Select the correct order of bond angle.

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(D) $1$. For $NO_2^+ > NO_2 > NH_4^+$: $NO_2^+$ is linear $(180^\circ)$,$NO_2$ is bent (approx $134^\circ$),and $NH_4^+$ is tetrahedral $(109.5^

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(D) $1$. For $NO_2^+ > NO_2 > NH_4^+$: $NO_2^+$ is linear $(180^\circ)$,$NO_2$ is bent (approx $134^\circ$),and $NH_4^+$ is tetrahedral $(109.5^\circ)$. Thus,the order is correct. $2$. For $COF_2 < COCl_2 < COBr_2$: As the size of the halogen atom increases,the steric repulsion between the halogen atoms increases,leading to an increase in the $X-C-X$ bond angle. Thus,the order is correct. $3$. For $H_2S < SF_2 < SCl_2$: The bond angle depends on the electronegativity of the central atom and the surrounding atoms. $H_2S$ (approx $92^\circ$) < $SF_2$ (approx $98^\circ$) < $SCl_2$ (approx $103^\circ$). Thus,the order is correct. Since all statements are correct,the answer is $D$.

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