For the following chemical species, (i) NO_2^- (ii) NO_2^+ (iii)...

The bond angles of the given species can be determined by finding the hybridization and the number of lone pairs on the central nitrogen atom.

For $\mathrm{NO_2^+}$, the central nitrogen atom has $2$ bond pairs and $0$ lone pairs. The hybridization is $sp$, resulting in a linear geometry with a bond angle of $180^\circ$.

For $\mathrm{NO_3^-}$, the central nitrogen atom has $3$ bond pairs and $0$ lone pairs. The hybridization is $sp^2$, resulting in a trigonal planar geometry with a bond angle of $120^\circ$.

For $\mathrm{NO_2^-}$, the central nitrogen atom has $2$ bond pairs and $1$ lone pair. The hybridization is $sp^2$. Due to lone pair-bond pair repulsion, the bond angle is slightly less than $120^\circ$ (approximately $115^\circ$).

For $\mathrm{NH_4^+}$, the central nitrogen atom has $4$ bond pairs and $0$ lone pairs. The hybridization is $sp^3$, resulting in a tetrahedral geometry with a bond angle of $109.5^\circ$.

Comparing the bond angles, we get $180^\circ > 120^\circ > 115^\circ > 109.5^\circ$.

Therefore, the decreasing order of bond angles is $\mathrm{NO_2^+} > \mathrm{NO_3^-} > \mathrm{NO_2^-} > \mathrm{NH_4^+}$, which corresponds to (ii) $>$ (iv) $>$ (i) $>$ (iii).

Answer: (ii) $>$ (iv) $>$ (i) $>$ (iii)