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Andhra Pradesh BIEAP AP Inter 1st Year Chemistry Study Material 4th Lesson States of Matter: Gases and Liquids Textbook Questions and Answers.

AP Inter 1st Year Chemistry Study Material 4th Lesson States of Matter: Gases and Liquids

Very Short Answer Questions

Question 1.
Name the different intermolecular forces experienced by the molecules of a gas.
Answer:
The different intermolecular forces experienced by the molecules of a gas are London (or) dispersion forces, Dipole-Dipole forces, Dipole- induced dipole forces, and hydrogen bonds.

Question 2.
State Boyle’s law. Give its mathematical expression.
Answer:
At constant temperature, the pressure of a given mass (fixed amount) of gas varies inversely with its volume. This is Boyle’s law.

• Mathematically it can be written as
  P ∝ \(\frac{1}{v}\) (At constant T and no.of moles (n))
  ⇒ Pv = \(\frac{k}{v}\) (constant).

Question 3.
State Charle’s law. Give its mathematical expression.
Answer:
At constant pressure the volume of a fixed mass of a gas is directly proportional to it’s absolute temperature. This is charle’s law.

• Mathematically it can be written as
  V ∝ T (At constant P and no.of moles (n))
  ⇒ V = kT
  ⇒ \(\frac{V}{T}\) = k (constant).

Question 4.
What are Isotherms?
Answer:
At constant temperature the curves which shows the relationship between variation of volume of a given mass of gas and pressure are called isotherms.

Question 5.
What is Absolute Temperature?
Answer:
It is also called thermodynamic temperature (or) Kelvin temperature. It is a temperature on the absolute (or) kelvin scale in which zero lies at – 273.16°C.
T = (t° C + 273.16) K

Question 6.
What are Isobars?
Answer:
The curves (or) graphs that can be drawn at constant pressure are called Isobars.
Eg : Graph drawn between volume and temperature.

Question 7.
What is Absolute Zero?
Answer:
It is the lowest temperature theoretically possible at which volume of a perfect gas is zero.

Question 8.
State Avogadro’s law.
Answer:
Equal volumes of all gases under the same conditions of temperature and pressure contains equal number of molecules
V ∝ n (mathematically)
v = kn

Question 9.
What are Isochores ?
Answer:
At constant volume a line on a graph showing the variation of temperature of a gas with its pressure is called Isochores.

• It is also called Isoplere.

Question 10.
What are S T P Conditions ?
Answer:
STP means Standard Temperature and Pressure conditions.

• Standard temperature is 0° C = 273 K
• Standard pressure is 1 atmosphere = 76 cm = 760 mm. of Hg.

At S.T.P. one mole of any gas occupy 22.4 lit. of volume.

Question 11.
What is Gram molar Volume ?
Answer:
The volume occupied by one gram molecular weight (or) one gram mole of an element (or) compound in the gaseous state is called gram molar volume.
(or)

• At STP one mole of any gas occupy 22.4 lit. of volume This is known as gram molar volume.

Question 12.
What is an Ideal gas ?
Answer:
A gas which obeys gas laws i.e. Boyle’s law, charle’s law and avagadro’s law exactly at all temperatures is called an ideal gas.

Question 13.
Why the gas constant ‘R’ is called Universal gas constant ?
Answer:
Gas constant ‘R’ is called universal gas constant because the value of ‘R‘ is same for all gases.

Question 14.
Why Ideal gas equation is called Equation of State ?
Answer:
Ideal gas equation is a relation between four variables (p, v, n, T) and it describes the state of any gas. Hence it is called equation of state.

Question 15.
Give the values of gas constant in different units.
Answer:
Gas constant ‘R’ has values in different units as follows.
R = 0.0821 lit. atm. k^-1 mol^-1
= 8.314 J. k^-1 mol^-1
= 1.987 (or) 2 cal. k^-1 mol^-1
= 8.314 × 10⁷ ergs. k^-1 mol^-1.

Question 16.
How are the density and molar mass of a gas related?
Answer:
Pv = n RT
Pv = \(\frac{w}{m}\) RT m
P = \(\left(\frac{w}{v}\right) \frac{R T}{M}\)
Molar mass M = \(\frac{\mathrm{dRT}}{\mathrm{P}}\) [∴ \(\frac{w}{v}\) = density(d)]
P = Pressure of gas
R = Universal gas constant
T = Temperature of gas in kelvins scale.

Question 17.
State Graham’s law of diffusion. (A.P. Mar. ‘16, ’14)
Answer:
The rate of diffusion of a given mass of gas at a given pressure and temperature is inversely proportional to the square root of its density
rate of diffusion r ∝ \(\frac{1}{\sqrt{d}}\).

Question 18.
Which of the gases diffuses faster among N₂, O₂ and CH₄? Why? (T.S. Mar. ‘15)
Answer:
CH₄ gas diffuse faster among N₂, O₂ and CH₄.
Reason : CH₄ (16) has low molecular weight than N₂ (28) and O₂ (32).

Question 19.
How many times methane diffuses faster than sulphurdioxide?
Answer:
According to Graham’s law of diffusion.

Hence methane gas diffuses 2 times faster than SO₂.

Question 20.
State Dalton’s law of Partial pressures. (Mar. ‘14)
Answer:
The total pressure exerted by a mixture of chemically non – reacting gases at given temperature and volume, is equal to the sum of partial pressures of the component gases.
P = P₁ + P₂ + P₃.

Question 21.
Give the relation between the partial pressure of a gas and its mole fraction.
Answer:
Partial pressure of a gas = mole fraction of the gas × Total pressure of the mixture of gases
Eg : Consider A and B in a container which are chemically non reaction.
∴ Partial pressure of A (P_A) = X_A × P_T
Partial pressure of B (P_B) = X_B × P_T
X_A = \(\frac{n_A}{n_A+n_B}\), X_B = \(\frac{n_B}{n_A+n_B}\)
X_A, X_B are mole fractions
P_T = Total pressure.

Question 22.
What is aqueous tension?
Answer:
The pressure exerted by the water vapour which is equilibrium with liquid water is called aqueous tension.
(or)
The pressure exerted by the saturated water vapour is called aqueous tension.

Question 23.
Give the two assumptions of Kinetic molecular theory of gases that do not hold good in explaining the deviation of real gases from ideal behaviour.
Answer:
The two assumptions of kinetic molecular theory of gases that do not hold good in explaining the deviation of real gases from ideal behaviour are

1. There is no force of attraction between the molecules of a gas.
2. Volume of the gas molecules is negligible when compared to the space occupied by the gas.

Question 24.
Give the Kinetic gas equation and write the terms in it.
Answer:
Kinetic gas equation is PV = \(\frac{1}{3} \mathrm{mnu}_{\mathrm{rms}}^2\)
P = Pressure of the gas
V = Volume of the gas
m = Mass of 1 mole of the gas
u_rms = RMS speed of the gas molecules.

Question 25.
Give an equation to calculate the kinetic energy of gas molecules.
Answer:
Kinetic energy for ‘n1 moles of gas is given by
K.E. = \(\frac{3}{2} \mathrm{nRT}\)
R = Universal gas constant
T = absolute temperature.

Question 26.
What is Boltzman’s constant ? Give its value.
Answer:
Boltzman’s constant is the gas constant per molecule.
Boltzman’s constant K = \(\frac{R}{N}\)
= 1.38 × 10^-16 erg/k. molecule
= 1.38 × 10^-23 J/k. molecule.

Question 27.
What is RMS speed ?
Answer:
The square root of mean of the squares of the speeds of all molecules of a gas is known as RMS speed (u_RMS)

Question 28.
What is Average speed ?
Answer:
The arithematic mean of speeds of gas molecules is known as average speed (u_av).

Question 29.
What is Most probable speed ?
Answer:
The speed possessed by the maximum number of molecules of the gas is known as most probable sPeed (u_mp).

Question 30.
What is the effect of temperature on the speeds of the gas molecules ?
Answer:
Temperature and speeds of the gases are directly related.
∴ By the rise of temperature the speeds of the gas molecules also increases.

Question 31.
What is the effect of temperature on the kinetic energy of the gas molecules ?
Answer:
According to the postulates of kinetic molecular theory of gases.
The kinetic energy of gas molecules is directly proportional to the absolute temperature.
K.E. ∝ T_abs

Question 32.
Give the ratio of RMS average and most probable speeds of gas molecules.
Solution:

Question 33.
Why RMS speed is taken in the derivation of Kinetic gas equation ?
Answer:
RMS speed is the mean of squares of speeds of all molecules of gas. Hence RMS speed, is taken into the derivation of kinetic gas equation.
PV = \(\frac{1}{3} m n u_{r m s}^2\)

Question 34.
What is Compressibility factor ?
Answer:
The ratio of the actual molar volume of a gas to the molar volume of a perfect gas under the same conditions is called compressibility factor.
Compressibility factor Z = \(\frac{\mathrm{PV}}{\mathrm{nRT}}\)
For a perfect gas Z = 1.

Question 35.
What is Boyle Temperature?
Answer:
The temperatue at which a real gas exibits ideal behaviour for a considerable range of pressure is called Boyle’s temperature.

Question 36.
What is critical temperature ? Give its value for CO₂.
Answer:
The temperature above which no gas can be liquified how ever high the pressure may be applied is called critical temperature.

• Critical temperature of CO₂ gas is 31.98° C.

Question 37.
What is critical Volume ?
Answer:
The volume occupied by one mole of gas at critical temperature and critical pressure is known as critical volume.

Question 38.
What is critical Pressure ?
Answer:
The pressure required to liquify a gas at critical temperature is known as critical pressure.

Question 39.
What are critical constants ?
Answer:
Critical temperatue (T_C), critical volume (V_C) and critical pressure (P_C) are called as critical constants.

Question 40.
Define vapour Pressure of a liquid.
Answer:
The pressure exerted by the vapour on the liquid surface. When it is in equilibrium with the liquid at a given temperature is known as vapour pressure of the liquid.

Question 41.
What are normal and standard boiling points ? Give their values for H₂O.
Answer:

• The boiling points at 1 atm. pressure are called normal boiling points.
• The boiling points at 1 bar pressure are called standard boiling points.
• For water normal boiling point is 100° C.
• For water standard boiling point is 99.6° C.

Question 42.
Why pressure Cooker is used for cooking food on hills ?
Answer:
At hill areas pressure cooker is used for cooking food because low atmospheric pressure is observed at high altitudes. At high altitudes liquids boil at low temperature. So water boils at low temperature on hills.

Question 43.
What is surface tension ?
Answer:
The force acting at right angles to the surface of the liquid along unit length of surface is called surface tension.

• Units : dynes / cm.

Question 44.
What is laminar flow of a liquid ?
Answer:
In liquids a regular gradation of velocity for layers in passing from one layer to the next observed. This flow of liquid is called Laminar flow.

Question 45.
What is coefficient of Viscosity ? Give its units.
Answer:
The force of friction required to maintain velocity difference of 1 cm. sec^-1 between two parallel layers of a liquid 1 cm. apart and each layer having an area 1cm² is called coefficient of viscosity.

• It is denoted by η
• F = η A \(\frac{\mathrm{du}}{\mathrm{dx}}\)
• Units : Poise : In CGS system 1 poise = 1g. cm^-1 sec^-1.

Short Answer Questions

Question 1.
State and explain Boyle’s law.
Answer:
Boyle’s law : At constant temperature, the volume of a given mass of gas is inversely proportional to pressure of the gas.
If ‘V’ is the volume of a given mass of the gas and its pressure, then the law can be written as
V ∝ \(\frac{1}{P}\) at constant temperature
V = \(\frac{k}{P}\)
∴ PV = k
or P₁V₁ = P₂V₂ = k.
Boyle’s law may also be stated as, “at constant temperature the product of the’ pressure and volume of a given mass of gas is constant.”

Question 2.
State and explain Charle’s law.
Answer:
Charles’ law : At constant pressure, the volume of a given mass of gas is directly proportional to absolute temperature.
V ∝ T at constant pressure
V = kT
\(\frac{V}{T}\) = k
Where V is the volume and T is the absolute temperature. V₁ and V₂ are the initial and final volumes of a given mass of the gas at the absolute temperatures T₁ and T₂ respectively at constant pressure.
\(\frac{V_1}{T_1}\) = k; \(\frac{V_2}{T_2}\) = k
or \(\frac{v_1}{T_1}\) = \(\frac{V_2}{T_2}\) = k

Question 3.
Derive Ideal gas equation. (T.S. Mar. ’16)
Answer:
Ideal gas equation : The combination of the gas laws leads to the development of an equation which relates to the four parameters volume, pressure, absolute temperature and number of moles. This equation is known as ideal gas equation.
In this Boyle’s law and Charles’ law combined together and an equation obtained is called the gas equation.
V ∝ \(\frac{1}{p}\) (Boyle’s law)
V ∝ T (Charles law)
V ∝ n (Avogadro’s law)
Combining above three laws, we can write
V ∝ \(\frac{1}{p}\) ∝ T ∝ n (or) V = R × \(\frac{1}{P}\) × T × n
(Or) PV = nRT
Where V = volume of the gas
P = pressure of the gas
n = no. of moles of gas
T = absolute temperature
R = Universal gas constant.

Question 4.
State and explain Graham’s law of Diffusion. (A.P. Mar.’16) (Mar.’13)
Answer:
Graham’s law of diffusion : At a given temperature and pressure, the rate of diffusion of a gas is inversely proportional to the square root of density, vapour density or molecular weight.

If r₁ and r₂ are the rates of diffusion of two gases d₁ and d₂ are their densities respectively, then
\(\frac{r_1}{r_2}\) = \(\sqrt{\frac{d_2}{d_1}}\)
This eqaution can be written as:

Comparison of the volumes of the gases that diffuse in the same time. Let V₁ and V₂ are the volumes of two gases that diffuse in the same time ‘t’.

When time of flow is same then : \(\frac{\mathrm{r}_1}{\mathrm{r}_2}\) = \(\frac{v_1}{v_2}\)
When volume is the same then : \(\frac{r_1}{r_2}\) = \(\frac{t_2}{t_1}\).

Applications:

• This principle is used in the separation of isotopes like U²³⁵ and U²³⁸.
• Molar mass of unknown gas can be determined by comparing the rate of diffusion of a known gas molecular mass.
• Ansil’s alarms which are used in coal mines to detect the explosive maršh gas works on the principle of diffusion.

Question 5.
State and explain Dalton’s law of Partial pressures.
Answer:
Dalton’s law of Partial pressures : The total pressure exerted by a mixture of chemically non-reacting gases at given temperature and volume, is equal to the sum of the partial pressures of the component gases.
Explanation: Consider a mixture of three gases ¡n a vessel. Let P₁, P₂, P₃ be the partial pressures of the three gases in the mixture. According to Dalton’s law of partial pressures, the pressure (P) of the gaseous mixture is at the same temperature.
P = P₁ + P₂ + P₃

Let n₁, n₂, n₃ be the number of moles of three gases respectively in the mixture. Let ‘V’ be the volume of the mixture of gases at T K temperature.
According to ideal gas equation,

∴ P₁ = x₁ P
In a similar way P₂ = x₂ P and P₃ = x₃P
∴ Partial pressure = mole fraction × total pressure

Question 6.
Deduce
(a) Boyle’s law and
(b) Charle’s law from Kinetic gas equation.
Answer:
a) Deduction of Boyle’s law:
Kinetic gas equation is PV = \(\frac{1}{3} m n u^2\)
PV = \(\frac{1}{3}\) mnu²
= \(\frac{1}{3}\) × \(\frac{2}{2}\) mnu²
= \(\frac{2}{3}\) × \(\frac{1}{2}\) mnu²
PV = \(\frac{2}{3}\) [Kinetic energy (KE)] [∵ KE_{‘n; moles} = \(\frac{1}{2}\) mnu²]
PV = \(\frac{2}{3}\) KT [∵ KE = KT According to kinetic theory]
PV = \(\frac{2}{3}\)KT
According to Boyle’s law T is constant
∴ PV = \(\frac{2}{3}\)(constant)
∴ PV = constant
(or)
P ∝ \(\frac{1}{V}\)
Hence Boyle’s law proved from kinetic gas equation.

b) Deduction of Charle’s law:
Kinetic gas equation is PV = \(\frac{1}{3}\) mnu²
PV = \(\frac{1}{3}\) mnu²
= \(\frac{2}{2}\) × \(\frac{1}{3}\) mnu²
= \(\frac{2}{3}\) × \(\frac{1}{2}\) mnu²
= \(\frac{2}{3}\)(KE) [Kinetic energy (KE) = \(\frac{1}{2}\) mnu²]
PV = \(\frac{2}{3}\)KT [According to kinetic theory KE = KT]
\(\frac{V}{T}\) = \(\frac{2}{3}\) × \(\frac{K}{P}\)
According to Charles law ‘P is constant
∴ \(\frac{V}{T}\) = constant (or) V ∝ T
Hence Charle’s law proved from kinetic gas equation.

Question 7.
Deduce
(a) Graham’s law and
(b) Dalton’s law from Kinetic gas equation.
Answer:
a) Graham’s law: At constant temperature and pressure the rate of diffusion of a gas is inversely proportional to the square root of its density, r ∝ \(\frac{1}{\sqrt{d}}\)
Deduction : Kinetic gas equation is
PV = \(\frac{1}{3}\) mnu² = \(\frac{1}{3}\) mu²
u² = \(\frac{3 P V}{M}\) = \(\frac{3}{d}\)
∴ u = \(\sqrt{\frac{3 \mathrm{P}}{d}}\)
At constant pressure u = k. \(\frac{1}{\sqrt{d}}\)
k = constant; or u ∝ \(\frac{1}{\sqrt{d}}\)
∴ The rate of diffusion of gases depends upon the velocity of the gas molecules.
So r ∝ \(\frac{1}{\sqrt{d}}\).
This is Graham’s law.

b) Dalton’s law of partial pressures : The total pressure exerted by a mixture of chemically non-reacting gases at given temperature and volume, is equal to the sum of the partial pressures of the component gases.
Deduction :
Consider a gas present in vessel of volume = V
no. of molecules = n₁
mass of each molecule = m₁
RMS velocity = u₁
According to kinetic gas equation the pressure of the gas. P₁ = \(\frac{1}{3} \frac{m_1 n_1 u_1^2}{v}\)
When this gas is replaced by another gas in the same vessel, P₂ = \(\frac{1}{3} \frac{m_2 n_2 u_2^2}{V}\)
When these two gases are mixed in the same vessel, the total pressure of the mixture is P = \(\frac{1}{3} \frac{m_1 n_1 u_1^2}{V}+\frac{1}{3} \frac{m_2 n_2 u_2^2}{V}\) = P₁ + P₂
∴ P = P₁ + P₂.
This is Dalton’s law of partial pressures.

Question 8.
Derive an expression for Kinetic Energy of gas molecules.
Answer:
Kinetic gas equation is PV = \(\frac{1}{3}\)mnu²
For one mole of gas ‘n’ the no.of molecules will be equal to Avagadro’s number ‘N’.
∴ m × N = ‘M’ (gram molar mass of the gas)
∴ PV = \(\frac{1}{3}\) Mu²
= \(\frac{2}{2}\) × \(\frac{1}{3}\) Mu²
= \(\frac{2}{3}\) × \(\frac{1}{2}\) Mu²
= \(\frac{2}{3}\) (K.E)
Ideal gas equation for 1 mole of gas is PV = RT
∴ \(\frac{2}{3}\) KE = RT
⇒ KE = \(\frac{3}{2}\) RT
For ‘n’ moles KE = \(\frac{3}{2}\) nRT

Question 9.
Define
(a) RMS
(b) average and
(c) most probable speeds of gas molecules. Give their interrelationship.
Answer:
a) RMS speed:
The square root of mean of the squares of the speeds of all molecules of a gas is known as RMS speed (u_RMS)

b) Average speed:
The arthematic mean of speeds of gas molecules is known as average speed (u_av).

c) Most probable speed:
The speed possessed by the maximum number of molecules of the gas is known as most probable speed (u_mp).
\(u_{m p} \sqrt{\frac{2 R T}{M}}\) = \(\sqrt{\frac{2 P V}{M}}\) = \(\sqrt{\frac{2 P}{d}}\)

Inter relationships : –

• u_mp : u_av : u_rms = \(\sqrt{\frac{2 R T}{M}}\) : \(\sqrt{\frac{8 R T}{\pi M}}\) : \(\sqrt{\frac{3 R T}{M}}\)
  = 1 : 1.128 : 1.224.
• u_av = 0.9213 × u_rms
• u_mp = 0.8166 × u_rms

Question 10.
Explain the physical significance of Vander Waals paramaters.
Answer:
Vander Waals equation : [P + \(\frac{a n^2}{\mathrm{~V}^2}\)] [V – nb] = nRT
Where P = Pressure of the gas
n = Number of moles of the gas
a, b = Vander Waals parameters (or) empirical parameters
V = Volume of the container
R = Gas constant
T = Absolute temperature
Units of ’a’: – bar lit^-2 mole^-2
Units of ’b’: – lit. mol^-1

Significance : –

• ‘a’ is the measure of magnitude of inter molecular forces (attractive) with in the gas and is independent of temperature and pressure. If ‘a’ value is high the gas can be easily liquified.
• ‘b’ is the effective volume of the gas molecule. It indicates the effective size of the gas molecules. If the value of ‘b’ is constant over a long range of temperature and pressure then the gas cannot be compressed easily.

Question 11.
What is Surface Tension of liquids ? Explain the effect of temperature on the surface tension of liquids.
Answer:
Surface tension property (γ): ‘It is defined as the force acting along the surface of a liquid at right angles to any line of 1 unit length.”
It is numerically and dimensionally equal to surface energy. It has dimensions kg. s^-2 and in SI unit Nm^-1.

If we consider a molecule in the bulk of the liquid it experiences equal intermolecular forces in all directions. Hence there is no net force acting on it. But a molecule at the surface has intermolecular forces from inside only. Therefore there is a net attractive force on it towards the interior of the liquid. Due to this the surface area of the liquid tends to minimise. That is the molecules experience a downward force and have more energy than the molecules in the bulk.
Surface tension decreases with increase of temperature because of increase in K,E. of molecules and decrease in intermolecular forces,
e.g.:

Question 12.
What is Vapour Pressure of liquids? How the Vapour Pressure of a liquid is related to its boiling point ?
Answer:
The pressure exerted by vapour of a liquid when it is in equilibrium with liquid is known as vapour pressure.
Effect of temperature : When the temperature of a liquid increased the average kinetic energy of molecules increases. This increase in kinetic energy overcomes the attractive forces between the liquid molecules so that liquid molecules rapidly escape into air. Thus rise in temperature raises the escaping tendency of molecules. Hence the vapour pressure of a liquid increases with increase in temperature.

The vapour pressure of a liquid increases with increase of temperature. This goes on until the critical temperature of the liquid is reached. Above the critical temperature liquid state does not exist when the vapour pressure of the liquid becomes equal to the external (atmospheric) pressure the liquid is said to be boiling and the temperature at which this happens is known as boiling point.

For water the boiling point is 100°C at 1. atm pressure. If the external pressure is reduced, the liquid boils at lower temperature. The boiling point of a liquid can be increased by increasing the external pressure.

Question 13.
Define Viscosity and Coefficient of Viscosity. How does the Viscosity of liquids varies with temperature.
Answer:
Viscosity : Viscosity is a measure of resistance to flow of liquids. This arises due to internal friction between layers of fluids as they slip past one another while liquid flows.

Coefficient of Viscosity:

The force of friction required to maintain velocity difference of 1 cm. sec^-1 between two parallel layers of a liquid 1 cm. apart and each layer having an area 1 cm² is called coefficient of viscosity.

• It is denoted by η
• F = ηA\(\frac{d u}{d x}\)
• Units : Poise : In CGS system 1 poise = 1g. cm^-1 sec^-1.
• Viscosity of liquids decrease with increase of temperature due to high kinetic energy of molecules that over come the inter molecular forces.

Long Answer Questions

Question 1.
Write notes on Intermolecular Forces.
Answer:
Intermolecular forces :
a) Ion – Dipole forces : Ion dipole forces are mainly important in aqueous solutions of ionic substances such as NaCl in which dipolar water molecules surround the ions.

Water molecules are polar and in them hydrogen atoms possess partial positive charges and oxygen atoms possess partial negative charges due to electronegativity difference between hydrogen and oxygen atoms. When ionic compounds like NaCl dissolve in water, they dissociate into component ions like Na⁺ and Cl^–. Now the water molecules orient in the presence of ions in such a way that the positive end of the dipole is near an anion and the negative end of the dipole is near a cation.

b) Dipole-Dipole forces : Neutral but polar molecules experience dipole-dipole forces. These are due to the electrical interactions among dipoles on neighbouring molecules. These forces are again attractive between unlike poles and repulsive between like poles and depend on the orientation of the molecules. The net force in a large collection of molecules results from many individual interactions of both types. The forces are generally weak and are significant only when the molecules are in close contact.

c) London dispersion forces : These forces result from the motion of electrons around atoms. Take, for example, atoms of helium. The electron distribution around a helium atom is for averaged over time spherically symmetrical. However, at a given instant the electron distribution in an atom may be unsymmetrical giving the atom a short – lived dipole moment. This instantaneous dipole on one atom can affect the electron distribution is neighbouring atoms and induce temporary dipoles in those neighbours. As a result, weak attractive forces develop known as London forces or dispersion forces. London forces are generally small. Their energies are in the range 1 – 10k J mol^-1.

d) Dipole – Induced Dipole forces : These forces are between polar molecules with permanent dipole moments and the molecules with no permanent dipole moment. Permanent dipole of the polar molecule induces dipole on the electrically neutral molecule by deforming into electronic cloud.
Magnitude of these forces depends on the magnitude of the dipole moment of permanent dipole and polarisatricity of neutral molecule. This interaction is proportional to \(\left(\frac{1}{r^2}\right)\), where r = distance between molecules.

Question 2.
State Boyle’s law, Charle’s law and Avogadro’s law and derive Ideal gas equation.
Answer:
Boyle’s law : At constant temperature, the volume of a given mass of gas is inversely proportional to pressure of the gas.
If ‘V is the volume of a given mass of the gas and ‘P’ its pressure, then the law can be written as V ∝ \(\frac{1}{P}\) at constant temperature
V = \(\frac{k}{P}\)
∴ PV = k or P₁V₁ = P₂V₂ = k

Boyle’s law may also be stated as, “at constant temperature the product of the pressure and volume of a given mass of gas is constant.”

Charles’ law: At constant pressure, the volume of a given mass of gas is directly proportional to absolute temperature.
V ∝ T at constant pressure
V = kT
\(\frac{\mathrm{T}}{\mathrm{T}}\) = k

Where V is the volume and T is the absolute temperature. V₁ and V₂ are the initial and final volumes of a given mass of the gas at the absolute temperatures T₁ and T₂ respectively at constant pressure.
\(\frac{V_1}{T_1}\) = k ; \(\frac{V_2}{T_2}\) = k
or \(\frac{V_1}{T_1}\) = \(\frac{V_2}{T_2}\) = k
Avogadro’s law : Equal volumes of all gases contain equal number of moles at constant temperature and pressure.
V ∝ n (pressure and temperature are constant).

Ideal gas equation : The combination of the above gas laws leads to the development of an equation which relates to the four parameters volume, pressure, absolute temperature and number of moles. This equation is known as ideal gas equation.

In this Boyle’s law and Charles’ law combined together and an equation obtained called the gas equation.
V ∝ \(\frac{1}{P}\) (Boyle’s law) ‘
V ∝ T (Charles law)
V ∝ n (Avogadro’s law)

Combining above three laws, we can write
V ∝ \(\frac{1}{P}\) ∝ T ∝ n (or) V = R × \(\frac{1}{P}\) × T × n
(Or) PV = nRT
Where V = volume of the gas
P = pressure of the gas
n = no. of moles of gas
T = absolute temperature
R = Universal gas constant

Question 3.
Write notes on diffusion of Gases.
Answer:
Diffusion : The property of gases to spread and occupy the available space is known as diffusion.

• It is a non – directional phenomenon.
  Effusion : The escape of a gas from high pressure region into space through a fine hole is called effusion.
• It is uni directional phenomenon.

Rate of diffusion : No. of molecules diffused per unit time is called rate of diffusion.

Graham’s law of diffusion : At a given temperature and pressure, the rate of diffusion of a gas is inversely proportional to the square root of density, vapour density or molecular weight.
r ∝ \(\frac{1}{\sqrt{d}}\) ; r ∝ \(\frac{1}{\sqrt{V D}}\) ; r ∝ \(\frac{1}{\sqrt{M}}\)
If r₁ and r₂ are the rates of diffusion of two gases d₁ and d₂ are their densities respectively, then
\(\frac{r_1}{r_2}\) = \(\sqrt{\frac{\mathrm{d}_2}{\mathrm{~d}_1}}\)
This equation can be written as :

Comparison of the volumes of the gases that diffuse in the same time. Let V₁, and V₂ are the volumes of two gases that diffuse in the same time t’.

When time of flow is same then \(\frac{r_1}{r_2}\) = \(\frac{V_1}{V_2}\)
When volume is the same then : \(\frac{r_1}{r_2}\) = \(\frac{t_2}{t_1}\)

Applications :

• This principle is used in the separation of isotopes like U²³⁵ and U²³⁸.
• Molar mass of unknown gas can be determined by comparing the rate of diffusion of a known gas molecular mass.
• Ansil’s alarms which are used in coal mines to detect the explosive marsh gas works on the principle of diffusion.

Question 4.
State and explain Dalton’s law of Partial Pressures.
Answer:
Dalton’s law of Partial pressures : The total pressure exerted by a mixture of chemically non-reacting gases at given temperature and volume, is equal to the sum of the partial pressures of the component gases.

Explanation: Consider a mixture of three gases in a vessel. Let P₁, P₂, P₃ be the partial pressures of the three gases in the mixture. According to Dalton’s law of partial pressures, the pressure (P) of the gaseous mixture is at the same temperature.
P = P₁ + P₂ + P₃
Let n₁, n₂, n₃ be the number of moles of three gases respectively in the mixture. Let V be the volume of the mixture of gases at T K temperature.
According to ideal gas equation,
P₁ = \(\frac{n_1 R T}{V}\) ; P₂ = \(\frac{\mathrm{n}_2 \mathrm{RT}}{\mathrm{V}}\) ; P₃ = \(\frac{\mathrm{n}_3 \mathrm{RT}}{\mathrm{V}}\)
∴ Total pressure of the mixture P = P₁ + P₂ + P₃
P = \(\frac{n_1 R T}{V}\) + \(\frac{n_2 R T}{V}\) + \(\frac{n_3 R T}{V}\)
P = \(\frac{R T}{V}\)(n₁ + n₂ + n₃)
Since nn₁ + n2n₂ + nn₃ = n

∴ P₁ = x₁ P
In a similar way P₂ = x₂ P and P₃ = x₃P
∴ Partial pressure = mole fraction × total pressure

Question 5.
Write the postulates of Kinetic Molecular Theory of Gases.
Answer:
Assumptions:

1. Gases are composed of minute particles called molecules. All the molecules of a gas are identical.
2. Gaseous molecules are always, at a random movement. The molecules are moving in all possible directions in straight lines with very high velocities. They keep on colliding against each other and against the walls of the vessel at very small intervals of time.
3. The actual volume occupied by the molecules is negligible when compared to the total volume occupied by the gas.
4. There is no appreciable attraction or repulsion between the molecules.
5. There is no loss of kinetic energy when the molecules collide with each other or with the wall of vessel. This is because the molecules are spherical and perfectly elastic in nature.
6. The pressure exerted by the gas is due to the bombardment of the molecules of the gas on the walls of the vessel.
7. The average kinetic energy of the molecules of the gas is directly proportional to the absolute temperature, Average K.E. ∝ T.
8. The force of gravity has no effect on the speed of gas molecules.

Boyle’s law : According to kinetic theory of gases, the pressure of a gas is due to collisions of gas molecules on the walls of the vessel. At a particular temperature the molecules make definite number of collisions with the walls of the vessel; When the volume of the vessel is reduced the molecules have to travel lesser distance only before making collisions on the walls. As a result the number of collisions per unit increases. The pressure then increases, i.e., the pressure increases when the volume is reduced at constant temperature. This explains Boyle’s law.

Charles’ law : According to kinetic theory of gases, the average kinetic energy of the molecules is directly proportional to the absolute temperature of the gas.
K.E. ∝ T
but K.E. = \(\frac{1}{2}\) mc²

As temperature increases, the velocity of the molecules also increases. As a result the molecules make more number of collisions against the walls of the vessel. This results in an increase of pressure if the volume is kept constant. If the volume is allowed to increase the number of collisions decrease due to the increased distance between the molecules and the walls of the vessel. The pressure then decreases. In other words, with rise of temperature, the volume should increase in order to keep the pressure constant.
V ∝ T at constant pressure.
This is Charles’ law.

Question 6.
Deduce gas laws from Kinetic gas equation.
Answer:
a) Deduction of Boyle’s law :
Kinetic gas equation is PV = \(\frac{1}{3}\) mnu²
PV = \(\frac{1}{3}\) mnu²
= \(\frac{1}{3}\) × \(\frac{2}{2}\) mnu²
= \(\frac{2}{3}\) × \(\frac{1}{2}\) mnu²
PV = \(\frac{2}{3}\)[Kinetic energy (KE)] [∵ KE_{‘n’ moles} = \(\frac{1}{2}\) mnu²]
PV = \(\frac{2}{3}\)KT [∵ KE = KT According to kinetic theory]
PV = \(\frac{2}{3}\)KT
According to Boyle’s law T is constant
∴ PV = \(\frac{2}{3}\)(constant)
∴ PV = constant
(or)
P ∝ \(\frac{1}{V}\)
Hence Boyle’s law proved from kinetic gas equation.

b) Deduction of Chartes law:
Kinetic gas equation is PV = \(\frac{1}{3}\) mnu²
PV = \(\frac{1}{3}\) mnu²
= \(\frac{2}{2}\) × \(\frac{1}{3}\) mnu²
= \(\frac{2}{3}\) × \(\frac{1}{2}\) mnu²
= \(\frac{2}{3}\) (KE) [Kinetic Energy (KE) = \(\frac{1}{2}\) mnu²]
PV = \(\frac{2}{3}\)KT [According to kinetic theory KE = KT]
\(\frac{V}{T}\) = \(\frac{2}{3}\) × \(\frac{K}{P}\)
According to Charles law P’ is constant
∴ \(\frac{V}{T}\) = constant (or) V ∝ T
Hence Charles law proved from kinetic gas equation.

c) Graham’s law: At constant temperature and pressure the rate of diffusion of a gas is inversely proportional to the square root of its density, r ∝ \(\frac{1}{\sqrt{d}}\)
Deduction: Kinetic gas equation is
Pv = \(\frac{1}{3}\) mnu² = \(\frac{1}{3}\) Mu²

At constant pressure u = k. \(\frac{1}{\sqrt{d}}\)
k = constant; or u ∝ \(\frac{1}{\sqrt{d}}\)
∵ The rate of diffusion of gases depends upon the velocity of the gas molecules.
So r ∝ \(\frac{1}{\sqrt{d}}\)
This is Graham’s law.

d) Dalton’s law of partial pressures : The total pressure exerted by a mixture of chemically non-reacting gases at given temperature and volume, is equal to the sum of the partial pressures of the component gases.

Deduction:
Consider a gas present in vessel of volume = V
no. of molecules = n₁
mass of each molecule = m₁
RMS velocity = u₁

According to kinetic gas equation the pressure of the gas. P₁ = \(\frac{1}{3} \frac{m_1 n_1 u_1^2}{V}\)
When this gas is replaced by another gas in the same vessel, p₂ = \(\frac{1}{3} \frac{m_2 n_2 \cdot u_2^2}{V}\)
When these two gases are mixed in the same vessel, the total pressure of the mixture is P = \(\frac{1}{3} \frac{m_1 n_1 u_1^2}{v}\) + \(\frac{1}{3} \frac{m_2 n_2 u_2^2}{V}\) = P₁ + P₂
∴ P = P₁ + P₂
This is Dalton’s law of partial pressures.

Question 7.
Explain Maxwell-Boltzmann distribution curves of molecular speeds and give the important conclusions. Discuss the effect of temperature on the distribution of molecular speeds.
Answer:
According to kinetic gas equation it was assumed that all the molecules in a gas have the same velocity. But it is not correct. When any two molecules collide exchange of energy takes place and hence their velocities keep on changing. At any instant few molecules may have zero velocity, a few molecules may be at high velocities and some may be with low velocities.

The distribution of speeds between different molecules were worked out by Maxwell by applying probability considerations.

If one plots a graph between fraction of molecules \(\frac{\Delta \mathrm{N}}{\mathrm{N}}\) vs velocity one gets distribution curve of the type.

These curves are shown at different temperatures T₁ T₂ (T₁ < T₂)

The graph reveals that

1. There are no molecules with zero velocity and only very few molecules possess the highest velocity.
2. The velocities of most of the molecules lie near a mean value.
3. As the temperature of the gas is increased, the curve becomes more flattened and shifts towards higher velocity. It means that at higher temperature the number of molecules possessing higher velocities is more than at lower temperature.

The peak point corresponds to the most probable velocity. It is the velocity possessed by maximum number of molecules.
The average velocity of the molecules is slightly higher than the most probable velocity. The RMS velocity is slightly higher than the average velocity.

Question 8.
Write notes on the behaviour of real gases and their deviation from ideal behavior.
Answer:
Real gases are also called non – ideal gases; A gas which does not obey ideal gas equation PV = nRT is called Real gas.

• Real gases show ideal behaviour at low pressure and high temperature.
  The deviation of real gas from ideal behaviour can be measured in terms compressibility factor (Z), which is the ratio of product PV and nRT. (i.e.,) Z = \(\frac{\mathrm{PV}}{\mathrm{nRT}}\)

For ideal gas, Z = 1 at all temperatures and pressures because PV = nRT. The graph of Z Vs P will be a straight line parallel to pressure axis. For gases which deviate from ideality, value of Z deviates from unity. At very low pressures all gases shown Z = 1 and behave as ideal gas. At low pressures, inter molecular forces are negligible hence show ideal behaviour. At high pressures all the gases have Z > 1. These are more difficult to compress. At intermediate pressures, most gases have Z < 1.

Question 9.
Derive the Vander Waals equation of state. Explain the importance of Vander Waal’s gas eqaution.
Answer:
Vander Waal’s equation of state : Vander Waal’s proposed an approximate equation of state which involves the intermolecular interactions that contribute to the deviations of a gas from perfect gas law. It may be explained as follows. The repulsive interactions between two molecules cannot allow them to come closer than a certain distance. Therefore, for the gas molecules the available volume for free travel is not the volume of the container V but reduced to an extent proportional to the number of molecules present and the volume of each exclude.

Therefore, in the perfect gas equation a volume correction is made by changing v to (v – nb). Here, ‘b’ is the proportionality constant between the reduction in volume and the amount of molecules present in the container. P = \(\frac{n R T}{V-n b}\) If pressure is low, the volume is large compared with the volume excluded by the molecules (V > > nb). The nb can be neglected in the denominator and the equation reduces to the perfect gas equation of state.

The effect of attractive interactions between molecules is to reduce the pressure that the gas exerts. The attraction experienced by a given molecule is proportional to the concentration n/V of molecules in that container. As the attractions slow down the molecules, the molecules strike the waals less frequently and strike with a weaker impact. Therefore, we can expect the reduction in pressure to be proportional to the square of the molar-concentration, one factor of n/V showing the reduction in frequency of collisions and the other factor the reduction in the strength of their impulse.

Reduction in pressure ∝ \(\left(\frac{n}{V}\right)^2\)
Reduction in pressure = a. \(\left(\frac{n}{v}\right)^2\),
Where a = the proportionality constant.
Vander Waals equation is

The equation is called Vander Waals equation of state.

The constants ‘a’ and ‘b’ known as Vander Waals parameters (or) empirical parameters. They depend on the nature of the gas independent of temperature.

Significance : –

• ‘a’ is the measure of magnitude of inter molecular forces (attractive) with in the gas and is independent of temperature and pressure. If ‘a’ value is high the gas can be easily liquified.
• ‘b’ is the effective volume of the gas molecule. It indicates the effective size of the gas molecules. If the value of ‘b’ is constant over a long range of temperature and pressure then the gas cannot be compressed easily.

Question 10.
Explain the principle underlying the liquefacation of gases.
Answer:
Liquifacation of gases can be done by decreasing the temperature and increasing the pressure.

Liquefaction of gases: Any gas, if it to be liquefied, it must be cooled below its critical temperature. A gas liquefies if it is cooled below its boiling point at given pressure. For example, chlorine at room pressure say 1 atmosphere can be liquefied by cooling it to – 34.0°C in a dry ice bath. For N₂ and O₂ that have very low boiling points -196°C and -183°C. Such simple technique is not possible. Then, to liquify such type of gases the technique based on intermolecular forces is used. It is as follows. If the velocities of molecules are reduced to such lower values that neighbours can attract each other by their interaction or intermolecular attractions, then the cooled gas will condense to a liquid.

For this, the molecules are allowed to expand into available volume without supplying any heat from outside. In this, the molecules have to overcome the attractions of their neighbours and in doing so, the molecules convert some of their kinetic energy into potential energy and now travel slowly. The average velocity decreases and therefore the temperature of the gas decreases and the gas cools down compared to its temperature before its expansion. For this the gas is allowed to expand through a narrow opening called throttle. This way of cooling of gas by expansion from high pressure side to low pressure is called Joule – Thomson effect.

Question 11.
Write notes on the following properties of liquids
(a) Vapour Pressure
(b) Surface Tension
(c) Viscosity.
Answer:
(a) Vapour Pressure : The pressure exerted by vapour of a liquid when it is in equilibrium with liquid is known as vapour pressure.
Effect of temperature : When the temperature of a liquid increased the average kinetic energy of molecules increases. This increase in kinetic energy overcomes the attractive forces between the liquid molecules so that liquid molecules rapidly escape into air. Thus rise in temperature raises the escaping tendency of molecules. Hence the vapour pressure of a liquid increases with increase in temperature.

The vapour pressure of a liquid increases with increase of temperature. This goes on until the critical temperature of the liquid is reached. Above the critical temperature liquid state does not exist when the vapour pressure of the liquid becomes equal to the external (atmospheric) pressure the liquid is said to be boiling and the temperature at which this happens is known as boiling point.

For water the boiling point is 100°C at 1. aim pressure. If the external pressure is reduced, the liquid boils at lower temperature. The boiling point of a liquid can be increased by increasing the external pressure.

b) Surface tension property (γ) : “It is defined as the force acting along the surface of a liquid at right angles to any line of 1 unit length.”
It is numerically and dimensionally equal to surface energy. It has dimensions kg. s^-2 and in SI unit Nm^-1

If we consider a molecule in the bulk of the liquid it experiences equal intermolecular forces in all directions. Hence there is no net force acting on it. But a molecule at the surface has intermolecular forces from inside only. Therefore there is a net attractive force on it towards the interior of the liquid. Due to this the surface area of the liquid tends to minimise. That is the molecules experience a downward force and have more energy than the molecules in the bulk.

Surface tension decreases with increase of temperature because of increase in K, E. of molecules and decrease in intermolecular forces.
e.g.:

c) Viscosity : Viscosity is a measure of resistance to flow of liquids. This arises due to internal friction between layers of fluids as they slip past one another while liquid flows.
Coefficient of Viscosity:
The force of friction required to maintain velocity difference of 1cm. sec^-1 between two parallel layers of a liquid 1 cm. apart and each layer having an area 1cm² is called coefficient of viscosity.

• It is denoted by η
• F = η A \(\frac{\mathrm{du}}{\mathrm{dx}}\)
• Units : Poise : In CGS system 1 poise = 1g. cm^-1 sec^-1.
• Viscosity of liquids decrease with increase of temperature due to high kinetic energy of molecules that over come the inter molecular forces.

Solved Problems

Question 1.
What will be the minimum pressure required to compress 500 dm³ of air at 1 bar to 200 dm³ at 30°C?
Solution:
Formula:
P₁y₁ = P₂y₂
P₁ = 1 bar
V₁ = 500 dm³
V₂ = 200 dm³
P₂ = ?
1 × 500 = P₂ × 200
P₂ = \(\frac{5}{2}\) = 2.5 bar.

Question 2.
A vessel of 120 mL capacity contains a certain amount of gas at 35°C and 1.2 bar pressure. The gas is transferred to another vessel of volume 180 mL at 35 °C. What would be its pressure ?
Solution:
Formula:
P₁V₁ = P₂V₂
P₁ = 1.2 bar
V₁ = 120 ml
V₂ = 180 ml
P₂ = ?
1.2 × 120 = P₂ × 180
P₂ = \(\frac{1.2 \times 12}{18}\)
= \(\frac{2.4}{3}\) = 0.8 bar

Question 3.
Using the equation of state pV = nRT; show that at a given temperature density of a gas is proportional to gas pressure p.
Solution:
Consider the equation of state .
PV = nRT
PV = \(\frac{w}{M} R T\)
P = \(\frac{W}{V} \times \frac{R T}{M}\)
P = \(\frac{\mathrm{dRT}}{\mathrm{M}}\) (∵ d = \(\frac{w}{V}\))
From the above relation
P ∝ d

Question 4.
At 0°C, the density of a certain oxide of a gas at 2 bar is same as that of dinitrogen at 5 bar. What is the molecular mass of the oxide?
Solution:

• Given two gases one is unknown oxide and another one is dinitrogen.
• Density of two gases is same
  

Question 5.
Pressure of 1 gm. of an ideal gas A at 27 °C is found to be 2 bar. When 2 g of another ideal gas B is introduced in the same flask at same temperature the pressure becomes 3 bar. Find a relationship between their molecular masses.
Solution:
Given
Weight of gas A = 1 gm
Weight of gas B = 2 gms
Molecular mass of A = M_A
Molecular mass of B = M_B
Pressure of A = P_A = 2 bar
Given Total pressure = 3 bar (P_A + P_B)
∴ P_B = 3 – 2 = 1 bar

Question 6.
The drain cleaner, Drainex contains small bits of aluminum which react with caustic soda to produce dihydrogen. What volume of dihydrogen at 20°C and one bar will be released when 0.15 g of aluminium reacts?
Solution:
Chemical equation is
2Al + 2 NaOH + 2H₂O → 2 NaAlO₂ + 3H₂
From the above equation
2 gram atom of Al liberates 3 moles of H₂ at NTP
2 × 27 gms Al liberates 3 × 22.4 lit.
0.15 gms of Al liberates?
= \(\frac{0.15 \times 3 \times 22.4}{2 \times 27}\)
= 0.1866 li.t = 186.6 ml
P₁ = 1.013 bar P₂ = 1 bar
V₁ = 186.6 ml V₂ = ?
T₁ = 273 K T₂ = 20° C = 293 K
Formulae:

Question 7.
What will be the pressure extracted by a mixture of 3.2 g of methane and 4.4 g of carbon dioxide contained in a 9 dm³ flask at 27 °c?
Solution:
Formula:
Given 3.2 gmš of CH₄
no.of moles of CH₄ = \(\frac{w t}{\text { GMW }}\) = \(\frac{3.2}{16}\) = 0.2
no.of moles of CO₂ = \(\frac{4.4}{44}\) = 0.1
∴ n = \(\mathrm{n}_{\mathrm{CH}_4}\) + \(\mathrm{n}_{\mathrm{CO}_2}\)
= 0.2 + 0.1 = 0.3
R = 8.314
T = 27°C = 300 K
V = 9 dm³
PV = nRT
P = \(\frac{n R T}{V}\)
= \(\frac{0.3 \times 8.314 \times 300}{9}\) = 83.14
= 83.14 × 10³ pa
= 83.14 × 10⁴ pa
∴ P = 8.314 × 10⁴ pa

Question 8.
What will be the pressure of the gaseous mixture when 0.5 L of H₂ at 0.8 bar and 2.0 L of dioxygen at 0.7 bar are introduced in a 1 L vessel at 27°C?
Solution:
Case – I
Hydrogen gas
P₁ = 0.8 bar
P₂ = ?
V₁ = 0.5 lit
V₂ = 1.0 lit
P₁y₁ = P₂V₂
P₂ = \(\frac{0.8 \times 0.5}{1}\)
P₂ = 0.4 bar
Partial pressure of H₂ = 0.4 bar. \(\left[\mathrm{P}_{\mathrm{H}_2}\right]\)

Case-II:
Oxygen gas
P₁ = 0.7 bar
V₁ = 2 lit
V₂ = 1.0 lit
P₂ = ?
P₁V₁ = P₂V₂
P₂ = \(\frac{P_1 V_1}{V_2}=\frac{0.7 \times 2}{1}\)
= 1.4 bar
Partial pressure of O₂ = 1.4 bar. \(\left[\mathrm{P}_{\mathrm{O}_2}\right]\)
∴ Total pressure = \(P_{\mathrm{H}_2}\) + \(P_{\mathrm{O}_2}\)
= 0.4 + 1.4 = 1.8 bar

Question 9.
Density of a gas is found to be 5.46 g/dm³ at 27 °c at 2 bar pressure. What will be its density at STP?
Solution:
d₁ = 5.46 gm/dm³
T₁ = 27° C = 300 K
P₁ = 2 bar
P₂ = 1.013 bar (STP)
T₂ = 273 K(STP)
d₂ = ?

Question 10.
34.05 mL of phosphorus vapour weighs 0.0625 g at 546 °c and 0.1 bar pressure. What is the molar mass of phosphorus ?
Solution:
P = 0.1 bar
W = 0.0625 gms
R = 0.083 bar dm³ k^-1 mol^-1
V = 34.05 × 10^-3 lit
T = 546°C = 819 K
Formula:
PV = nRT
PV = \(\frac{w}{M} R T\)
0.1 × 34.05 × 10^-3 = \(\frac{0.0625}{M}\) × 0.083 × 819
M = \(\frac{0.0625 \times 0.083 \times 819}{0.1 \times 34.05 \times 10^{-3}}\)
= \(\frac{0.0625 \times 83 \times 819}{34.05}\)
= 124.77 gm/mole.

Question 11.
A student forgot to add the reaction mixture to the round bottomed flask at 27 °C but instead he/she placed the flask on the flame. After a lapse of time, he realized his mistake and using a pyrometer he found the temperature of the flask was 477 °C. What fraction of air would have been expelled out ?
Solution:
Formula:
T₁ = 27° C – 300 K
T₂ = 477° C = 750 K
\(\frac{V_1}{T_1}\) = \(\frac{V_2}{T_2}\)
\(\frac{V_1}{300}\) = \(\frac{V_2}{750}\)
V₂ = \(\frac{750 \times \mathrm{V}_1}{300}\)
V₂ = 2.5 V₁
The volume of air expelled = V₂ – V₁
= 2.5V₁ – V₁
= 1.5V₁
Fraction of air expelled out
= \(\frac{1.5 \mathrm{~V}_1}{2.5 \mathrm{~V}_1}\) = \(\frac{1.5}{2.5}\) = \(\frac{15}{25}\) = \(\frac{3}{5}\)

Question 12.
Calculate the temperature of 4.0 mol of a gas occupying 5 dm³ at 3.32 bar.
(R = 0.083 bar dm³ k^-1 mol^-1)
Solution:
Formulae: —
P = 3.32 bar
V = 5 dm³
R = 0.083 bar dm³ k^-1 mol^-1
n = 4 moles
PV = nRT
T = \(\frac{\mathrm{PV}}{\mathrm{nR}}\)
= \(\frac{3.32 \times 5}{4 \times 0.083}\)
= \(\frac{16.6}{0.332}\)
= 50
∴ T = 50 k

Question 13.
Calculate the total number of electrons present in 1.4 g of dinitrogen gas.
Solution:
14 gms of N₂ gas contains
6.023 × 10²³ atoms
1.4 gms of N₂ gas contains
6.023 × 10²² atoms
Each ‘N’ atom contains 7 electrons.
∴ Number of electrons present in 1.4 gms of Nitrogen
= 6.023 × 10²² × 7
= 42.161 × 10²²
= 4.2161 × 10²³ electrons.

Question 14.
How much time would it take to distribute one Avogadro number of wheat grains, if 10¹⁰ grains are distributed each second ?
Solution:
Given that
10¹⁰ grains are distributed in each second i.e., one second Avagodro number means 6.023 × 10²³
6.023 × 10²³ grains distributed in ?
X seconds
x = \(\frac{6.023 \times 10^{23}}{10^{10}}\) = 6.023 × 10¹³ seconds
The time taken to distribute the one Avagadro number of grains
= \(\frac{6.023 \times 10^{13}}{60 \times 60 \times 24 \times 365}\)
= \(\frac{6.023 \times 10^{13}}{3.153 \times 10^7}\) = 1.909 × 10⁶ years.

Question 15.
Ammonia gas diffuses through a fine hole at the rate 0.5 lit min^-1. Under thé same conditions find the rate of diffusion of chlorine gas.
Solution:
Rate of diffusion of ammonia (r₁)
= 0.5 lit min^-1
Molecular weight of ammonia (M₁) = 17
Rate of diffusion of chlorine (r₂) = ?
Molecular weight of chlorine (M₂) = 71
According to Graham’s law of diffusion

∴ Rate of diffusion of Cl₂ = 0.245 lit/min.

Question 16.
Find the relative rates of diffusion of CO₂ and Cl₂ gases.
Solution:

Question 17.
If 150 mL carbon monoxide effused in 25 seconds, what volume of methane would diffuse in same time?
Solution:
Rate of diffusion of CO (r₁)

Molecular weight of CO (M₁) = 28
Rate of diffusion of methane (r₂)

Molecular weight of methane (M₂) = 16
According to Graham’s law of diffusion

Question 18.
Hydrogen chloride gas is sent into a 1oo metre tube from one end ‘A’ and ammonia gas from the other end ‘B’, under similar conditions. At what distance from ‘A’ will be the two gases meet?
Solution:

The two gases HCl and NH₃ diffuse into the pipe from the ends A and B respectively to meet at a point O as indicated by formation of white ring of NH₄Cl. If the distance AO is x meters, the distance OB will be (100 – x) metres.
According to Graham’s law of diffusion. Ration of rates of diffusion of HCl and NH₃ gas is given by
\(\frac{735 \mathrm{~mm} \times 101.3 \mathrm{kPa}}{1760 \mathrm{~mm}}\) = 98 k Pa
It means that the two gases meet at the point O such that the ratio of the distances from the end A to O and B to O is 0.68 : 1.00
∴ \(\frac{0.68}{1}\) = \(\frac{x}{(100-x)}\)
or 0.68 (100 – x) = x or 68 – 0.68 x = x
or 68 = x + 0.68 x or 68 = x (1 + 0.68)
= 1.68 x
x = \(\frac{68}{1.68}\) = 40.48 metres.
Hence, the two gases meet at a distance of 40.48 metres from the end ‘A’.

Question 19.
Calculate the total pressure in a mixture of 8 g of dioxygen and 4 g of dihydrogen confined in a vessel of 1 dm³ at 27°C. R = 0.083 bar dm³ K^-1 mol^-1.
Solution:
Formula:

Question 20.
Calculate the total pressure in a mixture of 3.5g of dinitrogen 3.0g of dihydrogen and 8.0g dioxygen confined in vessel of 5 dm³ at 27°C (R = 0. 083 bar dm³ k^-1 mol^-1)
Solution:
Formula:
V = 5 dm³

Question 21.
Pay load is defined as the difference between the mass of displaced air and the mass of the balloon. Calculate the pay load when a balloon of radius 10m, mass 100 kg is filled with helium at 1.66 bar at 27°C. (Density of air = 1.2 kg m^-3 and R = 0.083 bar dm³ k^-1 mol^-1).
Solution:
Formula :
r = 10 m
m = 100 kg
T = 27° C = 300 K
d = 1.22 kg/m³
Volume of the ballon = \(\frac{4}{3} \pi r^3\)
= \(\frac{4}{3} \times \frac{22}{7} \times 10^3\)
= 4190.5 m³
P = 1.66 bar
T = 300 K
V = 4190.5 m³
R = 0.083 bar dm³ k^-1 mol^-1
PV = nRT
n = \(\frac{\mathrm{PV}}{\mathrm{RT}}\) = \(\frac{1.66 \times 4190.5}{0.083 \times 10^{-3} \times 300}\)
= 2793.70 moles
= 0.083 × 10^-3 bar m³k^-1mol^-1
∴ Weight of 279370 moles of He
= 279390 × \(\frac{4}{1000}\)
= 1117.48 kg
Total weight of balloon = 100 + 1117.48
= 1217.48 kg
Maximum weight of He = V × d
=4190.5 × 1.2
= 5028.6 kg
∴ Payload = 5028.6 – 1219.48
= 3811.12 kg.

Question 22.
Calculate the volume occupied by 8.8 g of CO₂ at 31 .1°C and 1 bar pressure. R = 0.083 bar L K^-1 mol^-1.
Solution:
Formula:

Question 23.
2.9 g of a gas at 95 °C occupied the same volume as 0.184 g of dihydrogen at 17°C, at the same pressure. What is the molar mass of the gas?
Solution:
Given unknown gas and dihydrogen
For unknown gas
V₁ = V
n₁ = \(\frac{2.9}{\mathrm{~m}}\)
T₁ = 95° C = 368 K
P₁V₁ = n₁RT₁
P₁ = \(\frac{n_1 R T_1}{V_1}\)
= \(\frac{2.9}{m} \times \frac{R \times 368}{V}\)
For dihydrogen

Question 24.
A mixture of dihydrogen and dioxygen at one bar pressure contains 20% by weight of dihydrogen. Calculate the partial pressure of dihydrogen.
Solution:
Given 20% by wt of dihydrogen so 80% oxygen remained for dihydrogen
n = \(\frac{w t}{\text { GMwt }}\) = \(\frac{0.2}{2}\) = 0.1
For dioxygen
n = \(\frac{w t}{\text { GMwt }}\) = \(\frac{0.8}{32}\) = 0.025
mole fraction of H₂
= \(\frac{0.1}{0.1+0.025}\) = \(\frac{0.1}{0.125}\) = 0.8
Partial pressure of dihydrogen
= mole fraction of H₂ × P_total
= 0.8 × 1 = 0.8 bar

Question 25.
What would be the SI unit for the quantity \(\frac{\mathrm{PV}^2 \mathrm{~T}^2}{\mathrm{n}}\)?
Solution:
Given quantity \(\frac{\mathrm{PV}^2 \mathrm{~T}^2}{\mathrm{n}}\) = \(\frac{\mathrm{N} / \mathrm{m}^2\left(\mathrm{~m}^3\right)^2(\mathrm{~K})^2}{\text { mole }}\)
= N × mole^-1m⁴k²
∴ The given quantity has SI units Nm⁴k² mole^-1.

Question 26.
In terms of Charles’ law explain why — 273°C is the lowest possible temperature.
Solution:
According to Charles law if we put the value of t = -273°C
in the equation V_t = V₀ \(\left[\frac{273.15+t}{273.15}\right]\).
In this case the volume of the gas becomes zero.
V₀ = Volume at 0° C
V_t = Volume at t° C

• This means the gas will not exist
• In fact all gases liquified before this temperature.

Question 27.
Critical temperature for carbon dioxide and methane are 31.1°C and – 81.9°C respectively. Which of these has
stronger intermolecular forces and why?
Solution:
Given Critical temperatures of CO₂, CH₄
T_C (CO₂) = 31.1°C
T_C (CH₄) = -81.9°C

• The gas with highest critical temperature value can be easily liquified because of high inter molecular forces.
  ∴ T_C(CO₂) is very high.
  So CO₂ gas liquified easily.
• ‘He’ gas has low T value so it is highly difficult to liquify.

Question 28.
Air is cooled form 25°C to 0°C. Calculate the decrease in rms speed of the molecules.
Solution:

Question 29.
Find the rms, most probable and average speeds of SO₂ at 27°c.
Solution:

Question 30.
Find the RMS. average and most probable speeds of O₂ at 27°c.
Solution:

u_average = 0.9213 × u_rms
= 0.9213 × 4.835 × 10⁴
= 4.455 × 10⁴ cm/sec.
u_mp = 0.8166 × u_rms
= 3.948 × 10⁴ cm/sec.

Question 31.
Give the values of Gas constant ‘R’ in different units.
Answer:
R = 0.0821 lit. atm. K^-1 .mol^-1
= 8.314 J.K^-1. mole^-1
= 1.987 (or) 2 cal.K^-1.mol^-1
= 8.314 × 10⁷ erg.K^-1.mol^-1

Andhra Pradesh BIEAP AP Inter 1st Year Chemistry Study Material 3rd Lesson Chemical Bonding and Molecular Structure Textbook Questions and Answers.

AP Inter 1st Year Chemistry Study Material 3rd Lesson Chemical Bonding and Molecular Structure

Very Short Answer Questions

Question 1.
What is Octet rule?
Answer:
Every atom must possess 8 electrons in its outermost energy level for its stability. Atoms combine in two ways to get octets either by transfer of electrons (or) by mutual sharing of electrons. They can attain ns² np⁶ configuration.
The tendency of an atom to achieve eight electrons in its outermost shell is known as the octet rule.

Question 2.
Write Lewis dot structures for S and S^2-.
Answer:

• Lewis dot structure for ‘s’ is
  
  Electronic configuration — 1s² 2s² 2p⁶ 3s² 3p⁴
• Lewis dot structure for s^-2 is
  
  Electronic configuration of s^-2 is 1s² 2s² 2p⁶ 3s² 3p⁶

Question 3.
Write the possible resonance structures for SO₃.
Answer:
The resonance structures of SO₃ as follows

Question 4.
Predict the change, if any, in hybridization of Al atom in the following reaction
AlCl₃ + Cl^– → \(\mathrm{AlCl}_4^{-}\).
Answer:
In AlCl₃ Aluminium undergoes sp² hybridisation
In \(\mathrm{AlCl}_4^{-}\) Aluminium undergoes sp³ hybridisation

Question 5.
Which of the two ions Ca²⁺ or Zn²⁺ is more stable and why ?
Answer:

• Ca⁺² has electronic configuration 1s²2s²2p⁶3s²3p⁶. This configuration is noble gas (or) inert gas configuration.
• Zn⁺² has electonic configuration 1s²2s²2p⁶3s²3p⁶4s⁰3d¹⁰. This configuration is psuedo inert gas configuration.
  ∴ Ca⁺² is more stable than Zn⁺² ion.

Question 6.
Cl^– ion is more stable than Cl atom—Why ?
Answer:
The electronic configurations of Cl and Cl^– is :
Cl = 1s² 2s² 2p⁶ 3s² 3p⁵
Cl^– = 1s² 2s² 2p⁶ 3s² 3p⁶.
This electronic configuration clearly shows that chlorine atom has 7 electrons in the outermost orbit. Whereas chloride has a stable octet electronic configuration (3s² 3p⁶). After gaining one electron, chloride ion attains the electronic configuration of Argon. Hence chloride ion has greater stability than chlorine atom.

Question 7.
Why argon does not form Ar₂ molecule ?
Answer:
Ar₂ represents diatomic molecule. But Argon does not form diatomic molecule. So it cannot represented as ‘Ar₂‘.
Reason : Since Ar’ has only paired electrons with stable octet configuration. It cannot share its electrons with another Ar atom and does not form diatomic molecule.

Question 8.
What is the best possible arrangement of four bond pairs in the valence shell of an atom to minimise repulsions ?
Answer:
The best possible arrangement of four bond pairs in the valency shell of an atom to minimise repulsions is Tetrahedral. (Bond angle 109°.28′)
Eg. : Methane (CH₄).

Question 9.
If A and B are two different atoms when does AB molecule become Covalent ?
Answer:

1. If the difference in electronegativity values between A and B is less than 1.7, then covalent compound formation is possible (according to Allred – Rochow scale).
2. If A and B are sharing one or more electron pairs mutually then AB will be a covalent compound.

Question 10.
What is meant by localized orbitals?
Answer:
The molecular orbital with bonded electron cloud localised between the two nuclei of bonded atoms is called localized orbital, (or) The orbitals which are involved in bond formation are called localized orbitals.

Question 11.
How many Sigma and Pi bonds are present in
(a) C₂H₂ and
(b) C₂H4₄?
Answer:
a) C₂H₂
H – C ≡ C – H
C₂H₂ contains 3 – sigma bonds and 2 – pi bonds.

b) C₂H₄

C₂H₄ contains 5 – sigma bonds and 1-pi bond.

Question 12.
Is there any change in the hybridization of Boron and Nitrogen atoms as a result of the following reaction? BF₃ + NH₃ → F₃BNH₃
Answer:
a) Ammonia – Boron trifluoride formation (H₃N → BF₃):
Ammonia molecule contains Nitrogen atom with a lone pair of electrons (in sp³ orbital). BF₃ has ‘B’ atom with an incomplete octet (with a vacant P_z orbital). Therefore, nitrogen of ammonia donates its lone pair to Boron and thus forms coordinate covalent bond. During this bond formation, the sp³ orbital of nitrogen having a lone pair overlaps the vacant ‘p’ orbital of Boron. The equation corresponding to the reaction is written as follows :

b) Change in hybridised states N and B during [H₃N → BF₃] formation :
Boron in BF₃ undergoes (sp² hybridization with one vacant unhybrid ‘p’ orbital. This orbital also undergoes) hybridization in presence of NH₃ so that the hybridised state of ‘B’ changes from sp² to sp³. This vacant hybrid orbital is bonded to NH₃ through dative bond. During this process there is no change in the hybridized state of Nitrogen in NH₃.

Question 13.
Give reasons for the following.
a) Why H₂O Boiling point is more than H₂S
b) Why H₂O Boiling point is more than HF
Answer:
a) H₂O has high boiling point than H₂S
Reason:
In H₂O inter molecular hydrogen bonding is present where as in case of H₂S such bonding is absent.

b) H₂O has high boiling point than HF
Reason:
In H₂O and H₂S inter molecular hydrogen bonding is present but in H₂O the no. of hydrogen bonds are more than in HF.

Short Answer Questions

Question 1.
Explain Kossel-Lewis approach to Chemical bonding.
Answer:
Kossel – Lewis Theory : This theory was also called as electronic theory of valency (or) chemical bond theory.
Postulates of Kossel – Lewis Theory : Kossel explained the formation of electrovalent bond while Lewis explained the formation of covalent bond. Their explanation of valency is mainly based on the inertness of noble gases.

Postulates:

• The chemical inertness of noble gases is due to the presence of octet structure. Octet rule was stated as follows “An atom must possess eight electrons in the outermost energy level for its stability”.
• Even though ‘He’ has only two electrons in the valency shell, it is highly stable and chemically inert.
• Elements other than zero group are chemically reactive because of having less than 8 electrons in their outer most shells.
• Every atom try to acquire the Eight electron configuration (octet) in its outer most shell. This can be possible by losing (or) sharing (or) gaining electrons.
• According to Lewis the valency electrons are represented by dots. These are called Lewis symbols.
  
• Lewis dot structures can be used to calculate the group valency of the element.
  Eg:
  has four electrons.
  ∴ Valency of ‘c’ is ‘4’.

Question 2.
Write the general properties of Ionic Compounds.
Answer:

1. Physical state : Due to close packing of ions, ionic compounds are crystalline solids.
2. Melting and Boiling points : In ionic crystals the oppositely charged ions are bound by strong electrostatic force of attraction. To overcome these attractive force between ions, more thermal energy is required. Hence the melting and boiling points of ionic compounds are high.
3. Solubility : Ionic compounds are soluble in polar solvents like water, liquid ammonia etc., but are insoluble in non – polar solvents like benzene, carbon disulphide etc.,
4. Reactivity: Reactions between ionic compounds in aqueous solution are very fast due to strong attraction among ions.
   e.g. : When AgNO₃ solution is added to NaCl solution, a white precipitate of AgCl is formed.
   
5. Isomerism : Ionic bond is non-directional.
   So ionic compounds cannot exhibit isomerism.
6. Electrical conductivity : Ionic substances conduct electricity in molten state and in aqueous solution. The ionic compounds are, therefore, electrolytes.

Question 3.
State Fajan’s rules and give suitable examples.
Answer:
Fajan’s rules:

1. Ionic nature of the bond increases with increase in the size of cation, e.g.: The ionic nature increases in the order
   Li⁺ < Na⁺ < K⁺ < Rb^– < Cs⁺
2. The formation of ionic bond is favoured with the decrease of the size of anion.
   e.g.: CaF₂ is more ionic than CaI₂.
3. If the charge on cation (or) anion (or) both is less, then they can form ionic bonds, e.g.: The ionic nature increases in the order
   
4. Cations with inert gas configurations form ionic compounds while those cations with pseudo inert gas configurations favour covalent bond formation.
   e.g.: Na⁺ in Na⁺Cl^– has an inert gas configuration. So Na⁺Cl^– is ionic. But CuCl is more covalent because Cu⁺ has not acquired inert gas configuration in this compound, instead it has acquired pseudo inert gas configuration.
5. The cation with inert gas configuration is more stable, e.g.: Ca²⁺ is more stable than Zn²⁺ ion.
6. 

Question 4.
What is Octet rule ? Briefly explain its significance and limitations.
Answer:
Octet rule: Every atom must possess 8 electrons in its outermost energy level for its stability. Atoms combine in two ways to get octet either by transfer of electrons (or) by mutual sharing of electrons. They can attain ns² np⁶ configuration.
e.g.:

1. Na loose one electron to get Ne configuration by possessing 8 electrons.
   Na : 1 s² 2s² 2p⁶ 3s¹ and Na⁺ : 1 s² 2s² 2p⁶.
2. ‘Cl’ atom take one electron to get “Ar” configuration. Cl^– : 1s² 2s² 2p⁶ 3s² 3p⁶.
   Here Na⁺ and Cl^– ions obey octet rule.
3. In H₂O molecule oxygen obey octet rule.

It is therefore, concluded that s2p6 configuration in the outer energy level constitutes a structure of maximum stability and therefore, of minimum energy.

The atoms of all elements when enter into chemical combination try to attain noble gas configuration (i.e.,) they try to attain 8 electrons in their outermost energy level which is of maximum stability and hence of minimum energy.
The tendency of an atoms to achieve eight electrons in their outermost shell is known as OCTET RULE.

Octet rule was the basis of electronic theory of valency.

Limitations : There are 3 types of exceptions to the octet rule. These are mentional below.

• Central atoms containing incomplete octet.
  Eg : BeH₂, BCl₃ etc.,
• Molecules containing odd number of electrons.
  
• Central atoms possessing more than 8 electrons which istermed as expanded octet.
  Eg : SF₆, H₂SO₄ etc.,
• This theory does not explained about shape of molecules.
• This theory does not explained about the formation of noble gas compounds like XeF₂, XeOF₂ etc.,

Question 5.
Write the resonance structures for NO₂ and \(\mathrm{NO}_3^{-}\)
Answer:
Resonance structure of NO₂

Question 6.
Use Lewis symbols to show electron transfer between the following pairs of atoms to form cations and anions:
(a) K and S
(b) Ca and O
(c) Al and N.
Answer:
a) Between the atoms K and S

b) Between Ca and O

c) Between Al and N

Question 7.
Explain why H₂O has dipole moment while CO₂ does not have.
Answer:

• H₂O molecule is a polar molecule and it has un symmetrical structure i.e. Angular (or) V – shape
• CO₂ molecule is non polar and it is linear molecule.
  
• SO H₂O has dipolemoment (μ = 1 ,835D) and CO₂ does not have dipolemoment (μ = 0)

Question 8.
Define Dipole moment. Write its applications.
Answer:
Dipole moment : The product of magnitude of the charge and the distance between the two poles (bond length) is called dipole moment.

• Dipole moment μ = q × d
  q = charge
  d = bond length
• Units : Debye (D), 1 Debye = 3.34 × 10^-30 coulombs metres.

Applications : –

• Dipole moment is used to calculate the percentage of ionic character in a molecule.
• It is used to know the shape of the molecule.
• Symmetry (symmetrical (or) non symmetrical) of the molecule can be known by dipole moment.

Question 9.
Explain why BeF₂ molecule has zero dipole moment although the Be-F bonds are polar.
Answer:

• Even though Be-F bonds in BeF₂ are polar, the dipole moment of BeF₂ molecule is zero. Because BeF₂ has linear shape.
  
• Flere the vectrorial sum of the dipole moment of two Be-F bonds is zero. Hence dipole moment
  (μ) = 0.

Question 10.
Explain the structure of CH₄ molecule.
Answer:
Formation of Methane molecule :

1. The central atom of methane is carbon.
2. The electronic configuration of carbon in ground state is and on excitation it is During excitation the 2s pair splits and the electron jumps into the adjacent vacant 2p_z orbital.
3. The undergo sp³ hybridisation giving four equivalent sp³ hybridised orbitals.
4. Each sp³ hybrid orbital overlaps with the 1s orbitals of hydrogen forming \(\sigma_{s p^3-s}\) bond.
5. In case of methane four \(\sigma_{s p^3-s}\) bonds are formed. The bonds are directed towards the four corners of a regular tetrahedron. The shape of methane molecule is tetrahedral with a bond angle 109°28’.
   

Question 11.
Explain Polar Covalent bond with a suitable example.
Answer:
The covalent bond which is formed by the mutual sharing of electron pairs between two dissimilar atoms is called polar covalent bond.

• Dissimilar atoms means two atoms having different electronegativity values (or) atoms of different elements.
  Eg : HF, HCl, H₂O, CO₂ etc.,
  Formation of HCl :
  
• In the above example H and Cl are two atoms of different elements having different electro negativities.
• These two atoms (H, Cl) mutually share the electron pairs and form the polar covalent bond.

Question 12.
Explain the shape and bond angle In BCl₃ molecule in terms of Valence Bond Theory.
Answer:
Boron trichloride molecule formation :

1. The electronic configuration of ‘B’ in the ground state is
2. On excitation the configuration is Now there are three half filled orbitals are available for hybridisation.
3. Now sp² hybridisation takes place at boron atom giving three sp² hybrid orbitals.
4. Each of them with one unpaired electron forms a ‘σ’ bond with one chlorine atom. The overlapping is σ_{sp² – p} (Cl atom has the unpaired electron in 2p_z orbital). In boron trichloride there are three ‘σ’ bonds.
   

Question 13.
What are σ and π bonds ? Specify the differences between them.
Answer:
Definition of σ bond : “σ – bond is along the internuclear axis. It has a cylindrical symmetry”.
Definition of π bond : “A covalent bond formed by a sidewise overlap of ‘p’ orbitals of atoms that are already bonded through a σ – bond, and in which the electron clouds are present above and below the internuclear axis is known as π – bond”.

Sigma bond (σ)

1. A ‘σ’ bond is formed by the axial overlap of two half filled orbitals belonging to the valence shells of the two combining atoms.
2. The ‘σ’ bonding electron cloud is symmetric about the inter-nuclear axis.
3. It is strong bond since the extent of overlap is much.
4. It allows free rotation of atoms or groups about the bond.
5. It can exist independently.
6. It determines the shape of the molecule.
7. There can be only one ‘σ’ bond between two atoms.
8. Hybrid orbitaIs form only ‘σ’ bonds.

Pi — bond (π)

1. A π-bond is formed by the lateral overlap of orbitals.
2. The π-bonding electron cloud lies above and below the flame of the internuclear axis.
3. It is a weaker than ‘σ’ bond, since the extent of overlap is less.
4. π-bond restricts such free rotation.
5. It is formed only after a ‘σ’ bond is formed.
6. It does not determine the shape of the molecule.
7. There can be one or two π-bonds between the atoms.
8. Hybrid orbitals cannot form π-bonds.

Question 14.
Even though nitrogen in ammonia is in sp³ hybridization, the bond angle deviate from 109°28. Explain.
Answer:
In NH₃ molecule the central nitrogen atom shares its ‘3’ unpaired electrons with three hydrogen atoms to form 3σ bonds. Hence NH₃ molecule contains one lone pair, three bond pairs.

a) Because of the repulsion between the lone pair and the bond pairs the angle reduces to 107°.
b) According to VSEPR theory the geometry of the molecule is pyramidal with bond angle 107°.

Question 15.
Show how a double and triple bond are formed between carbon atoms in
(a) C₂H₄ and
(b) C₂H₂ respectively.
Answer:
Formation of double bond between the carbon atoms of C₂H₄ : –
C – Ground state electronic configuration 1s² 2s² 2p²

•  In ethylene two carbons undergo sp² hybridisation.
• One of sp² hybrid orbital of carbon overlaps with sp² hybrid orbital of another carbon atom to form C – C sigma bond.
• The two other sp² hybrid orbitals of each carbon overlap with ‘s’ orbital of hydrogen atoms to form C – H bonds.

The unhybridised orbital of one carbon atom overlap side wisely with the similar orbital to form weak π bond.

Formation of triple bond between the carbon atoms of C₂H₂ : –
C – Ground state electronic configuration 1s² 2s² 2p²

• In acetylene two carbons undergoes sp hybridisation.
• One of sp hybrid orbital of carbon overlaps with sp hybrid orbital of another carbon atom to form C – C sigma bond.
• another sp hybrid orbitals of each carbon overlap with s orbital of hydrogen atoms to form C – H bonds.
• The un hybridised orbitals of two carbon atom overlap side wisely to form two weak π bonds.
  

Question 16.
Explain the hybridization involved in PCl₅ molecule. (T.S. Mar. ’16, ’15)
Answer:
1) In PCl₅ the electron configuration of phosphorus is

2) Phosphorus undergoes sp³d – hybridisation by intermixing of one s-orbital [3s], three p – orbitals [3p_x, 3p_y, 3p_z] and one d – orbital.
These five hybrid orbitals overlap. The p_z orbitals of chlorine atoms forming five \(\sigma_{s p^3 d-s}\) bonds. Out of these five p – Cl bonds three are coplanar and the remaining two are in the axial position. There by PCl₅ acquires the trigonal bipyramidal shape. The molecule contains two bond angles 90° and 120°.

Question 17.
Explain the hybridization involved in SF₆ molecule.
Answer:
In this hybridisation one ‘s’ orbital, three ‘p’ orbitals and two ‘d’ orbitals of the excited atom combine to form six equivalent sp³d² hybrid orbitals.
e.g. : SF₆

These six sp³ d² hybrid orbitals overlap six 2p_z orbitals of fluorine atoms to form six \(\sigma_{s p^3 d^2}\) bonds. The directions of the bonds give an octa-hedral shape to the molecule. The bond angle is 90° or 180° & 90°.

Question 18.
Explain the formation of Coordinate Covalent bond with one example.
Answer:
Co-ordinate covalent bond (dative bond) is a special type of covalent bond. It is proposed by Sidgwick. It is formed by the sharing of electrons between two atoms in which both the electrons of the shared electron pair are contributed by one atom and the other atom nearly participates in sharing.

The bond is represented as (“→”) an arrow starting from the donar atom and directed towards the acceptor atom.

Examples :
1) Ammonia – Boron trifluoride H₃N : → BF₃
Ammonia combines with boron trifluoride to give ammonium boron trifluoride.

In ammonia nitrogen has a complete octet and also it has a lone pair of electrons. In BF₃ the boron atom has a total of six electrons after sharing with fluorine. Nitrogen donates the electron pair to boron to form a co-ordinate covalent bond between ammonia and boron trifluoride.

2) Ammonium ion (\(\mathrm{NH}_4^{+}\))

3) Hydronium ion (\(\mathrm{H}_3 \mathrm{O}^{+}\))

Properties of co-ordinate covalent bond :

1. The bond do not ionise in water.
2. The compounds are generally soluble in organic solvents and are sparingly soluble in water.
3. These compounds exhibit space isomerism because the bond is rigid and directional.
4. The bond is semipolar in nature – so their volatility lies in between covalent and ionic bonds.

Question 19.
Which hybrid orbitals are used by Carbon atoms in the following molecules?
(a) CH₃ -CH₃
(b) CH₃ – CH = CH₂
(c) CH₃ – CH₂ – OH
(d) CH₃ – CHO
Answer:
a) CH₃ -CH₃ (ethane)
The two carbons of ethane undergo sp³ hybridisation.

Carbon – (1) – undergoes sp² hyrbidisation
Carbon – (2) – undergoes sp² hyrbidisation
Carbon – (3) – undergoes sp³ hyrbidisation

Carbon (1) and (2) both undergo ‘sp³‘ hybridisation.

Question 20.
What is Hydrogen bond ? Explain the different types of Hydrogen bonds with examples. (A.P., T.S. Mar. ’16)
Answer:
Hydrogen bond is a weak electrostatic bond formed between partially positive charged hydrogen atom and an highly electronegative atom of the same molecule or another molecule.

Hydrogen bond is formed when the Hydrogen is bonded to small, highly electronegative atoms like F, O and N. A partial positive charge will be on hydrogen atom and partial negative charge on the electronegative atom.

The bond dissociation energy of hydrogen bond is 40 KJ/mole. Hydrogen bond is represented with dotted lines (—–). Hydrogen bond is stronger than Vander Waals’ forces and weaker than covalent bond.
Hydrogen bonding is of two types.

(1) Intermolecular hydrogen bond and
(2) Intramolecular hydrogen bond.

1) Intermolecular hydrogen bond :
If the hydrogen bond is formed between two polar molecules it is called intermolecular hydrogen bond, i.e., the hydrogen bond is formed between hydrogen atom of one molecule and highly electronegative atom of another molecule is known as intermolecular hydrogen bond.
Ex. : Water (H₂O) ; HF : NH₃ ; p – nitrophenol, CH₃COOH, ethyl alcohol etc.

Water molecule forms oi. associated molecule through intermolecular hydrogen bond. Due to molecular association water possess high boiling point.

m or p – nitrophenol :

2) Intramolecular hydrogen bond:
If the hydrogen bond is formed within the molecule it is known as intramolecular hydrogen bond.
Ex. : o – nitrophenol; o – hydroxy benzaldehyde.

Abnormal behaviour due to hydrogen bond:

1. The physical state of substance may alter. They have high melting and boiling points.
2. Ammonia has higher boiling point than HCl eventhough nitrogen and chlorine have same electronegativity values (3.0). Ammonia forms an associated molecule through intermolecular hydrogen bond.
3. p – hydroxy benzaldehyde have higher boiling point than o- hydroxy benzaldehyde. This is due to intermolecular hydrogen bonding in para isomer.
4. Ethyl alcohol is highly soluble in water due to association and co-association through intermolecular hydrogen bonding.

Question 21.
Explain the formation of H₂ molecule on the basis of Valence Bond theory.
Answer:
Postulates of valency bond theory:

1. Covalent bond is formed by the overlap of an half filled atomic orbital of one atom with an half filled atomic orbital of the other atom involved in the bond formation.
2. The electrons in these two orbitals involved in the overlap shall have opposite spins.
3. Greater the overlap stronger the bond formed.
4. The bonds are formed mostly in the direction in which the electron clouds are concentrated.
   

Formation of H₂ molecule:
Hydrogen molecule is formed due to overlaping of s – s orbitals. When two Hydrogen atoms come together, is orbitals of the Hydrogen atoms overlap to form a strong ‘σ” bond. This is σ_s-s’

Question 22.
Using Molecular Orbital Theory explain why the B₂ molecule is paramagnetic ?
Answer:
Boron electronic configuration is – 1s² 2s² 2p¹
The molecular orbital energy level sequence for B₂ is

• Bond order = \(\frac{6-4}{2}\) = \(\frac{2}{2}\) = 1
• In the above sequence unpaired electrons are present.
• Presence of unpaired electrons leads to paramagnetic nature.
  ∴ B₂ molecule is paramagnetic.

Question 23.
Write the important conditions necessary for linear combination of atomic orbitals.
Answer:

1. The molecular orbitals are formed when the atomic orbitals combine linearly (i.e.,) when the atoms approach each other. The no. of molecular orbitals resulting are equal to the no.of atomic orbitals combining.
2. Only such atomic orbitals which are of similar energies and symmetry with respect to the inter nuclear axis combine to form molecular orbitals.
3. The total no. of molecular orbitals produced will be numerically equal to the no. of combining orbitals.
4. The order of energies of bonding, anti bonding and non bonding orbitals can be written as bonding orbitals < non bonding orbitals < anti – bonding orbitals.

Question 24.
What is meant by the term Bond order? Calculate the bond orders in the following
(a) N₂
(b) O₂
(c) \(\mathrm{O}_2^{+}\) and
(d) \(\mathrm{O}_2^{-}\)
Answer:
Bond order : The half of the difference between the no.of bonding electrons and anti bonding electrons is known as bond order. ’
a) N₂ : Molecular orbital energy level sequence.

→ Bond Order = \(\frac{10-4}{2}\) = \(\frac{6}{2}\) = 3

b) O₂ : Molecular orbital energy level sequence.

Question 25.
Of BF₃ and NF₃, dipole moment is observed for NF₃ and not for BF₃. Why ?
Answer:
Of BF₃ and NF₃ dipole moment is observed for NF₃ and not for BF₃.
Reasons:

• BF₃ molecule is non polar and is a symmetrical molecule. Symmetrical molecules have zero dipole moment.
• NF₃ molecule is polar and it is a unsymmetrical molecule so it has dipole moment.
• BF₃ molecule has trigonal planar structure.
  NF₃ molecule has pyramidal shape.
• NF₃ has dipole moment µ = 0.8 × 10^-30 coloumb × meter.
  

Question 26.
Eventhough both NH₃ and NF₃ are Pyramidal, NH₃ has a higher dipole moment compared to NF₃. Why? (A.P. Mar.’16)
Answer:

• Both NH₃ and NF₃ molecules have pyramidal shape and in two molecules N atom has lone pair of electrons.
• Even though fluorine has more electronegativity than nitrogen the dipole moment of NH₃ is greater than that of NF₃
  µ (NH₃) = 4.9 × 10^-30 coloumbs × meter.
  µ (NF₃) = 0.8 × 10^-30 coloumbs × meter.
• In case of NH₃ the orbital dipole due to lone pair is in the same direction as the resultant dipole moment of N – H bonds.

Where as in case of NF₃ the orbital dipole is in the direction opposite to the resultant dipole
moment of the three N – F bonds.

Question 27.
How do you predict the shapes of the following molecules making use of VSEPR Theory ?
(a) XeF₄
(b) BrF₅
(c) ClF₃ and
(d) IC\(l_4^{-}\)
Answer:
According to VSEPR theory the shape of the molecule can be predicted by counting no.of electron pairs (bond pairs, lone pairs) around the central atom.
a) XeF₄:
In XeF₄ No.of bond pairs present are ‘4’.
No.of lone pairs present are ‘2’.
According to VSEPR theory shape of molecule is square planar (Actual shape octahedral).

b) BrF₅ :
In BrF₅ No.of bond pairs present are ‘5’
No.of lone pairs present are ‘1’
According to VSEPR theory shape of the molecule is square pyramid (actual shape octahedral)

c) ClF₃:

In ClF₃ No.of bond pairs present are ‘3’
No.of lone pairs present are ‘2’
According to VSEPR Theory shape of ClF₃ Molecule is T – shape (Actual shape TBP)

d) IC\(l_4^{-}\):
In IC\(l_4^{-}\) No.of bond pairs present are ‘4’
No.of lone pairs present are ‘2’
According to VSEPR Theory shape of ICl\(l_4^{-}\) is square planar
(Actual shape octahedral)

Long Answer Questions

Question 1.
Explain the formation of Ionic Bond with a suitable example.
Answer:
The electrostatic force that binds the oppositely charged ions which are formed by the transfer of electrons from one atom with low ionization potential to the other with high electron affinity is called ionic bond or electrovalent bond.
It is formed when the electronegativity difference between the two atoms is more tha 1.7.
Example :
Formation of sodium chloride in terms of orbital concept:
1) Na (Z = 11). The electronic configuration is 1s² 2s² 2p⁶ 3s¹.
This can be expressed as

2) Cl (Z = 17). The electronic configuration is = 1s² 2s² 2p⁶ 3s² 3p⁵
This can be expressed as

3) The configurations after the transfer of electrons forming ions can be expressed as:

In the formation of sodium chloride the 3s electron of sodium atom is transferred to the 3p₂ orbital of chlorine atom. The Na⁺ ion and Cl^– ion so formed are now bound by strong coulombic electrostatic forces of attraction forming sodium chloride.

Question 2.
Explain the factors favourable for the formation of Ionic Compounds.
Answer:
Factors favour the ionic bond formation:
a) Cation formation:

1. Lower ionization energy: Lower ionization energy of an atom greater is the ease of formation of cation
   e.g. : The ionization energy of sodium is 117.9 kcal/mole and that of potassium is 100 kcal/mole So K⁺ ion can readily form than Na⁺ ion.
2. Large size of the atom : Large atoms can easily lose the valence electrons. If the size large, the distance between the nucleus and the valence electrons is more and so the force of attraction is less. Therefore the electron can be removed easily from the atom forming cation.
3. Ion with lower charge: Small magnitude of charge favours the formation of ions easily.
   e.g. : The ease of ion formation increases in the order Na⁺ > Mg²⁺ > Al³⁺.
4. Cations with inert gas configuration : Ion possessing electronic configuration similar to zero group elements are more stable than those ions which do not have such configuration.
   eg.: Ca²⁺ (2, 8, 8) is more stable than Zn²⁺ (2, 8, 18) because the former has inert gas
   configuration.

b) Anion formation:

1. High electron affinity: If the electron affinity of an element is high its anion can be easily formed.
   e.g.: Cl + e^— → Cl^–
2. Smaller size of atom : Smaller the size of the atom lesser is the distance between the nucleus and the valence orbit. Hence the nuclear attraction on incoming electron is more. So the anion is readily formed.
3. Lower charge : Ions with lower charge are more readily formed than those with higher charge.
   e.g.: Cl^– > O ^2- > N^3-
4. The ions with inert gas electronic configuration are more readily formed than others with the same charge. .
   c) If the two bonded atoms differ by more than 1.70 in their EN values, the bond between them is ionic in nature.

Question 3.
Draw Lewis Structures for the following molecules.
(a) H₂S
(b) SiCl₄
(c) BeF₂
(d) HCOOH
Answer:
Lewis structures:

Question 4.
Write notes on
(a) Bond Angle
(b) Bond Enthalpy
(c) Bond length and
(d) Bond order.
Answer:
a) Bond angle : The angle between the orbitals containing bonding electron pairs around the central atom in a molecule (or) complex ion is known as Bond angle.

• it is expressed in degrees.
• It is determined experimentally by spectroscopic methods.
  Eg: In H₂O (H — 0 — H) bond angle is 104.5° .

b) Bond Enthalpy : The amount of energy required to break one mole of bonds of a particular type between two atoms in a gaseous state is known as Bond Enthalpy.
Units: KJ/Mole.
Eg: H – H bond enthalpy in hydrogen is 435.8 KJ/mole
H_2(g) → H_(g) + H_(g) ∆H = 435.8 KJ/mole

• In case of poly atomic molecules average bond enthalpy is used.
  

c) Bond length : The distance between the nuclei of the atoms in a molecule is known as bond length.

• Bond length is equal to the sum of the covalent radii of the two atoms that are bonded.
• Units of Bond length A° (or) cm (or) m (or) pm
• As the number of bonds between two atoms increases the bond length decrease.
  

d) Bond Order:
According to Lewis the bond order is given by the number of bonds between the two atoms in a covalent molecule.
Eg: Bond order of N₂ – 3
Bond order of O₂ – 2
Bond order of H₂ – 1

• In case of Iso electronic species and ions bond orders are same.
• Bond order is useful in predecting stabilities of molecules.
• Bond order increases bond enthalpy increases and Bond length decreases.

Question 5.
Give an account of VSEPR Theory and its applications.
Answer:
VSEPR theory was proposed by Sidgwick and Powell and later extended by Gillespie and Nyholm. It was developed by Ronald and Nyholm.
This theory explains the shapes of simple molecules having electron pairs bonded or non-bonded. The repulsions among the electron pairs present in the valence shell of the central atom decides the shape of the molecules.

According to this theory :

a) The shape of the molecule is determined by repulsions between all of the electron pairs present in the valency shell of central atom.
b) The electron pairs orient in space so as to have minimum repulsions among them.
c) The magnitude of repulsions between bonding pairs of electrons depends on the electronegativity difference between the central atom and the other atoms.
d) The order of repulsions between various electron pairs is lone pair – lone pair > lone pair – bond pair > bond pair – bond pair.
e) The repulsive forces between different bonds is of the order triple bond > double bond > single bond.
f) The shapes of molecules can be predicted as.

In NH₃ molecule the central nitrogen atom shares its ‘3’ unpaired electrons with three hydrogen atoms to form 3σ bonds. Hence NH₃ molecule contains one lone pair, three bond pairs.

a) Because of the repulsion between the lone pair and the bond pairs the angle reduces to 107°.
b) According to VSEPR theory the geometry of the molecule is pyramidal with bond angle 107°.

Question 6.
How do you explain the geometry of the molecules on the basis of Valence bond Theory ?
Answer:
Postulates of valency bond theory :

1. Covalent bond is formed by the overlap of an half filled atomic orbital of one atom with an half filled atomic orbital of the other atom involved in the bond formation.
2. The electrons in these two orbitals involved in the overlap shall have opposite spins.
3. Greater the overlap stronger the bond formed.
4. The bonds are formed mostly in the direction in which the electron clouds are concentrated.
   

Formation of H₂ molecule :
Hydrogen molecule is formed due to overlapping of s – s orbitals. When two Hydrogen atoms come together, 1s orbitals of the Hydrogen atoms overlap to form a strong “σ” bond. This is σ_s-s.

Formation of Cl₂ molecule :
The electronic configuration of Chlorine atom is It has one half filled 3p_z orbital. The p_z orbital of one chlorine atom overlaps the p_z orbital of the other chlorine atom and the two electrons of opposite spins pair up to form covalent bond. As the overlap along the internuclear axis is maximum a strong bond is formed. The bond is formed due to \(\sigma_{p-p}\) overlap.

Formation of O₂ molecule:
The electronic configuration of oxygen atom is It has two half filled 2p orbitals i.e., 2p_y and 2p_z.
The p_y orbital of one atom overlaps the p_y orbital of the second atom to form a ‘σ’ bond \(\sigma_{p_y}-p_y\).
The p_z orbital in the two atoms will be at right angles to the internuclear axis. These two can have lateral overlap. The electron density of the bonded pair is distributed in two banana like regions lying on either side of the internuclear axis. Thus the oxygen molecule has a double bond. The molecule has one σ_{p – p} and one π_{p – p} between the two atoms.

Question 7.
‘What do you understand by Hybridisation? Explain different types of hybridization involving s and p orbitals. (Mar. ’13)
Answer:
Hybridisation is defined as the process of mixing of atomic orbitals of nearly equal energy of an atom to give the same number of new set of orbitals of equal energy and shapes.
Depending on the number and nature of orbitals involving hybridisation it is classified into different types. If ‘s’ and ‘p’ atomic orbitals are involved three types are possible namely sp³, sp² and sp.

1. sp³ hybridisation : In this hybridisation one’s and three ‘p’ atomic orbitals of the excited atom combine to form four equivalent sp³ hybridised orbitals.
This hybridisation is known as tetrahedral or tetragonal hybridisation.

Each sp³ hybridised orbital possess 25% ‘s’ nature and 75% of ‘p’ nature. The shape of the molecule is tetrahedral with a bond angle 109°28′, e.g. : Formation of Methane molecule :

1. The central atom of methane is carbon.
2. The electronic configuration of carbon in ground state is and on excitation it is During excitation the 2s pair splits and the electron jumps into the adjacent vacant 2p_z orbital.
3. The undergo sp³ hybridisation giving four equivalent sp³ hybridised orbitals.
4. Each sp³ hybrid orbital overlaps with the 1s orbitals of hydrogen forming \(\sigma_{s p^3-s}\) bond.
5. In case of methane four \(\sigma_{s p^3}-s\) bonds are formed. The bonds are directed towards the four corners of a regular tetrahedron. The shape of methane molecule is tetrahedral with a bond angle 109°28′.
   

2. sp² hybridisation : In this hybridisation one ‘s’ and two p’ atomic orbitals of the excited atom
combine to form three equivalent sp² hybridised orbitals.
This hybridisation is also known as trigorial hybridisation. In sp² hybridisation each sp² hybrid orbital has 33.33% ‘s’ nature and 66.66% ‘p’ nature. The shape of the molecule is trigonal with a bond angle 120°.
E.g.: Boron trichioride molecule formation:

1. The electronic configuration of ‘B’ in the ground state is
2. On excitation the configuration is Now there are three half filled orbitals are available for hybridisation.
3. Now sp² hybridisation takes place at boron atom giving three sp² hybrid orbitals.
4. Each of them with one unpaired electron forms ‘σ’ bond with one chlorine atom. The overlapping is \(\sigma_{s p^2-p}\) (Cl atom has the unpaired electron in 2P_z orbital). In boron trichloride there are three ‘σ’ bonds.
   

3. sp hybridisation : In this hybridisation one ‘s and one ‘p’ atomic orbitals of the excited atom combine to form two equivalent sp hybridised orbitals.
This hybridisation is also known as diagonal hybridisation. In sp hybridisation each sp hybrid orbital has 50% ‘s’ character and 50% ‘p’ character. The shape of the molecule is linear or diagonal with a bond angle 180°.
Ex. : Beryllium chloride molecule formation:

1. Be atom has electronic configuration.
2. In ground state it has no half filled orbitais. On excitation the configuration becomes \(1 s^2 2 s^1 2 P_x^1\)\(2 p_y^0 2 p_z^0\).
3. Now sp hybridisation takes place at beryllium atom giving two sp hybrid orbitais. Each of them with one unpaired electron forms a ‘σ’ bond with one chlorine atom.
4. The overlaping is σ_sp-p (Cl atom has the unpaired electron in 2p_z orbital). In beryllium chloride there are two ‘σ’ bonds.
   

Question 8.
Write the salient features of Molecular Orbital Theory.
Answer:
Molecular orbital theory:
Hund and Mulliken.

1. Theory was proposed by
2. Atomic orbitals (AO) of the bonded atoms combine loose their identity to form molecular orbitals (MO).
3. The electrons in a molecule reside in molecular orbitals.
4. Molecular orbital is the region around the nuclei where the probability of finding electon is maximum (or) the wave function of a molecule.
5. The electrons of all the atoms in a molecule are revolving under the influence of all the nuclei in the molecule.
6. The molecular orbitals are formed when the atomic orbitals combine linearly.
7. The shape of the molecular orbitals depends on the shape of the atomic orbitals.
8. Each molecular orbital can accommodate two electrons with opposite spins.
9. The molecular orbitals are arranged in the increasing order of energy, and electrons are filled in the same order.
10. Hund’s rule of maximum multiplicity is to be followed while filling molecular orbitals.
11. Atomic orbitals with similar energy and symmetry can combine to give molecular orbitals.
12. Molecular orbitals with energy lower than A.O are known as bonding molecular orbital; while those with higher energy are known as anti bonding molecular orbitals. Those which are not involved in combination are called non bonding orbitals.
    
13. The order of energies of molecular orbital is : bonding < nonbonding < antibonding molecular orbitals.
14. The bonding orbitals are designated a σ and π.
15. The antibonding orbitals are designated σ* and π*.

Filling of electrons into molecular orbitals :
The sequence of energy levels of molecular orbitals is given by

sequence is valid for oxygen and other heavier elements.

This sequence is valid for lighter elements like B.CandN.

Question 9.
Give the Molecular Orbital Energy diagram of
(a) N₂ and
(b) O₂. Calculate the respec- five bond order. Write the magnetic nature of N₂ and O₂ molecules.
Answer:
Molecular orbital energy level diagram (MOED) of ‘N₂‘ ; Electronic configuration of nitrogen (z = 7) is 1s² 2s² 2p³. Since nitrogen atom has 7 electrons, the molecular orbitals of nitrogen molecule (N₂) has 14 electrons which are distributed as below :

• Bond order = \(\frac{8-2}{2}\) = 3 ( N ≡ N)
• Absence of unpaired electrons showed that N₂ molecule is diamagnetic.

MOED of O₂:
Electronic configuration of Oxygen (Z = 8) is 1s² 2s² 2p⁴. Since Oxygen atom has 8 electrons, the molecular orbitais of Oxygen molecule (O²) has 16 electrons, which are distributed as below:

• Bond order = \(\frac{10-6}{2}\) = 2 (O = O)
• Presence of two unpaired 6 electrons \(\left(\pi_{2 p_y^1}^{\star}, \pi_{2 p_z^1}^{\star}\right)\) showed that O₂ molecule is paramagnetic.

Solved Problems

Question 1.
Write the Lewis dot structure of CO molecule.
Solution:
Step 1. Count the total number of valence electrons of carbon and oxygen atoms. The outer (valence) shell configurations of carbon and oxygen atoms are: 2s² 2p² and 2s² 2p⁴, respectively. The valence electrons available are 4 + 6 = 10.

Step 2. The skeletal structure of CO is written as: C O

Step 3. Draw a single bond (one shared electron pair) between C and O and complete the octet on O, the remaining two electrons are the lone pair on C.

This does not complete the octet on carbon and hence we have to resort to multiple bonding (in this case a triple bond) between C and O atoms. This satisfies the octet rule condition for both atoms.

Question 2.
Write the Lewis structure of the nitrite ion, \(\mathrm{NO}_2^{-}\).
Solution:
Step 1. Count the total number of valence electrons of the nitrogen atom, the oxygen atoms and the additional one negative charge (equal to one electron).
N(2s² 2p³), O (2s² 2p⁴)
5 + (2 × 6) + 1 = 18 electrons

Step 2. The skeletal structure of \(\mathrm{NO}_2^{-}\) is written as: O N O

Step 3. Draw a single bond (one shared electron pair) between the nitrogen and each of the oxygen atoms completing the octets on oxygen atoms. This, however, does not complete the octet on nitrogen if the remaining two electrons constitute lone pair on it.

Hence we have to resort to multiple bonding between nitrogen and one of the oxygen atoms (in this case a double bond). This leads to the following Lewis dot structures.

Question 3.
Explain the structure of \(\mathrm{CO}_3^{2-}\) ion interms of resonance.
Solution:
The single Lewis structure based on the presence of two single bonds and one double bond between carbon and oxygen atoms is inadequate to represent the molecule accurately as it represents unequal bonds. According to the experimental findings, all carbon to oxygen bonds in CCO² are equivalent.

Therefore the carbonate ion is best described as a resonance hybrid of the canonical forms I, II, and III shown below.

Question 4.
Explain the structure of CO₂ molecule.
Solution:
The experimentally determined carbon to oxygen bond length in CO₂ is 115 pm. The lengths of a normal carbon to oxygen double bond (C = O) and carbon to oxygen triple bond(C ≡ O) are 121 pm and 110 pm respectively. The carbon-oxygen bond lengths in CO₂ (115 pm) lie between the values for C = O and C ≡ O. Obviously, a single Lewis structure cannot depict this position and it becomes necessary to write more than one Lewis structures and to consider that the structure of CO₂ is best described as a hybrid of the canonical or resonance forms I, II and III.

Additional Problems

Question 1.
The experimental dipole moment of HCl is 1.03D and its bond length (distance) is 1.27Å. Calculate the % of ionic character of HCl.
Answer:
Calculated dipole moment = q × d
= 4.8 × 10^-10 × 1.27 × 10^-8 cm
= 6.09 Debye
% of ionic character = \(\frac{\mu_{\text {ods }}}{\mu_{\text {calc }}}\) × 100
= \(\frac{1.03}{6.09}\) × 100
= 16.9%

Question 2.
The dipole moment of H₂S is 0.95D. Find the bond moment if the bond angle is 97° (Cos 48.5° = 0.662).
Answer:
\(\mu_{\text {obs }}\) = 2 (bond moment) \(\left(\cos \frac{\theta}{2}\right)\)
0.95 = 2 (bond moment) (Cos 48.5°)
0.95 = 2 × bond moment × 0.662
Bond moment = \(\frac{0.95}{2 \times 0.662}\) = 0.72D

Andhra Pradesh BIEAP AP Inter 1st Year Chemistry Study Material 2nd Lesson Classification of Elements and Periodicity in Properties Textbook Questions and Answers.

AP Inter 1st Year Chemistry Study Material 2nd Lesson Classification of Elements and Periodicity in Properties

Very Short Answer Questions

Question 1.
What is the difference in the approach between Mendeleev’s periodic law and the modern periodic law?
Answer:

• According to Mendeleev, the physical and chemical properties of elements are periodic functions of their atomic weights.
• According to modem periodic law, elements’ physical and chemical properties are periodic functions of their atomic numbers.

Question 2.
In terms of period and group, where would you locate the element with Z = 114?
Answer:
Element Z = 114 is present in 7^th period and IVA group (Group – 14)

Question 3.
Write the atomic number of the element, present in the third period and seven¬teenth group of the periodic table.
Answer:
The Element present in 3^rd period and Group – 17 (VIIA group) is chlorine (Cl). It’s atomic number is 17.

Question 4.
Which element do you think would have been named by
a) Lawrence Berkeley Laboratory .
b) Seaborg’s group
Answer:
a) Lawrence Berekeley Laboratory – Lanthanide
b) Seaborg’s group – Actinide (Transuranic element).

Question 5.
Why do elements in the same group have similar physical and chemical properties ?
Answer:
Elements in the same group have same no. of valency shell electrons and have similar outer elec¬tronic configuration so these have similar physical and chemical properties.

Question 6.
What are representative elements ? Give their valence shell configuration.
Answer:

• Representative elements are s and p-block elements except zero group.
• These have general electronic configuration ns^1-2np^1-5.

Question 7.
Justify the position of f-block elements in the periodic table.
Answer:
The two series of elements lanthanides and actinides have been grouped separately and placed at the bottom of the periodic table, though they belong to the sixth and seventh periods of third group (III B).

The justification for assigning one place to these elements has been given on the basis of their similar properties. The properties are so similar that elements from Ce to Lu can be considered as equivalent to one element. In case these elements are assigned different positions (i.e.,) arranged in order of their increasing atomic numbers, the symmetry of the whole arrangement would be disrupted. The same explanation can be given in the case of actinides.

Question 8.
An element ‘X’ has atomic number 34. Give its position in the periodic table.
Answer:
The element X with atomic no (z) = 34 has electronic configuration 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁴. Hence the element is present in 4,h period and 16th group (VIA group).

Question 9.
What factors impart characteristic properties to the transition elements ?
Answer:
Transition elements exhibits characteristic properties because

• The differentiating electron enters into penultimate d-subshell
• These elements have small size.
• These possess high effective nuclear charge.

Question 10.
Give the outer shells configuration of d-block and f-block elements.
Answer:

• The outer shell electronic configuration of d-block – elements is ns^1-2 (n-1)d^1-10
• The outer shell electronic configuration of f-block – elements is ns²(n-1)d^{0 (or) 1} (n – 2) f^1-14

Question 11.
State and give one example for Dobereiner’s law of triads and Newland’s law of octaves.
Answer:
1) Dobereiner law :

• According to Dobereiner in a traid (3 – elements) the atomic weight of the middle element is the arithmatic mean of the other two elements.
  

2) Newland’s law of octaves : According to Newlands, the elements arranged in the increasing order of atomic weights noticed that every eight element had properties similar to the first element. This relationship was just like every eight note that resembles the first in octaves of music.

Question 12.
Name the anomalous pairs of elements in the Mendaleev’s periodic table.
Answer:
In Mendeleev’s periodic table anamalous pairs are the elements whose atomic weights increasing order is reversed.
Eg:

Question 13.
How does atomic radius vary in a period and in a group ? How do you explain the variation ?
Answer:
In a period : Atomic radius decreases generally from left to right in a period.
Reason : In periods electrons are entered into same subshells.
In a group : Atomic radius increases generally from top to bottom in a group.
Reason : In groups electrons are entered into new subshells.

Question 14.
Among N^-3; O^-2, F^–, Na⁺, Mg⁺² and Al⁺³
a. What is common in them ?
b. Arrange them in the increasing ionic radii.
Answer:
Given ions are
N^-3, O^-2, F^–, Na⁺, Mg⁺² and Al⁺³.
a) The above ions have same number of electrons (All have 10 electrons). So these are called iso electronic species.
b) The increasing order of ionic radii among above ions is
Al⁺³ < Mg⁺² < Na⁺ < F^– < O^-2 < N^-3
Reason : – In case of iso electronic species as the nuclear charge increases ionic radii decreases.

Question 15.
What is the significance of the term isolated gaseous atom while defining the ionization enthalpy.
Hint: Requirement for comparison.
Answer:

• Isolated gaseous atom’s ionisation enthalpy is taken as reference value and it is required to compare this values to various ions of this elements and to compare this values with various elements.

Question 16.
Energy of an electron in the ground state of the hydrogen atom is – 2.18 × 10^-18J. Calculate the ionization enthalpy of atomic hydrogen in terms of J mol-1.
Answer:
Given that the energy of the electron in the ground state for hydrogen atom = – 2.18 × 10^-18 J.
For 1 mole of atoms is given by – 2.18 × 10^-18 J × 6.023 × 10²³
= -13.13 × 10⁵ J/Mole
∴ Ionisation enthalpy of hydrogen atom = 13.13 × 10⁵ J/Mole.

Question 17.
Ionization enthalpy, (IE₁) of O is less than that of N — explain.
Answer:

• Oxygen has electronic configuration 1s² 2s² 2p⁴
  
• Nitrogen has electronic configuration 1s² 2s² 2p³
  
• Nitrogen has half filled shell and is stable so more amount of energy is required to remove an electron, than in oxygen.
  Hence IE, of ‘O’ is less than that of ‘N’.

Question 18.
Which in each pair of elements has a more negative electron gain enthalpy?
a. O or Fb. F or Cl
Answer:
a)

has more negative electron gain enthalpy than that of

b)

has more negative electron gain enthalpy than that of

Question 19.
What are the major differences between metals and non-metals ?
Answer:
Metals

• These are generally in solid form (Except Hg)
• These are good conductors of heat and electricity.
• These have high m.pts and b.pts.
• Generally these are electropositive.
• These forms more ionic compounds.

Nón metals

• These may be solids (or) gases (or) liquids.
• These are not good conductors of heat and electricity.
• These have low m.pts and b.pts.
• Generally these are electronegative.
• These forms more covalent compounds.

Question 20.
Use the periodic table to identify elements.
a. With 5 electrons in the outer subshell
b. Would tend to lose two electrons
c. Would tend to gain two electrons.
Answer:
a) The elements possessing 5 electrons in the outer most shell are group 15 (V_A) elements.

• General outer electronic configuration is ns² np³ Eg. : N, P, As………

b) The elements tend to lose two electrons are Group — II elements.

• General outer electronic configuration is ns² Eg: Mg, Ca, Sr etc.

c) The elements tend to gain two electrons are Group – VI_A elements (16^th group).

• General outer electronic configuration is ns² np⁴ Eg : O, S, Se ………

Question 21.
Give the outer electronic configuration of s, p, d and f – block elements.
Answer:

Question 22.
Write the increasing order of the metallic character among the elements B, Al, Mg and K.
Answer:
Given elements are B, Al, Mg and K
The increasing order of metallic character is

Question 23.
Write the correct increasing order of non – metallic character for B, C, N, F and Si.
Answer:
Given elements are B, C, N. F and Si
The increasing order of non- metallic character is

Question 24.
Write the correct increasing order of chemical reactivity in terms of oxidizing property for N, O, Fand Cl.
Answer:
The correct increasing order of chemical reactivity in terms of oxidizing property for N, O, F and Cl is F > 0 > Cl > N.

Question 25.
What is electronegativity ? How is this useful in understanding the nature of elements?
Answer:
Electronegativity : The tendency of an element (or) atom to attract the shared pair of electrons towards itself in a molecule is called electronegativity.

• On the basis of electronegativity values nature of elements can be predicted. Higher electro-negativity values indicates that element is non metal and lower values indicates that the element is a metal.
• On the basis of electronegativity values bond nature also predicted (Ionic/covalent).

Question 26.
What is screening effect? How is it related to lE?
Answer:
The decrease of nuclear attraction on outer most shell electrons due to presence of inner energy electrons is called screening effect.

• As the screening effect increases LE. values decreases.

Question 27.
How are electronegativity and metallic & non-metallic characters related?
Answer:

• Greater the electronegativity values of an element indicates that non metallic nature and low metallic nature of that element.
• Lower the electronegativity value of an element indicates that low non metallic nature and high metallic nature of that element.

Electronegativity ∝ Non metallic nature
Electronegativity ∝ \(\frac{1}{\text { Metallic Nature }}\)

Question 28.
What is the valency possible to Arsenic with respect to oxygen and hydrogen?
Answer:

• The valency of Arsenic with respect to hydrogen is ‘3’
  Eg : AsH₃
• The valency of Arsenic with respect to oxygen is ‘5’
  Eg : As₂O₅

Question 29.
What is an amphoteric oxide? Give the formula of an amphoteric oxide formed by an element of group – 13.
Answer:
The oxide which contains both acidic as well s basic nature is called amphoteric oxide.

• The oxides reacts with both acids and bases and forms salts.
  Eq : Al₂O₃ is one of the amphoteric oxide formed by the Group — 13 element Aluminium.

Question 30.
Name the most electronegative element. Is it also having the highest electron gain enthalpy? Why or Why not?
Answer:
The most electronegative element is fluorine (F).

• It doesnot have high electron gain énthalpy.

Reasons : —

• Due to small size
• Due to high inter electronic repulsions.
• Chlorine has high electron gain enthalpý.

Question 31.
What is diagonal relation? Give one pair of elements, that have this relation.
Answer:
On moving diagonally across the periodic table the elements show certain similarities. An element of group in 2nd period has similar properties with second element of the next higher group in the 3rd period. This type of resemblance is called diagonal relationship. e.g. : (Li, Mg); (Be, Al); (B, Si)

Question 32.
How does the nature of oxides vary in the third period?
Answer:
In 3^rd period from left to right the oxide nature varies from high basic nature to high acidic nature.

• Basic nature gradually decreases and acidic nature gradually increases.
• 

Question 33.
Radii of iron atom and its ions follow Fe > Fe²⁺ > Fe³⁺ – explain.
Answer:
When the positive charge on the ion increases, the effective nuclear charge on the outer electrons increases.
Hence the ionic size decreases in the order Fe > F²⁺ > F³⁺.

Question 34.
IE₂ > IE₁ for a given element — why?
Answer:
IE₂ > IE₁ for a given element

Reason : —

• IE₁ means minimum amount of energy required to remove an electron from isolated neutral
  gaseous atom.
• IE₂ means minimum amount of energy required to remove an electron from uni positive ion.
• In case of unipositive ion nuclear attraction increases on outer most electrons than in isolated gaseous atom. So more amount of energy needed to remove an electron from unipositive ion.
  Hence IE₂ > → IE₁.

Question 35.
What is Ianthanide contraction? Give one of its consequences.
Answer:
Lanthanide contraction : Slow decrease in size of the atoms or ions among the lanthanides is known as lanthanide contraction.

Consequences due to lanthanide contraction:

1. Due to lanthanide contraction, the crystall structure and other properties of the elements become very closely similar.
2. Due to this, it becomes difficult to separate lanthanides from a mixture.

Question 36.
What is the atomic number of the element, having maximum number of unpaired 2p electrons ? To which group does it belong ?
Answer:

• The atomic number of the element having maximum no.of unpaired 2p electrons is 7′ (Z = 7)
• Element is nitrogen.
• Electronic configuration is 1s² 2s² 2p³ (3 unpaired 2p electrons)

Question 37.
Sodium is strongly metallic, while chlorine is strongly non-metallic – explain.
Answer:
Sodium is an alkali metal and it is present in group – I and it has the ability to lose the valency electron readily.

• It has high electropositive nature. So it has metallic nature.
• Chlorine is a halogen and it is placed in Group – 17 and it has the ability to gain the electron readily.
• It has high electronegative nature. So it has non metallic nature.

Question 38.
Why are zero group elements called noble gases or inert gases ?
Answer:

• Zero group elements has general outer electronic configuration ns² np⁶ (except for He).
• These contains stable octet configuration. So these are stable and chemically inert. Hence these are called inert gases.
• These elements neither lose nor gain electrons. Hence these are called ‘noble gases’.

Question 39.
Select in each pair, the one having lower ionization energy and explain the reason,
a. I and I^–
b. Br and K
c. Li and Li⁺
d. Ba and Sr
e. O and S
f. Be and B
g. N and O
Answer:
a) I^– has lower ionisation energy than I because of increase of size \(\mathrm{I}^{\ominus}\) ion than ‘I’.
b) K has lower ionisation energy than ‘Br’ because of low electronegative value of K (0.8) than ‘Br’ (2.8).
c) ‘Li” has lower ionisation energy than Li+ because of large size of ‘Li’ than Li⁺.
d) ‘S’ has lower ionisation energy than ‘O’ because of large size of ‘S’ than ‘O’.
e) ‘B’ has lower ionisation energy than ‘Be’ because ‘Be’ has completely filled electronic configuration (1s² 2s²).
f) ‘O’ has lower ionisation energy than ‘N’ because ‘N’ has half filled electronic configuration (1s² 2s² 2p³).

Question 40.
IE₁ of O < IE₁ of N but IE₂ of O > IE₂ of N – Explain.
Answer:

• ‘N’ has half filled electronic configuration (1s² 2s² 2p³)
  So IE₁ of O < IE₁ of ‘N’.
• O⁺ ion has half filled electronic configuration (1s² 2s² 2p³)
  So IE₂ of O > IE₁ of N.

Question 41.
Na⁺ has higher value of ionization energy than Ne, though both have same electronic configuration – Explain.
Answer:
Na⁺ has higher value of I.E. than Ne, though both have same electronic configuration.

Reason : –

• Both have electronic configuration 1s² 2s² 2p⁶
• In case of Na⁺ ion effective nuclear charge increases and size decreases than in ‘Ne’.

Question 42.
Which in each pair of elements has a more electronegative gain enthalpy ? Explain.
a. N or O
b. F or Cl
Answer:
a) Oxygen has high electronegative gain enthalpy than Nitrogen because ‘N’ has stable half filled electron configuration.
b) Chlorine (- 349 KJ /mole) has high electronegative gain enthalpy than Fluorine (- 328 KJ/mole) because ‘F’ has small size and more inter electronic repulsions.

Question 43.
Electron affinity of chlorine is more than that of fluorine – explain.
Answer:
Chlorine (- 349 KJ / mole) has high electronegative gain enthalpy than Fluorine (- 328 KJ/mole) because ‘F’ has small size and more inter electronic repulsions.

Question 44.
Which in each has higher electron affinity ?
a. F or Cl^–
b. O or O-
c. Na⁺ or F
d. F or F^–
Answer:
a) Fluorine has high electron affinity than Cl^– ion because of inert gas configuration of Cl^– ion.
b) Oxygen has high electron affinity than O^– because O^– has positive of 2nd electron affinity.
c) F has high electron affinity than Na⁺ because Na⁺ has inert gas configuration.
d) F has high electron affinity than F^– because F^– has inert gas configuration.

Question 45.
Arrange the following in order of increasing ionic radius :
a. Cl^–, P^-3, S^-2, F^–
b. Al⁺³, Mg⁺², Na⁺, O^-2, F^–
c. Na⁺, Mg⁺², K⁺
Answer:
a) The increasing order of ionic radius is F^– < Cl^– < S^-2 < P^-3
b) The increasing order of ionic radius is Al⁺³ < Mg⁺² < Na⁺ < F^– < O^-2
c) The increasing order of ionic radius is Mg⁺² < Na⁺ < K⁺

Question 46.
Mg⁺² is smaller than O^-2 in size, though both have same electronic configuration –
explain.
Answer:
Mg⁺² and O^-2 ions are iso electronic species.
In case of iso electronic species nuclear charge increases size of ion decreases. So Mg⁺² has small size than O^-2.

Question 47.
Among the elements B, Al, C and Si
a. Which has the highest first ionization enthalpy ?
b. Which has the most negative electron gain enthalpy ?
c. Which has the largest atomic radius ?
d. Which has the most metallic character ?
Answer:
a) Highest I.E. is possessed by the element carbon
b) Most negative gain enthalpy is for carbon (- 122 KJ/mole)
c) Large atomic radius is for Al (1.43 A)
d) Most metallic nature having element is ‘Al’.

Question 48.
Consider the elements N, P, O and S and arrange them in order of ;
a. Increasing first ionization enthalpy
b. Increasing negative electron gain enthalpy
c. Increasing non-metallic character
Answer:
a) Increasing first Ionisation energy order is S < P < O < N.
b) Increasing negative electron gain enthalpy order is N < P < O < S. .
c) Increasing non metallic nature order is P < N < S < O.

Question 49.
Arrange in given order :
a. Increasing EA :O, S and Se
b. Increasing IE₁ : Na, K and Rb
c. Increasing radius : I^–, I⁺ and I
d. Increasing electronegativity : F, Cl, Br, I
e. Increasing EA : F, Cl, Br, I
f. Increasing radius : Fe, Fe⁺², Fe⁺³
Answer:
a) Increasing order of electron affinity is O < Se < S.
b) Increasing order of IE₁ is Rb < K < Na.
c) Increasing order of radius is I⁺ < I < I^–
d) Increasing order of electronegativity is I < Br < F < Cl
e) Increasing order of electron affinity is I < Br < F < Cl
f) Increasing order of radius is Fe⁺³ < Fe⁺² < Fe

Question 50.
a. Name the element with highest ionization enthalpy
b. Name the family with highest value of ionization enthalpy.
c. Which element possesses highest electron affinity ?
d. Name unknown elements at the time of Mendeleef
e. Name any two typical elements.
Answer:
a) Highest I.E₁ possessing element is ‘Helium’.
b) The family that possess highest values of I.E is Noble gases (or) inert gases.
c) Highest electron affinity element is ‘Chlorine’.
d) Unknown elements at the time of mendeleef are Germanium (Eka silicon). Scandium (Eka Alu-minium), Gallium (Eka Boron).
e) 3^rd period elements are called typical elements
Eg : Al, Si, P, Na, Mg etc

Question 51.
a. Name any two bridge elements.
b. Name two pairs showing diagonal relationship.
c. Name two transition elements.
d. Name two rare earths.
e. Name two transuranic elements.
Answer:
a) Bridge elements 2nd period elements are called Bridge elements Eg : Be, B.
b) i) ‘Li’ diagonally relates with ‘Mg’,
ii) ‘Be diagonally relates with ‘Al’.
c) Scandium, Titanium, Vanadium, Chromium, etc., are examples of transition elements.
d) Lanthanides are called rare earths
Eg : Cerium (Ce), Prasodimium (Pr), Promethium (Pm).
e) Neptunium (Np), Californium (Cf), Fermium (Fm) are examples of transuranic elements.

Question 52.
On the basis of quantum numbers, justify that the 6^th period of the periodic table should have 32 elements.
Answer:
6^th period contains the subshells 6s, 4f, 5d, 6p
6s can accomodate two electrons (2 elements)
4f can accomodate 14 electrons (14 elements)
5d can accomodate 10 electrons (10 elements)
6p can accomodate 6 electrons (6 elements)

Total no.of electrons can accomodation 6^th period are 2 + 14 + 10 + 6 = 32
∴ 6^th period of periodic table contains 32 elements.

Question 53.
How did Moseley’s work on atomic numbers show that atomic number is a fundamental property better than atomic weight ?
Answer:
Mosley’s equation is

where v = frequency, Z = atomic number a, b = constants
A plot of \(\sqrt{v}\) against ‘Z’ gives a straight line.
However, no such relationship was obtained when the plot was drawn between frequency and the atomic mass. The atomic number of the elements, according to Mosley, stands for serial numbers of the elements in the periodic table. As the atomic number of the elements increase, the wavelengths of characteristic X – rays decrease. Mosley concluded that there is a fundamental quantity in an atom which increases in regular steps with increasing atomic number. The correlation between X – ray spectra and atomic number indicated that an element is characterized by its atomic number and not by atomic mass.

Question 54.
State modern periodic law. How many groups and periods are present in the long form of the periodic table ?
Answer:
Modern periodic law : – The physical and chemical properties of elements are periodic functions of their atomic numbers.

• In modern periodic table 7 periods and 18 groups are present.

Question 55.
Why f-block elements are placed below the main table ?
Answer:
In Lanthanides 4f – orbitals and in Actinides 5f – orbitals are filled. Since these elements have the same electronic configuration in the ultimate and penultimate shells they have similar properties. Hence they were placed at the bottom of the periodic table though they belongs to the sixth and seventh periods of IIIB groups.

Question 56.
Mention the number of elements present in each of the periods in the long form periodic table.
Answer:

Question 57.
Give the outer orbit general electronic configuration of
a. Noble gases
b. Representative elements
c. Transition elements
d. Inner transition elements
Answer:
Type of elements – General electronic configurations
a) Noble gases – ns² np⁶ (except ‘He’ which has 1s²)
b) Representative elements – ns^1-2 np^0-5
c) Transition elements – (n – 1)d^1-10ns^1-2
d) Inner transition elements – (n – 2) f^1-14(n – 1)d^{0, 1} ns²

Question 58.
Give any four characteristic properties of transition elements.
Answer:
Characteristic properties of elements:
a) They exhibit more than one oxidation state.
b) Most of the elements and their ions exhibit colour.
c) These elements and their compounds are good catalysis for various chemical processes.
d) They and their ions exhibit paramagnetic properties.
e) They form useful alloys.

Question 59.
What are rare earths and transuranic elements?
Answer:

1. Lanthanides are rare earths. In these elements the differentiating electron enters into 4f – orbital.
2. The elements present after Uranium are called Transurariic elements. All of these are radioactive and synthetic elements.

Question 60.
What is isoelectronic series? Name a series that will be isoelectronic with each of the following atoms or ions.
a. F^–
b. Ar
c. He
d. Rb⁺
Answer:
The species containing same no.of electrons are called Iso electronic species and this series is called Isoelectronic series.
a) F^– relating series
N^-3, O^-2, F^–, Ne, Na⁺, Mg⁺², Al⁺³
b) Ar relating series P^-3, S^-2, Cl^–, Ar, K⁺, Ca⁺²
c) ‘He’ relating series H^–, He, Li⁺, Be⁺²
d) Rb⁺ relating series
As^-3, Se^-2, Br^–, Kr, Rb⁺, Sr⁺²

Question 61.
Explain why cation is smaller and anion is larger in radii than their parent atoms.
Answer:

• Cation means a positively charged species which is formed by an atom (or) element when an electron is lost.
  M → M⁺ + \(\mathrm{e}^\Theta\)
• Cation has high effective nuclear charge and decrease in size observed.
  Hence cation has smaller radii.
• Anion means a negatively charged species which is formed by an atom (or) element when an electron is gained.
  M + e^– → \(\mathrm{M}^{\Theta}\).
• Anion has very low effective nuclear charge and increase in size observed.
  Hence anion has larger radii.

Question 62.
Arrange the second period elements in the increasing order of their first ionization enthalpies. Explain why Be has higher IE₁ than B.
Answer:
The increasing order of the I.E.s of 2^nd period elements are as follows.

• Due to presence of incompletely filled p – orbitals in Boron, its IE value is less.
• Due to presence of completely filled s² configuration in ‘Be’, it has higher IE value.

Question 63.
IE₁ of Na is less than that of Mg but IE₂ of Na is higher than that of Mg – explain.
Answer:
IE₁ of Na is less than that of ‘Mg’
Reason :-
Na — has electronic configuration [Ne] 3¹
Mg — has electronic configuration [Ne] 3s²
Mg has completely filled configuration so Mg has more IE₁ than Na.
IE₂ of Na is higher than that of Mg

• 
• Na⁺ has stable inert gas configuration so IE₂ of Na is very high
  
• By the lose of one electron from Mg⁺ ion forms Mg⁺² ion which is more stable so low amount of energy is required.
  ∴ IE₂ of Na is higher than Mg.

Question 64.
What are the various factors due to which the IE of the main group elements tends to decrease down a group ?
Answer:
The factors influencing on IE are

1. Atomic radius
2. Nuclear charge
3. Screening effect
4. Half filled, completely filled electronic configurations
5. peretrating power
   • In main group elements in a group IE decrease from top to bottom.

Reason: –
In groups from top to bottom size of elements increases hence IE values decreases.

Question 65.
The first ionization enthalpy values (in KJ mol^-1) of group 13 elements are :

How do you explain this deviation from the general trend ?
Answer:
The given IE, values (in KJ / mole) of group 13 are as follow

• In genaral in a group IE values decrease down a group but from the above values we observe that there is no smooth decrease down the group.
• The decrease from B to Al is due to increase of size.
• The observed discontinuity in the IE values of Al and Ga, and between In and Tl are due to inability of d- and f – electrons, which have low screening effect to compensate the increase in nuclear charge.

Question 66.
Would you expect the second electron gain enthalpy of oxygen as positive, more negative or less negative than the first ? Justify.
Answer:
2^nd gain enthalpy means energy released when an electron is added to uni negative ion.
O_(g) + e^– → \(\mathrm{O}_{(g)}^{-}\) + 141 KJ/mole
O_(g) + e^– → \(O_{(g)}^{-(2)}\) – 780 KJ/mole

• 2^nd gain enthalpy of oxygen is positive because O^– ion doesnot accept an electron readily and the entering electron have to over come the repulsive force.

Question 67.
What is the basic difference between the electron gain enthalpy and electropositivity?
Answer:

• Electron gain enthalpy means energy released when an electron is added to isolated neutral gaseous atom.
• The tendency to lose the electrons by an element is called electropositivity.
• Electron gain enthalpy is the measure of electronegativity.
• Electron gain enthalpy and electropositivity are inversely related.

Question 68.
Would you expect IE₁ for two isotopes of the same element to be the same or different ? Justify.
Answer:

• Isotopes means existance of same elements with different mass no.s
• The isotope with higher mass no. have low I.E value than the normal isotope.
• This is due to the less nuclear attraction on valency electrons in case of heavier nuclide.
• But overall, the IE values of the isotopes are nearly same and the difference in IE values is negligible.

Question 69.
Increasing order of reactivity among group-1 elements is Li < Na < K < Rb < Cs, where as among group-17 elements it is F > Cl > Br > I- explain.
Answer:
a) Increasing order of reactivity among group -1 elements is Li < Na < K < Rb < Cs

Explanation: –

• Group -1 elements are Alkali metals.
• Group -1 elements have the tendency to lose the electrons
• Group -1 elements forms ionic bonds readily by losing electrons
• Group – 1 elements good reducing agents.
• Electro positive character increases from top to bottom this is due to increase of size.

b) Increasing order of reactivity among group -17 elements is F > Cl > Br > I

Explanation: –

• These are halogens (Group – 17)
• These have high electronegativity due to small size.
• These have the tendency to gain electrons.
• In a group from top to bottom electronegativity decrease, due to increase of size.
• These are oxidising agents and forms ionic bonds by gaining electrons.

Question 70.
Assign the position of the element having outer electronic configuration.
a. ns²np⁴ for n = 3
b. (n – 1)d²ns² for n = 4
Answer:
a) ns²np⁴ for n = 3 .

• 3s²3p⁴ → element is sulphur
• Sulphur belongs to VIA group (Group – 16) and 3^rd period in periodic table,

b) (n – 1 )d² ns² for n = 4
3d² 4s² → element is Titanium

• Titanium belongs to IVB group (Group – 4) and 4^th period in the periodic table.

Question 71.
Predict the formulae of the stable binary compounds that would be formed by the combination of the following pairs of elements.
a. Li and O
b. Mg and N
c. Al and I
d. Si and O
e. p and Cl,
f. Element with atomic number 30 and Cl
Answer:
a) Stable binary compound formed between Li and O is Li₂O (Lithiumoxide).
b) Stable binary compound formed between Mg and N is Mg₃N₂ (Magnesium nitride).
c) Stable binary compound formed between Al and I is AlI₃ (Aluminium Iodide).
d) Stable binary compound formed between Si and I is SiO₂ (Silicondioxide).
e) Stable binary compound formed between P and Cl is PCl₃ and PCl₅ (Phosphorous trichloride and phosphorous pentain chloride).
f) Stable binary compound formed between element with At. NO – 30 and Cl is ZnCl₂ (Zinc chloride) [At. No – 30 – (Zn)].

Question 72.
Write a note on the variation of metallic nature in a group or in a period.
Answer:
Metals shows electropositive nature (i.e.,) loss of electrons and form positive ions.
Non – metals shows electronegative nature (i.e.,) gain of electrons and form negative ions.
Periodicity:
a) Down the group : Going down a group of the periodic table, the tendency to form positive ions, increases. That means there is an increase in metallic nature as the size of atom increases down the group.
b) Along a period : From left to right in a period, size of atom decreases. So there is decrease in metallic nature.

Question 73.
How does the covalent radius increase in group- 7 ?
Answer:

• Covalent radius increases in a group from top to bottom
• The increase of covalent radius in VIIA group elements as follows.
  

Question 74.
Which element of 3rd period has the highest IE₁ ? Explain the variation of IE₁ in this period.
Answer:

• In periods IE values increase from left to right
• Among 3^rd period elements Argon (Ar) [Z = 18] possess highest IE,sub>1 value.
• IE₁ values of 3^rd period elements given below.
  

Exceptions:

• IE, of ‘Mg’ is higher than ‘Al’ because ‘Mg’ has completely filled ‘s’ subshells.
• IE, of ‘p’ is higher than ‘s’ because ‘p’ has half filled ‘p’ subshells.

Question 75.
What is valency of an element ? How does it vary with respect to hydrogen in the third period?
Answer:
Valency : The combining capacity of an element with another element is called valency.

The number of hydrogen atoms (or) chlorine atoms (or) double the number of oxygen atoms, with which one atom of the element combine is also called valency.
∴ Valency = no. of hydrogens = no. of chlorine atoms
= 2 × no. of oxygen atoms present in the molecule.
e.g.:

Periodicity of valency:
1) Each period starts with valency ‘1’ and ends in ‘0’.

2) In a group valency is either equal to the group number (i.e., upto 4^th group) or is equal to (8 – group number) (i.e., from 5^th group onwards).
Significance :
Valency of an element is useful in writing the formulae of compounds.

Question 76.
What is diagonal relationship ? Give a pair of elements having diagonal relationship. Why do they show this relation ?
Answer:
On moving diagonally across the periodic table the elements show certain similarities. An element of group in 2nd period has similar properties with second element of the next higher group in the 3rd period. This type of resemblance is called diagonal relationship, e.g. : (Li, Mg); (Be, Al); (B, Si)

Diagonal relationship is due to similar sizes of atoms (or of ions) and similar electronegativities of the representative elements.
Diagonally similar elements posses the same polarizing power.
Polarizing power = \(\frac{\text { (ionic charge) }}{\text { (ionic radius) }^2}\)

Question 77.
What is lanthanide contraction ? What are its consequences ?
Answer:
Lanthanide contraction : Slow decrease in size of the atoms or ions among the lanthanides is known as lanthanide contraction.

Consequences due to lanthanide contraction:

1. Due to lanthanide contraction, the crystall structure and other properties of the elements become very closely similar.
2. Due to this, it becomes difficult to separate lanthanides from a mixture.

Question 78.
The first IP of lithium is 5.41 eV and electron affinity of Cl is – 3.61 eV. Calculate ΔH in kJ mol^-1 for the reaction : Li_(g) + Cl_(g) → \(\mathrm{Li}_{(\mathrm{g})}{ }^{+}\) + \(\mathrm{Cl}{ }_{(g)}^{-}\)
Answer:
Given reaction is

ΔH = ΔH₁ + ΔH₂
= 5.41 – 3.61
= 1.8 ev
= 1.8 × 9.65 × 10⁴ J/mole
= 17.37 × 10⁴ J/Mole
= 173.7 KJ/mole

Question 79.
How many Cl atoms can you ionize in the process Cl → Cl⁺ + e by the energy liberated for the process Cl + e → Cl^– for one Avogadro number of atoms. Given IP = 13.0 eV and EA = 3.60 eV. Avogadro number = 6 × 10²³
Answer:
Given
Cl_(g) + e^– → \(\mathrm{Cl}_{(\mathrm{g})}^{-}\) ΔH = – 3.6ev
1 – atom → Electron affinity = 3.6 ev
6.23 × 10²³ atoms → ?
6.23 × 10²³ × 3.6 = 21.6828 × 10²³
∴ For one avagadro no.of Cl atoms electron affinity = 21.6828 × 10²³ eV
Given 13 eV can ionise 1 atom of Cl
21.6828 × 10²³ eV ionise -?
\(\frac{21.6828 \times 10^{23}}{13}\) = 1.667 × 10²³ eV

Question 80.
The electron affinity of chlorine is 3.7 eV. How much energy in kcal is released when 2g of chlorine atoms is completely converted to Cl^– ions in the gaseous state ? (1 e V = 23.06 kcal)
Answer:
Given electron affinity of Cl = 3.7 ev
Cl + e^– → Cl^– ΔH = – 3.7ev
35.5 gms of CZ contains 6.023 × 10²³ atoms
2 gms of Cl contains ?
= \(\frac{2 \times 6.023 \times 10^{23}}{35.5}\) atoms
6.23 × 10²³ atoms can liberate 3.7 eV
\(\frac{2 \times 6.023 \times 10^{23}}{35.5}\) atom can liberate?
= \(\frac{2 \times 6.023 \times 10^{23} \times 3.7}{35.5 \times 6.023 \times 10^{23}}\) = \(\frac{2 \times 3.7}{35.5}\) = \(\frac{7.4}{35.5}\)
= 0.2084 eV
= 0.2084 × 23.06 kcal/mole
= 4.81 k.cal/mole

Long Answer Questions

Question 1.
Discuss the classification of elements by Mendaleev’
Answer:
The periodic classification of elements based on “atomic weights” was done by “Lothar Meyer” and “Mendeleev” independently.
Mendeleev’s periodic law : “The physical and chemical properties of elements and their compounds are a periodic function of their atomic weights”.

Mendeleev arranged the 65 elements in a periodic table. He did not blindly follow the atomic weight but gave more importance to their chemical properties in arranging them in the table.
Explanation of the periodic law : When the elements are arranged in the increasing order of their atomic weights, elements with similar properties appear again and again, at regular intervals. This is called, periodicity of properties.

Mendeleev’s table : Mendeleev introduced a periodic table containing the known 65 elements. In this table, while arranging the elements, he gave importance only to their atomic weights, but also to their physical and chemical properties. This table was defective in some responses. Then he introduced another table, after rectifying the defects of that table. lt is called, “short form of periodic table”. He named the horizontal rows as ‘periods’ and the vertical columns, as ‘groups’. It has in all ‘9’ groups, I to VIII and a ‘O’ group. The first ‘7’ groups were divided into A and B sub groups. There are ‘7’ periods in the table. The VIII group contains three triods, namely, (Fe, Co, Ni), (Ru, Rh, Pd) and (Os, Ir, Pt).

Mendeleev’s observations :

1. When the elements are arranged according to their atomic weights, they exhibit periodicity of properties.
2. Elements with similar chemical properties have nearly equal atomic weights: Iron (55.85),. Cobalt (58.94) and Nickel (58.69).
3. The group number corresponds to the valency of element in that group.
4. Most widely distributed elements like H,C, O, N, Si, S etc., have relatively low atomic weights.
5. The atomic weight of an element may be corrected if the atomic weights of the adjacent elements are known. The properties of an element are the average properties of the neighbouring elements.
6. The atomic weights of beryllium. Indium, Uranium etc. were corrected, based on this observation.

Merits of Mendeleev’s table:

1. Actually it formed the basis for the development of other modern periodic tables.
2. Mendeleeffs left some vacant spaces in his periodic table, for the unknown elements. But the predicted the properties of those elements. Later on, when these elements were discovered, they exactly fitted into those vacant places having properties, predicted by Mendeleev.
   Ex : E_ka – boron (Scandium), E_Ka – Silicon (germanium)
   E_Ka – aluminium (gallium) etc.
3. ‘O’ group elements were not known at the time of Mendeleeff. Later when they were discovered, they found a proper place in that table under ‘0’ group of elements. Similarly, the radioactive elements.
4. In case of these pairs of elements Tellurium – Iodine Argon – Potassium and Cobalt – Nickel, there is a reversal of the trend. The first element has higher atomic weight than the second one. These are called anomalous pairs.
   However, based on their atomic numbers, and chemical properties, this arrangement proves quite justified.

Draw-backs of Mendeleev’s periodic table :

1. Dissimilar elements were placed in the same group.
   Ex : The coinage mentals Cu, Ag and Au are placed along with the alkali metals K, Rb, Cs etc. in the I group. The only common property among them is that they are all univalent (Valency = 1).
2. The 14 rare earths having different atomic weights are kept in the same place.
3. Hydrogen could not be given a proper place, as it resembles alkali metals and halogens in its properties.

Question 2.
From a study of properties of neighbouring elements, the properties of an unknown element can be predicted – Justify with an example.
Answer:
From a study of adjacent elements and their compounds, it is possible to predict the characteristics of certain elements. These predictions were found to be very accurate. These predicted properties helped the future scientists in the discovery of unknown elements, e.g. : E_Ka Aluminium (E_Ka Al) (now known as Gallium); E_Ka Silicon (E_Ka Si) (now known as Germanium); E_Ka Boron (E_Ka B) (now known as Scandium).

Illustration : Following table shows a comparison of the properties predicted by’Mendeleeff’ for the elements and those found experimentally after their discovery.

Question 3.
Discuss the construction of long form periodic table.
Answer:
The elements are arranged in the long form of the periodic table in the increasing order of atomic numbers. ‘Neils Bohr’ constructed the long form of the periodic table based on electronic configuration of elements.
The important features of the long form of the periodic table are :
It consists seven horizontal rows called periods and 18 vertical columns which are classified into 16 groups only.

Periods : Every period starts with an alkali metal and ends with an inert gas. The first period consists of two elements only (H, He) and is called very short period. Second period consists 8 elements (Li to Ne) and is called first short period. The third period consists (Na to Ar) 8 elements and is called second short period.

Fourth period contains 18 elements (K to Kr) and is called first long period. Fifth period is the second long period with 18 elements (Rb to Xe).
Sixth period is the longest period with 32 elements which starts with Cs and ends with Rn. This period includes 14 lanthanides.
Seventh period is an incomplete period with 20 radioactive elements.
Groups : There are 16 groups in the long form of the periodic table (in transition elements three vertical columns are fused and designated as VIII group). These groups are IA, IIA, NIB, IVB, VB, VIB, VIIB, VIII, IB, MB, IMA, IVA, VA, VIA, VIIA and zero group.

The elements of IA, IIA, IIIA, IVA, VA, VIA, VIIA groups are called representative elements or normal elements. Elements of IB, MB, NIB, IVB, VB, VIB, VIIB and VIII groups have their ultimate and penultimate shell incomplete. These are called transition elements. MB elements have (n – 1) d¹⁰ ns² outermost electronic configuration.

Zero group elements have stable electronic configuration. These elements are called inert gases, noble gases. These elements have been grouped at the extreme right of the periodic table.
In this long periods have been expanded and short periods are broken to accommodate the transitional elements in the middle of the long period.

Lanthanides and actinides have been grouped separately and placed at the bottom of the periodic table.

Question 4.
Discuss the relation between the number of electrons filled into the sub energy levels of an orbit and the maximum number of elements present in a period.
Answer:
Elements have been accommodated in these periods according to the following scheme.
1st period : The first main energy shell (K – shell) is completed. As the maximum capacity of K- shell is of 2 electrons, it consists of only two elements, hydrogen (1s¹)and helium (1s²).
2nd period : The second main energy shell is completed (i.e.,) 2s and 2p are completed. It includes eight elements from Li (2s¹) to Ne (2s²2p⁶).
3rd period : The 3s and 3p energy shells are completed. It includes also eight elements from Na(3s¹) to Ar (3s² 3p⁶).
4th period : The 4s, 3d and 3p energy shells are completed. It includes 18 elements from K(4s¹) to Kr (3d¹⁰ 4s²4p⁶). It includes two s – block elements, ten d – block elements and six p – block – elements.
5th period : The 5s, 4d and 5p energy shells are completed. It includes 18 elements from Rb (5s) to Xe (4d¹⁰ 5s¹ 5p⁶). 1st also includes two s – block elements, ten d – block elements and six p – block elements.
6th period : The 6s, 4f, 5d and 6p energy shells are completed, (i.e.,) it includes 32 elements from Cs (6s¹) to Rn (4f¹⁴ 5d¹⁰ 6s²6p⁶). It consists of two s – block elements, ten d – block elements, six p – block elements and fourteen f – block elements.
As this period can accommodate only 18 elements in the table, 14 members of 4f – series are separately accommodated in a horizontal row below the periodic table.
7th period : This period at present consists 26 elements. The 7s, 5f and 6d are completed (i.e.,) twenty six elements. Seven of the elements from atomic numbers 106 to 112 have recently been reported.

Question 5.
Write an essay on s, p, d and f block elements .
Answer:
According to the electronic configuration of elements, the elements have been classified into four blocks. The basis for this classification is the entry of the differentiating electron into the subshell. They are classified into s, p, d and f blocks.
‘s’ block elements : If the differentiating electron enters into ‘s’ orbital, the elements belongs to ‘s’ block.

In every group there are two ‘s’ block elements. As an ‘s’ orbital can have a maximum of two electrons, ‘s’ block has two groups IA and IIA.
‘p‘ block elements : If the differentiating electron enters into ‘p’ orbital, the elements belongs to ‘p’ block.

‘p’ block contains six elements in each period. They are IIIA to VIIA and zero group elements. The electronic configuration of ‘p’ block elements varies from ns²np¹ to ns²np⁶.

‘d‘ block elements : It the differentiating electron enters into (n – 1) d – orbitals the elements belongs to’d’ block. These elements are in between ‘s’ and ‘p’ blocks. These elements are also known as transition elements. In these elements n and (n – 1) shells are incompletely filled. The general electronic configuration of’d’ block elements is (n – 1) d^1-10 ns^1-2. This block consists of IIIB to VIIB, VIII, IB and IIB groups.

‘f’ block elements : If the differentiating electron enters into ‘f’ orbitals of antipenultimate shell (n – 2) of atoms of the elements belongs to ‘f block. They are in sixth and seventh periods in the form of two series with 14 elements each. They are known as lanthanides and actinides and are arranged at the bottom of the periodic table. The general electronic configuration is (n – 2)f^1-14 (n – 1)d^{0 – 1} ns².

In these shells the last three shells (ultimate, penultimate and anti penultimate) are incompletely filled. Lanthanides belongs to 4f series. It contains Ce to Lu. Actinides belong to 5f series. It contains Th to Lr.

Advantages of this kind of classification :

As a result of this classification of elements were placed in correct positions in the periodic table. It shows a gradual gradation in physical and chemical properties of elements. The metallic nature gradually decreases and non – metallic nature gradually increases from’s1′ block to ‘p1 block. This classification gave a special place for radioactive elements.

Question 6.
Relate the electronic configuration of elements and their properties in the Classification of elements.
Answer:
The chemical properties of all elements depends upon the electronic configuration. Upon the basis of complete and incomplete electron shells and chemical properties, the elements are classified into four types. (Type -1, Type – II, Type – III and Type – IV).

Type – I(inert gas elements): All the elements with an electronic configuration ns² np⁶ including He belongs to this type, nth shell of those elements are completely filled.
The elements show chemical inertness due to completely filled shells and hence they have extra stability. Because of their stability, they are chemically inactive.
e.g. : He, Ne, Ar, Kr, Xe and Rn.

Type – II (Representative elements) : Except inert gases, the remaining elements of s and p – blocks are called representative elements. All the elements with an electronic configuration ns¹ to ns² np⁵ excluding. He comes under this type. These are atoms in which all except the outermost shells (n^th) are complete. Elements of this type enter into chemical combination by loosing, gaining or sharing electrons to get stable inert gas configuration. Many of the metals, all non – metals and metalloids come under this type. Chemically these elements are reactive.

Type – III (Transition elements): All the elements with an electronic configuration (n – 1) d^{1 – 9} ns^1or2 belongs to this type. Atoms in which the two outermost shells are incomplete. These elements show variable oxidation states, form complex ions and coloured ions. The electronic configuration of d – block elements is (n – 1) d^{1 – 10} ns^{1 – 2}. Small size, high nuclear charge and unpaired’d’ orbitals impart characteristic properties to be transition elements.

Type- IV (Inner transition elements): All the elements with an electronic configuration (n – 2) f^{1 – 14} (n -1) d^{0, 1} ns² belongs to this type. Atoms in which three outermost shells are incomplete. Lanthanides and Actinides belong to this type.

Question 7.
What is a periodic property ? How the following properties vary in a group and in a period ? Explain
a) Atomic radius
b) Electron gain enthalpy.
Answer:
Recurrence of similar properties of elements at definite regular intervals with increasing atomic number i.e., according to their electronic configurations is known as periodicity. Any property which is periodic in nature is called periodic property.
a) Atomic radius : The atomic radius decreases from left to right in a period. With an increase in the atomic number in a period the nuclear charge increases. As a result the effective nuclear charge over the outermost electrons increases, due to this the orbitals are pulled closer to the nucleus causing in a decrease in the atomic radius.

The atomic radius increases from top to the bottom in a group because – with an increase in the atomic number the electrons are added to new shells resulting an increase in the number of inner shells. Hence atomic radius increases from top to bottom in a group.

b) Electron gain enthalpy : Electron gain enthalpy increases in a period from left to right because the size of the atom decreases and the nature of the element changes from metallic to non – metallic nature when we move from left to right in a period.

Electron gain enthalpy decreases from top to bottom in a group because there is an increase in the atomic size. But the second element has greater electron gain enthalpy than the first element.
e.g. : Chlorine has more electron affinity value (- 348 kJ mol^-1) than Fluorine (-333 kJ mol^-1). It is because fluorine atom is smaller in size than chlorine atom. There is repulsion between the incoming electron and electrons already present in fluorine atom i.e., due to stronger inter electronic repulsions.

Question 8.
What is a periodic property ? How the following properties vary in a group and in a period ?
Explain
a. IE
b. EN
Answer:
a) IE : In groups and periods of the periodic table the ionization enthalpy values are periodically change depend upon the electronic configuration and size of elements.
In a group of elements ionization energy decreases from top to bottom because atomic radius increases.
In general, in a period the atomic size decreases. Because of this, the ionization energy increases across a period.

b) Electro negativity : Electronegativity increases from left to right in a period since the atomic: radii decreases and nuclear attraction increases.
In a group electronegativity decreases from top to bottom due to an increase in atomic radii and a decrease in nuclear attraction.

Question 9.
Write a note on
a. Atomic radius
b. Metallic radius
c. Covalent radius
Answer:
a) Crystal Radius (Atomic radius or Metallic radius): The term is used for metal atoms. A metallic crystal contains metal atoms in close packing. These metal atoms are considered spherical. They are supposed to touch each other in the crystal.
The crystal radius is half the internuclear distance between two adjacent atoms.
e.g.: The internuclear distance between two adjacent sodium atoms in a crystal of sodium metal is 3.72 A. So crystal radius of sodium is \(\frac{3.72}{2}\) = 1.86 A
For potassium it is 2.31 A.

b) Covalent radius : It is used generally for non – metals. Covalent radius is half the equilibrium distance between the nuclei of two atoms with a covalent bond.
Covalent radii of two atoms can be added to give internuclear distance between them.
e.g. : Covalent radius of H is 0.37A and for chlorine it is 0.99A. Hence internulear distance between H and Cl in HCl is 1.36A.

c) Vander Waals radius (Collision radius): The Vander Waals radius is half the equilibrium distance between the nuclei of two atoms bound by Vander Waals forces.
It is used for molecular substances in solid state and for inert gases.
e.g. : The Vander Waals radius for hydrogen is 1.2 A and that of chlorine is 1.80 A.
The Vander Waals radius of an atom is 40% larger than its covalent radius.

Question 10.
Define IE₁ and IE₂. Why is IE₂ > IE₁ for a given atom? Discuss the factors that effect IE of an element. (A.P. Mar. ’16)(T.S. Mar. ’13)
Answer:
1) Ionization energy is the amount of energy required to remove the most loosely held electron from isolated a neutral gaseous atom to convert it into gaseous ion. It is also known as first ionization energy because it is the energy required to remove the first electron from the atom.
It is denoted as I₁, and is expressed in electron volts per atom, kilo calories (or) kilo joules per mole.
M_(g) + I₁ → \(M_{(g)^{+}}\) + e^–
I₁ is first ionization potential.

2) The energy required to remove another electron from the unipositive ion is called the second ionization energy. It is denoted as I₂.
\(M_{(g)}^{+}\) + I₂ → \(\mathrm{M}_{(\mathrm{g})}{ }^{2+}\) + e^–

3) The second ionization potential is greater than the first ionization potential. On removing an electron from an atom, the unipositive ion formed will have more effective nuclear charge than the number of electrons. As a result the effective nuclear charge increases over the outermost electrons. Hence more energy is required to remove the second electron. This shows that the second ionization potential is greater than the first ionization potential.
For sodium, I₁ is 5.1 eV and I₂ is 47.3 eV.
I₁ < I₂ < I₃ ….. I_n

Factors affecting ionization potential:

1. Atomic radius : As the size of the atom increases the distance between the nucleus and the outermost electrons increases. So the effective nuclear charge on the outermost electrons decreases. In such a case the energy required to remove the electrons also decreases. This shows that with an increase in atomic radius the ionization energy decreases.

2. Nuclear charge : As the positive charge of the nucleus increases its attraction increases over the electrons. So it becomes more difficult to remove the electrons. This shows that the ionization energy increases as the nuclear charge increases.

3. Screening effect or shielding effect: In multielectron atoms, valence electrons are attracted by the nucleus as well as repelled by electrons of inner shells. The electrons present in the inner shells screen the electrons present in the outermost orbit from the nucleus. As the number of electrons in the inner orbits increases, the screening effect increases. This reduces the effective nuclear charge over the outermost electrons. It is called screening or shielding effect. With the increase of screening effect the ionization potential decreases. Screening efficiency of the orbitals falls off in the order s > p > d > f.

(Magnitude of screening effect) ∝ \(\frac{1}{(\text { Ionization enthalpy) }}\)
TREND IN A GROUP : The ionisation potential decreases in a group, gradually from top to bottom as the size of the elements increases down a group.
TREND IN A PERIOD : In a period from left to right I.P. value increases as the size of the elements decreases along the period.

Question 11.
How do the following properties change in group-1 and in the third period ? Explain with example.
a. Atomic radius
b. IE
c. EA
d. Nature of oxides
Answer:
a) Atomic radius

• In Group – 1 atomic radius from Li to Cs increases
  
  → In 3rd period from Na to Cl atomic radius decreases.
  

b) Ionization energy : In a group ionization energy values decreases with an increase in the size of the atom. In IA group Li is the element with highest ionization potential and Cs is the element with lowest ionization potential.

In third period the ionization potential increases from Na to Mg and then decreases at Al and increases upto P and decreases in case of S and then increases upto argon, i.e., Al has a lower value than Mg and S has a lower value than P, due to stable electronic configurations of Mg and P Among these elements argon has highest ionization potential.

c) Electron affinity :

• In 3^rd period E.A > from ‘Si’ to ‘P’ decreases and P to Cl increases.
  ‘Mg’ and ‘Ar’ has positive values.
• In 1^st group from Li to Cs electron affinity decreases due to increase of size.

d) Nature of oxides of elements : All IA group elements are alkali metals. Their oxides are basic in nature. They dissolve in water to give basic solutions which changes fed litmus blue.
e.g. : Na₂O, CaO , MgO etc.
Na₂O + H₂O → 2 NaOH
CaO + H₂O → Ca(OH)₂
MgO + H₂O → Mg (OH)₂
The basic nature of these oxides increases from top to bottom in the group. In a period basic properties decreases and acidic properties increases.

Question 12.
Define electron gain enthalpy. How it varies in a group and in a period? Why is the electron gain enthalpy of O or F is less negative than that of the succeeding element in the group ?
Answer:

1. Electron affinity is the amount of energy released when an electron is added to a neutral gaseous atom in its ground state. It is known as first Electron affinity EA,sub>1. It has -ve value.
   X_(g) + e → \(\mathrm{X}_{(\mathrm{g})}{ }^{-}\) + EA₁
2. Energy can be absorbed when an other electron is added to uni negative ion. It is because to overcome the repulsion between negative ion and electron. Hence second electron affinity, EA₂ has +ve value.
   \(X_{(g)}{ }^{-}\) + e → \(X_{(g)}^{2-}\) + EA₂
3. It can be indirectly determined from Born —Haber cycle. Its units are kJ/mole.
4. Trend in a group : Generally EA decreases down the group. In a group the second element has higher EA value than the first member.
   e.g. : EA of Cl is more than that of F
   Fluorine possess lower EA value than chlorine. It is due to inner electron repulsions in Fluorine.
5. Trend in a period : Generally EA value increases across a period, but some irregularities can be observed.

a) I A group elements possess low EA values than the corresponding III A elements, e.g. : Be has EA value zero, it is due to completely filled 2s² orbital.
b) VA elements have low EA values than that of VI A elements.
e.g. : Due to presence of half filled p – orbitals [2s² 2p³], nitrogen has lower EA value than that of oxygen.
c) For inert gases EA value is zero.
d) The element with the highest EA value is chlorine.

• The electron gain enthalpy of O and F is less negative than the succeeding element in the group because. These have small size and inter electronic repulsions are high in these elements
  O → 141 KJ / mole and S → 200 kJ/Mole
  F → 328 KJ / mole and Cl → 349 kJ/Mole

Question 13.
a. What is electronegativity ?
b. How does it vary in a group and in a period ?
Answer:
a) Electronegativity : ‘The tendency of the atom of an element to attract the shared electron pair(s) more towards itself in a hetronuclear diatomic molecule or in a polar covalent bond.” Measuring electronegativity Pauling scale : Pauling scale is based on the values of bond energy. The bond energy of a compound A – B is the average of bond energies of A – A and B – B molecules.
E_{A – B} = \(\frac{1}{2}\left(E_{A-A}+E_{B-B}\right)\)
But the experimental value of E_{A – B} is found to exceed the theoretical value. The difference is ∆.
∴ ∆ = E’_{A – B} – E_{A – B}
∆ indicates the polarity of the covalent bond. It is measured in k. cal. mol^-1.
Pauling gave the relation X_A – X_B = 0.208 × \(\sqrt{\Delta}\)
In S.l. units X_A – X_B = 0.1017 \(\sqrt{\Delta}\) where ∆ is measured in kJ/mole
X_A and X_B are the electronegativities of A and B. Pauling arbitrarily fixed 2.1 as the electronegativity value of Flydrogen and calculated the electronegativities of other elements. On Pauling scale Fluorine has the highest EN value 4.0.

From the values of the electronegativity of elements, the nature of the chemical bond formed can be understood. If two bonded atoms differ by 1.70 or more ion their EN values, the bond between them would be either 50% or more than 50% ionic in nature. Similarly if the difference in the EN values of the atoms is less than 1.70, the bond formed is more than 50% covalent in nature.

b) Variation in a group and period : Electronegativity increases from left to right in a period since the atomic: radii decreases and nuclear attraction increases. In a group electronegativity decreases from top to bottom due to an increase in atomic radii and a decrease in nuclear attraction.

Question 14.
Explain the following
a. Valency
b. Diagonal relation
c. Variation of nature of oxides in the Group -1
Answer:
a) Valency : The combining capacity of an element with another element is called valency. The number of hydrogen atoms (or) chlorine atoms (or) double the number of oxygen atoms, with which one atom of the element combine is also called valency.
∴ Valency = no. of hydrogens
= no. of chlorine atoms
= 2 × no. of oxygen atoms present in the molecule

Periodicity of valency :
1) Each period starts with valency’11 and ends in ‘O’.
e.g.:

2) In a group valency is either equal to the group number (i.e., upto 4^th group) or is equal to (8 – group number) (i.e., from 5^th group onwards).
Significance :
Valency of an element is useful in writing the formulae of compounds.

b) Diagonal relation : On moving diagonally across the periodic table the elements show certain similarities. An element of group in 2nd period has similar properties with second element of the next higher group in the 3rd period. This type of resemblance is called diagonal relationship,
e.g. : (Li, Mg); (Be, Al); (B, Si)

c) Nature of oxides of elements : All IA group elements are alkali metals. Their oxides are basic in nature. They dissolve in water to give basic solutions which changes red litmus blue.
e.g. : Na₂O, CaO , MgO etc.
CaO + H₂O → Ca(OH)₂
MgO + H₂O → Mg (OH)₂
The basic nature of these oxides increases from top to bottom in the group. In a period basic properties decreases and acidic properties increases.

Solved Problems

Question 1.
What would be the IUPAC name and symbol for the element with atomic number 120?
Solution:
The roots for 1, 2 and 0 are un, bi and nil, respectively. Hence, the symbol and the name respectively are Ubn and unbinilium.

Question 2.
How would you justify the presence of 18 elements in the 5t” period of the Periodic Table ?
Solution:
When n = 5, l = 0, 1, 2, 3. The order in which the energy of the available orbitals 4d, 5s and 5p increases is 5s < 4d < 5p. The total number of orbitals available are 9. The maximum number of electrons that can be accommodated is 18; and therefore 18 elements are there in the 5^th period.

Question 3.
The elements Z = 117 and 120 have not yet been discovered. In which family / group would you place these elements and also give the electronic configuration in each case.
Solution:
we see from that element with Z = 117, would belong to the halogen family (Group 17) and the electronic configuration would be [Rn] 5f¹⁴6d¹⁰7s²7p⁵. The element with Z = 120, will be placed in Group 2 (alkaline earth metals), and will have the electronic configuration [Uuo]8s²

Question 4.
Considering the atomic number and position in the periodic table, arrange the following elements in the increasing order of metallic character: Si, Be, Mg, Na, P.
Solution:
Metallic character increases down a group and decreases along a period as we move from left to right. Hence the order of increasing metallic character is : P < Si < Be < Mg < Na.

Question 5.
Which of the following species will have the largest and the smallest size ?
Mg, Mg²⁺, Al, Al³⁺.
Solution:
Atomic radii decrease across a period. Cations are smaller than their parent atoms. Among isoelectronic species, the one with the larger positive nuclear charge will have a smaller radius.
Hence the largest species is Mg; the smallest one is Al³⁺.

Question 6.
The first ionization enthalpy (∆_iH) values of the third period elements, Na, Mg and Si are respectively 496, 737 and 786 kJ mol^-1. Predict whether the first ∆_iH value for Al will be more close to 575 or 760 kJ mol^-1 ? Justify your answer.
Solution:
It will be more close to 575 kJ mol^-1. The value for Al should be lower than that of Mg because of effective shielding of 3p electrons from the nucleus by 3s-electrons.

Question 7.
Which of the following will have the most negative electron gain enthalpy and which the least negative ? P, S, Cl, F. Explain your answer.
Solution:
Electron gain enthalpy generally becomes more negative across a period as we move from left to right. Within a group, electron gain enthalpy becomes less negative down a group. However, adding an electron to the 2p-orbital leads to greater repulsion than adding an electron to the larger 3p-orbital. Hence the element with most negative electron gain enthalpy is chlorine; the one with the least negative electron gain enthalpy is phosphorous.

Question 8.
Using the Periodic Table, predict the formulas of compounds which might be formed by the following pairs of elements;
(a) silicon and bromine
(b) aluminium and sulphur.
Solution:
(a) Silicon is group 14 element with a valence of 4; bromine belongs to the halogen family with a valency of 1. Hence the formula of the compound formed would be SiBr₄.
(b) Aluminium belongs to group 13 with a valence of 3; sulphur belongs to group 16 elements with a valence of 2. Hence, the formula of the compound formed would be Al₂S₂.

Question 9.
Are the oxidation state and covalency of Al in [AICI(H₂O)₅]²⁺ same ?
Solution:
No. The oxidation state of Al is +3 and the covalency is 6.

Question 10.
Show by a chemical reaction with water that Na₂O is a basic oxide and Cl₂O₇ is an acidic oxide.
Solution:
Na₂O with water forms a strong base Whereas Cl₂O₇ forms strong acid.
Na₂O + H₂O → 2NaOH
Cl₂O₇ + H₂O → 2HClO₄
Their basic or acidic nature can be qualitatively tested with litmus paper.

Andhra Pradesh BIEAP AP Inter 1st Year Chemistry Study Material 1st Lesson Atomic Structure Textbook Questions and Answers.

AP Inter 1st Year Chemistry Study Material 1st Lesson Atomic Structure

Very Short Answer Questions

Question 1.
What is the charge, mass, and charge-to-mass ratio of an electron?
Answer:

• Charge of an electron = – 1.602 × 10^-19 coloumbs (or) -4.8 × 10^-19 esu
• Mass of an electron = 9.1 × 10^-28 gms
• Charge to mass ratio of an electron \(\left(\frac{\mathrm{e}}{\mathrm{m}}\right)\) i.e., specific charge = 1.758 × 10¹¹ coloumbs/kg

Question 2.
Calculate the charge of one mole of electrons.
Answer:
One electron has charge – 1.602 × 10^-19 coloumbs.
One mole of electrons has charge -6.023 × 10²³ × 1.602 × 10^-19
= 9.648846 × 10⁴ = 96488.5 coloumbs.

Question 3.
Calculate the mass of one mole of electrons.
Answer:
Mass of electron = 9.1 × 10-31 kg (or) 9.1 × 10^-28 gms.
One mole of electrons has mass 6.023 × 10²³ × 9.1 × 10^-31 = 54.8 × 10^-8 = 5.48 × 10^-7 kg.

Question 4.
Calculate the mass of one mole of protons. ”
Answer:
One proton has mass 1.672 × 10^-27 kg
One mole protons has mass 6.023 × 10²³ × 1.672 × 10^-27
= 10.0704 × 10^-4
= 1.00704 × 10^-3 kg.

Question 5.
Calculate the mass of one mole of neutrons.
Answer:
One neutron has mass 1.675 × 10^-27 kg
One mole neutrons has mass 6.023 × 10²³ × 1.675 × 10^-27
= 10.088 × 10^-4
= 1.0088 × 10^-3 kg.

Question 6.
How many neutrons and electrons are present in the nuclei of ₆C¹³, ₈O¹⁶, ₁₂Mg²⁴, ₂₆Fe⁵⁶ and ₃₈Sr⁸⁸.
Answer:

Question 7.
What is a black body ?
Answer:
The body which is perfect absorber and emmiter of all type of radiations incident on it is called a black body.

Question 8.
Which part of electromagnetic spectrum does Balmer series belong?
Answer:
Balmer series (n = 2) belongs to visible region of electromagnetic spectrum.

Question 9.
What is an atomic orbital?
Answer:
In an atom, the region around the nucleus where the probability of finding the electron is maximum is known as atomic orbital.

• From the value of magnitude of square of wave function (|\(\psi^2\)|) this region can be predicted.

Question 10.
When an electron is transferred in hydrogen atom from n = 4 orbit to n=5 orbit to which spectral series does this belong?
Answer:

• By the absorption of energy electron jumps from n = 4 orbit to n = 5 orbit.
• The electron present in n = 5 orbit emitts energy and return to n = 4 orbit. Hence the spectral lines series obtained in Brackett series (IR region)

Question 11.
How many “p” electrons are present in sulphur atom?
Answer:
Sulphur has electronic configuration – 1s² 2s² 2p⁶ 3s² 3p⁴
∴ Sulphur has 10 ‘p’ electrons.

Question 12.
What are the values of principal quantum number (n) and azimuthal quantum number (l) for a 3d electron?
Answer:
For a 3d – electron principal quantum number (n) = 3 and
Azimuthal quantum number (l) = 2.

Question 13.
What is the complete symbol for the atom with the given atomic number (Z) and atomic mass (A)?
I) Z = 4, A = 9 ;
II) Z = 17, A = 35 ;
III) Z = 92, A = 233.
Answer:
I) Z = 4, A = 9 Complete symbol is ₄Be⁹
II) Z = 17, A = 35 Complete symbol is ₁₇Cl³⁵
III) Z = 92, A = 233 complete symbol is ₉₂U²³³.

Question 14.
Draw the shape of \(\mathrm{d}_{\mathrm{z}^2}\) orbital.
Answer:

Question 15.
Draw the shape of d_{x² – y²} orbital.
Answer:

Question 16.
What is the frequency of radiation of wavelength 600nm?
Answer:
Formula:
\(v=\frac{c}{\lambda}\)
= \(\frac{3 \times 10^8}{6 \times 10^{-7}}\)
= \(\frac{1}{2} \times 10^{15}\)
= 0.5 × 10¹⁵ = 5 × 10¹⁴ sec^-1
λ = 600 nm
= 600 × 10^-9 m
= 6 × 10^-7 m
C = 3 × 10⁸ m/sec.

Question 17.
What is Zeeman effect?
Answer:
The splitting up of spectral lines in presence of strong external magnetic field is called as Zeeman effect.

Question 18.
What is Stark effect?
Answer:
The splitting of spectral lines in presence of strong electric field is called as Stark effect.

Question 19.
To which element does the following electronic configuration correspond?
I) 1s²2s² 2p⁶3s²3p¹
II) 1s²2s²2p⁶3s²3p⁶
III) 1s²2s²2p⁵
IV) 1s²2s²2p².
Answer:
I) 1s²2s² 2p⁶ 3s² 3p¹ (Atomic no. (Z) = 13) – Aluminium.
II) 1 s²2s²2p⁶3s²3p⁶(Atomic no. (Z) = 18) – Argon.
III) 1s²2s²2p⁵ (Atomic no. (Z) = 9) – Fluorine.
IV) 1s²2s²2p² (Atomic no. (Z) = 6) – Carbon.

Question 20.
Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 4000 A. What is the threshold frequency (\(v_0\))?
Answer:
Formula:
hv = hv₀ + \(\frac{1}{2} m v^2\)
hv = hv₀ + \(\frac{1}{2} m(0)^2\)
hv = hv₀
λ = 4000 A
= 4 × 10³ × 10^-10 = 4 × 10^-7 m.
V = 0
C = 3 × 10⁸ m/sec.
⇒ v = v₀
∴ v = \(\frac{\mathrm{C}}{\lambda}\) = \(\frac{3 \times 10^8}{4 \times 10^{-7}}\) = \(\frac{3}{4}\) × 10¹⁵
= 0.75 × 10¹⁵
= 7.5 × 10¹⁴ sec^-1

Question 21.
Explain Pauli’s exclusion principle.
Answer:
Pauli’s exclusion principle:
According to this principle
“No two electrons in an atom can have the same set of four quantum numbers”. This can also be stated as “only two electrons may exist in the same orbital and these electrons must have opposite spins”.

This means that the two electrons can have the same value of three quantum numbers n, l and m_l but have the opposite spin quantum number Ex : Consider ‘K’ shell of the atom having two electrons

Question 22.
What is Aufbaus principle ?
Answer:
Aufbau’s principle:
This principle states
“In the ground state of the atoms, the orbitals are filled in order of their increasing energies”. In other words electrons first occupy the lowest energy orbital available to them and enter into higher energy orbitals only after the lower energy orbitals are filled.
The order in which the orbitals are filled as follows :
1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 4f < 5d < 6p < 7s

Question 23.
What is Hund’s rule ?
Answer:
Hund’s rule: This rule deals with the filling of electrons in degenerate orbitals. It states “Pairing of electrons in the orbitals belonging to the same subshell (p, d or f) does not take place until each orbital belonging to that subshell has got one electron each (i.e.,) all the orbitals are singly occupied”.
Since there are three ’p’, five ‘d’ and seven ‘f’ orbitals, therefore the pairing of electrons will start in the p, d and f orbitals with the entry of 4^th, 6^th and 8^th electrons respectively.
Ex : ‘₈O’ electronic configuration is

Question 24.
Explain Heisenberg’s uncertainty principle.
Answer:
Heisenberg uncertainty principle : “Simultaneous and exact determination of the position and momentum of a sub-atomic particle, like electron moving with high speed is impossible.”
If Δx and Δp represents the uncertainties in the position and momentum respectively. Then according to Heisenberg
Δx. Δp ≥ \(\frac{\mathrm{h}}{4 \pi}\) ——- (1)

The product of uncertainties in position (Δx) and momentum (Δp) of an electron cannot be less than \(\frac{h}{4 \pi}\). It can be equal or greater than \(\frac{h}{4 \pi}\).
Since momentum = mass x velocity, the equation (1) can be written as
Δx × m (Δv) ≥ \(\frac{\mathrm{h}}{4 \pi}\) = Δx × Δv ≥ \(\frac{\mathrm{h}}{4 \pi \mathrm{m}}\)
If the position is determined accurately Δx = 0 and Δv = ∝. That means the inaccuracy in measuring the velocity is ∝. If velocity is determined accurately Δv = 0 and Δx = ∝.

Question 25.
What is the wavelength of an electron moving with a velocity of 2.0 × 10⁷m/s ?
Answer:
Formulae:
λ = \(\frac{h}{m v}\)
= \(\frac{6.625 \times 10^{-34}}{9.1 \times 10^{-31} \times 2 \times 10^7}\)
= 0.3640 × 10^-34 × 10⁺²⁴
= 0.3640 × 10¹⁰ m
= 0.3640 A
h = 6.625 × 10^-34 J.Sec
m = 9.1 × 10^-31 kg
V = 2.0 × 10⁷ m/sec.

Question 26.
An atomic orbital has n = 2, what are the possible values of l and m_l?
Answer:
For n = 2, l values are 0, 1
For l = 0 → m_l = 0
For l = 1 → m_l = -1, 0, +1.

Question 27.
Which of the following orbitals are possible? 2s, 1p, 3f, 2p.
Answer:
2s, 2p orbitals are possible among 2s, 1p, 3f, 2p and 1 p, 3f orbitals are not possible.

Question 28.
The static electric charge on the oil drop is – 3.2044 × 10^-19 C. How many electrons are present on it?
Answer:
Given static electric charge on oil drop = – 3.2044 × 10^-19 C
Charge of electron = – 1.602 × 10^-19 C.
Number of electrons present =

Question 29.
Arrange the following type of radiation in increasing order of frequency:
(a) X – rays
(b) visible radiation
(c) microwave radiation and
(d) radiation from radio waves.
Answer:
Increasing order of frequency of given radiations is
Radio waves < Micro waves < Visible radiation < X – rays.

Question 30.
How many electrons in an atom may have n = 4 and m_s = +1/2 ?
Answer:
For n = 4 → l values are 0, 1, 2, 3
l = 0 → s contains 1 electron with m_s = + 1/2
l = 1 → p contains 3 electron with m_s = + 1/2
l = 2 → d contains 5 electron with m_s = + 1/2
l = 3 → f contains 7 electron with m_s = + 1/2
∴ Total no.of electrons with ms = +1/2 for n = 4
= 1 + 3 + 5 + 7 = 16.

Question 31.
How many sub-shells are associated with n = 5 ?
Answer:
For n = 5
l values are 0, 1, 2, 3, 4
l = 0 → s – orbital
l = 1 → p — orbital
l = 2 → d – orbital
l = 3 → f – orbital
l = 4 → g — orbital
→ Five subshells are associated with n = 5.

Question 32.
Explain the particle nature of electromagnetic radiation.
Answer:

• According to earlier days concepts light was supposed to be made of particles. This assumption was made by Newton in his corpuscular theory. He called the particles as corpuscales.
• The particle nature of light explains the black body radiations and photo electric effect satisfactorily.
• The particle nature of light could not satisfactorily explains the phenomenon of diffraction and Interferance.

Question 33.
Explain the significance of Heisenberg’s Uncertainty principle.
Answer:
Significance of Uncertainty Principle:

1. This principle rules out the existence of definite paths or trajectories of electrons and other similar particles.
2. This principle is significant only for motion of microscopic objects, and is negligible for that of macroscopic objects.
3. In dealing with milligram size or heavier objects, the associated uncertainties are hardly of any real consequence.

Question 34.
What series of lines are observed in hydrogen spectra?
Answer:
The series of lines observed in hydrogen spectra are

Additional Answer Questions

Question 35.
How many newtrons and electrons are present in the nuclei of \({ }_6 C^{13}\), \({ }_8 \mathrm{O}^{16}\), \({ }_{12} \mathrm{Mg}^{24}\), \({ }_{26} \mathrm{Fe}^{56}\), \({ }_{38} \mathrm{Sr}^{88}\)
Answer:

Additional Problems

Question 36.
Calculate the wave no. and wave length of first line of lyman series.
Answer:

Question 37.
Calculate the wave no. of and wave length of first line of Balmer series.
Answer:

Short Answer Questions

Question 38.
What is the wavelength of light emitted when the electron in a hydrogen atom undergoes transition from an energy level with n = 5 to an energy level with n = 3 ?
Answer:
Formulae:

R = 1,09,677 cm^-1
n₁ = 3
n₂ = 5.
\(\bar{v}\) = 7799.25 cm^-1
λ = \(\frac{1}{\bar{v}}\) = \(\frac{1}{7799.25}\) = 1.2821 × 10^-4 cm

Question 39.
An atom of an element contains 29 electrons and 35 neutrons. Deduce

1. the number of protons and
2. the electronic configuration of the element.

Answer:
Given no.of electrons ’29’, ∴ Z = 29

1. So, no.of protons = 29
2. Electronic configuration of the element (Z = 29)
   = 1s²2s²2p²3s²3p⁶4s¹3d¹⁰ [anamalous electronic configuration]

Question 40.
Explain giving reasons, which of the following sets of quantum numbers are not possible.
a) n = 0, l = 0, m_l = 0, m_s = +\(\frac{1}{2}\)
b) n = 1, l = 0, m_l = 0, m_s = –\(\frac{1}{2}\)
c) n = 1, l = 1, m_l = 0, m_s = +\(\frac{1}{2}\)
d) n = 2, l = 1, m_l = 0, m_s = +\(\frac{1}{2}\)
e) n = 3, l = 3, m_l = – 3, m_s = +\(\frac{1}{2}\)
f) n = 3, l = 1, m_l = 0, m_s = +\(\frac{1}{2}\)
Answer:
Following set of quantum numbers are not possible.
a) n = 0, l = 0, m_l = 0, m_s = +\(\frac{1}{2}\)
Reason:
‘n’ is principal quantum number, whose values are from 1 to n. The value of ‘n’ never equal to zero. But given n = 0.

c) n = 1, l = 1, m_l = 0, m_s = +\(\frac{1}{2}\)
Reason:
Values of ‘l’ are from 0 to (n – 1).
If n = 1 then the value of ‘l’ is zero not equal to ‘1’.

e) n = 3, l = 3, m_l = – 3, m_s = +\(\frac{1}{2}\)
Reason:
If n = 3, possible values of ‘l’ are 0, 1, 2, but not equal to ‘3’.

Question 41.
Show that the circumference of the Bohr orbit for the hydrogen atom is an integral multiple of the de Broglie wavelength associated with the electron revolving around the orbit.
Answer:
Consider the Bohr’s angular momentum equation.
mvr = \(\frac{\mathrm{nh}}{2 \pi}\)
i.e The angular momentum of an electron is integral multiple of ‘\(\frac{\mathbf{h}}{2 \pi}\)‘
mvr = \(\frac{\mathrm{nh}}{2 \pi}\)
2πr = \(\frac{\mathrm{nh}}{\mathrm{mv}}\)
According to de-Broglie’s wavelength λ = \(\frac{h}{m v}\)
2πr = \(n\left(\frac{h}{m v}\right)\)
2πr = nλ.
Thus the circumference of the Bohr orbit is integral multiple of de-Broglie’s wave length.

Question 42.
The longest wavelength doublet absorption transition is observed at 589.0 and 589.6 nm. Calculate the frequency of each transition and energy difference between two excited states.
Answer:
Given largest wave length doublet absorption transition is observed at 589.0 and 589.6 nm.
∴ \(v_1=\frac{c}{\lambda_1}\)
= \(\frac{3 \times 10^8}{589 \times 10^{-9}}\) = 0.005093 × 10⁺⁷
= 5.093 × 10¹⁴ sec^-1
λ₁ = 589 × 10^-9 m
∴ \(v_2=\frac{c}{\lambda_2}\)
= \(\frac{3 \times 10^8}{589.6 \times 10^{-9}}\)
= 0.005088 × 10¹⁷
= 5.088 × 10¹⁴ sec^-1
λ₁ = 5.089 × 10^-9 m
Energy difference between two states = h[\(v_1\) – h\(v_2\)]
= h[\(v_1\) – \(v_2\)]
= 6.625 × 10^-34[5.093 × 10^-14 – 5.088 × 10¹⁴]
= 6.625 × 10^-34 × 0.005 × 10¹⁴
= 0.0331 × 10^-20 J.

Question 43.
What are the main features of quantum mechanical model of an atom?
Answer:
Important features of quantum mechanical model of atom:

1. The energy of electrons in an atom is quantized (it can only have certain specific values).
2. The existence of quantized electronic energý levels is a direct result of the wave like properties at electrons and are allowed solution at Schrodinger wave equation.
3. All the information about the electron in an atom is contained in its orbital wave function ‘Ψ’ and quantum mechanics makes it possible to extract this information from “Ψ’.
4. The path of the electron can never be determined accurately. Therefore, we find only the probability of the electron at different points in space, around an atom.
5. The probability of finding an electroñat a point within an atom is proportional to the square of the orbital wave function i.e., \(|\Psi|^2\) at that point. \(|\Psi|^2\) is known as probability density and is always positive. From the value of \(|\Psi|^2\) at different points with in the atom, it is possible to predict the region around the nucleus where electron will most probably be found.

Question 44.
What is a nodal plane? How many nodal planes are possible for 2p – and 3d – orbitals?
Answer:
The plane at which the probability of finding the electron is zero is called as nodal plane.

• For 2p orbitaIs one nodal plane is possible for each ‘p’ orbital.
• For 3d orbitais two nodal planes are possible for each ‘d’ orbital.

Question 45.
The Lyman series occurs between 91.2 nm and 121.6 nm, the Balmer series occurs between 364.7 nm and 656.5 nm and the Paschen series occurs between 820.6 nm and 1876 nm. Identify the spectral regions to which these wavelengths correspond?
Answer:
In electromagnetic spectrum,
a) 91.2 – 121.6 nm (Lyman senes) corresponds to u.v. region.
b) 364.7 – 656.5 nm (Balmer series) corresponds to visible region.
c) 820.6 – 1876 nm (Paschen series) corresponds to I.R. region.

Question 46.
How are the quantum numbers n, l, m_l, for hydrogen atom obtained?
Answer:
Electronic configuration of hydrogen is 1s¹
For ns¹
Principal quantum no. (n) = 1
Azimuthal quantum no. (l) = 0
Magnetic quantum no. (m_l) = 0
and Spin quantum no. (m_s) = + 1/2

Question 47.
A line in Lyman series of hydrogen atom has a wavelength of 1.03 × 10^-7 m. What is the initial energy level of the electron?
Answer:
Given λ= 1.03 × 10^-7 m = 1.03 × 10^-5 cm
n₂ = 1 (for Lyman series)
We have, R = 109677 cm^-1

⇒ n₂ = 3 (i.e.,) original energy level of electron is 3.

Question 48.
If the position of the electron is measured within an accuracy of ±0.002 nm. Calculate the uncertainty in the momentum of the electron.
Answer:
Formulae:
Δx × Δp = \(\frac{h}{4 \pi}\)
Δp = \(\frac{h}{\Delta x \times 4 \pi}\)
= \(\frac{6.625 \times 10^{-34}}{4 \times 3.14 \times 2 \times 10^{-12}}\)
= 0.2637 × 10^-22
= 2.637 × 10^-23 J/m.
Δx = 0.002 nm
= 2 × 10^-3 × 10⁹ m
= 2 × 10^-12 m
h = 6.625 × 10^-34 J.sec.
∴ Uncertainity in momentum of electron = 2.637 × 10^-23 J/m.

Question 49.
If the velocity of the electron is 1.6 × 10⁶ m/s^-1. Calculate de Brogue wavelength associated with this electron.
Answer:
Formulae:
λ = \(\frac{\mathrm{h}}{\mathrm{mv}}\)
= \(\frac{6.625 \times 10^{-34}}{9.1 \times 10^{-31} \times 1.6 \times 10^6}\)
= 0.455 × 10^-9 m
= 0.455 nm.
v = 1.6 × 10⁶ m/sec
h = 6.625 × 10^-34 J.sec
m = 9.1 × 10^-31 Kg.

Question 50.
Explain the difference between emission and absorption spectra. (A.P. Mar. ‘15)
Answer:
Emission spectrum

1. It is produced by analysing the radiant energy emitted by an excited substance.
2. It consists of bright lines on dark back ground.
3. Produced due to the emission of energy by electrons.
4. Emission spectra contains bright ineson dark back ground.

Absorption spectrum

1. It is produced when white light is passed through a substance and the transmitted light is analysed by a spectrograph.
2. It consists of dark lines on bright background.
3. Produced due to the adsorption of energy by electrons.
4. Absorption spectra contains dark lines on bright back ground.

Question 51.
The quantum numbers of electrons are given below. Arrange them in order of increasing energies.
a) n = 4, l = 2, m_l = -2, m_s = +\(\frac{1}{2}\)
b) n = 3, l = 2, m_l = -1, m_s = –\(\frac{1}{2}\)
c) n = 4, l = 1, m_l = 0, m_s = +\(\frac{1}{2}\)
d) n = 3, l = 1, m_l = -1, m_s = –\(\frac{1}{2}\)
Answer:
a) n = 4, l = 2, → 4d
b) n = 3, l = 2, → 3d
c) n = 4, l = 1, → 4p
d) n = 3, l = 1, → 3p
∴ 3p < 3d < 4p < 4d
According (n + l) values
Hence d < b < c < a is order of increasing energy.

Question 52.
The work function for Cesium atom is 1.9 eV. Calculate the threshold frequency of the radiation. If the Cesium element is irradiated with a wavelength of 500 nm, calculate the kinetic energy of the ejected photoelectron?
Answer:
Case-I
Photo electric effect equation is
hv = hv₀ + 1/2 mv²
w = hv₀

Case-II
Photo electric effect equation is
E = \(\frac{\mathrm{hc}}{\lambda}\)
= \(\frac{6.625 \times 10^{-34} \times 3 \times 10^8}{5 \times 10^{-7}}\)
= \(\frac{19.878 \times 10^{-26}}{5 \times 10^{-7}}\)
= 3.9756 × 10^-19 J.

Given work function hv₀ = 1.9 ev
= 1.9 × 1.602 × 10^-19J.

Kinetic Energy (KE) = \(\frac{1}{2} m v^2\)
From Photo electric effect
\(\frac{1}{2} m v^2\) = hv – hv₀
K.E. = 3.9756 × 10^-19 × 1.602 × 10^-19
= 3.9756 × 10^-19 – 3.0438 × 10^-19 = 0.9318 × 10^-19 = 9.318 × 10^-20J.

Question 53.
Calculate the wavelength for the emission transition if it starts from the orbit having radius 1.3225 nm and ends at 211.6 pm. Name the series to which this transition belongs and the region of the spectrum.
Answer:
Given the radius of orbit from which it started = 1.35225 × 10^-9 m = 1 3.225Å
In general radius of orbit = 0.529 × n²Å
n² = \(\frac{13.225}{0.529}\) = 25
n² = 25 ⇒ n = 5
Given that the
Radius of orbit at which the transition ended = 211.6 pm
= 211.6 × 10^-12m
= 2.116A
Similarly as above
n² = \(\frac{2.116}{0.529}\) = 4
n² = 4 ⇒ n = 2
∴ transition takes place from n = 5 to n = 2 level
∴ spectral lines are obtained in Balmer series (visible region)

Question 54.
Explain the difference between orbit and orbital.
Answer:
Orbit

1. A circular path which is present around the nucleus in which electrons revolve is called as orbit.
2. Orbits are circular and are non directional paths.
3. The maximum no.of electrons in any orbit is given by the formula 2n² (n = orbit number).

Orbital

1. The 3 – dimension space where the probability of finding the electron is maximum around the nucleus is called as orbital.
2. These have definite shape and these are directional except’s orbital.
3. Each orbital can occupy a maximum of two electrons.

Question 55.
Explain photoelectric effect.
Answer:
The ejection of electrons from a metal surface, when the radiations of suitable frequency strikes the metal surface is called photoelectric effect.

Explanation using Einstein’s quantum theory:

1) To explain photoelectric effect, Einstein utilised Quantum theory.
2) When a photon strikes metal surface, it uses some part of its energy to eject the electron from the metal atom. The remaining part of the total energy is given to the ejected electrons in the form of kinetic energy.
Hence we can write hv = W + KE ⇒ hv = hv₀ + \(\frac{1}{2} m_e v^2\)
where hv = energy of photon,
v₀ = Threshold frequency,
m_e = mass of electron
W = energy required to overcome the attractive forces on the electron in the metal (work function)
KE = kinetic energy of ejected electron,
V = Velocity of ejected electron.

3) If a photon of sufficient energy struck the metal surface and could eject the electron. But if a photon has insufficient energy, it cannot eject the electron from the metal.
eg. : A photon of violet light [high frequency] can eject the electrons from the surface of potassium but a photon of red light [low frequency] cannot eject the electrons.

Question 56.
Explain Rutherford’s nuclear model of an atom. What are its drawbacks?
Answer:
Rutherford’s Planetary model: Rutherford drew some conclusions regarding the structure of atom.

1. Most of the space in the atom is empty (as most of the α – particles passed through the foil undeflected).
2. A few — positive charges were deflected. The deflection must be due to enormous repulsive force showing that the positive charge of the atom is not spread throughout the atom as Thomson predicted. The positive charge is concentrated in a very small volume. Which is responsible for the deflection of α — particles.

On the basis of the above observations. Rutherford proposed the nuclear model. According to his model.

1. The positive charge in the atom is concentrated in the small dense portion, called the NUCLEUS.
2. The nucleus is surrounded by the electrons that move around it in circular paths called the ORBITS. Thus Rutherford’s model resembles the solar system.
3. Electrons and the nucleus are held together by electrostatic forces of attraction.

Drawbacks of Rutherford model:

1. Rutherford’s atomic model of an atom is like a small scale solar system. This similarity suggests that electrons should move around the nucleus in well defined orbits. However, when a body is moving, it undergoes acceleration. According to electromagnetic theory, charged particles, when accelerated, should emit radiation. Therefore, an electron in an orbit will emit radiation, thus the orbit will continue to shrink. But this does not happen. Thus Rutherford’s model cannot explain the stability of the atom.

2. If we assume that electrons as stationary around the nucleus, the electrostatic attraction between the nucleus and the electrons would pull the electrons towards the nucleus to form a miniature version of Thomson’s model.

3. Rutherford model does not explain the electronic structure of the atom i.e., how the electrons are distributed around the nucleus and what are the energies of these electrons.
Before studying further developments that lead to the formulation of various atomic models, it is necessary to study about light and its nature.

Question 57.
Explain briefly the Planck’s quantum theory.
Answer:
The postulates. of Planck’s quantum theory are
a) The emission of radiation is due to vibrations of charged particles (electrons) in the body.
b) The emission is not continuous but in discrete packets of energy called quanta. This emitted radiation propagates in the form of waves.
c) The energy (E) associated with each quantum for a particular radiation of frequency V is given by E = hv, Here ‘h’ is Planck’s constant.
d) A body can emit or absorb either one quantum (hv) of energy or some whole number multiple of it. Thus energy can be emitted or absorbed as hv, 2hv, 3hv etc., but not fractional values. This is called quantisation of energy.
e) The emitted radiant energy is propagated in the form of waves.
f) Values of Planck’s constant in various units:
h = 6.6256 × 10^-27 erg.sec (or) g cm²s^-1
= 6.6256 × 10^-34J.s (or) kg m²s^-1 = 1.58 × 10^-34 cal.s.
Success of Planck’s quantum theory: This theory successfully explains the black body radiations. A black body is a perfect absorber and also a perfect radiator of radiations.

Question 58.
What are the postulates of Bohr’s model of hydrogen atom? Discuss the importance of this model to explain various series of line spectra In hydrogen atom. (T.S. Mar. ‘16)(A.P. Mar.’15. ‘13)
Answer:
Niels Bohr quantitatively gave the general features of hydrogen atom structure and it’s spectrum. His theory is used to evaluate several points in the atomic structure and spectra.
The postulates of Bohr atomic model for hydrogen as follows

Postulates : –

• The electron in the hydrogen atom can revolve around the nucleus in a circular path of fixed radius and energy. These paths are called orbits (or) stationary states. These circular orbits are concentric (having same center) around the nucleus.
• The energy of an electron in the orbit does not change with time.
• When an electron moves from lower stationary state to higher stationary state absorption of energy takes place.
• When an electron moves from higher stationary state to lower stationary state emission of energy takes place.
• When an electronic transition takes place between two stationary states that differ in energy by ΔE is given by
  ΔE = E₂ – E₁ = hv
  ∴ The frequency of radiation absorbed (or) emitted v = \(\frac{E_2-E_1}{h}\) E₁ and E₂ are energies of lower, higher energy states respectively.
• The angular momentum of an electron is given by mvr = \(\frac{\mathrm{nh}}{2 \pi}\)
  An electron revolve only in the orbits for which it’s angular momentum is integral multiple of \(\frac{\mathrm{h}}{2 \pi}\)

Line spectra of hydrogen, Bohr’s Theory:

• In case of hydrogen atom line spectrum is observed and this can be explained by using Bohr’s Theory.
• According to Bohr’s postulate when an electronic transition takes place between two stationary states that differ in energy is given by
  
  
• In case of absorption spectrum n_f > n_i → energy is absorbed (+ve) energy is absorbed (+Ve)

• In case of emission spectrum n_i > n_f → energy is emitted (- Ve)
• Each spectral line in absorption (or) emission spectrum associated to the particular transition in hydrogen atom
• In case of large no.of hydrogen atoms large no.of transitions possible they rsults in large no.of spectral lines.
  The series of lines observed in hydrogen spectra are
  

Question 59.
Explain the success of Bohr’s theory for hydrogen atom.
Answer:
Succes of Bohrs Theory for hydrogen atom:

• Bohr’s theory gave the information about the principal quantum number. Principal quantum number represents the stationary states. (n = 1, 2, 3 integral numbers)
• Bohr’s theory gave the information about the radius of the stationary states (or) orbits.
  r = 0.529 × n² A (or)
  r = 52.9 × n² pm
  \(\left[r=\frac{n^2 h^2}{4 \pi^2 m e^2}\right]\) (for hydrogenation)
• This theory gave the information about the energy of the electron of particular stationary state.
  E_n = -R_H \(\left[\frac{1}{n^2}\right]\)n = 1, 2, 3,……
  R_H = Ryd berg constant
  = 1,09,677 cm^-1.
• This theory explained the line spectra of hydrogenation.
• This theory can also be applicable to the ions containing only one electron. Eg. : He⁺, Li⁺², Be⁺³….
• This theory can also gave information about velocity of electrons moving in the orbits.

Question 60.
What are the consequences that lead to the development of quantum mechanical model of an atom?
Answer:
Consequences that lead to development of quantum mechanical model of an atom are as follows.

• Clàssical mechanics successfully explained the motion of macro scopic objects.
  Eg: Falling stone, Planets etc.,.
• Classical mechanics failed to explain the motion of microscopic objects like electrons, atoms, molecules etc.
• Classical mechanics ignores the concept of dual behaviour of matter and especially for sub atomic particles
  Quantum mechanics:
  The Branch of science deals with the dual behaviour of matter is called quantum mechanics.
• This deals with the motions of microscopic objects like electron.

Important features of quantum mechanical model of atom:

1. The energy of electrons in an atom is quantized (it can only have certain specific values).
2. The existence of quantized electronic energy levels is a direct result of the wave like properties at electrons and are allowed solution at schrodinger wave equations.
3. All the information about the electron in an atom is contained in its orbital wave function ‘\(\Psi^{\prime}\) and quantum mechanics makes it possible to extract this information from ‘\(\Psi^{\prime}\).
4. The path of the electron can never be determined accurately. Therefore, we find only the probability of the electron at different points in space, around an atom.
5. The probability of finding an electron at a point within an atom is proportional to the square of the orbital wave function i.e., \(|\Psi|^2\) at that point. \(|\Psi|^2\) is known as probability density and is always positive. From the value of \(|\Psi|^2\) at different points with in the atom, it is possible to
   predict the region around the nucleus where electron will most probably be found.

Question 61.
Explain the salient features of quantum mechanical model of an atom.
Answer:
Important features of quantum mechanical model of atom:

1. The energy of electrons in an atom is quantized (it can only have certain specific values).
2. The existence of quantized electronic energy levels is a direct result of the wave like properties at electrons and are allowed solution at schrodinger wave equations.
3. All the information about the electron in an atom is contained in its orbital wave function ‘\(\Psi^{\prime \prime}\) and quantum mechanics makes it possible to extract this information from \(\Psi^{\prime \prime}\).
4. The path of the electron can never be determined accurately. Therefore, we find only the probability of the electron at different points in space, around an atom.
5. The probability of finding an electron at a point within an atom is proportional to the square of the
   orbital wave function i.e., \(|\Psi|^2\) at that point. \(|\Psi|^2\) is known as probability density and is always positive. From the value of \(|\Psi|^2\) at different points with in the atom, it is possible to predict the region around the nucleus where electron will most probably be found.

Question 62.
What are the limitations of Bohr’s model of an atom?
Answer:
Limitations:

1. Spectra of multielectron atoms: Bohr’s theory could explain the spectra of Hydrogen and single electron species like He⁺, Li²⁺, Be³⁺, but it fails to explain the spectra of multielectron atoms.
2. Fine structure: It fails to explain this fine structure of Hydrogen atom.
3. Splitting up of spectral lines : The theory fails to explain Zeeman effect and Stark effect.
   The splitting up of spectral lines when an atom is subjected to strong magnetic field is called Zeeman effect.
   The splitting up of spectral lines when an atom is subjected to strong electric field is called Stark effect.
4. Flat model: Bohr’s theory gives a flat model of the orbits. Bohr’s theory predicts definite orbits for electrons considering them as particles. But according to de Brogue electron has both wave nature and particle nature. Bohr’s theory cannot explain this dual role.
5. It fails to support the uncertainty principle proposed by Heisenberg.
6. It could not explain the ability of atoms to form molecules by chemical bonds.

Question 63.
What are the evidences in favour of dual behaviour of electron?
Answer:

• The particle nature of light explains the phenomenon of blackbody radiations and photo electric effect but it could not explain about wave nature of light.
• Wave nature of light explains the phenomenon of interference and diffraction.
• So, light has dual nature i.e it behaves as a wave (or) as a stream of particles.
• According to de-Broglie, light has dual behaviour i.e both particle and wave nature.
  de-Broglies gave the following relationship
  λ = \(\frac{\mathrm{h}}{\mathrm{mv}}\) = \(\frac{h}{p}\)
  λ = wave length
  P = momentum
• Heisen bergs uncertainty principle also a consequence of dual behaviour of matter and radiation.
  Statement :— It is impossible to determine simultaneously, the exact momentum and exact position of a small particle like lectron.
  Δx × Δp ≥ \(\frac{h}{4 \pi}\)
  Δx = uncertainty in position
  Δp = uncertainty in momentum

Question 64.
How are the quantum numbers n, l and m_l, arrived at? Explain the significance of these quantum numbers. (A.P. Mar. ‘16)(T.S. Mar. ‘15, ‘14)
Answer:

• In general a large no.of orbitals are possible in an atom.
• These orbitaIs are distinguished by their size, shape and orientation.
• An orbital of smaller size means there is more chance to find electron near the nucleus.
• Atomic orbitals are precisely distinguished by quantum numbers. Each orbital is designated by three major quantum numbers.

1) Principal quantum number (n)
2) Azimuthal quantum number (l)
3) Magnetic quantum number (m)

1) Principal quantum number : The principal quantum number was introduced by Neils Bohr. It reveals the size of the atom (main energy levels). With increase in the value of ‘n’ the distance between the nucleus and the orbit also increases.
It is denoted by the letter ‘n’. It can have any simple integer value 1, 2, 3, ……. but not zero. These are also termed as K, L, M, N etc.
The radius and energy of an orbit can be determined basing on ”n” value.
The radius of n^th orbit is r_n = \(\frac{n^2 h^2}{4 \pi^2 m e^2}\)
The energy of n^th orbet is E_n = \(\frac{-2 \pi^2 m e^4}{n^2 h^2}\)

2) Azimuthal quantum number: It was proposed by Sommerfeld. it is also known as angular momentum quantum number or subsidiary quantum number.
it indicates the shapes of orbitals. It is denoted by ‘l’. The values of ‘l’ depend on the values of ‘n’, ‘l’ has values ranging from ‘0 to (n – 1) i.e.. l = 0, 1, 2,….. (n – 1). The maximum number of electrons present in the subshells s, p, d, f are 2, 6, 10, 14 respectively.

3) Magnetic quantum number: It was proposed by Lande. It shows the orientation of the orbitals in space. ‘p’ — orbital has three orientations. The orbital oriented along the x-axis is called p_x orbital, along the y-axis is called p_y -orbital and along the z-axis is called p_z orbital. In a similar way d – orbital has five orientations. They are d_xy, d_yz, d_zx, d_{x² – y²} and d_z². It is denoted by ‘m’. Its values depends on azimuthal quantum number, ‘m’ can have all the integral values from -l to +l including zero. The total number of ‘m’ values are (2l + 1).

Question 65.
Explain the dual behaviour of matter. Discuss its significance to microscopic particles like electrons.
Answer:

• The particle nature of light explains the phenomenon of blackbody radiations and photo electric effect but it couldnot explain about wave nature of light.
• Wave nature of light explains the phenomenon of interference and diffraction.
• So, light has dual nature i.e it behaves as a wave (or) as a stream of particles.
• According to de-Broglie light has dual behaviour i.e both particle and wave nature.
  de-Broglies gave the following relationship
  λ = \(\frac{\mathrm{h}}{\mathrm{mv}}\) = \(\frac{h}{p}\)
  λ = wavelength
  P = momentum
• Heisen bergs uncertainty principle also a consequence of dual behaviour of matter and radiation.
  Statement :— It is impossible to determine simultaneously, the exact momentum and exact position of a small particle like electron.
  Δx × Δp ≥ \(\frac{h}{4 \pi}\) = uncertainty in position
  Δx = uncertainty in position
  Δp = uncertainty in momentum

Significance of Uncertainty Principle:

1. This principle rules out the existence of definite paths or trajectories of electrons and other similar
   particles.
2. This principle is significant only for motion of microscopic objects and is negligible for that of macroscopic objects.
3. In dealing with milligram size or heavier objects, the associated uncertainties are hardly of any real consequence.

Question 66.
What are various ranges of electromagnetic radiation ? Explain the characteristics of electromagnetic radiation.
(or)
Explain diagrammatically the boundary surfaces for three 2p orbitais and five 3d
Answer:
Electromagnetic radiation : When electrically charged particle is accelerated alternating electric and magnetic fields are produced and transmitted. These fields are transmitted in the form of waves called electromagnetic waves or electromagnetic radiation.

Important Characteristics of a wave:

1) These are produced by oscillating charged particles in a body.
2) These radiations can pass through vacuum also. So medium for transmission is not required.
3) Velocity (c) : It is defined as the linear distance travelled by the wave in one second.
Units : cm sec^-1 (or) m see^-1
All kinds of electromagnetic waves have the same velocity.
(3 × 10⁸ m sec^-1 or 3 × 10¹⁰ cm sec^-1)

4) Wavelength (λ) : It is defined as the distance between any two successive crests or troughs of wavez.
Units:A ; m ; cm ; nm or pm 1 A° = 10^-10 m
1 nm = 10^-9 m = 10^-7 cm
1 pm = 10^-12m.

5) Frequency (v): It is defined as the number of waves passing through a point in one second.
Units: Hertz (Hz); cycles sec^-1 or sec^-1.
v = \(\frac{c}{\lambda}\)

6) Wave number (\(\bar{v}\)): It is defined as the number of waves present in one unit length. It is equal
to the reciprocal of the wavelength.
\(\bar{v}=\frac{1}{\lambda}=\frac{v}{c}\)
Relation between wavelength and frequency :
c = v × λ ⇒ v = \(\frac{c}{\lambda}\)

7) Amplitude (A) is the height of the crest (or) depth of through of a wave, It determines intensity (or) brightness of the wave.

Question 67.
Define atomic orbital. Explain the shapes of s, p and d orbitals with the help of diagrams.
Answer:
Atomic orbital : A three dimensional space around the nucleus in an atom. Where the probability of finding an electron is maximum (i.e.,) \(\Psi^2\) is maximum is called an atomic orbital.

Shapes of orbitals:

a) s – orbital : Boundary surface diagram for ‘s’ orbital is spherical in shape, ‘s – orbitals are spherically symmetric (i.e.,) the probability of finding the electron at a given distance is equal in all directions.

b) p – orbitals : p – orbital consists of two sections called lobes that are either side of the plane that passes through the nucleus. The size, shape and energy of the three orbitals are identical. They differ only in the orientation. These are mutually perpendicular to each other and oriented along x, y and z axes. Each p-orbital is of dumb-bell shape.

c) d – orbitals: Five d-orbitals are designated as d_xy, d_yz, d_zx, d_{x² – y²} and d_z². The shapes of first four d – orbitals are similar to each other of double dumb-bell whereas that of the fifth one dd_z² is different from others, but all five d-orbitals are equivalent in energy.

Question 68.
Illustrate the reasons for the stability of completely filled and half filled subshells.
Answer:
Chromium and Copper shows anamalous electronic configurations
Cr – [Ar] 4s¹ 3d⁵
Cu – [Ar] 4¹ 3d¹⁰

• Cr — gets half filled 3d— shell electronic configuration.
• Cu — gets full filled 3d— shell electronic configuration.
• Half filled and full filled subshells are more stable than others.
  Causes of Stability of Completely filled and Half filled Sub-shells
  The completely filled and half filled sub-shells are stable due to the following reasons :

1. Symmetrical distribution of electrons : It is well known that symmetry leads to stability. The completely filled or half filled subshells have symmetrical distribution of electrons in them and are therefore more stable. Electrons in the same subshell (here 3d) have equal energy but different spatial distribution, Consequently, their shielding of one another is relatively small and the electrons are more strongly attracted by the nucleus.


2. Exchange Energy: The stabilizing effect arises whenever two or more electrons with the same spin are present in the degenerate orbitals of a subshell. These electrons tend to exchange their positions and the energy released due to this exchange is called exchange energy. The number of exchanges that can take place is maximum when the subshell is either half filled or completely filled.
As a result the exchange energy is maximum and so is the stability.

The extra stability of half-filled and completely filled subshell is due to:

1. relatively small shielding,
2. smaller coulombic repulsion energy and
3. larger exchange energy.

Question 69.
Explain emission and absorption spectra. Discuss the general description of line spectra in hydrogen atom.
Answer:
Emission spectrum:

1. It is produced by analysing the radiant energy emitted by an excited substance.
2. It consists of bright lines on dark background.
3. Produced due to the emission of energy by electrons.
4. Emission spectra contains bright lines on dark background.

Absorption spectrum:

1. It is produced when white light is passed through a substance and the transmitted light is analysed by a spectrograph.
2. It consists of dark lines on bright background.
3. Produced due to the adsorption of energy by electrons.
4. Absorption spectra contains dark lines on bright background.
   Line spectra of Hydrogen — Bohrs Theory:
   • In case of hydrogen atom line spectrum is observed and this can be explained by using Bohr’s Theory.
   • According to Bohrs postulate when an electronic transition takes place between two stationary states that differ in energy is given by
     ΔE = E_f – E_i
     E_f = final orbit energy
     E_i = initial orbit energy

In terms of wave numbers

• In case of absorption spectrum n_f > n_i → energy is absorbed (+ Ve)
• In case of emission spectrum n_i > n_f → energy is emitted (- Ve)
• Each spectral line in absorption (or) emission spectrum associated to the particular transition in hydrogen atom
• In case of large no.of hydrogen atoms large no.of transitions possible they results in large no.of spectral lines.

The Spectral Lines for Atomic Hydrogen

Solved Problems

Question 1.
Calculate the num Notons, neutrons and electrons species?
Solution:
In this case, \({ }_{35}^{80} \mathrm{Br}\), Z = 35, 80, species is neutral
Number of protons = number of electrons = Z = 35
Number of neutrons = 80 – 35 = 45, Mass number (A) = number of protons (Z) + number of neutrons (n).

Question 2.
The number of electrons, protons and neutrons in a species are equal to 18, 16 and 16 respectively. Assign the proper symbol to the species.
Solution:
The atomic number is equal to number of protons =16. The element is sulphur (S).
Atomic mass number = number of protons + number of neutrons = 16 + 16 = 32
Species is not neutral as the number of protons is not equal to electrons. It is anion (negatively charged) with charge equal to excess electrons = 18 – 16 = 2.
Symbol is \(\frac{32}{16} \mathrm{~s}^{2-}\)
Note: Before using the notation find out whether the speclés is a neutral atom, acation or an anion, If it is a neutral atom, Atomic number (Z) = number of protons in the nucleus of an atom = number of electrons in a neutral atom is valid, i.e., number of protons = number of electrons = atomic number. If the species is an ion, determine whether the number of protons are larger (cation, positive ion) or smaller (anion, negative ion) than the number of electrons. Number of neutrons is always given by A-Z, whether the species is neutral or ion.

Question 3.
The Vividh Bharati station of All India Radio, Delhi, broadcasts on a frequency of 1,368 kHz (kilo hertz). Calculate the wavelength of the electro-magnetic radiation emitted by transmitter. Which part of the electromagnetic specturm does it belong to?
Solution:
The wavelength, λ, is equal to c/v. where c is the speed of electromagnetic radiation in vacuum and v is the frequency. Substituting the given values, we have

Question 4.
The wavelength range of the visible spectrum extends from violet (400 nm) to red (750 nm). Express these wave
lengths in frequencies (Hz). (1 nm = 10^-9 m).
Solution:
Using c = v λ frequency of violet light.
V = \(\frac{\mathrm{c}}{\mathrm{v}}=\frac{3.00 \times 10^8 \mathrm{~ms}^{-1}}{400 \times 10^{-9} \mathrm{~m}}\)
= 7.50 × 10¹⁴ Hz
Frequency of red light
v = \(\frac{c}{v}=\frac{3.00 \times 10^8 \mathrm{~ms}^{-1}}{750 \times 10^{-9} \mathrm{~m}}\) = 4.00 × 10¹⁴ Hz
The range of visible spectrum is from
4.0 × 10¹⁴ to 7.5 × 10¹⁴Hz in terms of frequency units.

Question 5.
Calculate
(a) wave number and
(b) frequency of yellow radiation having wavelength 5000 A.
Solution:
(a) Calculation of wavenumber \((\bar{v})\)
λ = 5800 A = 5800 × 10^-8 cm
= 5800 × 10^-10m
\((\bar{v})\) = \(\frac{1}{\lambda}=\frac{1}{5800 \times 10^{-10} \mathrm{~m}}\)
= 1.724 × 10⁶m^-1
= 1.724 × 10⁴ cm^-1

Question 6.
Calculate energy of one mole of photons of radiation whose frequency is 5 × 10¹⁴ Hz.
Solution:
Energy (E) of one photon is given by the expression
E = hv .
h = 6.626 × 10^-34 J s
v = 5 × 10¹⁴ s^-1 (given)
E = (6.626 × 10^-34J s) × (5 × 10¹⁴ s^-1)
= 3.313 × 10^-19 J
Energy of one mole of photons
= (3.313 × 10^-19J) × (6.022 × 10²³ mol^-1)
= 199.51 kJ mol^-1.

Question 7.
A 100 watt bulb emits monochromatic light of wavelength 400 nm. Calculate the number of photons emitted per second by the bulb.
Solution:
Power of the bulb = 100 watt
= 100 J s^-1
Energy of one photon E = hv = hc/λ.
= \(\frac{6.626 \times 10^{-34} \mathrm{Js} \times 3 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}}{400 \times 10^{-19} \mathrm{~m}}\)
= 4.969 × 10^-19J
Number of photons emitted
\(\frac{100 \mathrm{~J} \mathrm{~s}^{-1}}{4.969 \times 10^{-19} \mathrm{~J}}\) = 2.012 × 10²⁰ s^-1

Question 8.
When electromagnetic radiation of wavelength 300 nm falls on the surface of sodium, electrons are emitted with a kinetic energy of 1.68 × 10⁵ J mol^-1. What is the minimum energy needed to remove an electron from sodium? What is the maximum wavelength that will cause a photoelectron to be emitted?
Solution:
The energy (E) of a 300 nm photon is given by
hv = hc/λ.
= \(\frac{6.626 \times 10^{-34} \mathrm{Js} \times 3.0 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}}{300 \times 10^{-9} \mathrm{~m}}\)
= 6.626 × 10^-19 J
The energy of one mole of photons
= 6.626 × 10^-19 J × 6.022 × 10²³ mol^-1
= 3.99 × 10⁵ mol^-1
The minimum energy needed to remove one mole of electrons from sodium
= (3.99 – 1.68) 10⁵ J mol^-1
= 2.31 × 10⁵ J mol^-1
The minimum energy for one electron
= \(\frac{2.31 \times 10^5 \mathrm{~J} \mathrm{~mol}^{-1}}{6.022 \times 10^{23} \text { elelctrons } \mathrm{mol}^{-1}}\)
= 3.84 × 10^-19J
This corresponds to the wavelength
∴ λ = \(\frac{\mathrm{hc}}{\mathrm{E}}\)
= \(\frac{6.626 \times 10^{-34} \mathrm{~J} \mathrm{~s} \times 3.0 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}}{3.84 \times 10^{-9} \mathrm{~J}}\)
= 517 nm
(This corresponds to green light)

Question 9.
The threshold frequency v₀ for a metal is 7.0 × 10¹⁴ s^-1. Calculate the kinetic energy of an electron emitted when radiation of frequency v = 1.0 × 10¹⁵ s^-1 hits the metal.
Solution:
According to Einstein’s equation
Kinetic energy = 1/2 m_ev² = h(v – v₀)
= (6.626 × 10^-34 J s)
(1.0 × 10¹⁵ s^-1 – 7.0 × 10¹⁴ s^-1)
= (6.626 × 10^-34 J s)
(10.0 × 10¹⁴s^-1 – 7.0 × 10¹⁴s^-1)
= (6.626 × 10^-34 J s)
(3.0 × 10¹⁴s^-1) = 1.988 × 10^-19 J

Question 10.
What are the frequency and wave-length of a photon emitted during a transition from n = 5 state to the n = 2 state in the hydrogen atom?
Solution:
Since n₁ = 5 and n_f = 2, this transition gives rise to a spectral line in the visible region of the Balmer series. From ΔE = \(\mathrm{R}_{\mathrm{H}}\left(\frac{1}{\mathrm{n}_i^2}-\frac{1}{\mathrm{n}_{\mathrm{f}}^2}\right)\)
= 2.18 × 10^-18J\(\left(\frac{1}{n_i^2}-\frac{1}{n_f^2}\right)\)
ΔE = 2.18 × 10^-18J\(\left(\frac{1}{5^2}-\frac{1}{2^2}\right)\)
= -4.58 × 10^-19 J.
It is an emission energy.
The frequency of the photon (taking energy in terms of magnitude) is given by

Question 11.
Calculate the energy associated with the first orbit of He⁺. What is the radius this orbit?
Solution:

The radius of the orbit is given by

Question 12.
What will be the wavelength of a ball of mass 0.1 kg moving with a velocity of 10 m s^-1?
Solution:
According to de Brogue λ = \(\frac{h}{m V}=\frac{h}{p}\)
λ = \(\frac{\mathrm{h}}{\mathrm{mv}}\) = \(\frac{\left(6.626 \times 10^{-34} \mathrm{Js}\right)}{(0.1 \mathrm{~kg})\left(10 \mathrm{~m} \mathrm{~s}^{-1}\right)}\)
= 6.626 × 10^-34m (J = kg m² s^-2)

Question 13.
The mass of an electron is 9.1 × 10^-31 kg. If its K.E. is 3.0 × 10^-25 J, calculate its wavelength.
Solution:

Question 14.
Calculate the mass of a photon with wavelength 3.6 A.
Solution:
λ = 3.6 A = 36 × 10^-10 m
Velocity of photon = velocity of light
m = \(\frac{\mathrm{h}}{\lambda v}=\frac{6.626 \times 10^{-34} \mathrm{Js}}{\left(3.6 \times 10^{-10} \mathrm{~m}\right)\left(3 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}\right)}\)
= 6.135 × 10^-29kg

Question 15.
A microscope using suitable photons is employed to locate an electron in an atom within a distance of 0.1 A. What
is the uncertainty involved in the measurement of its velocity?
Solution:
∆x ∆p = \(\frac{h}{4 \pi}\) or ∆x m∆v = \(\frac{h}{4 \pi}\)
∆v = \(\frac{h}{4 \pi \Delta x m}\)
∆v = \(\frac{6.626 \times 10^{-34} \mathrm{Js}}{4 \times 3.14 \times 0.1 \times 10^{-10} \mathrm{~m} \times 9.1 \times 10^{31} \mathrm{kc}}\)
= 0.579 × 10⁷ m s^-1 (1 J = 1 kg m² s^-2)
= 5.79 × 10⁶ ms^-1

Question 16.
A golf ball has a mass of 40g and a speed of 45 m/s. If the speed can be measured within accuracy of 2%, calculate the uncertainty in the position.
Solution:
The uncertainty in the speed is 2%, i.e.,
4 × \(\frac{2}{100}\) = 0.9 ms^-1
Using the λ = \(\frac{\mathrm{h}}{\mathrm{mv}}=\frac{\mathrm{h}}{\mathrm{p}}\)
Δx = \(\frac{h}{4 \pi m \Delta v}\)
= \(\frac{6.626 \times 10^{-34} \mathrm{Js}}{4 \times 3.14 \times 40 \mathrm{~g} \times 10^{-3} \mathrm{~kg} \mathrm{~g}^{-1}\left(0.9 \mathrm{~ms}^{-1}\right)}\)
= 1.46 × 10^-33 m
This is nearly \(\sim\) 10¹⁸ times smaller than the diameter of a typical atomic nucleus. As mentioned earlier for large particles, the uncertainty principle sets no meaningful limit to the precision of measurements.

Question 17.
What is the total number of orbitals associated with the principal quantum number n = 3?
Solution:
For n = 3, the possible values of 1 are 0, 1 and 2. Thus there is one 3s orbital (n = 3, l = 0 and m_l = 0); there are three 3p orbitals
(n = 3, 1 = 1 and m_l = -1, 0, +1); there are five 3d orbitais (n = 3, l = 2 and m_l = -2, -1, 0, +1, +2).
Therefore, the total number of orbitals is 1 + 3 + 5 = 9
The same value can also be obtained by using the relation; number of orbitals = n²,
i.e. 3² = 9

Question 18.
Using s, p, d, f notations, describe the orbital with the following quantum numbers
(a) n = 2, l = 1,
(b) n = 4, l = 0,
(c) n = 5, l = 3,
(d) n = 3, l = 2
Solution:

Question 19.
Calculate its wave length of 1^st line in Balmer series of hydrogen spectrum.
Solution:
Ryd berg’s equation \(\bar{v}\) = \(\frac{1}{\lambda}\) = \(\mathrm{R}_{\mathrm{H}}\left[\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right]\)
For the 1^st line of Balmer series
n₁ = 2, n₂ = 3
R = 1,09,677 cm^-1
\(\bar{v}\) = \(\frac{1}{\lambda}\) = 1,09,677\(\left[\frac{1}{2^2}-\frac{1}{3^2}\right]\)
= 1,09,677 × \(\frac{5}{36}\)
Wave no. \((\bar{v})\) = 15232.9 cm^-1
Wave length λ = \(\frac{1}{\bar{v}}\) = \(\frac{1}{15232.9}\) = 6.5 × 10^-5 cm^-1

Question 20.
Calculate the shortest wave length in lyman series of hydrogen spectrum (R_H = 1,09,677 cm^-1).
Solution:
To calculate shortest wave length

Question 21.
What is the maximum no.of emission lines when the excited electron of a ‘H’ atom in n =6 drops to ground state.
Solution:
The no. of spectral lines found when an electron return from n^th orbit to ground state.
= \(\frac{n(n-1)}{2}\) = \(\frac{6(6-1))}{2}\) = \(\frac{30}{2}\) = 15

Question 22.
Calculate the longest wavelength transition in the paschen series of He⁺.
Solution:

Question 23.
The no. of waves in the forth Bohr’s orbit of hydrogen is
a) 3
b) 4
c) 9
d) 12
Solution:

Question 24.
It the speed of the electron in 1^st Bohr’s orbit of hydrogen is x, then the speed of the electron in the 3^rd orbit of hydrogen is
Solution:
Given
Velocity of electron in 1^st Bohr’s orbit of hydrogen = x
Velocity of electron in 3^rd Bohr’s orbit of hydrogen = \(\frac{x}{n}\) = \(\frac{x}{3}\)

Question 25.
The ratio of radii of the fifth orbits of He⁺ and Li⁺² will be
a) 2 : 3
b) 3 : 2
c) 4 : 1
d) 5 : 3
Solution:
Z₂(li)=3
Z₁ (He) = 2
\(\frac{r_1}{r_2}\) = \(\frac{Z_2}{Z_1}\) = \(\frac{3}{2}\) = 3 : 2

Question 26.
What is the lowest value of ‘n’ that allows ‘g’ orbitals to exist ?
Solution:
The lowest value of ‘n that allows ‘g’ orbitals to exist is ‘5’.

Question 27.
What is the orbital angular momentum of a d-electron
Solution:
Orbital angular momentum = \(\sqrt{l(l+1)} \frac{\mathrm{h}}{2 \pi}\)
For a d electron l = 2
= \(\sqrt{2(2+1)} \frac{h}{2 \lambda}\)
= \(\frac{\sqrt{6 h}}{2 \lambda}\)

Question 28.
What is the total spin and magnetic moment of an atom with atomic number 7’?
Solution:
Z = 7 (Nitrogen)
Electronic configuration is 1s² 2s² 2p³ (in ground state)

Question 29.
The quantum number of electrons are given below. Arrange in order of increasing energies.

Solution:
a) 4d
b) 3d
c) 4p
d) 3d
e) 3p
f) 4p
∴ Increasing order of energy
e < b = d < c = f < a

Question 30.
If the value of n + l = 7 then what should be the increasing order of energy of the possible subshells.
Solution:
Given
n + l = 7

∴ The increasing order of energy
4f < 5d < 6p < 7s
(According Aufbau principle)

Question 31.
Which of the following sets of quantum number is not permitted?
a) n = 3, l = 3, m = +1, s = +\(\frac{1}{2}\)
b) n = 3, l = 3, m = +2, s = –\(\frac{1}{2}\)
c) n = 3, l = 1, m = +2, s = –\(\frac{1}{2}\)
d) n = 3, l = 0, m = 0, s = +\(\frac{1}{2}\)
Solution:
Only d is permitted i.e., 3s¹
In a, b n = l but n always > l
in c m = + 2 is not permitted
Because l = 1,’m’ has -1, 0, +1 values only.

Question 32.
Ground state electronic configuration of nitrogenators can be represented as

Solution:
(a) and (d) are correct representations.

Question 33.
Which of the following is possible
a) 3f
b) 4d
c) 2d
d) 3p
Solution:
4d and 3p are possible.

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• Chapter 8 Ecology and Environment Notes

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AP Intermediate 1st Year Botany Notes

• Chapter 1 The Living World Notes
• Chapter 2 Biological Classification Notes
• Chapter 3 Science of Plants – Botany Notes
• Chapter 4 Plant Kingdom Notes
• Chapter 5 Morphology of Flowering Plants Notes
• Chapter 6 Modes of Reproduction Notes
• Chapter 7 Sexual Reproduction in Flowering Plants Notes
• Chapter 8 Taxonomy of Angiosperms Notes
• Chapter 9 Cell: The Unit of Life Notes
• Chapter 10 Biomolecules Notes
• Chapter 11 Cell Cycle and Cell Division Notes
• Chapter 12 Histology and Anatomy of Flowering Plants Notes
• Chapter 13 Ecological Adaptation, Succession and Ecological Services Notes

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AP Intermediate 1st Year Zoology Important Questions with Answers Chapter Wise 2022

Intermediate 1st Year Zoology Important Questions Chapter Wise 2022

• Chapter 1 Diversity of Living World Important Questions
• Chapter 2 Structural Organisation in Animals Important Questions
• Chapter 3 Animal Diversity-I: Invertebrate Phyla Important Questions
• Chapter 4 Animal Diversity-II: Phylum Chordata Important Questions
• Chapter 5 Locomotion and Reproduction in Protozoa Important Questions
• Chapter 6 Biology in Human Welfare Important Questions
• Chapter 7 Type Study of Periplaneta Americana Important Questions
• Chapter 8 Ecology and Environment Important Questions

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AP Intermediate 1st Year Zoology Study Material Pdf Download | Jr Inter 1st Year Zoology Textbook Solutions

• Chapter 1 Diversity of Living World
• Chapter 2 Structural Organisation in Animals
• Chapter 3 Animal Diversity-I: Invertebrate Phyla
• Chapter 4 Animal Diversity-II: Phylum Chordata
• Chapter 5 Locomotion and Reproduction in Protozoa
• Chapter 6 Biology in Human Welfare
• Chapter 7 Type Study of Periplaneta Americana (Cockroach)
• Chapter 8 Ecology and Environment

TS AP Inter 1st Year Zoology Weightage Blue Print 2022-2023

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TS AP Inter 1st Year Zoology Weightage to Objectives

TS AP Inter 1st Year Zoology Weightage to Form of Questions

TS AP Inter 1st Year Zoology Weightage to Content: Units

Intermediate 1st Year Zoology Syllabus

TS AP Inter 1st Year Zoology Syllabus

Unit I Zoology – Diversity of Living World

• 1.1 What is life?
• 1.2 Nature, Scope & meaning of zoology
• 1.3 Branches of Zoology
• 1.4 Need for classification- Zoos as tools for the study of taxonomy
• 1.5 Basic principles of Classification: Biological system of classification- (Phylogenetic classification only)
• 1.6 Levels of Hierarchy of classification
• 1.7 Nomenclature – Bi & Trinominal
• 1.8 Species concept
• 1.9 Kingdom Animalia
• 1.10 Biodiversity – Meaning and distribution (Genetic diversity, Species diversity, Ecosystem diversity(alpha, beta, and gamma), other attributes of biodiversity, the role of biodiversity, threats to biodiversity, methods of conservation, IUCN Red data books, Conservation of wildlife in India – Legislation, Preservation, Organisations, Threatened species.

Unit II Structural Organization in Animals

• 2.1 Levels of organization, Multicellularity: Diploblastic & Triploblastic conditions.
• 2.2 Asymmetry, Symmetry: Radial symmetry, and Bilateral symmetry (Brief account giving one example for each type from the representative phyla)
• 2.3 Acoelomates, Pseudocoelomates, and Eucoelomates: Schizo & Entero coelomates (Brief account of the formation of coelom)
• 2.4 Tissues: Epithelial, Connective, Muscular, and Nervous tissues. (make it a little more elaborative)

Unit III Animal Diversity – I: Invertebrate Phyla
General Characters – Strictly restrict to 8 salient features only
Classification up to Classes with two or three examples – Brief account only

• 3.1 Porifera
• 3.2 Cnidaria
• 3.3 Ctenophora
• 3.4 Platyhelminthes
• 3.5 Nematoda
• 3.6 Annelida (Include Earthworm as a type study strictly adhering to NCERT textbook)
• 3.7 Arthropoda
• 3.8 Mollusca
• 3.9 Echinodermata
• 3.10 Hemichordata

Unit IV Animal Diversity – II: Phylum: Chordata
General Characters – Strictly restrict to 8 points only Classification up to Classes – Brief account only with two or three examples

• 4.1 Sub phylum: Urochordata
• 4.2 Sub phylum: Cephalochordata
• 4.3 Sub phylum: Vertebrata
• 4.4 Super Class: Agnatha
• 4.4.1 Class Cyclostomata
• 4.5 Super Class: Gnathostomata
• 4.5.1 Super Class: Pisces
• 4.5.2 Class: Chondrichthyes
• 4.5.3 Class: Osteichthyes
• 4.6 Tetrapoda
• 4.6.1 Class: Amphibia (Include Frog as a type study strictly adhering to NCERT textbook)
• 4.6.2 Class: Reptilia
• 4.6.3 Class: Aves
• 4.6.4 Class: Mammalia

Unit V Locomotion & Reproduction in Protozoa

• 5.1 Locomotion: Definition, types of locomotor structures pseudopodia (basic idea of pseudopodia without going into different types), flagella & cilia (Brief account giving two examples each)
• 5.2 Flagellar & Ciliary movement – Effective & Recovery strokes in Euglena, Synchronal & Metachronal movements in Paramecium
• 5.3 Reproduction: Definition, types. Asexual Reproduction: Transverse binary fission in Paramecium & Longitudinal binary fission in Euglena. Multiple fission, Sexual Reproduction.

Unit VI Biology & Human Welfare

• 6.1 Parasitism and parasitic adaptation
• 6.2 Health and disease: introduction (follow NCERT) Life cycle, Pathogenicity, Treatment & Prevention (Brief account only)
  1. Entamoeba histolytica 2. Plasmodium vivax 3. Ascaris lumbricoides 4. Wuchereria bancrofti
• 6.3 A brief account of pathogenicity, treatment & prevention of Typhoid, Pneumonia, Common cold, & Ringworm.
• 6.4 Drugs and Alcohol abuse

Unit VII Type study of Periplaneta Americana

• 7.1 Habitat and habits
• 7.2 External features
• 7.3 Locomotion
• 7.4 Digestive system
• 7.5 Respiratory system
• 7.6 Circulatory system
• 7.7 Excretory system
• 7.8 Nervous system – sense organs, the structure of ommatidium.
• 7.9 Reproductive system

Unit VIII Ecology & Environment

• 8.1 Organisms and Environment: Ecology, population, communities, habitat, niche, biome, and ecosphere (definitions only)
• 8.2 Ecosystem: Elementary aspects only Abiotic factors – Light, Temperature & Water (Biological effects only), Ecological adaptations
• 8.3 Population interactions
• 8.4 Ecosystems: Types, Components, Lake ecosystem
• 8.5 Food chains, Food web, Productivity and Energy flow in Ecosystem, Ecological pyramids – Pyramids of numbers, biomass, and energy.
• 8.6 Nutrition cycling – Carbon, Nitrogen, & Phosphorous cycles (Brief account)
• 8.7 Population attributes Growth, Natality, and Mortality, Age distribution, and Population regulation.
• 8.8 Environmental issues

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AP Intermediate 1st Year Botany Study Material Pdf Download | Jr Inter 1st Year Botany Textbook Solutions

Unit I Diversity in the Living World

• Chapter 1 The Living World
• Chapter 2 Biological Classification
• Chapter 3 Science of Plants – Botany
• Chapter 4 Plant Kingdom

Unit II Structural Organisation in Plants – Morphology

• Chapter 5 Morphology of Flowering Plants

Unit III Reproduction in Plants

• Chapter 6 Modes of Reproduction
• Chapter 7 Sexual Reproduction in Flowering Plants

Unit IV Plant Systematics

• Chapter 8 Taxonomy of Angiosperms

Unit V Cell Structure and Function

• Chapter 9 Cell: The Unit of Life
• Chapter 10 Biomolecules
• Chapter 11 Cell Cycle and Cell Division

Unit VI Internal Organisation of Plants

• Chapter 12 Histology and Anatomy of Flowering Plants

Unit VII Plant Ecology

• Chapter 13 Ecological Adaptation, Succession and Ecological Services

TS AP Inter 1st Year Botany Weightage Blue Print 2022-2023

TS AP Inter 1st Year Botany Weightage 2022-2023 | TS AP Inter 1st Year Botany Blue Print 2022

Intermediate 1st Year Botany Syllabus

TS AP Inter 1st Year Botany Syllabus

Unit I Diversity in the Living World (30 Periods)

Chapter 1 The Living World
What is living? Diversity in the living world; Taxonomic categories and taxonomical aids.

Chapter 2 Biological Classification
Five kingdom classifications – Monera, Protista, Fungi, Plantae and Animalia, Three domains of life (six kingdom classification), Viruses, Viroids, Prions & Lichens.

Chapter 3 Science of Plants – Botany
Origin, Development, Scope of Botany and Branches of Botany.

Chapter 4 Plant Kingdom
Sailent features, classification, and alternation of generations of the plants of the following groups – Algae, Bryophytes, Pteridophytes, Gymnosperms, and Angiosperms.

Unit II Structural Organisation in Plants – Morphology (20 Periods)

Chapter 5 Morphology of flowering plants
Vegetative: Parts of a typical Angiosperimic plant; Vegetative morphology and modifications – Root, stem, and leaf – types; venation, Phyllotaxy.
Reproductive: Inflorescence – Racemose, Cymose, and special types (in brief).
Flower: Parts of a flower and their detailed description; Aestivation, Placentation.
Fruits: Types – True, False, and parthenocarpic fruits.

Unit III Reproduction in Plants (25 Periods)

Chapter 6 Modes of Reproduction
Asexual reproduction, binary fission, sporulation, budding, fragmentation, vegetative propagation in plants, sexual reproduction, in brief, Overview of the angiosperm life cycle.

Chapter 7 Sexual Reproduction in Flowering Plants
Stamen, Microsporangium/pollen grain, Pistil, megasporangium (ovule) and embryo sac; development of male and female gametophytes. Pollination types, agents, outbreeding devices, and pollen- pistil interaction Double Fertilization; Post fertilization events; Development of endosperm and embryo; development of seed, the structure of Dicotyledonous and Monocotyledonous seeds, Significance of fruit and seed Special modes Apomixis, parthenocarpy, polyembryony.

Unit IV Plant Systematics (10 Periods)

Chapter 8 Taxonomy of Angiosperms
Introduction. Types of Systems of classification (In brief). Semitechnical description of a typical flowering plant description of families: Fabaceae, Solanaceae, and Liliaceae.

Unit V Cell Structure and Function (35 Periods)

Chapter 9 Cell – The Unit of Life
Cell-Cell theory and cell as the basic unit of life-overview of the cell. Prokaryotic cells, Ultra Structure of Plant cell (structure in detail and functions in brief). Cell membrane, Cell wall, Cell organelles; Endoplasmic reticulum, Mitochondria, Plastids, Ribosomes, Golgi bodies, Vacuoles, Lysosomes, Microbodies, Centrosome and centriole, Cilia, Flagella, Cytoskeleton, and Nucleus. Chromosomes: Number, structural organization; Nucleosome.

Chapter 10 Biomolecules
Structure and function of Proteins, Carbohydrates, Lipids, and Nucletic acids.

Chapter 11 Cell Cycle and Cell Division
Cell Cycle, Mitosis, Meiosis – significance.

Unit VI Internal Organisation of Plants (25 Periods)

Chapter 12 Histology and Anatomy of Flowering Plants
Tissues – types, structure, and functions: Meristematic; Permanent tissues- Simple and Complex tissues. Tissue systems – Types, structure, and function: Epidermal, Ground, and Vascular tissue systems. Anatomy of Dicotyledonous and Monocotyledonous plant-Root, stem, and leaf. Secondary growth in Dicot stem and Dicot root.

Unit VII Plant Ecology (12 Periods)

Chapter 13 Ecological Adaptations, Succession, and Ecological Services
Introduction, Plant communities and Ecological adaptations: Hydrophytes, Mesophytes, and Xerophytes. Plant succession. Ecological services – Carbon fixation, Oxygen release, and pollination (In brief).

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AP Inter 1st Year Study Material Pdf

• Intermediate 1st Year Maths 1A Textbook Solutions
• Intermediate 1st Year Maths 1B Textbook Solutions
• AP Inter 1st Year Physics Study Material
• AP Inter 1st Year Chemistry Study Material
• AP Inter 1st Year Botany Study Material
• AP Inter 1st Year Zoology Study Material
• AP Inter 1st Year Civics Study Material
• AP Inter 1st Year Commerce Study Material
• AP Inter 1st Year Accountancy Study Material
• AP Inter 1st Year Economics Study Material
• AP Inter 1st Year Telugu Study Material
• AP Inter 1st Year English Study Material
• AP Inter 1st Year Sanskrit Study Material
• AP Inter 1st Year Hindi Study Material

AP Inter 1st Year Important Questions

• Inter 1st Year Maths 1A Important Questions
• Inter 1st Year Maths 1B Important Questions
• AP Inter 1st Year Physics Important Questions
• AP Inter 1st Year Chemistry Important Questions
• AP Inter 1st Year Botany Important Questions
• AP Inter 1st Year Zoology Important Questions
• AP Inter 1st Year Civics Important Questions
• AP Inter 1st Year Commerce Important Questions
• AP Inter 1st Year Accountancy Important Questions
• AP Inter 1st Year Economics Important Questions

AP Inter 1st Year Notes

• Inter 1st Year Maths 1A Formulas
• Inter 1st Year Maths 1B Formulas
• AP Inter 1st Year Physics Notes
• AP Inter 1st Year Chemistry Notes
• AP Inter 1st Year Botany Notes
• AP Inter 1st Year Zoology Notes
• AP Inter 1st Year Civics Notes
• AP Inter 1st Year Commerce Notes
• AP Inter 1st Year Economics Notes

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