Why transition metals show variable oxidation states — explain with examples

Question

Why do transition metals show variable oxidation states? Explain with examples. Which transition metal shows the maximum number of oxidation states?

(NCERT Class 12, Chapter 8)

Solution — Step by Step

Why This Works

The closeness of $3d$ and $4s$ energy levels is the fundamental reason. In s-block elements, the energy gap between orbitals is large — Na always loses just 1 electron. In transition metals, multiple electrons from different subshells can participate, depending on the chemical environment.

The most stable oxidation state is determined by factors like exchange energy (stabilization from half-filled or fully-filled $d$ orbitals), electronegativity of the bonding partner, and lattice/hydration energy of the resulting ion. For instance, Mn$^{2+}$ ($3d^5$, half-filled) is very stable, while Mn$^{7+}$ ($3d^0$) exists only with highly electronegative elements like O and F (as in $\text{KMnO}_4$).

Alternative Method

You can also use electrode potentials to predict stable oxidation states. If $E°(\text{M}^{3+}/\text{M}^{2+})$ is very positive, M$^{3+}$ is a strong oxidising agent and M$^{2+}$ is the stable form. If $E°$ is very negative, M$^{2+}$ is easily oxidised to M$^{3+}$.

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