To determine the hybridisation of the central atom, we calculate its steric number (number of $\sigma$ bonds + number of lone pairs):

• $\text{NO}_2^-$: The central nitrogen atom forms $2$ $\sigma$ bonds with oxygen atoms (one single and one double bond) and possesses $1$ lone pair . The steric number is $2 + 1 = 3$, which corresponds to $sp^2$ hybridisation.
• $\text{NH}_3$: The central nitrogen atom forms $3$ $\sigma$ bonds with hydrogen atoms and has $1$ lone pair . The steric number is $3 + 1 = 4$, which gives $sp^3$ hybridisation.
• $\text{BF}_3$: The central boron atom forms $3$ $\sigma$ bonds with fluorine atoms and has $0$ lone pairs . The steric number is $3 + 0 = 3$, corresponding to $sp^2$ hybridisation.
• $\text{NH}_2^-$: The central nitrogen atom forms $2$ $\sigma$ bonds with hydrogen atoms and possesses $2$ lone pairs (due to the extra electron). The steric number is $2 + 2 = 4$, resulting in $sp^3$ hybridisation.
• $\text{H}_2\text{O}$: The central oxygen atom forms $2$ $\sigma$ bonds with hydrogen atoms and has $2$ lone pairs . The steric number is $2 + 2 = 4$, leading to $sp^3$ hybridisation.

Therefore, both central atoms in the pair $\text{BF}_3$ and $\text{NO}_2^-$ exhibit $sp^2$ hybridisation.