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My understanding of basicity and reducing character:

  • Reducing character is the ability of a substance to reduce something else. So it itself must get oxidized. Since $\ce{BiH3}$ has a large radius, it will loose hydrogen atoms easily. Hence I conclude $\ce{BiH3}$ has more reducing character than $\ce{NH3}$.
  • Basicity is how easily can the compound give up hydrogen atoms/electrons.

Now I end up thinking basicity and reducing character are the same thing. And that $\ce{BiH3}$ should have more basicity than $\ce{NH3}$. But, this is not the corrrect answer.

So what's the difference between reducing character and basicity?

On mostly all Google searches I find that reducing character is directly proportional to basicity. But that doesn't say why they are different in the first place.

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    $\begingroup$ They have, like, nothing in common, unless you wanna talk about obscure Usanovich's theory. $\endgroup$
    – Mithoron
    Mar 13, 2018 at 19:53
  • $\begingroup$ @Mithoron That's the thing I don't understand. Why aren't they the same. I write why I think they are the same in the question. $\endgroup$ Mar 13, 2018 at 19:57
  • $\begingroup$ Well, then it sounds like Usanovich, and also why his theory wasn't successful. Reasoning in vein of BiH3 barely holds to its protons and looses them in oxidation would be much better on your part. $\endgroup$
    – Mithoron
    Mar 13, 2018 at 21:15

2 Answers 2

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Basicity is how easily can the compound give up hydrogen atoms.

Isn't this the definition for acidity instead? (assuming your "hydrogen atoms" = proton) Basicity instead refers to how easily a compound can accept protons on its lone pair, not give them up.

$\ce{BiH3}$ is a weak base, because the lone pair on bismuth is dispersed over a large surface area, as compared to that on the nitrogen in ammonia. (details here) Being diffused, the lone pair is unable to attack a proton, and hence, is a weak base.

$\ce{BiH3}$ is a better reducing agent, because on getting oxidised (losing a hydrogen), the negative charge formed on the bismuth atom is dispersed over a much larger surface area. Hence, the negative charge is stabilized much more on the bismuth atom than it would be stabilized on the nitrogen atom in ammonia.

Notice that the dispersion of the lone pair makes it more diffused, hence, less basic. While dispersion of the excess negative charge stabilizes the specifies formed after loss of a hydrogen, hence, a better reducing agent.

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  • $\begingroup$ +1. I think the key word is polarizability (when you talk about surface area).. $\endgroup$
    – user43021
    Mar 14, 2018 at 2:41
  • $\begingroup$ @santimirandarp Thanks for your comment, but I don't get it exactly. Are you talking about the polarizability of the bismuth and the nitrogen atoms? How exactly does that fit in here? (I am more used to seeing polarizability in applications of Fajan's rule. So I may be missing something) $\endgroup$ Mar 14, 2018 at 3:57
  • $\begingroup$ if the negative charge lies on the bismuth, then how come the hydrogen is responsible for the reducing character (if it doesnt even have the electrons)? $\endgroup$
    – Ankush
    Aug 13, 2022 at 15:28
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Your confusion seem to stem from the fact that getting hydrogen is reduction (so losing hydrogen is oxidation) and getting electrons is reduction (so losing electrons being oxidation) as well. So you must have concluded that greater reducing character means greater basicity as basicity is ability to donate electrons.

But here we are considering hydrides of pnictogens that involves lone pair of electrons as well as hydrogens.

$\ce {NH3}$ is most basic as it's lone pair of electrons is fairly well localised in the 2s orbital which is very small in size and thus could concentrate the charge density into a tiny volume. Such a high charge density of lone pair makes it quite easy to be donated and hence making it most basic. But on the other hand, $\ce {NH3}$ has the least reducing character in group 15 because nitrogen's p orbitals are also compact due to it's small size. This compact size means p orbitals are localised fairly well too and the H orbitals will very well overlap with that of N to form a very strong bond in $\ce {NH3}$. This high bond strength makes it difficult to lose H and hence $\ce {NH3}$ being least reductive.

Going to bismuth, as elucidated in the other answer, it has such a large size that it's lone pair is too diffused to be concentrated in a suitable volume for donation which makes it least basic. But the same factor of size and diffused orbitals make the overlap of its orbitals with that of H orbitals quite poor. So $\ce {BiH3}$ can easily lose hydrogen and hence has the greatest reducing character.

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