Decreasing order of bond angle is

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(A) To determine the bond angle,we analyze the hybridization and geometry of each molecule:$1$. $BeCl_2$: It has $sp$ hybridization and a linear geome

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$BeCl_2 > NO_2 > SO_2$

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(A) To determine the bond angle,we analyze the hybridization and geometry of each molecule:$1$. $BeCl_2$: It has $sp$ hybridization and a linear geometry with a bond angle of $180^{\circ}$.$2$. $NO_2$: It has $sp^2$ hybridization with one unpaired electron on the central nitrogen atom. The bond angle is approximately $134^{\circ}$.$3$. $SO_2$: It has $sp^2$ hybridization with one lone pair on the central sulfur atom. The lone pair-bond pair repulsion is greater than the single electron-bond pair repulsion,leading to a smaller bond angle of approximately $119^{\circ}$.Therefore,the decreasing order of bond angle is $BeCl_2 (180^{\circ}) > NO_2 (134^{\circ}) > SO_2 (119^{\circ})$.

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