a) Account for the following:
i) Mn shows the highest oxidation state of +7 with oxygen but with fluorine it shows a highest oxidation state of +4.
ii) Zirconium and hafnium exhibit similar properties.
iii) Transition metals act as catalysts.
b) Complete the following equations:
i) 2MnO₂ + 4KOH + O₂ →(Δ)
ii) Cr₂O₇²⁻ + 14H⁺ + 6I⁻ →
i) Mn shows the highest oxidation state of +7 with oxygen but with fluorine it shows a highest oxidation state of +4.
ii) Zirconium and hafnium exhibit similar properties.
iii) Transition metals act as catalysts.
b) Complete the following equations:
i) 2MnO₂ + 4KOH + O₂ →(Δ)
ii) Cr₂O₇²⁻ + 14H⁺ + 6I⁻ →
Solution
(a)(i) Oxygen reaches Mn(+7) with only four O atoms because it forms multiple pπ–dπ bonds (MnO); fluorine bonds singly, and seven F atoms around Mn are sterically/energetically impossible — the highest fluoride is MnF.
(ii) The lanthanoid contraction leaves Zr and Hf with virtually identical radii; same group configuration + same size → nearly identical properties.
(iii) Variable oxidation states (reactive intermediates) and adsorbing surfaces with vacant d-orbitals lower activation energies — hence their wide catalytic use.
(b)(i)
(ii)
(ii) The lanthanoid contraction leaves Zr and Hf with virtually identical radii; same group configuration + same size → nearly identical properties.
(iii) Variable oxidation states (reactive intermediates) and adsorbing surfaces with vacant d-orbitals lower activation energies — hence their wide catalytic use.
(b)(i)
(ii)
CBSE Class 12 Board · 2016 · All India · Subjective