Question

The number of unpaired d-electrons in \([ \text{Co}(\text{H}_2\text{O})_6 ]^{3+}\) is ______

• A. 4
• B. 2
• C. 0
• D. 1

Answer 1

The Correct Option is C

The complex \([ \text{Co(H}_2\text{O)}_6 ]^{3+}\) consists of a \(\text{Co}^{3+}\) ion, which has a \(d^6\) electronic configuration. In an octahedral field created by the water ligands, the splitting of the \(d\)-orbitals leads to the \(t_{2g}\) and \(e_g\) orbitals. As cobalt in this state is in a low-spin configuration (due to the relatively strong ligand field of water), all six electrons pair up in the \(t_{2g}\) set, resulting in no unpaired electrons.

\[ \text{at } (0,0) \, t_{2g}: \quad (0,-0.3); \, (1.2,-0.3); \, (0.2,-0.3) \text{ circle (0.05); } \, (0.6,-0.3) \text{ circle (0.05); } \, (1.0,-0.3) \text{ circle (0.05);} \] \[ \text{at } (2,0) \, e_g: \quad (2.2,-0.3); \, (3.2,-0.3); \, (2.2,-0.3) \text{ circle (0.05); } \, (2.6,-0.3) \text{ circle (0.05);} \]
Thus, there are no unpaired electrons in \([ \text{Co(H}_2\text{O)}_6 ]^{3+}\).

Answer 2

Step 1: Identify the oxidation state and electronic configuration of cobalt
In the complex \([ \text{Co}(\text{H}_2\text{O})_6 ]^{3+}\), cobalt exists in the +3 oxidation state.
Atomic number of cobalt (Co) = 27.
Electronic configuration of neutral Co atom = [Ar] 3d⁷ 4s².

For Co³⁺, three electrons are removed (two from 4s and one from 3d):
\[ \text{Co}^{3+} = [\text{Ar}]\,3d^6. \]

Step 2: Nature of the ligand
In this complex, the ligands are six water molecules (\(\text{H}_2\text{O}\)).
Although water is generally a weak field ligand, in the case of Co³⁺ (a small, highly charged ion), the ligand field strength increases significantly due to the high effective nuclear charge. This can lead to pairing of electrons within the 3d orbitals, resulting in a low-spin (inner orbital) complex.

Step 3: Crystal field splitting and electron arrangement
For an octahedral complex, the five d orbitals split into two sets:
- \(t_{2g}\) (lower energy) → three orbitals
- \(e_g\) (higher energy) → two orbitals

For a low-spin \(d^6\) configuration, all six electrons occupy the lower-energy \(t_{2g}\) orbitals, with complete pairing:
\[ t_{2g}^6\,e_g^0. \]

Step 4: Determine number of unpaired electrons
Since all six d-electrons are paired in the \(t_{2g}\) orbitals, there are:
\[ \text{Number of unpaired electrons} = 0. \]

Step 5: Magnetic behavior
As all electrons are paired, \([ \text{Co}(\text{H}_2\text{O})_6 ]^{3+}\) is a diamagnetic complex.

Final answer

0