Question

Identify correct order for repulsion between electron pair present in valence shell of central atom of molecule?

• A. \(\text{Bp-Bp} > \text{Lp-Bp} > \text{Lp-Lp}\)
• B. \(\text{Lp-Lp} > \text{Lp-Bp} > \text{Bp-Bp}\)
• C. \(\text{Lp-Bp} > \text{Bp-Bp} > \text{Lp-Lp}\)
• D. \(\text{Lp-Lp} > \text{Bp-Bp} > \text{Lp-Bp}\)

Hint:

Lone pairs sit closer to the nucleus and spread out more, so they repel most.

Answer 1

The Correct Option is B

Step 1: Reasoning:
Think about the size of the electron cloud. A lone pair is pulled by one nucleus only, so its cloud is fat. A bonding pair is stretched between two nuclei, so it is thin.

Step 2: Result:
Fat clouds repel each other most. So Lp-Lp is highest, Lp-Bp is in the middle and Bp-Bp is lowest.

Step 3: Real Examples:
In water the two lone pairs squeeze the O-H bonds, reducing the angle from $109.5^{\circ}$ to about $104.5^{\circ}$. This confirms the order and the answer (B).

Final Answer:
Option (B) gives the correct repulsion order. \[ \boxed{\text{(B) } \text{Lp-Lp} > \text{Lp-Bp} > \text{Bp-Bp}} \]