Valence Bond Theory, Hybridisation and Geometry Class 12 Chemistry | Coordination Compounds

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Student-friendly explanation

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.

How to write this in exams

1
Start with the exact idea
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
Then show how to use it
Find metal oxidation state. Write d-electron configuration of the metal ion. Identify ligand strength as strong or weak field. Decide whether electron pairing occurs. Select hybridisation from coordination number and available orbitals. State geometry and magnetic character.
3
Add one concrete 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.
4
Avoid this incomplete 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.

Definition

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.

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.

Memory hook

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

Examples and method

Worked example

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

Method to apply

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

Diagram support

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.

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-.

Avoid common mistakes

Common confusion

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.

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.

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.

Quick check

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

dsp2 hybridisation is commonly associated with square planar geometry.

Answer writing and exam use

1-mark answer

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-mark answer

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. 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. [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-mark answer

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

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Valence Bond Theory, Hybridisation and Geometry

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