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Coordination Compounds - Valence Bond Theory (VBT)

Grade 12ICSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The central metal atom or ion provides a number of empty orbitals for the formation of coordinate bonds with suitable ligand orbitals. This number is equal to the coordination number of the metal.

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The empty atomic orbitals (ss, pp, and dd) of the metal ion undergo hybridization to yield a set of equivalent hybrid orbitals of definite geometry such as octahedral, tetrahedral, or square planar.

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A coordinate bond is formed by the overlap of a vacant hybrid orbital of the metal ion with a filled orbital of the ligand containing a lone pair of electrons.

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Inner orbital complexes (low spin) involve the (n−1)d(n-1)d orbitals, usually occurring with strong field ligands like CN−CN^- or COCO. These result in d2sp3d^2sp^3 hybridization for coordination number 6.

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Outer orbital complexes (high spin) involve the ndnd orbitals, usually occurring with weak field ligands like F−F^- or Cl−Cl^-. These result in sp3d2sp^3d^2 hybridization for coordination number 6.

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Magnetic property: If the complex contains one or more unpaired electrons, it is paramagnetic. If all electrons are paired, it is diamagnetic.

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Coordination Number 4 can lead to sp3sp^3 (Tetrahedral) or dsp2dsp^2 (Square Planar) geometry.

📐Formulae

μ=n(n+2) BM\mu = \sqrt{n(n+2)} \text{ BM}

where n=number of unpaired electrons, BM = Bohr Magneton\text{where } n = \text{number of unpaired electrons, BM = Bohr Magneton}

CN=4  ⟹  sp3 (Tetrahedral) or dsp2 (Square Planar)CN = 4 \implies sp^3 \text{ (Tetrahedral) or } dsp^2 \text{ (Square Planar)}

CN=6  ⟹  d2sp3 (Inner Octahedral) or sp3d2 (Outer Octahedral)CN = 6 \implies d^2sp^3 \text{ (Inner Octahedral) or } sp^3d^2 \text{ (Outer Octahedral)}

💡Examples

Problem 1:

Discuss the hybridization, geometry, and magnetic property of the complex ion [Co(NH3)6]3+[Co(NH_3)_6]^{3+}. (Atomic number of Co=27Co = 27)

Solution:

  1. Oxidation state of CoCo is +3+3. Electronic configuration of Co3+Co^{3+} is [Ar]3d6[Ar] 3d^6.
  2. NH3NH_3 is a strong field ligand, causing the pairing of 3d3d electrons.
  3. The six electrons in 3d3d orbitals pair up, leaving two 3d3d, one 4s4s, and three 4p4p orbitals vacant.
  4. These 6 orbitals undergo d2sp3d^2sp^3 hybridization.
  5. Six pairs of electrons from NH3NH_3 ligands are donated into these hybrid orbitals.

Explanation:

Since (n−1)d(n-1)d orbitals are used, it is an inner orbital complex. The hybridization is d2sp3d^2sp^3, giving an octahedral geometry. All electrons are paired, so the complex is diamagnetic (μ=0\mu = 0).

Problem 2:

Predict the geometry and magnetic behavior of [NiCl4]2−[NiCl_4]^{2-}. (Atomic number of Ni=28Ni = 28)

Solution:

  1. Oxidation state of NiNi is +2+2. Configuration of Ni2+Ni^{2+} is [Ar]3d8[Ar] 3d^8.
  2. Cl−Cl^- is a weak field ligand and cannot cause pairing of electrons.
  3. To accommodate 4 ligands, the 4s4s and three 4p4p orbitals hybridize to form sp3sp^3 hybrid orbitals.
  4. Two unpaired electrons remain in the 3d3d subshell.

Explanation:

The hybridization is sp3sp^3, resulting in a tetrahedral geometry. Since there are n=2n=2 unpaired electrons, the complex is paramagnetic with a magnetic moment μ=2(2+2)=8≈2.83 BM\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83 \text{ BM}.

Valence Bond Theory (VBT) - Revision Notes & Key Formulas | ICSE Class 12 Chemistry