In the linear `I_3^(-)` (triiodide ion), the central iodine atom contains

Options

A. two unpaired electrons
B. no unshared pair of electrons
C. four unshared pairs of electrons
D. three unshared pairs of electrons

Answer

D

Explanation

The formation of triiodide ion can be conceptualized as follows:
`I_2+I^(-)rarrI_3^(-)`
or `:overset(..)underset(..)I-overset(..)underset(..)I:+:overset(..)underset(..)I:^(-)rarr[:overset(..)underset(..)I-overset(..)underset(..)I:larr:overset(..)underset(..)I:]`
Calculation of formal charge (FC) shows that
for I on ends, `FC=7-(6+1)=0`
for I in middle, `FC=7-(6+2)=-1`
Note that the negative charge on `I^(-)` is nullified in `I_3^(-)` by the positive formal charge of the coordinate covalent bond or we can say that it is transferred to the central iodine. Also note that the central iodine atom in `I_3^(-)` has an expanded valence shell with 3 lone pairs of electrons. There is a shortcut method. Considering the negative charge on central I atom, we have `7+1=8` valence electrons for central I. Two of these are used up to form two covalent bonds leaving 3 lone pairs.