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: Understanding the Concept:
VSEPR theory says electron pairs around the central atom arrange themselves to minimise repulsion. The strength of repulsion depends on how much space each pair occupies.

Step 2: Compare the Pairs:
A lone pair (Lp) is held by only one nucleus. It spreads out more and takes more space around the central atom.
A bond pair (Bp) is shared by two nuclei, so it is pulled along the bond axis and is more compact.

Step 3: Rank the Repulsions:
Lp-Lp repulsion is strongest because both clouds are large.
Lp-Bp repulsion is in the middle.
Bp-Bp repulsion is weakest because both clouds are compact.
\[ \text{Lp-Lp} > \text{Lp-Bp} > \text{Bp-Bp} \]

Step 4: Check the Other Options:
Option (A) is exactly reversed. Options (C) and (D) place Bp-Bp above or below Lp-Bp in a way that breaks the pattern; they do not follow the lone pair spreading argument. This order is also why bond angles shrink in \(\text{NH}_3\) and \(\text{H}_2\text{O}\).

Final Answer:
The correct order is Lp-Lp > Lp-Bp > Bp-Bp, option (B). \[ \boxed{\text{(B) } \text{Lp-Lp} > \text{Lp-Bp} > \text{Bp-Bp}} \]