The geometry with respect to the central atom | vedclass

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Question

(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

$planar, pyramidal, pyramidal$

Vedclass · Answer

(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.

List-$I$	List-$II$
$a. NH_3$	$i$. Square pyramidal
$b. ClF_3$	$ii$. Trigonal bipyramidal
$c. PCl_5$	$iii$. Trigonal pyramidal
$d. BrF_5$	$iv$. $T$-shape

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

Similar Questions

Which of the following pairs is isostructural?

The number of species below that have two lone pairs of electrons on their central atom is ..... . (Round off to the nearest integer)
$SF_{4}, BF_{4}^{-}, ClF_{3}, AsF_{3}, PCl_{5}, BrF_{5}, XeF_{4}, SF_{6}$

Match List-$I$ with List-$II$ :
List-$I$ List-$II$
$a. NH_3$ $i$. Square pyramidal
$b. ClF_3$ $ii$. Trigonal bipyramidal
$c. PCl_5$ $iii$. Trigonal pyramidal
$d. BrF_5$ $iv$. $T$-shape

Choose the correct answer from the options given below :

What are the structures and shapes of $SF_4$ and $ClF_3$?

Which of the following statements is incorrect?

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