A) sp hybridized
B) \[s{{p}^{2}}\] hybridized
C) sp and \[s{{p}^{2}}\] hybridized
D) sp, \[s{{p}^{2}}\] and \[s{{p}^{3}}\] hybridized
Correct Answer: C
Solution :
\[\begin{matrix} N\equiv \underset{sp}{\mathop{C}}\, \\ {} \\ N\equiv \underset{sp}{\mathop{C}}\, \\ \end{matrix}\,>\underset{s{{p}^{2}}}{\mathop{C}}\,=\underset{s{{p}^{2}}}{\mathop{C}}\,<\begin{matrix} \overset{sp\,\,\,\,\,\,\,\,\,\,\,}{\mathop{C\equiv N}}\, \\ {} \\ \underset{sp\,\,\,\,\,\,\,\,\,\,\,\,\,}{\mathop{C\equiv N}}\, \\ \end{matrix}\]
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