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Question:

(a) Given reasons for the following: (i) Mn³⁺ is a good oxidizing agent. (ii) E°M²⁺/M values are not regular for first row transition metals (3d series). (iii) Although F is more electronegative than O, the highest Mn fluoride is MnF₄, whereas the highest oxide is Mn₂O₇. (b) Complete the following equations: (i) 2CrO₄²⁻ + 2H⁺ → (ii) KMnO₄ Δ →

Solution:

a) i. Mn³⁺ is a good oxidizing agent because the +2 oxidation state of Mn is more stable than +3. Mn³⁺ is reduced to Mn²⁺.

ii. Because values of ionization energy first and second are not regular for first row transition metals of 3d series.

iii. Oxygen can form multiple bonds while fluorine can form single bonds.

b) i) 2CrO₄²⁻ + 2H⁺ → Cr₂O₇²⁻ + H₂O

ii) 2KMnO₄ → K₂MnO₄ + MnO₂ + O₂