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