To determine the type of hybrid orbitals used by the chlorine atom in ${\mathrm{ClO}}_2^{-}$, we first need to find the steric number of the central atom (chlorine).

Step 1: Calculate the total number of valence electrons.
Chlorine (Cl) is in Group 17, so it has 7 valence electrons.
Oxygen (O) is in Group 16, so it has 6 valence electrons.
The ion has a -1 charge, meaning there is an additional electron.
Total valence electrons = $7\text{ (for Cl)}+2\times 6\text{ (for 2 O atoms)}+1\text{ (for -1 charge)}=7+12+1=20$ electrons.

Step 2: Draw the Lewis structure.
Place Cl as the central atom. Connect the oxygen atoms to Cl with single bonds. This uses 4 electrons (2 bonds).
Remaining electrons = $20-4=16$.
Distribute these 16 electrons as lone pairs to satisfy the octets of the oxygen atoms first. Each oxygen needs 6 more electrons (3 lone pairs). This uses $2\times 6=12$ electrons.
Remaining electrons = $16-12=4$.
Place these 4 electrons as lone pairs on the central chlorine atom. This means Cl has 2 lone pairs.

Step 3: Determine the steric number.
Steric number = (Number of sigma bonds around the central atom) + (Number of lone pairs on the central atom).
In ${\mathrm{ClO}}_2^{-}$, there are 2 sigma bonds (Cl-O single bonds) and 2 lone pairs on the central Cl atom.
Steric number = $2\text{ (sigma bonds)}+2\text{ (lone pairs)}=4$.

Step 4: Determine the hybridization.
A steric number of 4 corresponds to $s{p}^3$ hybridization.

Option Analysis:

• A) $s{p}^3$: This is correct. A steric number of 4 (2 bond pairs + 2 lone pairs) leads to $s{p}^3$ hybridization. The molecular geometry is bent, but the electron geometry is tetrahedral.
• B) $s{p}^2$: This would correspond to a steric number of 3.
• C) $sp$: This would correspond to a steric number of 2.
• D) none of these: This is incorrect as $s{p}^3$ is the correct hybridization.

Correct Answer: (A)