Question

$N{a}_2{B}_4{O}_7.10H_{2}O\underset{\Delta }{\overset{720{}^{o}C}{\to }}2NaB{O}_2+\left(X\right)+10H_{2}O$

$\left(X\right)+Ca{F}_2+H_{2}S{O}_4⟶\left(Y\right)+CaS{O}_4+H_{2}O$

$\left(Y\right)+NaB{H}_4⟶\left(Z\right)+NaB{F}_4$

$\left(Z\right)\underset{N{H}_3(1:2)}{\overset{\text{High temperature}}{\to }}\underset{\text{Inorganic benzene}}{B_3{N}_3{H}_6}$

Select the correct statements regarding compound $\left(X\right),\left(Y\right)$ and $\left(Z\right)$.

• A. Compound $\left(X\right)$ is $B_2{O}_3$ and compound $\left(Z\right)$ contains two $3C-2e$ bonds.
• B. Compound $\left(X\right)$ is $B_2{O}_3$ and compound $\left(Y\right)$ contains one $3C-2e$ bonds.
• C. Compound $\left(Y\right)$ is Lewis acid and stable due to $p\pi -p\pi$ back bonding.
• D. Compound $\left(Z\right)$ reacts with excess of $N{H}_3$ at high temperature gives inorganic benzene.

Answer 1

Answer: (A), (C), (D)

The correct options are
A Compound $\left(X\right)$ is $B_2{O}_3$ and compound $\left(Z\right)$ contains two $3C-2e$ bonds.
C Compound $\left(Y\right)$ is Lewis acid and stable due to $p\pi -p\pi$ back bonding.
D Compound $\left(Z\right)$ reacts with excess of $N{H}_3$ at high temperature gives inorganic benzene.

Borax on heating loses its water of crystallisation and then decomposes into $(NaB{O}_2+B_2{O}_3)$, the borax beads.
$N{a}_2{B}_4{O}_7.10H_{2}O\underset{\Delta }{\overset{720{}^{o}C}{\to }}2NaB{O}_2+\underset{\left(X\right)}{B_2{O}_3}+10H_{2}O$

$\underset{\left(X\right)}{B_2{O}_3}+Ca{F}_2+H_{2}S{O}_4⟶\underset{\left(Y\right)}{B{F}_3}+CaS{O}_4+H_{2}O$

$\underset{\left(Y\right)}{B{F}_3}+NaB{H}_4⟶\underset{\left(Z\right)}{B_2{H}_6}+NaB{F}_4$

$\underset{\left(Z\right)}{B_2{H}_6}\underset{N{H}_3(1:2)}{\overset{\text{High temperature}}{\to }}\underset{\text{Inorganic benzene}}{B_3{N}_3{H}_6}$



The central boron atom in $B{F}_3$ is electron-deficient. Boron has an empty pure unhybrid $2p$ orbital. Both boron and fluorine belong to the same period. So the lone pairs on the fluorine atoms are perfectly poised for $p\pi -p\pi$ backbonding. This makes $B{F}_3$ the weakest Lewis acid among boron trihalides.