A) \[{{24}^{o}}44'\]
B) \[{{29}^{o}}16'\]
C) \[{{19}^{o}}22'\]
D) \[{{9}^{o}}44'\]
Correct Answer: D
Solution :
When carbon is bonded to four other atoms, the angle between any pair of bonds \[={{109}^{o}},28'\] (tetrahedral angle) but the ring of cyclobutane is square with four angles of \[{{90}^{o}}\]. So, deviation of the bond angle (angle strain) in cyclobutane \[={{109}^{o}}28'-{{90}^{o}}/2\] \[={{19}^{o}}28'/2={{9}^{o}}=44'\]
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