(a) In gallium, due to poor shielding of valence electrons by the intervening $3 d$ electrons. The nuclear charge becomes effective, thus, atomic radius decreases and hence, the ionisation enthalpy of gallium is higher than that of aluminium.

(b) Due to small size of boron, the sum of its first three ionisation enthalpies is very high. This prevent it to form +3 ions and force it to form only covalent compound. That's why boron does not exist as $\mathrm{B}^{3+}$ ion.

(c) Aluminium forms $\left[\mathrm{AlF}_6\right]^{3-}$ ion because of the presence of vacant $d$-orbitals so it can expand its coordination number from 4 to 6 . In this complex, Al undergoes $s p^3 d^2$ hybridisation.

On the other hand, boron does not form $\left[\mathrm{BF}_6\right]^{3-}$ ion, because of the unavailability of $d$-orbitals as it cannot expand its coordination number beyond four. Hence, it can form $\left[\mathrm{BF}_4\right]^{-}$ion (sp ${ }^3$ hybridisation).

(d) Due to inert pair effect, Pb in +2 oxidation state is more stable than in +4 oxidation state hence $\mathrm{PbX}_2$ is more stable than $\mathrm{PbX}_4$.

(e) Due to inert pair effect, tendency to form +2 ions increases down the group, hence $\mathrm{Pb}^{2+}$ is more stable than $\mathrm{Pb}^{4+}$. That's why $\mathrm{Pb}^{4+}$ acts as an oxidising agent while $\mathrm{Sn}^{2+}$ is less stable than $\mathrm{Sn}^{4+}$ and hence $\mathrm{Sn}^{2+}$ acts as a reducing agent.

$\underset{\text { Reducing agent }}{\mathrm{Sn}^{2+}} \longrightarrow \mathrm{Sn}^{4+}+2 \mathrm{e}^{-}$

$\underset{\text { Oxidising agent }}{\mathrm{Pb}^{4+}}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pb}^{2+}$

(f) Electron gain enthalpy of Cl is more negative than electron gain enthalpy of fluorine because when an electron is added to $F$, the added electron goes to the smaller $n=2$ quantum level and suffers significant repulsion from other electrons present in this level. For $n=3$ quantum level (in Cl ) the added electron occupies a larger region of space and the electron-electron repulsion is much less.

(g) Due to inert pair effect, TI in +1 oxidation state is more stable than that of +3 oxidation state. Therefore, $\mathrm{TI}\left(\mathrm{NO}_3\right)_3$ acts as an oxidising agent.

(h) Property of catenation depends upon the atomic size of the element. Down the group, size increases and electronegativity decreases, thus the tendency to show catenation decreases. As the size of C is much smaller than Pb , therefore, carbon show property of catenation but lead does not show catenation.

(i) Unlike other boron halides, $\mathrm{BF}_3$ does not hydrolyse completely. However, it form boric acid and fluoroboric acid. This is because the HF first formed reacts with $\mathrm{H}_3 \mathrm{BO}_3$.

$\begin{aligned} & \left.\mathrm{BF}_3+3 \mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{H}_3 \mathrm{BO}_3+3 \mathrm{HF}\right] \times 4 \\ & \left.\mathrm{H}_2 \mathrm{BO}_3+4 \mathrm{HF} \longrightarrow \mathrm{H}^{+}\left[\mathrm{BF}_4\right]^{-}+3 \mathrm{H}_2 \mathrm{O}\right] \times 3 \\ & 4 \mathrm{BF}_3+3 \mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{H}_3 \mathrm{BO}_3+3\left[\mathrm{BF}_4\right]^{-}+3 \mathrm{H}^{+}\end{aligned}$

(j) In graphite, C is $s p^2$ hybridised. Carbon due to its smallest size and highest electronegativity among group 14 elements has strong tendency to form $p \pi-p \pi$ multiple bonds while silicon due to its larger size and less electronegativity has poor ability to form $p \pi-p \pi$ multiple bonds. That's why the element silicon does not form a graphite like structure.

(i) $$\mathop {N{a_2}{B_4}{O_7}}\limits_{Borax\,(A)} + 2HCl + 5{H_2}O \to 2NaCl + \mathop {4{H_3}B{O_3}}\limits_{Orthoboric\,acid\,(X)} $$

(ii) $${H_3}B{O_3}\buildrel {\Delta ,\,370\,K} \over \longrightarrow \mathop {HB{O_2}}\limits_{Metaboric\,acid} + {H_2}O$$

(iii) $$4HB{O_2}\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{ - {H_2}O}^{\Delta > 370\,K}} \mathop {[{H_2}{B_4}{O_7}]}\limits_{Tetraboric\,acid} \mathrel{\mathop{\kern0pt\longrightarrow} \limits_{Heat}^{{\mathop{\rm Re}\nolimits} d}} \mathop {2{B_2}{O_3} + {H_2}O}\limits_{Boron\,trioxide\,(Z)} $$

(i) $$4B{F_3} + 3LiAl{H_4} \to \mathop {2{B_2}{H_6}}\limits_{Diborane\,(X)} + 3LiF + 3Al{F_3}$$

(ii) $$\mathop {{B_2}{H_6}}\limits_{(X)} + 6{H_2}O \to \mathop {2{H_3}B{O_3}}\limits_{Orthoboric\,acid} + 6{H_2}$$

(iii) $$\mathop {{B_2}{H_6}}\limits_{(X)} + 3{O_2}\buildrel \Delta \over \longrightarrow {B_2}{O_3} + 3{H_2}O$$

A. $\rightarrow$ (5)

B. $\rightarrow$ (3)

C. $\rightarrow$ (4)

D. $\rightarrow(1,2)$

A. $\mathrm{BF}_4^{-}$Tetrahedral shape $s p^3$ hybridisation regular geometry.

B. $\mathrm{AlCl}_{3^{-}}$Octet not complete of Al , act as Lewis acid.

C. $\mathrm{SnO}^3 \mathrm{Sn}^{2+}$ can show +4 oxidation state.

D. $\mathrm{PbO}_2$ Oxidation state of Pb in $\mathrm{PbO}_2$ is +4 . Due to inert pair effect $\mathrm{Pb}^{4+}$ is less stable than $\mathrm{Pb}^{2+}$, acts as strong oxidising agent.

A. $\rightarrow$ (3)

B. $\rightarrow$ (4)

C. $\rightarrow$ (1)

D. $\rightarrow$ (5)

E. $\rightarrow$ (2)

A. $\mathrm{BH}_3$ is unstable forms diborane $\mathrm{B}_2 \mathrm{H}_6$ by 3 centre -2 electron bond show banana bond.

B. Gallium with low melting point and high boiling point makes it useful to measure high temperatures.

C. Borax is used as a flux for soldering metals for heat, scratch resistant coating in earthernwares.

D. Alumino silicate used as catalyst in petrochemical industries.

E. Quartz, is a crystalline form of silica.