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