Which of the following ions forms coloured solution?

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: Sort the ions by their d-electron count, because a coloured aqueous ion must permit a d-d transition. Colourless ions are those that are either d⁰ or d¹⁰.
Step 2: Ti⁴⁺ has lost all its 3d and 4s electrons giving d⁰; Zn²⁺ and Cu⁺ both reach the stable d¹⁰ shell. All three cannot absorb visible light through d-d jumps, so their solutions are colourless.
Step 3: Fe³⁺ keeps five 3d electrons (d⁵), an intermediate, partly filled configuration. Absorption of visible light promotes a d electron to a higher-energy d level, and the transmitted light gives the solution its yellow-brown colour.
Result: The coloured ion is Fe³⁺, alternative (ii).

Which of the following ions forms coloured solution?

Show Hint

The Correct Option is B

Solution and Explanation

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