Question

Explain the following:
(i) Gallium has higher ionization enthalpy than aluminum.
(ii) Boron does not exist as $B^{3+}$ion.
(iii) Aluminum forms $[Al{F}_6{]}^{3-}$ ion but boron does not form $[B{F}_6{]}^{3-}$ ion.
(iv) $Pb{X}_2$ is more stable than $Pb{X}_4$
(v) $P{b}^{4+}$ acts as an oxidizing agent, but $S{n}^{2+}$ acts as a reducing agent.
(vi) Electron gain enthalpy of chlorine is more negative as compared to fluorine.
(vii) $TI(N{O}_3{)}_3$ acts as an oxidizing agent.
(viii) Carbon shows catenation property but lead does not.
(ix) $B{F}_3$ does not hydrolyze.
(x) Why does the element silicon, not form a graphite-like structure whereas carbon does?

Answer 1

(i) The ionization enthalpy value of Ga is higher than Al due to the inability of d- and f- electrons, which have a screening effect to compensate for the increase in nuclear charge.
(ii) Boron has small size and sum of ${\Delta }_i{H}_1+{\Delta }_i{H}_2+{\Delta }_i{H}_3$ is very high. Boron does not form $B^{3+}$ ion therefore, give covalent compounds.
(iii) $Al$ has vacant 3d-orbitals and can expand its co-ordination no and forms $[Al{F}_6{]}^{3-}$. On the other hand, Boron does not have orbitals and can not form $[B{F}_6{]}^{3-}$ and cannot expand its covalence beyond $4$ and thus, gives$\left[B{F}_4\right].$
(iv) Due to the inert pair effect, a $+2$ oxidation state is more stable than a $+4$ oxidation state.
(v) $P{b}^{4+}$ by gaining $2$ electrons change into $P{b}^{2+}$ which is more stable due to inert pair effect. $S{n}^{2+}$ is less stable than $S{n}^{4+}$ by losing electrons. Therefore, $P{b}^{4+}$ acts as an oxidizing agent while $S{n}^{2+}$ acts as a reducing agent.
(vi) The size of F atom is very small and incoming electron feels interelectronic repulsion and electron gain enthalpy of F is less negative as compared to Cl.
(vii) Due to the inert pair effect, Tl is more stable in $+1$ oxidation state than that of $3+$ oxidation state. Therefore, $Tl(N{O}_3{)}_3$ acts as a strong oxidizing agent.
(viii) Carbon atoms have the tendency to link with another through covalent bonds to form chains and rings. This property is called catenation. This is because $C-C$ bonds are very strong. Down the group, the size increases, and electronegativity decreases, and thereby, the tendency to show catenation decreases. This can be clearly seen from bond enthalpy values. The order of catenation is $C>>Si>GE\approx Sn.$ Lead does not show catenation.
(ix) $B{F}_3$ does not hydrolyze completely. It forms boric acid and fluroboric acid this is because the HF formed reacts with $H_{3}B{O}_3$.

$B{F}_3+3H_{2}O\to H_{3}B{O}_3+3HF\times 4$

$H_{3}B{O}_3+3HF\to H^{+}[B{F}_4{]}^{-}+3H_{2}O\times 3$

$4B{F}_3+3H_{2}O\to H_{3}B{O}_3+3[B{F}_4{]}^{-}+3H^{+}$
(x) In graphite, carbon is $s{p}^3$ hybridized. Carbon has a tendency to form multiple $p\pi -p\pi$ bonds due to its small size and highest electronegativity in group $14$. Silicon due to its large size and less electronegativity cannot form multiple bonds. Thus, silicon cannot form a graphite-like structure.