Among all the given options, XeF₂ has more than one lone pair, i.e. 3.

SO₂ has one lone pair.

SiF₄ and CH₄ has no lone pairs.

BF₃ forms complex halides of the type [B${\mathrm{F}}_4^{-}$] in which B atom has C.N. 4, it cannot extend its C.N. beyond 4 due to the non-availability of d-orbitals in its configuration. Hence, B${\mathrm{F}}_6^{3-}$ ion (sp³d² hybridisation) is not formed. On the other hand, Al can extend its C.N. beyond 4 due to the availability of d-orbitals in its configuration.

1 mL of CS₂ weighs 2.63 gm

10 mL of CS₂ weighs 26.3 gm

$\underset{76 \mathrm{gm}}{\underset{12 + (2 \times 32)}{{\mathrm{CS}}_2}} + {\mathrm{O}}_2 \to \underset{67.2 \mathrm{L}}{\underbrace{\underset{22.4 \mathrm{L}}{{\mathrm{CO}}_2} +\underset{44.8 \mathrm{L}}{ 2{\mathrm{SO}}_2}}}$

$\because$ 76 gm of CS₂ will yield 67.2 L of a mixture of CO₂ and SO₂ at STP.

$\therefore$ 26.3 gm of CS₂ would yield $\frac{67.2}{76} \times 26.3$= 23.25 L

According to molecular orbital theory, the energy level of the given molecules are

${\mathrm{C}}_2^{2-} \to \mathrm{\sigma }1{\mathrm{s}}^2 {\mathrm{\sigma }}^{*}1{\mathrm{s}}^2 \mathrm{\sigma }2{\mathrm{s}}^2 {\mathrm{\sigma }}^{*}2{\mathrm{s}}^2 \mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^2 \mathrm{\pi }2{\mathrm{p}}_{\mathrm{y}}^2 \mathrm{\sigma }2{\mathrm{p}}_{\mathrm{z}}^2$

Bond Order : $\frac{1}{2}[10 - 4] = 3$

${\mathrm{He}}_2^{+} \to \mathrm{\sigma }1{\mathrm{s}}^2 {\mathrm{\sigma }}^{*}1{\mathrm{s}}^1$

Bond Order : $\frac{1}{2}\left[2 - 1\right] = \frac{1}{2} = 0.5$

 NO $\to \mathrm{\sigma }1{\mathrm{s}}^2 {\mathrm{\sigma }}^{*}1{\mathrm{s}}^2 \mathrm{\sigma }2{\mathrm{s}}^2 {\mathrm{\sigma }}^{*}2{\mathrm{s}}^2 \mathrm{\sigma }2{\mathrm{p}}_{\mathrm{z}}^2 \mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^2 \mathrm{\pi }2{\mathrm{p}}_{\mathrm{y}}^2 {\mathrm{\pi }}^{*}2{\mathrm{p}}_{\mathrm{x}}^1$

Bond Order : $\frac{1}{2}\left[10 - 5\right] = 2.5$

${\mathrm{O}}_2^{-} \to \mathrm{\sigma }1{\mathrm{s}}^2 {\mathrm{\sigma }}^{*}1{\mathrm{s}}^2 \mathrm{\sigma }2{\mathrm{s}}^2 {\mathrm{\sigma }}^{*}2{\mathrm{s}}^2 \mathrm{\sigma }2{\mathrm{p}}_{\mathrm{z}}^2 \mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^2 \mathrm{\pi }2{\mathrm{p}}_{\mathrm{y} }^2{\mathrm{\pi }}^{*}2{\mathrm{p}}_{\mathrm{x}}^2 {\mathrm{\pi }}^{*}2{\mathrm{p}}_{\mathrm{y}}^1$

Bond Order : $\frac{1}{2}\left[10 - 7\right] = 1.5$

So, the correct order of their increasing bond order is ${\mathrm{He}}_2^{+} < {\mathrm{O}}_2^{-} < \mathrm{NO}\hspace{0.17em}<\hspace{0.17em}{\mathrm{C}}_2^{2-}$

H₃PO₂ is a good reducing agent as it contains two P-H bonds and thus reduces AgNO₃ to Ag.

4AgNO₃ + H₃PO₂ + 2H₂O $\to$ 4Ag$\downarrow$  + H₃PO₄ + 4HNO₃

Assuming that reaction is started with 1 mol of A and at equilibrium x moles of A have reacted,

$\underset{1-\mathrm{x} \mathrm{mol}}{\underset{1 \mathrm{mol}}{\mathrm{A}}} \to \underset{\mathrm{x} \mathrm{mol} (\mathrm{at} \mathrm{equilibrium})}{\underset{0 \left(\mathrm{initially}\right)}{\mathrm{B}}}$

K_p = $\frac{{\mathrm{p}}_{\mathrm{B}}}{{\mathrm{p}}_{\mathrm{A}}}= 4$

    = $\frac{\mathrm{x}}{1-\mathrm{x}} = 4$

 x = 0.8

According to Bohr's theory,

r = $\frac{{\mathrm{n}}^2{\mathrm{h}}^2}{4{\mathrm{\pi }}^2{\mathrm{mZe}}^2\mathrm{k}}$ and E_n = $\frac{-{\mathrm{k}}^{2}2{\mathrm{\pi }}^2{\mathrm{mZ}}^2{\mathrm{e}}^4}{{\mathrm{n}}^2{\mathrm{h}}^2}$

Therefore, E_n ^{$\propto \frac{1}{{\mathrm{n}}^2}$}and r $\propto$n²

H₃PO₂, H₃PO₃ and H₃PO₄ contain one, two and three ionisable hydrogen atoms respectively.

H₃PO₂ $\rightleftharpoons$H⁺ + H₂P${\mathrm{O}}_2^{-}$

H₃PO₃ $\rightleftharpoons$ H⁺ + H₂P${\mathrm{O}}_3^{-}$ $\rightleftharpoons$ H⁺ + HP${\mathrm{O}}_3^{2-}$

H₃PO₄ $\rightleftharpoons$ H⁺ + H₂P${\mathrm{O}}_4^{-}$ $\rightleftharpoons$ H⁺ + HP${\mathrm{O}}_4^{2-}$

                                    $\rightleftharpoons$ H⁺ + P${\mathrm{O}}_4^{3-}$

But there is very little difference in acidity. In H₃PO₂, H₃PO₃ and H₃PO₄, P is sp³ hybridised in all. Therefore, all are tetrahedral.

H₃BO₃ is a very weak monobasic acid (K_a= 5.6 $\times$ 10^-10) but it does not act as a proton donor. It behaves as a Lewis acid i.e., it accepts a pair of electrons from OH^- ion of H₂O.

H - OH + B(OH)₃ $\to$ [B(OH)₄]^- + H⁺

The electronic configuration of (X) can be written as

X = 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p³

So element (X) has completely filled s orbital and d orbitals and half filled p orbitals.