The number of unpaired electrons in \( Cu^{2+} \) is

Question

Match List-I with List-II
\[ \begin{array}{|c|c|} \hline \text{List-I (Species)} & \text{List-II (Electronic distribution)} \\ \hline \text{(A) Cr}^{+2} & \text{(I) } 3d^8 \\ \text{(B) Mn}^{+} & \text{(II) } 3d^3 4s^1 \\ \text{(C) Ni}^{+2} & \text{(III) } 3d^4 \\ \text{(D) V}^{+} & \text{(IV) } 3d^5 4s^1 \\ \hline \end{array} \]

Answer 1

Step 1: Number of unpaired electrons in an ion is found from its d-electron count. For \(Cu^{2+}\) the d-count is \(10-1=9\).
Step 2: Fill nine electrons into five d-orbitals following Hund's rule: \(\uparrow\downarrow\;\uparrow\downarrow\;\uparrow\downarrow\;\uparrow\downarrow\;\uparrow\).
Step 3: Only the last orbital carries a lone (unpaired) electron.
Step 4: Therefore \(Cu^{2+}\) (\(3d^{9}\)) has 1 unpaired electron, which also explains its paramagnetism and blue colour.

The number of unpaired electrons in \( Cu^{2+} \) is

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The Correct Option is C

Solution and Explanation

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