Subjective Type

Arrange the following in order of decreasing bond angles, giving reasons (i) $$CH_{4}, NH_{3}, H_{2}O, BF_{3}, C_{2}H_{2}$$ (ii) $$NH_{3}, NH_{2}^{-}, NH_{4}^{+}$$.

Solution

(i) $$C_{2}H_{2} (180^\circ) > CH_{4} (109.28^\circ) > BF_{3} (120^\circ) > NH_{3} (107^\circ) > H_{2}O (104.5^\circ)$$
(ii) $$NH_{4}^{+} > NH_{3} > NH_{2}^{-}$$
This is because all of them involve $$sp^{3}$$ hybridization. The number of lone pair of electrons present on N-atom are 0, 1 and 2 respectively. Greater the number of lone pairs, greater are the repulsion on the bond pairs and hence smaller is the angle.


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