Find oxidation state, coordination number and geometry of [Co(NH₃)₆]³⁺

Question

For the complex ion $[\text{Co}(\text{NH}_3)_6]^{3+}$ , find:

The oxidation state of cobalt
The coordination number of cobalt
The geometry of the complex

Solution — Step by Step

Let the oxidation state of Co be $x$ .

$\text{NH}_3$ is a neutral ligand (oxidation state = 0). The overall charge of the complex ion is $+3$ .

Setting up the equation:

x + 6(0) = +3

x = +3

The oxidation state of cobalt is +3, written as Co(III).

The coordination number is the number of ligand atoms directly bonded to the central metal atom.

$\text{NH}_3$ is a monodentate ligand — it binds through its nitrogen atom (one donor atom per ligand). There are 6 NH₃ ligands, each contributing 1 bond.

\text{Coordination number} = 6 \times 1 = \mathbf{6}

For a coordination number of 6, the geometry is almost always octahedral — the 6 ligands arrange themselves at the corners of a regular octahedron around the central metal atom.

Co(III) with 6 NH₃ ligands forms a regular octahedral complex. The bond angles are all 90° (adjacent ligands) and 180° (opposite ligands).

Co has atomic number 27. Co³⁺ configuration: [Ar] $3d^6$ (loses 3 electrons from 4s and 3d).

NH₃ is a strong field ligand (appears high in the spectrochemical series). In a strong field, the 3d electrons pair up:

3d^6 \text{ with strong field: } t_{2g}^6 e_g^0

All 6 electrons are paired → diamagnetic complex.

Hybridisation: $d^2sp^3$ (inner orbital complex — uses $3d$ , $4s$ , and $4p$ orbitals).

Why This Works

The IUPAC rules for finding oxidation state follow simple algebra: sum of oxidation states = overall charge. Since NH₃ is neutral and doesn’t carry a charge, all the positive charge comes from cobalt.

The octahedral geometry for CN = 6 follows from VSEPR theory and the geometrical arrangement that minimises repulsion among 6 identical ligands — an octahedron achieves equal 90° angles between all adjacent ligands.

Common Mistake

Students often confuse oxidation state with coordination number. Oxidation state is determined by charge; coordination number is determined by number of bonds from ligands. In $[\text{Co}(\text{NH}_3)_6]^{3+}$ , the oxidation state is +3 (from charge calculation) and the coordination number is 6 (from counting ligand bonds). These are both 6 and 3 respectively — not the other way around.

For NEET and JEE: $[\text{Co}(\text{NH}_3)_6]^{3+}$ is a classic example of an inner orbital ( $d^2sp^3$ ) octahedral complex. Contrast it with $[\text{CoF}_6]^{3-}$ — F⁻ is a weak field ligand, giving outer orbital ( $sp^3d^2$ ) hybridisation and a paramagnetic complex with unpaired electrons.

Question

Solution — Step by Step

Why This Works

Common Mistake

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