Question

Which of the following statement is correct regarding $I_3^{-}$ ion?

• 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 two 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

The molecules assume different geometry based on the number of bond pairs/lone pairs in the valence shell of the central atom. In trigonal bipyramidal geometry, lone pairs prefer to occupy the equatorial positions to ensure minimal repulsion and maximum stability.

Steric number of $I_3^{-}$ = $\frac{1}{2}(7+2+1)=5$, which means that there are 2 bond pairs and 3 lone pairs and the hybridisation is $s{p}^{3}d$ giving it a trigonal bipyramdal geometry. In this geometry, the lone pairs must be placed in an equatorial position to reduce the net repulsions and thus, in this molecule the 3 lone pairs are placed in the equatorial positions and the two bond pairs are placed in the axial positions giving it a linear shape.