The geometry with respect to the central atom | vedclass

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(B) $1$. In $N(SiH_3)_3$,the nitrogen atom has a lone pair which is donated into the vacant $3d$-orbital of the silicon atom,forming a $p\pi-d\pi$ bac

Vedclass · Accepted Answer

$planar, pyramidal, pyramidal$

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(B) $1$. In $N(SiH_3)_3$,the nitrogen atom has a lone pair which is donated into the vacant $3d$-orbital of the silicon atom,forming a $p\pi-d\pi$ back-bonding. This makes the nitrogen atom $sp^2$ hybridized,resulting in a $planar$ geometry.$2$. In $Me_3N$ (trimethylamine),the carbon atom does not have vacant $d$-orbitals,so no back-bonding occurs. The nitrogen atom remains $sp^3$ hybridized with a lone pair,resulting in a $pyramidal$ geometry.$3$. In $(SiH_3)_3P$,the phosphorus atom is larger and has a lower electronegativity compared to nitrogen. The lone pair on phosphorus is stereochemically active and does not participate in significant back-bonding with silicon. Thus,it retains a $pyramidal$ geometry.

The geometry with respect to the central atom of the molecules $N(SiH_3)_3$,$Me_3N$,and $(SiH_3)_3P$ respectively are

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