Question

Incorrectly matched pair is:

• A. $XeO_3$ - pyramidal
• B. $XeF_4$ - tetrahedral
• C. $XeF_6$ - disorted octahedral
• D. $XeOF_4$ - square pyramidal

Answer 1

Answer: (B)

$XeF_4$ - tetrahedral

Answer 2

In $XeF _{6}$ there are $6$ bond pairs and one lone pair. Thus the shape of $XeF _{6}$ is distorted octahedral.

In $XeOF _{4}, Xe$ is $sp ^{3} d$ hybridized and the shape is square pyramidal.

In $XeO _{3}$, $Xe$ is $sp ^{3}$ hybridized and the shape is trigonal pyramidal.

In $XeF _{4}, Xe$ is $sp ^{3} d ^{2}$ hybridized. It has $4$ bond pairs and two lone pairs. Thus, the shape is square planar.