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.
(a) What are silicones? State the uses of silicones.
(b) What are boranes ? Give chemical equation for the preparation of diborane.
(a) Silicones are a group of organosilicon polymers, which have $\left(R_2 \mathrm{SiO}\right)$ as a repeating unit. These may be linear silicones, cyclic silicones and cross-linked silicones.
These are prepared by the hydrolysis of alkyl or aryl derivatives of $\mathrm{SiCl}_4$, like $R \mathrm{SiCl}_3, R_2 \mathrm{SiCl}_2$, and $R_3 \mathrm{SiCl}$ and polymerisation of alkyl or aryl hydroxy derivatives obtained by hydrolysis.
Uses
These are used as sealant, greases, electrical insulators and for water proofing of fabrics. These are also used in surgical and cosmetic plants.
(b) Boron forms a number of covalent hydrides with general formulae $\mathrm{B}_n \mathrm{H}_{n+4}$ and $\mathrm{B}_n \mathrm{H}_{n+6}$. These are called boranes. $\mathrm{B}_2 \mathrm{H}_6$ and $\mathrm{B}_4 \mathrm{H}_{10}$ are the representative compounds of the two series respectively.
Preparation of Diborane
It is prepared by treating boron trifluoride with $\mathrm{LiAlH}_4$ in diethyl ether.
$$4 \mathrm{BF}_3+3 \mathrm{LiAlH}_4 \longrightarrow 2 \mathrm{~B}_2 \mathrm{H}_6+3 \mathrm{LiF}+3 \mathrm{AlF}_3$$
On industrial scale it is prepared by the reaction of $\mathrm{BF}_3$ with sodium hydride.
$$2 \mathrm{BF}_3+6 \mathrm{NaH} \xrightarrow{450 \mathrm{~K}} \mathrm{~B}_2 \mathrm{H}_6+6 \mathrm{NaF}$$