Question

The electronic configuration of \(\text{Co}^{3+}\) metal ion is........

• A. \(3d^5 4s^1\)
• B. \(3d^6 4s^0\)
• C. \(3d^4 4s^2\)
• D. \(3d^3 4s^2 4p^1\)

Hint:

For all transition metal ions (\(\text{M}^{2+}\), \(\text{M}^{3+}\)), always remove \(4s\) electrons before \(3d\) electrons.
Neutral \(\text{Co} = 3d^7 4s^2\).
Loss of 3 electrons: 2 from \(4s\) and 1 from \(3d \rightarrow 3d^6 4s^0\).

Answer 1

The Correct Option is B

Concept:
The electronic configuration of transition elements follows the Aufbau principle and Hund's rule.
During cation formation in transition metals, electrons are removed first from the outermost \(s\)-orbital (highest principal quantum number \(n\)) before removing electrons from the inner \((n-1)d\) subshell.

Step 1: Ground-State Configuration of Neutral Cobalt:
Cobalt (\(\text{Co}\)) has atomic number \(Z = 27\).
Using the noble gas core of argon (\(Z = 18\)), the ground-state electronic configuration of a neutral cobalt atom is:
\[ \text{Co}: 1s^2 2s^2 2p^6 3s^2 3p^6 3d^7 4s^2 = [\text{Ar}] \, 3d^7 4s^2 \]

Step 2: Ionization to the Trivalent State (\(\text{Co}^{3+}\)):
To form the \(\text{Co}^{3+}\) ion, three valence electrons must be removed from the neutral atom.
The \(4s\) electrons are at a higher principal quantum level (\(n = 4\)) than the \(3d\) electrons (\(n = 3\)).
Consequently, the two electrons occupying the \(4s\) orbital are removed first:
\[ \text{Co}^{2+}: [\text{Ar}] \, 3d^7 4s^0 \] The third electron is then removed from the \(3d\) subshell:
\[ \text{Co}^{3+}: [\text{Ar}] \, 3d^6 4s^0 \]

Step 3: Verification with Options:
The configuration obtained for \(\text{Co}^{3+}\) is \(3d^6 4s^0\), which matches option (B).
Final Answer:
The electronic configuration of \(\text{Co}^{3+}\) is \(3d^6 4s^0\), corresponding to option (B).