# Oxidation state of d block elements [closed]

Why d block elements do not show +1 oxidation state ? Only copper shows +1 , whereas oxidation number starts from +3 in scandium .

• Are you sure about that? imgur.com/L5UZNx9 – Nilay Ghosh Sep 7 '20 at 5:11
• @nilay Ghosh yes, I'm pretty sure , you may also check that . – SUDEEPA GUPTA Sep 7 '20 at 6:13
• And I'm pretty sure you're wrong. d block elements show even negative oxidation states, also zero, +1 and pretty much whatever you'd wish. – Mithoron Sep 7 '20 at 13:34

In general, transition metals do not have a common +1 oxidation state because they have a $$\ce{ns^2}$$ valence shell. It would be more stable for metals to lose 2 electrons instead of 1. There are exceptions to this rule. Group 11 elements, Copper, Silver and Gold, do have a common +1 oxidation state. This is because instead of a $$\ce{nd^9ns^2}$$ configuration, they have a $$\ce{nd^{10}ns^1}$$ configuration. As such, they readily lose only 1 electron to form a stable cation with a +1 charge.

I would not be looking at Group 12 elements since strictly speaking, they are not transition elements. However, even if they are part of the discussion, only mercury has a common +1 oxidation state in the form of $$\ce{Hg^{2+}2}$$.

Cotton and Wilkinson state that

1. $$\ce{Cr(I)}$$ exists in the ion $$\ce{[Cr(bipy)3]+}$$
2. $$\ce{Mn(I)}$$ exist in $$\ce{Mn(CO)5Cl}$$ and $$\ce{K5Mn(CN)6}$$
3. $$\ce{Fe(I)}$$ exists in the ion $$\ce{[Fe(H2O)5NO]^{2+}}$$*
4. $$\ce{Co(I)}$$ exists in the ion $$\ce{[Co(bipy)_{3}]^+}$$ and in the molecule $$\ce{CoBr(PR_3)_3}$$

Ref.: F. A. Cotton, G. Wilkinson, Advanced Inorganic Chemistry, Interscience Publ., 3rd Ed. 1972

*This complex, however, is no longer considered a metal(I) compound, although iron(I) is still known. Another answer explores that case in more detail.

Tl, dr: There is more to getting iron(I) than meets the eye. The aquo-nitrosyl complex doesn't do it. To get there we need a nonaqueous complex formed under strongly reducing conditions.

As noted in a comment to another answer, this question is not as cut and dried as it may appear, especially in the case of iron. As explained in this answer, it was long thought that the "brown ring" iron-nitrosyl complex was iron(I) bound to $$\ce{NO^+}$$, but in 2002 (three decades after the Cotton-Wilkinson text) spectroscopic studies revealed that it actually consists of iron(III) bound and antiferromagnetically coupled to $$\ce{NO^-}$$. The reader may want to digress to an addendum at the bottom, in which iron is rendered so highly electropositive that this outcome is really no surprise.

Here the case of iron is explored in more detail, using the work of Keilwerth et al. [1]. Keilwerth et al. performed extensive spectroscopic and electrochemical studies of a series of iron complexes with nitrous oxide and tris[2-(3-mesitylimidazol-2-ylidene)ethyl]- amine ($$\text{TIMEN}^{\text{Mes}}$$), which have the chemical formula $$\ce{(TIMEN^{\text{Mes}})FeNO(CH3CN)^q}$$ with varying charge states $$q$$. Initially the complex is formed with $$q=+3$$. When this tri-cation is reduced -- by zinc, with one electron, by magnesium, with two electrons, or by sodium amalgam, with three electrons -- the nitrisyl group in the resulting reduced species is identified as antiferromagnetically coupled $$\ce{NO^-}$$, with the iron oxidation state being decreased from +3 to +2 to +1 as the number of accepted electrons increases. Thus the sodium-reduced species, with three added electrons, is found to indeed contain iron(I), bonded to $$\ce{NO^-}$$ rather than $$\ce{NO^+}$$. As an aside, and consistent with the addendum, it takes a reducing agent as strong as magnesium even to reach iron(II).

Reference

1. Martin Keilwerth, Johannes Hohenberger, Frank W. Heinemann, Jörg Sutter, Andreas Scheurer, Huayi Fang, Eckhard Bill, Frank Neese, Shengfa Ye, and Karsten Meyer, "A Series of Iron Nitrosyl Complexes {Fe−NO}6−9 and a Fleeting {Fe− NO}10 Intermediate en Route to a Metalacyclic Iron Nitrosoalkane", J. Am. Chem. Soc. 2019, 141, 17217−17235. pdf

We usually think of "highly electropositive elements" as alkali or alkaline earth metals, plus a few others such as aluminum and some rare earths. But iron is actually not far behind. It is actually a quite electropositive element for which even the +2 oxidation state is a low one in an electronegative environment. What is often called $$\ce{FeO}$$ is actually not a pure iron(II) oxide, but a nonstoichiometric compound containing a significant amount of iron(III). Moreover, when cooled to ambient conditions this oxide is only meta-stable, tending to decompose to form $$\ce{Fe3O4}$$. Under certain naturally occurring conditions, iron(II) can even displace hydrogen from water, again forming $$\ce{Fe3O4}$$ instead of $$\ce{FeO}$$. And of course, rusting produces a combination of $$\ce{Fe3O4}$$ and various iron(III) compounds. Given these properties, we should realize that to stabilize iron into a low oxidation state requires low-electronegativity ligands and strongly reducing conditions. We see this in the main body of the answer.