The shape of a moleculs is determined by electron pair repulsions in the valence shell A.lp occupies a larger space than a bp because it is not shared by two nuclei The the lp lp repulsion is greater than the lp-lp repulsion, which in trun is greater the lp-lp repulsion. The presence of lp causes distortion of bond angles hence, a daviation from an ideal shape THe extent of distortion depends upon the orientation of the lp's around the central atom In a trigonal bipyramid, the lp's occupy equatorial positions than the apical ones In `AB_(n)` type molecules, as the `EN` of A increases, the bp's come closer and the repulsion between them increases. On the other hand, as `EN` of `B` increases, the lp s get farther and repulsion decreases
The shape of which of the following molecules will not be distored ?

Options

A. `BrF_(3)`
B. `CIF_(3)`
C. `XeF_(4)`
D. `XeF_(6)`

Answer

C

Explanation

For (a) and (b) `: :overset(..)Br F_(3) , :overset(..)CIF_(3)`
`SN =lp + bp = 2 + 3 =5`
Hybridisation `= sp^(3) d` Geometry = Tbp Shape = T -shape
For (c ) `:overset(..)XeF_(4), SN = lp + bp = 2 + 4 =6`
Hybridisation `=sp^(3) d^(2)` Geometry `=OH` Shape = Square planar
For (d) `: overset(..)XeF_(6),SN =lp + lp = 1 + 6 =7`
Hybridisation `sp^(3) d^(3)` Geometry = Pbp Shape = Distorted `OH` .