A) \[ds{{p}^{2}},\] square planar
B) \[s{{p}^{3}}{{d}^{2}},\] octahedral
C) \[s{{p}^{3}},\] tetrahedral
D) \[s{{p}^{3}}{{d}^{2}},\] square planar
Correct Answer: D
Solution :
[d]The hybridisation of a compound can be calculated using formula. |
\[H=\frac{1}{2}[V+MO\pm \text{anion}/\text{cation}]\] |
where, H= Hybridised orbitals |
V= Valence |
MO = Number of monoatomic atoms |
For \[Xe{{F}_{4}}\] |
\[H=\frac{1}{2}[8+4]=6\] |
\[\Rightarrow \] \[s{{p}^{3}}{{d}^{2}}\] |
It has 4 bond pairs and 2 lone pair, where 2 lone pair occupies axial positions and four F occupies equatorial positions. So, its geometry is square planar and not octahedral. |
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