Question

The complex \([\mathrm{Co(NH_3)_6]^{3+}\) is:}

• A. \(sp^3d^2\) hybridised and paramagnetic
• B. \(sp^3d^2\) hybridised and diamagnetic
• C. \(d^2sp^3\) hybridised and paramagnetic
• D. \(d^2sp^3\) hybridised and diamagnetic

Hint:

\([\mathrm{Co(NH_3)_6}]^{3+}\) is a classic low-spin \(d^6\) octahedral complex and therefore diamagnetic.

Answer 1

The Correct Option is D

Concept: The hybridisation and magnetic behaviour of coordination compounds can be determined from the electronic configuration of the central metal ion and the strength of the ligand.

Step 1: Determine oxidation state of cobalt.
Since ammonia is neutral, \[ x+6(0)=+3 \] \[ x=+3 \] Therefore metal ion is \[ \mathrm{Co^{3+}} \]

Step 2: Write electronic configuration.
Atomic number of cobalt: \[ Z=27 \] \[ \mathrm{Co} = [Ar]3d^74s^2 \] \[ \mathrm{Co^{3+}} = [Ar]3d^6 \]

Step 3: Pair the electrons.
For \(\mathrm{Co^{3+}}\), ligand field splitting is sufficiently large. Electrons pair up in lower \(d\)-orbitals. Thus configuration becomes low-spin.

Step 4: Determine hybridisation and magnetic nature.
Two vacant \(3d\) orbitals become available. Hence hybridisation is \[ d^2sp^3 \] All electrons become paired. Therefore the complex is diamagnetic. \[ \boxed{d^2sp^3 \text{ hybridised and diamagnetic}} \]