A) \[Xe{{F}_{4}},s{{p}^{3}}\]
B) \[Xe{{F}_{2}},sp\]
C) \[Xe{{F}_{2}},s{{p}^{3}}d\]
D) \[Xe{{F}_{4}},s{{p}^{2}}\]
Correct Answer: C
Solution :
[c] Hybridisation in each case is \[Xe{{F}_{4}},s{{p}^{3}}{{d}^{2}},Xe{{F}_{2}},s{{p}^{3}}d.\]
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