A) \[PC{{l}_{5}},I{{F}_{5}},Xe{{O}_{2}}{{F}_{2}}\]
B) \[B{{F}_{3}},PC{{l}_{3}},Xe{{O}_{3}}\]
C) \[S{{F}_{4}},Xe{{F}_{4}},CC{{l}_{4}}\]
D) \[Cl{{F}_{3}},XeO{{F}_{2}},Xe{{F}_{3}}^{+}\]
Correct Answer: D
Solution :
[d] \[Cl{{F}_{3}}\xrightarrow{{}}Hybridisation\] \[=3+\frac{1}{2}[7-3]=5(s{{p}^{3}}d)\] \[XeO{{F}_{2}}\xrightarrow{{}}Hybridisation\] \[=3+\frac{1}{2}[8-4]=5(s{{p}^{3}}d)\] \[XeF_{3}^{-}\xrightarrow{{}}Hybridisation\] \[=3+\frac{1}{2}[8-3-1]=5(s{{p}^{3}}d)\] All molecules have \[s{{p}^{3}}d\] hybridization and 2 lone pairs. Hence all have identical (T-shape).
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