Crystal field theory

A second way to look at the same complex

Valence bond theory had to be told which ligands are strong. It also could not explain colour. We now use a simpler picture that does both.

Crystal field theory makes three assumptions.

• The ligands are point negative charges. A neutral ligand such as water points its negative end at the metal ion.
• The bond between metal and ligand is the attraction of opposite charges. No orbitals overlap.
• Only the five d orbitals of the metal ion matter, and they are repelled by the negative charges.

The five d orbitals

In a free ion, all five d orbitals have the same energy. They differ in shape.

• Three of them, $d_{xy}$, $d_{yz}$ and $d_{xz}$, have lobes between the axes.
• Two of them, $d_{x^2-y^2}$ and $d_{z^2}$, have lobes along the axes.

Now surround the ion with six ligands. First imagine that their charge is spread evenly over a sphere. Every d orbital is repelled by the same amount, so all five rise together and stay equal.

Then place the ligands where they really are: one on each end of the three axes, which makes an octahedron. Now the orbitals are no longer equal.

Octahedral splitting

• The $d_{x^2-y^2}$ and $d_{z^2}$ lobes point straight at the ligands. An electron there is strongly repelled. These two orbitals rise. They are called the $e_g$ set.
• The $d_{xy}$, $d_{yz}$ and $d_{xz}$ lobes point between the ligands. An electron there is repelled less. These three orbitals are lower. They are called the $t_{2g}$ set.

The gap between the two sets is the crystal field splitting, written $\Delta_o$ ("delta o", for octahedral).

The average energy of the five orbitals does not change. Two orbitals go up and three go down, so:

• each $e_g$ orbital rises by $0.6\,\Delta_o$,
• each $t_{2g}$ orbital falls by $0.4\,\Delta_o$.

Check: $2 \times 0.6 = 1.2$ and $3 \times 0.4 = 1.2$. The rises and falls cancel.

Filling the orbitals: high spin and low spin

Electrons go into the lower $t_{2g}$ orbitals first, one in each orbital, with parallel spins. This is Hund's rule. The first three electrons ($d^1$, $d^2$, $d^3$) have no choice.

The fourth electron has two choices:

• It can go to an $e_g$ orbital. This costs energy $\Delta_o$.
• It can pair up in a $t_{2g}$ orbital. This costs the pairing energy, $P$, the extra repulsion of two electrons in one orbital.

The electron takes the cheaper path.

• If $\Delta_o < P$ (a weak field), the electron goes to $e_g$. We get a high spin complex with many unpaired electrons.
• If $\Delta_o > P$ (a strong field), the electron pairs up. We get a low spin complex with few unpaired electrons.

Only $d^4$, $d^5$, $d^6$ and $d^7$ ions in an octahedral field have this choice. For $d^1$ to $d^3$ and for $d^8$ to $d^{10}$ there is just one way to fill the orbitals.

Here are the four cases:

• $d^4$: high spin $t_{2g}^3 e_g^1$ with 4 unpaired electrons. Low spin $t_{2g}^4$ with 2.
• $d^5$: high spin $t_{2g}^3 e_g^2$ with 5 unpaired electrons. Low spin $t_{2g}^5$ with 1.
• $d^6$: high spin $t_{2g}^4 e_g^2$ with 4 unpaired electrons. Low spin $t_{2g}^6$ with 0.
• $d^7$: high spin $t_{2g}^5 e_g^2$ with 3 unpaired electrons. Low spin $t_{2g}^6 e_g^1$ with 1.

Start with $d^6$ and $\Delta_o$ below $P$, then raise $\Delta_o$ past $P$. Watch the electrons jump from $e_g$ into $t_{2g}$ and the number of unpaired electrons drop.

What decides the size of the splitting

The ligand. Ligands are ranked by the splitting they cause. This list is the spectrochemical series, from weak to strong:

The left end holds the weak field ligands and the right end the strong field ones. This is why $\text{CN}^-$ forces pairing and $\text{F}^-$ does not.

The metal ion. A higher charge on the metal ion gives a larger splitting, so $\text{Fe}^{3+}$ complexes split more than $\text{Fe}^{2+}$ complexes. Ions of the 4d and 5d rows give much larger splitting than 3d ions. So second and third row complexes are nearly always low spin.

Tetrahedral complexes

Put four ligands at the corners of a tetrahedron. No ligand points straight at a d orbital. But the three orbitals $d_{xy}$, $d_{yz}$ and $d_{xz}$ lie a little closer to the ligands than the other two. So the order turns upside down:

• the two $e$ orbitals ($d_{z^2}$ and $d_{x^2-y^2}$) are lower, at $-0.6\,\Delta_t$,
• the three $t_2$ orbitals are higher, at $+0.4\,\Delta_t$.

Only four ligands are present, and they do not point at the orbitals. So the gap is small:

For the same metal and the same ligand, the tetrahedral gap is less than half of the octahedral one. It is nearly always smaller than $P$. So tetrahedral complexes are almost always high spin. Low spin tetrahedral complexes are not found with ordinary ligands.

A square planar complex splits the five orbitals into four levels. The $d_{x^2-y^2}$ orbital is the highest. An ion with eight d electrons and a strong field leaves it empty, and all the others are paired. That is why $[\text{Ni(CN)}_4]^{2-}$ is diamagnetic.

Crystal field stabilisation energy

Electrons in the lower set are more stable than in a free ion. The total gain in energy, measured from the average level, is the crystal field stabilisation energy, CFSE. For an octahedral complex:

Here $n_{t_{2g}}$ and $n_{e_g}$ are the numbers of electrons in each set. Here are the values for high spin ions:

• $d^1$: $-0.4\,\Delta_o$. $d^2$: $-0.8\,\Delta_o$. $d^3$: $-1.2\,\Delta_o$.
• $d^4$: $-0.6\,\Delta_o$. $d^5$: $0$. $d^6$: $-0.4\,\Delta_o$. $d^7$: $-0.8\,\Delta_o$.
• $d^8$: $-1.2\,\Delta_o$. $d^9$: $-0.6\,\Delta_o$. $d^{10}$: $0$.

For low spin ions, $d^4$ gives $-1.6\,\Delta_o$, $d^5$ gives $-2.0\,\Delta_o$, $d^6$ gives $-2.4\,\Delta_o$ and $d^7$ gives $-1.8\,\Delta_o$.

A fuller count also adds the pairing energy for each extra pair. Here we count the splitting only.

A worked example. The ion $[\text{Cr(H}_2\text{O)}_6]^{3+}$ has $d^3$ and $\Delta_o = 208\ \text{kJ/mol}$.

1. The three electrons sit alone in the $t_{2g}$ orbitals: $n_{t_{2g}} = 3$ and $n_{e_g} = 0$.
2. $\text{CFSE} = -0.4 \times 3 \times 208\ \text{kJ/mol} = -249.6\ \text{kJ/mol}$.
3. The sign is negative because the electrons are more stable than in the free ion.

Practise with new numbers

Try it in Python

The program fills the five orbitals for any number of electrons, any splitting and any pairing energy. It then prints the unpaired electrons, the CFSE and the moment.

Common mistakes

• Putting $e_g$ below $t_{2g}$ in an octahedron. The orbitals pointing at the ligands are the high ones.
• Swapping 0.4 and 0.6. The $t_{2g}$ set is at $-0.4\,\Delta_o$. The $e_g$ set is at $+0.6\,\Delta_o$.
• Drawing a low spin tetrahedral complex. The gap is too small to pay for pairing.
• Asking whether a $d^2$ or a $d^9$ ion is high spin. Only $d^4$ to $d^7$ have the choice.
• Mixing up $P$ and $\Delta_o$. Pairing happens when the gap is larger than the pairing energy.

Remember these

• Octahedral: $t_{2g}$ at $-0.4\,\Delta_o$ and $e_g$ at $+0.6\,\Delta_o$. Tetrahedral: $e$ at $-0.6\,\Delta_t$ and $t_2$ at $+0.4\,\Delta_t$.
• $\Delta_t = \tfrac{4}{9}\Delta_o$, so tetrahedral complexes are high spin.
• $\Delta_o > P$: low spin. $\Delta_o < P$: high spin. Only for $d^4$ to $d^7$.
• Spectrochemical series: $\text{I}^-$ weakest, CO among the strongest.
• $\text{CFSE} = (-0.4\,n_{t_{2g}} + 0.6\,n_{e_g})\,\Delta_o$.

Check yourself

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

1. Why do the $d_{z^2}$ and $d_{x^2-y^2}$ orbitals go up in an octahedral complex?

2. A $d^5$ ion has $\Delta_o$ smaller than $P$. How many unpaired electrons does it have, and what is the CFSE?

3. For the same ion, a ligand that is stronger on the spectrochemical series is used. What changes?

4. Why is a tetrahedral complex nearly always high spin?