The lanthanoid contraction

Two metals that behave as one

Zirconium is element 40. Hafnium is element 72. A hafnium atom has 32 more protons and 32 more electrons than a zirconium atom, so you would expect a much bigger atom. It is not bigger. The two are so alike that hafnium was found only in 1923, long after zirconium. Zirconium minerals usually carry a little hafnium, about one or two parts in a hundred, and chemists find it hard to take out.

Let us see why, starting with what we would expect.

What we expect down a group

Going down a group, a new shell is added, and the atoms grow. The first move, from the 3d to the 4d series, does this.

Titanium has a radius of $147\ \text{pm}$ and zirconium has $160\ \text{pm}$, a gain of $13\ \text{pm}$. Vanadium to niobium gains $11\ \text{pm}$, and chromium to molybdenum gains $10\ \text{pm}$. Then the next move, from 4d to 5d, should add another shell and give another increase of about $13\ \text{pm}$. It does not. Hafnium is $159\ \text{pm}$, tantalum is the same as niobium, and tungsten is the same as molybdenum. Something shrank the 5d atoms back. That something sits between the two series, in the lanthanoids.

The contraction

Between lanthanum and hafnium come the fourteen lanthanoids. Along them the radius of the 3+ ion falls steadily.

The ion $\text{La}^{3+}$ has a radius of $103.2\ \text{pm}$ and $\text{Lu}^{3+}$ has $86.1\ \text{pm}$. The fall is $17.1\ \text{pm}$, which is about $1.2\ \text{pm}$ at each step. Each step is tiny. Fourteen steps add up. This steady shrinking is the lanthanoid contraction. (Other tables give $106\ \text{pm}$ and $85\ \text{pm}$ for the same two ions. The scale is a little different, but the fall is about the same.)

The atoms also shrink, in a less regular way. Europium and ytterbium are exceptions. They give only two electrons to the metallic bond, and so their atoms are larger than their neighbours.

The cause, from the electrons

What changes from one lanthanoid to the next? The nucleus gains one proton, and one electron joins. The new electron goes into the 4f subshell, which lies inside the 5s and 5p electrons.

The outer electrons feel the nucleus, but only after the inner electrons have cancelled a part of its charge. This is called shielding. The charge that the outer electrons feel is the effective nuclear charge. If the new 4f electron shielded the outer electrons perfectly, it would cancel the new proton exactly. The effective charge would stay the same, and the ion would not shrink.

It does not shield perfectly. The f orbitals have complicated shapes, and their electrons do not stay neatly between the nucleus and the outer electrons. So a little of each new proton is felt on the outside. The outer electrons are pulled in a little more at each step.

Let us count it. The rules of John Slater give a rough way to do so. For a 5s or 5p electron of $\text{La}^{3+}$:

• the 28 electrons of shells 1, 2 and 3 cancel $1.00$ each, which is $28$
• the 18 electrons of shell 4 cancel $0.85$ each, which is $15.3$
• the 7 other electrons of shell 5 cancel $0.35$ each, which is $2.45$

The total shielding is $45.75$, so the effective charge is $57 - 45.75 = 11.25$. In the next element, a proton is added and a 4f electron cancels $0.85$ of it. The effective charge rises by only $0.15$. After fourteen steps it is $11.25 + 14 \times 0.15 = 13.35$. A smaller ion follows, because the radius goes roughly as one over the effective charge:

The measured value is $86.1\ \text{pm}$. The count is crude, because it gives every inner electron the same $0.85$, and a real 4f electron shields worse than that. But the idea is clear. A small shortfall in shielding at each step adds up to a large contraction over fourteen steps.

Try $s = 1$. The model radius stays at $103.2\ \text{pm}$ for every element, and there is no contraction at all. So the contraction exists only because the shielding is less than perfect.

Effect 1: the 4d and 5d metals have the same size

Now we can explain zirconium and hafnium. Going from zirconium to hafnium we add a shell, which would increase the radius by about $13\ \text{pm}$. But we also add 14 electrons to the 4f subshell, and they pull the atom in at each step. The two effects cancel almost exactly.

The slider shows that a pull-in of about $0.93\ \text{pm}$ per 4f electron makes hafnium as small as zirconium. That is close to the $1.2\ \text{pm}$ per step that we found for the ions.

What follows from the same radii?

• Chemistry. Zr and Hf, Nb and Ta, Mo and W behave alike. They form the same kinds of compounds with the same formulas. They occur together in nature. Niobium and tantalum are found in the same ore, and hafnium comes with zirconium.
• Density. The mass of a mole nearly doubles from Zr ($91.2\ \text{g/mol}$) to Hf ($178.5\ \text{g/mol}$), but the volume stays the same. So hafnium is almost twice as dense, $13.3\ \text{g/cm}^3$ against $6.51\ \text{g/cm}^3$.
• Ionisation enthalpy. The 5d metals hold their electrons more tightly, because the outer electrons feel a bigger effective charge. The first ionisation enthalpies are $652\ \text{kJ/mol}$ for Nb and $761\ \text{kJ/mol}$ for Ta, and $684\ \text{kJ/mol}$ for Mo and $770\ \text{kJ/mol}$ for W. This makes the 5d metals less reactive. Gold and platinum are the extreme case.

Effect 2: the hydroxides get less basic

The ion gets smaller from La to Lu, and its charge stays at $3+$. So the charge on a given volume rises. Charge divided by radius gives the number: $\text{La}^{3+}$ has $29.1$ charges per nanometre and $\text{Lu}^{3+}$ has $34.8$. That is a rise of about 20 per cent.

A small, strongly charged ion pulls electrons from the oxygen of the hydroxide ion. The bond between the metal and oxygen becomes more covalent and less ionic. A bond that is more covalent does not break so easily to release $\text{OH}^-$. So the hydroxide $\text{Ln(OH)}_3$ is a weaker base for lutetium than for lanthanum. $\text{La(OH)}_3$ is the strongest base and $\text{Lu(OH)}_3$ is the weakest.

The same trend appears elsewhere. The ions form stronger complexes, and the salts are hydrolysed a little more, towards lutetium.

Effect 3: the elements are hard to separate

Look at the figures. Two neighbouring ions differ in radius by about $1\ \text{pm}$ out of about $90\ \text{pm}$. They have the same charge. Their solubilities, their precipitation points and their reactions are almost the same. A method that picks out one lanthanoid also takes some of its neighbours.

The old method was fractional crystallisation. A salt was crystallised, redissolved and crystallised again, thousands of times. The modern methods are ion exchange and solvent extraction.

In ion exchange, the mixture of $\text{Ln}^{3+}$ ions is put on a column of resin beads. A solution of a complexing agent, such as citrate, is washed through. The small ions form stronger complexes, so they spend more of their time in solution and move down faster. The lutetium comes out first and lanthanum last. The order follows the size, and it is the same trend as the one in the hydroxides.

There are two easy cases. Cerium can be oxidised to $\text{Ce}^{4+}$, which behaves nothing like a 3+ ion. Europium can be reduced to $\text{Eu}^{2+}$, and $\text{EuSO}_4$ is insoluble like $\text{SrSO}_4$, while the sulfates of the 3+ ions are soluble. Both use a different oxidation state, and that is the only way the chemistry changes.

Yttrium is a second example. Its $\text{Y}^{3+}$ ion has a radius of $90.0\ \text{pm}$. That is almost the same as $\text{Ho}^{3+}$ at $90.1\ \text{pm}$. So yttrium is found in the same minerals as the heavy lanthanoids, and it is separated by the same methods.

Practise with new numbers

Try it in Python

The program finds the steps of the contraction and the charge per nanometre. Then it tests the shielding model that you used above.

Common mistakes

• Saying the contraction happens because the 4f electrons are far from the nucleus. They are close in, and they shield badly. The outer electrons are pulled in.
• Thinking the atom gains a shell at each step. The shell is the same, and only the nuclear charge changes.
• Saying the basic strength rises from La to Lu. It falls, because the smaller ion pulls the oxygen harder.
• Forgetting that the effect is the sum of fourteen small steps. The change per step is only about $1\ \text{pm}$.
• Claiming that cerium and europium are as hard to separate as the others. Their other oxidation states make them easy.

Remember these

• The lanthanoid contraction is the steady fall of radius from La to Lu, 14 steps and about $17\ \text{pm}$ in total for the 3+ ions.
• Cause: the 4f electrons shield each other and the outer electrons poorly, so the effective nuclear charge rises at each step.
• Effects: Zr and Hf (and Nb and Ta, and Mo and W) are the same size and behave alike. The 5d metals are far denser. The hydroxides $\text{Ln(OH)}_3$ get less basic from La to Lu.
• Separation: neighbouring ions are almost identical, so it takes ion exchange or solvent extraction. Ce and Eu are easier because of $\text{Ce}^{4+}$ and $\text{Eu}^{2+}$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. What is the lanthanoid contraction, and which subshell is responsible?

2. The radius of $\text{Gd}^{3+}$ is $93.8\ \text{pm}$ and of $\text{Ho}^{3+}$ is $90.1\ \text{pm}$. Holmium is three places after gadolinium. What is the average fall at each of these steps?

3. Explain in two sentences why tantalum is as big as niobium although it has 32 more electrons.

4. Which is the stronger base, $\text{Pr(OH)}_3$ or $\text{Er(OH)}_3$? Give the reason.

5. In an ion-exchange separation of $\text{Lu}^{3+}$ and $\text{La}^{3+}$, which one leaves the column first?