When $\mathrm{BC}_3$ is treated with water, it hydrolyses and forms $\left[\mathrm{B}[\mathrm{OH}]_4\right]^{-}$ only whereas $\mathrm{Ald}_3$ in acidified aqueous solution forms $\left[\mathrm{Al}\left[\mathrm{H}_2 \mathrm{O}\right]_6\right]^{3+}$ ion. Explain what is the hybridisation of boron and aluminium in these species?
In trivalent state, most of the compounds being covalent are hydrolysed in water, e.g., $\mathrm{BCl}_3$ on hydrolysis in water form $\left[\mathrm{B}(\mathrm{OH})_4\right]^{-}$ species, the hybridisation state of B is $s p^3$.
$$\begin{gathered} \mathrm{BCl}+3 \mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{B}(\mathrm{OH})_3+3 \mathrm{HCl} \\ \mathrm{B}(\mathrm{OH})_3+\mathrm{H}_2 \mathrm{O} \longrightarrow\left[\mathrm{B}(\mathrm{OH})_4\right]^{-}+\mathrm{H}^{+} \end{gathered}$$
$\mathrm{AlCl}_3$ in acidified aqueous solution form octahedral $\left[\mathrm{Al}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$ ion. In this complex, the $3 d$ orbitals of Al are involved and the hybridisation state of Al is $s p^3 d^2$
$$\mathrm{AlCl}_3+\mathrm{H}_2 \mathrm{O} \xrightarrow{\mathrm{HCl}}\left[\mathrm{Al}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}+3 \mathrm{Cl}^{-}(\mathrm{aq})$$
Aluminium dissolves in mineral acids and aqueous alkalies and thus shows amphoteric character. A piece of aluminium foil is treated with dilute hydrochloric acid or dilute sodium hydroxide solution in a test tube and on bringing a burning matchstick near the mouth of the test tube, a pop sound indicates the evolution of hydrogen gas. The same activity when performed with concentrated nitric acid, reaction doesn't proceed. Explain the reason.
Aluminium being amphoteric in nature dissolves both in acids and alkalies evolving $\mathrm{H}_2$ gas which burns with a pop sound.
$2 \mathrm{Al}+6 \mathrm{HCl} \longrightarrow 2 \mathrm{AlCl}_3+3 \mathrm{H}_2$
$2 \mathrm{AI}+\mathrm{NaOH}+2 \mathrm{H}_2 \mathrm{O} \longrightarrow \underset{\text { Sodium meta aluminate }}{2 \mathrm{NaAlO}_2}+3 \mathrm{H}_2$
But when Al is treated with conc. $\mathrm{HNO}_3$, a thin protective layer of $\mathrm{Al}_2 \mathrm{O}_3$ is formed on its surface which prevents further action.
$2 \mathrm{AI}+6 \mathrm{HNO}_3 \longrightarrow \mathrm{Al}_2 \mathrm{O}_3+6 \mathrm{NO}_2+3 \mathrm{H}_2 \mathrm{O}$
Explain the following.
(a) Gallium has higher ionisation enthalpy than aluminium.
(b) Boron does not exist as $\mathrm{B}^{3+}$ ion.
(c) Aluminium forms $\left[\mathrm{AlF}_6\right]^{3-}$ ion but boron does not form $\left[\mathrm{BF}_6\right]^{3-}$ ion.
(d) $\mathrm{Pb} X_2$ is more stable than $\mathrm{PbX}_4$.
(e) $\mathrm{Pb}^{4+}$ acts as an oxidising agent but $\mathrm{Sn}^{2+}$ acts as a reducing agent.
(f) Electron gain enthalpy of chlorine is more negative as compared to fluorine.
(g) $\mathrm{Tl}\left(\mathrm{NO}_3\right)_3$ acts as an oxidising agent.
(h) Carbon shows catenation property but lead does not.
(i) $\mathrm{BF}_3$ does not hydrolyse.
(j) Why does the element silicon, not form a graphite like structure whereas carbon does
(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.
Identify the compounds $A, X$ and $Z$ in the following reactions.
$$A+2 \mathrm{HC}+5 \mathrm{H}_2 \mathrm{O} \longrightarrow 2 \mathrm{NaC}+X$$
$X \xrightarrow[370 \mathrm{~K}]{\Delta} \mathrm{HBO}_2 \underset{>370 \mathrm{~K}}{\stackrel{\Delta}{\longrightarrow}} Z$
(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) $$4HB{O_2}\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{ - {H_2}O}^{\Delta > 370\,K}} \mathop {[{H_2}{B_4}{O_7}]}\limits_{Tetraboric\,acid} \mathrel{\mathop{\kern0pt\longrightarrow} \limits_{Heat}^{{\mathop{\rm Re}\nolimits} d}} \mathop {2{B_2}{O_3} + {H_2}O}\limits_{Boron\,trioxide\,(Z)} $$
Complete the following chemical equations.
$$\begin{gathered} \mathrm{Z}+3 \mathrm{LiAlH}_4 \longrightarrow X+3 \mathrm{LiF}+3 \mathrm{AlF}_3 \\ X+6 \mathrm{H}_2 \mathrm{O} \longrightarrow Y+6 \mathrm{H}_2 \\ 3 X+3 \mathrm{O}_2 \xrightarrow{\Delta} \mathrm{B}_2 \mathrm{O}_3+3 \mathrm{H}_2 \mathrm{O} \end{gathered}$$
(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) $$\mathop {{B_2}{H_6}}\limits_{(X)} + 3{O_2}\buildrel \Delta \over \longrightarrow {B_2}{O_3} + 3{H_2}O$$