Question

The $I_3^{-}$ ion has:

• A. Five equatorial lone pairs on the central $I$ atom and two axial bonding pairs in a trigonal bipyramidal arrangement
• B. Five equatorial lone pairs on the central $I$ atom and two axial bonding pairs in a pentagonal bipyramidal arrangement
• C. Three equatorial lone pairs on the central $I$ atom and two axial bonding pairs in a trigonal bipyramidal arrangement
• D. Two equatorial lone pairs on the central $I$ atom and three axial bonding pairs in a trigonal bipyramidal arrangement

Answer 1

The correct option is C Three equatorial lone pairs on the central $I$ atom and two axial bonding pairs in a trigonal bipyramidal arrangement

Hybridisation of $I_3^{-}=\frac{1}{2}(7+2+1)=5=s{p}^{3}d$
So the geometry of the molecule is trigonal bipyramidal.
Number of bond pairs in $I_3^{-}=2$
So, Number of lone pairs in $I_3^{-}=5-2=3$
So the shape of the molecule is linear and there are three equatorial lone pairs on the central $I$ atom and two axial bonding pairs in a trigonal bipyramidal arrangement.