The sum of lone pairs present on the central atom of the interhalogen IF₅ and IF₇ is _________

Solution

Chemistry ma’am · 2 steps

Start with iodine's 7 valence electrons, take away one for each I–F bond, and pair up what is left.

1. Count on iodine. Iodine is in group 17, so it has 7 valence electrons. Each I–F single bond uses one of them, and the leftover electrons pair up as lone pairs.

   lone pairs on I = (7 - number of F atoms)/2

   Molecule	Electrons used in bonds	Electrons left on I	Lone pairs on I	Shape
   IF₅	5	7 - 5 = 2	1	square pyramidal (sp³d²)
   IF₇	7	7 - 7 = 0	0	pentagonal bipyramidal (sp³d³)

2. So. The sum of lone pairs on the central iodine atoms is 1 + 0 = 1.

The common slip: 2. Counting the 2 leftover electrons on iodine in IF₅ as two lone pairs gives 2. Two electrons make only one pair.

The idea to keep. For a central halogen, lone pairs = (7 − number of bonds) ÷ 2. In XFₙ with odd n, everything left over always pairs up neatly.

Answer from the NTA final answer key. Solution by PrepWiser.