A) \[{{H}_{2}}O\]
B) \[C{{H}_{4}}\]
C) \[BC{{l}_{3}}\]
D) \[N{{H}_{3}}\]
Correct Answer: C
Solution :
Key Idea: Hybridisation \[=\frac{1}{2}\] [no. of electron in valence shell + no. of monovalent atoms charge on cation + charge on onion] [a] \[{{H}_{2}}O\] \[H=\frac{1}{2}\,(6+2+0-0)\] \[=\frac{8}{4}=4\] \[\therefore \] \[s{{p}^{3}}\] hybridisation [b] \[C{{H}_{4}}\] \[H=\frac{1}{2}\,(4+4+0-0)\] \[=\frac{8}{4}=4\] \[\therefore \] \[s{{p}^{3}}\] hybridisation [c] \[BC{{l}_{3}}\] \[H=\frac{1}{2}(3+3+0-0)\] \[=\frac{6}{2}=3\] \[\therefore \] \[s{{p}^{2}}\] hybridisation. [d] \[N{{H}_{3}}\] \[H=\frac{1}{2}(5+3+0-0)\] \[=\frac{8}{2}=4\] \[\therefore \] \[s{{p}^{3}}\] hybridisation. \[\therefore \] [c] is correct answer.
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