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.