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Write Brief Answer · Q3

Q.In CH₄, NH₃ and H₂O, the central atom undergoes sp³ hybridisation - yet their bond angles are different. Why?

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✓ Free question

Step 1. All three central atoms (C in CH₄, N in NH₃, O in H₂O) have four electron domains, so all three ARE sp³ hybridised, with an underlying tetrahedral electron-pair geometry.

Step 2. CH₄'s carbon has ZERO lone pairs — all four domains are bond pairs, so only the weakest repulsion (b.p.–b.p.) is present, and the angle sits at the ideal, undistorted tetrahedral value, 109°28'.

Step 3. NH₃'s nitrogen has ONE lone pair and three bond pairs — the stronger lone pair–bond pair repulsion pushes the three N-H bonds slightly closer together, compressing the angle to 107°18'.

Step 4. H₂O's oxygen has TWO lone pairs and only two bond pairs — now BOTH lone pair–lone pair (the strongest) and lone pair–bond pair repulsions act to compress the two O-H bonds further, giving the smallest angle of the three, 104°35'.

Step 5. So even though all three molecules share the same sp³ hybridisation, the progressively increasing lone-pair count (0 → 1 → 2) explains the progressively shrinking bond angle.

✓Final answer

The bond angles shrink because lone pairs repel more strongly than bond pairs (l.p.–l.p. > l.p.–b.p. > b.p.–b.p.), and the lone-pair count rises from 0 (CH₄) to 1 (NH₃) to 2 (H₂O).

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