Correct order regarding bond angle in the following species?

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${NH}_{4}^{+} > {NH}_{3} > {NH}_{2}^{-}$ ( $H \hat{N} H$ bond angle)

${NO}_{2}^{+} > {NO}_{3}^{-} > {NO}_{2}^{-}$ ( $O \hat{N} O$ bond angle)

${OCl}_{2} > {OF}_{2} > {OH}_{2}$ ( $X \hat{O} X$ ) (X = terminal atom)

${BCl}_{3} > {PCl}_{3} > {NCl}_{3}$ $(Cl \hat{X} Cl)$ CX = central atom)

answer is A, B, D.

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Detailed Solution

Due to the presence of 3 bond pairs and 1 lone pair of electrons in ${NH}_{3}$ , ${NH}_{4}^{+}$ will have a greater bond angle than ${NH}_{3}$ .
Since ${NO}_{2}^{+}$ only has 2 bonds and no lone pair is available, the bond angle is 180 degrees, hence ${NO}_{2}^{+}$ should have a bigger bond angle than ${NO}_{2}^{-}$ . Whereas in the case of ${NH}_{2}$ , the angle is less than 180 degrees since there is a lone pair with a bond pair. Thus, ${NO}_{2}^{-}$ has a smaller bond angle than ${NO}_{2}^{+}$ .
Compared to ${BF}_{3}$ , ${BCl}_{3}$ has a greater bond angle around the B atom. Compared to Cl, F is more electronegative. Consequently, the $B - F$ bond's bond a pair of electrons is, while the bond pair of electrons in the $B - Cl$ bond are more near B and less away from Cl.