The Trans Effect: Predicting the Product of Square Planar Substitution

Inorganic Chemistry · Reaction Mechanism

The Trans Effect: Predicting the Product of Square Planar Substitution

A synthesis question in coordination chemistry usually reduces to one thing — knowing which ligand directs the incoming group to the position opposite itself.

BSc & MSc · Inorganic Chemistry · Concept

The short answer: In square planar complexes, certain ligands strongly labilise the position trans to themselves. Arranging synthesis steps so the right ligand is present at the right time is how specific geometric isomers are made deliberately rather than as a mixture.

What the effect is

In a square planar complex, a ligand influences how easily the ligand trans to it can be replaced. Some ligands do this powerfully; others barely at all. The consequence is that the position of substitution can be controlled by choosing the order in which ligands are introduced.

This is a kinetic effect — it concerns rate of substitution, not thermodynamic stability. That distinction matters, and it is what separates the trans effect from the related trans influence, which is a ground-state structural property.

The trans series

CO ≈ CN− ≈ C2H4 > PR3 ≈ H− > NO2− > I− > Br− > Cl− > NH3 > OH− > H2O

The order is worth knowing at least approximately, since the whole topic depends on comparing two ligands in a complex and deciding which exerts the stronger effect.

Why some ligands are strong directors

Two contributions combine, and being able to separate them is what higher-level questions ask for.

  • The σ contribution. A strong σ donor pushes electron density toward the metal along its own axis, weakening the bond directly opposite it. Hydride and phosphines act largely this way.
  • The π contribution. A π-acceptor ligand withdraws electron density from the metal, which stabilises the five-coordinate transition state formed when a new ligand adds. Carbon monoxide, cyanide and alkenes act mainly this way, and they sit at the top of the series because the effect is large.
The π explanation is about the transition state, not the ground state. That is why the trans effect is kinetic. A question asking why CO ranks above hydride despite hydride being the stronger σ donor is answered by pointing to transition-state stabilisation.

Using it to make a specific isomer

The classic application is preparing the cis and trans isomers of a diamminedichloroplatinum complex from different starting materials.

Making the cis isomer

Start from the tetrachloro complex and add ammonia twice. Chloride has a stronger trans effect than ammonia. After the first ammonia replaces one chloride, the remaining chlorides direct the second ammonia to a position trans to a chloride — which places the two ammonias cis to each other.

Making the trans isomer

Start instead from the tetraammine complex and add chloride twice. After the first chloride enters, its strong trans effect directs the second chloride to the position opposite it, putting the two chlorides trans and therefore the two ammonias trans as well.

Same two products, opposite geometry, decided entirely by which ligand was in place when the second substitution happened. Working through this reasoning is a very common exam question, and the answer must explain the ordering rather than simply state the outcome.

Trans effect versus trans influence

Trans effectTrans influence
NatureKineticThermodynamic and structural
ObservableRate of substitution trans to the ligandBond length and stretching frequency trans to it
Main originσ donation plus π acceptanceσ donation only

Because the trans influence is purely a σ effect, the two orderings differ — most obviously for π-acceptor ligands, which rank high in the trans effect but not correspondingly high in trans influence. Being asked to distinguish them is a reliable discriminating question.

Frequently asked questions

Does the trans effect apply to octahedral complexes?

It is far less important there. Octahedral substitution is usually dissociative, so the mechanism through which the trans effect operates does not dominate. The topic is essentially a square planar one.

Why are square planar substitutions associative?

Because the geometry leaves the metal accessible above and below the plane, so an incoming ligand can approach and form a five-coordinate intermediate without a ligand leaving first.

Is the ligand exerting the effect itself replaced?

No. It stays in place and labilises the ligand opposite it. That is exactly what makes it useful for directing synthesis.

How reliable is the series?

Reliable enough for exam prediction, but it is an ordering assembled from many observations rather than a single measured scale. Ligands close together in the series may swap order depending on the metal and conditions.

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