Both N(SiH_3)_3 and NH(SiH_3)_2 compounds hav | vedclass

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(B) In $NH(SiH_3)_2$,the lone pair on the $N$ atom is involved in back bonding with two empty $3d$-orbitals of two silicon atoms,whereas in $N(SiH_3)_

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$N-Si$ bond length in $NH(SiH_3)_2 > N-Si$ bond length in $N(SiH_3)_3$

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(B) In $NH(SiH_3)_2$,the lone pair on the $N$ atom is involved in back bonding with two empty $3d$-orbitals of two silicon atoms,whereas in $N(SiH_3)_3$,it is involved with three empty $3d$-orbitals of three silicon atoms.Due to the distribution of the lone pair over fewer silicon atoms in $NH(SiH_3)_2$,the extent of $ppi-dpi$ back bonding is greater in $NH(SiH_3)_2$ compared to $N(SiH_3)_3$.Greater back bonding strength results in a shorter $N-Si$ bond length. Therefore,the $N-Si$ bond length in $NH(SiH_3)_2$ is less than that in $N(SiH_3)_3$.Thus,the statement that the $N-Si$ bond length in $NH(SiH_3)_2 > N-Si$ bond length in $N(SiH_3)_3$ is incorrect.

Both $N(SiH_3)_3$ and $NH(SiH_3)_2$ compounds have a trigonal planar skeleton. Which of the following statements is incorrect regarding these compounds?

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