0
$\begingroup$

We know that iron(II,III) oxide has the formula $\ce{Fe3O4}$ which is a combination of iron(II) oxide and iron(III) oxide with formulas $\ce{FeO}$ and $\ce{Fe2O3}$ respectively.

So why isn't this the case in $\ce{Pb3O4}$? Why can't we call it lead(II,III) oxide just like the iron example?

On the other hand, lead(II,IV) oxide should have the formula $\ce{Pb2O3}$ which is the same with lead(III) oxide.

Lead(II)oxide: $\ce{PbO}$ and lead(IV) oxide: $\ce{PbO2}$.

Improve this question
$\endgroup$
11
  • 5
    $\begingroup$ Because lead (III) is not a thing. $\endgroup$
    – Ivan Neretin
    Commented Mar 17, 2022 at 14:54
  • $\begingroup$ @IvanNeretin According to Wikipedia, the oxidation states of lead are $−4, −2, −1, +1, +2, +3$, and $+4$; with $+2$ and $+4$ being dominant. $\endgroup$
    – user85778
    Commented Mar 17, 2022 at 14:58
  • $\begingroup$ These are not combinations of 2 oxides, but they are rather single oxides where metal atoms alternate their oxidation numbers ( II,III for Fe, II,IV for Pb ), often not exactly in stoichiometric ratios of small integers. This can be determined from X-ray crystallographic analysis. $\ce{Pb3O4}$ is AFAIK actually $\ce{Pb2[PbO4]}$ $\endgroup$
    – Poutnik
    Commented Mar 17, 2022 at 14:59
  • 4
    $\begingroup$ The name of both cases is based on what is known, not what is liked. // Chemistry SE site strongly recommends plain text titles for index/search reasons and due possible displaying issues in question lists. $\endgroup$
    – Poutnik
    Commented Mar 17, 2022 at 15:10
  • 2
    $\begingroup$ lead(II,IV) oxide should have the formula Pb2O3 It would be true if Pb(II)/Pb(IV) ratio were 1:1, but it is 2:1. There is $\ce{Pb2^{II}Pb^{IV}O4}$ versus $\ce{Fe^{II}Fe2^{III}O4}$ $\endgroup$
    – Poutnik
    Commented Mar 17, 2022 at 15:19

2 Answers 2

Reset to default
3
$\begingroup$

@Poutnik gave us the Wikipedia link which basically contains the answer:

Lead tetroxide ("red lead"), a mixed oxide with formula $\ce{Pb3O4}$, may be thought of as lead(II) ortho-plumbate(IV), $\ce{[Pb^2+]2[PbO4]^4−}$. Lead sesquioxide, $\ce{Pb2O3}$, is also known, and has the structure lead(II) meta-plumbate(IV), $\ce{[Pb^{2+}][PbO3]^2−}$.

Since $\ce{Pb3O4}$ contains $\ce{Pb(II)}$ and $\ce{Pb(IV)}$, it is lead(II,IV) oxide. Magnetite is actually $\ce{[Fe^{2+}][(Fe^{3+})2O4]}$, so it is iron(II,III) oxide.

Improve this answer
$\endgroup$
1
$\begingroup$

Both crystal structure and periodic characteristics favor the formulation $\ce{Pb^{II}_2Pb^{IV}O4}$.

Below is a picture of the crystal structure from Wikipedia.

enter image description here

Source

This structure has at least two features that would be consistent with a lead(II,IV) identification:

  1. One third of the lead atoms have higher coordination number (six) to the electronegative oxygen and the rest have lower coordination number (two). The more highly coordinated lead atoms would naturally fit with a higher oxidation state, and $\ce{Pb^{II}_2Pb^{IV}O4}$ fits with the ratio of high to low coordinate lead atoms.

  2. The six-coordinate lead atoms noted above have the same coordination as they do in the pure lead (IV) oxide $\ce{PbO2}$. Note the contrast here with iron, where six-coordination is typical for oxidation states +2 and +3.

In addition, $p$-block elements — unlike transition metals — strongly favor paired-electron structures, which means even-$Z$ elements favor even oxidation states and odd-$Z$ elements favor odd ones. Where the oxidation state has the "wrong" parity it's usually accompanied by homonuclear bonding which allows the electrons to be paired without being polarized (like peroxides with their oxygen-oxygen bond or lead(III) in $\ce{(CH3)3Pb–Pb(CH3)3}$). No such homonuclear bonding is evident in $\ce{Pb3O4}$, so the lead atoms ($Z=82$) would be assigned even oxidation states to fit the general trend.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.