Question

The electronic configuration \( 1s^{2}\, 2s^{2}\, 2p^{6}\, 3s^{2}\, 3p^{6}\, 3d^{10} \) represents:

• A. $Cu^{+}$
• B. $Cu^{2+}$
• C. $Ni^{2+}$
• D. $Ni$

Hint:

Transition metals lose $4s$ electrons before $3d$ electrons during ionization.

Answer 1

The Correct Option is A

Step 1: Concept
Determine the total number of electrons in the given configuration.
Step 2: Analysis
Total electrons $= 2+2+6+2+6+10 = 28$. Neutral $Cu$ ($Z=29$) has 29 electrons ($3d^{10} 4s^{1}$). $Cu^{+}$ loses one electron from $4s$.
Step 3: Conclusion
$Cu^{+}$ has 28 electrons with the configuration $[Ar] 3d^{10}$.
Final Answer: (A)