Question

The hybridisation of the underlined atom is not affected in which of the following changes?

• A. $\underset{–}{P}{H}_3+H^{+}\to P{H}_4^{+}$
• B. $\underset{–}{B}{F}_3+H_{2}S\to F_{3}B\leftarrow H_{2}S$
• C. $H_3\underset{–}{B}{O}_3\to HB{O}_2+H_{2}O$
• D. $2H\underset{–}{Cl}{O}_2\to HClO+HCl{O}_3$

Answer 1

Answer: (A), (D)

The correct options are
A $\underset{–}{P}{H}_3+H^{+}\to P{H}_4^{+}$
D $2H\underset{–}{Cl}{O}_2\to HClO+HCl{O}_3$

Let us discuss each of the given reactions in detail.
$\left(A\right)$ $\underset{–}{P}{H}_3+H^{+}\to P{H}_4^{+}$
The hybridization of P remains the same, i.e., $s{p}^3$ hybridization.
In $P{H}_3$, $P$ atom contains $3$ bond pairs of electrons and one lone pair of electrons.
In $P{H}_4^{+}$, $P$ atom contains $4$ bond pairs of electrons.

$\left(B\right)$ $\underset{–}{B}{F}_3+H_{2}S\to H_{2}S\to B{F}_3$
The hybridization of $B$ changes from $s{p}^2$ to $s{p}^3$.
In $B{F}_3$, $B$ contains $3$ bond pairs and zero lone pair of electrons.
In $H_{2}S\to B{F}_3$, $B$ contains $4$ bond pairs of electrons and zero lone pair of electrons.

$\left(C\right)$ $H_3\underset{–}{B}{O}_3\to HB{O}_2+H_{2}O$
The hybridization of $B$ changes from $s{p}^2$ to $sp$.
In $H_{3}B{O}_3$, B contains three bond pairs of electrons and zero lone pairs of electrons.
In $HB{O}_2$, B contains two bonding domains.

(D) $2H\underset{–}{Cl}{O}_2\to HClO+HCl{O}_3$
The hybridization changes from $s{p}^3$ to no hybridization.
In $HCl{O}_2$, $Cl$ has $2$ bonding domains and $2$ lone pairs of electrons. It undergoes $s{p}^3$ hybridization.
In $HCl{O}_3$, $Cl$ has $3$ bond pairs of electrons and $1$ lone pair of electrons and undergoes $s{p}^3$ hybridization.

Hence, the first and fourth reactions do not undergo any change in the hybridisation of the underlined atom.