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.

  • $\begingroup$ They have, like, nothing in common, unless you wanna talk about obscure Usanovich's theory. $\endgroup$ – Mithoron Mar 13 '18 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. $\endgroup$ – SmarthBansal Mar 13 '18 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 '18 at 21:15

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.

| improve this answer | |
  • $\begingroup$ +1. I think the key word is polarizability (when you talk about surface area).. $\endgroup$ – user43021 Mar 14 '18 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) $\endgroup$ – Gaurang Tandon Mar 14 '18 at 3:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.