A) \[s{{p}^{2}}\] and \[s{{p}^{2}}\]
B) \[s{{p}^{3}}\] and \[sp\]
C) \[sp\] and \[s{{p}^{2}}\]
D) \[sp\] and \[sp\]
Correct Answer: C
Solution :
sp and \[s{{p}^{2}}\] \[N\equiv \overset{sp}{\mathop{\underset{1}{\mathop{C}}\,}}\,-\underset{2}{\overset{s{{p}^{2}}}{\mathop{CH}}}\,=\underset{3}{\mathop{C{{H}_{2}}}}\,\]
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