A) \[sp,s{{p}^{3}}\]
B) \[s{{p}^{2}},sp\]
C) \[s{{p}^{2}},s{{p}^{2}}\]
D) \[sp,sp\]
Correct Answer: B
Solution :
\[\overset{1}{\mathop{C}}\,{{H}_{3}}-\underset{s{{p}^{2}}}{\mathop{\underset{\uparrow }{\overset{2}{\mathop{C}}}\,H}}\,=\underset{sp}{\mathop{\underset{\uparrow }{\overset{3}{\mathop{C}}}\,}}\,=\overset{4}{\mathop{C}}\,H-\overset{5}{\mathop{C}}\,{{H}_{3}}\]
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