Question

$I_2$ has less solubility in water and its solubility increases on adding KI solution. When KI and $I_2$ reacts then a species 'X' is formed by which solubility of $I_2$ increases.

Hybridization of anionic part of 'X' is :

• A. $s{p}^2$
• B. $s{p}^3$
• C. $s{p}^{3}d$
• D. $s{p}^3{d}^2$

Answer 1

The correct option is C $s{p}^{3}d$

$KI\left(aq\right)+I_2\left(aq\right)\to K^{+}\left(aq\right)+I^{3-}\left(aq\right)$

Consider the central I atom in $I_3^{-}$. It has seven valence electrons (group 17) two of which are used to for the I: I bonds; one electron is added for the charge:

lp = 1/2[7 -2 +1] = 3 lp This is an AX2E3 system and is based on trigonal bipyramidal coordination ($s{p}^{3}d$ hybridization).

Lone pairs are found to adopt the equatorial sites (less steric repulsion) and so the terminal I atoms adopt the two axial sites. The ion $I_3^{-}$ is predicted and found to be linear.

Hence option C is correct.