Is ICl2- nonpolar or polar?

Nonpolar. Its dipoles all cancel.

Never really heard of `"ICl"_2^(-)`, but since it's more probable than `"ICl"^(2-)`...

To draw the Lewis structure, each halogen contributes `7` valence electrons, and the charge contributes `1`. So we have `7+7+7+1 = 22` valence electrons.

Hence, we can distribute `6` on each `"Cl"` and `2` per single bond for a total of `6+6+2+2 = 16`, putting the remaining `6` on iodine.

The hypothetical VSEPR-predicted structure would look like this:

Since the electron geometry was trigonal bipyramidal (`5` electron groups), the molecular geometry is triatomic linear.

(Taking away atoms from a trigonal bipyramidal molecular geometry, you would get, in order, see-saw, T-shaped, triatomic linear, then diatomic linear.)

Since:

• The molecule has two identical non-central atoms.
• The structure is linear, giving dipoles that are opposite in direction to each other.
• The three lone pairs of electrons are `120^@` away from each adjacent one, a rotationally-symmetric configuration; so, the lone-pair-bonding-pair repulsions sum to cancel out as well.

...it doesn't matter what the electronegativity difference is between `"Cl"` and `"I"`; the dipoles all cancel out to give a net dipole moment of `\mathbf(0)` in all directions.

Therefore, `"ICl"_2^(-)` is projected to be nonpolar.

NOTE: Although iodine is less electronegative, it has to hold the `-1` formal charge (but it would have a `+1` oxidation state, while each `"Cl"` holds a `-1` oxidation state and a `0` formal charge).

Since the only way to rework formal charges is to form a double bond using one of `"Cl"`'s lone pairs (giving a `-2` formal charge to iodine and `+1` to chlorine), it's most favorable as it is now.