The geometry with respect to the central atom | vedclass

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Question

(B) $1$. In $N(SiH_3)_3$,the lone pair on $N$ is donated into the empty $d$-orbitals of $Si$ ($p\pi-d\pi$ back bonding). This makes the $N$ atom $sp^2

Vedclass · Accepted Answer

planar,pyramidal,pyramidal

Vedclass · Answer

(B) $1$. In $N(SiH_3)_3$,the lone pair on $N$ is donated into the empty $d$-orbitals of $Si$ ($p\pi-d\pi$ back bonding). This makes the $N$ atom $sp^2$ hybridized,resulting in a planar geometry.$2$. In $Me_3N$ (trimethylamine),there is no back bonding. The $N$ atom is $sp^3$ hybridized with one lone pair,resulting in a pyramidal geometry.$3$. In $(SiH_3)_3P$,the central atom $P$ is larger than $N$. The energy gap between the $P$ lone pair and the empty $d$-orbitals of $Si$ is large,making back bonding insignificant. Thus,it retains a pyramidal geometry.

List-$I$	List-$II$
$A$. $BF_3$	$II$. Trigonal planar
$B$. $ClF_3$	$III$. $T$-shape
$C$. $NH_3$	$IV$. Trigonal pyramidal
$D$. $NH_4^+$	$I$. Tetrahedral

The geometry with respect to the central atom of the following molecules are: $N(SiH_3)_3;\, Me_3N;\, (SiH_3)_3P$

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Which of the following pairs of species have identical shapes?

$A$ molecule of the type $AX_5$ has square pyramidal geometry. Hence,the number of lone pairs on '$A$' is:

Match the following:
List-$I$ List-$II$
$A$. $BF_3$ $II$. Trigonal planar
$B$. $ClF_3$ $III$. $T$-shape
$C$. $NH_3$ $IV$. Trigonal pyramidal
$D$. $NH_4^+$ $I$. Tetrahedral

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