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 anion] (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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