Question

Amongst the following which one will have maximum 'lone pair- lone pair' electron repulsions?

• A. ClF₃
• B. IF₅
• C. SF₄
• D. XeF₂

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
According to VSEPR theory, electron pairs around a central atom arrange themselves to minimize repulsion. The magnitude of repulsion follows the order: lone pair - lone pair (lp-lp) > lone pair - bond pair (lp-bp) > bond pair - bond pair (bp-bp). The number of lp-lp repulsions depends on the number of lone pairs and their spatial arrangement.
Step 2: Detailed Explanation:
Let's analyze the valence shell electronic environment of each central atom:
1. $ClF_3$: Chlorine (Group 17) has 7 valence electrons. It forms 3 bond pairs with Fluorine and has 2 lone pairs. Geometry is T-shaped. Number of lp-lp interactions at $90^\circ$ is 1.
2. $IF_5$: Iodine has 7 valence electrons. It forms 5 bond pairs and has 1 lone pair. Geometry is square pyramidal. There are no lp-lp interactions as there is only one lone pair.
3. $SF_4$: Sulfur (Group 16) has 6 valence electrons. It forms 4 bond pairs and has 1 lone pair. Geometry is see-saw. No lp-lp interactions.
4. $XeF_2$: Xenon (Group 18) has 8 valence electrons. It forms 2 bond pairs and has 3 lone pairs. In its trigonal bipyramidal electron geometry, the 3 lone pairs occupy the equatorial positions at $120^\circ$ to each other. This results in 3 sets of lp-lp interactions (though at $120^\circ$, they are still the most significant in the molecule). Compared to the other options, $XeF_2$ has the highest number of lone pairs.
Step 3: Final Answer:
$XeF_2$ has the maximum number of lone pairs (3), leading to the maximum lp-lp repulsion.