Question

Which is the correct order of increasing number of unpaired electrons in the following ions?
\( A = \text{Cr}^{2+} \, (Z = 24), \quad B = \text{Cu}^{2+} \, (Z = 29), \quad C = \text{Ni}^{2+} \, (Z = 28), \quad D = \text{Fe}^{3+} \, (Z = 26) \)

• A. B < C < A < D
• B. D < C < A < B
• C. B < C < D < A
• D. C < B < D < A

Hint:

To determine the number of unpaired electrons, write the electron configuration and count the unpaired electrons in the highest energy orbitals.

Answer 1

The Correct Option is A

Step 1: Write the electronic configuration for each ion.
We need to write the electron configurations for each of the ions.
For Cr\(^{2+}\) (\(Z = 24\)): \[ \text{Cr} = [Ar] 3d^5 4s^1, \quad \text{Cr}^{2+} = [Ar] 3d^4 \]
In Cr\(^{2+}\), there are 4 electrons in the 3d orbitals. Hence, 4 unpaired electrons.
For Cu\(^{2+}\) (\(Z = 29\)):
\[ \text{Cu} = [Ar] 3d^{10} 4s^1, \quad \text{Cu}^{2+} = [Ar] 3d^9 \]
In Cu\(^{2+}\), there are 1 unpaired electron in the 3d orbitals.
For Ni\(^{2+}\) (\(Z = 28\)):
\[ \text{Ni} = [Ar] 3d^8 4s^2, \quad \text{Ni}^{2+} = [Ar] 3d^8 \]
In Ni\(^{2+}\), there are 2 unpaired electrons in the 3d orbitals.
For Fe\(^{3+}\) (\(Z = 26\)):
\[ \text{Fe} = [Ar] 3d^6 4s^2, \quad \text{Fe}^{3+} = [Ar] 3d^5 \]
In Fe\(^{3+}\), there are 5 unpaired electrons in the 3d orbitals.

Step 2: Order the ions based on the number of unpaired electrons.
- Cr\(^{2+}\) has 4 unpaired electrons.
- Cu\(^{2+}\) has 1 unpaired electron.
- Ni\(^{2+}\) has 2 unpaired electrons.
- Fe\(^{3+}\) has 5 unpaired electrons.
The correct order of increasing unpaired electrons is:
\[ \text{Cu}^{2+} < \text{Ni}^{2+} < \text{Cr}^{2+} < \text{Fe}^{3+} \]
which corresponds to option (A).