I read this in a textbook:

$\ce{Bi(V)}$ is very unstable and is a good oxidizing agent.

Why does it happen that way? Is it because in $\pu{+5}$ oxidation state $\ce{Bi}$ pulls in more electrons and hence gets reduced fast or there's a different concept?

Also then $\ce{BiCl5}$ should be more stable than $\ce{BiCl3}$ because it's getting the electrons that it needs, which is not true.

  • 1
    $\begingroup$ In one word, it's due to inert pair effect. If you want to go deeper, you can pull in Relativistic effect to ultimately conclude the cause of inert pair effect and explain the reason of stability of $+3$ oxidation state rather than $+5$ oxidation state. $\endgroup$
    – Soumik Das
    Feb 16 '19 at 9:01
  • 2
    $\begingroup$ I require a proper explanation for this. I know inert pair effect is affecting this, but how? That's the main question. $\endgroup$
    – Aashi
    Feb 16 '19 at 9:18
  • $\begingroup$ As far as I can tell BiCl5 doesn't exist. So "not very stable" was a serious understatement. It would be significantly stronger oxidant then elemental chlorine which marks some serious problem with it's hypothetical synthesis. $\endgroup$
    – Mithoron
    Feb 16 '19 at 17:25
  • $\begingroup$ @Mithoron why would it be a strong oxidant? Is the reasoning in my question right? $\endgroup$
    – Aashi
    Feb 16 '19 at 18:32
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    $\begingroup$ Well, your inert pair effect makes the pair harder to use - one needs stronger oxidant to oxidate Bi (III) to Bi (V) and therefore Bi (V) compounds are stronger oxidants themselves. $\endgroup$
    – Mithoron
    Feb 16 '19 at 23:15

The reason behind this is the mainly inert pair effect .

In $\ce{BiCl3}$, due to the much electronegativity difference between $\ce{Bi}$ and $\ce{Cl}$, Chlorine atom forms bonds with almost pure $p$ orbitals of $\ce{Bi}$, and the lone pairs on $\ce{Bi}$ are of almost pure $s$ character. Thus, $\ce{Bi}$ atom doesn't actually utilise much of its $s$ orbital electrons in forming the bonds which is energetically much more preferable. ( If it seems non-obvious, recall Bent's rule which says that more electronegative atoms prefer to form bonds with orbitals with more $p$ - character.)

On the other hand, if $\ce{Bi}$ is present in $\ce{BiCl5}$, it has no other option other than utilising its $s$ orbitals for bonding. The equatorial bonds will mainly consist $s$ character and $p_x, p_y$ orbitals will also participate in forming the equatorial bonds. So, $s$ character in the equatorial bonds will be slightly lesser than $33$% (again due to Bent's rule). And the axial bonds will consist of mainly $p_z $ and $d_{z^2}$ orbitals and little bit of $s$ mixing may also be there as it is spherically symmetric. So, overall central atom $\ce{Bi}$ has to use its $s$ electrons in bonding which is difficult energetically.

Now, the participation of $s$ electrons is difficult in bonding is due to the relativistic contraction of $s$ orbitals. It becomes closer to the nucleus and more stable in case of heavy elements like $\ce{Au, Hg, Tl, Pb, Bi}$ etc. Hence, the central atom needs to pay more energy to involve those stable electrons in bonding. That's why, the $s$ electrons become kind of inert and thus these heavy elements don't prefer higher oxidation states to avoid participation of these inner $s$ orbitals in bonding.

  • $\begingroup$ This contraction is relative to the 2nd period and 3rd period elements,I presume? $\endgroup$ Feb 16 '19 at 10:38
  • $\begingroup$ @YUSUFHASAN Yes ! $\endgroup$
    – Soumik Das
    Feb 16 '19 at 11:02

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