A) \[sp-s{{p}^{2}}\]
B) \[s{{p}^{3}}-s{{p}^{3}}\]
C) \[sp-s{{p}^{3}}\]
D) \[s{{p}^{2}}-s{{p}^{3}}\]
Correct Answer: D
Solution :
[d] \[\overset{1}{\mathop{C}}\,{{H}_{2}}=\overset{2}{\mathop{C}}\,H-\overset{3}{\mathop{C}}\,{{H}_{2}}-\overset{4}{\mathop{C}}\,{{H}_{2}}-\overset{5}{\mathop{C}}\,\equiv CH\] Hybridisation at \[{{C}_{2}}=s{{p}^{2}}\] Hybridisation at \[{{C}_{3}}=s{{p}^{3}}\] Thus, the \[{{C}_{2}}-{{C}_{3}}\] bond is of \[s{{p}^{2}}\cdot s{{p}^{3}}\]type.
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