Concept Breakdown
1. TITLE
VSEPR Repulsion Order — Explained in 2 minutes for NEET
2. CONCEPT EXPLANATION
The Valence Shell Electron Pair Repulsion (VSEPR) Theory helps predict molecular shapes based on the idea that electron pairs repel each other and try to get as far apart as possible. However, not all electron pairs repel equally.
The Golden Rule: Repulsion strength follows this order:
lp–lp > lp–bp > bp–bp.
Why does this happen?
Think of a Lone Pair (lp) like a passenger on a crowded local train taking up extra space because they aren't holding onto anything—they are localized only on the central atom and spread out. In contrast, a Bonding Pair (bp) is shared between two nuclei (like a shared cab), so it is more constrained and takes up less space.
Because lone pairs are "fatter" and occupy more space, they push the bonding pairs closer together, squeezing the bond angle to be smaller than the ideal geometry.
Common Misconception:
Students often think that if the hybridization is the same (e.g., $sp^3$), the bond angle must be identical. This is false; lone pairs distort the angle.
3. MCQ
Question:
Consider three molecules: Methane ($CH_4$), Ammonia ($NH_3$), and Water ($H_2O$). All involve $sp^3$ hybridization of the central atom. Which option correctly identifies the bond angle trend and the valid reason for the deviation from the ideal $109.5^\circ$ angle?
A) $H_2O > NH_3 > CH_4$; because lone pairs repel less than bonding pairs.
B) $CH_4 = NH_3 = H_2O$; because they all have the same tetrahedral geometry.
C) $CH_4 > NH_3 > H_2O$; because the repulsion order is lp-lp > lp-bp > bp-bp.
D) $NH_3 > CH_4 > H_2O$; because nitrogen is less electronegative than oxygen.
4. ANSWER + EXPLANATION
Correct Answer: (C)
Explanation:
The bond angle decreases as the number of lone pairs increases due to the repulsion order lp–lp > lp–bp > bp–bp.
- $CH_4$ (0 lp): Ideal tetrahedral angle of $109.5^\circ$ because there are only bp-bp repulsions.
- $NH_3$ (1 lp): The single lone pair exerts lp-bp repulsion, pushing the N-H bonds closer to $107^\circ$.
- $H_2O$ (2 lp): The two lone pairs create intense lp-lp repulsion (the strongest type), squeezing the H-O-H angle significantly further to $104.5^\circ$.
Option B is a "trap" because while the electron arrangement (geometry) is tetrahedral for all, the molecular shape and angles change due to lone pair distortion.
5. KEY TAKEAWAY BOX
Remember:
Lone pairs are localized on a single atom and occupy more space than bonding pairs, causing them to compress bond angles significantly ($H_2O < NH_3 < CH_4$).
6. NCERT REFERENCE
NCERT Class 11, Chapter 4, Page 113
"NCERT defines this as: 'The repulsive interaction of electron pairs decrease in the order: Lone pair (lp) – Lone pair (lp) > Lone pair (lp) – Bond pair (bp) > Bond pair (bp) – Bond pair (bp)'".
7. CTA
Full chapter explanation → Chemical Bonding & Molecular Structure Complete Guide