Ionisation Enthalpies

The Overall Trend

Ionisation enthalpy rises across each transition series from left to right, because nuclear charge keeps increasing as the inner $d$ orbitals are progressively filled. However, the successive ionisation enthalpies of a transition series do not climb nearly as steeply as they do across a period of non-transition elements — the rise along a transition series is comparatively gentle. This is true of the first ionisation enthalpy in particular; the second and third ionisation enthalpies of successive elements, by contrast, increase much more sharply along the series than the first does.

Why the First-Ionisation-Enthalpy Trend Is Irregular

The first ionisation enthalpy of the 3$d$ metals does not rise smoothly, and although this irregularity has little chemical significance in itself, it can be traced to how removing one electron shifts the relative energies of the $4s$ and $3d$ orbitals. Recall that $d$-block elements lose their $ns$ electrons before their $(n-1)d$ electrons when forming ions. Across the 3$d$ series, nuclear charge increases from scandium to zinc, but each additional electron goes into the inner $3d$ subshell — and these $3d$ electrons shield the $4s$ electrons from the growing nuclear charge somewhat more effectively than outer-shell electrons shield one another. As a result, atomic radius contracts only slowly, and ionisation enthalpy correspondingly rises only slightly across the 3$d$ series.

Second and Third Ionisation Enthalpies

Once an ion carries two or more positive charges, it typically has a pure $d^{n}$ configuration with no $4s$ electrons remaining. For such ions, one $d$ electron does not shield another from the nuclear charge very effectively (because the $d$ orbitals point in different directions), so the second ionisation enthalpy is expected to increase fairly steadily as effective nuclear charge rises. That steady increase, however, breaks at $\text{Mn}^{2+}$ (for the second ionisation enthalpy) and at $\text{Fe}^{3+}$ (for the third), because in both cases the ion being formed has the particularly stable $d^{5}$ configuration.

The Exchange-Energy Explanation

The value of an ionisation enthalpy for a $d^{n}$ configuration reflects three contributions: the attraction of each electron toward the nucleus, the repulsion between electrons, and the exchange energy. Exchange energy is approximately proportional to the total number of possible pairs of electrons with parallel spin occupying a set of degenerate orbitals, and it stabilises the electronic state — the larger the exchange energy, the more stable (and so the harder to ionise) the configuration.

When several electrons occupy a set of degenerate orbitals, Hund's rule tells us that the lowest-energy arrangement is the one with the maximum possible number of singly occupied orbitals, all with parallel spin. On ionising such a configuration, some of that exchange stabilisation can be lost — and losing exchange energy makes ionisation easier — except exactly at $d^{5}$ and $d^{10}$ (and, by the same reasoning, at $d^{6}$ when one electron is removed to give $d^{5}$ back), where no exchange energy is lost on that particular ionisation step.

This explains the specific anomalies noted above:

• $\text{Mn}^{+}$ has the configuration $3d^{5}4s^{1}$, while $\text{Cr}^{+}$ has $3d^{5}$. Because there is no loss of exchange energy going from $\text{Mn}$ to $\text{Mn}^{+}$ in the way there is for $\text{Cr} \to \text{Cr}^{+}$, the ionisation enthalpy of $\text{Mn}^{+}$ comes out lower than that of $\text{Cr}^{+}$.
• Similarly, $\text{Fe}^{2+}$ has configuration $3d^{6}$ while $\text{Mn}^{2+}$ has $3d^{5}$; the ionisation enthalpy of $\text{Fe}^{2+}$ is lower than that of $\text{Mn}^{2+}$ — equivalently, the third ionisation enthalpy of iron is lower than that of manganese.

Ionisation Enthalpy and the $+2$ Oxidation State …