Question

Using valence bond theory, predict: (a) Hybridisation of central metal atom of the complex \([Co(NH_3)_6]^{3+}\)
(b) Its shape and magnetic behaviour
(c) Whether it is a high spin or a low spin complex
[Atomic number: $CO$ = 27]

Hint:

Use the valence bond theory to determine geometry from hybridisation: - \(sp^3\): tetrahedral, - \(dsp^2\): square planar, - \(d^2sp^3\)/\(sp^3d^2\): octahedral. For low/high spin, check whether the ligand is strong or weak field.

Answer 1

Step 1: Determine the oxidation state of cobalt.
In \([Co(NH_3)_6]^{3+}\), NH₃ is a neutral ligand. Therefore, the oxidation state of Co is +3.
Step 2: Find the electronic configuration.
Atomic number of Co = 27
Electronic configuration of Co: \([Ar]\,3d^7\,4s^2\)
Co³⁺: \([Ar]\,3d^6\)
Step 3: Pairing of electrons and hybridisation
NH₃ is a strong field ligand (though borderline), so it causes pairing of electrons in the 3d orbitals: \[ \text{3d orbital: } \uparrow\downarrow\,\uparrow\downarrow\,\uparrow\downarrow\,\_\,\_ \quad (\text{all 6 electrons paired}) \] The hybridisation involves: - Two 3d orbitals - One 4s orbital - Three 4p orbitals \[ \text{Hybridisation: } d^2sp^3 \]
Step 4: Shape and magnetic behaviour
- \(d^2sp^3\) hybridisation gives an octahedral geometry.
- All electrons are paired, so the complex is diamagnetic. Step 5: High spin or low spin
- As pairing occurs due to the strong ligand field of NH₃, this is a low spin complex.