Geometrical isomerism

A drug and its useless twin

Cisplatin, $[\text{Pt}(\text{NH}_{3})_{2}\text{Cl}_{2}]$, is a drug used against some cancers. Its twin has the same atoms and the same bonds, and it does not work as a drug. The only difference is where the two chloride ligands sit. In cisplatin they are next to each other. In the twin they are opposite each other.

You know a similar idea from Class 11. But-2-ene has cis and trans forms because a double bond cannot rotate. In a complex the reason is different. The ligands sit at the corners of a fixed shape. No turning can move a ligand to another corner without breaking a bond.

Two words describe the positions.

• Cis: two ligands at neighbouring corners, $90^\circ$ apart as seen from the metal.
• Trans: two ligands at opposite corners, $180^\circ$ apart.

Isomers that differ in this way are geometrical isomers. They are a kind of stereoisomer.

Four ligands round the metal

In a tetrahedral complex every corner is next to every other corner. There is no cis and trans. A tetrahedral complex such as $[\text{NiCl}_{2}\text{Br}_{2}]^{2-}$ has only one arrangement.

A square planar complex has its four ligands in one plane, at the corners of a square. Now each ligand has two neighbours and one opposite partner. Look at $[\text{Pt}(\text{NH}_{3})_{2}\text{Cl}_{2}]$.

The two forms differ in more than looks.

• In cis, the two Pt-Cl bond dipoles point at an angle of $90^\circ$. They add up, and the molecule is polar.
• In trans, the two Pt-Cl bond dipoles point in opposite directions. They cancel, and the dipole moment is zero.

So the dipole moment tells the two apart. The cis form is polar, so it dissolves better in water than the trans form.

Set the angle to $90^\circ$ and then to $180^\circ$. The same two bonds give a large dipole in one case and none in the other.

Practise with new numbers

Here is how many geometrical isomers each square planar type has.

• $\text{MA}_{4}$ and $\text{MA}_{3}\text{B}$: one each. Every arrangement can be turned into every other.
• $\text{MA}_{2}\text{B}_{2}$: two, cis and trans.
• $\text{MA}_{2}\text{BC}$: two. The two A ligands are cis or trans.
• $\text{MABCD}$: three. Whichever ligand sits opposite A is B, C or D.
• $\text{M}(\text{AB})_{2}$, with an unsymmetrical bidentate ligand: two.

Werner used this counting. He found two forms of $[\text{Pt}(\text{NH}_{3})_{2}\text{Cl}_{2}]$ and no more. A tetrahedral arrangement allows only one. A square planar arrangement allows two. So platinum(II) is square planar here.

Six ligands round the metal

In an octahedral complex each ligand has four neighbours at $90^\circ$ and one opposite partner at $180^\circ$.

Werner used counting again. Two forms are known for $[\text{Co}(\text{NH}_{3})_{4}\text{Cl}_{2}]^{+}$, one violet and one green. An octahedron gives two arrangements of two chloride ligands (cis and trans). A flat hexagon would give three, and so would a trigonal prism. Only the octahedron matches the experiment.

• $\text{MA}_{4}\text{B}_{2}$ has a cis and a trans form, as in the top row.
• $\text{MA}_{3}\text{B}_{3}$ has a fac and a mer form, as in the bottom row. In fac (facial) the three A ligands sit at the corners of one triangular face, all $90^\circ$ apart. In mer (meridional) they lie in one plane through the metal, like a meridian on a globe. Two of them are then $180^\circ$ apart.

Other octahedral types have more isomers. $\text{MA}_{4}\text{BC}$ has two, and $\text{MA}_{3}\text{B}_{2}\text{C}$ has three. $\text{MA}_{2}\text{B}_{2}\text{C}_{2}$ has five, and $\text{MABCDEF}$ has fifteen.

Counting by trans pairs

Listing every drawing is slow. Here is a faster way for monodentate ligands. An octahedron has three pairs of opposite corners. Write the six ligands as three trans pairs. Two arrangements with the same three pairs are the same geometrical isomer.

Try $\text{MA}_{3}\text{B}_{3}$. The pairs are either AB, AB, AB (that is fac) or AA, AB, BB (that is mer). There are two isomers.

Now try $\text{MA}_{2}\text{B}_{2}\text{C}_{2}$. The possible sets of trans pairs are:

• AA, BB, CC
• AA, BC, BC
• BB, AC, AC
• CC, AB, AB
• AB, AC, BC

That is five geometrical isomers. The last one has no two identical ligands trans, and it also exists as a mirror-image pair. The next lesson explains why.

Step through the eight types. The count stays small for the first six, then climbs to fifteen. The orange bar is higher than the blue one for two types. Those are the types with mirror-image forms.

A chelate ring needs two cis corners

A bidentate ligand such as ethane-1,2-diamine, written en, is a short chain. It cannot reach across the metal, so it must take two cis corners. So $[\text{Co}(\text{en})_{2}\text{Cl}_{2}]^{+}$ can have its two chloride ligands cis or trans. It cannot have a trans en. That is the key idea for the next lesson.

Try it in Python

The program tries every way of placing the ligands on the corners. It then counts an arrangement as new only if no turn of the shape makes it match one already seen.

Common mistakes

• Counting a turned drawing as a new isomer. Turn the shape in your head. If one can be turned into the other, they are one isomer.
• Looking for cis and trans in a tetrahedral complex. Every corner is next to every other corner, so there is none.
• Using cis and trans for $\text{MA}_{3}\text{B}_{3}$. Use fac and mer.
• Believing $\text{MA}_{5}\text{B}$ has several isomers. All six corners are alike, so B has only one place to go.
• Placing a chelate ring across the metal. The ring spans two cis corners only.

Remember these

• Geometrical isomers have the same bonds in a different arrangement in space.
• Cis is $90^\circ$ and trans is $180^\circ$. Tetrahedral complexes have no geometrical isomers.
• Square planar: $\text{MA}_{2}\text{B}_{2}$ 2, $\text{MA}_{2}\text{BC}$ 2, $\text{MABCD}$ 3.
• Octahedral: $\text{MA}_{4}\text{B}_{2}$ 2, $\text{MA}_{3}\text{B}_{3}$ 2 (fac, mer), $\text{MA}_{2}\text{B}_{2}\text{C}_{2}$ 5, $\text{MABCDEF}$ 15.
• To count: write three trans pairs and count the different sets.

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 geometrical isomers has the square planar complex $[\text{Pt}(\text{NH}_{3})(\text{py})\text{ClBr}]$? Here py is pyridine, a ligand that bonds through one atom.

2. Why can a tetrahedral complex $\text{MA}_{2}\text{B}_{2}$ not have cis and trans forms?

3. Which isomer of $\text{MA}_{3}\text{B}_{3}$ has no pair of identical ligands trans to each other?

4. Which is more polar, cis or trans $[\text{Pt}(\text{NH}_{3})_{2}\text{Cl}_{2}]$, and why?

5. Why does $[\text{Co}(\text{en})_{2}\text{Cl}_{2}]^{+}$ have no isomer with a trans en ligand?