Fajans' rules for Transition Metals and Lanthanides

We may recall that Fajans' Rules do not fully determine the balance between ionic and covalent bonding. As described in this answer, lithium and magnesium hydrides show more ionic character than would be expected based on the usual electronegativity and Fajans' rules. Although the cations as such are relatively small and the electronegativity difference seems not great enough, their diffuse vacant orbitals overlap poorly with hydrogen. Hence the bonding defaults to more ionic character, which is evidenced by their crystal structures (the metal ions are octahedrally coordinated) and their reactivity (readily hydrolyzed by water with hydrogen formation).

Lanthanaides show a similar overlap problem with most nonmetals. The $4f$ valence orbitals overlap poorly with nonmetals because of the mismatch between their three angular nodes and the nonmetal atom orbitals ($s$ and $p$) having zero or one. The $5d$ and $6s$ orbitals have better angular overlap, but with low effective nuclear charge they are diffuse and high-energy.

Lanthanides and actinides can be made to bond more covalently by going to high oxidation states. Among lanthanides we identify praesodymium(V) covalently bonded with nitrogen[1], and among actinides we have the familiar covalently bonded uranyl cation $\ce{U^{VI}O2^{2+}}$.

Reference

1. Boggiano, Andrew C.; Studvick, Chad M.; Roy Chowdhury, Sabyasachi; Niklas, Julie E.; Tateyama, Haruko; Wu, Hongwei; Leisen, Johannes E.; Kleemiss, Florian; Vlaisavljevich, Bess; Popov, Ivan A.; La Pierre, Henry S. (July 2025). "Praseodymium in the formal +5 oxidation state". Nature Chemistry. 17 (7): 1005–1010. doi:10.1038/s41557-025-01797-w. ISSN 1755-4349.

Fajans’ rules say that a cation’s polarizing power increases when its charge is large and its radius is small. A convenient way to express this is

$$\text{polarizing power} \propto \frac{Z_\text{eff}}{r^2}$$

for a given charge state, where $Z_\text{eff}$ is the effective nuclear charge at the cation surface and $r$ is the ionic radius.

For the first-row transition metals in the $M^{2+}$ state (Sc–Zn), the ions are essentially $[\text{Ar}]\,3d^n$. The outermost electrons that define the size of the ion are the $3d$ electrons. As you move across the period, the nuclear charge $Z$ increases by $+1$ each step and you add one more $3d$ electron. Electrons in the same $3d$ shell shield each other poorly (small increment in screening per added $3d$ electron), so $Z_\text{eff}$ experienced by the $3d$ shell increases substantially from left to right. Because that $3d$ shell is the outer shell, the increase in $Z_\text{eff}$ pulls it in strongly, causing a significant decrease in $r$ across the series. With charge fixed at $+2$, $\dfrac{Z_\text{eff}}{r^2}$ rises rapidly, so the polarizing power of $M^{2+}$ increases quickly and the bonding becomes more covalent going from early to late 3d metals for a given anion.

For the lanthanides in the $Ln^{3+}$ state, the situation is different. The ions are approximately $[\text{Xe}]\,4f^n$, with an outer closed $5s^2 5p^6$ shell and an inner, partially filled $4f$ shell. As you go from La$^{3+}$ to Lu$^{3+}$, you add protons to the nucleus and fill the inner $4f$ orbitals. The $4f$ electrons are very core-like: they are radially contracted, lie mostly inside the $5s$ and $5p$ shells, and shield outer electrons poorly. So for the outer $5s$ and $5p$ electrons that control the ionic radius, each step adds $+1$ to $Z$ and one more inner $4f$ electron that only partially screens this extra nuclear charge. This does increase $Z_\text{eff}$ on the outer shell, but more gradually per step than in the 3d case, and the principal quantum number and character of the outer shell (5s/5p) do not change along the series.

Because of that, the lanthanide contraction (the decrease in $r$ from La$^{3+}$ to Lu$^{3+}$) is relatively modest compared with the contraction of the 3d $M^{2+}$ ionic radii across a single period. The radius $r$ of $Ln^{3+}$ decreases, and $\dfrac{Z_\text{eff}}{r^2}$ does increase, but the fractional change is smaller than for the $M^{2+}$ 3d ions. Consequently, the polarizing power of $Ln^{3+}$ increases more slowly across the series than that of $M^{2+}$ across the 3d row.

In terms of penetration and shielding, the key point is that $s > p > d > f$ in penetration. The $3d$ electrons in $M^{2+}$ are the outer electrons and shield each other poorly, so the increase in $Z_\text{eff}$ directly contracts the outer shell strongly from one element to the next. The $4f$ electrons in $Ln^{3+}$ also shield poorly, but they are inner electrons; their poor shielding only slowly increases $Z_\text{eff}$ at the outer $5s/5p$ shell, giving a smaller stepwise contraction. This is why, using Fajans’ rules, 3d $M^{2+}$ ions show a rapid increase in polarizing power and covalent character across the period, whereas $Ln^{3+}$ ions show a slower, more gradual increase across the lanthanide series.