# ICl2 è non polare o polare?

Nonpolar. Its dipoles all cancel.

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

To draw the Struttura di Lewis, each halogen contributes 7 elettroni di valenza, and the charge contributes 1. So we have 7+7+7+1 = 22 elettroni di valenza.

Hence, we can distribute 6 su ciascun "Cl" e 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:

Poiché l' geometria elettronica was trigonal bipyramidal (5 electron groups), the geometria molecolare 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.)

Da:

• The molecule has two identical non-central atoms.
• La struttura è lineare, 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 elettronegatività difference is between "Cl" e "I"; the dipoles all cancel out to give a net dipole moment of mathbf(0) in all directions.

Perciò, "ICl"_2^(-) is projected to be non polare.

NOTA: Although iodine is meno electronegative, it ha 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.