## Is ICl2- nonpolar or polar?

Answer:

Nonpolar. Its dipoles all cancel.

Never really heard of \(\mathrm{ICl}_{2}^{-}\), but since it’s more probable than \(\mathrm{ICl}^{2-} \ldots\)

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:

IMG

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

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 0 in all directions.

Therefore, \(\mathrm{ICl}_{2}^{-}\) 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.