Correct order of bond angle, $∠ X O X (X = H, O$ of $F$ ) is:

H_{2} O > O_{3} > F_{2} O

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O_{3} > H_{2} O > F_{2} O

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F_{2} O > H_{2} O > O_{3}

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O_{3} > F_{2} O > H_{2} O

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Solution

The correct option is B $O_{3} > H_{2} O > F_{2} O$
Due to less lone pair-bond pair repulsion, $O_{3}$ has bond angle greater than $H_{2} O$ & $F_{2} O$ . Among, $H_{2} O$ & $F_{2} O$ , bond angle of $H_{2} O$ is greater than that of $F_{2} O$ because Fluorine has a higher electronegativity than Hydrogen and thus faces more repulsion from the lone pair which decreases its bond angle.

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