Number of molecules/ions from the following in which the central atom is involved in sp3 hybridization is ________. NO₃^–, BCl₃, ClO₂^–, ClO₃

The correct answer is 2.

To determine the hybridization of the central atom in each of the given species, we can determine the steric number (SN) using the valence shell electron pair repulsion (VSEPR) theory formula:
$\mathrm{SN}=\frac{1}{2}\left(V+M-C+A\right)$
where:
- $V$ is the number of valence electrons on the central atom,
- $M$ is the number of monovalent atoms (like Cl, H) bonded to the central atom,
- $C$ is the cationic charge, and
- $A$ is the anionic charge.

The steric number determines the hybridization of the central atom as follows:
- SN = 2 corresponds to ^sp hybridization
- SN = 3 corresponds to ^sp2 hybridization
- SN = 4 corresponds to ^sp3 hybridization

Let us analyze each species:

1. NO₃^– (Nitrate ion):
- Central atom: Nitrogen (N), which has 5 valence electrons (V = 5).
- Oxygen is a divalent atom, so the number of monovalent atoms M = 0.
- Anionic charge A = 1.
- Cationic charge C = 0.
Calculating the steric number:
$\mathrm{SN}=\frac{1}{2}\left(5+0-0+1\right)=3$
Since the steric number is 3, the hybridization of N in NO₃^– is sp².

2. BCl₃ (Boron trichloride):
- Central atom: Boron (B), which has 3 valence electrons (V = 3).
- Chlorine is monovalent, so M = 3.
- C = 0, A = 0.
Calculating the steric number:
$\mathrm{SN}=\frac{1}{2}\left(3+3-0+0\right)=3$
Since the steric number is 3, the hybridization of B in BCl₃ is sp².

3. ClO₂^– (Chlorite ion):
- Central atom: Chlorine (Cl), which has 7 valence electrons (V = 7).
- Oxygen is divalent, so M = 0.
- Anionic charge A = 1.
- C = 0.
Calculating the steric number:
$\mathrm{SN}=\frac{1}{2}\left(7+0-0+1\right)=4$
Since the steric number is 4, the hybridization of Cl in ClO₂^– is sp³ (tetrahedral geometry with 2 bond pairs and 2 lone pairs, resulting in a bent shape).

4. ClO₃ (Chlorine trioxide radical):
- Central atom: Chlorine (Cl), which has 7 valence electrons.
- It forms three double bonds with three oxygen atoms, which uses 6 of its valence electrons.
- This leaves 1 unpaired electron on the chlorine atom.
- The molecule adopts a pyramidal geometry where the three bond pairs and the single unpaired electron occupy four tetrahedrally oriented orbitals.
Therefore, the hybridization of Cl in ClO₃ is sp³.

Among the given species, the central atom is involved in sp³ hybridization in 2 of them: ClO₂^– and ClO₃.