The correct order of bond angle is

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(A) Let us analyze each option:$A$: In $NCl_3$ and $NBr_3$,the central atom is the same $(N)$. As the electronegativity of the surrounding atom decrea

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$NCl_3 > NBr_3$

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(A) Let us analyze each option:$A$: In $NCl_3$ and $NBr_3$,the central atom is the same $(N)$. As the electronegativity of the surrounding atom decreases $(Cl > Br)$,the bond pair-bond pair repulsion decreases,leading to a smaller bond angle. Thus,$NCl_3 > NBr_3$ is correct.$B$: In hydrides of group $16$ $(H_2O, H_2S, H_2Se, H_2Te)$,as the size of the central atom increases,the bond angle decreases. Thus,$H_2S > H_2Te$ is the correct order.$C$: Both $N_3^-$ and $I_3^-$ are linear molecules with a bond angle of $180^{\circ}$. Thus,$N_3^- = I_3^-$.$D$: In $NO_3^-$ ($sp^2$ hybridized,$120^{\circ}$) and $NO_2^-$ ($sp^2$ hybridized with one lone pair,$ NO_2^-$.Therefore,the correct statement is $A$.

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