Oxidation States of Transition Elements

A Great Variety of Oxidation States

The single most distinctive chemical feature of transition metals is their variety of oxidation states. Manganese, for example, shows every state from +2 to +7.

Why? Because the $ns$ and $(n-1)d$ electrons have similar energies, so both sets can take part in bonding. An element can use just its $ns$ electrons (giving lower states like +2) or also involve $d$ electrons (giving higher states).

Contrast with main-group elements, which usually show one or two oxidation states differing by 2. Transition metals show many states differing by 1 (e.g. Mn: +2, +3, +4, +5, +6, +7).

Trends in Oxidation States

Maximum oxidation state: rises to a maximum in the middle of the 3d series. The highest is +7 for manganese ($3d^5 4s^2$: all seven outer electrons can be involved). After Mn the maximum falls, because the increasing nuclear charge holds the $d$ electrons more tightly.

Where do the highest states appear? Almost always in oxides and fluorides — the most electronegative, small ligands that can stabilise high oxidation states (e.g. $\text{MnO}_4^-$, $\text{Cr}_2\text{O}_7^{2-}$, $\text{CrO}_3$, $\text{Mn}_2\text{O}_7$). Fluorine and oxygen, being highly electronegative, can support a metal in a very high oxidation state.

The +2 state becomes more and more stable (relative to higher states) across the series. From Mn onwards, the +2 ion is the most common because the 4s electrons are easily lost and the remaining $d$ electrons are increasingly hard to remove.

[JEE Tip] "Why is the highest oxidation state shown in the oxide or fluoride?" — because O and F are highly electronegative and small, able to form strong bonds that stabilise the metal's high oxidation state. This is a recurring exam answer.

Relative Stability of Oxidation States

Some pairs of ions have the same $d$-configuration but very different behaviour — and the reason is the drive toward a stable $d^3$, $d^5$, or $d^{10}$ configuration:

• Cr$^{2+}$ is a strong reducing agent ($d^4$); it readily oxidises to Cr$^{3+}$ ($d^3$), which is especially stable in an octahedral field.
• Mn$^{3+}$ is a strong oxidising agent ($d^4$); it readily gains one electron to give Mn$^{2+}$ ($d^5$), the stable half-filled configuration.

Both Cr$^{2+}$ and Mn$^{3+}$ are $d^4$, yet one is reducing and the other oxidising — decided by which stable configuration they move toward.

Similarly:

• Mn$^{2+}$ ($d^5$) is especially stable (half-filled), so manganese resists oxidation beyond +2 in many situations.
• Cu$^+$ disproportionates in water ($2\text{Cu}^+ \rightarrow \text{Cu}^{2+} + \text{Cu}$) because the hydration energy and stability favour Cu$^{2+}$.

Key Point: To compare the stability or redox behaviour of two transition-metal ions, look at the configuration they move toward — anything heading to $d^5$ or $d^{10}$ is favoured.

Solved Examples

Example 1: Why so many oxidation states?

Why do transition metals show a variety of oxidation states?

Solution: The $ns$ and $(n-1)d$ electrons are close in energy, so both can take part in bonding. An element may lose only its $ns$ electrons (lower state) or additionally some $d$ electrons (higher states), giving many possible oxidation states differing by 1.

Example 2: Highest oxidation state in the 3d series

Which element shows the highest oxidation state in the 3d series, and what is it?

Solution: Manganese, with a maximum of +7 (as in $\text{MnO}_4^-$), because all seven of its $3d^5 4s^2$ electrons can be involved.

Example 3: Cr2+ reducing vs Mn3+ oxidising

Both Cr$^{2+}$ and Mn$^{3+}$ are $d^4$. Why is Cr$^{2+}$ a reducing agent but Mn$^{3+}$ an oxidising agent?

Solution:

• Cr$^{2+}$ ($d^4$) is easily oxidised to Cr$^{3+}$ ($d^3$), which is very stable in solution — so Cr$^{2+}$ acts as a reducing agent.
• Mn$^{3+}$ ($d^4$) is easily reduced to Mn$^{2+}$ ($d^5$), a stable half-filled configuration — so Mn$^{3+}$ acts as an oxidising agent.

Example 4: Why high states in oxides/fluorides

Why is the highest oxidation state of a metal shown in its oxide or fluoride?

Solution: Oxygen and fluorine are small and highly electronegative, forming strong bonds that stabilise the metal in its highest oxidation state (e.g. Mn in $\text{Mn}_2\text{O}_7$, $\text{MnO}_4^-$).

Example 5: Stronger reducing agent

Which is a stronger reducing agent, Cr$^{2+}$ or Fe$^{2+}$, and why?

Solution: Cr$^{2+}$ is the stronger reducing agent. Cr$^{2+}$ ($d^4$) is readily oxidised to the very stable Cr$^{3+}$ ($d^3$). Fe$^{2+}$ ($d^6$) oxidises to Fe$^{3+}$ ($d^5$); although $d^5$ is stable, the change for Cr is more favourable, making Cr$^{2+}$ the stronger reducing agent.

Example 6: Stability of +2 across the series

How does the stability of the +2 oxidation state change across the 3d series?

Solution: The +2 state becomes increasingly stable from left to right (especially from Mn onward), because the 4s electrons are lost easily while the remaining $d$ electrons become progressively harder to remove.

Example 7: Disproportionation of Cu+

Why is Cu$^+$ unstable in aqueous solution?

Solution: Cu$^+$ disproportionates: $2\text{Cu}^+ \rightarrow \text{Cu}^{2+} + \text{Cu}$. The high hydration enthalpy of Cu$^{2+}$ (more than compensating its higher second ionisation enthalpy) makes Cu$^{2+}$ + Cu more stable than 2 Cu$^+$ in water.

Example 8: Mn2+ stability

Why are Mn$^{2+}$ compounds more stable to oxidation than Fe$^{2+}$ compounds?

Solution: Mn$^{2+}$ is $3d^5$ — a stable half-filled subshell — so it resists oxidation to Mn$^{3+}$ ($d^4$). Fe$^{2+}$ is $3d^6$ and is readily oxidised to the stable Fe$^{3+}$ ($d^5$). Hence Mn$^{2+}$ is more stable toward oxidation.

Example 9: Maximum oxidation state of chromium

What is the maximum oxidation state of chromium, and in what species does it appear?

Solution: +6, as in the dichromate ($\text{Cr}_2\text{O}_7^{2-}$) and chromate ($\text{CrO}_4^{2-}$) ions and $\text{CrO}_3$. Chromium ($3d^5 4s^1$) can involve up to six electrons.

Example 10: Predict the stable state

A 3d-series element has the configuration $[\text{Ar}]\,3d^3 4s^2$. What is its likely maximum oxidation state?

Solution: With 3 + 2 = 5 available electrons (3d and 4s), the maximum oxidation state is +5 (this is vanadium, which shows +5 in $\text{VO}_4^{3-}$ / $\text{V}_2\text{O}_5$).