A. Account for the following:

(i) Mn shows the highest oxidation state of +7 with oxygen but with fluorine it shows the highest oxidation state of +4.

(ii) Zirconium and Hafnium exhibit similar properties.

(iii) Transition metals act as catalysts.

B. Complete the following equations:

(i)

(ii)

a) (i) Mn shows the highest oxidation state of + 7 with oxygen but with fluorine it shows the highest oxidation state of + 4 due to the following reasons:

⇒ In covalent compounds, fluorine can form only single bond with metal. Hence with fluorine, Mn shows +4 oxidation state.

⇒ On the other hand, oxygen forms multiple bonds with metal. Hence with oxygen, Mn shows +7 oxidation state.

(ii) Zirconium and Hafnium exhibit similar properties due to lanthanoid contraction.

Note: The regular decrease (contraction) in the atomic and ionic radii of lanthanoids with increasing atomic number is called lanthanoid contraction.

(iii) Transition metals act as catalysts due to the following reasons:

⇒ The catalytic properties of the transition elements are probably due to the presence pf unpaired electrons in their incomplete d-orbitals and hence possess the capacity o absorb and re-emit wide range of energies.

⇒ This makes the required energy of activation available.

⇒ According to activated complex theory (intermediate compound formation), due to variable oxidation states, transition metals combine with the reactants to form an intermediate called activated complex.

⇒ This activated complex finally decomposes to form the products regenerating the catalyst.

⇒ Transition metals provides a large surface area with free valencies on which the reactants are adsorbed. As a result, the rate of reaction increases.

b) (i) 2MnO₂ + + 4KOH + O₂→ 2K₂MnO₄ + 2H₂O

Potassium Manganate

(ii) Cr₂O₇^2- + 14H⁺ + 6I^-→ 2Cr³⁺ + 3I₂ + 7H₂O

Explanation:

Cr₂O₇^2- + 14H⁺ + 6e^-→ 2Cr³⁺ + 7H₂O

2I^-→ I₂ + 2e^-

By multiplying the second reaction by 3, we get

Cr₂O₇^2- + 14H⁺ + 6I^-→ 2Cr³⁺ + 7H₂O

6I^-→ 3I₂ + 6e^-

(by adding both the reactions)

Cr₂O₇^2- + 14H⁺ + 6I^-→ 2Cr³⁺ + 3I₂ + 7H₂O