Electronic configuration and oxidation states

The two rows under the table

Look at a periodic table. Below the main block sit two rows of fourteen elements, set apart from the rest. Why are they pulled out?

Each row fills one subshell that holds fourteen electrons. In the d-block the electrons enter the $(n-1)d$ subshell. Here they enter the $(n-2)f$ subshell, which lies one shell deeper. So these elements are called the inner transition elements, or the f-block.

• The first row is the lanthanoids. They fill the 4f subshell. The fourteen elements run from cerium (atomic number 58) to lutetium (71). Lanthanum (57) is placed with them, because its chemistry matches theirs.
• The second row is the actinoids. They fill the 5f subshell. They run from thorium (90) to lawrencium (103), with actinium (89) placed with them. Lesson 3 is about this row.

The older name for the first row is lanthanides, and you will also hear rare earths. The IUPAC name ends in "oid" because an ending "ide" belongs to negative ions. The name "rare earths" is misleading. Cerium is about as common in the Earth's crust as copper. Promethium is the one lanthanoid with no stable isotope.

How the 4f subshell fills

Xenon has 54 electrons. The lanthanoids sit after xenon, so every configuration starts with $[\text{Xe}]$. Two electrons go into 6s. Then the 4f subshell fills, and from time to time one electron sits in 5d instead. The general form is:

Here are the atoms. Look for the three that do not follow the simple order.

• La: $[\text{Xe}]\,5d^1\,6s^2$
• Ce: $[\text{Xe}]\,4f^1\,5d^1\,6s^2$
• Pr to Eu: $4f^3$, $4f^4$, $4f^5$, $4f^6$ and $4f^7$, each with $6s^2$
• Gd: $[\text{Xe}]\,4f^7\,5d^1\,6s^2$
• Tb to Yb: $4f^9$ up to $4f^{14}$, each with $6s^2$
• Lu: $[\text{Xe}]\,4f^{14}\,5d^1\,6s^2$

Why do Ce, Gd and Lu put an electron into 5d? In cerium, the 4f and 5d levels are almost at the same energy, so one electron stays in 5d. In gadolinium the 4f subshell is exactly half full, with seven electrons, and that arrangement is unusually stable. Adding the next electron to 5d keeps it that way. In lutetium the 4f subshell is full, and the new electron has nowhere to go except 5d.

The same reasoning explains europium and ytterbium. Europium has $4f^7\,6s^2$, with a half-full subshell. Ytterbium has $4f^{14}\,6s^2$, with a full one.

The 3+ ion

When a lanthanoid becomes an ion, the 6s electrons go first, and then the 5d electron, or one 4f electron. All fifteen elements reach the same pattern:

So $\text{La}^{3+}$ has $4f^0$ and $\text{Gd}^{3+}$ has $4f^7$. $\text{Lu}^{3+}$ has $4f^{14}$. This rule is much simpler than the atoms, and it is the one to remember.

Unpaired electrons and magnetism

Seven f orbitals hold fourteen electrons. By Hund's rule the first seven electrons go into separate orbitals with parallel spins. The next seven then pair up. So the number of unpaired electrons rises from 0 to 7, and then falls back to 0.

Look at the shape of the graph. Two ions with $n$ and $14 - n$ electrons have the same number of unpaired electrons. $\text{Ce}^{3+}$ ($4f^1$) and $\text{Yb}^{3+}$ ($4f^{13}$) each have one. $\text{Pr}^{3+}$ and $\text{Tm}^{3+}$ each have two.

The spin-only moment, in Bohr magnetons (BM), is

where $n$ is the number of unpaired electrons. Every ion with unpaired electrons is paramagnetic. $\text{La}^{3+}$ and $\text{Lu}^{3+}$ have none, so they are diamagnetic.

Now a warning. For most lanthanoid ions the measured moment is not the spin-only value. For $\text{Dy}^{3+}$ the spin-only moment is $5.92\ \text{BM}$, but the measured value is about $10.6\ \text{BM}$. Why? The 4f electrons sit deep inside the ion, under the filled 5s and 5p subshells. The surroundings hardly touch them. So the orbital motion of the electrons still adds to the magnetism. In the d-block the surrounding ligands "quench" this orbital part, and the spin-only formula works well. For the 4f electrons it does not. The ion $\text{Gd}^{3+}$ is the exception that works, because its half-full subshell has no net orbital motion.

Colour

The same deep position explains the colours. A 4f electron jumps from one f orbital to another, and the energy gap is small and sharp. Because the surroundings hardly change the gap, the colour of a lanthanoid ion is nearly the same in every compound. Ions with $n$ and $14 - n$ electrons have similar colours.

• Colourless: $\text{La}^{3+}$ ($4f^0$), $\text{Lu}^{3+}$ ($4f^{14}$), $\text{Gd}^{3+}$ ($4f^7$), $\text{Ce}^{3+}$ ($4f^1$), $\text{Yb}^{3+}$ ($4f^{13}$)
• Green: $\text{Pr}^{3+}$ ($4f^2$) and, paler, $\text{Tm}^{3+}$ ($4f^{12}$)
• Lilac and pink: $\text{Nd}^{3+}$ ($4f^3$) is lilac and $\text{Er}^{3+}$ ($4f^{11}$) is pink
• Yellow: $\text{Sm}^{3+}$ ($4f^5$), $\text{Dy}^{3+}$ ($4f^9$)
• Pale pink: $\text{Eu}^{3+}$ ($4f^6$) and $\text{Tb}^{3+}$ ($4f^8$)

An ion with empty or full 4f cannot jump, so it is colourless. $\text{Gd}^{3+}$ has a half-full subshell, and its first f-f jump needs ultraviolet light, so it is colourless too.

Oxidation states

The usual state of every lanthanoid is +3. Why? Take the energy needed to remove electrons. The first three ionisation enthalpies are low, because the 6s and 5d electrons are far out. The fourth electron has to come from the 4f subshell, which is held much more tightly. The energy released when a 3+ ion is hydrated, or fits into a crystal, repays the cost of the third electron, but it does not repay a fourth.

Two other states appear, but only where the electrons left in 4f make a specially stable subshell, empty, half full or full.

• +4: $\text{Ce}^{4+}$ has $4f^0$. $\text{Tb}^{4+}$ has $4f^7$.
• +2: $\text{Eu}^{2+}$ has $4f^7$. $\text{Yb}^{2+}$ has $4f^{14}$.

Look at the third ionisation enthalpy. Europium needs $2404\ \text{kJ/mol}$ and ytterbium needs $2415\ \text{kJ/mol}$ to lose a third electron, because that would break a half-full or full subshell. Gadolinium needs only $1990\ \text{kJ/mol}$ and lutetium $2022\ \text{kJ/mol}$, because they are losing the last electron outside a stable subshell. This is why Eu and Yb stop at +2 and Gd and Lu stop at +3.

The pattern is a guide and not a law. $\text{Sm}^{2+}$ ($4f^6$), $\text{Pr}^{4+}$ and $\text{Dy}^{4+}$ also exist in some compounds, and the graph shows how close they come to a stable count.

What the states mean in practice

An ion that is out of its usual +3 state wants to return.

• $\text{Ce}^{4+}$ is a strong oxidising agent. Its standard potential for $\text{Ce}^{4+}/\text{Ce}^{3+}$ is $+1.74\ \text{V}$, high enough to oxidise water, though the reaction is slow. So cerium(IV) salts such as ceric ammonium sulfate are used in titrations.
• $\text{Eu}^{2+}$, $\text{Yb}^{2+}$ and $\text{Sm}^{2+}$ are strong reducing agents. They give up one electron to become the stable 3+ ion.

Europium(II) has a size close to strontium, and $\text{EuSO}_4$ is as insoluble as $\text{SrSO}_4$. The next lesson uses this fact for separation.

Practise with new numbers

Try it in Python

The program counts the 4f electrons in each 3+ ion and finds the unpaired electrons and the spin-only moment. Then it looks for the 2+ and 4+ ions with an empty, half-full or full 4f subshell.

Common mistakes

• Counting 4f electrons as $Z - 58$ for the 3+ ion. The rule is $Z - 57$, so $\text{Ce}^{3+}$ has one and $\text{La}^{3+}$ has none.
• Taking the unpaired number above seven as $n$. After seven the extra electrons pair up. $4f^9$ has five unpaired electrons, not nine.
• Writing gadolinium as $4f^8\,6s^2$. The atom is $4f^7\,5d^1\,6s^2$.
• Saying every lanthanoid shows +4 because cerium does. Only cerium and terbium are common, and a few others in special compounds.
• Using the spin-only moment for every lanthanoid ion. The orbital part is large for most of them.

Remember these

• The 3+ ion is $[\text{Xe}]\,4f^{n}$ with $n = Z - 57$. The atoms $\text{Ce}$, $\text{Gd}$ and $\text{Lu}$ have an electron in 5d.
• Unpaired electrons: $n$ up to seven, then $14 - n$. The spin-only formula works well for $\text{Gd}^{3+}$ only.
• Colour comes from f-f jumps. It hardly depends on the ligand. Empty or full 4f gives no colour.
• The usual state is +3. The +4 state of Ce and Tb and the +2 state of Eu and Yb leave $4f^0$, $4f^7$ or $4f^{14}$.
• $\text{Ce}^{4+}$ is an oxidising agent. $\text{Eu}^{2+}$, $\text{Yb}^{2+}$ and $\text{Sm}^{2+}$ are reducing agents.

Check yourself

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

1. Write the electronic configuration of $\text{Nd}^{3+}$ and give its number of unpaired electrons. The atomic number of neodymium is 60.

2. Why do $\text{Pr}^{3+}$ and $\text{Tm}^{3+}$ have similar colours?

3. Why is europium found in the +2 state in some compounds while gadolinium is not?

4. A cerium(IV) solution turns slowly from yellow to colourless when left. What has happened to the cerium?

5. The measured moment of $\text{Dy}^{3+}$ is far above its spin-only value. Why is the spin-only formula poor for a lanthanoid ion?