Question:hard

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

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Colour in transition metal ions is generally due to partially filled d-orbitals. Completely filled or completely empty d-orbitals usually produce colourless ions.
Updated On: Jun 29, 2026
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Solution and Explanation

Step 1: Melting point comparison (Fe vs Cu).
Fe ($[Ar]3d^6 4s^2$) has more unpaired $d$-electrons than Cu ($[Ar]3d^{10}4s^1$). More unpaired electrons contribute to stronger metallic bonds, so more energy is needed to melt Fe. Hence Fe has the higher melting point.
Step 2: Colour in aqueous solution ($Ti^{3+}$ vs $Sc^{3+}$).
$Ti^{3+}$ has configuration $[Ar]3d^1$: one $d$-electron is present, enabling $d$-$d$ electronic transitions that absorb visible light and produce colour. $Sc^{3+}$ has an empty $d$-subshell with no $d$-electrons, so no $d$-$d$ transitions occur and it is colourless. Hence $Ti^{3+}$ is coloured in aqueous solution.
Step 3: Third ionisation enthalpy (Cr vs Zn).
$Zn^{2+}$ has the completely filled $3d^{10}$ configuration, which is exceptionally stable. Removing the third electron disrupts this stability, requiring very high energy. $Cr^{2+}$ ($3d^4$) has no such special stability, so its third ionisation enthalpy is lower. Hence Zn has the higher third ionisation enthalpy. \[ \boxed{(I)\ Fe \quad (II)\ Ti^{3+} \quad (III)\ Zn} \]
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