Question:hard

Describe, giving the reason, which one of the following pairs has the property indicated :
(I) Fe or Cu – higher melting point
(II) $\mathrm{Ti^{3+}}$ or $\mathrm{Sc^{3+}}$ – coloured in aqueous solution
(III) Cr or Zn – higher third ionisation enthalpy

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More unpaired d-electrons → higher mp; d¹ → coloured; breaking d¹⁰ → high IE.
Updated On: Jun 16, 2026
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Solution and Explanation

(I) Fe or Cu, higher melting point.
Melting point of a metal depends on how strongly its atoms grip each other, and that grip is stronger when more unpaired d-electrons take part in the metallic bonding. Iron has several unpaired d-electrons, so its bonding is strong, while copper has a filled $\mathrm{3d^{10}}$ set with no unpaired d-electrons helping to bond, so its grip is weaker. Therefore iron melts at a higher temperature than copper.

(II) Ti³⁺ or Sc³⁺, coloured in water.
A transition metal ion shows colour when it has at least one electron in the d-orbitals that can jump from a lower to a higher d-level (a d to d transition) by absorbing visible light. $\mathrm{Ti^{3+}}$ has a $\mathrm{3d^1}$ arrangement, so it has that one electron available and appears coloured (purple). $\mathrm{Sc^{3+}}$ has an empty $\mathrm{3d^0}$ set, no d-electron to jump, so it is colourless. So $\mathrm{Ti^{3+}}$ is the coloured one.

(III) Cr or Zn, higher third ionisation enthalpy.
The third ionisation enthalpy is the energy to pull a third electron away. For chromium, removing the third electron gives $\mathrm{Cr^{3+}}$ with a stable half-spread $\mathrm{3d^3}$, which is reached fairly easily. For zinc, the third electron has to be torn out of the very stable, completely filled $\mathrm{3d^{10}}$ shell, which resists strongly. So zinc has the higher third ionisation enthalpy.

Final answer: Fe (higher melting point), $\mathrm{Ti^{3+}}$ (coloured), Zn (higher third ionisation enthalpy).
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