$\mathrm{BCl}_3$ 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.
$$2 \mathrm{BF}_3+6 \mathrm{NaH} \xrightarrow{450 \mathrm{~K}} \mathrm{~B}_2 \mathrm{H}_6+6 \mathrm{NaF}$$
A compound (A) of boron reacts with $\mathrm{NMe}_3$ to give an adduct (B) which on hydrolysis gives a compound (C) and hydrogen gas. Compound (C) is an acid. Identify the compounds A, B, and C. Give the reactions involved.
Since, compound $(A)$ of boron reacts with $\mathrm{NMe}_3$ to form an adduct $(B)$ thus, compound $(A)$ is a Lewis acid. Since, adduct $(B)$ on hydrolysis gives an acid $(C)$ and hydrogen gas, therefore, $(A)$ must be $\mathrm{B}_2 \mathrm{H}_6$ and $(C)$ must be boric acid.
$\underset{\text { Diborane }(A)}{\mathrm{B}_2 \mathrm{H}_6}+\underset{\text { Adduct }(B)}{2 \mathrm{NMe}_3} \longrightarrow \underset{\text { Adduct }(B)}{2 \mathrm{BH}_3 \mathrm{NMe}_3}$
$\underset{(B)}{\mathrm{BH}_3 \cdot \mathrm{NMe}_3}+3 \mathrm{H}_2 \mathrm{O} \longrightarrow \underset{\text { Boric acid (C) }}{\mathrm{H}_3 \mathrm{BO}_3}+\mathrm{NMe}_3+6 \mathrm{H}_2$
A non-metallic element of group 13, used in making bullet proof vests is extremely hard solid of black colour. It can exist in many allotropic forms and has unusually high melting point. Its trifluoride acts as Lewis acid towards ammonia.
The element exihibits maximum covalency of four. Identify the element and write the reaction of its trifluoride with ammonia. Explain why does the trifluoride act as a Lewis acid?
The only non-metallic element of group 13 is boron. It is an extremely hard substance and is used in making bullet proof vests. It exists in many allotropy forms and usually high melting point. Since $B$ has only $s$ and $p$-orbitals but no $d$-orbitals. The maximum covalency of boron is 4 .
In trivalent state, the number of electrons around the central atom in a molecule will be six as in case of $\mathrm{BF}_3$. Such electron deficient molecules have tendency to accept a pair of electron to achieve stable electronic configuration and behave as Lewis acid. $\mathrm{BF}_3$ easily accepts lone pair of electron from $\mathrm{NH}_3$.
