Question 2

1. $NO_2^+$: The central Nitrogen atom has $sp$ hybridization. It is linear with a bond angle of $180^\circ$.
2. $NO_2$: Nitrogen has $sp^2$ hybridization with one odd electron. The repulsion from a single electron is less than that from a lone pair or a bond pair, leading to an angle of approximately $134^\circ$ (greater than $120^\circ$).
3. $NO_3^-$: Nitrogen has $sp^2$ hybridization with three bonding domains and no lone pairs. It is trigonal planar with a bond angle of $120^\circ$.
4. $NO_2^-$: Nitrogen has $sp^2$ hybridization with one lone pair. Lone pair - bond pair repulsion reduces the ONO angle to approximately $115^\circ$ (less than $120^\circ$).

Comparing the angles: $115^\circ (NO_2^-) < 120^\circ (NO_3^-) < 134^\circ (NO_2) < 180^\circ (NO_2^+)$. Thus, the order is $NO_2^- < NO_3^- < NO_2 < NO_2^+$.