Question

Answer the following: \[ (a)(i)\;\text{Describe giving reason which one of the following pairs has the property indicated:} \] \[ (I)\;Fe\text{ or }Cu-\text{higher melting point} \] \[ (II)\;Ti^{3+}\text{ or }Sc^{3+}-\text{coloured in aqueous solution} \] \[ (III)\;Cr\text{ or }Zn-\text{higher third ionisation enthalpy} \] \[ (a)(ii)\;\text{Write the ionic equations for the oxidizing action of }MnO_4^-\text{ in acidic medium with:} \] \[ (I)\;Fe^{2+}\text{ ion} \] \[ (II)\;I^-\text{ ion} \]

Hint:

Colour in transition metal ions usually requires partially filled \(d\)-orbitals. In acidic medium, \(MnO_4^-\) is reduced to \(Mn^{2+}\).

Answer 1

Concept:
Transition metals show characteristic properties due to the presence of partially filled \(d\)-orbitals. Important properties such as melting point, colour, ionisation enthalpy and oxidising behaviour depend on: number of unpaired electrons metallic bonding electronic configuration stability of oxidation states

Step 1: Compare Fe and Cu for higher melting point.
Iron has more unpaired \(d\)-electrons than copper. Due to more unpaired electrons, metallic bonding in iron is stronger. Stronger metallic bonding requires more heat energy to break. Therefore, iron has a higher melting point than copper. Hence: \[ Fe\text{ has higher melting point than }Cu. \]

Step 2: Compare Ti^3+

and Sc^3+

for colour.
Titanium has atomic number \(22\). \[ Ti:[Ar]3d^24s^2 \] \[ Ti^{3+}:[Ar]3d^1 \] So \(Ti^{3+}\) has one unpaired \(d\)-electron. Therefore, it can show \(d-d\) transition and is coloured. Scandium has atomic number \(21\). \[ Sc:[Ar]3d^14s^2 \] \[ Sc^{3+}:[Ar]3d^0 \] Since \(Sc^{3+}\) has no \(d\)-electron, \(d-d\) transition is not possible. Therefore, \(Sc^{3+}\) is colourless. Hence: \[ Ti^{3+}\text{ is coloured in aqueous solution.} \]

Step 3: Compare Cr and Zn for higher third ionisation enthalpy.
Zinc has electronic configuration: \[ Zn:[Ar]3d^{10}4s^2 \] After losing two electrons: \[ Zn^{2+}:[Ar]3d^{10} \] This is a completely filled and highly stable configuration. The third ionisation would require removal of an electron from stable \(3d^{10}\) configuration. Therefore, the third ionisation enthalpy of zinc is very high. Hence: \[ Zn\text{ has higher third ionisation enthalpy than }Cr. \]

Step 4: Oxidising action of MnO_4^-

with Fe^2+

in acidic medium.
In acidic medium, permanganate ion is reduced to \(Mn^{2+}\). \[ MnO_4^-\rightarrow Mn^{2+} \] Ferrous ion is oxidised to ferric ion. \[ Fe^{2+}\rightarrow Fe^{3+} \] Balanced ionic equation: \[ MnO_4^-+5Fe^{2+}+8H^+ \rightarrow Mn^{2+}+5Fe^{3+}+4H_2O \]

Step 5: Oxidising action of MnO_4^-

with I^-

in acidic medium.
Iodide ion is oxidised to iodine. \[ I^- \rightarrow I_2 \] Permanganate is reduced to \(Mn^{2+}\). Balanced ionic equation: \[ 2MnO_4^-+10I^-+16H^+ \rightarrow 2Mn^{2+}+5I_2+8H_2O \] Hence: \[ Fe,\quad Ti^{3+},\quad Zn \] \[ MnO_4^-+5Fe^{2+}+8H^+ \rightarrow Mn^{2+}+5Fe^{3+}+4H_2O \] \[ 2MnO_4^-+10I^-+16H^+ \rightarrow 2Mn^{2+}+5I_2+8H_2O \]