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

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(D) In compound $(I)$,the lone pair of nitrogen is involved in $p\pi-d\pi$ back-bonding with the empty $d$-orbitals of silicon. This reduces the avail

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$II > I$

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(D) In compound $(I)$,the lone pair of nitrogen is involved in $p\pi-d\pi$ back-bonding with the empty $d$-orbitals of silicon. This reduces the availability of the lone pair for $p\pi-p\pi$ back-bonding with boron. In compound $(II)$,carbon has no empty $d$-orbitals,so the lone pair of nitrogen is fully available for $p\pi-p\pi$ back-bonding with boron. Therefore,the $\pi$-bond strength between $B$ and $N$ is greater in $(II)$ than in $(I)$.

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

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