The angle strain in cyclobutane is

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(D) In an ideal $sp^3$ hybridized carbon atom,the bond angle is $109^{\circ} 28^{\prime}$.In cyclobutane,the carbon atoms form a square,so the interna

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$9^{\circ} 44^{\prime}$

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(D) In an ideal $sp^3$ hybridized carbon atom,the bond angle is $109^{\circ} 28^{\prime}$.In cyclobutane,the carbon atoms form a square,so the internal bond angle is $90^{\circ}$.The total deviation from the tetrahedral angle is $109^{\circ} 28^{\prime} - 90^{\circ} = 19^{\circ} 28^{\prime}$.According to Baeyer's strain theory,the angle strain is half of this deviation.$	ext{Angle strain} = \frac{1}{2} 	imes (109^{\circ} 28^{\prime} - 90^{\circ}) = \frac{19^{\circ} 28^{\prime}}{2} = 9^{\circ} 44^{\prime}$.

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