Question

The repulsive interaction between different types of electron pairs increases in the order of

• A. bond pair - bond pair \(<\) lone pair - bond pair \(<\) lone pair - lone pair
• B. lone pair - lone pair \(<\) lone pair - bond pair \(<\) bond pair - bond pair
• C. lone pair - bond pair \(<\) lone pair - lone pair \(<\) bond pair - bond pair
• D. lone pair - lone pair \(<\) bond pair - bond pair \(<\) lone pair - bond pair

Hint:

In VSEPR theory, lone pairs always exert stronger repulsion than bond pairs because lone pair electrons are localized closer to the central atom.

Answer 1

The Correct Option is A

Step 1: Recall VSEPR thinking.
Electron pairs around a central atom push each other apart. How hard they push depends on whether each pair is a lone pair or a bonding pair.

Step 2: Compare how much room each takes.
A bonding pair is tied between two nuclei, so its cloud is pulled thin and takes little space. A lone pair sits on one atom only, so its cloud is fat and spreads wider.

Step 3: Rank the pushes.
Two thin bond pairs repel the least. A fat lone pair against a thin bond pair repels harder. Two fat lone pairs repel the hardest of all.

Step 4: Answer.
So the rising order is \[ \boxed{\text{bp-bp} < \text{lp-bp} < \text{lp-lp}} \] which is option A.