Among all the given statements, statement b is correct. Other statements can be corrected as-

a. For ${\mathrm{NH}}_4^{+}$, number of valance electrons = 10

For CH₄, number of valance electrons = 10.

They both are isoelectronic species.

c. O - Cl - O does not follow octet rule.

d. O^- in O₃ is sp^2- hybridised.

AlCl₃ is the salt of weak base and strong acid which renders H⁺ ions on hydrolysis. Thus, aqueous solution of AlCl₃ is acidic in nature.

Al³⁺ + 3H₂O $\rightleftharpoons$Al(OH)₃ + 3H⁺

On the other hand, NaCl is the salt of strong acid and strong base which does not undergo hydrolysis.

Noble gases make up a group of chemical elements with similar properties. These include helium, neon, argon, krypton, xenon and radon. 

Among all the given statements, statement c is incorrect. It can be corrected as valence shells of noble gases are complete, thus, they do not show tendency to gain electrons. Therefore, they have zero electron affinity.

For most substances, increasing the pressure when a system is in equilibrium between liquid and solid phases will increase the phase transition temperature. Ice melts at a temperature lower than its usual melting point because ice is less denser than water.

Among all the given pairs, second pair has sp²- hybrid orbitals.

SO₂ is trigonal planar. It has 3 sp²- hybrid orbitals.

CH₄ is tetrahedral. It has 4 sp³- hybrid orbitals.

SO₃ and C₂H₄ are trigonalplanar and have 3 sp²- hybrid orbitals.

BF₃ is trigonal planar. It also has 3 sp²- hybrid orbitals.

SF₄ is irregular tetrahedral. It has 5 sp³d hybrid orbitals.

${\mathrm{I}}_3^{-}$L is linear and has 5 sp³d hybrid orbitals.

Work done in isothermal reversible process is 

w = -2.303 nRT ${\log }_{10}\left(\frac{{\mathrm{p}}_1}{{\mathrm{p}}_2}\right)$

Given, n = 1; R = 2 cal K^-1 mol^-1

T = (27 + 273) K = 300 K

p₁ = 2 atm; p₂ = 10 atm

Therefore,

w = -2.303 $\times$ 1 $\times$ 2 $\times$ 300 ${\log }_{10}\left(\frac{2}{10}\right)$ 

w = + 965.54 cal

For isothermal change, $∆\mathrm{U} = 0$

Now, from first law of thermodynamics,

q = $∆$U - w = 0 - 965.84 cal

q = -965.84 cal

Beryllium or Be differs in properties from other elements of its own group but shows resemblance with aluminium because of its smaller ionic radius and high polarising power. Polarising powers of Be²⁺ and Al³⁺ ions are almost the same thus, they show similarities in their properties.

Number of waves = $\frac{\mathrm{Number} \mathrm{of} \mathrm{waves}}{\mathrm{de}- \mathrm{Brogile} \mathrm{wavelength} \mathrm{of} \mathrm{electron}}$

But, $\mathrm{\lambda } = \frac{\mathrm{h}}{\mathrm{mv}}$ 

$\therefore$ $\mathrm{\lambda } = \frac{2\mathrm{\pi }3 \times \mathrm{\mu }}{\mathrm{h}} = \frac{2\mathrm{\pi }}{\mathrm{n}} \times \mathrm{mvr}$

$\because \mathrm{mvr} = \frac{\mathrm{nh}}{2\mathrm{\pi }} = \frac{2\mathrm{\pi }}{\mathrm{h}} \times \frac{\mathrm{nh}}{2\mathrm{\pi }}$

mvr = n

For second orbit, n = 2. Therefore, number of waves = 2.

Due to gravity, barbell is lifed and hence work is done.

Given that,

p₁ = p₁T₁ = t; V₁ = v

p₂ = $\frac{p}{2}$; V₂ = $\frac{v}{2}$

T = ?

By using Boyle's law

$\frac{p_1{V}_1}{T_1} = \frac{p_2{V}_2}{T_2}$

$\frac{p \times V}{t} = \frac{p \times V}{2 \times 2 \times T_2}$

$T_2 = \frac{t}{4}$