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