Ligands and coordination number

The groups round the metal

Every ligand has at least one atom with a lone pair of electrons. That atom is the donor atom. It shares its lone pair with the metal, and a coordinate bond forms. In $[\text{Co}(\text{NH}_{3})_{6}]^{3+}$ the donor atom of each ammonia is nitrogen. In $[\text{Fe}(\text{CN})_{6}]^{4-}$ it is the carbon of each cyanide ion.

A ligand can be:

• a neutral molecule, such as $\text{NH}_{3}$, $\text{H}_{2}\text{O}$ or $\text{CO}$;
• a negative ion, such as $\text{Cl}^{-}$, $\text{CN}^{-}$ or $\text{OH}^{-}$;
• rarely, a positive ion, such as the nitrosonium ion $\text{NO}^{+}$.

Denticity: how many donor atoms touch the metal

Some ligands have one donor atom. Others have two or more, and they can bond through several at once. The number of donor atoms that a ligand uses is its denticity.

• A monodentate ligand has one donor atom: $\text{NH}_{3}$, $\text{H}_{2}\text{O}$, $\text{Cl}^{-}$, $\text{CN}^{-}$, $\text{CO}$, $\text{OH}^{-}$.
• A bidentate ligand has two. Ethane-1,2-diamine, written en, is $\text{H}_{2}\text{N}\text{-}\text{CH}_{2}\text{-}\text{CH}_{2}\text{-}\text{NH}_{2}$ and uses its two nitrogen atoms. The oxalate ion, $\text{C}_{2}\text{O}_{4}^{2-}$, uses two oxygen atoms. The glycinate ion, $\text{H}_{2}\text{N}\text{-}\text{CH}_{2}\text{-}\text{COO}^{-}$, uses one nitrogen and one oxygen.
• A polydentate ligand has more. The ion of EDTA, ethylenediaminetetraacetate, has six donor atoms: two nitrogen atoms and four oxygen atoms. It is hexadentate.

A ligand with two or more donor atoms wraps round the metal like a claw. The ring that closes is a chelate ring, and the ligand is a chelating ligand. The word comes from the Greek word for a claw. A five-membered ring or a six-membered ring is the most stable. A single EDTA ion can take a calcium ion out of hard water, because it grips from six sides.

Ligands with two different donor atoms

Some ligands have two different donor atoms, but they use only one of them at a time. They are ambidentate ligands:

• The nitrite ion, $\text{NO}_{2}^{-}$, can bond through the nitrogen atom or through an oxygen atom.
• The thiocyanate ion, $\text{SCN}^{-}$, can bond through the sulphur atom or through the nitrogen atom.

Do not confuse the two ideas. A chelating ligand uses two or more donor atoms at once. An ambidentate ligand has a choice, but it takes only one. Each choice makes a different compound, and the names show which atom is used.

The coordination number

The coordination number of a metal is the number of donor atoms bonded to it. It is not the number of ligands. The two are equal only when every ligand is monodentate.

• $[\text{Co}(\text{NH}_{3})_{6}]^{3+}$ has 6 ligands and a coordination number of 6.
• $[\text{Ni}(\text{en})_{3}]^{2+}$ has 3 ligands, each with two donor atoms, so the coordination number is 6.
• $[\text{Fe}(\text{EDTA})]^{-}$ has one ligand and a coordination number of 6.
• $[\text{Ag}(\text{CN})_{2}]^{-}$ has 2 ligands and a coordination number of 2.

The common values are 2, 4 and 6. A large metal ion with a high charge can hold more ligands. Large ligands crowd one another, so they give smaller numbers. Chloride ions give $[\text{CoCl}_{4}]^{2-}$ with 4, while the smaller water molecules give $[\text{Co}(\text{H}_{2}\text{O})_{6}]^{2+}$ with 6.

Set 2 monodentate and 2 bidentate ligands. There are four ligands, and the bar is 6 blocks long. Now add the EDTA ion. The bar runs far past the grey line, and no such complex is found. Real EDTA complexes have the EDTA ion alone, or with one small extra ligand.

Practise with new numbers

The shape follows the coordination number

The donor atoms arrange themselves round the metal as far from one another as they can.

• Coordination number 2 gives a linear complex. $[\text{Ag}(\text{NH}_{3})_{2}]^{+}$ is the diamminesilver(I) ion in Tollens' reagent.
• Coordination number 4 gives a tetrahedral complex or a square planar one. $[\text{Ni}(\text{CO})_{4}]$ is tetrahedral. $[\text{PtCl}_{4}]^{2-}$ is square planar.
• Coordination number 6 gives an octahedral complex. This is the most common shape. $[\text{Co}(\text{NH}_{3})_{6}]^{3+}$ is one.

The same metal can choose either shape for 4. The bonding lessons later in this topic explain why.

Try it in Python

The program counts ligands and donor atoms for five complexes. It keeps a table that holds the charge and the number of donor atoms of each ligand.

Homoleptic and heteroleptic complexes

A complex with only one kind of ligand is homoleptic. $[\text{Co}(\text{NH}_{3})_{6}]^{3+}$ and $[\text{Fe}(\text{CN})_{6}]^{4-}$ are homoleptic. A complex with more than one kind of ligand is heteroleptic. $[\text{Co}(\text{NH}_{3})_{4}\text{Cl}_{2}]^{+}$ and $[\text{Pt}(\text{NH}_{3})_{2}\text{Cl}_{2}]$ are heteroleptic.

The charge on the complex and the oxidation state

The charge on a complex ion is the sum of the oxidation state of the metal and the charges of all the ligands:

Here $x$ is the oxidation state of the metal. We want $x$, so we rearrange: $x = \text{charge} - (\text{sum of the ligand charges})$.

Take $[\text{Fe}(\text{CN})_{6}]^{4-}$. The ligands are six cyanide ions, each $-1$. So $x + 6(-1) = -4$, and $x = +2$. The iron is iron(II).

Take $[\text{Co}(\text{NH}_{3})_{6}]^{3+}$. Ammonia is neutral, so the ligands add nothing. $x = +3$.

Take $[\text{Cu}(\text{NH}_{3})_{4}]\text{SO}_{4}$, the deep blue compound of the first page. Sulphate outside the brackets has a charge of $-2$, so the complex ion is $+2$. Ammonia is neutral. So $x = +2$. The copper is copper(II), as it was before the ammonia came.

Take $\text{K}_{3}[\text{Fe}(\text{C}_{2}\text{O}_{4})_{3}]$. Three $\text{K}^{+}$ ions mean the complex ion has a charge of $-3$. Each oxalate ion is $-2$. So $x + 3(-2) = -3$, and $x = +3$.

Zero is possible too. In $[\text{Ni}(\text{CO})_{4}]$ every ligand is neutral, and the complex has no charge. So $x = 0$.

Start with the charge on the complex at $-3$ and six ligands with a charge of $-1$ each. The metal is at $+3$. Now change the number of ligands to 4. The oxidation state drops to $+1$. Try making the ligands neutral. The oxidation state then equals the charge on the complex.

Practise with new numbers

Try it in Python

The program turns the rule into a function. It takes the charge of the complex and the ligands, and it gives the oxidation state of the metal.

Common mistakes

• Counting ligands, not donor atoms. $[\text{Ni}(\text{en})_{3}]^{2+}$ has 3 ligands but a coordination number of 6.
• Adding the ligand charges to the charge of the complex. You must subtract them.
• Forgetting the counter ions. In $\text{K}_{3}[\text{Fe}(\text{C}_{2}\text{O}_{4})_{3}]$ the charge of the complex comes from the three $\text{K}^{+}$ ions.
• Treating the oxidation state as the coordination number. In $[\text{Ni}(\text{CO})_{4}]$ the first is 0 and the second is 4.
• Calling a chelating ligand ambidentate. A chelating ligand uses all its donor atoms at once.

Remember these

• A ligand donates a lone pair. Denticity is the number of donor atoms it uses.
• Coordination number = the number of donor atoms bonded to the metal.
• 2 gives linear, 4 gives tetrahedral or square planar, 6 gives octahedral.
• Charge on the complex = $x$ + the ligand charges.
• Ambidentate: two donor atoms, one used at a time. Chelating: several at once.

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 coordination number of $[\text{Ni}(\text{en})_{2}\text{Cl}_{2}]$?

2. Find the oxidation state of copper in $[\text{CuCl}_{4}]^{2-}$.

3. The nitrite ion can bond through nitrogen or through oxygen. Is it a chelating ligand?

4. A student says that $[\text{Fe}(\text{EDTA})]^{-}$ has a coordination number of 1. What is wrong?