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d-and f-Block Elements - Oxides and Oxoanions of Metals

Grade 12CBSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Transition metals form oxides of the general composition MOMO, M2O3M_2O_3, M3O4M_3O_4, MO2MO_2, M2O5M_2O_5, and MO3MO_3.

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As the oxidation state of the metal increases, the ionic character of the oxide decreases and the acidic character increases. For example, MnOMnO is basic, Mn2O3Mn_2O_3 is amphoteric, and Mn2O7Mn_2O_7 is acidic.

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Potassium dichromate (K2Cr2O7K_2Cr_2O_7) is prepared from chromite ore (FeCr2O4FeCr_2O_4). The process involves fusion with sodium carbonate followed by acidification.

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The chromate (CrO42−CrO_4^{2-}) and dichromate (Cr2O72−Cr_2O_7^{2-}) ions are interconvertible in aqueous solution depending upon the pH: CrO42−CrO_4^{2-} is stable in alkaline medium, while Cr2O72−Cr_2O_7^{2-} is stable in acidic medium.

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Potassium permanganate (KMnO4KMnO_4) is prepared by the alkaline oxidative fusion of pyrolusite ore (MnO2MnO_2) to form K2MnO4K_2MnO_4 (green), which is then oxidized electrolytically or by chlorine to KMnO4KMnO_4 (purple).

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In acidic medium, KMnO4KMnO_4 acts as a powerful oxidizing agent where MnMn is reduced from +7+7 to +2+2 state, accepting 55 electrons.

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In neutral or weakly alkaline medium, KMnO4KMnO_4 is reduced to MnO2MnO_2, with the oxidation state changing from +7+7 to +4+4.

📐Formulae

2CrO42−+2H+→Cr2O72−+H2O2CrO_4^{2-} + 2H^+ \rightarrow Cr_2O_7^{2-} + H_2O

Cr2O72−+2OH−→2CrO42−+H2OCr_2O_7^{2-} + 2OH^- \rightarrow 2CrO_4^{2-} + H_2O

Cr2O72−+14H++6e−→2Cr3++7H2OCr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O

MnO4−+8H++5e−→Mn2++4H2OMnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O

2MnO2+4KOH+O2→2K2MnO4+2H2O2MnO_2 + 4KOH + O_2 \rightarrow 2K_2MnO_4 + 2H_2O

MnO4−+2H2O+3e−→MnO2+4OH−MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-

💡Examples

Problem 1:

Calculate the molecular mass of Potassium Permanganate (KMnO4KMnO_4) given Atomic Masses: K=39K = 39, Mn=55Mn = 55, O=16O = 16.

Solution:

The molecular mass is the sum of the atomic masses of all atoms in the formula: K=1×39=39K = 1 \times 39 = 39 Mn=1×55=55Mn = 1 \times 55 = 55 O4=4×16=64O_4 = 4 \times 16 = 64 Calculation: 3955+64158\begin{array}{r} 39 \\ 55 \\ + 64 \\ \hline 158 \end{array}

Explanation:

Summing the standard atomic weights of Potassium, Manganese, and four Oxygen atoms gives the molar mass of 158 g/mol158\text{ g/mol}.

Problem 2:

What is the effect of increasing pH on a solution of Potassium dichromate?

Solution:

When pH is increased (adding OH−OH^-), the orange dichromate ion converts to the yellow chromate ion: Cr2O72−+2OH−→2CrO42−+H2OCr_2O_7^{2-} + 2OH^- \rightarrow 2CrO_4^{2-} + H_2O

Explanation:

The equilibrium shifts towards the formation of the chromate ion (CrO42−CrO_4^{2-}) in basic conditions, causing a color change from orange to yellow.

Problem 3:

Write the balanced ionic equation for the reaction between acidified MnO4−MnO_4^- and Fe2+Fe^{2+} ions.

Solution:

The reduction half-reaction is: MnO4−+8H++5e−→Mn2++4H2OMnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O The oxidation half-reaction is: 5Fe2+→5Fe3++5e−5Fe^{2+} \rightarrow 5Fe^{3+} + 5e^- Overall reaction: MnO4−+8H++5Fe2+→Mn2++5Fe3++4H2OMnO_4^- + 8H^+ + 5Fe^{2+} \rightarrow Mn^{2+} + 5Fe^{3+} + 4H_2O

Explanation:

In acidic medium, Permanganate oxidizes Ferrous (Fe2+Fe^{2+}) ions to Ferric (Fe3+Fe^{3+}) ions while itself getting reduced to Mn2+Mn^{2+}.