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Valence Bond Theory, Hybridisation and Geometry

Valence Bond Theory explains bonding in coordination compounds by assuming that the central metal ion uses suitable hybrid orbitals to accept electron pairs donated by ligands, producing definite geometries such as octahedral, square planar, or tetrahedral.

In one line: VBT geometry follows the metal's prepared orbitals: sp3 makes tetrahedral, dsp2 makes square planar, six hybrids make octahedral.

Practice This Concept

Understand it

  • In VBT, the metal ion provides vacant hybrid orbitals and ligands donate lone pairs.
  • The type of hybridisation is linked to coordination number and geometry.
  • Coordination number 6 commonly gives octahedral geometry with d2sp3 or sp3d2 hybridisation.
  • Coordination number 4 can give tetrahedral geometry with sp3 hybridisation or square planar geometry with dsp2 hybridisation.
  • Strong field ligands may cause pairing of d electrons and form inner orbital complexes, while weak field ligands often do not cause pairing and form outer orbital complexes.

Example

[Ni(CN)4]2- is square planar and diamagnetic because CN- is a strong field ligand and causes pairing, allowing dsp2 hybridisation. [NiCl4]2- is tetrahedral and paramagnetic because Cl- is a weak field ligand and generally does not cause pairing, leading to sp3 hybridisation.

Rule to remember

Common VBT links: sp3 = tetrahedral, dsp2 = square planar, d2sp3 or sp3d2 = octahedral.

Magnetic moment is linked to unpaired electrons; more unpaired electrons means paramagnetic behaviour.

See it worked out

Follow the example line by line, then use the same method on new questions.

Worked example

Predict geometry and magnetic nature of [Ni(CN)4]2-.

  1. 1Let oxidation state of Ni be x. x + 4(-1) = -2, so x = +2.
  2. 2Ni2+ is d8.
  3. 3CN- is a strong field ligand, so electrons pair up and one 3d orbital becomes available for dsp2 hybridisation.
  4. 4The complex is square planar and diamagnetic because no unpaired electrons remain.

Method you can reuse

  1. 1Find metal oxidation state.
  2. 2Write d-electron configuration of the metal ion.
  3. 3Identify ligand strength as strong or weak field.
  4. 4Decide whether electron pairing occurs.
  5. 5Select hybridisation from coordination number and available orbitals.
  6. 6State geometry and magnetic character.

Visual tip

Orbital box diagrams can help show pairing and hybrid orbital formation, but a full orbital diagram is not always required for naming or geometry questions.

Check yourself

Answer the question in your head before opening the answer.

Try it yourself

Which geometry is commonly associated with dsp2 hybridisation in a coordination compound?

Think first, then show answer
dsp2 hybridisation is commonly associated with square planar geometry.

Common wrong answer

A realistic wrong answer is calling [Ni(CN)4]2- tetrahedral because its coordination number is 4, while ignoring strong-field pairing and dsp2 hybridisation.

Don't mix this up

Students often assign square planar geometry to every coordination number 4 complex. Coordination number 4 may be tetrahedral or square planar depending on metal ion, oxidation state, and ligand field strength.

Exam tip

For VBT questions, first find oxidation state and d-electron count, then judge ligand strength and possible pairing before assigning hybridisation and magnetic behaviour.

Write it in the exam

Each point is worth about one mark. Write point 1 for a 1-mark question and keep adding points as the marks go up.

How CBSE asks it

Questions commonly ask for hybridisation, geometry, inner or outer orbital nature, number of unpaired electrons, and magnetic behaviour for complexes such as [Ni(CN)4]2-, [NiCl4]2-, [Fe(CN)6]4-, and [CoF6]3-.
  1. 1

    Enough for 1 mark

    Valence Bond Theory explains bonding in coordination compounds by assuming that the central metal ion uses suitable hybrid orbitals to accept electron pairs donated by ligands, producing definite geometries such as octahedral, square planar, or tetrahedral.

  2. 2

    Add for 2 marks

    Example: [Ni(CN)4]2- is square planar and diamagnetic because CN- is a strong field ligand and causes pairing, allowing dsp2 hybridisation. [NiCl4]2- is tetrahedral and paramagnetic because Cl- is a weak field ligand and generally does not cause pairing, leading to sp3 hybridisation.

  3. 3

    Add for 3 marks

    In VBT, the metal ion provides vacant hybrid orbitals and ligands donate lone pairs.

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