If a reaction is exothermic, $∆$H will be -ve and in increasing order, $∆$S will be +ve thus, at this condition, $∆$G is negative according to following equation:
$∆$G = $∆$H - T$∆$S

$∆$G = -ve.

Hence, for spontaneous reaction $∆$G must be -ve

B₂H₆ is electron deficient molecule because boron atom has three half filled orbitals in excited state. The structure of B₂H₆ is represented as follows :

In it two electrons of a B - H bond are involved in formation of three centre bond, these bonds are represented as dotted lines.

Heat of neutralisation of strong acid and strong base is -57.33 kJ. MgO is weak base while HCl is strong acid, so the heat of neutralisation of MgO and HCl is lower to -57.33 kJ because MgO requires some heat in ionisation, then net released amount of heat is decreased.

(i) NO (g) + $\frac{1}{2}{\mathrm{O}}_2$(g) $\stackrel{{\mathrm{K}}_1}{\rightleftharpoons }$ NO₂(g)

 So, K₁ = $\frac{\left[{\mathrm{NO}}_2\right]}{\left[\mathrm{NO}\right] [{\mathrm{O}}_2{]}^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}$          ...(1)

(ii) 2NO₂ (g) $\stackrel{{\mathrm{K}}_2}{\rightleftharpoons }$ 2NO + O₂ (g)

So, K₂ = $\frac{[\mathrm{NO}{]}^2 [{\mathrm{O}}_2]}{\left[{\mathrm{NO}}_2\right]}$             ...(2)

From equation (1)

${\mathrm{K}}_1^2 = \frac{[{\mathrm{NO}}_2{]}^2}{\left[{\mathrm{NO}}_2\right]}$  or

$\frac{1}{{\mathrm{K}}_1^2} = \frac{\left[\mathrm{NO}{]}^{2 }\right[{\mathrm{O}}_2]}{[{\mathrm{NO}}_2{]}^2}$              ...(3)

from equation (2) and (3)

K₂ = $\frac{1}{{\mathrm{K}}_1^2}$

Among all the given options, IF₃ has trigonal planar geometry.

PCl₃ has pyramidal geometry.

NH₃ has pyramidal geometry.

BF₃ has trigonal planar geometry

In Hall and Heroult process

2Al₂O₃ $\to$ 4Al + 3O₂

4C + 3O₂ $\to$ 2CO₂ + 2CO $\uparrow$

2Al₂O₃ + 4C $\to$4Al + 2CO₂ + 2CO

Only for removal of CO₂, following equation is possible

2Al₂O₃ + 3C $\to$4Al + 3CO₂

Molecular weight of Carbon = 12

Molecular weight of Aluminium = 27

$\therefore$ For Carbon/ C = 3 x 12 = 36

    For Aluminium/ Al = 4 x 27 = 108

$\because$ For 108 gm of Al, 36 gm of C is required in above reaction

$\therefore$ For 270 gm of Al require amount of C = $\frac{36}{108} \times 270 =90 \mathrm{gm}$

The spontaneity of reaction is based upon the negative value of $∆$G, $∆$G is based upon T, $∆$S and $∆$H according to the following equation (Gibbs-Helmholtz equation):

$∆\mathrm{G} = ∆\mathrm{H} - \mathrm{T}∆\mathrm{S}$

If the magnitude of $∆$H - T$∆$S is negative, then the reaction is spontaneous.

When T$∆$S > $∆$H and $∆$H and $∆$S are +ve, then $∆$G is negative.

In alkaline solution, KMnO₄ is reduced to MnO₂ (colourless).

$$\begin{gathered} 2{\mathrm{KMnO}}_4 + 2{\mathrm{H}}_2\mathrm{O} \to 2{\mathrm{MnO}}_2 + 2\mathrm{KOH} \\ \underline{ + 3\left[\mathrm{O}\right]\mathrm{Ki} + 3\left[\mathrm{O}\right] \to {\mathrm{KIO}}_{3 } } \\ 2{\mathrm{KMnO}}_4 + 2{\mathrm{H}}_2\mathrm{O} + \mathrm{KI} \to 2{\mathrm{MnO}}_2 + 2\mathrm{KOH} + {\mathrm{KIO}}_3 \end{gathered}$$

Hence, two moles of KMnO₄ are reduced by one mole of KI.

According to Fajan's rule, lower the size of cation higher will be its polarising power and higher will be covalent character.

$\therefore$ Polarising power $\propto$ $\frac{1}{\mathrm{size} \mathrm{of} \mathrm{action}}$

Covalent character $\propto$Polarising power

So, the correct order is NaCl < LiCl < BeCl₂

After decay of radioactive element, the last element Pb is obtained which belongs to group 14 of periodic table.