Question

(i) Account for the following :

(I) \(\mathrm{MnO}\) is basic while \(\mathrm{Mn_2O_7}\) is acidic.

(II) Iron has higher enthalpy of atomization than that of copper.

(III) \(\mathrm{Mn^{3+}}\) is a stronger oxidising agent than \(\mathrm{Cr^{3+}}\).

(ii) How do you prepare potassium dichromate from sodium chromate ? Write balanced chemical equation for each step.

Hint:

Link oxide nature to oxidation state, count unpaired d electrons, compare d4 to d5 and d3 to d4, then acidify sodium chromate and add KCl.

Answer 1

Step 1: (i)(I) Acidic and basic character of the oxides:
In MnO, manganese is in the +2 state. The metal ion is large and the bond has an ionic nature, so the oxide is basic.
In \(\mathrm{Mn_2O_7}\), manganese is in the +7 state. The small, highly charged metal makes the bonds covalent, so the oxide is acidic and gives permanganic acid with water.
In general, the acidic character of a transition metal oxide rises as the oxidation state rises.

Step 2: (i)(II) Enthalpy of atomization:
The strength of metallic bonding depends on the number of unpaired d electrons that take part in bonding.
Iron (\(3d^6 4s^2\)) has four unpaired d electrons plus the 4s electrons. Copper (\(3d^{10} 4s^1\)) has no unpaired d electron.
So iron has stronger metallic bonds and needs more energy to atomize.

Step 3: (i)(III) Oxidising power of \(\mathrm{Mn^{3+}}\):
\(\mathrm{Mn^{3+}}\) (\(3d^4\)) gains one electron and becomes \(\mathrm{Mn^{2+}}\) (\(3d^5\)), a stable half-filled configuration. The value of \(E^\circ\) is about +1.57 V.
\(\mathrm{Cr^{3+}}\) (\(3d^3\)) already has the stable \(t_{2g}^3\) arrangement. Gaining an electron to make \(\mathrm{Cr^{2+}}\) (\(3d^4\)) is not favourable, and \(E^\circ\) is about -0.41 V.
So \(\mathrm{Mn^{3+}}\) is the stronger oxidising agent.

Step 4: (ii) Sodium chromate to potassium dichromate:
Sodium chromate is obtained from chromite ore by fusion with sodium carbonate in air:
\[ \mathrm{4FeCr_2O_4 + 8Na_2CO_3 + 7O_2 \rightarrow 8Na_2CrO_4 + 2Fe_2O_3 + 8CO_2} \]
Step A: Acidify the sodium chromate solution with sulphuric acid. It turns orange as sodium dichromate forms.
\[ \mathrm{2Na_2CrO_4 + H_2SO_4 \rightarrow Na_2Cr_2O_7 + Na_2SO_4 + H_2O} \]
Step B: Treat the sodium dichromate solution with potassium chloride. Potassium dichromate is less soluble and crystallises out.
\[ \mathrm{Na_2Cr_2O_7 + 2KCl \rightarrow K_2Cr_2O_7 + 2NaCl} \]

Final Answer:
(I) Low oxidation state gives ionic, basic MnO; +7 gives covalent, acidic \(\mathrm{Mn_2O_7}\). (II) Iron has more unpaired d electrons. (III) \(\mathrm{Mn^{3+}}\) changes to the stable \(d^5\) \(\mathrm{Mn^{2+}}\). (ii) Acidify \(\mathrm{Na_2CrO_4}\), then add KCl.