Students get through AP Inter 2nd Year Chemistry Important Questions Lesson 6(a) Group-15 Elements which are most likely to be asked in the exam.

AP Inter 2nd Year Chemistry Important Questions Lesson 6(a) Group-15 Elements

Very Short Answer Questions

Question 1.
Nitrogen molecule is highly stable – Why? (IPE 2014)
Answer:
The nitrogen Molecule is more stable because in between two nitrogen atoms of N2, a triple bond is present. To break this triple bond high energy is required (941.4KJ/mole).

Question 2.
Why are the compounds of bismuth more stable in +3 oxidation state?
Answer:
Bismuth compounds are more stable in +3 oxidation state because ‘Bi’ exhibits +3 stable oxidation state instead of +5 due to inert pair effect.

AP Inter 2nd Year Chemistry Important Questions Chapter 6(a) Group-15 Elements

Question 3.
What is allotropy ? Explain the different allotropic forms of phosphorus. (IPE 2016(TS))
Answer:
Allotropy: The existence of an element in different physical forms having similar chemical properties is called allotropy.
→ Allotropes of ‘P’: → White ‘P’ (or) Yellow P.

  • Red P
  • Scarlet ‘P’
  • Violet P
  • α – black P’
  • β – black ‘P’.

White phosphorus :

  • It is poisonous and insoluble in water and soluble in carbon disulphide and glows in dark.
  • It is a translucent white waxy solid.
  • It dissolves in boiling NaOH solution and gives PH3.
    AP Inter 2nd Year Chemistry Important Questions Chapter 6(a) Group-15 Elements 1
  • It is more reactive than other solid phases.
  • Bond angle is 60° and it readily catches fire,

Red phosphorus :
AP Inter 2nd Year Chemistry Important Questions Chapter 6(a) Group-15 Elements 2

  • Red P possesses iron grey lustre.
  • In it odour less, non poisonous and insoluble in water as well as CS2.
  • Red P is much less reactive than white P.

Black P’:

  • α – Black P : It is formed when red P is heated in a sealed tube 803K.
  • β – Black P : It is prepared by heating white P at 473 K under high pressure.

Question 4.
Explain the difference in the structures of white and red phosphorus.
Answer:
White ‘P’ molecule has tetrahedral structure (discrete molecule). Discrete ‘P’ molecules are held by vander waal’s forces.
AP Inter 2nd Year Chemistry Important Questions Chapter 6(a) Group-15 Elements 3
Red ‘P’ is polymeric consisting of chains of P4 tetrahedron linked together through covalent bonds.
AP Inter 2nd Year Chemistry Important Questions Chapter 6(a) Group-15 Elements 4

Question 5.
What is inert pair effect ?
Answer:
Inert pair effect: The reluctance of ns pair of electrons to take part in bond formation is called inert pair effect.

  • Bi exhibits +3 oxidation state instead of +5 due to inert pair effect.

AP Inter 2nd Year Chemistry Important Questions Chapter 6(a) Group-15 Elements

Question 6.
How do calcium phosphide and heavy water react ?
Answer:
Calcium phosphide reacts with heavy water to form Deutero phosphine.
Ca3P2 + 6D2O → 3 Ca (OD)2 + 2PD3

Question 7.
Ammonia is a good complexing agent – Explain with an example.
Answer:
NH3 is a lewis base and it donates electron pair to form dative bond with metal ions. This results in the formation of complex compound.
Eg :
AP Inter 2nd Year Chemistry Important Questions Chapter 6(a) Group-15 Elements 5

Question 8.
NO is paramagnetic in gaseous state but diamagnetic in liquid and solid states – Why ?
Answer:
In gaseous state NO2 exists as a Monomer and contains one unpaired electron but in solid state it dimerises to N2O4 so it doesnot contain unpaired electron.
Hence NO2 is para magnetic in geseous state but diamagnetic in solid state.

Question 9.
Iron becomes passive in cone. HNO3 – Why ?
Answer:
Iron becomes passive in cone. HNO3 due to formation of a passive film of oxide on the surface of iron.

Question 10.
Give the neutral oxides of nitrogen.
Answer:
Nitrous oxide (N2O) and Nitric oxide (NO) are neutral oxides of nitrogen.

Question 11.
Give the paramagnetic oxides of nitrogen.
Answer:
Nitric oxide (NO) and nitrogen dioxide (NO2) are the paramagnetic oxides of nitrogen due to the presence of odd number of electrons.

Question 12.
Why is white phosphorus is more reactive than red phosphorus ?
Answer:
In white phosphorus the P – P – P bond angle is 60° hence the bonds are in strain. As a result the bonds are broken easily. In red phosphorus the bond angle is 120°, hence it is stable and less reactive.

AP Inter 2nd Year Chemistry Important Questions Chapter 6(a) Group-15 Elements

Question 13.
What happens when white phosphorus is heated with conc. NaOH solution in an inert atmosphere of CO2 ?
Answer:
When white phosphorus heated with con. NaOH solution in an inert atmosphere of CO2 forms PH3.
P4 + 3NaoH + 3H2O → PH3 + 3NaH2PO2.

Question 14.
Give the uses of
a) nitric acid and
b) ammonia.
Answer:
Nitric acid : It is used in the manufacture of fertilizers, explosives, nitro glycerine, nitro toluence etc.,
Ammonia : It is used to produce fertilizers like urea liquid NH3 is used as a refrigerant.

Question 15.
Give the disproportionation reaction of H3PO3.
Answer:
Orthophosphoric acid (H3PO3) on heating disproportionates to give orthophosphoric acid and phosphine.
4H3PO3 → 3H3PO4 + PH3.

Question 16.
Nitrogen exists as diatomic molecule and phosphorus as P4 – Why ?
Answer:
Nitrogen exists as diatomic molecule :

  • Nitrogen has small size and high electronegativity and nitrogen atom forms Pπ – Pπ multiple bonds with itself (triplebond). So it exists as a discrete diatomic molecule in elementary state.

Phosphorus exists as tetra atomic molecule :

  • Phosphorus has large size and less electronegative and it forms P-P single bonds. So it exists as tetra atomic i.e., P4.

Question 17.
Arrange the hydrides of group – 15 elements in the increasing order of basic strength and decreasing order of reducing character.
Answer:

  • Increasing order of basic strength of Group – 15 elements hydrides is
    BiH3 < SbH3 < AsH3 < PH3 < NH3.
  • Decreasing order of reducing character of Group – 15 elements hydrides is
    BiH3 > SbH3 > AsH3 > PH3 > NH3.

Question 18.
PH3 is a weaker base than NH3 – Explain.
Answer:
PH3 is a weaker base than NH3.

  • In NH3 nitrogen atom undergoes sp3 hybridisation and due to small size it has high electron density than in’P of PH3.
  • Due to large size of ‘P’ atom and availability of large surface area, lone pair of electron spread in PH3. Hence PH3 is weaker base than NH3.

Question 19.
A mixture of Ca3P2 and CaC2 is used in making Holme’s signal – Explain.
Answer:
A mixture of Ca3P2 and CaC2 is used in Holme’s signal. This Mixture containing containers are pierced and thrown in the sea, when the gas is evolved bum and serve as a signal.
The spontaneous combustion of PH3 is the technical use of Holme’s signal.

AP Inter 2nd Year Chemistry Important Questions Chapter 6(a) Group-15 Elements

Question 20.
Which chemical compound is formed in the brown ring test of nitrate ions ?
Answer:
In the brown ring test of nitrate salts a brown ring is formed. It’s formula is [Fe(H2O)5NO]+2.

Question 21.
Why does NH3 act as a Lewis base ?
Answer:
Nitrogen atom in NH3 has one lone pair of electrons with is available for donation. Therefore, it acts as a Lewis base.

Question 22.
Why does NO2 dimerise ?
Answer:
NO2 contains of odd number of electrons. It behaves as a typical odd molecule. On dimerisation, it is converted to stable N2O4 molecule with even number of electrons.

Question 23.
Why does PCl3 fume in moisture ?
Answer:
PCl3 hydrolyses in the presence of moisture giving fumes of HCl.
PCl3 + 3H2O → H3PO3 + 3HCl

Question 24.
Are all the five bonds in PCl5 molecule equivalent ? Justify your answer.
Answer:
PCl5 has a trigonal bipyramidal structure and the three equatorial P-Cl bonds are equivalent. While the two axial bonds are different and longer than equatorial bonds.

Question 25.
How is nitric oxide (NO) prepared ?
Answer:
Nitric oxide (NO) is prepared by the action of dilute nitric acid on copper.
AP Inter 2nd Year Chemistry Important Questions Chapter 6(a) Group-15 Elements 6

Question 26.
Explain the following .
a) reaction of alkali with red phosphorus.
b) reaction between PCl3 and H2O.
Answer:
a) Red phosphorus reacts with alkalies slowly and forms phosphine and hypophosphite.
P4 + 3NaOH + 3 H2O → PH3 + 3 NaH2 PO2
b) PCl3 is hydrolysed by water and gives H3PO3
AP Inter 2nd Year Chemistry Important Questions Chapter 6(a) Group-15 Elements 7

AP Inter 2nd Year Chemistry Important Questions Chapter 6(a) Group-15 Elements

Question 27.
Write the oxidation states of phosphorus in solid PCl5.
Answer:
In solid state PCl5 exists as an ionic solid [PCl4)+ [PCl6]

  • ‘P’ exhibits +5 oxidation state in [PCl4)+ and [PCl6]

Question 28.
Illustrate how copper metal can give different products on reaction with HNO3.
Answer:
AP Inter 2nd Year Chemistry Important Questions Chapter 6(a) Group-15 Elements 8

Question 29.
H3PO2 is a good reducing agent – Explain with an example.
Answer:
In H3PO2, two H-atoms are bonded directly to P-atom which imparts reducing character to the acid.

Question 30.
NH3 forms hydrogen bonds but PH3 does not – Why ?
Answer:
NH3 forms hydrogen bonds but PH3 does not.
Reason : Ammonia forms hydrogen bonds because it is a polar molecule and N-H bond is highly polar. Nitrogen has more electronegativity than phosphorus. In case of PH3, P-H bond polarity decreases.

Question 31.
PH3 has lower boiling point than NH3. Why ?
Answer:
Unlike NH3, PH3 molecules are not associated through inter molecular hydrogen bonding. That is why the boiling point of PH3 is lower than NH3.

Long Answer Questions

Question 1.
How is ammonia manufactured by Haber’s process ? Explain the reactions of ammonia with
a) ZnSO4(aq)
b) CuSO4(aq)
c) AgCl(s) (TS Mar. ’17 IPE 2015(AP))
Answer:
In Haber process ammonia is directly synthesised from elements (nitrogen and hydrogen). The principle involved in this is
AP Inter 2nd Year Chemistry Important Questions Chapter 6(a) Group-15 Elements 9
This is a reversible exothermic reaction.
According to Le Chatelier’s principle favourable conditions for the better yield of ammonia are low temperature and high pressure. But the optimum conditions are :
Temperature : 720k
Pressure : 200 atmospheres
Catalyst: Finely divided iron in the presence of molybdenum (Promoter).
Procedure : A mixture of nitrogen and hydrogen in the volume ratio 1 : 3 is heated to 725 – 775K at a pressure of 200 atmospheres is passed over hot finely divided iron mixed with small amount of molybdenum as promotor. The gases coming out of the catalyst chamber consists of 10 – 20% ammonia. Gases are cooled and compressed, so that ammonia gas is liquified and the uncondensed gases are sent for recirculation.
AP Inter 2nd Year Chemistry Important Questions Chapter 6(a) Group-15 Elements 10
a) Aq. ZnSO4 reacts with ammonia aqueous solution to form white ppt of Zinc hydroxide.
AP Inter 2nd Year Chemistry Important Questions Chapter 6(a) Group-15 Elements 11

AP Inter 2nd Year Chemistry Important Questions Chapter 6(a) Group-15 Elements

Question 2.
How is nitric acid manufactured by Ostwald’s process ? How does it react with the following ? (A.P. Mar. ’17)
a) Copper
b) Zn
c) S8
d) P4
Answer:
Ostwald’s process : Ammonia, mixed with air in 1 : 7 or 1 : 8, when passed over a hot platinum gauze catalyst, is oxidised to NO mostly (about 95%).
The reaction is
AP Inter 2nd Year Chemistry Important Questions Chapter 6(a) Group-15 Elements 12
The liberated heat keeps the catalyst hot. The ’NO’ is cooled and is mixed with oxygen to give the dioxide, in large empty towers (oxidation chamber). The product is then passed into warm water, under pressure in the presence of excess of air, to give HNO3.
4NO2 + O2 + 2H2O → 4HNO3.
The acid formed is about 61% concentrated.

a) Copper reacts with dil. HNO3 and cone. HNO3 and liberates Nitric Oxide and Nitrogen dioxide respectively
3 Cu + 5 HNO3 (dil) → 3 Cu (NO3)2 + 2 NO + 4 H2O
Cu + 4 HNO3 (conc.) → Cu (NO3)2 + 2 NO2 + 2 H2O

b) Zn reacts with dil. HNO3 and Cone. HNO3 and liberates Nitrous oxide and Nitrogen
dioxide respectively.
4 Zn + 10 HNO3 (dil) → 4 Zn (NO3)2 + 5 H2O + N2O.
Zn + 4 HNO3 (conc.) → Zn (NO3)2 + 2 H2O + 2 NO2

c) S8 reacts with cone, nitric acid to form Sulphuric acid, NO2 gas.
S8 + 48 HNO3 → 8 H2SO4 + 48 NO2 + 16 H2O
d) P4 reacts with cone, nitric acid to form phosphoric acid and NO2 gas.
P4 + 20 HNO3 → 4 H3PO4 + 20 NO2 + 4 H2O