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Question

The hydrogen bond strength in

(i) O—H.....O

(ii) O—H.....Cl

(iii) O—H.....N

will follow the order :

This question was previously asked in
MH SET 2020 Paper 2 Philosophy Question Paper (27-Dec-2020)
The correct answer is

(i) > (iii) > (ii)

The strength of a hydrogen bond X–H···Y depends on how electronegative the acceptor atom Y is and how compact its lone pair is. A small, highly electronegative acceptor with a concentrated lone pair (F, O, N) forms strong hydrogen bonds, while a large acceptor such as chlorine, whose lone pairs are spread over a bigger 3p orbital, forms only weak ones even though its Pauling electronegativity is high.

Step 1 — O—H.....O: oxygen is the most electronegative of the three acceptors and is small, so this hydrogen bond is the strongest of the set. Step 2 — O—H.....N: nitrogen is also a small second-period atom with a compact lone pair but is less electronegative than oxygen, so this interaction comes next. Step 3 — O—H.....Cl: chlorine is a third-period atom; its diffuse lone pairs overlap poorly with the hydrogen, so this hydrogen bond is the weakest.

This gives (i) > (iii) > (ii). Ranking the chlorine acceptor above oxygen and nitrogen, treating the O and Cl acceptors as equal, or placing the Cl acceptor above O ignores the large size and diffuse lone pairs of chlorine, which make it the poorest hydrogen-bond acceptor of the three.

(i) > (iii) > (ii) option is correct.

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