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I have noted down the available positive oxidation states of the first row of transition elements (on the Periodic Table) from the respective Wikipedia articles of the elements.

$\ce{Sc} - 3, 2,1$

$\ce{Ti} - 4, 3, 2, 1 $

$\ce{V} - 5, 4, 3, 2, 1$

$\ce{Cr} - 6, 5, 4, 3, 2, 1$

$\ce{Mn} - 7, 6, 5, 4, 3, 2, 1$

$\ce{Fe} - 6,5,4,3,2,1 $

$\ce{Co} - 5,4,3,2,1 $

$\ce{Ni}- 4,3,2,1$

$\ce{Cu} - 4,3,2,1$

$\ce{Zn} - 2,1$

  • Why does $\ce{Fe}$ show not show $+7$ or $+8$ oxidation state?
  • After $\ce{Fe}$ from $\ce{Co}$ why is there a decrease in number of possible oxidation states? Why don't $\ce{Fe,Co,Ni,Cu,Zn}$ show higher oxidation states though they have valence electrons left in $3d$ subshell?
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    $\begingroup$ For a full picture, you need to add the possible negative oxidation states. $\endgroup$
    – Jan
    Sep 29, 2016 at 13:29

2 Answers 2

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Regarding the maximum oxidation state achievable, there are two competing trends. Both have to do with ionisation energies.

In going to a higher oxidation state, the additional IEs are always balanced against the release in energy in the formation of stronger ionic/covalent bonds. However, if the additional IE is too large, then the favourable bonding will not be sufficient to bring out the higher oxidation state. A simple example is sodium, which cannot adopt the +2 oxidation state because its second IE is simply too large, even though the lattice energy of a hypothetical $\ce{NaCl2}$ crystal is larger than that of $\ce{NaCl}$.

Firstly, the maximum oxidation state is limited by the number of valence electrons available. This is exactly analogous to the case of sodium; therefore, manganese does not exhibit the +8 oxidation state, because its eighth IE involves ionisation from the 3p subshell, much lower in energy than either the 3d or 4s. This therefore accounts for the increase in the maximum oxidation state going from Sc to Mn.

Secondly, the ionisation energies generally increase going across the 3d block, because of an increase in effective nuclear charge. This is the factor which leads to the decrease in the maximum oxidation state after Mn. For example, the seventh IE of Fe is larger than that of Mn; therefore, Fe does not exhibit the +7 oxidation state.

You can look up the IEs in any relevant textbook. However, note that since the +2 oxidation state is very common in the later half of the 3d block, it is usually the third and higher IEs that are more important in this discussion. So, looking at the trend of first/second IEs won't get you anywhere.

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  • $\begingroup$ But that's not a good enough reason.You are justifying using IE.But what if you did'nt have access to I.E. values.Then how would you justify the sudden decrease in available oxidation states after $\ce{Fe}$ just by using the electronic configurations ? $\endgroup$
    – user14857
    Sep 29, 2016 at 7:39
  • $\begingroup$ You don't need to look up the IE values to know that there is an increasing trend. $\endgroup$
    – orthocresol
    Sep 29, 2016 at 7:39
  • $\begingroup$ You did not understand my question.Your answer does not explain why is there a decrease in available oxidation states.Say,why doesnt Fe show +7 oxidation state,why doesnt Co show +6 oxidation state though electrons are left in 3d? I know you can just look up I.E. values but that is not a reason. $\endgroup$
    – user14857
    Sep 29, 2016 at 7:41
  • $\begingroup$ Because to get cobalt to the +6 oxidation state, you need to remove six electrons. Six electrons are too hard to remove in cobalt, because there is a large effective nuclear charge, and the ionisation energies are large, despite the electrons still being in the "valence" 3d orbital. In fact, that's exactly what I wrote in my answer. $\endgroup$
    – orthocresol
    Sep 29, 2016 at 7:45
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    $\begingroup$ "Iron does not exhibit the +7 oxidation state". Lu et al., doi.org/10.1039%2FC6CP06753K. But yes, we would not see iron(VII) in normally encountered laboratory settings. $\endgroup$
    – Oscar Lanzi
    Oct 16, 2022 at 22:22
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In some cases what's anomalous is not only the number but the environment in which the oxidation state is realized. We might expect high oxidation states to be realized by combining the element with oxide or fluoride ligands, but for instance one of the most widely used lead(IV) compounds for many years was $\ce{Pb(C2H5)4}$.

Palladium(V) offers an even more extreme example. Silicon as such has low electronegativity, but the molecular orbital structure of silyl ligands makes them more electronegative and metals bonded to such ligands are considered to be at the positive end of such bonds. Palladium is combined with five silyl ligands to give the $\ce{Pd(V)}$ species identified in Ref. 1.

Reference

  1. Shimada, Shigeru; Li, Yong-Hua; Choe, Yoong-Kee; Tanaka, Masato; Bao, Ming; Uchimaru, Tadafumi (2007). "Multinuclear palladium compounds containing palladium centers ligated by five silicon atoms". Proceedings of the National Academy of Sciences. 104 (19): 7758–7763. https://doi.org/10.1073/pnas.0700450104.
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