the earth’s surface.
• Acid – rain is more in industrial areas.

Chemical equations involved:-
NO₂ + NO₃ → N₂O₅
N₂O₅ + H₂O + 2HNO₃
CO₂ + H₂O→ H₂CO₃
SO₃ + H₂O → H₂SO₄
The pH of acid rain is lessthan 5.6.

Acid rains are harmful because

1. The life of old buildings will be considerable reduced.
2. The pH of the soil changes affecting its fertility.
3. Ammonium salts formed by acid rains can be seen as atmospheric haze.
4. Ammonium salts in rain drops result in wet decomposition.
5. Acid rains are harmful to agriculture, trees, plants etc.
6. Due to acid rains aquatic life disturbed.
7. Due to acid rains the historical monuments like Tajmahal are damaged.

d) Eutrophication : Water present in ponds and lakes becomes over nutritious when organic sub-stances from agriculture and industry are thrown into it. It can support the luxuriant growth of algae and thus the lakes and ponds become marshy. This phenomenon is called Eutrophication.

Question 3.
Green Chemistry is to avoid environmental pollution. Explain.
Answer:
The ways of using the knowledge and the principles of chemistry and other sciences to develop methods to reduce as far as possible the pollution of the environment are known as green chemistry.
Ex : In the dry cleaning of clothes, earlier (CCl₂ = CCl₂) tetrachloroethane was used. This com-pound contaminates the ground water and is a suspected carcinogen. Therefore, using this com-pound is replaced by a process in which liquefied carbon dioxide with a suitable detergent is used. This would not pollute ground water much. Now-a-days hydrogen peroxide is used for bleaching clothes in laundries. This gives better results and decreases the consumption of water.
Ex : Synthesis of ethanol is now commercially prepared by one step by the oxidation of ethane in presence of catalyst.

Thus green chemistry is a cost effective approach and it involves reduction in material, energy consumption and waste greneration.

Andhra Pradesh BIEAP AP Inter 1st Year Chemistry Study Material 11th Lesson The p-Block Elements – Group 14 Textbook Questions and Answers.

AP Inter 1st Year Chemistry Study Material 11th Lesson The p-Block Elements – Group 14

Very Short Answer Questions

Question 1.
Discuss the variation of oxidation states in the group -14 elements.
Answer:

• The common oxidation states exhibited by group-14 elements are +4 and +2.
• Carbon exhibits negative oxidation states.
• Heavier elements exhibits +2 oxidation state.
• The tendency to show +2 oxidation state increases in the order Ge < Sn < pb.
• pb exhibits +2 oxidation state as stable state because of inert pair effect.

Question 2.
How the following compounds behave with water
a) BCl₃
b) CCl₄.
Answer:
a) BCl₃ reacts with water (hydrolysis) to form boric acid.

b) CCl₄ does not undergo hydrolysis due to lack of d-orbitals in the central atom ‘C1 and due to its highly non polar nature, CCl₄ does not acts as Lewis acid.

Question 3.
Are BCl₃ and SiCl₄ electron-deficient compounds ? Explain.
Answer:

• BCl₃ and SiCl₄ are electron-deficient compounds.
• These two compounds behave as Lewis acids.
• These compounds are electron pair acceptors.
• The following reacts support the electron deficiency of BCl₃ and SiCl₄.
  

Question 4.
Give the hybridization of carbon in
a) CO₃^-2
b) diamond
c) graphite
d) fullerene
Answer:
a) In C0₃^-2 carbon atom undergoes sp² hybridisation.
b) In Diamond carbon atom undergoes sp³ hybridisation.
c) In Graphite carbon atom undergoes sp² hybridisation.
d) In Fullerenes carbon atom undergoes sp² hybridisation.

Question 5.
Why is’CO’poisonous ? [T.S. Mar. 16]
Answer:
‘CO’ gas is highly poisonous because it has the ability to form a stable complex with haemoglobin.

Carboxy haemoglobin is 300 times more stable than oxyhaemoglobin.

Question 6.
What is allotropy ? Give the crystalline allotropes of carbon. [Mar. 13]
Answer:

• The phenomenon of existence of an element in different physical forms having similar (or) same chemical properties is called allotropy.
• Crystalline allotropes of carbon are
  a) Diamond
  b) Graphite
  c) Fullerenes.

Question 7.
Classify the following oxides as neutral, acidic, basic or amphoteric,
a) CO
b) B₂O₃
c) SiO₂
d) CO₂
e) Al₂O₃
f) PbO₂
g) Tl₂O₃
Answer:
a) CO is neutral oxide.
b) B₂O₃ is acidic oxide.
c) SiO₂ is acidic oxide.
d) CO₂ is acidic oxide.
e) Al₂O₃ is amphoteric oxide.
f) PbO₂ is amphoteric oxide.
g) Tl₂O₃ is basic oxide.

Question 8.
Name any two manmade silicates.
Answer:
Glass and cement are man made silicates.

Question 9.
Write the outer electron configuration of group -14 elements.
Answer:
The general outer most electronic configuration of group – 14 elements is ns²np².
1) Carbon – [He] 2s²2p²
2) Silicon – [Ne] 3s²3p²
3) Germanium – [Ar] 3d¹⁰4s²4p²
4) Tin – [Kr] 4d¹⁰ 5s² 5p²
5) Lead – [Xe] 4f⁴ 5d¹⁰ 6s² 6p²

Question 10.
How does graphite function as a lubricant ?
Answer:
Graphite is soft and it has layer lattice. In this structure one of the layer slided over another due to weak Vander Waal’s force of attraction. These layers are slippery. Hence it is greasy and function as lubricant.

Question 11.
Graphite is a good conductor – explain.
Answer:
In graphite carbon undergoes sp² hybridisation. Each carbon forms three a – bonds with three neighbouring carbon atoms. Fourth electron forms TC – bond and it is delocalised. Due to the presence of these moving (or) free electrons graphite acts as good conductor.

Question 12.
Explain the structure of silica.
Answer:

• Silica is a giant molecule with 3 – dimensional structure.
• In Silica eight membered rings are formed with alternate Silicon and Oxygen atoms.
• Each ‘Si’ is tetrahedrally surrounded by four Oxygen atoms.
• Each Oxygen at the vertex of the tetrahedron is shared by two Silicon atoms.
• In SiO₂ Silicon atom undergoes sp³ hybridisation.

Question 13.
What is ‘Synthesis gas’?
Answer:

• Water gas is also called as synthesis gas.
• It is a mixture of CO and H₂.
• It s prepared by passing steamover hot coke.
• It is used for the synthesis of methanol and a number of hydrocarbons. Hence it is called synthesis gas.

Question 14.
What is producer gas?
Answer:

• Producer gas is mixture of CO and N₂.
• It is prepared by passing air over hot coke.

Question 15.
Diamond has high melting point – Explain.
Answer:

• In Diamond each carbon undergoes sp³ hybridisation and it is surrounded by four other carbon atoms with strong a – bonds tetrahedrally.
• The C – C bond energy in diamond is very high and it has 3 – dimensional structure.
• Due to these reasons diamond has high melting point.
• It has melting point 4200 K.

Question 16.
Give the use of CO₂ in photosynthesis.
Answer:
The process of converting the atmospheric CO₂ into Carbohydrates by green plants is known as ‘photosynthesis’.
In Photosynthesis CO₂ changes to carbohydrates such as glucose.

Question 17.
How does CO₂ increase the green house effect ?
Answer:

• Green plants absorbs CO₂ gas for photosynthesis and releases O₂ gas.
• Due to deforestation, decomposition of lime stone and burning of fossil fuels CO₂ concentration is increased in atmosphere.
• The increase of CO₂ level disturbs the O₂ – CO₂ balance in the atmosphere and it is responsible for green house effect (or) global warming.

Question 18.
What are silicones ?
Answer:

• Silicones are the organo Silicon polymers containing R₂ SiO – repeating unit.
• These are synthetic compounds containing Si – O – Si.
• Linkage preparation : These are formed by the hydrolysis of chlorosilanes.

Question 19.
Give the uses of silicones.
Answer:
Uses of silicones :

• These are used in surgical and cosmetic plants.
• These are used in preparation of silicone rubbers.
• These are used as sealoant, greases etc.
• These are used in preparing water proof clothes and papers.
• These are used as insulators.
• These are used in paints and enamels.

Question 20.
What is the effect of water on tin ?
Answer:

• Tin metal reacts with steam to form tin dioxide and dihydrogen gas.
• In this reaction steam is decomposed.
  Sn + 2H₂O SnO₂ + 2H₂

Question 21.
Write an account of SiCl₄.
Answer:

• Silicon tetrachloride (SiCl₄) is also called as tetra chloro silico methane.
• SiCl₄ can acts as Lewis acid due to availability of 3d orbital in ‘Si’.
• SiCl₄ undergoes hydrolysis due to presence of vacant 3d-orbital. Here water molecules forms dative bonds with empty 3d-orbitals of Siratom.

Uses :

• SiCl₄ and NH₃ mixture used to produce smoke screens.
• Ultra pure Silicon is used to make transistors.
• SiO₂ prepared from SiCl₄ used in epoxypaints, resis etc..

Question 22.
SiO₂ is a solid while CO₂ is a gas – explain.
Answer:

• Silica (SiO₂) has giant molecular structure.
• In SiO₂ ‘Si’ undergoes sp³ hybridisation.
•  It is a 3 – dimension structure in which each ‘Si’ atom is tetrahedrally surrounded by four oxygen atoms.
• Hence it exists as solid compound.
• CO₂ has linear structure.
• In CO₂ ‘C’ undergoes sp hybridisation.
• In between CO₂ molecule weak Vander Waal’s forces are present.
• In CO₂ molecule two double bonds are present.
•  Hence CO₂ exists as a gas.

Question 23.
Write the use of ZSM – 5.
Answer:

1. ZSM – 5 is a zeolite.
2. It is used to convert alcohols directly into gasoline.

Question 24.
What is the use of dry ice ?
Answer:

1. Solid CO₂ is called as dry ice.
2. It is used as refrigirent for frozen food and ice – creams.

Question 25.
How is water gas prepared ?
Answer:
Water gas is prepared by passing superheated steam over hot coke.

Question 26.
How is producer gas prepared ?
Answer:
Producer gas is prepared by passing air over white not coke.

Question 27.
C-C bond length in graphite is shorter than C-C bond length in diamond – explain.
Answer:

1. In graphite each carbon undergoes sp² hybridisation and hence bond length is 1.42 A° (or) 141.5 pm.
2. Graphite has hexagonal layer like lattice. It is a 2-dimensional structure.
3. In diamond each carbon undergoes sp3 – hybridisation and hence bond length is 1.54 A° (or) 154 pm.
4. Diamond has regular tetrahedral giant polymeric structure. It is a 3-dimensional structure.

Question 28.
Diamond is used as precious stone – explain.
Answer:

• Diamonds are used as precious stones.
• Diamonds are clear, colourless form of pure carbon.
• These are hardest substances occurring naturally.
• The weight of diamond expressed in carats.
  1 carat = 200 mg.

Question 29.
Carbon never shows co-ordination number greater than four while other members of carbon family show co-ordination number as high as six – explain.
Answer:
Carbon never shows co-ordination number greater than four because of absence of d-orbitals in carbon atom.

The other members of carbon family show co-ordination number as high as six because of availability of d – orbitals.

Question 30.
Producer gas is less efficient fuel than water gas – explain.
Answer:

1. Producer gas has calorific value 5439.2 KJ/m³
2. Water gas has calorific value 13000 KJ/m³.
3. Due to high calorific value of watergas, it is more efficient fuel than producer gas (or) producer gas is less efficient than watergas.

Question 31.
SiF₆^-2 is known while SiCl₆^-2 is not. Explain. [A.P. Mar. 16]
Answer:
SiF₆^-2 is known while SiCl₆^-2 is not because

• Si⁺⁴ has small size so it cannot be accomodate six large chloride ions.
• The interaction between lone pairs of Cl^– ion and Si⁺⁴ is not very strong.

Short Answer Questions

Question 1.
Explain the difference in properties of diamond and graphite on the basis of their structure.
Answer:
Diamond
a) Each carbon is sp³ hybridised.
b) Each carbon is bonded to 4 other carbons tetrahedrally.
c) It has a 3 dimensional structure.
d) C – C bond length is 1.54 Å and bond angle is 109° 28′.
e) Carbon atoms are firmly held with strong covalent bonds.
f) Diamond is very hard.
g) Density = 3.5 g/cc.
h) Graphite is a conductor due to the presence of free electrons.
i) It is transparent to light and X-rays. It has high refractive index (2.45).

Graphite
a) Each carbon is sp² hybridised.
b) Each carbon is bonded to 3 other carbon atoms to form hexagonal rings. It has sheet like structure.
c) It has a 2 dimensional structure.
d) C – C bond length in hexagonal rings is 1.42 A° and bond angle is 120°.
e) The distance between two adjacent layers is 3.35 A°. These layers are held by weak Vander Waal’s forces.
f) Graphite is soft.
g) Density – 2.2g/cc.
h) Diamond is an insulator due to the absence of free electrons.
i) It has layer, lattice. The layers are slippery. Hence it is greasy.

Question 2.
Explain the following.
a) PbCl₂ reacts with Cl₂ to give PbCl₄
b) PbCl₄ is unstable to heat,
c) Lead is not known to form PbI₄.
Answer:
a) PbCl₂ + Cl₂ → PbCl₂
But PbCl₄ is unstable than PbCl₂. Because compounds of lead in +2 oxidation state are stable than +4 oxidation state.

b) PbCl₄ is unstable to heat:

• In PbCl₄ lead exhibits +4 oxidation state.
• he compounds of lead in +2 oxidation state are stable than +4 oxidation state. Hence PbCl₄ is unstable to heat.

c) Lead is not known to form PbI₄:

• Pb – I bond formed initially during the reaction does not release enough energy to unpair the 6s electrons.
• Lead compounds in +2 state are stable than +4 state.
  Due to inert pair effect Pb exhibits stable +2 oxidation state.

Question 3.
Explain the following :
a) Silicon is heated with methyl chloride at high temperature in presence of copper.
b) SiO₂ is treated with HF.
c) Graphite is a Lubricant
d) Diamond is an abrasive.
Answer:
a)

• Methyl chloride reacts with silicon at high temperature in presence of copper catalyst to form various types of methyl substituted chlorosilane of formula MeSiCl₃, Me₂SiCl₂, Me₃SiCl with small amount Me₄Si.
• Hydrolysis of dimethyl dichloro silane followed by condensation polymerisation forms a straight chain polymer (silicone) (Me = CH₃ – group)
  

b) SiO₂ is treated with HF to form SiF₄. Which on hydrolysis to form H₄SiO₄ and H₂SiF₆.
SiO₂ + 4HF → SiF₄ + 2H₂O
SiF₄ + 4H₂O → H₄SiO₄ + 2H₂SiF₆

c) Graphite is soft and it has layer lattice. In this structure one of the layer slided over another due to weak Vander Waal’s force of attraction. These layers are slippery. Hence it is greasy and function as lubricant.

d) The covalent bonds in diamond are very strong and difficult to break, Hence diamond is used as an abrasive for sharpening hard tpols, in making dyes and in the manufacturing of tungsten filaments etc.

Question 4.
What do you understand by
a) Allotropy
b) Inert pair effect
c) Catenation.
Answer:
a) Allotropy :
The phenomenon of existence of an element in different physical forms having similar (or) same chemical properties is called allotropy.
Crystalline allotropes of carbon are
a) Diamond
b) Graphite.

b) Inert pair effect : The reluctance of ‘ns’ pair of electrons to take part in bond formation is known as inert pair effect.
(or)
The occurrence of oxidation states two units less than the group oxidation states is known as inert pair effect.
Eg : Lead exhibits +2 oxidation state as stable oxidation state due to inert pair effect. (Instead of +4 state).

c) Catenation : The phenomenon of self linkage of atoms among themselves to form long chains (or) rings is called as catenation.
Carbon has highest catenation tendency due to its high bond energy (348 KJ/mole).

Question 5.
If the starting material for the manufacturing of Silicons is RSiCi3. Write the structure of the product formed.
Answer:
When RSiCl₃ type of compound is used for the manufacturing of silicones a cross – linked silicon is formed.
Eg : When Methyl trichloro silane (CH₃SiCl₃) undergoes hydrolysis to give monomethyl silane triol. This undergoes polymerisation to form a very complex cross-linked polymer (Silicone).

Question 6.
Write a short note on Zeolites.
Answer:
Zeolites : The three dimensional structure contain no metal ions. If some of the Si⁺⁴ in them is replaced by Al⁺³ and an additional metal ion an infinite three dimensional lattice is formed. Replacements of one or two silicon atoms in [Si₂O₈]^-2n form zeolites. Zeolites act as ion exchanges and as molecular sieves. The structures of zeolites permit the formation of cavities of different sizes. Water molecules and a variety of other molecules like NH₃, CO₂ and ethanol can be trapped in these cavities. Then zeolites serve as molecular sieves. They trap Ca⁺² ions from hard water and replace them by Na⁺² ions.
Uses of Zeolites :
Zeolites are used as catalysts in petrochemical industries for cracking of hydrocarbons and in Isomerisation.
Zeolite ZSM – 5 is used to convert alcohols directly into gasoline.

Question 7.
Write a short note on Silicates.
Answer:
Silicates : Many building materials are silicates.
Ex : Granites, slates, bricks and cement. Ceramics and glass are also silicates. The Si – O bonds in silicates are very strong. They do not dissolve in any of the common solvents nor do they mix with other substances readily.
The silicates can be divided into six types. They are mentioned here.

1. Orthosilicate or Nesosilicates : Their general formula may be M₂¹¹ (SiO₄).
   Ex : Willemite Zn₂ (SiO₄).
2. Pyrosilicates or sorosilicates or Disilicates : These contain SiO₇^-6 units. Pyrosilicates are rare.
   Ex: Thorveitite Ln₂ (Si₂O₇).
3. Chain silicates : They have the units (SiO₃)_n^2n-
   Ex : Spodumene LiA/ (SiO₃)₂.
   Amphiboles are one type chain silicates. Generally double chains are formed in them.
4. Cyclic silicates : They are silicates having ring structures. They may be formed of general formula (SiO₃)_n^2n-. Rings containing three, four, six and eight tetrahedral units are known. But rings with three and six are the most common.
   Ex : Beryl Be₃Al₂ (Si₆O₁₈)
5. Sheet silicates : When SiO₄ units share three corners the structure formed is an inifinite two dimensional sheet. The empirical formula (Si₂O₅)_n^2n-. These compounds appear in layer struc-tures. They can be cleaved.
   Ex : Kaolin Al₂ (OH)₄ Si₂O₅.
6. Frame work silicates or three dimensional silicates : Sharing all the four corners of a SiO₄ tetrahedron results in three dimensional lattice of formula SiO₂.
   Ex : Quartz, Tridymite; Cristobalite; Feldspar and ultramarine (Na₈ [Al₆ Si₆ O₂₄] S₂), zeolites.

Question 8.
What are Silicones ? How are they obtained ?
Answer:

• Silicones are the organo silicon polymers containing R₂ SiO – repeating unit.
• These are synthetic compounds containing Si – 0 – Si.
  Preparation : These are formed by the hydrolysis of chlorosilanes.
• Methyl chloride reacts with Silicon at high temperature in presence of copper catalyst to form various types of methyl substituted chlorosilane of formula MeSiCl₃, Me₂SiCl₂, Me₃SiCl with small amount Me₄Si
• Hydrolysis of dimethyl dichloro silane followed by condensation polymerisation forms a straight chain polymer (Silicone) (Me = CH₃ – group).
  

Question 9.
Write a short note on Fullerene.
Answer:
Fullerenes :

• Fullerenes are one type of crystalline allotropes of carbon.
• These are formed by heating graphite in an electric arc in presence of inertgases such as Helium (or) Argon.
• These have smooth structure without dangling bonds. Hence Fullerenes are the only pure forms of carbon.
• C₆₀ molecule is called Buck minster fullerene and it’s shave was like a soccer ball.
• C₆₀ contains twenty 6 – membered rings and twelve 5 – membered rings.
  
•  In C₆₀ 6 – membered rings can combine with 5 (or) 6 – mem-bered rings while 5 – membered rings only combine with 6 – membered rings.
• In Fullerens each carbon undergoes sp2 hybridisation.
• Fullere has aromatic nature due to delocalisation of electrons in unhybrid p-orbitals.
• The C – C bond length in these compound lies between single and double bond lengths.
• Spherical fullerene are also named as bucky balls.
• The ball shaped molecule has 60 vertices.
• The C – C bond distances are 1,43A° and 1.38A° respectively.

Question 10.
Why SiO₂ does not dissolve in water ?
Answer:
Silica (SiO₂) is a non reactive compound in it’s normal state.

1. This non reactivity is due to very high Si – O bond enthalpy.
2. Silica is a giant molecule with 3-dimensional structure.
3. In Silica each silicon atom is covalently bonded in a tetrahedral manner to four oxygen atoms
4. Hence SiO₂ is insoluble in water.
5. But slightly dissolves at high pressures when heated.

Question 11.
Why is diamond hard ?
Answer:
In diamond, each carbon undergoes sp3 hybridization. A carbon atom is bound to four carbon atoms, arranged in a tetrahedral symmetry, with single bonds. A three dimensional arrangement of the tetrahedral structures result in giant molecule. The bond energy is very high (348 kJ mol^-1). It is very difficult to break the bonds. So, diamond is hard.

Question 12.
What happens when the following are heated
a) CaCO₃
b) CaCO₃ and SiO₂
c) CaCO₃ and excess of coke.
Answer:
CaCO₃ up on heating gives Quick lime.


Quick lime (CO) with silica gives Calcium silicate.


Coke reacts with Quick lime and form Carbides.

Question 13.
Why does Na₂CO₃ solution turn into a suspension, when saturated with CO₂ gas.
Answer:
An aq. solution of Na₂CO₃ when saturated with CO₂, gives Sodium bicarbonate (NaHCO₃).
Na₂CO₃ + H₂O + CO₂ → 2NaHCO₃
NaHCO₃ is less soluble compared to Sodium carbonate, hence suspension is formed.

Question 14.
What happens when
a) CO₂ is passed through slaked lime
b) CaC₂ is heated with N₂.
Answer:
a) Slaked lime, Ca(OH)₂ is turned milky on passing CO₂ with the formation of insoluble calcium
carbonate Ca(OH)₂ + CO₂ → CaCO₃ + H₂O on passing more ‘CO₂‘, CaCO₃ is converted into Calcium bicarbonate.
CaCO₃ + H₂O + CO₂ → Ca(HCO₃)₂
b) CaC₂ on heating with N2 gives calcium cyanamide.

Question 15.
Write a note on the anomalous behaviour of carbon in the group -14.
Answer:
Carbon shows ariamalous behaviour in group – 14 elements. The following facts support that anamalous behaviour.
Except carbon all other elements of group -14 has available d-orbitals and can expand octet in valency shell.
Carbon occurs in free state but not the other elements of this group.

• Maximum covalency of carbon is four but for silicon is six.
• C – C bond energy is very high (348 kJ/Mde).
• Carbon can form multiple bonds with C, O, S, etc.
• Hydrocarbons are more stable thermally than silanes.

Long Answer Questions

Question 1.
What are Silicones ? How are they prepared ? Give one example. What are their uses?
Answer:

• Silicones are the organo silicon polymers containing R₂ SiO – repeating unit.
• These are synthetic compounds containing Si – 0 – Si.
  Preparation : These are formed by the hydrolysis of chlorosilanes.
• Methyl chloride reacts with Silicon at high temperature in presence of copper catalyst to form various types of methyl substituted chlorosilane of formula MeSiCl₃, Me₂SiCl₂, Me₃SiCl with small amount Me₄Si
• Hydrolysis of dimethyl dichloro silane followed by condensation polymerisation forms a straight chain polymer (Silicone) (Me = CH₃ – group).
  

Uses of Silicones:

• These are used in surgical and cosmetic plants.
• These are used in preparation of Silicons rubbers.
• These are used as sealoant, greases etc.
• These are used in preparing water proof clothes and papers.
• These are used as insulators.
• These are used in paints and enamels.

Question 2.
Explain the structure of Silica. How does it react with
a) NaOH and
b) HF.
Answer:

• Silica is a giant molecule with 3 – dimensional structure.
• In Silica eight membered rings are formed with alternate Silicon and Oxygen atoms.
• Each ‘Si’ is tetrahedrally surrounded by four Oxygen atoms.
• Each Oxygen at the vertex of the tetrahedron is shared by two Silicon atoms.
• In SiO₂ Silicon atom under goes sp³ hybridisation.
  

a) Silica reacts with NaOH and forms Sodium Silicate (Na₂SiO₃)
SiO₂ + 2 NaOH → Na₂SiO₃ + H₂O

b) SiO₂ is treated with HF to form SiF₄. Which on hydrolysis to form H₄SiO₄ and H₂SiF₆.
SiO₂ + 4HF → SiF₄ + 2H₂O
SiF₄ + 4H₂O → H₄SiO₄ + 2H₂SiF₆

Question 3.
Write a note on the allotropy of carbon.
Answer:
The property of an element to exist in two or more physical forms due to difference in the arrangement of atoms is called Allotropy. Allotropes have more or less similar chemical properties but different physical properties.

Diamond, graphite and fullerenes are the crystalline allotropes of carbon.
Structure of Diamond : In diamond, each carbon atom bonded to four carbon atoms situated tetrahydrally around it.
In diamond, each carbon atom is in sp³ hybridisation and is linked to four carbon atoms by single covalent bonds.
C – C bond distance in diamond is 1.54 A°.
bond angle in diamond is 109° 28″.
Uses of Diamond :

1. Diamonds are used as precious stones for jewellery because of their ability to reflect light.
2. Diamonds are used for cutting glass and drilling rocks due to their remarkable hardness.

Structure of Graphite : Graphite consists of a series of layers in which hexagonal rings made up of carbon atoms.
In Graphite, each carbon atom undergo sp² hybridisation and forms three covalent bonds with three other carbon atoms.

The fourth electron present in the pure ‘p orbitaI which is unhybridised. The electron become Free Electron.
The C — C bond length in graphite is 1.42 A°.

The distance between the two layers in graphite is 3,4 A°.
These layers are held together by Vander Waal’s forces which are weak.
Graphite is a layer lattice structure.
Uses of Graphite:

1. Graphite is used as a lubricant.
2. It is used in the manufacturing of lead pencils.
3. It is used in the manufacturing of Electrodes and Refractory crucibles.

Fullerenes :

• Fullerenes are one type of crystalline allotropes of carbon.
• These are formed by heating graphite in an electric arc in presence of inertgases such as Helium (or) Argon.
• These have smooth structure without dangling bonds. Hence Fullerenes are the only pure forms of carbon.
• C₆₀ molecule is called Buck minster fullerene and it’s shave was like a soccer ball.
• C₆₀ contains twenty 6 – membered rings and twelve 5 – membered rings.
• In C₆₀ 6 – membered rings can combine with 5 (or) 6 – membered rings while 5 – membered rings only combine with 6 – membered rings.
•  In Fullerens each carbon undergoes sp² hybridisation.
• Fullere has aromatic nature due to delocalisation of electrons in unhybrid p-orbitals.
• The C – C bond length in these compound lies between single and double bond lengths.
• Spherical fullerene are also named as bucky balls.
• The ball shaped molecule has 60 vertices.
• The C – C bond distances are 1.43A° and 1.38A° respectively.
• Amorphous allotropes of carbon are coal, coke, animal charcoal wood charcoal, lamp black, carbon black, gas carbon, petroleum coke and sugar charcoal.

Question 4.
Write a note on
a) Silicates
b) Zeolites
c) Fullerenes.
Answer:
a) Silicates : Many building materials are silicates.
Ex : Granites, slates, bricks and cement. Ceramics and glass are also silicates. The Si – O bonds in silicates are very strong. They do not dissolve in any of the common solvents nor do they mix with other substances readily.

The silicates can be divided into six types. They are mentioned here.

1. Orthosilicate or Nesosilicates : Their general formula may be M₂¹¹ (SiO₄).
   Ex : Willemite Zn₂ (SiO₄).
2. Pyrosilicates or sorosilicates or Disilicates : These contain SiO₇^-6 units. Pyrosilicates are rare.
   Ex : Thorveitite Ln₂ (Si₂O₇).
3. Chain silicates : They have the units (SiO₃)_n^2n-
   Ex : Spodumene LiAl (SiO₃)₂.
   Amphiboles are one type chain silicates. Generally double chains are formed in them.
4. Cyclic silicates : They are silicates having ring structures. They may be formed of general formula (SiO₃)_n^2n-. Rings containing three, four, six and eight tetrahedral units are known. But rings with three and six are the most common.
   Ex : Beryl Be₃Al₂ (Si₆O₁₈)
5. Sheet silicates : When SiO₄ units share three corners the structure formed is an inifinite two dimensional sheet. The empirical formula (Si₂O₅)_n^2n-. These compounds appear in layer struc-tures. They can be cleaved.
   Ex : Kaolin Al₂ (OH)₄ Si₂O₅.
6. Frame work silicates or three dimensional silicates : Sharing all the four corners of a SiO₄ tetrahedron results in three dimensional lattice of formula SiO₂.
   Ex : Quartz, Tridymite; Cristobalite; Feldspar and ultramarine (Na₈ [Al₆ Si₆ O₂₄] S₂), zeolites.

b) Zeolites : The three dimensional structure contain no metal ions. If some of the Si⁺⁴ in them is replaced by Al⁺³ and an additional metal ion an infinite three dimensional lattice is formed. Replacements of one or two silicon atoms in [Si₂O₈]^-2n form zeolites. Zeolites act as ion ex-changes and as molecular sieves. The structures of zeolites permit the formation of cavities of different sizes. Water molecules and a variety of other molecules like NH₃, CO₂ and ethanol can be trapped in these cavities. Then zeolites serve as molecular sieves. They trap Ca⁺² ions from hard water and replace them by Na⁺² ions.

Uses of Zeolites :

• Zeolites are used as catalysts in petrochemical industries for cracking of hydrocarbons and in Isomerisation.
• Zeolite ZSM – 5 is used to convert alcohols directly into gasoline.

c) Fullerenes :

• Fullerenes are one type of crystalline allotropes of carbon.
• These are formed by heating graphite in an electric arc in presence of inertgases such as Helium (or) Argon.
• These have smooth structure without dangling bonds. Hence Fullerenes are the only pure forms of carbon. .
• C₆₀ molecule is called Buck minster fullerene and it’s shave was like a soccer ball.
• C₆₀ contains twenty 6 – membered rings and twelve 5 – mem- bered rings.
  
• In C₆₀. 6 – membered rings can combine with 5 (or) 6 – membered rings while 5 – membered rings only combine with 6 – membered rings.
• In Fullerens each carbon undergoes sp² hybridisation.
• Fullere has aromatic nature due to delocalisation of electrons in unhybrid p-orbitals.
• The C – C bond length in these compound lies between single and double bond lengths.
• Spherical fullerene are also named as bucky balls.
• The ball shaped molecule has 60 vertices.
• The C – C bond distances are 1.43A° and 1.38A° respectively.

Solved Problems

Question 1.
Select the member(s) of group 14 that

1. forms the most acidic dioxide
2. is commonly found in +2 oxidation state
3. used as semiconductor.

Solution:

1. Carbon
2. lead
3. Silicon and germanium

Question 2.
[SiF₆]^2- is known whereas [SiCl₆]^2- not. Give possible reasons.
Solution:
The main reasons are :

1. Six large chloride ions cannot be accommodated around Si⁴⁺ due to limitation of its size.
2. Interaction between lone pair of chloride ion and Si⁴⁺ is not very strong.

Question 3.
Diamond is covalent, yet it has high melting point. Why ?
Solution:
Diamond has a three – dimensional network involving strong C – C bonds, which are very difficult to break and in turn has high melting point.

Question 4.
What are Silicones ?
Solution:
Simple Silicones consist of chains in which alkyl or phenyl groups occupy the remaining bonding positions on each silicon. They are hydrophobic (water repellant) in nature.

Andhra Pradesh BIEAP AP Inter 1st Year Chemistry Study Material 10th Lesson The p-Block Elements – Group 13 Textbook Questions and Answers.

AP Inter 1st Year Chemistry Study Material 10th Lesson The p-Block Elements – Group 13

Very Short Answer Questions

Question 1.
Discuss the pattern of variation in the oxidation states of Boron to Thallium.
Answer:

• Boron exhibits – 3 oxidation state due to its small size and non-metalic nature.
• Aluminium exhibits +3 oxidation state.
• Gallium, Indium and Thallium exhibits both +1 and +3 oxidation states.
• In Thallium +1 oxidation state is more stable than +3 due to inert pair effect.

Question 2.
How do you explain higher stability of TlCl₃ ?
Answer:
[TlCl₃ is not stable because Tl doesnot exist in Tl³ stable state.] [TlCl is stable because of inertpair effect Tl⁺¹ is stable].

Question 3.
Why does BF₃ behave as a Lewis acid ?
Answer:
BF₃ is a electron deficient molecule. It has the tendency to accept an electron pair. Electron pair acceptors are Lewis acids. Hence BF₃ behave as a Lewis acid.

Question 4.
Is boric acid a protic acid ? Explain.
Answer:
Boric acid is a weak mono basic acid. In Boric acid plannar BO₃ units are joined by hydrogen bonds. It has layer structure (polymeric). Hence it is not a protic acid. It does not give up a proton.

Question 5.
What happens when boric acid is heated ?
Answer:
Boric acid when heated above 370 K forms meta boric acid. This on heating forms Boric oxide.

Question 6.
Describe the shapes of BF₃ and BH₄^–. Assign the hybridization of boron in these species.
Answer:

• Shape of BF₃ molecule is Trigonal planar
  Hybridisation of ‘B’ in BF₃ is sp²
• Shape of BH₄^– molecule is Tetrahedral
  Hybridisation of ‘B’ in BH₄^– is sp³

Question 7.
Explain why atomic radius of Ga is less than that of ‘Al’.
Answer:
In Gallium penultimate shell contains 10-d electrons. Due to this 10-d electrons shielding effect becomes poor on outer most electrons. So nuclear charge increased in Gallium. Hence atomic radius of Ga is lessthan that of ‘Al’.

Question 8.
Explain inert pair effect.
Answer:
The reluctance of ‘ns’ pair of electrons to take part in bond formation is called inert pair effect.
(or)
The occurrence of oxidation states two unit lessthan the group oxidation states is called inert pair effect.
Eg : In Group – 13 Tl exhibits +1 oxidation state instead of +3 oxidation state due to inert pair effect.

Question 9.
Write balanced equations for
a) BF₃ + LiH →
b) B₂H₆ + H₂O →
c) NaH + B₂H₆ →
d) H₃BO₃
e) B₂H₆ + NH₃
Answer:
a) 2BF₃ + 6LiH → B₂H₆ + 6 LiF
b) B₂H₆ + 6H₂O → 2B(OH)₃ + 6H₂
c) B₂H₆ + 2NaH → 2NaBH₄
d) H₃BO₃ HBO₂ B₂O₃
e) B₂H₆ + 6NH₃ → 3[BH₂(NH₃)₂]⁺ (BH₄)^– 2B₃N₃H₆ + 12H₂

Question 10.
Why is boric acid polymeric ?
Answer:
Boric acid has layer like lattice. In this structure planar BO₃ units are joined by hydrogen bonds and forms a polymeric structure.

Question 11.
What is the hybridization of B in diborane and borazine ?
Answer:

• In diborane ‘B’ hybridisation is sp³
• In Borazine ‘B’ hybridisation is sp²

Question 12.
Write the electronic configuration of group – 13 elements.
Answer:
General outer electronic configuration of Group – 13 elements is ns²np¹

• B – 1s²2s²2p¹
• Al – [Ne] 3s²3p¹
• Ga – [Ar] 3d¹⁰4s²4p¹
• In – [Kr] 4d¹⁰5s²5p¹
• Tl – [Xe] 5d¹⁰ 6s²6p¹

Question 13.
Give the formula of borazine. What is its common name ?
Answer:

• The formula of borazine is B₃N₃H₆.
• It’s common name is “In organic benzene” because it is – iso structural with benzene.

Question 14.
Give the formulae of
a) Borax
b) Colemanite.
Answer:
a) Formula of Borax is Na₂B₄O₇. 10H₂O
b) Formula of Colemanite is Ca₂B₆O₁₁.5H₂O

Question 15.
Give two uses of aluminium.
Answer:
Uses of Aluminium :

• Aluminium is used in packing.
• Aluminium is used in utensil making.
• Aluminium alloys are used in shaping of pipes, tubes, wires etc.
• Aluminium alloys are used in making air craft bodies.

Question 16.
What happens when
a) LiAlH₄ and BCl₃ mixture in dry ether is warmed and
b) Borax is heated with H₂SO₄ ?
Answer:
a) When LiAlH₄ and BCl₃ mixture is warmed in dry ether diborane is formed.
4BF₃ + 3 LiAlH₄ → 2B₂H₆ + 3 LiF + 3 AlF₃
b) Borax is heated with H2S04 then boric acid is formed
Na₂B₄O₇ + H₂SO₄ + 5H₂O → Na₂SO₄ + 4H₃BO₃

Question 17.
Sketch the structure of Orthoboric acid.
Answer:

Question 18.
Write the structure of AlCl₃ as a climer.
Answer:

Question 19.
Metal borides (having ¹⁰B) are used as protective shield – Why ?
Answer:
Boron – 10 (¹⁰B) has the capacity to absorb neutrons. Hence metal borides (having ¹⁰B) are used as protective shields in nuclear industry.

Short Answer Questions

Question 1.
Write reactions to justify amphoteric nature of aluminium.
Answer:

• Amphoteric nature means having acidic as well as basic nature.
• Aluminium reacts with both mineral acids as well as aqueous alkalis.

a) Reaction with mineral acid :
‘Al’ reacts with dil.HCl and liberates hydrogen gas.
2Al + 6HCl → 2AlCl₃ + 3H₂ ↑

b) Reaction with aqueous alkali :
‘Al’ reacts with aqueous alkali (NaOH) and liberates hydrogen gas.
2Al + 2NaOH + 6H₂O → Na₂[Al(OH)₄] + 3H₂ ↑

Question 2.
What are electron deficient compounds ? Is BCl₃ an electron deficient species ? Explain.
Answer:
These are the compounds in which the available no.of valence electrons is lessthan the number required for normal covalent bond formation (or) for writting the Lewis structure of the molecule.

• These compounds are electron pair acceptors and acts as Lewis acids.
• BCl₃ is an electron deficient compound.
• In BCl₃ ‘B’ contains only six electrons instead of eight electrons.
• BCl₃ has the tendency to accept an electron pair and acts as Lewis acid.
  Eg : Formation of BCl₃ . NH₃ :-
  BCl₃ accepts an electron pair from NH₃ and forms the compound BCl₃.NH₃ (Tetrahedral)
  

Question 3.
Suggest reasons why the B – F bond lengths in BF₃ (130 pm) and BF₄^– (143 pm) differ.
Answer:
About BF₃ :

• In BF₃ the central atom ‘B’ contains three bond pairs in the valency shell.
• ‘B’ under goes sp² hybridisation.
• Shape of the molecule is trigonal planar.

About BF₄^– :

• In BF₄^– the central atom ‘B’ contains four bond pairs in the valency shell.
• ‘B’ under goes sp³ hybridisation.
• Shape of the molecule is tetrahedral.
• The above reasons suggent that the difference in bond lengths of BF₃ (130 pm) and BF₄^– (143 pm).

Question 4.
B – Cl bond has a bond moment. Explain why BCl₃ molecule has zero dipolemoment.
Answer:

• B – Cl bond is a polar bond so it has bond moment.
• BCl₃ molecule is non-polar because of its symmetrical structure. (Trigonal planar structure)
• Symmetrical molecules has zero dipole moment.
  ∴ Dipole moment of BCl₃ (μ) = 0

Question 5.
Explain the structure of boric acid.
Answer:

• Boric acid has a layer lattice.
• In this layer lattice planar BO₃ units are joined by hydrogen bonds.
• The structure of Boric acid is polymeric as shown in following figure.
  
• In the above structure dotted lines represents the hydrogen bonds

Question 6.
What happens when
a) Borax is heated strongly
b) Boric acid is added to water
c) Aluminium is heated with dilute NaOH
d) BF₃ is treated with ammonia
e) Hydrated alumina is treated with aq.NaOH solution.
Answer:
a) Borax on heating first loses water molecules and forms sodium tetraborate. This on further heating forms a mixture of sodium metaborate and boric an hydride. This mixture is solidifies into glass like substance.

b) Boric acid is added to water, boric acid accepts a hydroxyl ion from water.
B(OH)₃ + 2H₂O → [B(OH)₄]^– + H₃O⁺

c) Aluminium is heated with dilute NaOH, sodium metaluminate is formed with the liberation of hydrogen gas.
2Al + 2NaOH → 2NaAlO₂ + H₂ ↑

d) BF₃ is treated with NH₃ an addition compound BF₃. NH₃ is formed. BF₃ accepts an electron pair from NH3 and forms a dative bond.
BF₃ + → [BF₃ ← NH₃] → [BF₃.NH₃]

e) Hydrated Alumina is treated with aq.NaOH to form sodium metaluminate.
Al₂O₃.2H₂O + 2NaOH_(aq) → 2NaAlO_(aq) + 3H₂O

Question 7.
Give reasons
a) Conc.HNO₃ can be transported in aluminium container.
b) A mixture of dil. NaOH and aluminium pieces is used to open drain.
c) Aluminium alloys are used to make aircraft body.
d) Aluminium utensils should not be kept in water overnight.
e) Aluminium wire is used to make transmission cables.
Answer:
a) Cone. HNO₃ can be transported in Aluminium containers because Al is passive towards Conc.HNO₃ due to the formation of thin layer of Al₂O₃ on the surface.

b) A mixture of dil.NaOH and aluminium pieces is used to open drain because it acts as cleaning agent.
2Al + 2NaOH → 2NaAlO₂ + H₂

c) Aluminium alloys are used to make air craft body because it is a light metal, soft, malleable, ductile and tenacious. It shows resistance to atmosphere corrosion.

d) Aluminium utensils should not be kept in water overnight because Aluminium reacts with water and liberates hydrogen gas and heat. It makes colour dissolving and sometimes Aluminium compounds are toxic in nature.

e) Aluminium wire used to make transmission cables because of it’s good conductivity (electrical) and resistance to atmospheric corrosion.

Question 8.
Explain why the electronegativity of Ga, In and Tl will not vary very much.
Answer:

1. Ga, In and Tl have the electro negativity values 1.6, 1.7 and 1.8 respectively.
2. In Ga, In and Tl the d-electrons (d¹⁰) in penultimate shell do not shield the outer most electrons from nuclear attraction effectively.
3. The reason for the above fact is.the shielding effect of various electrons in the orbitals follows the order
   s > p > d > f.
4. Hence the outer electrons are held more firmly by the nucleus. Because of this, atoms with d- electrons in the penultimate shell (d¹⁰) are smaller in size. Ga, In and Tl has same number of penultimate shell electrons.
5. So, Ga, In and Tl will not vary very much in their electronegativities.

Question 9.
Explain Borax bead test with a suitable example. [T.S. Mar. 16] [Mar. 13]
Answer:
Borax bend test: This test is useful in the identification of basic radicals in qualitative analysis. On heating borax swells into a white, opaque mass of anhydrous sodium tetra borate. When it is fused, borax glass is obtained. Borax glass is sodium meta borate and B203. The boric anhydride, B203, combined with metal oxides to form metal metaborates as coloured beads. The reactions are as follows :

Question 10.
Explain the structure of diborane. [A.P. Mar. 16] [A.P. & T.S. Mar. 15]
Answer:
Diborane is an electron deficient compound. It has ’12’ valency electrons for bonding purpose instead of ’14’ electrons.

In diborane each boron atom undergoes sp3 hybridization out of the four hybrid orbitals one is vacant.

Each boron forms two, σ – bonds (2 centred – 2 electron bonds) bonds with two hydrogen atoms by overlapping with their ‘1s’ orbital.

The remaining hybrid orbitals of boran used for the formation of B-H-B bridge bonds.

In the formation of B-H-B bridge, half filled sp3 hybrid ofbital of one boron atom and vacant sp³ hybrid orbital of second boron atom overlap with 1s orbital of H-atom.

These three centred two electron bonds are also called as banana bonds. These bonds are present above and below the plane of BH₂ units.

Diborane contains two coplanar BH₂ groups. The four hydrogen atoms are called terminal hydrogen atoms and the remaining two hydrogens are called bridge hydrogen atoms.

Bonding in diborane, Each B atom uses sp³ hybrids for bonding. Out of the four sp³ hybrids on each B atom, one is without an electron shown in broken lines. The terminal B-H bonds are normal 2-centre-2- electron bonds but the two bridge bonds are 3-centre-2- electron bonds. The 3-centre-2-electron bridge bonds are also referred to as banana bonds.

Question 11.
Explain the reactions of aluminium with acids.
Answer:
Reactions of ‘Al’ with adds :
i) DiS. (or) cone. HCl dissolves Al and gives H₂.

ii) a) Dil.H₂SO₄ liberates H₂.

b) Cone. H₂SO₄ dissolves the metal ‘Al’ and gives SO₂.

iii) a) Very dil.HNO₃ is reduced to NH₄ NO₃ by Al.

b) Cone. HNO₃ makes ‘Al’ passive due to the formation of a thin film of oxide layer on the metal surface.

Question 12.
Write a short note on the anamalous behaviour of boron in group – 13.
Answer:

• Among Group – 13 elements ‘B’ is only the non metal.
• ‘B’ forms only covalents compounds.
• ‘B’ doesnot displaces hydrogen from acids.
• ‘B’ shows diagonal relationship with ‘Si’.
• ‘B’ forms acidic oxide where as other elements of group forms amphoteric oxides and basic oxides.
• ‘B’ has only two electrons in the penultimate shell.
• ‘B’ has covalency ‘4’ where as other elements has covalency of maximum ‘6’.

Question 13.
Aluminium reacts with dil.HNO₃ but not with conc.HNO₃ – explain.
Answer:

• Dilute HNO₃ reacts with Aluminium slowly and forms aluminium nitrate and ammonium nitrate.
  8Al + 30 HNO₃ → 8 Al(NO₃)₃ + 3NH₄NO₃ + 9H₂O
• Aluminium doesnot react with cone. HNO₃.

Reasons :

• Aluminium is passive towards cone. HNO₃ due to the formation of thin film of Al₂O₃ layer on the surface.
• Because of this passivity between Al and conc.HNO₃, conc.HNO₃ is transported in Aluminium containers.

Question 14.
Give two methods of preparation of diborane.
Answer:

1. In industries diborane is prepared by the reaction between boron tri fluoride and lithium hydride.
   
2. Boron trichloride and hydrogen mixture subjected to silent electric discharge at low pressure to from diborane.
   2BCl₃ + 6H₂ → B₂H₆ + 6HCl
3. Boron trichloride undergo reduction with LiAlH₄ to form diborane.
   

Question 15.
How does diborane react with
a) H₂O
b) CO
c) N(CH₃)₃ ?
Answer:
a) Diborane reacts with water to form boric acid and hydrogen.
B₂H₆ + 6H₂O → 2H₃BO₃ + 6H₂↑

b) Diborane reacts with CO at 100° C and 20 atm. pressure to form borane carbonyl.

c) Diborane reacts with N(CH₃)₃ and form a adduct.
B₂H₆ + 2N(CH₃)₃ → 2BH₃.N(CH₃)₃ (adduct)
Reactions b, c are cleavage reactions.

Question 16.
Al₂O₃ is amphoteric – explain with suitable reactions.
Answer:

• Amphoteric oxides are the oxides which possess both acidic as well as basic nature.
  Al₂O₃ possess both acidic as well as basic behaviour.
• Al₂O₃ react with both acids as well as bases to produce salts and water.
  Supporting reactions for amphoteric nature of Al₂O₃
  a) With acids:
  
  b) With bases:
  

Question 17.
Na₂B₄O₇ + Cone. H₂SO₄ → A > B (Green edged flame) Identify A and B
Hint: A = H₃BO₃ B = (C₂H₅)₃ BO₃.
Answer:
Na₂B₄O₇ + Cone. H₂SO₄ + 5H₂O → 4H₃BO₃ A) B) 4(C₂H₅)₃BO₃

• ‘A’ is H₃BO₃
• ‘B’ is (C₂H₅)₃ BO₃.

Long Answer Questions

Question 1.
How are borax and boric acid prepared ? Explain the action of heat on them.
Answer:
Preparation of Borax:
Boric acid on heating first forms tetraboric acid. This on reaction with sodium hydroxide to form borax.

Preparation of Boric acid : Borax is treated with conc.H₂SO₄ boric acid is formed.
Na₂B₄O₇ + H₂SO₄ + 5H₂O → 4H₃BO₃ + Na₂SO₄.

Heating of Borax :
Borax on heating first loses the water molecules and forms sodium tetraborate. This on further heating forms a mixture of sodium meta borate and boric anhydride.
Na₂B₄O₇.10H₂O Na₂B₄O₇ imgg 2 2 NaBO₂ + B₂O₃.

Heating of Boric acid :
Boric acid on heating forms boric anhydride. The reaction depends on the temperature used.

Question 2.
How is diborane prepared ? Explain its structure.
Answer:
In industries diborane is prepared by the reaction between boran tri fluoride and lithium hydride.

Boron trichioride and hydrogen mixture subjected to silent electric discharge at low pressure to from diborane.
2BCl₃ + 6H₂ → B₂H₆ + 6HCl
Boron trichloride undergo reduction with LiAlH₄ to form diborane.

Diborane is an electron deficient compound. It has ’12’ valency electrons for bonding purpose instead of ’14’ electrons.

In diborane each boron atom undergoes sp3 hybridization out of the four hybrid orbitals one is vacant.

Each boron forms two, σ – bonds (2 centred – 2 electron bonds) bonds with two hydrogen atoms by overlapping with their ‘1s’ orbital.

The remaining hybrid orbitals of boran used for the formation of B-H-B bridge bonds.

In the formation of B-H-B bridge, half filled sp3 hybrid ofbital of one boron atom and vacant sp³ hybrid orbital of second boron atom overlap with 1s orbital of H-atom.

These three centred two electron bonds are also called as banana bonds. These bonds are present above and below the plane of BH₂ units.

Diborane contains two coplanar BH₂ groups. The four hydrogen atoms are called terminal hydrogen atoms and the remaining two hydrogens are called bridge hydrogen atoms.

Bonding in diborane, Each B atom uses sp³ hybrids for bonding. Out of the four sp³ hybrids on each B atom, one is without an electron shown in broken lines. The terminal B-H bonds are normal 2-centre-2- electron bonds but the two bridge bonds are 3-centre-2- electron bonds. The 3-centre-2-electron bridge bonds are also referred to as banana bonds.

Question 3.
Write any two methods of preparation of diborane. How does it react with
a) Carbon monoxide and
b) Ammonia ?
Answer:
Preparation of diborane:
In industries diborane is prepared by the reaction between boroh tri fluoride and lithium hydride.

Boron trichloride and hydrogen mixture subjected to silent electric discharge at low pressure to from diborane.
2BCl₃ + 6H₂ → B₂H₆ + 6HCl
Boron trichloride undergo reduction with LiAlH₄ to form diborane.

a) Reaction with carbon monoxide :
Diborane reacts with CO at 100° C and 20 atm. pressure to form borane carbonyl.

b) Reaction with ammonia
Diborane reacts with ammonia at 120° C first forms B₂H₆.2NH₃ (or) [BH₂(NH₃)₂]⁺[BH₄]^– and on further heating forms borazole (or) borazine. Which is also called as “In organic benzene”. It has iso structural with benzene. Flence it is named as “Inorganic Benzene”.

Solved Problems

Question 1.
Standard electrode potential values, E^Θ for Al³⁺ / Al is – 1.66 V and that of Tl³⁺ / Tl is + 1.26 V. Predict about the formation of M³⁺ ion in solution and compare the electropositive character of the two metals.
Solution:
Standard electrode potential values for two half cell reactions suggest that aluminium has high tendency to make Al³⁺ (aq) ions, whereas Tl³⁺ is not only unstable in solution but is a powerful oxidising agent also. Thus Tl⁺ is more stable in solution than Tl³⁺. Aluminium being able to form +3 ions easily, is more electropositive than thallium.

Question 2.
White fumes appear around the bottle of anhydrous aluminium chloride. Give reason.
Solution:
Anhydrous aluminium chloride is partially hydrolysed with atmospheric moisture to liberate HCl gas. Moist HCl appears white in colour.

Question 3.
Boron is unable to form \(\mathrm{BF}_6^{3-}\) ion. Explain.
Solution:
Due to non – availability of d orbitals, boron is unable to expand its octet. Therefore, the maximum covalence of boron cannot exceed 4.

Question 4.
Why is boric acid considered as a weak acid ?
Solution:
Because it is not able to release H⁺ ions on its own. It receives OH⁺ ions from water molecule to complete its octet and in turn releases H⁺ ions.

Andhra Pradesh BIEAP AP Inter 1st Year Chemistry Study Material 9th Lesson The s-Block Elements Textbook Questions and Answers.

AP Inter 1st Year Chemistry Study Material 9th Lesson The s-Block Elements

Very Short Answer Questions

Question 1.
Give reasons for the diagonal relationship observed in the periodic table.
Answer:

• Diagonal relationship is due to similar sizes of atoms (or) ions
• Diagonal relationship is due to similar electro negativities of the respective elements. Diagonally similar elements possess same polarising power.
• Polarizing Power = \(\frac{\text { ionic charge }}{\text { (ionic radius) }^2}\)

Question 2.
Write completly the electronic configurations of K and Rb.
Answer:
The electronic configuration of ‘K’ is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s²
The electronic configuration of ‘Rb’ is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s¹.

Question 3.
Lithium salts are mostly hydrated. Why ?
Answer:
Hydration enthalpy of Li⁺ ion is very high. It has very high degree of hydration. So ‘Li’ salts are mostly hydrated.
Eg : LiCl . 2H₂O.

Question 4.
Which of the alkali metals shows abnormal density ? What is the order of the varia-tion of density among the IA group elements.
Answer:

• ‘K’ has abnormal density among alkalimetals due to high inter atomic distances in crystal lattice.
• The order of the variation of density among the IA group elements as follows.
  Li < Na > K < Rb < Cs.

Question 5.
Lithium reacts with water less vigorously than sodium. Give your reasons.
Answer:
Lithium reacts with water less vigorously than sodium.

Reasons :

• Lithium has small size.
• Lithium has very high hydration energy.

Question 6.
Lithium Iodide is the most covalent among the alkali metal halides. Give the reasons.
Answer:
Lithium iodide is the most covalent among the alkalimetal halides.

Reasons :

• The polarising capability of lithium ion is high.
• Li⁺ ion has very small size.
• Li⁺ ion has high tendency to distort electron cloud around the iodide ion.

Question 7.
In what respects lithium hydrogen carbonate differs from other alkali metal hydrogen carbonates.
Answer:
Lithium hydrogen carbonate cannot exist in solid form but remaining alkali metal hydrogen carbonates exist as solids.

Question 8.
Write the complete electronic configurations of any two alkaline earth metals.
Answer:

1. The electronic configuration of ‘Mg’ is 1s² 2s² 2p⁶ 3s²
2. The electronic configuration of ‘Ca’ is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s².

Question 9.
Tell about the variation of m.pts., and b.pts among the alkaline earth metals.
Answer:

1. The m.pts and b.pts of alkaline earth metals are higher than corresponding alkali metals due to smaller sizes.
2. Due to low I.P. values the variation of m.pts and b.pts among alkaline earth metals is not sys-tematic.

Question 10.
What are the characterstic colours imparted by the HA elements ?
Answer:
Elements – Imparted colours towards flame
Calcium – Brick red
Strontium – Crimson red
Barium – Apple green
Beryllium – No colour
Magnesium – No colour

Question 11.
What happens when magnesium metal is burnt in air ?
Answer:
Magnesium metal burns with dazzling brilliance in air to give MgO and Mg₃N₂.
2 Mg + O₂ → 2 MgO
3 Mg + N₂ → Mg₃N₂.

Question 12.
Lithium carbonate is not so stable to heat as the other alkali metal carbonates. Explain.
Answer:
Lithium carbonate is not so stable to heat because Lithium has very small size and it polarises the large CO₃^-2 ion which leads to the formation of more stable Li₂O and CO₂.
As the electro positive character increases down the group, the stability of carbonats increase.

Question 13.
Write a balanced equation for the formation of ammoniated IIA metal ions from the metals in liquid ammonia.
Answer:
Alkaline earth metals dissolve in liquid ammonia to give deep blue black solutions forming ammoniated ions.
M + (x + y) NH₃ → [M(NH₃)_x]²⁺ + 2 [e(NH₃)_y]^–
From the above solutions, ammoniates [M(NH₃)_x]²⁺ can be recovered.

Question 14.
The fluorides of alkaline earth metals are relatively less soluble than their respective chlorides in water. Why ?
Answer:
Because of their high lattice energies fluorides of alkaline earth metals are relatively less soluble than their respective chlorides in water.

Question 15.
What happens when hydrated Mg (NO₃)₂ is heated ? Give the balanced equation.
Answer:
When hydrated Mg(NO₃)₂ is heated, it first loses the six water molecules and on further heating
decomposes to give the oxide.
2 Mg (NO₃)₂ → 2 MgO + 4NO₂ + O₂.

Question 16.
Why does the solubility of alkaline earth metal hydroxides in water increases down the group ?
Answer:
Among alkaline earth metal hydroxides, the anion being common the cationic radius will influence the lattice enthalpy. Since lattice enthalpy decreases much more than the hydration enthalpy with increasing ionic size, the solubility increases as we go down the group.

Question 17.
Why does the solubility of alkaline earth metal Carbonates and sulphates in water decrease down the group ?
Answer:
The size of anions being much larger compared to cations, the lattice enthalpy will remain almost constant within a particular group. Since the hydration enthalpies decrease down the group, solubility will decrease as found for alkaline earth metal carbonates and sulphates.

Question 18.
Write the average composition of Portland cement.
Answer:
Composition of port land cement is
Cao – 50 – 60%
Sio₂ – 20 – 25%
Al₂O₃ – 5 – 10%
Mgo – 2 – 3%
Fe₂O₃ – 1 – 2% and
SO₂ – 1 – 2%

Question 19.
Why is gypsum added to cement ?
Answer:
Gypsum is added to cement to slow down the process of setting of the cement and to get sufficiently hardened cement.

Question 20.
Why are alkali metals not found in the free state in nature ? [Mar. 13]
Answer:
Alkali metals are not found in the free state in nature because they readily lose their valency electron to form M⁺ ion (a nonvalent ion).

Question 21.
Potassium carbonate cannot be prepared by Solvay process. Why ?
Answer:
Potassium carbonate cannot be prepared by solvay process because potassium bi carbonate is more soluble and to be precipitated by the addition of ammonium bi carbonate to a saturated solution of potassium chloride.

Question 22.
Describe the important uses of caustic soda.
Answer:
Uses:

• It is used in petrol refining
• It is used in the purification of bauxite.
• It is used in manufacturing of soap, paper.
• It is used in manufacturing of antificial silk.
• It is used in manufacturing of so many chemically.
• It is used in textile industries for mercerising cotton fabrics.
• It is used in preparation of pure fats and oils.
• It is used in as laboratory reagent.

Question 23.
Describe the important uses of sodium carbonate.
Answer:
Uses:

• Na₂CO₃ is used in the manufacturing of glass.
• Na₂CO₃ is used in the manufactuing of borax, caustic soda.
• Na₂CO₃ is used in paper, paints and textile industries.
• Na₂CO₃ is used in softening of water.
• Na₂CO₃ is used in laundries.
• Na₂CO₃ is used an important laboratory reagent both in qualitative and quantitative analysis.

Question 24.
Describe the important uses of quick line.
Answer:
Uses:

• Quick lime is used in the purification of sugar.
• Quick lime is used in the manufacture of dyestuffs.
• Quick lime is used in the manufacture of Na₂CO₃ from NaOH.
• It is an important material for manufacturing of cement and it is the cheapest form of alkali.

Question 25.
Draw the structures of

1. BeCl₂ (vapour) and
2. BeCl₂ (Solid).

Answer:

1. In vapour phase BeCl₂ forms a bridged dimer which disociates into monomer at high temperatures (around 1200 k)
   Cl – Be – Cl
2. In solid state BeCl₂ has a chain structure.
   

Question 26.
Describe the importance of Plaster of Paris.
Answer:

1. Plaster of paris has an important property of setting with water.
2. It forms a hard solid in 5 to 15 min. When it is mixed with suitable quantity of water.
3. It is majorly used in building industry and as well as plasters.
4. It is used in the bone fractures (or) sprain conditions.
5. It is used in dentistry.
6. It is used in manufacturing status and busts.

Question 27.
Which of the alkaline earth metal carbonates is thermally the most stable ? Why ?
Answer:
Among Alkaline earth metal carbonates BaCO₃ is thermally most stable.
Reason :
As the cationic size increases thermal stability also increases. So BaCO₃ is most stable thermally.

Question 28.
Write balanced equations for the reactions between

1. Na₂O₂ and water
2. K₂O and water.

Answer:

1. Na₂O₂ + 2H₂O → 2 NaOH + H₂O₂
2. K₂O + H₂O → 2 KOH.

Short Answer Questions

Question 1.
Alkali metals and their salts impart characteristic colours to an oxidizing flame. Explain the resonizing flame. Explain the reason.
Answer:
Alkali metals and their salts impart characterstic colours to an oxidizing flame.

Reasons :
The heat from the flame excites the outer most orbital electron to a higher energy level. When the excited electron emitts the radiation and comes back to the ground state. This falls in the visible region.

Question 2.
What makes caesium and potassium useful as electrodes in photoelectric cells ?
Answer:

• Alkali metals can be detected by the respective flame tests and can be determined by the flame photo metry (or) atomic absorption spectroscopy.
• Alkali metals when irradiated with light, the light energy absorbed may be sufficient to make an atom lose electron.
• This makes caesium and potassium useful as electrodes in photo electric cells.

Question 3.
Write a short note on the reactivity of alkali metals towards air.
Answer:

• The alkali metals forms their oxides in presence of dry air and tarnished.
• These oxides reacts with moisture to form hydroxides.
• They burn vigorously in oxygen and forms oxides.
  • Lithium forms Lithium monoxide.
  • Sodium forms monoxide with limited supply of oxygen and peroxide with excess of oxygen.
  • Other metals of this group forms super oxides. The super oxide ion (O₂^–) is stable only in presence of large cations.

Reactions :
4Li + O₂ → 2 Li₂O
4Na + O₂ (Limited) → 2 Na₂O₂
2Na + O₂ (Excess) → Na₂O₂
K + O₂ (Excess) → KO₂
Lithium shows a different character. It reacts directly with nitrogen of air and forms Li₃N (Lithium nitride).

Question 4.
Give any two uses for each of the following metals.

1. Lithium
2. Sodium.

Answer:

1. Uses of Lithium : –

• ‘Li’ metal is used to make alloys.
  Eg: 1) Lithium with lead forms an alloy which is used for making white metal bearings for motor engines.
  2) Lithium with aluminium forms alloys which are used to make air craft parts.
• ‘Li’ metal is used in thermo nuclear reactions.
• ‘Li’ metal is used to make electro chemical cells.

2) Uses of sodium : –

• Sodium metal is used to make Na and Pb alloy needed to make TEL. this TEL (tetra ethyl lead) is used as antiknock additives to petrol.
• Liquid ‘Na’ metal is used as a coolant in fast breeder nuclear reactors.
• ‘Na’ metal is used in the manufacturing of rabber.

Question 5.
Give an account of properties of washing soda.
Answer:
Properties of washing soda : –

• Na₂CO₃ is a white crystalline solid.
• Na₂CO₃ exists as a decahydrate Na₂CO₃. 10H₂O which is called washing soda.
• Na₂CO₃ is readily soluble in water.
• Na₂CO₃ (deca hydrate) when heated it loses the water molecules and forms monohydrate. This monohydrate on heating above 373 K it forms anhydrous form which is called soda ash, a white powder.
  Reactions : –
  
• Na₂CO₃ (aq) solution is alkaline (basic) in nature Because it under goes anionic hydrolysis. (PH > 7).
  CO₃^-2 + H₂O → HCO₃^– + OH^–

Question 6.
Mention some uses of sodium carbonate.
Answer:
Uses :

• Na₂CO₃ is used in the manufacturing of glass.
• Na₂CO₃ is used in the manufacturing of borax, caustic soda.
• Na₂CO₃ is used in paper, paints and textile industries. .
• Na₂CO₃ is used in softening of water.
• Na₂CO₃ is used in laundries.
• Na₂CO₃ is used an important laboratory reagent both in qualitative and quantitative analysis.

Question 7.
How do you obtain pure sodium chloride from a crude sample ?
Answer:

1. Crude sodium chloride is obtained by the crystallisation of brine solution.
2. Crude sodium chloride contains sodium sulphate, calcium sulphate, calcium chloride and magnesium chloride.
3. CaCl₂ and MgCl₂ are the impurities in the Crude NaCl because these absorb moisture easily from the atmosphere.
4. Pure sodium chloride’s obtained by dissolving Crude NaCl in minimum amount of water and filtered to remove insoluble impurities.
5. This solution is saturated with HCl gas. Then crystals of pure NaCl are separated out.
6. Ca and Mg chloride are more soluble in solution than NaCl and these remained in the solution.

Question 8.
What do you know about Castner – Kellner process ? Write the principle involved in it.
Answer:

• Castner – Kellner process is a commercial method used for the preparation of sodium hydroxide.
• In this process sodium hydroxide is prepared by the electrolysis of sodium chloride in Castner – Kellner cell.
• Brine solution is electrolysed using a mercury cathode and a carbon anode.
• Sodium metal is formed at cathode and it combine with mercury to form sodium amalgam. Chlorine gas is evolved at anode.
• The amalgam is treated with water to form, sodium hydroxide.
  Cell Reactions :
  2NaCl → 2Na⁺ + 2Cl^–
  2Na⁺ + 2e^– 2Na – amalgam
  2Cl^– → Cl₂ + 2e^–
  2Na – amalgam + 2H₂O → 2NaOH + 2Hg + H₂
• This process is also called as mercury cathode process.

Question 9.
Write a few applications of caustic soda.
Answer:
Uses :

• It is used in petrol refining
• It is used in the purification of bauxite.
• It is used in manufacturing of soap, paper.
• It is used in manufacturing of antificial silk.
• It is used in manufacturing of so many chemically.
• It is used in textile industries for mercerising cotton fabrics.
• It is used in preparation of pure fats and oils.
• It is used in as laboratory reagent.

Question 10.
Give an account of the biological importance of Na⁺ and K⁺ ions.
Answer:

• Na⁺ ions participate in the transmission of nerve signals.
• Na⁺ ions regulates the flow of water accross cell membranes.
• Na⁺ ions responsible for transport of sugars and amino acids into cells.
• K⁺ ions are useful in activating enzymes.
• K⁺ ions participate in the oxidation of glucose to produce ATP.
• K⁺ along with Na+ responsible for the transmission of nerve signals.

Question 11.
Mention the important uses of Mg metal.
Answer:

1. Magnesium forms so many useful alloys with Al, Zn, Mn and Sn.
2. Mg – Al alloys are useful in air – craft construction.
3. Mg powder and ribbon is used in flash powders bulbs.
4. Mg is used in incendiary bombs and signals.
5. Milk of magnesice is used as antacid in medicine.
6. MgCO₃ is the main ingradient in tooth pastes.

Question 12.
Show that Be(OH)₂ is amphoteric in nature.
Answer:

1. Be(OH)₂ is amphoteric in nature. This can be evidented by the following reactions.
2. Be(OH)₂ reacts with both acids and alkalis.
   Be(OH)₂ + 2OH^– [Be(OH)₄]^2- (Beryllation)
   Be(OH)₂ + 2HCl + 2H₂O → [Be(OH)₄]Cl₂
3. Hence Be(OH)₂ is amphoteric in nature.

Question 13.
Write a note on anomalous behaviour of beryllium.
Answer:
Anomalous characters of Be:
As was already discussed in the earlier sections, the first element shows some differences from the properties of the other elements in the group. Be differs from the other alkaline earth metals because of its small size and high electronegativity. Be differs from the other elements in the following aspects.

1. Be compounds are predominantly covalent due to its high polarizing power and its salts are readily Hydrolyzed.
2. Be is not easily affected by dry air and does not decompose water at ordinary temperature.
3. Be is an amphoteric metal. It dissolves in alkali solutions forming beryllates.
4. Be SO₄ is soluble in water whereas the sulphates of Ca, Sr and Ba are not soluble.
5. Be and its salts do not respond to flame test while Ca, Sr and Ba give characteristic flame colours.
6. Be forms many complexes while the heavier elements do not show a great tendency to form complexes.
7. Be has a maximum covalency of 4 while other can have a maximum covalency of 6.

Question 14.
Be shows diagonal relationship with Al. Discuss.
Answer:

1. ‘Be’ shows diagonal relation ship with ‘Al’.
2. The ionic radius of Be⁺² is nearly same as that of Al⁺³ so ‘Be’ resembles ‘Al’ in several ways.
3. Al, Be both not readily reacts with acids. This is due to the presence of an oxide film on the surface of metal.
4. Al(OH)₃, Be(OH)₂ both similarly dissolves in excess of alkali and forms Beryllate ion [Be(OH)₃]^2- and Aluminate [Al(OH)₄] ions respectively.
5. The chlorides of Be, Al have bridged chloride structures in vapour phase.
6. Both the chlorides of Be, Al used as strong Lewis acids.
7. Both the chlorides of Be, Al used in Friedal craft catalysts.
8. Be, Al ions have strong tendency to form complexes.

Question 15.
What is Plaster of Paris ? Write a short note on it. [T.S. Mar. 16]
Answer:
Plaster of paris is the hemi hydrate of CaSO₄ with formula CaSO₄. \(\frac{1}{2}\)H₂O.
Preparation: –
Plaster of paris is obtained by heating gypsum at 393 K.

1. If temperature is used greater than 393 K then an hydrous CaSO₄ is formed which is called ‘dead burnt plaster’.
2. Plaster of paris has an important property of setting with water.
3. It forms a hard solid in 5 to 15 min. when it is mixed with suitable quantity of water.
4. It is majorly used in building industry and as well as plasters.
5. It is used in the bone fractures (or) sprain conditions.
6. It is used in dentistry.
7. It is used in manufacturing status and busts.

Question 16.
In what ways lithium shows similarities to magnesium in its chemical behaviour ?
Answer:
Diagonal relationship of Li : In the periodic table an element of a group in the 2^nd period shows similar properties with the second element of the next group in the third period. This relation is known as diagonal relationship. For examples. Lithium and Magnesium show diagonal relationship. The elements show that diagonal relationship have similar polarizing powers, electronegativities, nature of the compounds. The diagonal similarity may be due to the effects of size and charge. For example, charge per unit area.
Lithium shows similarity to Magnesium in the following respects.
a) Lithium is slow to react with water. Magnesium decomposes water only in the hot condition.
2Li + 2H₂O → + H₂;
Mg + 2H₂O → Mg(OH)₂ + H₂

b) Lithium combines directly with N2 forming nitride.
6Li + N₂ → 2Li₃N

c) Both Lithium and Magnesium give only monoxides Li₂O, MgO.

d) Lithium chloride is deliquescent like MgCl₂, LiCl undergoes hydrolysis to a smaller extent in hot water in a similar way to MgCl₂.

e) Due to their covalent nature, the halides Lithium and Magnesium are soluble in organic solvents.

f) Both Li⁺ and Mg⁺² are highly hydrated.

g) The Carbonates, Phosphates and Fluorides of both Li and Mg are sparingly soluble in water.

h) Lithium alkyls (Li⁺ R^–) are chemically similar to Grignard reagents in organic synthesis.

Question 17.
When an alkali metal dissolves in liquid ammonia the solution can acquire different colours. Explain the reasons for this type of colour change.
Answer:

• The alkali metals dissolve in liquid NH3 and gives deep blue solutions. These are conducting in nature.
• The blue colour of the solution is due to the ammoniated electrons which absorbs energy in the visible region of light and thus imparts blue colour to the solution.
• These solutions are paramagnetic and on standing liberate hydrogen resulting in the formation of amide.
  M + (x + y) NH₃ → [M(NH₃)_x]⁺ + [e(NH₃)_y]^–
  M_(am)⁺ + e^– + NH₃ → MNH_2(am) + 1/2 H_2(g)
• In concentrated solution the blue colour Changes to bronze.colour on warming and becomes did magnetic.

Question 18.
What happens when

1. Sodium metal is dropped in water ?
2. Sodium metal is heated in a free supply of air ?
3. Sodium peroxide dissolves in water ?

Answer:

1. Sodium metal when dropped in water it reacts with water vigourously and liberates H₂ gas.
   2Na + 2H₂O → 2NaOH + H₂
2. Sodium metal is heated in free supply of air to form sodium peroxide.
   2Na + O₂ → Na₂O₂ (sodium peroxide)
3. Sodium peroxide dissolves in water and forms NaOH and hydrogen peroxide.
   Na₂O₂ + 2H₂O → 2NaOH + H₂O₂

Question 19.
States as to why
i) An aqueous solution of Na₂CO₃^– is alkaline ;
ii) Alkali metals are prepared by the electrolysis of their fused chlorides ?
Answer:
i) An aqueous solution of Na₂CO₃ is alkaline. This is due to anionic (CO₃^-2) hydrolysis.
CO₃^-2 + H₂O → HCO₃^-2 + OH^–
∴ PH > 7. So the solution is alkaline in nature.

ii) Chemically alkali metals are highly reactive and they are placed at top in the electro chemical series.
∴ Common methods of extraction of the metals are not applicable for the alkali metals. So electrolytic reduction of their used chlorides is the possible method for extracting alkali metals.
Eg: ‘Na’ metal obtained from fused ‘NaCl’.

Question 20.
How would you explain the following observations ?

1. BeO is almost insoluble but BeSO₄ is soluble in water ?
2. BaO is soluble but BaSO₄ is insoluble in water ?

Answer:

1. BeO has amphoteric nature and the solubility in water is low because of its covalent nature.
   BeSO₄ is soluble in water. This is due to greater hydration energy of Be⁺² ion.
2. BaO is soluble in water because of its ionic nature.
   BaSO₄ is insoluble in water because of low hydration energy of Ba⁺² ion.

Long Answer Questions

Question 1.
Justify the inclusion of alkali metals in the same group of the periodic table with reference to the following :
i) Electronic configuration
ii) Reducing nature
iii) Oxides and hydroxides.
Answer:
i) Electronic configuration : All the alkali metals have one valence electron, ns¹.

ii) Reducing nature :

• Alkali metals are strong reducing agents.
• ‘Li’ is most powerful reducing agent and ‘Na’ is poor reducing agent. .
  The standard electrode potential (E°) is the measure of reducing power.
• ‘Li’ has highest hydration enthalpy. It has high negative S.E.P. (E°) hence it has high reducing power.

iii) Oxides and hydroxides : –

• The alkali metals forms their oxides in presence of dry air and tarnished.
• These oxides reacts with , .noisture to form hydroxides.
• They burn vigorously in oxygen and forms oxides.
  1. Lithium forms lithium monoxide.
  2. Sodium forms monoxide with limited supply of oxygen and peroxide with’ excess of oxygen
  3. Other metals of this group forms super oxides. The super oxide ion (O₂^–) is stable only in presence of large cations.

Reactions:
4Li + O₂ → 2 Li₂O
4Na + O₂ (Limited) → 2 Na₂O
2 Na + O₂ (Excess) → Na₂O₂
K + O₂ (Excess) → KO₂
Lithium shows a different character. It reacts directly with nitrogen of air and forms Li₃N (Lithium nitride).

• Alkali metal oxides easily hydrolysed by water to form the hydroxides.
  Monoxide : M₂O + H₂O → 2MOH
  Peroxide : M₂O₂ + 2H₂O → 2MOH + H₂O₂
  Superoxide : 2MO₂ + 2H₂O → 2MOH + H₂O₂ + O₂
• Oxides, peroxides are colourless whereas superoxides are coloured because of their para mag-netic property.
• Hydroxides are white crystalline solids.
• Hydroxides are strong bases and dissolved freely in water and evolve much heat.

Question 2.
Write an essay on the differences between lithium and other alkali metals.
Answer:
Anomalous properties of Lithium : In the periodic table some representative elements of different series show similarities, known as diagonal relationship. Li, of the alkali metals, show such a similarity with Mg of II group. That means it differs from alkali metals. Some of the important abnormal characters of lithium are given below.
a) Lithium is hard metal while other alkali metals are soft and can be cut with the knife. Its melting point and boiling point are high.

b) Lithium directly unites with N₂ while no other alkali metal combines directly.
6Li + N₂ → 2Li₃N

c) Lithium element forms a carbide on direct combination. Group IA elements do not form directly. But all these elements are known to give carbides.

d) The solubilities of Lithium Hydroxide (LiOH), Lithium Carbonate (Li₂CO₃), Lithium Phosphate (Li₃PO₄) and Lithium Fluoride (LiF), are very less compared to the high solubilities of the other alkali metal compounds.

e) Lithium Hydroxide is a weaker alkali than the alkali metal Hydroxides. Basic nature of other alkali metal Hydroxides is more than Li(OH). Because of this Lithium Hydroxide Carbonate, nitrates are thermally unstable.

Question 3.
Discuss the preparation and properties of sodium carbonate.
Answer:
Preparation :

• Sodium carbonate is prepared by solvay process.
• In this process sodium chloride reacts with ammonium bicarbonate and gets precipitated the low soluble sodium bicarbonate.
• The lather is prepared by passing CO₂ into a concentrated solution of NaCl saturated with ammonia. Here (NH₄)₂CO₃ followed by NH₄ HCO₃ are formed.
  Chemical equations involved :
  2NH₃ + H₂O + CO₂ → (NH₄)₂ CO₃
  (NH₄)₂ CO₃ + H₂O + CO₂ → 2NH₄HCO₃
  NH₄HCO₃ + NaCl → NH₄Cl + NaHCO₃
• The separated NaHC03 Crystals heated to get Na₂CO₃.
  2NaHCO₃ → Na₂CO₃ + H₂O + CO₂
• In this process ammonia is regenerated by Ca(OH)₂.
  2NH₄Cl + Ca(OH)₂ → 2NH₃ + CaCl₂ + H₂O

Properties of washing soda :

• Na₂CO₃ is a white crystalline solid.
• Na₂CO₃ exists as a decahydrate Na₂CO₃. 10H₂O which is called washing soda.
• Na₂CO₃ is readily soluble in water.
• Na₂CO₃ (decahydrate) when heated it loses the water molecules and forms monohydrate. This monohydrate on heating above 373 K it forms anhydrous form which is called soda ash, a white powder.
  Reactions : –
  
• Na₂CO₃ (aq) solution is alkaline (basic) in nature Because it under goes anionic hydrolysis. (PH > 7).
  CO₃^-2 + H₂O → HCO₃^– + OH^–

Uses :

• Na₂CO₃ is used in the manufactuing of glass.
• Na₂CO₃ is used in the manufactuing of borax, caustic soda.
• Na₂CO₃ is used in paper, paints and textile industries.
• Na₂CO₃ is used in softening of water.
• Na₂CO₃ is used in laundries.
• Na₂CO₃ is used an important laboratory reagent both in qualitative and quantitative analysis.

Question 4.
Discuss the similarities between alkaline earth metals and gradation in the following aspects.
i) Electronic configuration
ii) Hydration enthalpies
iii) Nature of oxides and hydroxides.
Answer:
i) Electronic configuration :
The general electronic configuration of alkaline earth metals represented by [noble gas] ns².

ii) Hydration Enthalpies :

• The hydration enthalpies of alkaline earth metal ions decrease with increase in ionic size down the group.
  Be⁺² > Mg⁺² > Ca⁺² > Sr⁺² > Ba⁺²
• Hydration enthalpies of these elements ions are larger than those of alkali metal ions.
  Eg : MgCl₂ MgCl₂ . 6H₂O
  CaCl₂ CaCl₂ . 6H₂O

iii) Nature of oxides and hydroxides : –

• Alkaline earth metals forms oxides of type MO.
• These are formed by burning in oxygen.
• BeO is amphoteric and covalent in nature whereas other oxides are ionic and basic in nature. Other oxides i.e., except BeO forms hydroxides with water.
  Eg : MgO + H₂O → Mg(OH)₂
• The solubility, thermal stability and the basic character of these hydroxides increase with increase of atomic no. from Mg(OH)₂ to Ba(OH)₂. H
• These hydroxides are less basic, less stable than alkali metal hydroxide.
• Be(OH)₂ is amphoteric in nature. This can be evidented by the following reactions.
• Be(OH)₂ reacts with both acids and alkalis.
  Be(OH)₂ + 2OH^– [Be(OH)₄]^2- (Beryllation)
  Be(OH)₂ + 2HCl + 2H₂O → [Be(OH)₄]Cl₂
• Hence Be(OH)₂ is amphoteric in nature.

Question 5.
Discuss on;
i) Carbonates
ii) Sulphates and
iii) Nitrates of alkaline earth metals.
Answer:
i) Carbonates :

• Alkaline earth metals forms MCO₃ type carbonates.
• These Carbonates are insoluble in water.
• The solubility of these carbonates in water decreases as the atomic no.of the element increases.
•  The thermal stability increases with increasing cationic size.
• These carbonates decompose on heating to give CO₂
  CaCO₃ Cao + CO₂
• BeCO₃ is unstable and kept only in at atmosphere of CO₂.

ii) Sulphates :

• Alkaline earth metals forms MSO₄ type sulphates.
• These are white solids and are stable to heat.
• BeSO₄ and MgSO₄ are readily soluble in water due to high hydration enthalies of Be²⁺, Mg²⁺.
• The solubility decrease from CaSO₄ to BaSO₄.

iii) Nitrates :

• Alkaline earth metals forms M(NO₃)₂ type Nitrates.
• These are formed by the reaction of carbonates in dil.HNO₃.
• Mg(N03)2 crystallises with six water molecules and Ba(NO₃)₂ is an hydrous.
• All of these decompose on heating to give the respective oxides.
  2M(NO₃)₂ → 2MO + 4NO₂ + O₂
  M = Be, Mg, Ca, Sr, Ba.

Question 6.
What are the common physical and chemical features of alkali metals ?
Answer:
Physical features :

• Alkali metals are silvery white, soft and light metals.
• These elements have low density which increases down the group from Li to Cs. (one exception density of K < density of Na).
• The m.pts, b.pts of alkali metals are low.
• Alkali metals and their salts impart characterstic colours to an oxidizing flame.

Reasons :
The heat from the flame excities the outer most orbital electron to a higher energy level. When the excited electron emitts the radiation and comes back to the ground state. This falls in the visible region.

• Alkali metals can be detected by the respective flame tests and can be determined by the flame photo metry (or) atomic absorption spectroscopy.
• Alkali metals when irradiated with light, the light energy absorbed may be sufficient to make an atom lose electron. .
• This makes calsium and potassium useful as electrodes in photo electric cells.

Chemical features : –
i) Reactivity towards Air : —

• The alkali metals forms their oxides in presence of dry air and tarnished.
• These oxides reacts with moisture to form hydroxides.
• They burn vigorously in oxygen and forms oxides.
  1. Lithium forms Lithium monoxide.
  2. Sodium forms monoxide with limited supply of oxygen and peroxide with excess of oxygen.
  3. Other metals of this group forms super oxides. The super oxide ion (O₂^–) is stable only in presence of large cations.
• Reasons :
  4Li + O₂ → 2 Li₂O
  4Na + O₂ (Limited) → 2 Na₂O
  2Na + O₂ (Excess) → Na₂O₂
  K + O₂ (Excess) → KO₂
• Lithium shows a different character. It reacts directly with nitrogen of air and forms Li₃N (Lithium nitride).

ii) Reactivity with H₂: Alkali metals react with H₂ directly at 300 – 600° C and form hydrides. The reaction can be written as follows :

Where M = Li, Na, K, Rb or Cs. These hydrides are ionic in nature. Their ionic nature increases with the metalic nature of alkali metals.

iii) Reactivity with halogens : All the alkali’ metals react with halogens to give the binary compounds. The chemical reactivity in the alkali metals increases with increase in atomic number.
2M + X₂ → 2MX (where M is any alkali metal)
All the metal halides are ionic compounds.

iv) Reactivity with water : The alkali metals decompose water vigorously and liberate hydrogen gas! The chemical reactivity of these metals increases as the atomic number increases. The metal hydroxides are formed.
2M + 2H₂O → 2MOH + H₂
Where M = any one of the alkali metals.

ii) Reducing nature :

• Alkali metals are strong reducing agents.
• ‘Li1 is most powerful reducing agent and ‘Na’ is poor reducing agent.
• The standard electrode potential (E°) is the measure of reducing power.
• ‘Li’ has highest hydration enthalpy. It has high negative S.E.P. (E°) hence it has high reducing power.
• The alkali metals dissolves in liquid NH3 and gives deep blue solutions. These are conducting in nature.
• The blue colour of the solution is due to the ammoniated electrons which absorbs energy in the visible region of light and thus imparts blue colour to the solution.
• These solutions are paramagnetic and on standing Liberate hydrogen resulting in the formation of amide.
  M + (x + y) NH₃ → [M(NH₃)_x]⁺ + [e(NH₃)_y]^–
  M_(am)⁺ + e^– + NH₃ → MNH_2(am) + 1/2 H_2(g)
• In concentrated solution the blue colour Changes to bronze.colour on warming and becomes did magnetic.

Question 7.
Discuss the general characterstics and gradation in properties of alkaline earth metals.
Answer:
The general characterstics and gradation in properties of alkaline earth metals follows.
i) Oxides and hydroxides :

• Alkaline earth metals forms oxides of type Mo.
• These are formed by burning in oxygen.
• BeO is amphoteric and covalent in nature where as other oxides are ionic and basic in nature.
• Other oxides i.e., except Beo forms hydroxides with water.
  Eg : MgO + H₂O → Mg(OH)₂
• The solubility, thermal stability and the basic character of these hydroxides increase with increase of atomic no. from Mg(OH)₂ to Ba(OH)₂.
• These hydroxides are less basic, less stable than alkali metal hydroxide.
• Be(OH)₂ is amphoteric in nature. This can be evidented by the following reactions.
• Be(OH)₂ reacts with both acids and alkalis.
  Be(OH)₂ + 2OH^– [Be(OH)₄]^2- (Beryllation)
  Be(OH)₂ + 2HCl + 2H₂O → [Be(OH)₄]Cl₂
• Hence Be(OH)₂ is amphoteric in nature.

ii) Halides:-

• These forms MX₂ type halides.
• Except Be – halides, all other halides of these metals are ionic.
• Be – halides are covalent and soluble in organic solvents.
• In vapour phase BeCl₂ forms a bridged dimer which disociates into monomer at high temperatures (around 1200 k)
  Cl – Be – Cl
• In solid state BeCl₂ has a chain structure.
  
• The tendency to form halide hydrates gradually decreases – down the group.
  Eg : MgCl₂, 8H₂O, CaCl₂, 6H₂O, BaCl₂. 2H₂O.
• Ca, Sr and Ba halides, dehydration can be done by heating.
• Fluorides are less soluble than the chlorides due to their high lattice energies.

i) Carbonates :

• Alkaline earth metals forms MCO₃ type carbonates.
• These Carbonates are insoluble in water.
• The solubility of these carbonates in water decreases as the atomic no.of the element increases.
• The thermal stability increases with increasing cationic size.
• These carbonates decompose on heating to give CO₂
  CaCO₃ Cao + CO₂
• BeCO₃ is unstable and kept only in at atmosphere of CO₂.

ii) Sulphates:

• Alkaline earth metals forms MSO₄ type sulphates.
• These are white solids and are stable to heat.
• BeSO₄ and MgSO₄ are readily soluble in water due to high hydration enthalpies of Be²⁺, Mg²⁺.
• The solubility decrease from CaSO₄ to BaSO₄.

iii) Nitrates:

• Alkaline earth metals forms M(NO₃)₂ type Nitrates.
• These are formed by the reaction of carbonates in dil.HNO₃.
• Mg(NO₃)₂ crystallises with six water molecules and Ba(NO₃)₂ is an hydrous.
• All of these decompose on heating to give the respective oxides.
  2M(NO₃)₂ → 2MO + 4NO₂ + O₂
  M = Be, Mg, Ca, Sr, Ba.

Question 8.
Discuss the various reactions that occur in the solvay process. [A.P. Mar. 16]
Answer:
Preparation:

• Sodium carbonate is prepared by solvay process.
• In this process sodium chloride reacts with ammonium bicarbonate and gets precipitated the low soluble sodium bicarbonate. ‘
• The lather is prepared by passing CO₂ in to a concentrated solution of NaCl saturated with ammonia. Here (NH₄)₂CO₃ followed by NH₄ HCO₃ are formed.
• Chemical equations involved:
  2NH₃ + H₂O + CO₂ → (NH₄)₂ CO₃
  (NH₄)₂ CO₃ + H₂O + CO₂ → 2NH₄HCO₃
  NH₄HCO₃ + NaCl → NH₄Cl + NaHCO₃
• The separated NaHCO₃ Crystals heated to get Na₂CO₃.
  2NaHCO₃ → Na₂CO₃ + H₂O + CO₂
• In this process ammonia is regenerated by Ca(OH)₂.
  2NH₄Cl + Ca(OH)₂ → 2NH₃ + CaCl₂ + H₂O

Question 9.
Starting with sodium chloride how would you proceed to prepare
i) Sodium metal
ii) Sodium hydroxide
iii) Sodium peroxide
iv) sodium carbonate.
Answer:
i) Fused NaCl on electrolysis gives sodium metal.
2NaCl → 2Na⁺ + 2Cl^–
2Na⁺ + 2e^– → 2Na (Cathode)
2Cl^– → Cl₂ + 2e^– (anode)

ii)

• Castner-Kellner process is a commercial method used for the preparation of sodium hydroxide.
• In this process sodium hydroxide is prepared by the electrolysis of sodium chloride in Castner – Kellner cel..
• Brine solution is electrolysed using a mercury cathode and a carbon anode.
• Sodium metal is formed at cathode and it combine with mercury to form sodium amalgam. Chlorine gas is evolved at anode.
• The amalgam is streated with water to form sodium hydroxide.
  Cell Reactions:
  2NaCl → 2Na⁺ + 2Cl^–
  2Na⁺ + 2e^– 2Na – amalgam
  2Cl^– → Cl₂ + 2e^–
  2Na-amalgam + 2H₂O → 2NaOH + 2Hg + H₂
• This process is also called as mercury cathode process.

iii) The obtained Na – metal reacts with excess of oxygen to form sodium peroxide.
2Na + O₂ → Na₂O₂ (Sodium peroxide.)

iv) Preparation:

1. Sodium carbonate is prepared by solvay process.
2. In this process sodium chloride reacts with ammonium bicarbonate and gets precipitated the low soluble sodium bicarbonate.
3. The lather is prepared by passing CO₂ into a concentrated solution of NaCl saturated with ammonia. Here (NH₄)₂CO₃ followed by NH₄ HCO₃ are formed.

Question 10.
What happens when
i) Magnesium is burnt in air ?
ii) Quick lime is heated with silica
iii) Chlorine reacts with slaked lime
iv) calcium nitrate is strongly heated.
Answer:
i) Magnesium burns with dazzling brilliance in air to give MgO and Mg₃N₂.
2Mg + O₂ →2MgO
3Mg + N₂ → Mg₃N₂

ii) Quick lime heated with silica to form calcium silicate
Cao + SiO₂ → CaSio₃

iii) Slaked lime reacts with chlorine gas to form bleaching powder.

iv) Calcium nitrate on strong heating to form respective oxide
2Ca(NO₃)₂ → 2CaO + 4NO₂ + O₂.

Question 11.
Explain the significance of sodium, potassium, magnesium and calcium in biological fluids.
Answer:
Biological importance of Na, K.

• Na⁺ ions participate in the transmission of nerve signals.
• Na⁺ ions regulates the flow of water accross cell membranes.
• Na⁺ ions responsible for transport of sugars and amino acids into cells.
• K⁺ ions are useful in activating enzymes.
• K⁺ ions participate in the oxidation of glucose to produce ATP.
• K⁺ along with Na⁺ responsible for the transmission of nerve signals.

Biological importance of Mg and Ca :
Role of Mg²⁺ in biology :

1. Mg²⁺ ions are concentrated in animal cells.
2. Enzymes like “phosphohydrolases1 and ‘Phospho transferases’ contain Mg²⁺ ions. These enzymes participate in ATP reactions and release energy in the process. Mg²⁺ forms a complex with ATP.
3. Mg²⁺ is a constituent of chlorophyll, the green component of plants.

Role of Ca⁺²:
About 99% of body calcium is present in bones and teeth. It also plays important roles in neuromuscular function, interneuronal transmission, cell membrance integrity and blood coaqulation.

The calcium concentration in plasma is regulated at about 100 mg/Lit. It is maintained by two hormones, calcitonin and parathyroid hormone. Ca²⁺ ion are necessary for muscle contraction.

Question 12.
Write few lines about cement ?
Answer:

• Cement is an important building material.
• It is also called portland cement.
• Cement is obtained by combining a material rich in lime, CaO with other material such as clay which contains Sio₂ along with the oxides of Al, Fe and Mg.
• Composition of port land cement is
  Cao – 50 – 60%
  Sio₂ – 20 – 25%
  Al₂O₃ – 5 – 10%
  Mgo – 2 – 3%
  Fe₂O₃ – 1 – 2%
  and SO₂ – 1 – 2%
• For a good quality of cement the ratio of SiO₂ to Al₂O₃ is between 2.5 and 4 and the ratio of lime (Co) to the total of the oxides of SiO₂, Al₂O₃ and Fe₂O₃ is as close as ‘2’.
• The raw materials used for the manufacture of cement are lime stone and clay.
  Clay + lime cement clinker.
• This cement clinker mixed with 2 – 3% by wt. of gypsum to form cement.
• Important ingradients in portland cement are
  Ca₂SiO₄ – 26%, Ca₃SiO₅ – 51 % and Ca₃Al₂O₆ – 11 %

Setting of Cement: –

• Cement mixed with water to give a hard mass i.e setting of cement takes place.
• This is due to the hydration of molecules of the cement.
• The purpose of adding gypsum is to slow down the process of setting and to get sufficiently hardness.

Uses:

• It is used in concrete and rein forced concrete.
• It is used in plastering.
• It is used in construction of bridges, dams and buildings.

Solved Problems

Question 1.
What is the oxidation state of K in KO₂ ?
Solution:
The superoxide species is represented as O₂^–; since the compound is neutral, the oxidation state of potassium is +1.

Question 2.
The E^Θ for Cl₂ / Cl^– is + 1.36, for I₂/I^– is + 0.53, for Ag⁺/Ag is + 0.79, Na⁺/Na is – 2.71 and for Li⁺/Li is – 3.04. Arrange the following ionic species in decreasing order of reducing strength : I^–, Ag, Cl^–, Li, Na
Solution:
The order is Li > Na > I^– > Ag > Cl^–.

Question 3.
Why is KO₂ paramagnetic ? [T.S. Mar. 16]
Solution:
The superoxide O₂^– is paramagnetic because of one unpaired electron in π*2p molecular orbital.

Question 4.
Why does the solubility of alkaline earth metal hydroxides in water increases down the group ?
Solution:
Among alkaline earth metal hydroxides, the anion being common the cationic radius will influence the lattice enthalpy. Since lattice enthalpy decreases much more than the hydration enthalpy with increasing ionic size, the solubilit y increases as we go down the group.

Question 5.
Why does the solubility of alkaline earth metal carbonates and sulphates in water decrease down the group ?
Solution:
The size of anions being much larger compared to cations, the lattice enthalpy will remain almost constant within a particular group. Since the hydration enthalpies decrease down the group, solubility will decrease as found for alkaline earth metal carbonates and sulphates.

Andhra Pradesh BIEAP AP Inter 1st Year Chemistry Study Material 8th Lesson Hydrogen and its Compounds Textbook Questions and Answers.

AP Inter 1st Year Chemistry Study Material 8th Lesson Hydrogen and its Compounds

Very Short Answer Questions

Question 1.
The three isotopes of hydrogen differ in their rates of reaction. Give the reasons.
Answer:
The three isotopes of hydrogen differ in their rates of reactions due to their different enthalpies of bond dissociation.

Question 2.
Why is dihydrogen used in the welding of high melting metals?
Answer:
Dihydrogen is used in welding high melting metals because the atomic hydrogens which are formed by the dissociation of dihydrogen are recombined on the surface of the metal to be welded and produce a high temperature of 4000 K.

Question 3.
Describe one method of producing high purity hydrogen.
Answer:
Highly pure hydrogen is obtained by the electrolysis of hot aq.Ba(OH)₂ solution between nickel electrodes. The purity of hydrogen is about more than 99.95%.

Question 4.
Explain the term “SYNGAS”.
Answer:
The mixture of CO and H₂ which is used for the synthesis of methanol and a number of hydrocar-bons is called “SYNGAS”. It is also called as synthesis gas.
Preparation :

This reaction is called ‘Coal gasification’.

Question 5.
What is meant by coal gasification ? Explain with relevant, balanced equation.
Answer:
The process of producing synthesis gas (SYNGAS) by using coal at 1270K temperature is called coal gasification.
Balanced equation :

Question 6.
Define the term Hydride. How many categories of hydrides are known ? Name them.
Answer:
The binary compounds of hydrogen formed by the other elements except noble gases are called hydrides.
Hydrides are categorised into three types.

1.  Ionic hydrides
2. Covalent hydrides
3. Metallic hydrides

Question 7.
The unusual property of water in condensed phase leads to its high heat of vapourization. What is that property ?
Answer:
In water inter molecular hydrogen bonding is present. Due to this unusual property water has high Freezing point. Boiling point and high heat of vapourization.

Question 8.
During photosynthesis, water is oxidised to O₂. Which element is reduced ?
Answer:
During photosynthesis, the element reduced is carbon.

Oxidation state change from +4 to 0 [Reduction]

Question 9.
What do you mean by autoprotolysis ? Give the equation to represent the autoprotolysis of water.
Answer:
Water has the ability to behave as an acid as well as base. It behaves as an amphoteric substance. The self-ionising property of water is called auto protolysis.
The equation that represent the auto protolysis of water is as follows.

Question 10.
Water behaves as an amphoteric substance in the Bronsted sense. How do you explain ?
Answer:
According to Bronsted proton donor is acid and acceptor is base. Water has the ability to behave as an acid as well as base. So it is an amphoteric substance.
Eg :

1. Water acts as acid with ammonia.
   H₂O_(l) + NH_3(aq) ⇌ OH^Θ_(aq) + \(\mathrm{NH}_{4 \text { (aq) }}^{\oplus}\)
   Here water is proton donor.
2. Water acts as base with H₂S.
   H₂O_(l) + H₂S_(aq) ⇌ H₃O^⊕_(aq) + HS^Θ_(aq)
   Here water is proton acceptor.

Short Answer Questions

Question 1.
The boiling points of NH₃ ; H₂O and HF are higher than those of hydrides of the subsequent members of the group. Give your reasons.
Answer:
The boiling points of NH₃, H₂O and HF are higher than those of hydrides of the subsequent mem-bers of the group.
Reasons :

• NH₃, H₂O and HF are electron rich hydrides and these have 1, 2 and 3 lone pairs respectively.
• Due to the presence of lonepairs on the high electronegative elements results in the formation of hydrogen bond [i.e. intermolecular hydrogen bonding].
• Due to the formation of hydrogen bond association of molecules takes place. Hence these hydrides has high boiling points.

Question 2.
Discuss the position of hydrogen in the periodic table on the basis of its electronic configuration.
Answer:
Hydrogen is the simplest element with one electron and proton. Its electronic configuration is 1s. The configuration is responsible for its dual nature. It behaves both like alkali metals and halogens. So it can be placed along with alkali metals (IA) or along with halogens (VIIA).
Points in support of placing it in IA group :
a) Like alkali metals hydrogen also has a single electron in the outer shell 1s¹,
b) Its ability to form hydrated unipositive ion, H⁺ (aq).
c) It is quite reasonable to start the periodic table with an element having the least atomic number (Z = 1).

Points in support of placing it in VIIA group :
a) Hydrogen is a gas like fluorine or chlorine.

b) It can form diatomic molecule like halogens.

c) It has a tendency of gaining an electron and attains a stable electronic configuration of He forming H⁺ ion like halogens.
At the same time it should be noted that hydrogen has not such a great tendency to lose electron like alkali metals and gain an electron like halogens. In view of this it is difficult to assign any definite position to hydrogen. Sometimes it is placed in IA group and sometime with VIIA group.

Question 3.
How is the electronic configuration of hydrogen suitable for its chemical reactions ?
Answer:
Electronic configuration of hydrogen is 1s¹.

• The atomic hydrogen obtained from dihydrogen by the treatment with UV rays does combine with allmost all the elements.
• This atomic hydrogen successfully complete the reactions,
  a) By the loss of one electron to give H^⊕.
  b) By the gain of one electron to form H^Θ and
  c) By the sharing of electrons to form a single covalent bond.
  Eg : 1) H_2(g) + F_2(g) → 2HF_(g)
  2) 2Li_(s) + H_2(g) → 2LiH

Question 4.
What happens when dihydrogen reacts with
a) Chlorine and
b) Sodium metal. Explain.
Answer:
a) Reaction of dihydrogen with chlorine : Hydrogen reacts with chlorine to form hydrogen chloride gas. This reaction occurs in presence of sun light.

b) Reaction of dihydrogen with sodium metal : Hydrogen reacts with highly reactive metal like sodium and forms sodium hydride. The reaction occur at a high temperature.
2Na_(s) + H_2(g) → 2NaH_(s)

Question 5.
Write a note on heavy water.
Answer:
Deuterium oxide (D₂O) is known as heavy water.
Peparation : Heavy water is obtained by the exhaustive electrolysis of water.

The physical properties like molecular mass, melting point, Boiling point etc., of D₂O are heavier than watet. But dielectric constant and solubility are low for heavy water when compared to water.

Chemical properties :
a) Heavy water reacts with calcium carbide and forms Deuteroacetylene.
CaC₂ + 2D₂O → C₂D₂ + Ca(OD)₂

b) Heavy water reacts with sulphur trioxide and forms Deutero sulphuric acid.
SO₃ + D₂O → D₂SO₄

c) Heavy water reacts with Aluminium carbide and forms Deutero methane.
Al₄C₃ + 12D₂O → 3CD₄ + 4Al (OD)₃
Uses :

• It is used as a moderator in nuclear reactors to decrease the speed of neutrons.
• It is used to study the reaction mechanism in exchange reactions.
• It is used for the preparation of deuterium and deuterium compounds.

Question 6.
Name the isotopes of hydrogen. What is the ratio of masses of these isotopes ?
Answer:
Hydrogen has three isotopes.

1. Protium (₁H¹)
2. Deuterium (₁H² (or) D)
3. Tritium (₁H³ or T)

Tritium is radioactive isotope.

• The ratio of masses of these isotopes is 1 : 2 : 3 respectively for protium, deuterium and Tritium.
• Protium has no neutrons, deuterium has one neutron and tritium has two neutrons.

Question 7.
What is water – gas shift reaction? How can the production of dihydrogen be increased by this reaction?
Answer:
The mixture of CO and H₂ is called water gas. It is also called ‘Syngas’.
Water-gas shift reaction : When carbon monoxide of the syngas mixture reacted in presence of Iron chromate as catalyst then the reaction is called as water-gas shift reaction.
By using this reaction dihydrogen production can be increased.

In this reaction CO₂ gas is removed by using sodiumarsenite solution by scrubbing.

Question 8.
Complete and balance the following reactions :

Answer:

Question 9.
What is the nature of the hydrides formed by elements of 13 group ?
Answer:
Group 13 elements are belongs to p-block of periodic table.
Generally p-block elements forms covalent (or) molecular hydrides.
These molecular hydrides are further classified into 3 types

1. Electron deficient hydrides
2. Electron – precise hydrides
3. Electron – rich hydrides

Group – 13 elements forms electron deficient hydrides. Electron deficient hydrides are the molecular hydrides which have less no.of electrons to write the lewis structure.
Eg : Diborane (B₂H₆)

These hydrides acts as Lewis acids. Lewis acids accepts electron pairs and forms co-ordinate covalent bonds with donors.

Question 10.
Discuss the principle and the method of softening of hard water by synthetic, ion- exchange resins.
Answer:
Synthetic ion – exchange resins method :

• In present days, this method is mostly used for softening the hard water by using synthetic resins.
• This method is more useful than zeolite (or) permutit process.

Principle : Formation of de-ionised water (or) de mineralised water by passing the water successively through cation exchange and anion exchange resins.

De-ionised water means, water which is free from all soluble mineral salts.

Process : The de-ionization of water can be done in two steps by this process.
Step – I: Cation exchange process.
Step – II: Anion exchange process.

Step – I: Cation exchange process

In this process synthetic resins used are – So₃H group containing large organic molecule. (R-SO₃H)
Here R = organic group (or) resin anion.

• At first the synthetic resin converted to RNa by reacting it with NaCl.
• This resin i.e; RNa exchanges Ca⁺² and Mg⁺² ions of hard water and the softening of water takes place.
  2RNa(s) + \(\mathrm{M}_{(\mathrm{aq})}^{+2}\) → R₂M_(s) + 2\(\mathrm{Na}_{(\mathrm{aq})}^{+}\)
• The resin can be regenerated by using aq.NaCl solution.
• In this step H⁺ ions exchanges Na⁺, Ca⁺², Mg⁺² and proton formation takes place.
  2RH_(s) + \(\mathrm{M}_{(\mathrm{aq})}^{+2}\) ⇌ MR_2(s) + 2\(\mathrm{H}_{(\mathrm{aq})}^{+}\)

Step – II: Anion exchange process :

• In this process resins used are RNH₃OH i.e., a basic compound.
• The OH^– ions of resin exchanges the’anions Cl^–, SO₄^-2, HCO₃^– etc., and OH^Θ ion formation takes place.
  
• The H⁺ ions and OH^– ions obtained in the two steps are get neutralised to form de-ionised water.
  \(\mathrm{H}_{(\mathrm{aq})}^{+}+\mathrm{OH}_{(\mathrm{aq})}^{-} \longrightarrow \mathrm{H}_2 \mathrm{O}_{(l)}\)
• The cation exchange resins and anion exchange resins are regenerated by treatment with dil.acid and base respectively.

Question 11.
Write a few lines on the utility of hydrogen as a fuel. [Mar. 13]
Answer:
Hydrogen as a fuel :
The heat of combustion of hydrogen is high i.e about 242kj/mole. Hence hydrogen is used as industrial fuel.

• The energy released by the combustion of dihydrogen is more than the petrol (3 times).
• Hydrogen is major constituent in fuel gases like coal gas and water gas.
• Hydrogen is also used in fuel cells for the generation of electric power.
• 5% dihydrogen is used in CNG for running four-wheeler vehicles:
• By hydrogen economy principle the storage and transportation of energy in the form of liquid (or) gaseous state. Here energy is transmitted in the form of dihydrogen and not as electric power.

Question 12.
A 1% solution of H₂O₂ is provided to you. What steps do you take to prepare pure H₂O₂ from it ?
Answer:
To obtain pure H₂O₂ from the provided 1% H₂O₂ the following steps are involved.

Step -1:
The provided 1% H₂O₂ solution is carefully evoparated on a water bath under reduced pressure by distillation.
Here approximately 30% H₂O₂ solution is obtained.

Step – II :
The obtained solution in the above step is heated in a distillation flask at a low pressure of 15mm. Here approximately 85% H₂O₂ solution is obtained.

Step – III :
The above sample (obtained in the step – II) crystallised by freezing and pure H₂O₂ is obtained (≅ 100%).

Question 13.
Mention any three uses of H₂O₂ in modern times.
Answer:
Uses of H₂O₂ in modern times :

• H₂O₂ is used in modern times in Green chemistry to control the pollution.
  Eg : It is used in the treatment of domestic and industrial effluents and in the oxidation of cyanides.
• H₂O₂ is used in manufacturing of Sodium perborate and sodium per carbonates. These are used in high quality detergents.
• H₂O₂ is used in certain food products and in certain pharmaceuticals.
• H₂O₂ is used as bleaching agent to bleach paper pulp, leather etc.
• H₂O₂ is used as an antiseptic and in hair bleach.

Long Answer Questions

Question 1.
Write an essay on the commercial preparation of dihydrogen. Give balanced equations.
Answer:
Commercial methods of preparation of dihydrogen
i) From Hydrocarbons : When hydrocarbons undergo reaction with steam at high temperatures in presence of catalyst liberates hydrogen gas.
Eg : C₃H _8(g) + 3H2°(g) 3CO + 7H₂

ii) By the electrolysis of water : Acidified (or) alkaline water undergo electrolysis using platinum electrodes liberates hydrogen gas.

Highly pure hydrogen is liberated by the electrolysis of hot aq. Ba(OH)₂ by using ‘Ni’ electrodes.

iii) By Nelson’s Process : Hydrogen gas is obtained as a Byproduct by the electrolysis of brine solution. This process is mainly used for the manufacturing of NaOH, \(\mathrm{Cl}_2^{-}\) gas.
Cell Reactions : –
2NaCl → 2Na⁺ + 2Cl^–
2Cl^– → Cl₂ + 2e^– (Anode)
2H₂O + 2e^– → H₂ + 2OH^– (Cathode)
2Na⁺ + 2OH^– → 2 NaoH

iv) By water gas shift reaction : Water gas is the mixture of CO and H₂. It is also called as syngas. The process of producing syngas from coal is called as coal gasification.

The dihydrogen production can be increased by the reaction of steam with syngas in presence of catalyst.

This reaction is called as water gas shift reaction.

Question 2.
Illustrate the chemistry of dihydrogen by its reaction with
i) N₂
ii) Metal ions and metal oxides and
iii) Organic compounds. How is dihydrogen used in the manufacture of chemicals ?
Answer:
i) Reaction with N₂: Dihydrogen reacts with nitrogen in presence of iron catalyst to form ammonia.

This process is called as Haber’s process. Here temperature used is around 700K and pressure used is 200 atm.

ii) a) Reaction with metal ions : Hydrogen reduces metal ions in aqueous solution into metals.
H_2(g) + \(\mathrm{Pd}_{(\mathrm{aq})}^{+2}\) → Pd_(s) + \(2 \mathrm{H}_{\text {(aq) }}^{+}\)

b) Reaction with metal oxides : Hydrogen reduces metal oxides into metals.
WO₃ + 3H₂ → W + 3H₂O

iii) Reaction with organic compounds :
a) Vegetable oils undergoes hydrogenation in presence of ‘Ni’ catalyst and forms vanaspathi.
b) Alkenes undergo hydro formylation and forms aldehydes. These aldehydes undergo reduction to form alcohols.
CH₂ = CH₂ + CO + H₂ → CH₃ – CH₂CHO (Aldehyde)
CH₃CH₂CHO + H₂ → CH₃CH₂CH₂OH (Alcohol )

Use of Dihydrogen in the manufacture of chemicals
Dihydrogen is used in the manufacturing of industrial cherfiicals like methanol, ammonia, hydrogen chloride etc.

Question 3.
Explain, with suitable examples, the following :

1. electron deficient
2. electron – precise and
3. electron – rich hydrides.

Answer:
Molecular hydrides (or) covalent hydrides are formed by the p-block elements.

These molecular hydrides are divided into three types.

1. Electron deficient hydrides
2. Electron – precise hydrides
3. Electron – rich hydrides

1) Electron – deficient hydrides : These are the molecular hydrides in which the available no.of valency electrons is less than the number-required for normal covalent bond formation.
(or) .
These are the molecular hydrides in which the available valency electrons are lessthan the required for writting the Lewis structure of the molecule.
Eg : (AlH₃)_n, B₂H₆ etc.
These hydrides acts as Lewis acids i.e. electron pair acceptors. These forms dative bond with donors.

2) Electron – precise hydrides : These are the molecular hydrides in which all the valency electrons of the central atom are involved in bond formation.
(or)
These are the molecular hydrides which contains the required no.of valency electrons to write the Lewis structure of the molecule.
Eg : Group 14 elements forms this type of hydrides CH₄, C₂H₆ etc., are examples of this type.
These have tetrahedral geometry.

3) Electron – rich hydrides :
These are the molecular hydrides in which the valency electrons on the central atom are more than that are required for bond formation.
(or)

• These are the molecular hydrides in which the available valency electrons are more than the required for writting the Lewis structure of the molecule.
• These hydrides contains lone pairs on central atoms.
  Eg :
  These hydrides have high boiling points than those of the hydrides of the subsequent members of group because of hydrogen bond formation.

Question 4.
Write in brief on
i) ionic hydrides
ii) interstitial hydrides.
Answer:
i) Ionic hydrides : These are also called as saline hydrides (or) salt like hydrides.
* These are the hydrides formed by combining di hydrogen with s-block elements (Electro posi-tive elements).
* These are stoichiometric compounds.
Eg : LiH, NaH, CaH₂ etc.
NaH is formed by the direct union of Na and H₂

Physical properties :

• These hydrides are crystalline.
• These hydrides are non volatile and non – conducting in solidstate.
• These conducts electricity in moltenstate.
• These have high melting points.

Chemical properties :
These hydrides on electrolysis, liberates dihydrogen gas at anode.
2H^– → H_2(g) + 2e^– [Anode]
Lithium hydride is used in the synthesis of other useful hydrides like LiAlH₄ and LiBH₄.
8LiH + Al₂CL₆ → 2LiAlH₄ + 6LiCl
2LiH + B₂H₆ → 2LiBH₄
2NaH + B₂H₆ → 2NaBH₄
These hydrides react with acids and water vigorously and liberate dihydrogen.
LiH + H₂O → LiOH + H₂ ↑

ii) Interstitial hydrides : These are the hydrides formed by the reaction of hydrogen with d-block and f-block elements. These are also called as metallic hydrides.
Eg : CrH, CrH₂, ZnH₂, ThH₂

• The hydrogen of the metallic hydride occupy the intersticies of metallic lattice. Hence these are called as Interstitial hydrides.
• These are non-stiochiometric compounds
  Eg : TiH_1.5-1.8 LaH_2.87 etc-
• In these hydrides law of constant composition does not found.
• Metals of group 7, 8 and 9 do not form hydrides and in group 6 only chromium forms hydrides.
• The conductivity of these hydrides is less than the parent metals.
• The formation of metallic hydrides and their capacity to release hydrogen at high temperature are utilised in the purification of H₂.
• Some metals can accomodate a very large volume of hydrogen and acts as storage media.

Question 5.
Explain any four of the chemical properties of water.
Answer:
i) Hydrolysis Reaction : The chemical interaction of a compound with water is called as hydrolysis. -» Water has high hydrating ability because of high dielectric constant.
Covalent as well as ionic compounds undergo hydrolysis.
Eg: P₄O_10(s) + 6H₂O_(l) → 4H₃PO_4(aq) (orthophosphoric acid)
NCl₃ + 3H₂O → NH₃ + 3HOCl
SiCl₄ + 2H₂O → SiO₂ + 4HCl