How many lone pairs of electrons are on the I atom in IF4+? Question 23 options: 0 1 2 3
The correct answer and explanation is:
The correct answer is: 1 lone pair
Explanation (Approximately 300 words):
To determine the number of lone pairs of electrons on the iodine (I) atom in IF₄⁺, we need to use principles from valence shell electron pair repulsion (VSEPR) theory and electron counting.
Step 1: Determine the number of valence electrons
Iodine (I) is in group 17 of the periodic table and has 7 valence electrons. Each fluorine (F) atom forms a single bond with iodine, and there are 4 fluorine atoms, contributing 4 bonding pairs (or 8 electrons).
However, because the species is a positively charged ion (IF₄⁺), one electron is removed, reducing the total electron count by 1.
So the total electron count on iodine becomes:
7 (from I) − 1 (positive charge) = 6 valence electrons available for bonding/lone pairs
Step 2: Count the bonding pairs
Each F atom forms a single bond with I, accounting for 4 bonding pairs, which equals 8 bonding electrons. These electrons are shared between iodine and fluorine, and they count toward the electron domain around iodine.
Now, subtract the 4 bonding pairs (8 electrons) from the available 6 valence electrons on iodine:
But wait—this isn’t quite right. Let’s reframe this using electron domains, which VSEPR theory uses.
Step 3: Apply VSEPR theory
Iodine has:
- 4 bonded fluorine atoms = 4 bonding pairs
- Total valence electron pairs around iodine = 5 electron domains (since 6 valence electrons gives 3 pairs; but 4 are used in bonding, and 1 remains as a lone pair)
So, iodine in IF₄⁺ has:
- 4 bonding pairs (shared with fluorine)
- 1 lone pair of electrons
This makes the molecular geometry seesaw (5 electron domains: 4 bonds + 1 lone pair), derived from a trigonal bipyramidal electron geometry.