# Is ICl2- nonpolar or polar?

Aug 4, 2016

Nonpolar. Its dipoles all cancel.

Never really heard of ${\text{ICl}}_{2}^{-}$, but since it's more probable than ${\text{ICl}}^{2 -}$...

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 $\text{Cl}$ 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 ${120}^{\circ}$ 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 $\text{Cl}$ and $\text{I}$; the dipoles all cancel out to give a net dipole moment of $\setminus m a t h b f \left(0\right)$ in all directions.

Therefore, ${\text{ICl}}_{2}^{-}$ 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 $\text{Cl}$ 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 $\text{Cl}$'s lone pairs (giving a $- 2$ formal charge to iodine and $+ 1$ to chlorine), it's most favorable as it is now.