Which of the following pair of compounds exhibits same colour in aque…

The correct answer is Option (2) → VOCl₂, CuCl₂

Transition metal ions are coloured because a d electron absorbs visible light and jumps to a higher d level; the colour seen is the complementary colour. The colour of an aquated ion therefore depends mainly on its d electron configuration, so first find the oxidation state and the d count of the metal in each compound:

• \( \ce{VOCl2} \): V + (−2) + 2(−1) = 0, so V is +4. V (Z = 23) is [Ar]3d³4s²; losing 4 electrons gives \( \ce{V^4+} \): 3d¹, blue.

• \( \ce{CuCl2} \): Cu is +2. Cu (Z = 29) is [Ar]3d¹⁰4s¹; losing 2 electrons gives \( \ce{Cu^2+} \): 3d⁹, blue.

• \( \ce{FeCl2} \): Fe is +2. Fe (Z = 26) is [Ar]3d⁶4s²; \( \ce{Fe^2+} \): 3d⁶, green.

• \( \ce{MnCl2} \): Mn is +2. Mn (Z = 25) is [Ar]3d⁵4s²; \( \ce{Mn^2+} \): 3d⁵, pink.

So the pair with the same colour (blue) in aqueous solution is VOCl₂ and CuCl₂.

Why the other options are incorrect

• Option (1) FeCl₂, CuCl₂ — Incorrect. \( \ce{Fe^2+} \) solutions are green, while \( \ce{Cu^2+} \) solutions are blue.

• Option (3) VOCl₂, FeCl₂ — Incorrect. Blue \( \ce{V^4+} \) against green \( \ce{Fe^2+} \).

• Option (4) VOCl₂, MnCl₂ — Incorrect. Blue \( \ce{V^4+} \) against pale pink \( \ce{Mn^2+} \).

NCERT (2026-27 edition):
Formation of Coloured Ions – Table 4.8: Colours of Some of the First Row (aquated) Transition Metal Ions