Valence bond theory

A puzzle about two complexes of iron

Take two complex ions of iron(III). Both have six ligands in an octahedron, and both have the same charge: $[\text{Fe(CN)}_6]^{3-}$ and $[\text{FeF}_6]^{3-}$.

Hold a sample of each near a strong magnet. The cyanide complex is pulled only a little. The fluoride complex is pulled much harder. The measured magnetic moments are $1.73\ \text{BM}$ and $5.92\ \text{BM}$. BM stands for the Bohr magneton, the unit of magnetic moment.

The metal ion is the same and the shape is the same. So why do the magnets disagree? Let us find out.

What Fe3+ brings to the complex

An iron atom has the configuration $[\text{Ar}]\,3d^6\,4s^2$. To form $\text{Fe}^{3+}$ it loses the two 4s electrons and one 3d electron. That leaves $3d^5$.

By Hund's rule the five electrons sit in the five 3d orbitals one each, with parallel spins. So a free $\text{Fe}^{3+}$ ion has five unpaired electrons. The 4s, 4p and 4d orbitals are all empty.

The main idea of valence bond theory

In Class 11 you used valence bond theory for molecules. Atoms overlap their orbitals, and the orbitals may mix, or hybridise, to give bonds in the right directions. Complexes follow the same plan.

1. The ligand has a lone pair of electrons.
2. The metal ion has empty orbitals of the right energy.
3. The ligand gives its pair into an empty orbital. This forms a coordinate bond.
4. The empty orbitals first hybridise. The new orbitals are equal in energy and point to the corners of a shape.
5. The number of hybrid orbitals equals the coordination number.

So the metal ion must have as many empty orbitals as it has ligands. The 4s and 4p orbitals are always available. If more are needed, the ion uses 3d orbitals or 4d orbitals.

Hybrid orbitals and shapes

• Coordination number 4, tetrahedral: $sp^3$, made from one 4s and three 4p orbitals.
• Coordination number 4, square planar: $dsp^2$, made from one 3d, one 4s and two 4p orbitals.
• Coordination number 6, octahedral: $d^2sp^3$, made from two 3d, one 4s and three 4p orbitals. Or $sp^3d^2$, made from one 4s, three 4p and two 4d orbitals.

Why two kinds for the octahedron? In $d^2sp^3$ the d orbitals belong to the inner shell, 3d. We call this an inner orbital complex. In $sp^3d^2$ the d orbitals belong to the outer shell, 4d. We call this an outer orbital complex.

Inner or outer: the ligand decides

Look at the five 3d electrons of $\text{Fe}^{3+}$. To make $d^2sp^3$, the ion needs two empty 3d orbitals. It has none, because all five orbitals hold one electron each.

Cyanide. The $\text{CN}^-$ ion is a strong field ligand. It pushes the 3d electrons together, so they pair up. Two pairs and one single electron fit into three orbitals. Two 3d orbitals are now empty. With the 4s and three 4p orbitals they make six $d^2sp^3$ hybrids. Each takes a pair from a cyanide ion. One unpaired electron remains, so $\mu = 1.73\ \text{BM}$. This is an inner orbital complex. Chemists also call it a low spin complex.

Fluoride. The $\text{F}^-$ ion is a weak field ligand. It leaves the electrons alone. All five 3d orbitals stay half full, so no 3d orbital is free. The ion uses 4s, three 4p and two 4d orbitals, which gives $sp^3d^2$. Five unpaired electrons remain, so $\mu = 5.92\ \text{BM}$. This is an outer orbital complex, or a high spin complex.

That solves the puzzle. Strong ligands pair the electrons and give few unpaired electrons. Weak ligands do not.

Strong field ligands include $\text{CN}^-$, CO, $\text{NH}_3$ and en. Weak field ligands include $\text{F}^-$, $\text{Cl}^-$, $\text{Br}^-$, $\text{I}^-$ and $\text{H}_2\text{O}$. Valence bond theory has to be told which is which. It gives no reason. The next lesson does.

Magnetic behaviour

• A substance with only paired electrons is diamagnetic. A magnet pushes it away, very weakly.
• A substance with unpaired electrons is paramagnetic. A magnet pulls it in.

The size of the pull is the magnetic moment. When the moment comes only from the spins, we use the spin-only formula:

Here $n$ is the number of unpaired electrons. The values to remember:

• $n = 1$: $1.73\ \text{BM}$
• $n = 2$: $2.83\ \text{BM}$
• $n = 3$: $3.87\ \text{BM}$
• $n = 4$: $4.90\ \text{BM}$
• $n = 5$: $5.92\ \text{BM}$

A measured moment therefore tells you $n$, and $n$ tells you which hybridisation the complex uses.

Three complexes of nickel

Nickel gives a good test of the theory, because one metal makes three different cases. Take the $\text{Ni}^{2+}$ ion, which is $3d^8$.

• $[\text{NiCl}_4]^{2-}$. Chloride is weak. The eight 3d electrons stay as they are, with two unpaired. The ion uses one 4s and three 4p orbitals: $sp^3$, tetrahedral. The complex has two unpaired electrons, so $\mu = 2.83\ \text{BM}$.
• $[\text{Ni(CN)}_4]^{2-}$. Cyanide is strong. The two unpaired 3d electrons pair up, which empties one 3d orbital. That orbital, the 4s and two 4p orbitals give $dsp^2$, square planar. All electrons are paired, so the complex is diamagnetic.
• $[\text{Ni(CO)}_4]$. Here nickel has an oxidation state of zero. The atom is $3d^8\,4s^2$. Carbon monoxide is a strong ligand. The 4s electrons move into 3d, which becomes $3d^{10}$. All are paired, and the 4s orbital is empty. With three 4p orbitals it gives $sp^3$, tetrahedral, and the complex is diamagnetic.

The same metal gives a tetrahedral or a square planar shape, and the magnet tells which one.

Try $n = 6$ with a weak ligand, then with a strong one. A $d^6$ ion with a strong ligand has no unpaired electrons. Then raise $n$ to 4 and compare again.

Practise with new numbers

Try it in Python

The program applies the steps of this lesson: unpaired electrons, hybridisation, shape and moment.

Where valence bond theory falls short

• It cannot say why some ligands are strong and others are weak. We have to be told.
• It gives no reason for the colour of complexes.
• It treats the magnetic moment as a count of unpaired electrons and nothing more.
• It cannot explain why $[\text{Cu(NH}_3)_4]^{2+}$ is square planar, without moving a 3d electron to a 4p orbital by hand.
• It says nothing about how stable a complex is.

Crystal field theory, in the next lesson, answers most of these questions.

Common mistakes

• Writing $sp^3d^2$ for every octahedral complex. A strong ligand with a $d^4$, $d^5$ or $d^6$ ion gives $d^2sp^3$.
• Using the 3d count of the atom. Count the electrons of the ion. Iron is $3d^6$ as an atom and $3d^5$ as $\text{Fe}^{3+}$.
• Calling $d^2sp^3$ an outer orbital complex. The 3d orbitals are the inner ones.
• Using the moment formula with the number of electrons. Use only the unpaired electrons.
• Forgetting that CO and $\text{CN}^-$ are strong. They almost always force pairing.

Remember these

• The ligand gives a lone pair. The metal ion gives an empty hybrid orbital.
• Octahedral: $d^2sp^3$ (inner orbital, low spin) or $sp^3d^2$ (outer orbital, high spin). Tetrahedral: $sp^3$. Square planar: $dsp^2$.
• A strong ligand pairs the electrons. A weak ligand does not.
• $\mu = \sqrt{n(n+2)}\ \text{BM}$, with $n$ the number of unpaired electrons.
• Diamagnetic: no unpaired electrons. Paramagnetic: at least one.

Check yourself

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

1. How many empty orbitals must a metal ion provide for the complex $[\text{Co(NH}_3)_6]^{3+}$, and which ones are used?

2. A complex has a spin-only moment of $3.87\ \text{BM}$. How many unpaired electrons does it have?

3. Why is $[\text{FeF}_6]^{3-}$ called an outer orbital complex?

4. What is the hybridisation and magnetic behaviour of $[\text{Ni(CN)}_4]^{2-}$?