Account for the following observations.
(a) $\mathrm{AlCl}_3$ is a Lewis acid.
(b) Though fluorine is more electronegative than chlorine yet $\mathrm{BF}_3$ is a weaker Lewis acid than $\mathrm{CI}_3$.
(c) $\mathrm{PbO}_2$ is a stronger oxidising agent than $\mathrm{SnO}_2$.
(d) The +1 oxidation state of thallium is more stable than its +3 state.
(a) In $\mathrm{AlCl}_3, \mathrm{Al}$ has only six electrons in its valence shell. It is an electron deficient species. Therefore, it acts as a Lewis acid (electron acceptor).
(b) In $\mathrm{BF}_3$ boron has a vacant $2 p$-orbital and fluorine has one $2 p$-completely filled unutilised orbital. Both of these orbitals belong to same energy level therefore, they can overlap effectively and form $p \pi-p \pi$ bond. This type of bond formation is known as back bonding.
While back bonding is not possible in $\mathrm{BCl}_3$, because there is no effective overlapping between the $2 p$-orbital of boron and $3 p$-orbital of chlorine. Therefore, electron deficiency of B is higher in $\mathrm{BCl}_3$ than that of $\mathrm{BF}_3$. That's why $\mathrm{BF}_3$ is a weaker Lewis acid than $\mathrm{BCl}_3$.
(c) In $\mathrm{PbO}_2$ and $\mathrm{SnO}_2$, both lead and tin are present in +4 oxidation state. But due to stronger inert pair effect, $\mathrm{Pb}^{2+}$ ion is more stable than $\mathrm{Sn}^{2+}$ ion. In other words, $\mathrm{Pb}^{4+}$ ions i.e., $\mathrm{PbO}_4$ is more easily reduced to $\mathrm{Pb}^{2+}$ ions than $\mathrm{Sn}^{4+}$ ions reduced to $\mathrm{Sn}^{2+}$ ions. Thus, $\mathrm{PbO}_2$ acts as a stronger oxidising agent than $\mathrm{SnO}_2$.
(d) $\mathrm{TI}^{+}$is more stable than $\mathrm{TI}^{3+}$ because of inert pair effect.
When aqueous solution of borax is acidified with hydrochloric acid, a white crystalline solid is formed which is soapy to touch. Is this solid acidic or basic in nature ? Explain.
When an aqueous solution of borax is acidified with HCl boric acid is formed.
$$\mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7+2 \mathrm{HCl}+5 \mathrm{H}_2 \mathrm{O} \longrightarrow 2 \mathrm{NaCl}+\underset{\text { Boric acid }}{4 \mathrm{H}_3 \mathrm{BO}_3}$$
Boric acid is a white crystalline solid. It is soapy to touch because of its planar layered structure. Boric acid is a weak monobasic acid. It is not a protonic acid but acts as a Lewis acid by accepting electrons from a hydroxyl ion.
$$\mathrm{B}(\mathrm{OH})_3+2 \mathrm{HOH} \longrightarrow\left[\mathrm{B}(\mathrm{OH})_4\right]^{-}+\mathrm{H}_3 \mathrm{O}^{+}$$
Three pairs of compounds are given below. Identify that compound in each of the pairs which has group 13 element in more stable oxidation state. Given reason for your choice. State the nature of bonding also.
(a) $\mathrm{TlCl}_3, \mathrm{TlCl}$
(b) $\mathrm{AlCl}_3, \mathrm{AlCl}$
(c) $\mathrm{InCl}_3, \mathrm{InCl}$
(a) TICl more stable than $\mathrm{TICl}_3$ due to inert pair effect. $\mathrm{TICl}_3$ is less stable and covalent in nature but TICl is more stable and ionic in nature.
(b) Due to absence of $d$-orbitals, Al does not show inert pair effect. Hence, its most stable oxidation state is +3 . Thus, $\mathrm{AlCl}_3$ is much more stable than AlCl . Further, in the solid or the vapour state, $\mathrm{AlCl}_3$ covalent in nature but in aqueous solutions, it ionises to form A $\mathrm{Al}^{3+}(a q)$ and $\mathrm{Cl}^{-}(a q)$ ions.
(c) Due to inert pair effect, indium exists in both +1 and +3 oxidation states out of which + 3 oxidation state is more stable than +1 oxidation state. In other words, $\mathrm{InCl}_3$ is more stable than InCl . Being unstable, In Cl undergoes disproportionation reaction.
$$3 \mathrm{InCl}(a q) \longrightarrow 2 \ln (s)+\ln ^{3+}(a q)+3 \mathrm{Cl}^{-}(a q)$$
$\mathrm{BCl}_3$ exists as monomer whereas $\mathrm{AlCl}_3$ is dimerised through halogen bridging. Give reason. Explain the structure of the dimer of $\mathrm{AlCl}_3$ also.
Boron halides do not exist as dimer due to small size of boron atom which makes it unable to accommodate four large sized halide ions. $\mathrm{AlCl}_3$ exists as dimer. Al makes use of vacant $3 p$-orbital by coordinate bond i.e., Al atoms complete their octet by forming dimers.
Boron fluoride exists as $\mathrm{BF}_3$ but boron hydride doesn't exist as $\mathrm{BH}_3$. Give reason. In which form does it exist ? Explain its structure.
Due to $p \pi-p \pi$ back bonding, the lone pair of electrons of $F$ is donated to the $B$-atom. This delocalisation reduces the deficiency of electrons on B thereby increasing the stability of $\mathrm{BF}_3$ molecule.
Due to absence of lone pair of electrons on H -atom, this compensation does not occur in $\mathrm{BH}_3$. In other words, electron deficiency of B stays and hence to reduce its electron deficiency, $\mathrm{BH}_3$ dimerises to form $\mathrm{B}_2 \mathrm{H}_6$. In $\mathrm{B}_2 \mathrm{H}_6$, four terminal hydrogen atoms and two boron atoms lie in one plane. Above and below this plane there are two bridging H -atoms. The four terminal $\mathrm{B}-\mathrm{H}$ bonds are regular while the two bridge $(\mathrm{B}-\mathrm{H}-\mathrm{B})$ bonds are three centre- two electron bonds.