Correct order of bond angle in $C H_{2} F_{2}$ is:

H^C H > H^C F > F^C F

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F^C F > H^C F > H^C H

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H^C H > F^C F > H^C F

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H^C F > F^C F > H^C H

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Solution

The correct option is A $F^C F > H^C F > H^C H$
In ${C H}_{2} F_{2}$ , the two Flourine atoms repel each other. So $F^C F$ angle is more. Two $^{'} H^{'}$ atoms are less repulsive.
So $H^C H$ bond angle is minimum and $H^C F$ bond angle is intermediate between them.
So $F^C F > H^C F > H^C H$

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