Question:medium

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

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Transition metals lose $4s$ electrons before $3d$ electrons during ionization.
Updated On: Apr 8, 2026
  • $Cu^{+}$
  • $Cu^{2+}$
  • $Ni^{2+}$
  • $Ni$
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The Correct Option is A

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