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A Level · Cambridge (CIE) · 9701

Describe cis/trans isomerism and polarity in complexes

Identify geometrical isomers in square-planar and octahedral complexes and deduce whether bond-dipole vectors cancel.

Section 28 · Chemistry of transition elements

Chemistry of transition elements · Cambridge International AS & A Level Chemistry 9701 · 28.4

A Level; Cambridge International Chemistry 9701, 2028–2030 syllabus, Topic 28. Use the first-row transition elements titanium to copper and the species explicitly named in the syllabus. Where redox potentials or stability constants are needed, use the values supplied in the question. Colour descriptions can depend on conditions and concentration. Written practice is self-checked

Before you begin

  • Use electron configurations, oxidation numbers, ionic equations, equilibrium expressions and the AS redox and bonding models.

Work on paper. Try each question before revealing help, and explain why your method works.

Quick readiness check

In trans-[Pt(NH₃)₂Cl₂], where are the two Cl ligands?

What you will learn

  • Recognise cis and trans arrangements in named complexes
  • Describe geometrical isomerism for square-planar and octahedral examples
  • Deduce overall polarity from ligand arrangement and symmetry

Cis and trans describe relative ligand positions

In square-planar [Pt(NH₃)₂Cl₂], cis has the two Cl ligands adjacent; trans has them opposite. The two forms have the same formula and connectivity but different spatial arrangements.

Octahedral complexes such as [Co(NH₃)₄(H₂O)₂]²⁺ can also show cis/trans arrangements. State the geometry first, then locate identical ligands relative to the central metal.

Use vector cancellation to infer polarity

Bond dipoles can cancel when identical ligands occupy opposite, symmetry-related positions. Thus the trans form of [Pt(NH₃)₂Cl₂] has cancelling bond-dipole contributions and is non-polar in the usual molecular model; its cis form has a net dipole and is polar.

Apply the same symmetry reasoning to named octahedral examples: opposite identical ligands can cancel, while adjacent unlike orientations generally leave a net dipole. Do not infer polarity from formula alone; draw or reason from the three-dimensional arrangement.

Pause and explain

Why is the cis-[Pt(NH₃)₂Cl₂] arrangement polar in the vector model?

Put the idea to work

Worked example

Compare cis- and trans-[Pt(NH₃)₂Cl₂] in polarity.

Show the worked solution
  1. In cis, the two Cl ligands are adjacent.
  2. In trans, they are opposite.
  3. Compare whether bond-dipole contributions cancel by symmetry.

Answer The cis isomer is polar because the ligand dipoles do not cancel. The trans isomer is non-polar in the usual vector model because opposite positions cancel the dipole contributions.

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From guided practice to a new situation

Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.

1 · Guided

Name the geometrical isomer when two identical ligands occupy adjacent positions in an octahedral complex.

Give me a hint

Use the relative positions of the identical ligands.

Compare my reasoning
  1. Adjacent positions correspond to cis.
  2. Opposite positions correspond to trans.
  3. Keep the stated octahedral geometry.

The arrangement is cis.

Look for these in your work

  • Relative positions identified
  • Term selected correctly
  • Geometry retained
2 · Independent

A trans square-planar complex has two pairs of identical ligands opposite each other. Predict polarity using the bond-dipole model.

Give me a hint

Compare opposite vector contributions pair by pair.

Compare my reasoning
  1. Identify each pair of opposite ligands.
  2. Opposing bond-dipole contributions cancel.
  3. Conclude no net dipole in this symmetric arrangement.

The trans arrangement is non-polar in the usual bond-dipole vector model because the opposite contributions cancel.

Look for these in your work

  • Symmetry identified
  • Cancellation reasoned
  • Conclusion stated
3 · Transfer

A learner says ‘trans always means non-polar.’ Give a safer method for a complex not yet classified.

Give me a hint

Polarity depends on the full arrangement and ligand dipoles, not the label alone.

Compare my reasoning
  1. Draw or identify the three-dimensional ligand positions.
  2. Consider the direction and relative size of bond dipoles.
  3. Check whether their vector sum is zero.

Determine the three-dimensional arrangement, consider the bond-dipole vectors, and test whether their sum cancels. The word trans alone is not enough for every possible ligand set.

Look for these in your work

  • Geometry used
  • Dipoles considered
  • Vector sum checked

Common mistakes

  • Calling cis ligands opposite
  • Using a flat drawing without identifying 3D positions
  • Inferring polarity only from total charge
  • Assuming every trans complex is non-polar regardless of ligand identity
Recall without your notes

What can you explain now?

In trans-[Pt(NH₃)₂Cl₂], where are the two Cl ligands?

Compare with the explanation

Opposite each other

Trans describes opposite positions in the square plane.

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Make it stick

How do cis/trans positions change geometry, and how can symmetry help decide whether a complex is polar?

If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.

Keep learning Chemistry

Course scope and assessment

Four hundred and forty-two original lessons map the published learning outcomes across AS Chemistry Topics 1–22 and A Level Topics 23–37, with guided, independent and transfer practice and explained answers. Every numbered subject-content topic and listed outcome tag now has a mapped lesson sequence. Topic 32 maps all 12 hydroxy-compound outcome tags in 11 lessons; Topic 33 maps all 15 carboxylic-acid and derivative outcome tags in 16 lessons; Topic 34 maps all 22 nitrogen-compound outcome tags in 20 lessons; Topic 35 maps all 12 polymerisation outcome tags in 16 lessons; Topic 36 maps all four organic-synthesis outcome tags in 14 lessons; Topic 37 maps all 22 chromatography and NMR outcome tags in 18 lessons. Cambridge's 2028–2030 Version 1 syllabus states that there are no significant changes which affect teaching from the preceding edition; the routes are aligned to that edition. This is 37 of the 37 numbered subject-content topics. Full cumulative exam assessment, Paper 3 laboratory competence, Paper 5 investigation practice and human subject-teacher review remain outstanding. Written practice is self-checked and does not certify exam readiness.

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