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 $\mathrm{NH}_3$.
A tetravalent element forms monoxide and dioxide with oxygen. When air is passed over heated element ( 1273 K ), producer gas is obtained. Monoxide of the element is a powerful reducing agent and reduces ferric oxide to iron.
Identify the element and write formulae of its monoxide and dioxide. Write chemical equations for the formation of producer gas and reduction of ferric oxide with the monoxide.
Producer gas is a mixture of CO and $\mathrm{N}_2$, therefore, the tetravalent element is carbon and its monoxide and dioxide are CO and $\mathrm{CO}_2$ respectively.
$2 \mathrm{C}(\mathrm{s})+\underbrace{\mathrm{O}_2(\mathrm{~s})+4 \mathrm{~N}_2(g)}_{\text {Air }} \xrightarrow{1273 \mathrm{~K}} \underbrace{2 \mathrm{CO}(g)+4 \mathrm{~N}_2(g)}_{\text {Producer gas }}$
The carbon monoxide is a strong reducing agent and reduces ferric oxide to iron.
$$\mathrm{Fe}_2 \mathrm{O}_3(\mathrm{~s})+3 \mathrm{CO}(\mathrm{g}) \xrightarrow{\Delta} 2 \mathrm{Fe}(\mathrm{s})+3 \mathrm{CO}_2(\mathrm{~g})$$