Compare pi-bond strength between B and N in t | vedclass

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(D) In compound $(I)$,the lone pair of electrons on the $N$ atom is partially donated to the empty $d$-orbitals of $Si$ through $p\pi-d\pi$ back-bondi

Vedclass · Accepted Answer

$II > I$

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(D) In compound $(I)$,the lone pair of electrons on the $N$ atom is partially donated to the empty $d$-orbitals of $Si$ through $p\pi-d\pi$ back-bonding.This decreases the electron density available for $p\pi-p\pi$ back-bonding with the empty $p$-orbital of $B$.In compound $(II)$,$C$ has no empty $d$-orbitals,so the lone pair on $N$ is fully available for $p\pi-p\pi$ back-bonding with $B$.Therefore,the $\pi$-bond strength is greater in $(II)$,i.e.,$II > I$.

List-$I$	List-$II$
$(A)$ $C_{2}H_{4}$	$(I)$ $3\sigma$ bonds, $2\pi$ bonds
$(B)$ $C_{2}H_{2}$	$(II)$ $3\sigma$ bonds, $1$ lone pair
$(C)$ $CH_{4}$	$(III)$ $4\sigma$ bonds
$(D)$ $NH_{3}$	$(IV)$ $5\sigma$ bonds, $1\pi$ bond

Compare $\pi$-bond strength between $B$ and $N$ in the following compounds: $(I)$ $((CH_3)_3Si)_2N^{-}BH_2$ and $(II)$ $((CH_3)_3C)_2N^{-}BH_2$.

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