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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 inter molecular forces experienced by the molecules of a gas are London (or) dispersion forces, Dipole – Dipole forces, Dipole- induced dipole forces, hydrogen bond.

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 it’s 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 8th Lesson Hydrogen and its Compounds Textbook Questions and Answers.

AP Inter 1st Year Chemistry Study Material 8th Lesson Hydrogen and its Compounds

Very Short Answer Questions

Question 1.
The three isotopes of hydrogen differ in their rates of reaction. Give the reasons.
Answer:
The three isotopes of hydrogen differ in their rates of reactions due to their different enthalpies of bond dissociation.

Question 2.
Why is dihydrogen used in welding of high melting metals ?
Answer:
Dihydrogen is used in welding of high melting metals because the atomic hydrogens which are formed by the dissociation of dihydrogen are recombined on the surface of the metal to be welded and produce the high temperature of 4000 K.

Question 3.
Describe one method of producing high purity hydrogen.
Answer:
Highly pure hydrogen is obtained by the electrolysis of hot aq.Ba(OH)₂ solution between nickel electrodes. The purity of hydrogen is about more than 99.95%.

Question 4.
Explain the term “SYNGAS”.
Answer:
The mixture of CO and H₂ which is used for the synthesis of methanol and a number of hydrocar-bons is called “SYNGAS”. It is also called as synthesis gas.
Preparation :

This reaction is called ‘Coal gasification’.

Question 5.
What is meant by coal gasification ? Explain with relevant, balanced equation.
Answer:
The process of producing synthesis gas (SYNGAS) by using coal at 1270K temperature is called coal gasification.
Balanced equation :

Question 6.
Define the term Hydride. How many categories of hydrides are known ? Name them.
Answer:
The binary compounds of hydrogen formed by the other elements except noble gases are called hydrides.
Hydrides are categorised into three types.

1.  Ionic hydrides
2. Covalent hydrides
3. Metallic hydrides

Question 7.
The unusual property of water in condensed phase leads to its high heat of vapourization. What is that property ?
Answer:
In water inter molecular hydrogen bonding is present. Due to this unusual property water has high Freezing point. Boiling point and high heat of vapourization.

Question 8.
During photosynthesis, water is oxidised to O₂. Which element is reduced ?
Answer:
During photosynthesis, the element reduced is carbon.

Oxidation state change from +4 to 0 [Reduction]

Question 9.
What do you mean by autoprotolysis ? Give the equation to represent the autoprotolysis of water.
Answer:
Water has the ability to behave as an acid as well as base. It behaves as an amphoteric substance. The self-ionising property of water is called auto protolysis.
The equation that represent the auto protolysis of water is as follows.

Question 10.
Water behaves as an amphoteric substance in the Bronsted sense. How do you explain ?
Answer:
According to Bronsted proton donor is acid and acceptor is base. Water has the ability to behave as an acid as well as base. So it is an amphoteric substance.
Eg :

1. Water acts as acid with ammonia.
   H₂O_(l) + NH_3(aq) ⇌ OH^Θ_(aq) + \(\mathrm{NH}_{4 \text { (aq) }}^{\oplus}\)
   Here water is proton donor.
2. Water acts as base with H₂S.
   H₂O_(l) + H₂S_(aq) ⇌ H₃O^⊕_(aq) + HS^Θ_(aq)
   Here water is proton acceptor.

Short Answer Questions

Question 1.
The boiling points of NH₃ ; H₂O and HF are higher than those of hydrides of the subsequent members of the group. Give your reasons.
Answer:
The boiling points of NH₃, H₂O and HF are higher than those of hydrides of the subsequent mem-bers of the group.
Reasons :

• NH₃, H₂O and HF are electron rich hydrides and these have 1, 2 and 3 lone pairs respectively.
• Due to the presence of lonepairs on the high electronegative elements results in the formation of hydrogen bond [i.e. intermolecular hydrogen bonding].
• Due to the formation of hydrogen bond association of molecules takes place. Hence these hydrides has high boiling points.

Question 2.
Discuss the position of hydrogen in the periodic table on the basis of its electronic configuration.
Answer:
Hydrogen is the simplest element with one electron and proton. Its electronic configuration is 1s. The configuration is responsible for its dual nature. It behaves both like alkali metals and halogens. So it can be placed along with alkali metals (IA) or along with halogens (VIIA).
Points in support of placing it in IA group :
a) Like alkali metals hydrogen also has a single electron in the outer shell 1s¹,
b) Its ability to form hydrated unipositive ion, H⁺ (aq).
c) It is quite reasonable to start the periodic table with an element having the least atomic number (Z = 1).

Points in support of placing it in VIIA group :
a) Hydrogen is a gas like fluorine or chlorine.

b) It can form diatomic molecule like halogens.

c) It has a tendency of gaining an electron and attains a stable electronic configuration of He forming H⁺ ion like halogens.
At the same time it should be noted that hydrogen has not such a great tendency to lose electron like alkali metals and gain an electron like halogens. In view of this it is difficult to assign any definite position to hydrogen. Sometimes it is placed in IA group and sometime with VIIA group.

Question 3.
How is the electronic configuration of hydrogen suitable for its chemical reactions ?
Answer:
Electronic configuration of hydrogen is 1s¹.

• The atomic hydrogen obtained from dihydrogen by the treatment with UV rays does combine with allmost all the elements.
• This atomic hydrogen successfully complete the reactions,
  a) By the loss of one electron to give H^⊕.
  b) By the gain of one electron to form H^Θ and
  c) By the sharing of electrons to form a single covalent bond.
  Eg : 1) H_2(g) + F_2(g) → 2HF_(g)
  2) 2Li_(s) + H_2(g) → 2LiH

Question 4.
What happens when dihydrogen reacts with
a) Chlorine and
b) Sodium metal. Explain.
Answer:
a) Reaction of dihydrogen with chlorine : Hydrogen reacts with chlorine to form hydrogen chloride gas. This reaction occurs in presence of sun light.

b) Reaction of dihydrogen with sodium metal : Hydrogen reacts with highly reactive metal like sodium and forms sodium hydride. The reaction occur at a high temperature.
2Na_(s) + H_2(g) → 2NaH_(s)

Question 5.
Write a note on heavy water.
Answer:
Deuterium oxide (D₂O) is known as heavy water.
Peparation : Heavy water is obtained by the exhaustive electrolysis of water.

The physical properties like molecular mass, melting point, Boiling point etc., of D₂O are heavier than watet. But dielectric constant and solubility are low for heavy water when compared to water.

Chemical properties :
a) Heavy water reacts with calcium carbide and forms Deuteroacetylene.
CaC₂ + 2D₂O → C₂D₂ + Ca(OD)₂

b) Heavy water reacts with sulphur trioxide and forms Deutero sulphuric acid.
SO₃ + D₂O → D₂SO₄

c) Heavy water reacts with Aluminium carbide and forms Deutero methane.
Al₄C₃ + 12D₂O → 3CD₄ + 4Al (OD)₃
Uses :

• It is used as a moderator in nuclear reactors to decrease the speed of neutrons.
• It is used to study the reaction mechanism in exchange reactions.
• It is used for the preparation of deuterium and deuterium compounds.

Question 6.
Name the isotopes of hydrogen. What is the ratio of masses of these isotopes ?
Answer:
Hydrogen has three isotopes.

1. Protium (₁H¹)
2. Deuterium (₁H² (or) D)
3. Tritium (₁H³ or T)

Tritium is radioactive isotope.

• The ratio of masses of these isotopes is 1 : 2 : 3 respectively for protium, deuterium and Tritium.
• Protium has no neutrons, deuterium has one neutron and tritium has two neutrons.

Question 7.
What is water – gas shift reaction? How can the production of dihydrogen be increased by this reaction?
Answer:
The mixture of CO and H₂ is called water gas. It is also called ‘Syngas’.
Water-gas shift reaction : When carbon monoxide of the syngas mixture reacted in presence of Iron chromate as catalyst then the reaction is called as water-gas shift reaction.
By using this reaction dihydrogen production can be increased.

In this reaction CO₂ gas is removed by using sodiumarsenite solution by scrubbing.

Question 8.
Complete and balance the following reactions :

Answer:

Question 9.
What is the nature of the hydrides formed by elements of 13 group ?
Answer:
Group 13 elements are belongs to p-block of periodic table.
Generally p-block elements forms covalent (or) molecular hydrides.
These molecular hydrides are further classified into 3 types

1. Electron deficient hydrides
2. Electron – precise hydrides
3. Electron – rich hydrides

Group – 13 elements forms electron deficient hydrides. Electron deficient hydrides are the molecular hydrides which have less no.of electrons to write the lewis structure.
Eg : Diborane (B₂H₆)

These hydrides acts as Lewis acids. Lewis acids accepts electron pairs and forms co-ordinate covalent bonds with donors.

Question 10.
Discuss the principle and the method of softening of hard water by synthetic, ion- exchange resins.
Answer:
Synthetic ion – exchange resins method :

• In present days, this method is mostly used for softening the hard water by using synthetic resins.
• This method is more useful than zeolite (or) permutit process.

Principle : Formation of de-ionised water (or) de mineralised water by passing the water successively through cation exchange and anion exchange resins.

De-ionised water means, water which is free from all soluble mineral salts.

Process : The de-ionization of water can be done in two steps by this process.
Step – I: Cation exchange process.
Step – II: Anion exchange process.

Step – I: Cation exchange process

In this process synthetic resins used are – So₃H group containing large organic molecule. (R-SO₃H)
Here R = organic group (or) resin anion.

• At first the synthetic resin converted to RNa by reacting it with NaCl.
• This resin i.e; RNa exchanges Ca⁺² and Mg⁺² ions of hard water and the softening of water takes place.
  2RNa(s) + \(\mathrm{M}_{(\mathrm{aq})}^{+2}\) → R₂M_(s) + 2\(\mathrm{Na}_{(\mathrm{aq})}^{+}\)
• The resin can be regenerated by using aq.NaCl solution.
• In this step H⁺ ions exchanges Na⁺, Ca⁺², Mg⁺² and proton formation takes place.
  2RH_(s) + \(\mathrm{M}_{(\mathrm{aq})}^{+2}\) ⇌ MR_2(s) + 2\(\mathrm{H}_{(\mathrm{aq})}^{+}\)

Step – II: Anion exchange process :

• In this process resins used are RNH₃OH i.e., a basic compound.
• The OH^– ions of resin exchanges the’anions Cl^–, SO₄^-2, HCO₃^– etc., and OH^Θ ion formation takes place.
  
• The H⁺ ions and OH^– ions obtained in the two steps are get neutralised to form de-ionised water.
  \(\mathrm{H}_{(\mathrm{aq})}^{+}+\mathrm{OH}_{(\mathrm{aq})}^{-} \longrightarrow \mathrm{H}_2 \mathrm{O}_{(l)}\)
• The cation exchange resins and anion exchange resins are regenerated by treatment with dil.acid and base respectively.

Question 11.
Write a few lines on the utility of hydrogen as a fuel. [Mar. 13]
Answer:
Hydrogen as a fuel :
The heat of combustion of hydrogen is high i.e about 242kj/mole. Hence hydrogen is used as industrial fuel.

• The energy released by the combustion of dihydrogen is more than the petrol (3 times).
• Hydrogen is major constituent in fuel gases like coal gas and water gas.
• Hydrogen is also used in fuel cells for the generation of electric power.
• 5% dihydrogen is used in CNG for running four-wheeler vehicles:
• By hydrogen economy principle the storage and transportation of energy in the form of liquid (or) gaseous state. Here energy is transmitted in the form of dihydrogen and not as electric power.

Question 12.
A 1% solution of H₂O₂ is provided to you. What steps do you take to prepare pure H₂O₂ from it ?
Answer:
To obtain pure H₂O₂ from the provided 1% H₂O₂ the following steps are involved.

Step -1:
The provided 1% H₂O₂ solution is carefully evoparated on a water bath under reduced pressure by distillation.
Here approximately 30% H₂O₂ solution is obtained.

Step – II :
The obtained solution in the above step is heated in a distillation flask at a low pressure of 15mm. Here approximately 85% H₂O₂ solution is obtained.

Step – III :
The above sample (obtained in the step – II) crystallised by freezing and pure H₂O₂ is obtained (≅ 100%).

Question 13.
Mention any three uses of H₂O₂ in modern times.
Answer:
Uses of H₂O₂ in modern times :

• H₂O₂ is used in modern times in Green chemistry to control the pollution.
  Eg : It is used in the treatment of domestic and industrial effluents and in the oxidation of cyanides.
• H₂O₂ is used in manufacturing of Sodium perborate and sodium per carbonates. These are used in high quality detergents.
• H₂O₂ is used in certain food products and in certain pharmaceuticals.
• H₂O₂ is used as bleaching agent to bleach paper pulp, leather etc.
• H₂O₂ is used as an antiseptic and in hair bleach.

Long Answer Questions

Question 1.
Write an essay on the commercial preparation of dihydrogen. Give balanced equations.
Answer:
Commercial methods of preparation of dihydrogen
i) From Hydrocarbons : When hydrocarbons undergo reaction with steam at high temperatures in presence of catalyst liberates hydrogen gas.
Eg : C₃H _8(g) + 3H2°(g) 3CO + 7H₂

ii) By the electrolysis of water : Acidified (or) alkaline water undergo electrolysis using platinum electrodes liberates hydrogen gas.

Highly pure hydrogen is liberated by the electrolysis of hot aq. Ba(OH)₂ by using ‘Ni’ electrodes.

iii) By Nelson’s Process : Hydrogen gas is obtained as a Byproduct by the electrolysis of brine solution. This process is mainly used for the manufacturing of NaOH, \(\mathrm{Cl}_2^{-}\) gas.
Cell Reactions : –
2NaCl → 2Na⁺ + 2Cl^–
2Cl^– → Cl₂ + 2e^– (Anode)
2H₂O + 2e^– → H₂ + 2OH^– (Cathode)
2Na⁺ + 2OH^– → 2 NaoH

iv) By water gas shift reaction : Water gas is the mixture of CO and H₂. It is also called as syngas. The process of producing syngas from coal is called as coal gasification.

The dihydrogen production can be increased by the reaction of steam with syngas in presence of catalyst.

This reaction is called as water gas shift reaction.

Question 2.
Illustrate the chemistry of dihydrogen by its reaction with
i) N₂
ii) Metal ions and metal oxides and
iii) Organic compounds. How is dihydrogen used in the manufacture of chemicals ?
Answer:
i) Reaction with N₂: Dihydrogen reacts with nitrogen in presence of iron catalyst to form ammonia.

This process is called as Haber’s process. Here temperature used is around 700K and pressure used is 200 atm.

ii) a) Reaction with metal ions : Hydrogen reduces metal ions in aqueous solution into metals.
H_2(g) + \(\mathrm{Pd}_{(\mathrm{aq})}^{+2}\) → Pd_(s) + \(2 \mathrm{H}_{\text {(aq) }}^{+}\)

b) Reaction with metal oxides : Hydrogen reduces metal oxides into metals.
WO₃ + 3H₂ → W + 3H₂O

iii) Reaction with organic compounds :
a) Vegetable oils undergoes hydrogenation in presence of ‘Ni’ catalyst and forms vanaspathi.
b) Alkenes undergo hydro formylation and forms aldehydes. These aldehydes undergo reduction to form alcohols.
CH₂ = CH₂ + CO + H₂ → CH₃ – CH₂CHO (Aldehyde)
CH₃CH₂CHO + H₂ → CH₃CH₂CH₂OH (Alcohol )

Use of Dihydrogen in the manufacture of chemicals
Dihydrogen is used in the manufacturing of industrial cherfiicals like methanol, ammonia, hydrogen chloride etc.

Question 3.
Explain, with suitable examples, the following :

1. electron deficient
2. electron – precise and
3. electron – rich hydrides.

Answer:
Molecular hydrides (or) covalent hydrides are formed by the p-block elements.

These molecular hydrides are divided into three types.

1. Electron deficient hydrides
2. Electron – precise hydrides
3. Electron – rich hydrides

1) Electron – deficient hydrides : These are the molecular hydrides in which the available no.of valency electrons is less than the number-required for normal covalent bond formation.
(or) .
These are the molecular hydrides in which the available valency electrons are lessthan the required for writting the Lewis structure of the molecule.
Eg : (AlH₃)_n, B₂H₆ etc.
These hydrides acts as Lewis acids i.e. electron pair acceptors. These forms dative bond with donors.

2) Electron – precise hydrides : These are the molecular hydrides in which all the valency electrons of the central atom are involved in bond formation.
(or)
These are the molecular hydrides which contains the required no.of valency electrons to write the Lewis structure of the molecule.
Eg : Group 14 elements forms this type of hydrides CH₄, C₂H₆ etc., are examples of this type.
These have tetrahedral geometry.

3) Electron – rich hydrides :
These are the molecular hydrides in which the valency electrons on the central atom are more than that are required for bond formation.
(or)

• These are the molecular hydrides in which the available valency electrons are more than the required for writting the Lewis structure of the molecule.
• These hydrides contains lone pairs on central atoms.
  Eg :
  These hydrides have high boiling points than those of the hydrides of the subsequent members of group because of hydrogen bond formation.

Question 4.
Write in brief on
i) ionic hydrides
ii) interstitial hydrides.
Answer:
i) Ionic hydrides : These are also called as saline hydrides (or) salt like hydrides.
* These are the hydrides formed by combining di hydrogen with s-block elements (Electro posi-tive elements).
* These are stoichiometric compounds.
Eg : LiH, NaH, CaH₂ etc.
NaH is formed by the direct union of Na and H₂

Physical properties :

• These hydrides are crystalline.
• These hydrides are non volatile and non – conducting in solidstate.
• These conducts electricity in moltenstate.
• These have high melting points.

Chemical properties :
These hydrides on electrolysis, liberates dihydrogen gas at anode.
2H^– → H_2(g) + 2e^– [Anode]
Lithium hydride is used in the synthesis of other useful hydrides like LiAlH₄ and LiBH₄.
8LiH + Al₂CL₆ → 2LiAlH₄ + 6LiCl
2LiH + B₂H₆ → 2LiBH₄
2NaH + B₂H₆ → 2NaBH₄
These hydrides react with acids and water vigorously and liberate dihydrogen.
LiH + H₂O → LiOH + H₂ ↑

ii) Interstitial hydrides : These are the hydrides formed by the reaction of hydrogen with d-block and f-block elements. These are also called as metallic hydrides.
Eg : CrH, CrH₂, ZnH₂, ThH₂

• The hydrogen of the metallic hydride occupy the intersticies of metallic lattice. Hence these are called as Interstitial hydrides.
• These are non-stiochiometric compounds
  Eg : TiH_1.5-1.8 LaH_2.87 etc-
• In these hydrides law of constant composition does not found.
• Metals of group 7, 8 and 9 do not form hydrides and in group 6 only chromium forms hydrides.
• The conductivity of these hydrides is less than the parent metals.
• The formation of metallic hydrides and their capacity to release hydrogen at high temperature are utilised in the purification of H₂.
• Some metals can accomodate a very large volume of hydrogen and acts as storage media.

Question 5.
Explain any four of the chemical properties of water.
Answer:
i) Hydrolysis Reaction : The chemical interaction of a compound with water is called as hydrolysis. -» Water has high hydrating ability because of high dielectric constant.
Covalent as well as ionic compounds undergo hydrolysis.
Eg: P₄O_10(s) + 6H₂O_(l) → 4H₃PO_4(aq) (orthophosphoric acid)
NCl₃ + 3H₂O → NH₃ + 3HOCl
SiCl₄ + 2H₂O → SiO₂ + 4HCl

ii) Formation of hydrogen : Water can be reduced to dihydrogen by reacting with highly elec-tropositive metals.
2Ma_(s) + 2H₂O_(l) → 2NaOH_(aq) + H_2(g)
Water is an important source of dihydrogen.

iii) Photosynthesis reaction : During photo synthesis reaction water gets oxidised to O₂.
6CO_2(g) + 6H₂O_(l) → C₆H₁₂O_6(aq) + 6O_2(g)

iv) Hydrates formation : By the association of water molecules of different types hydrated salts formed from the crystallisation of salts.
Eg : BaCl₂.2H₂O, CuSO₄. 5H₂O

v) Amphoteric Nature : Water has the ability to act as an acid as well as base. It is an amphoteric substance.
Eg: i) H₂O_(l) + NH_3(aq) ⇌ OH^–_(aq) + NH₄⁺_(aq)
In this reaction H₂O acts as Bronsted acid.
ii) H₂O_(l) + H₂S_(aq) ⇌ H₃O⁺_(aq) + HS^–_(aq)
In this reaction H₂O acts as Bronsted base.

Question 6.
Explain the terms hard water and soft water. Write a note on the
i) ion-exchange method and
ii) calgon method for the removal of hardness of water.
Answer:
Hard water : Water does not give lather readily with soap is called hard water.
Hard water contains hardness. This hardness is due to presence of Ca, Mg soluble salts.
Presence of Ca, Mg – bicarbonates causes temporary hardness.
Presence of Ca, Mg – chlorides, sulphates causes permanent hardness.

Soft water: Water which give lather immediately with soap is called soft water.
(or)
Water which is free from soluble salts of Ca, Mg is called soft water.

i) Ion – Exchange method :
This method is useful to remove the permanent hardness of water.

• This method is also named as permutil (or) zeolite process.
• Permutit is the artificial zeolite, j.e sodium aluminium orthosilicate. (Na₂Al₂Si₂O₈xH₂O (or) NaAlSiO₄)
• Permutit is written in short form as Naz.
• When permutit is added to hard water, the following ion-exchange reactions takes place.
  
• Caz₂ and Mgz₂ are called as exhausted permutit. These are regenerated to permutit by the treatment with brine solution.
  \(\mathrm{Caz}_{2_{(5)}}+2 \mathrm{Na}_{(\mathrm{aq})}^{+} \longrightarrow 2 \mathrm{Naz}_{(\mathrm{s})}+\mathrm{Ca}_{(\mathrm{aq})}^{+2}\)

ii) Calgon process : [A.P. Mar. 16]

• Calgon is sodium hexametaphosphate. [Na₆P₆O₁₈ (or) (NaPO₃)₆]
• Calgon does not precipitate the Ca (or) Mg – salts but removes Ca⁺² and Mg⁺² ions from water. -4 The removal of Ca⁺² (or) Mg⁺² ions from water may takes place either by adsorption (or) by complex formation.
• Reactions :
  

Question 7.
Write the chemical reaction to justify that hydrogen peroxide can function as on oxidizing as well as reducing agent. [T.S. Mar. 16]
Answer:

• H₂O₂ has the ability to function as an oxidising agent as well as reducing agent in both acid and alkaline solutions.
• In H₂O₂ oxidation state of oxygen is -1. It oxidised to O₂. Here H₂O₂ is reductant.
• H₂O₂ can be reduced to H₂O (or) OH^–. Here H₂O₂ is oxidant.

1. Oxidising action in acidic medium :
   \(2 \mathrm{Fe}_{(a q)}^{+2}+2 \mathrm{H}_{(\mathrm{aq})}^{+}+\mathrm{H}_2 \mathrm{O}_{2_{(\text {aq })}} \longrightarrow 2 \mathrm{Fe}_{(\text {aq })}^{+3}+2 \mathrm{H}_2 \mathrm{O}_{(l)}\)
2. Reducing action in acidic medium :
   \(2 \mathrm{MnO}_4^{-}+6 \mathrm{H}^{+}+5 \mathrm{H}_2 \mathrm{O}_2 \longrightarrow 2 \mathrm{Mn}^{+2}+8 \mathrm{H}_2 \mathrm{O}+5 \mathrm{O}_2\)
3. Oxidising action in basic medium :
   2Fe⁺² + H₂O₂ → 2Fe⁺³ + 2OH^–
4. Reducing action in basic medium :
   \(2 \mathrm{MnO}_4^{-}+3 \mathrm{H}_2 \mathrm{O}_2 \longrightarrow 2 \mathrm{MnO}_2+3 \mathrm{O}_2+2 \mathrm{H}_2 \mathrm{O}+2 \mathrm{OH}^{-}\)

Question 8.
Complete and balance the following chemical reactions :
i) PbS_(s) + H₂O₂ →
ii) \(\mathrm{MnO}_{4_{(a q)}}^{-}\) + H₂O_2(aq) →
iii) CaO_(s) + H₂O_(g) →
(iv) Ca₃N_2(s) + H₂O_(l) →
Classify the above into
a) hydrolysis
b) redox and
c) hydration reactions.
Answer:

iii) CaO_(s) + H₂O_(g) → Ca(OH)₂
(iv) Ca₃N_2(s) + H₂O_(l) → 3Ca(OH)₂ + 2NH₃
(iii) and (iv) reactions are hydrolysis reactions

Question 9.
Discuss, with relevant chemical equations, various methods of preparing hydrogen peroxide. Which of these methods is useful to prepare D₂O₂ ?
Answer:
Preparations of H₂O₂:
i) From Acidified BaO₂ removing the excess of water by evaporation under reduced pressure gives H₂O₂.
BaO₂ . 8H₂O_(s) + H₂SO_4(aq) → BaSO_4(s) + H₂O_2(aq) + 8H₂O_(l)

ii) Auto oxidation method : H₂O₂ is prepared industrially by the auto oxidation of 2-ethyl anthraquinol.

iii) Electrolysis of peroxo disulphuric acid : H₂O₂ is obtained by the electrolysis of 50% H₂SO₄ solution, peroxo disulphuric acid obtained undergo hydrolysis to form H₂O₂.

Preparation of D₂O₂: D₂O₂ is obtained by the reaction of K₂S₂O₈ with heavy water. This is similar to above method (iii).
K₂S₂O_8(s) + 2D₂O_(l) → 2KDSO_4(aq) + D₂O_2(l)

Question 10.
In how many ways can you express the strength of H₂O₂ ? Calculate the strength of 15 volume solution of H₂O₂ in g/l. Express this strength in normality and molarity.
Answer:
Strength of H₂O₂ can be expressed majorly in two ways.
H₂O₂ strength can be expressed in terms of i) Molarity ii) Normality.
Problem :
Solution:
15 volume H₂O₂ solution means 1 lit, of H₂O₂
will give 15 lit, of O₂ at STP

22.4 lit, of 02 produced from 68 gms of H₂O₂
15 lit, of 02 produced from x gms of H₂O₂
x = \(\frac{68 \times 15}{22.4}\) = 45.53 gms. of H₂O₂
∴ Strength of 15 volume H₂O₂ = 45.53 gms/lit.
= 4.5% H₂O₂

→ Molarity of the 15 volume H₂O₂ solution
= \(\frac{\mathrm{Wt}}{\mathrm{GMW} \times 1 \text { lit. }}\)
= \(\frac{45.53}{34 \times 1}\) = 1.339M.
→ Normality of the 15 volume H₂O₂ solution
= \(\frac{W t}{\text { G.E.W } \times 1 \text { lit. }}=\frac{45.53}{17 \times 1}\) = 2.678 N.

Solved Problems

Question 1.
Comment on the reactions of dihydro-gen with

1. chlorine
2. sodium and
3. copper (II) oxide.

Solution:

1. Hydrogen reacts with chlorine to form hydrogen chloride. An electron pair is shared between H and Cl leading to the formation of a covalent molecule.
2. Hydrogen is reduced by sodium to form NaH. An electron is transferred from Na to H leading to the formation of an ionic compound, Na⁺ H .
3. Hydrogen reduces copper(ll) oxide to copper and itself gets oxidised to H₂O.

Question 2.
H₂O has a higher boiling point than that of H₂S. Give reasons.
Solution:
On the basis of molecular mass of H₂O. Its boiling point is expected to be lower than that of H₂S. However, due to higher electronegativity of O, the magnitude of hydrogen bonding in H₂O will be quite appreciable. Hence, the boiling point of H₂O will be higher than that of H₂S.

Question 3.
How many hydrogen – bonded water molecule(s) are associated in CuSO₄. 5H₂O ?
Solution:
Only one water molecule, which is outside the brackets (coordination sphere), is hydrogen-bonded. The other four molecules of water are coordinated.

Question 4.
Calculate the strength of 10 volume solution of hydrogen peroxide.
Solution:
10 volume solution of H₂O₂ means that 1L of this H₂O₂ solution will give 10 L of oxygen at STP.

On the basis of above equation 22.4 L of O₂ is produced from 68 g H₂O₂ at STP 10 L of O₂ at STP is produced from \(\frac{68 \times 10}{22.4}\) g = 30.36g
≈ 30g H₂O₂
Therefore, strength of H₂O₂ in 10 volume H₂O₂ solution = 30.36 g/L = 3% H₂O₂ solution.

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 outer most 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.
The tendency of an atom to achieve eight electrons in their outer most shell is known as 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 9th Lesson The s-Block Elements Textbook Questions and Answers.

AP Inter 1st Year Chemistry Study Material 9th Lesson The s-Block Elements

Very Short Answer Questions

Question 1.
Give reasons for the diagonal relationship observed in the periodic table.
Answer:

• Diagonal relationship is due to similar sizes of atoms (or) ions
• Diagonal relationship is due to similar electro negativities of the respective elements. Diagonally similar elements possess same polarising power.
• Polarizing Power = \(\frac{\text { ionic charge }}{\text { (ionic radius) }^2}\)

Question 2.
Write completly the electronic configurations of K and Rb.
Answer:
The electronic configuration of ‘K’ is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s²
The electronic configuration of ‘Rb’ is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s¹.

Question 3.
Lithium salts are mostly hydrated. Why ?
Answer:
Hydration enthalpy of Li⁺ ion is very high. It has very high degree of hydration. So ‘Li’ salts are mostly hydrated.
Eg : LiCl . 2H₂O.

Question 4.
Which of the alkali metals shows abnormal density ? What is the order of the varia-tion of density among the IA group elements.
Answer:

• ‘K’ has abnormal density among alkalimetals due to high inter atomic distances in crystal lattice.
• The order of the variation of density among the IA group elements as follows.
  Li < Na > K < Rb < Cs.

Question 5.
Lithium reacts with water less vigorously than sodium. Give your reasons.
Answer:
Lithium reacts with water less vigorously than sodium.

Reasons :

• Lithium has small size.
• Lithium has very high hydration energy.

Question 6.
Lithium Iodide is the most covalent among the alkali metal halides. Give the reasons.
Answer:
Lithium iodide is the most covalent among the alkalimetal halides.

Reasons :

• The polarising capability of lithium ion is high.
• Li⁺ ion has very small size.
• Li⁺ ion has high tendency to distort electron cloud around the iodide ion.

Question 7.
In what respects lithium hydrogen carbonate differs from other alkali metal hydrogen carbonates.
Answer:
Lithium hydrogen carbonate cannot exist in solid form but remaining alkali metal hydrogen carbonates exist as solids.

Question 8.
Write the complete electronic configurations of any two alkaline earth metals.
Answer:

1. The electronic configuration of ‘Mg’ is 1s² 2s² 2p⁶ 3s²
2. The electronic configuration of ‘Ca’ is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s².

Question 9.
Tell about the variation of m.pts., and b.pts among the alkaline earth metals.
Answer:

1. The m.pts and b.pts of alkaline earth metals are higher than corresponding alkali metals due to smaller sizes.
2. Due to low I.P. values the variation of m.pts and b.pts among alkaline earth metals is not sys-tematic.

Question 10.
What are the characterstic colours imparted by the HA elements ?
Answer:
Elements – Imparted colours towards flame
Calcium – Brick red
Strontium – Crimson red
Barium – Apple green
Beryllium – No colour
Magnesium – No colour

Question 11.
What happens when magnesium metal is burnt in air ?
Answer:
Magnesium metal burns with dazzling brilliance in air to give MgO and Mg₃N₂.
2 Mg + O₂ → 2 MgO
3 Mg + N₂ → Mg₃N₂.

Question 12.
Lithium carbonate is not so stable to heat as the other alkali metal carbonates. Explain.
Answer:
Lithium carbonate is not so stable to heat because Lithium has very small size and it polarises the large CO₃^-2 ion which leads to the formation of more stable Li₂O and CO₂.
As the electro positive character increases down the group, the stability of carbonats increase.

Question 13.
Write a balanced equation for the formation of ammoniated IIA metal ions from the metals in liquid ammonia.
Answer:
Alkaline earth metals dissolve in liquid ammonia to give deep blue black solutions forming ammoniated ions.
M + (x + y) NH₃ → [M(NH₃)_x]²⁺ + 2 [e(NH₃)_y]^–
From the above solutions, ammoniates [M(NH₃)_x]²⁺ can be recovered.

Question 14.
The fluorides of alkaline earth metals are relatively less soluble than their respective chlorides in water. Why ?
Answer:
Because of their high lattice energies fluorides of alkaline earth metals are relatively less soluble than their respective chlorides in water.

Question 15.
What happens when hydrated Mg (NO₃)₂ is heated ? Give the balanced equation.
Answer:
When hydrated Mg(NO₃)₂ is heated, it first loses the six water molecules and on further heating
decomposes to give the oxide.
2 Mg (NO₃)₂ → 2 MgO + 4NO₂ + O₂.

Question 16.
Why does the solubility of alkaline earth metal hydroxides in water increases down the group ?
Answer:
Among alkaline earth metal hydroxides, the anion being common the cationic radius will influence the lattice enthalpy. Since lattice enthalpy decreases much more than the hydration enthalpy with increasing ionic size, the solubility increases as we go down the group.

Question 17.
Why does the solubility of alkaline earth metal Carbonates and sulphates in water decrease down the group ?
Answer:
The size of anions being much larger compared to cations, the lattice enthalpy will remain almost constant within a particular group. Since the hydration enthalpies decrease down the group, solubility will decrease as found for alkaline earth metal carbonates and sulphates.

Question 18.
Write the average composition of Portland cement.
Answer:
Composition of port land cement is
Cao – 50 – 60%
Sio₂ – 20 – 25%
Al₂O₃ – 5 – 10%
Mgo – 2 – 3%
Fe₂O₃ – 1 – 2% and
SO₂ – 1 – 2%

Question 19.
Why is gypsum added to cement ?
Answer:
Gypsum is added to cement to slow down the process of setting of the cement and to get sufficiently hardened cement.

Question 20.
Why are alkali metals not found in the free state in nature ? [Mar. 13]
Answer:
Alkali metals are not found in the free state in nature because they readily lose their valency electron to form M⁺ ion (a nonvalent ion).

Question 21.
Potassium carbonate cannot be prepared by Solvay process. Why ?
Answer:
Potassium carbonate cannot be prepared by solvay process because potassium bi carbonate is more soluble and to be precipitated by the addition of ammonium bi carbonate to a saturated solution of potassium chloride.

Question 22.
Describe the important uses of caustic soda.
Answer:
Uses:

• It is used in petrol refining
• It is used in the purification of bauxite.
• It is used in manufacturing of soap, paper.
• It is used in manufacturing of antificial silk.
• It is used in manufacturing of so many chemically.
• It is used in textile industries for mercerising cotton fabrics.
• It is used in preparation of pure fats and oils.
• It is used in as laboratory reagent.

Question 23.
Describe the important uses of sodium carbonate.
Answer:
Uses:

• Na₂CO₃ is used in the manufacturing of glass.
• Na₂CO₃ is used in the manufactuing of borax, caustic soda.
• Na₂CO₃ is used in paper, paints and textile industries.
• Na₂CO₃ is used in softening of water.
• Na₂CO₃ is used in laundries.
• Na₂CO₃ is used an important laboratory reagent both in qualitative and quantitative analysis.

Question 24.
Describe the important uses of quick line.
Answer:
Uses:

• Quick lime is used in the purification of sugar.
• Quick lime is used in the manufacture of dyestuffs.
• Quick lime is used in the manufacture of Na₂CO₃ from NaOH.
• It is an important material for manufacturing of cement and it is the cheapest form of alkali.

Question 25.
Draw the structures of

1. BeCl₂ (vapour) and
2. BeCl₂ (Solid).

Answer:

1. In vapour phase BeCl₂ forms a bridged dimer which disociates into monomer at high temperatures (around 1200 k)
   Cl – Be – Cl
2. In solid state BeCl₂ has a chain structure.
   

Question 26.
Describe the importance of Plaster of Paris.
Answer:

1. Plaster of paris has an important property of setting with water.
2. It forms a hard solid in 5 to 15 min. When it is mixed with suitable quantity of water.
3. It is majorly used in building industry and as well as plasters.
4. It is used in the bone fractures (or) sprain conditions.
5. It is used in dentistry.
6. It is used in manufacturing status and busts.

Question 27.
Which of the alkaline earth metal carbonates is thermally the most stable ? Why ?
Answer:
Among Alkaline earth metal carbonates BaCO₃ is thermally most stable.
Reason :
As the cationic size increases thermal stability also increases. So BaCO₃ is most stable thermally.

Question 28.
Write balanced equations for the reactions between

1. Na₂O₂ and water
2. K₂O and water.

Answer:

1. Na₂O₂ + 2H₂O → 2 NaOH + H₂O₂
2. K₂O + H₂O → 2 KOH.

Short Answer Questions

Question 1.
Alkali metals and their salts impart characteristic colours to an oxidizing flame. Explain the resonizing flame. Explain the reason.
Answer:
Alkali metals and their salts impart characterstic colours to an oxidizing flame.

Reasons :
The heat from the flame excites the outer most orbital electron to a higher energy level. When the excited electron emitts the radiation and comes back to the ground state. This falls in the visible region.

Question 2.
What makes caesium and potassium useful as electrodes in photoelectric cells ?
Answer:

• Alkali metals can be detected by the respective flame tests and can be determined by the flame photo metry (or) atomic absorption spectroscopy.
• Alkali metals when irradiated with light, the light energy absorbed may be sufficient to make an atom lose electron.
• This makes caesium and potassium useful as electrodes in photo electric cells.

Question 3.
Write a short note on the reactivity of alkali metals towards air.
Answer:

• The alkali metals forms their oxides in presence of dry air and tarnished.
• These oxides reacts with moisture to form hydroxides.
• They burn vigorously in oxygen and forms oxides.
  • Lithium forms Lithium monoxide.
  • Sodium forms monoxide with limited supply of oxygen and peroxide with excess of oxygen.
  • Other metals of this group forms super oxides. The super oxide ion (O₂^–) is stable only in presence of large cations.

Reactions :
4Li + O₂ → 2 Li₂O
4Na + O₂ (Limited) → 2 Na₂O₂
2Na + O₂ (Excess) → Na₂O₂
K + O₂ (Excess) → KO₂
Lithium shows a different character. It reacts directly with nitrogen of air and forms Li₃N (Lithium nitride).

Question 4.
Give any two uses for each of the following metals.

1. Lithium
2. Sodium.

Answer:

1. Uses of Lithium : –

• ‘Li’ metal is used to make alloys.
  Eg: 1) Lithium with lead forms an alloy which is used for making white metal bearings for motor engines.
  2) Lithium with aluminium forms alloys which are used to make air craft parts.
• ‘Li’ metal is used in thermo nuclear reactions.
• ‘Li’ metal is used to make electro chemical cells.

2) Uses of sodium : –

• Sodium metal is used to make Na and Pb alloy needed to make TEL. this TEL (tetra ethyl lead) is used as antiknock additives to petrol.
• Liquid ‘Na’ metal is used as a coolant in fast breeder nuclear reactors.
• ‘Na’ metal is used in the manufacturing of rabber.

Question 5.
Give an account of properties of washing soda.
Answer:
Properties of washing soda : –

• Na₂CO₃ is a white crystalline solid.
• Na₂CO₃ exists as a decahydrate Na₂CO₃. 10H₂O which is called washing soda.
• Na₂CO₃ is readily soluble in water.
• Na₂CO₃ (deca hydrate) when heated it loses the water molecules and forms monohydrate. This monohydrate on heating above 373 K it forms anhydrous form which is called soda ash, a white powder.
  Reactions : –
  
• Na₂CO₃ (aq) solution is alkaline (basic) in nature Because it under goes anionic hydrolysis. (PH > 7).
  CO₃^-2 + H₂O → HCO₃^– + OH^–

Question 6.
Mention some uses of sodium carbonate.
Answer:
Uses :

• Na₂CO₃ is used in the manufacturing of glass.
• Na₂CO₃ is used in the manufacturing of borax, caustic soda.
• Na₂CO₃ is used in paper, paints and textile industries. .
• Na₂CO₃ is used in softening of water.
• Na₂CO₃ is used in laundries.
• Na₂CO₃ is used an important laboratory reagent both in qualitative and quantitative analysis.

Question 7.
How do you obtain pure sodium chloride from a crude sample ?
Answer:

1. Crude sodium chloride is obtained by the crystallisation of brine solution.
2. Crude sodium chloride contains sodium sulphate, calcium sulphate, calcium chloride and magnesium chloride.
3. CaCl₂ and MgCl₂ are the impurities in the Crude NaCl because these absorb moisture easily from the atmosphere.
4. Pure sodium chloride’s obtained by dissolving Crude NaCl in minimum amount of water and filtered to remove insoluble impurities.
5. This solution is saturated with HCl gas. Then crystals of pure NaCl are separated out.
6. Ca and Mg chloride are more soluble in solution than NaCl and these remained in the solution.

Question 8.
What do you know about Castner – Kellner process ? Write the principle involved in it.
Answer:

• Castner – Kellner process is a commercial method used for the preparation of sodium hydroxide.
• In this process sodium hydroxide is prepared by the electrolysis of sodium chloride in Castner – Kellner cell.
• Brine solution is electrolysed using a mercury cathode and a carbon anode.
• Sodium metal is formed at cathode and it combine with mercury to form sodium amalgam. Chlorine gas is evolved at anode.
• The amalgam is treated with water to form, sodium hydroxide.
  Cell Reactions :
  2NaCl → 2Na⁺ + 2Cl^–
  2Na⁺ + 2e^– 2Na – amalgam
  2Cl^– → Cl₂ + 2e^–
  2Na – amalgam + 2H₂O → 2NaOH + 2Hg + H₂
• This process is also called as mercury cathode process.

Question 9.
Write a few applications of caustic soda.
Answer:
Uses :

• It is used in petrol refining
• It is used in the purification of bauxite.
• It is used in manufacturing of soap, paper.
• It is used in manufacturing of antificial silk.
• It is used in manufacturing of so many chemically.
• It is used in textile industries for mercerising cotton fabrics.
• It is used in preparation of pure fats and oils.
• It is used in as laboratory reagent.

Question 10.
Give an account of the biological importance of Na⁺ and K⁺ ions.
Answer:

• Na⁺ ions participate in the transmission of nerve signals.
• Na⁺ ions regulates the flow of water accross cell membranes.
• Na⁺ ions responsible for transport of sugars and amino acids into cells.
• K⁺ ions are useful in activating enzymes.
• K⁺ ions participate in the oxidation of glucose to produce ATP.
• K⁺ along with Na+ responsible for the transmission of nerve signals.

Question 11.
Mention the important uses of Mg metal.
Answer:

1. Magnesium forms so many useful alloys with Al, Zn, Mn and Sn.
2. Mg – Al alloys are useful in air – craft construction.
3. Mg powder and ribbon is used in flash powders bulbs.
4. Mg is used in incendiary bombs and signals.
5. Milk of magnesice is used as antacid in medicine.
6. MgCO₃ is the main ingradient in tooth pastes.

Question 12.
Show that Be(OH)₂ is amphoteric in nature.
Answer:

1. Be(OH)₂ is amphoteric in nature. This can be evidented by the following reactions.
2. Be(OH)₂ reacts with both acids and alkalis.
   Be(OH)₂ + 2OH^– [Be(OH)₄]^2- (Beryllation)
   Be(OH)₂ + 2HCl + 2H₂O → [Be(OH)₄]Cl₂
3. Hence Be(OH)₂ is amphoteric in nature.

Question 13.
Write a note on anomalous behaviour of beryllium.
Answer:
Anomalous characters of Be:
As was already discussed in the earlier sections, the first element shows some differences from the properties of the other elements in the group. Be differs from the other alkaline earth metals because of its small size and high electronegativity. Be differs from the other elements in the following aspects.

1. Be compounds are predominantly covalent due to its high polarizing power and its salts are readily Hydrolyzed.
2. Be is not easily affected by dry air and does not decompose water at ordinary temperature.
3. Be is an amphoteric metal. It dissolves in alkali solutions forming beryllates.
4. Be SO₄ is soluble in water whereas the sulphates of Ca, Sr and Ba are not soluble.
5. Be and its salts do not respond to flame test while Ca, Sr and Ba give characteristic flame colours.
6. Be forms many complexes while the heavier elements do not show a great tendency to form complexes.
7. Be has a maximum covalency of 4 while other can have a maximum covalency of 6.

Question 14.
Be shows diagonal relationship with Al. Discuss.
Answer:

1. ‘Be’ shows diagonal relation ship with ‘Al’.
2. The ionic radius of Be⁺² is nearly same as that of Al⁺³ so ‘Be’ resembles ‘Al’ in several ways.
3. Al, Be both not readily reacts with acids. This is due to the presence of an oxide film on the surface of metal.
4. Al(OH)₃, Be(OH)₂ both similarly dissolves in excess of alkali and forms Beryllate ion [Be(OH)₃]^2- and Aluminate [Al(OH)₄] ions respectively.
5. The chlorides of Be, Al have bridged chloride structures in vapour phase.
6. Both the chlorides of Be, Al used as strong Lewis acids.
7. Both the chlorides of Be, Al used in Friedal craft catalysts.
8. Be, Al ions have strong tendency to form complexes.

Question 15.
What is Plaster of Paris ? Write a short note on it. [T.S. Mar. 16]
Answer:
Plaster of paris is the hemi hydrate of CaSO₄ with formula CaSO₄. \(\frac{1}{2}\)H₂O.
Preparation: –
Plaster of paris is obtained by heating gypsum at 393 K.

1. If temperature is used greater than 393 K then an hydrous CaSO₄ is formed which is called ‘dead burnt plaster’.
2. Plaster of paris has an important property of setting with water.
3. It forms a hard solid in 5 to 15 min. when it is mixed with suitable quantity of water.
4. It is majorly used in building industry and as well as plasters.
5. It is used in the bone fractures (or) sprain conditions.
6. It is used in dentistry.
7. It is used in manufacturing status and busts.

Question 16.
In what ways lithium shows similarities to magnesium in its chemical behaviour ?
Answer:
Diagonal relationship of Li : In the periodic table an element of a group in the 2^nd period shows similar properties with the second element of the next group in the third period. This relation is known as diagonal relationship. For examples. Lithium and Magnesium show diagonal relationship. The elements show that diagonal relationship have similar polarizing powers, electronegativities, nature of the compounds. The diagonal similarity may be due to the effects of size and charge. For example, charge per unit area.
Lithium shows similarity to Magnesium in the following respects.
a) Lithium is slow to react with water. Magnesium decomposes water only in the hot condition.
2Li + 2H₂O → + H₂;
Mg + 2H₂O → Mg(OH)₂ + H₂

b) Lithium combines directly with N2 forming nitride.
6Li + N₂ → 2Li₃N

c) Both Lithium and Magnesium give only monoxides Li₂O, MgO.

d) Lithium chloride is deliquescent like MgCl₂, LiCl undergoes hydrolysis to a smaller extent in hot water in a similar way to MgCl₂.

e) Due to their covalent nature, the halides Lithium and Magnesium are soluble in organic solvents.

f) Both Li⁺ and Mg⁺² are highly hydrated.

g) The Carbonates, Phosphates and Fluorides of both Li and Mg are sparingly soluble in water.

h) Lithium alkyls (Li⁺ R^–) are chemically similar to Grignard reagents in organic synthesis.

Question 17.
When an alkali metal dissolves in liquid ammonia the solution can acquire different colours. Explain the reasons for this type of colour change.
Answer:

• The alkali metals dissolve in liquid NH3 and gives deep blue solutions. These are conducting in nature.
• The blue colour of the solution is due to the ammoniated electrons which absorbs energy in the visible region of light and thus imparts blue colour to the solution.
• These solutions are paramagnetic and on standing liberate hydrogen resulting in the formation of amide.
  M + (x + y) NH₃ → [M(NH₃)_x]⁺ + [e(NH₃)_y]^–
  M_(am)⁺ + e^– + NH₃ → MNH_2(am) + 1/2 H_2(g)
• In concentrated solution the blue colour Changes to bronze.colour on warming and becomes did magnetic.

Question 18.
What happens when

1. Sodium metal is dropped in water ?
2. Sodium metal is heated in a free supply of air ?
3. Sodium peroxide dissolves in water ?

Answer:

1. Sodium metal when dropped in water it reacts with water vigourously and liberates H₂ gas.
   2Na + 2H₂O → 2NaOH + H₂
2. Sodium metal is heated in free supply of air to form sodium peroxide.
   2Na + O₂ → Na₂O₂ (sodium peroxide)
3. Sodium peroxide dissolves in water and forms NaOH and hydrogen peroxide.
   Na₂O₂ + 2H₂O → 2NaOH + H₂O₂

Question 19.
States as to why
i) An aqueous solution of Na₂CO₃^– is alkaline ;
ii) Alkali metals are prepared by the electrolysis of their fused chlorides ?
Answer:
i) An aqueous solution of Na₂CO₃ is alkaline. This is due to anionic (CO₃^-2) hydrolysis.
CO₃^-2 + H₂O → HCO₃^-2 + OH^–
∴ PH > 7. So the solution is alkaline in nature.

ii) Chemically alkali metals are highly reactive and they are placed at top in the electro chemical series.
∴ Common methods of extraction of the metals are not applicable for the alkali metals. So electrolytic reduction of their used chlorides is the possible method for extracting alkali metals.
Eg: ‘Na’ metal obtained from fused ‘NaCl’.

Question 20.
How would you explain the following observations ?

1. BeO is almost insoluble but BeSO₄ is soluble in water ?
2. BaO is soluble but BaSO₄ is insoluble in water ?

Answer:

1. BeO has amphoteric nature and the solubility in water is low because of its covalent nature.
   BeSO₄ is soluble in water. This is due to greater hydration energy of Be⁺² ion.
2. BaO is soluble in water because of its ionic nature.
   BaSO₄ is insoluble in water because of low hydration energy of Ba⁺² ion.

Long Answer Questions

Question 1.
Justify the inclusion of alkali metals in the same group of the periodic table with reference to the following :
i) Electronic configuration
ii) Reducing nature
iii) Oxides and hydroxides.
Answer:
i) Electronic configuration : All the alkali metals have one valence electron, ns¹.

ii) Reducing nature :

• Alkali metals are strong reducing agents.
• ‘Li’ is most powerful reducing agent and ‘Na’ is poor reducing agent. .
  The standard electrode potential (E°) is the measure of reducing power.
• ‘Li’ has highest hydration enthalpy. It has high negative S.E.P. (E°) hence it has high reducing power.

iii) Oxides and hydroxides : –

• The alkali metals forms their oxides in presence of dry air and tarnished.
• These oxides reacts with , .noisture to form hydroxides.
• They burn vigorously in oxygen and forms oxides.
  1. Lithium forms lithium monoxide.
  2. Sodium forms monoxide with limited supply of oxygen and peroxide with’ excess of oxygen
  3. Other metals of this group forms super oxides. The super oxide ion (O₂^–) is stable only in presence of large cations.

Reactions:
4Li + O₂ → 2 Li₂O
4Na + O₂ (Limited) → 2 Na₂O
2 Na + O₂ (Excess) → Na₂O₂
K + O₂ (Excess) → KO₂
Lithium shows a different character. It reacts directly with nitrogen of air and forms Li₃N (Lithium nitride).

• Alkali metal oxides easily hydrolysed by water to form the hydroxides.
  Monoxide : M₂O + H₂O → 2MOH
  Peroxide : M₂O₂ + 2H₂O → 2MOH + H₂O₂
  Superoxide : 2MO₂ + 2H₂O → 2MOH + H₂O₂ + O₂
• Oxides, peroxides are colourless whereas superoxides are coloured because of their para mag-netic property.
• Hydroxides are white crystalline solids.
• Hydroxides are strong bases and dissolved freely in water and evolve much heat.

Question 2.
Write an essay on the differences between lithium and other alkali metals.
Answer:
Anomalous properties of Lithium : In the periodic table some representative elements of different series show similarities, known as diagonal relationship. Li, of the alkali metals, show such a similarity with Mg of II group. That means it differs from alkali metals. Some of the important abnormal characters of lithium are given below.
a) Lithium is hard metal while other alkali metals are soft and can be cut with the knife. Its melting point and boiling point are high.

b) Lithium directly unites with N₂ while no other alkali metal combines directly.
6Li + N₂ → 2Li₃N

c) Lithium element forms a carbide on direct combination. Group IA elements do not form directly. But all these elements are known to give carbides.

d) The solubilities of Lithium Hydroxide (LiOH), Lithium Carbonate (Li₂CO₃), Lithium Phosphate (Li₃PO₄) and Lithium Fluoride (LiF), are very less compared to the high solubilities of the other alkali metal compounds.

e) Lithium Hydroxide is a weaker alkali than the alkali metal Hydroxides. Basic nature of other alkali metal Hydroxides is more than Li(OH). Because of this Lithium Hydroxide Carbonate, nitrates are thermally unstable.

Question 3.
Discuss the preparation and properties of sodium carbonate.
Answer:
Preparation :

• Sodium carbonate is prepared by solvay process.
• In this process sodium chloride reacts with ammonium bicarbonate and gets precipitated the low soluble sodium bicarbonate.
• The lather is prepared by passing CO₂ into a concentrated solution of NaCl saturated with ammonia. Here (NH₄)₂CO₃ followed by NH₄ HCO₃ are formed.
  Chemical equations involved :
  2NH₃ + H₂O + CO₂ → (NH₄)₂ CO₃
  (NH₄)₂ CO₃ + H₂O + CO₂ → 2NH₄HCO₃
  NH₄HCO₃ + NaCl → NH₄Cl + NaHCO₃
• The separated NaHC03 Crystals heated to get Na₂CO₃.
  2NaHCO₃ → Na₂CO₃ + H₂O + CO₂
• In this process ammonia is regenerated by Ca(OH)₂.
  2NH₄Cl + Ca(OH)₂ → 2NH₃ + CaCl₂ + H₂O

Properties of washing soda :

• Na₂CO₃ is a white crystalline solid.
• Na₂CO₃ exists as a decahydrate Na₂CO₃. 10H₂O which is called washing soda.
• Na₂CO₃ is readily soluble in water.
• Na₂CO₃ (decahydrate) when heated it loses the water molecules and forms monohydrate. This monohydrate on heating above 373 K it forms anhydrous form which is called soda ash, a white powder.
  Reactions : –
  
• Na₂CO₃ (aq) solution is alkaline (basic) in nature Because it under goes anionic hydrolysis. (PH > 7).
  CO₃^-2 + H₂O → HCO₃^– + OH^–

Uses :

• Na₂CO₃ is used in the manufactuing of glass.
• Na₂CO₃ is used in the manufactuing of borax, caustic soda.
• Na₂CO₃ is used in paper, paints and textile industries.
• Na₂CO₃ is used in softening of water.
• Na₂CO₃ is used in laundries.
• Na₂CO₃ is used an important laboratory reagent both in qualitative and quantitative analysis.

Question 4.
Discuss the similarities between alkaline earth metals and gradation in the following aspects.
i) Electronic configuration
ii) Hydration enthalpies
iii) Nature of oxides and hydroxides.
Answer:
i) Electronic configuration :
The general electronic configuration of alkaline earth metals represented by [noble gas] ns².

ii) Hydration Enthalpies :

• The hydration enthalpies of alkaline earth metal ions decrease with increase in ionic size down the group.
  Be⁺² > Mg⁺² > Ca⁺² > Sr⁺² > Ba⁺²
• Hydration enthalpies of these elements ions are larger than those of alkali metal ions.
  Eg : MgCl₂ MgCl₂ . 6H₂O
  CaCl₂ CaCl₂ . 6H₂O

iii) Nature of oxides and hydroxides : –

• Alkaline earth metals forms oxides of type MO.
• These are formed by burning in oxygen.
• BeO is amphoteric and covalent in nature whereas other oxides are ionic and basic in nature. Other oxides i.e., except BeO forms hydroxides with water.
  Eg : MgO + H₂O → Mg(OH)₂
• The solubility, thermal stability and the basic character of these hydroxides increase with increase of atomic no. from Mg(OH)₂ to Ba(OH)₂. H
• These hydroxides are less basic, less stable than alkali metal hydroxide.
• Be(OH)₂ is amphoteric in nature. This can be evidented by the following reactions.
• Be(OH)₂ reacts with both acids and alkalis.
  Be(OH)₂ + 2OH^– [Be(OH)₄]^2- (Beryllation)
  Be(OH)₂ + 2HCl + 2H₂O → [Be(OH)₄]Cl₂
• Hence Be(OH)₂ is amphoteric in nature.

Question 5.
Discuss on;
i) Carbonates
ii) Sulphates and
iii) Nitrates of alkaline earth metals.
Answer:
i) Carbonates :

• Alkaline earth metals forms MCO₃ type carbonates.
• These Carbonates are insoluble in water.
• The solubility of these carbonates in water decreases as the atomic no.of the element increases.
•  The thermal stability increases with increasing cationic size.
• These carbonates decompose on heating to give CO₂
  CaCO₃ Cao + CO₂
• BeCO₃ is unstable and kept only in at atmosphere of CO₂.

ii) Sulphates :

• Alkaline earth metals forms MSO₄ type sulphates.
• These are white solids and are stable to heat.
• BeSO₄ and MgSO₄ are readily soluble in water due to high hydration enthalies of Be²⁺, Mg²⁺.
• The solubility decrease from CaSO₄ to BaSO₄.

iii) Nitrates :

• Alkaline earth metals forms M(NO₃)₂ type Nitrates.
• These are formed by the reaction of carbonates in dil.HNO₃.
• Mg(N03)2 crystallises with six water molecules and Ba(NO₃)₂ is an hydrous.
• All of these decompose on heating to give the respective oxides.
  2M(NO₃)₂ → 2MO + 4NO₂ + O₂
  M = Be, Mg, Ca, Sr, Ba.

Question 6.
What are the common physical and chemical features of alkali metals ?
Answer:
Physical features :

• Alkali metals are silvery white, soft and light metals.
• These elements have low density which increases down the group from Li to Cs. (one exception density of K < density of Na).
• The m.pts, b.pts of alkali metals are low.
• Alkali metals and their salts impart characterstic colours to an oxidizing flame.

Reasons :
The heat from the flame excities the outer most orbital electron to a higher energy level. When the excited electron emitts the radiation and comes back to the ground state. This falls in the visible region.

• Alkali metals can be detected by the respective flame tests and can be determined by the flame photo metry (or) atomic absorption spectroscopy.
• Alkali metals when irradiated with light, the light energy absorbed may be sufficient to make an atom lose electron. .
• This makes calsium and potassium useful as electrodes in photo electric cells.

Chemical features : –
i) Reactivity towards Air : —

• The alkali metals forms their oxides in presence of dry air and tarnished.
• These oxides reacts with moisture to form hydroxides.
• They burn vigorously in oxygen and forms oxides.
  1. Lithium forms Lithium monoxide.
  2. Sodium forms monoxide with limited supply of oxygen and peroxide with excess of oxygen.
  3. Other metals of this group forms super oxides. The super oxide ion (O₂^–) is stable only in presence of large cations.
• Reasons :
  4Li + O₂ → 2 Li₂O
  4Na + O₂ (Limited) → 2 Na₂O
  2Na + O₂ (Excess) → Na₂O₂
  K + O₂ (Excess) → KO₂
• Lithium shows a different character. It reacts directly with nitrogen of air and forms Li₃N (Lithium nitride).

ii) Reactivity with H₂: Alkali metals react with H₂ directly at 300 – 600° C and form hydrides. The reaction can be written as follows :

Where M = Li, Na, K, Rb or Cs. These hydrides are ionic in nature. Their ionic nature increases with the metalic nature of alkali metals.

iii) Reactivity with halogens : All the alkali’ metals react with halogens to give the binary compounds. The chemical reactivity in the alkali metals increases with increase in atomic number.
2M + X₂ → 2MX (where M is any alkali metal)
All the metal halides are ionic compounds.

iv) Reactivity with water : The alkali metals decompose water vigorously and liberate hydrogen gas! The chemical reactivity of these metals increases as the atomic number increases. The metal hydroxides are formed.
2M + 2H₂O → 2MOH + H₂
Where M = any one of the alkali metals.

ii) Reducing nature :

• Alkali metals are strong reducing agents.
• ‘Li1 is most powerful reducing agent and ‘Na’ is poor reducing agent.
• The standard electrode potential (E°) is the measure of reducing power.
• ‘Li’ has highest hydration enthalpy. It has high negative S.E.P. (E°) hence it has high reducing power.
• The alkali metals dissolves in liquid NH3 and gives deep blue solutions. These are conducting in nature.
• The blue colour of the solution is due to the ammoniated electrons which absorbs energy in the visible region of light and thus imparts blue colour to the solution.
• These solutions are paramagnetic and on standing Liberate hydrogen resulting in the formation of amide.
  M + (x + y) NH₃ → [M(NH₃)_x]⁺ + [e(NH₃)_y]^–
  M_(am)⁺ + e^– + NH₃ → MNH_2(am) + 1/2 H_2(g)
• In concentrated solution the blue colour Changes to bronze.colour on warming and becomes did magnetic.

Question 7.
Discuss the general characterstics and gradation in properties of alkaline earth metals.
Answer:
The general characterstics and gradation in properties of alkaline earth metals follows.
i) Oxides and hydroxides :

• Alkaline earth metals forms oxides of type Mo.
• These are formed by burning in oxygen.
• BeO is amphoteric and covalent in nature where as other oxides are ionic and basic in nature.
• Other oxides i.e., except Beo forms hydroxides with water.
  Eg : MgO + H₂O → Mg(OH)₂
• The solubility, thermal stability and the basic character of these hydroxides increase with increase of atomic no. from Mg(OH)₂ to Ba(OH)₂.
• These hydroxides are less basic, less stable than alkali metal hydroxide.
• Be(OH)₂ is amphoteric in nature. This can be evidented by the following reactions.
• Be(OH)₂ reacts with both acids and alkalis.
  Be(OH)₂ + 2OH^– [Be(OH)₄]^2- (Beryllation)
  Be(OH)₂ + 2HCl + 2H₂O → [Be(OH)₄]Cl₂
• Hence Be(OH)₂ is amphoteric in nature.

ii) Halides:-

• These forms MX₂ type halides.
• Except Be – halides, all other halides of these metals are ionic.
• Be – halides are covalent and soluble in organic solvents.
• In vapour phase BeCl₂ forms a bridged dimer which disociates into monomer at high temperatures (around 1200 k)
  Cl – Be – Cl
• In solid state BeCl₂ has a chain structure.
  
• The tendency to form halide hydrates gradually decreases – down the group.
  Eg : MgCl₂, 8H₂O, CaCl₂, 6H₂O, BaCl₂. 2H₂O.
• Ca, Sr and Ba halides, dehydration can be done by heating.
• Fluorides are less soluble than the chlorides due to their high lattice energies.

i) Carbonates :

• Alkaline earth metals forms MCO₃ type carbonates.
• These Carbonates are insoluble in water.
• The solubility of these carbonates in water decreases as the atomic no.of the element increases.
• The thermal stability increases with increasing cationic size.
• These carbonates decompose on heating to give CO₂
  CaCO₃ Cao + CO₂
• BeCO₃ is unstable and kept only in at atmosphere of CO₂.

ii) Sulphates:

• Alkaline earth metals forms MSO₄ type sulphates.
• These are white solids and are stable to heat.
• BeSO₄ and MgSO₄ are readily soluble in water due to high hydration enthalpies of Be²⁺, Mg²⁺.
• The solubility decrease from CaSO₄ to BaSO₄.

iii) Nitrates:

• Alkaline earth metals forms M(NO₃)₂ type Nitrates.
• These are formed by the reaction of carbonates in dil.HNO₃.
• Mg(NO₃)₂ crystallises with six water molecules and Ba(NO₃)₂ is an hydrous.
• All of these decompose on heating to give the respective oxides.
  2M(NO₃)₂ → 2MO + 4NO₂ + O₂
  M = Be, Mg, Ca, Sr, Ba.

Question 8.
Discuss the various reactions that occur in the solvay process. [A.P. Mar. 16]
Answer:
Preparation:

• Sodium carbonate is prepared by solvay process.
• In this process sodium chloride reacts with ammonium bicarbonate and gets precipitated the low soluble sodium bicarbonate. ‘
• The lather is prepared by passing CO₂ in to a concentrated solution of NaCl saturated with ammonia. Here (NH₄)₂CO₃ followed by NH₄ HCO₃ are formed.
• Chemical equations involved:
  2NH₃ + H₂O + CO₂ → (NH₄)₂ CO₃
  (NH₄)₂ CO₃ + H₂O + CO₂ → 2NH₄HCO₃
  NH₄HCO₃ + NaCl → NH₄Cl + NaHCO₃
• The separated NaHCO₃ Crystals heated to get Na₂CO₃.
  2NaHCO₃ → Na₂CO₃ + H₂O + CO₂
• In this process ammonia is regenerated by Ca(OH)₂.
  2NH₄Cl + Ca(OH)₂ → 2NH₃ + CaCl₂ + H₂O

Question 9.
Starting with sodium chloride how would you proceed to prepare
i) Sodium metal
ii) Sodium hydroxide
iii) Sodium peroxide
iv) sodium carbonate.
Answer:
i) Fused NaCl on electrolysis gives sodium metal.
2NaCl → 2Na⁺ + 2Cl^–
2Na⁺ + 2e^– → 2Na (Cathode)
2Cl^– → Cl₂ + 2e^– (anode)

ii)

• Castner-Kellner process is a commercial method used for the preparation of sodium hydroxide.
• In this process sodium hydroxide is prepared by the electrolysis of sodium chloride in Castner – Kellner cel..
• Brine solution is electrolysed using a mercury cathode and a carbon anode.
• Sodium metal is formed at cathode and it combine with mercury to form sodium amalgam. Chlorine gas is evolved at anode.
• The amalgam is streated with water to form sodium hydroxide.
  Cell Reactions:
  2NaCl → 2Na⁺ + 2Cl^–
  2Na⁺ + 2e^– 2Na – amalgam
  2Cl^– → Cl₂ + 2e^–
  2Na-amalgam + 2H₂O → 2NaOH + 2Hg + H₂
• This process is also called as mercury cathode process.

iii) The obtained Na – metal reacts with excess of oxygen to form sodium peroxide.
2Na + O₂ → Na₂O₂ (Sodium peroxide.)

iv) Preparation:

1. Sodium carbonate is prepared by solvay process.
2. In this process sodium chloride reacts with ammonium bicarbonate and gets precipitated the low soluble sodium bicarbonate.
3. The lather is prepared by passing CO₂ into a concentrated solution of NaCl saturated with ammonia. Here (NH₄)₂CO₃ followed by NH₄ HCO₃ are formed.

Question 10.
What happens when
i) Magnesium is burnt in air ?
ii) Quick lime is heated with silica
iii) Chlorine reacts with slaked lime
iv) calcium nitrate is strongly heated.
Answer:
i) Magnesium burns with dazzling brilliance in air to give MgO and Mg₃N₂.
2Mg + O₂ →2MgO
3Mg + N₂ → Mg₃N₂

ii) Quick lime heated with silica to form calcium silicate
Cao + SiO₂ → CaSio₃

iii) Slaked lime reacts with chlorine gas to form bleaching powder.

iv) Calcium nitrate on strong heating to form respective oxide
2Ca(NO₃)₂ → 2CaO + 4NO₂ + O₂.

Question 11.
Explain the significance of sodium, potassium, magnesium and calcium in biological fluids.
Answer:
Biological importance of Na, K.

• Na⁺ ions participate in the transmission of nerve signals.
• Na⁺ ions regulates the flow of water accross cell membranes.
• Na⁺ ions responsible for transport of sugars and amino acids into cells.
• K⁺ ions are useful in activating enzymes.
• K⁺ ions participate in the oxidation of glucose to produce ATP.
• K⁺ along with Na⁺ responsible for the transmission of nerve signals.

Biological importance of Mg and Ca :
Role of Mg²⁺ in biology :

1. Mg²⁺ ions are concentrated in animal cells.
2. Enzymes like “phosphohydrolases1 and ‘Phospho transferases’ contain Mg²⁺ ions. These enzymes participate in ATP reactions and release energy in the process. Mg²⁺ forms a complex with ATP.
3. Mg²⁺ is a constituent of chlorophyll, the green component of plants.

Role of Ca⁺²:
About 99% of body calcium is present in bones and teeth. It also plays important roles in neuromuscular function, interneuronal transmission, cell membrance integrity and blood coaqulation.

The calcium concentration in plasma is regulated at about 100 mg/Lit. It is maintained by two hormones, calcitonin and parathyroid hormone. Ca²⁺ ion are necessary for muscle contraction.

Question 12.
Write few lines about cement ?
Answer:

• Cement is an important building material.
• It is also called portland cement.
• Cement is obtained by combining a material rich in lime, CaO with other material such as clay which contains Sio₂ along with the oxides of Al, Fe and Mg.
• Composition of port land cement is
  Cao – 50 – 60%
  Sio₂ – 20 – 25%
  Al₂O₃ – 5 – 10%
  Mgo – 2 – 3%
  Fe₂O₃ – 1 – 2%
  and SO₂ – 1 – 2%
• For a good quality of cement the ratio of SiO₂ to Al₂O₃ is between 2.5 and 4 and the ratio of lime (Co) to the total of the oxides of SiO₂, Al₂O₃ and Fe₂O₃ is as close as ‘2’.
• The raw materials used for the manufacture of cement are lime stone and clay.
  Clay + lime cement clinker.
• This cement clinker mixed with 2 – 3% by wt. of gypsum to form cement.
• Important ingradients in portland cement are
  Ca₂SiO₄ – 26%, Ca₃SiO₅ – 51 % and Ca₃Al₂O₆ – 11 %

Setting of Cement: –

• Cement mixed with water to give a hard mass i.e setting of cement takes place.
• This is due to the hydration of molecules of the cement.
• The purpose of adding gypsum is to slow down the process of setting and to get sufficiently hardness.

Uses:

• It is used in concrete and rein forced concrete.
• It is used in plastering.
• It is used in construction of bridges, dams and buildings.

Solved Problems

Question 1.
What is the oxidation state of K in KO₂ ?
Solution:
The superoxide species is represented as O₂^–; since the compound is neutral, the oxidation state of potassium is +1.

Question 2.
The E^Θ for Cl₂ / Cl^– is + 1.36, for I₂/I^– is + 0.53, for Ag⁺/Ag is + 0.79, Na⁺/Na is – 2.71 and for Li⁺/Li is – 3.04. Arrange the following ionic species in decreasing order of reducing strength : I^–, Ag, Cl^–, Li, Na
Solution:
The order is Li > Na > I^– > Ag > Cl^–.

Question 3.
Why is KO₂ paramagnetic ? [T.S. Mar. 16]
Solution:
The superoxide O₂^– is paramagnetic because of one unpaired electron in π*2p molecular orbital.

Question 4.
Why does the solubility of alkaline earth metal hydroxides in water increases down the group ?
Solution:
Among alkaline earth metal hydroxides, the anion being common the cationic radius will influence the lattice enthalpy. Since lattice enthalpy decreases much more than the hydration enthalpy with increasing ionic size, the solubilit y increases as we go down the group.

Question 5.
Why does the solubility of alkaline earth metal carbonates and sulphates in water decrease down the group ?
Solution:
The size of anions being much larger compared to cations, the lattice enthalpy will remain almost constant within a particular group. Since the hydration enthalpies decrease down the group, solubility will decrease as found for alkaline earth metal carbonates and sulphates.

Students can go through AP Inter 1st Year Physics Notes 14th Lesson Kinetic Theory will help students in revising the entire concepts quickly.

AP Inter 1st Year Physics Notes 14th Lesson Kinetic Theory

→ A molecule moves along a straight line between two successive collisions and the average straight distance covered between two successive collisions is called the mean free path of the molecules.

→ Boyle’s Law states that, at constant temperature, the volume of a given mass of a gas is inversly proportional
to the pressure. V ∝ \(\frac{1}{p}\) (constant temperature)

→ Charle’s laws :

1. At constant pressure, the volume of a given mass of a gas is directly proportional to the absolute temperature of the gas V ∝ T (P constant).
2. At constant volume, the pressure of a given mass of a gas is directly proportional to the absolute temperature of the gas. P ∝ T (V constant)

→ Dalton’s law of partial pressures : It states that the total pressure exerted by a mixture of non-reactive ideal gases is equal to sum of the partial pressures which each would exert, if it alone occupied the same volume at the given temperature.

→ Root mean square speed is defined as the square root of the mean of the squares of the random velocities of the individual molecules of a gas.
V_rms = \(\sqrt{\frac{3 K_B \mathrm{~T}}{\mathrm{~m}}}\)

Students can go through AP Inter 1st Year Physics Notes 13th Lesson Thermodynamics will help students in revising the entire concepts quickly.

AP Inter 1st Year Physics Notes 13th Lesson Thermodynamics

→ Zero law of thermodynamics states that, two systems in thermal equilibrium with a third system seperately are in thermal equilibrium with each other.

→ The zeroth law of thermodynamics led us to the con¬cept of temperature.

→ Reversible process: A process that can be retraced back in the opposite direction in such a way that the system passes through the same states as in the direct process, and finally the system and the surroundings return to their original states, with no other change any wherelse in the Universe is called a reversible process

→ Irreversible process : A process that cannot be retraced back in the opposite direction is called as Irreversible process.

→ Thermodynamics is the branch of Physics dealing with the inter relation between heat and mechanical energy.

→ Thermodynamics is applicable only when the system is T in equilibrium.

→ Temperature is a thermal condition of a substance and d measures its relative hotness.

→ The mathematical representation of zeroth law of thermodynamics f (P, V, T) = 0.

→ The energy transferred by a non-mechanical way is called heat.

→ The mechanical equivalent J is the amount of mechanical work to be done to produce unit quantity of heat.
J = \(\frac{W}{Q}\) its value in CCS system is 4.2 × 10⁷ erg/cal
In S.l. system J is 1.

→ First law of thermodynamics : The heat supplied to the system is equal to the sum of the increase in internal energy of the system and the external work done by the system.
dQ = dU + dW or dQ = dU + PdV

→ First law of thermodynamics is only another statement of the Law of Conservation of Energy.

→ Quasi – static process: An infinitesimally slow process in which at each and every intermediate stage, the system remains in thermal and mechanical equilibrium with the surroundings through out the entire process.

→ Cyclic process : A process in which the system after passing through various states pressure, volume and temperature changes) returns to its initial state is called a cyclic process.

→ In isobaric process the pressure is constant while in isochoric process the volume is constant.

→ Carnot engine is a reversible engine operating between two temperatures T₁ (source) and T₂ (sink). The efficiency of a cannot engine is given by
η = 1 – \(\frac{T_2}{T_1}\)

→ C_p is always greater than C_v.
∴ C_p – C_v = R and \(\frac{C_p}{C_v}\) = y
For mono atomic gas γ = \(\frac{5}{3}\), for diatomic gas γ = \(\frac{7}{5}\)
For tnatomic gas γ = \(\frac{4}{3}\)

→ Isothermal change: The changes of pressure and volume of a gas at constant temperature with exchange of heat is called isothermal changes.
PV = Constant

→ Workdone by an ideal gas during isothermal process
W = RT log_e \(\frac{V_2}{V_1}\) or 2.303 RT log₁₀\(\left[\frac{V_2}{V_1}\right]\)

→ Adiabatic change : The changes in pressure and volume of a gas resulting a change in temperature without exchange of heat in an isolated system is called adiabatic changes.

→ In adiabatic process
\(P_1 V_1^\gamma=P_2 V_2^\gamma, T_1 V_1^{\gamma-1}=T_2 V_2^{\gamma-1}, T_1^\gamma P_1^{1-\gamma}=T_2^\gamma P_2^{1-\gamma}\)

→ Workdone in an adiabatic change W = \(\frac{\mu \mathrm{R}}{\gamma-1}\left(\mathrm{~T}_1-\mathrm{T}_2\right)\)

→ Clausius statement of second law : Heat cannot itself flow from cold body to hot body.

→ Kelvin’s Statement : It is impossible to derive continuous supply of energy in cooling a body below the coldest of its surroundings.

→ Latent heat (L): The quantity of heat absorbed or liberated during the change of state by unit mass of substance, without any change in temperature is called Latent heat.
L = \(\frac{Q}{M}\)

→ Unit of 1′: Joule / kg
Dimension of L : L = \(\frac{Q}{M}\) = [L² T^-2]

→ Latent heat of fusion of ice L_ice = 80 cal/gm
= 0.335 × 10⁶ J kg^-1
Latent heat of steam L_steam = 540 cal/gm
= 2.26 × 10⁶ J kg^-1

Students can go through AP Inter 1st Year Physics Notes 12th Lesson Thermal Properties of Matter will help students in revising the entire concepts quickly.

AP Inter 1st Year Physics Notes 12th Lesson Thermal Properties of Matter

→ Temperature is a relative measure indicating the degree of hotness (or) coldness of a body.

→ Temperature is a macroscopic property of a body (or) system. It is a scalar quantity.

→ Heat is a form of energy transferred between two sys-tems by virtue of temperature difference.

→ The instrument used to measure temperature is called thermometer.

→ Relation between Celsius, Fahremheit, Reasumer and Kelvin Scales is \(\frac{c-0}{100}=\frac{F-32}{180}=\frac{R-0}{80}=\frac{K-273}{100}\)

→ The atoms are arranged in regular order in the form of a lattice in solids.

→ The interatomic force of attraction depends on the dis-tance between them.

→ As the temperature increases, the amplitudes of the vibrations of the atoms increases.

→ When a solid is heated its length, area and volume increase.

→ The increase in length, per unit length, per 1°C rise of temperature is called, the coefficient of linear expansion
α_l = \(\frac{\Delta l}{l \times \Delta \mathrm{T}} /{ }^{\circ} \mathrm{C}\)

→ The increase in area, per unit area, per 1°C rise of temperature is called, the coefficient of areal expansion, α_A = \(\frac{\Delta \mathrm{a}}{\mathrm{a} \times \Delta \mathrm{t}} /{ }^{\circ} \mathrm{C}\)

→ The increase in volume, per unit volume, per 1°C rise of temperature is called, the coefficient of volume expansion, α_v = \(\frac{\Delta v}{v \times \Delta t} /{ }^{\circ} \mathrm{C}\)

→ α_l = α_A = α_v = 1 : 2 : 3 (or) \(\frac{\alpha_l}{1}: \frac{\alpha_{\mathrm{A}}}{2}: \frac{\alpha_{\mathrm{v}}}{3}\)

→ The change in temperature of a substance, when a given quantity of heat is absorbed, Then it is called Heat capacities S = \(\frac{\Delta Q}{\Delta T}\).

→ The amount of heat absorbed (Or) rejected to change the temperature of unit mass in called specific heat capacity of the substance s = \(\frac{s}{m}=\frac{1}{m} \frac{\Delta Q}{\Delta T}\).

→ Molar specific heat is defined as the amount of heat absorbed (or) rejected to change the temperature of one mole of a substance C = \(\frac{s}{\mu}=\frac{1}{\mu} \frac{\Delta Q}{\Delta T}\).

→ The phenomenon of refreezing is called regelation.

→ The quantity of heat absorbed (or) liberated during the change of state by unit mass of substance, with out any change in temperature is called latent heat.

→ The latent heat of fusion (L_f) is the heat per unit mass required to change a substance from solid into liquid at the same temperature and pressure.

→ The latent heat of vaporisation (L_v) is the heat per unit mass required to change a substance from the liquid to the vapour state without change in the temperature and pressure.

→ Transfer of heat takes place in three modes from high temperature regions to low temperature regions in matter. These modes are conduction, convection and radiation.

→ Conduction of heat is possible in matter by transfer of energy among the molecules through collisions. The bulk of the material does not move while the molecules vibrate about their mean position and collide with each other.

→ The ability of conducting heat varies from material to material. It is measured by a quantity known as coefficient thermal conductivity.

→ Convection is the energy transfer by means of bulk movement of the fluid-

→ Convection is of two types

1. natural convection
2. forced convection.

→ Natural convection occurs when fluid with different density values moves under gravity.

→ Forced convection takes place when a fluid is allowed to pass over a body with different temperature forcibly.

→ Thermal radiation does not require a material medium.

→ Every body emits thermal radiation at all temperatures above absolute zero and exchanges heat energy with surroundings. This is known as Prevost theory.

→ The energy flux emitted by unit surface area of a radiating body is known as emissive power. Its units is Jm²s^-1 or Wm^-2. Dimensional formula is [MT^-3].

→ The ratio of energy flux absorbed in certain time to total energy flux incident on the body in the same time is known as absorptive power α . α can not be greater than 1.
For a black body absorptive power at all wave lengths is 1.

→ The ratio of emissivity of a body to its absorptivity is same for all bodies at a given temperature and equals to the emissivity of a black body at the same temperature. This is KirchhofFs law of radiation.

→ Emissive power of a black body is proportional to the forth power of the absolute temperature. P = σAT^-4. P emissive power, σ Stefan’s constant and equal to 5.67 × 10^-8 W/m⁴ k⁴ this can be written for a body (which is not a black body) as follows :
P = e_λ σ AT⁴ where e_λ is the emissivity of the body.

→ Newton’s law of cooling : The rate of loss of heat is directly proportional to the difference in temperature between the body and its surroundings provided the temperature difference is small.
\(\frac{-\mathrm{dQ}}{\mathrm{dt}}\) = α (T_B – T_S)

Students can go through AP Inter 1st Year Physics Notes 11th Lesson Mechanical Properties of Fluids will help students in revising the entire concepts quickly.

AP Inter 1st Year Physics Notes 11th Lesson Mechanical Properties of Fluids

→ The property by virtue of which a liquid opposes the relative motion between its layers is called viscosity.

→ If the velocity of any point in the flow is independent of time is called streamline flow.

→ The velocity during the streamline flow is less than a particular velocity called critical velocity.

→ If the streamlines are parallel, the flow is called laminar flow.

→ Pressure difference at different depths in a fluid cause an upward thrust on an object immersed in the fluid. This upward thrust is called buoyant force. Buoyant force is equal to the weight of the fluid displaced by the immersed body.
An object floats on a fluid if its density is less than that of the fluid.
Fraction of the volume of the object submerged in a fluid is equal to the ratio of density of the object to density of the fluid.

→ Study of moving fluids is called fluid dynamics. Flow of fluids is of two types

1. Streamline flow
2. turbulent flow.

→ Fluid flow may be

1. steady or non-steady,
2. rotational or irrotational,
3. compressible or incompressible and iv. viscous or non-viscous.

→ A line in a fluid on which, a tangent drawn at every point represents the direction of velocity of the fluid is called a streamline.
The density of streamlines at a point is proportional to the magnitude of the fluid at that point. If streamlines coincide with the trajectories of the flowing fluid particles, the flow is called a steady flow.
Bundle of streamlines is known as a tube of flow.

→ If the velocity of the fluid at a point varies with time the flow is called turbulent flow.

→ The minimum velocity of the fluid at which the flow changes from streamline to turbulent flow is called critical velocity.

→ For a steady flowing fluid in a pipe the product of cross sectional area A and velocity of the fluid v at a point is constant.
AV = constant.
This is known as equation of continuity.

→ The motion of objects in flowing fluid can be understood by Bernoulli theorem. It is stated like this:
For a steady flow of a non-viscous, incompressible fluid sum of the pressure energys kinetic energy and the potential energy per unit volume remains constant at all points. The equation is given by
P + ρgh + \(\frac{1}{2}\) ρ v² = constant.

→ The frictional force between the successive layers of a flowing fluid is called viscous force. This frictional force hampers the flow of the fluid.
The property of a fluid which opposes the relative motion between, different layers is called viscosity.

→ Viscous force F = –\(\eta \mathrm{A}\left(\frac{\Delta \mathrm{v}}{\Delta \mathrm{x}}\right)\)

→ Stokes law: The viscous force F acting on a smooth spherical body .which is moving in a fluid is given by
F = 6 πη r v
h coefficient of viscosity of the fluid r is the radius of the spherical body, v the velocity of the body in the fluid.

Students can go through AP Inter 1st Year Physics Notes 10th Lesson Mechanical Properties of Solids will help students in revising the entire concepts quickly.

AP Inter 1st Year Physics Notes 10th Lesson Mechanical Properties of Solids

→ The property of a body by virtue of which it resists the deformation forces and regains its original shape when the deformation forces are removed is called elasticity.

→ The restoring force per unit area developed inside the body is called stress.
∴ Stress = \(\frac{\text { Restoring force }}{\text { Area }}=\frac{R}{A}\)

→ The maximum value of stress within which a body completely regains its original condition when the deforming forces are removed is called its elastic limit.

→ Hooke’s law: Within the proportionality limit, the stress is directly proportional to the strain produced in a body. Stress × strain = constant within elastic limit.

→ The ratio of increase in length of a wire to its original length is called tensile or longitudinal or linear strain.
Linear strain = \(\frac{\Delta \mathrm{l}}{\mathrm{L}}=\frac{\mathrm{e}}{\mathrm{L}}\)

→ The ratio of the relative displacement between two layers to the distance between them is called shearing strain, θ.

→ The volume strain or bulk strain in the ratio of change in volume to its original volume.
i. e., Bulk strain = \(\frac{\Delta V}{V}\).

→ Within elastic limit of a body, the ratio of linear stress to linear strain is called Youngs modulus of elasticity Y. Y = \(\frac{F}{A} / \frac{e}{L}=\frac{F}{A} \cdot \frac{L}{e}\)

→ Within the elastic limit of a body, the ratio of shearing stress to shearing strain is called rigidity modulus of elasticity (G). .
∴ Rigidity modulus G = \(\frac{F / A}{\theta}=\frac{F}{A \theta}\)

→ Within the elastic limit of a body, the ratio of bulk stress to bulk strain is called bulk modulus of elasticity B.
PV = –\(\frac{\mathrm{PV}}{\Delta \mathrm{V}}\)

→ When a body is subjected to continuous elastic straining for some time, it appears to have lost its elastic nature temporary which it regains after taking sufficient rest. This behaviour is called elastic fatigue.

→ The ratio of lateral contractional strain to longitudinal elongational strain for the same stress is called Poisson’s ratio σ.

→ Shearing strain = 2 × longitudinal strain

→ Bulk strain = 3 × longitudinal strain

→ Hooke’s law is valid only in the linear part of stress – strain curve.

→ The young’s modulus and shear modulur are relevant only for solids. Bulk modulus is relevant for solids, liquids and gases.

→ Metals have larger values of young’s modulus than alloys and elastomes.

→ Examples of ductile metals are, Copper, Aluminium, lead, gold, Examples of brittle metals are ; Glass ceramic

→ Strain energy is the energy stored in a body due to its deformation.

→ Stress is not a vector quantity.

Students can go through AP Inter 1st Year Physics Notes 9th Lesson Gravitation will help students in revising the entire concepts quickly.

AP Inter 1st Year Physics Notes 9th Lesson Gravitation

→ There are four basic forces in nature, they are
a) Gravitational force
b) Electromagnetic force
c) Strong nuclear force
d) Weak nuclear force

→ Newton’s Universal Law of Gravitation : Every particle of matter in the universe attracts every other particle with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres.
F ∝ \(\frac{\mathrm{m}_1 \mathrm{~m}_2}{\mathrm{~d}^2}\); F = \(\frac{\mathrm{Gm}_1 \mathrm{~m}_2}{\mathrm{~d}^2}\)

→ Gravitational field : A mass particle modifies space around it in some way and through which it interacts with other mass particles.

→ Gravitational field travels with speed of light in vacuum.

→ Black hole is dense object into which other objects could fall but out of which no object or even light could ever come out.

→ A non-accelerating frame is called inertial frame of refer-enceAn accelerating frame is called non-inertial frame of reference.

→ Inertial mass (m₁) of an object and its gravitational mass (mg) are equal.

→ On the basis of principle of equivalence, the gravitational and inertial masses are equal.

→ Acceleration due to gravity at a height h
g_h = \(\frac{G M}{(R+h)^2}\) = g(1 – \(\frac{\mathrm{2h}}{\mathrm{R}}\))

→ The value of decreases with depth g_d = g(1 – \(\frac{\mathrm{d}}{\mathrm{R}}\))

→ The variation of g with latitude of the place on earth g_Φ = g – Rw² cos² Φ.

→ ‘g’ value vary with latitude due to non-spherical nature of the earth.

→ Due to local conditions also the value of g will vary from place to place on the surface of the earth.

→ The orbital velocity of a projected body is the velocity with which it revolves in the orbit.
v₀ = \(\sqrt{\mathrm{gR}}=\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}\) = 7.92 kms^-1

→ The escape velocity of a body is the minimum velocity that should be given in order that it may overcome the earth’s attraction and go into the space.
v_e = \(\sqrt{2 \mathrm{gR}}=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}\) = 11.2 kms^-1

→ The escape velocity of a satellite from an orbit is \(\sqrt{2}\) times the orbital velocity.

→ The geostationary satellites are in specified orbits at a height of about 36,000 km above the earth.

→ Law of orbits : All planets move in ellipitcal orbits with the sun situated at one of the foci.

→ Law of areas: The line that joins any planet to the sun sweeps equal areas in equal internals of time.

→ Law of periods : The square of the time period of revolution of a planet is proportional to the cube of the semi-major axis of the ellipse tracedout by the planet
T² ∝ R³

→ The gravitational potential energy is given by
V = \(\frac{-\mathrm{Gm}_1 \mathrm{~m}_2}{\mathrm{r}}\)

→ Time period of polar satellites is 100 minutes.

→ Polar satellites are low altitude satellites, but they go around the poles of the earth in a north- south direction.