Find the pair with sp2 hybridisation of the central molecule

The correct option is BF₃ and NO₂⁻.

To determine the hybridisation of the central atom in a molecule or ion, we can use the steric number (SN) formula:
$$\mathrm{Steric\; Number\; (SN)}=\frac{1}{2}\left[V+M-C+A\right]$$
where:
V = number of valence electrons of the central atom
M = number of monovalent atoms/groups attached to the central atom
C = charge on the cation
A = charge on the anion

Let us calculate the steric number for both species in the correct option to verify their hybridisation:

1. Boron Trifluoride (BF₃):
Here, the central atom is Boron (B).
The number of valence electrons of Boron (V) = 3 (since B belongs to group 13).
The number of monovalent Fluorine atoms attached (M) = 3.
There is no charge on the molecule, so C = 0 and A = 0.
Calculating the steric number:
$$\mathrm{SN}=\frac{1}{2}\left[3+3-0+0\right]=\frac{6}{2}=3$$
A steric number of 3 corresponds to sp² hybridisation with a trigonal planar geometry.

2. Nitrite Ion (NO₂⁻):
Here, the central atom is Nitrogen (N).
The number of valence electrons of Nitrogen (V) = 5 (since N belongs to group 15).
Oxygen is a divalent atom, so it is not counted in the number of monovalent atoms (M = 0).
The anion has a charge of -1, so A = 1 and C = 0.
Calculating the steric number:
$$\mathrm{SN}=\frac{1}{2}\left[5+0-0+1\right]=\frac{6}{2}=3$$
A steric number of 3 corresponds to sp² hybridisation (with one lone pair and two bond pairs, yielding a bent/angular shape).

Since the central atoms of both BF₃ and NO₂⁻ are sp² hybridised, this pair is the correct answer.