Question

The electronic configuration of Cu2+ (Z = 29) is:

• A. [Ar] \(3d^{8}\,4s^{1}\)
• B. [Ar] \(3d^{7}\,4s^{2}\)
• C. [Ar] \(3d^{9}\)
• D. [Ar] \(3d^{6}\,4s^{2}\,4p^{1}\)

Hint:

For transition metals, ionization removes \(ns\) electrons before \( (n-1)d \) electrons.

Answer 1

The Correct Option is C

Step 1: Start from neutral Cu.
Cu atom: [Ar] \(3d^{10}4s^{1}\).
Step 2: Remove electrons for Cu$^{2+$.}
Electrons are removed first from \(4s\) then \(3d\):
Cu$^{2+}$: \([Ar]\,3d^{9}\).
Step 3: Conclusion.
Therefore, the configuration is [Ar] \(3d^{9}\).