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Among $N{{H}_{3}}$, ${{H}_{2}}O$ and HF, which would you expect to have the highest magnitude of hydrogen bonding and why?


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Last updated date: 20th Jun 2024
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Answer
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Hint: As we know that hydrogen bond is present in between those atoms that have a high electron affinity. It mainly depends on the amount of hydrogen atom present and the electronegativity of the other atom. It is found that these mainly belong to the second row elements in the periodic table like nitrogen, oxygen and fluorine.

Complete answer:
- As we know that electronegativity is the tendency of an atom to attract the shared pair of electrons towards itself. And electron affinity is the energy released when an electron is attached to a neutral atom.
- In the periodic table we can see that electronegativity decreases across a period and increases down the group. So, we can say fluorine is a more electronegative atom than oxygen, nitrogen and fluorine.
Hence, we can write the increasing order of electronegativity as: N- According to this order, HF should have the highest magnitude of hydrogen bonding, but is not true. Because the number of hydrogen atoms also matters. We can see that in the case of water there are sufficient hydrogen atoms present for the oxygen to form hydrogen bonds.
Whereas, in case of HF there is only one hydrogen atom.
- We can see that in ammonia, lone pairs are present on nitrogen, through which a hydrogen bond is formed, but as it is found to be present in limited amounts.
- So, we can write the order of highest magnitude of hydrogen bonding as:
${{H}_{2}}O$ > HF >$N{{H}_{3}}$
- Hence, we can conclude that ${{H}_{2}}O$ have highest magnitude of hydrogen bonding

Note: - We should not get confused in between the two types of hydrogen bonding that are intermolecular and intramolecular hydrogen bonding. As, intermolecular takes place between different molecules is intermolecular.
- Whereas, intramolecular bonding which takes place in the same molecule by the different parts.