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VSEPR – bond angles
Concepts tested here
- bond-angle-ch4-nh3-h2o
All Questions
2016 1 question
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Consider the molecules $CH_4$, $NH_3$ and $H_2O$. Which of the given statements is false?
Bond angle: $CH_4$ > $NH_3$ > $H_2O$.
Lone pairs are more diffused than bond pairs. According to VSEPR theory, repulsion order: lp–lp > lp–bp > bp–bp.
Therefore the H–O–H bond angle in $H_2O$ (104.5°) is smaller than the H–C–H bond angle in $CH_4$ (109°28'), so the first statement is false.
