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CBSE Class 12 Chemistry 2013 Solved Paper

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Question : 22 of 30
Marks: +1, -0
How would you account for the following?
(i) Transition metals exhibit variable oxidation states.
(ii) Zr(Z=40)\mathrm{Zr}(Z=40) and Hf(Z=72)\mathrm{Hf}(Z=72) have almost identical radii.
(iii) Transition metals and their compounds act as catalyst.
OR
Complete the following chemical equations:
(i) Cr2O72−+6Fe2++14H+→\mathrm{Cr}_2\mathrm{O}_7^{2-}+6\mathrm{Fe}^{2+}+14\mathrm{H}^{+} \rightarrow
(ii) 2CrO42−+2H+→2\mathrm{CrO}_4^{2-}+2\mathrm{H}^{+} \rightarrow
(iii) 2MnO4−+5C2O42−+16H+→2\mathrm{MnO}_4^{-}+5\mathrm{C}_2\mathrm{O}_4^{2-}+16\mathrm{H}^{+} \rightarrow
Solution:  
(i) Cr2O72−+6Fe2++14H+→6Fe3+\mathrm{Cr}_2\mathrm{O}_7^{2-}+6\mathrm{Fe}^{2+}+14\mathrm{H}^{+} \rightarrow 6\mathrm{Fe}^{3+} +2Cr3++7H2O+2\mathrm{Cr}^{3+}+7\mathrm{H}_2\mathrm{O}
(ii) 2CrO42−+2H+→Cr2O72−+H2O2\mathrm{CrO}_4^{2-}+2\mathrm{H}^{+} \rightarrow \mathrm{Cr}_2\mathrm{O}_7^{2-}+\mathrm{H}_2\mathrm{O}
(iii) 2MnO4−+5C2O42−+16H+→2Mn2+2\mathrm{MnO}_4^{-}+5\mathrm{C}_2\mathrm{O}_4^{2-}+16\mathrm{H}^{+} \rightarrow 2\mathrm{Mn}^{2+} +10CO2+8H2O+10\mathrm{CO}_2 +8\mathrm{H}_2\mathrm{O}
OR
(i) Transition metals shows variable oxidation state because they have tendency to loose electron from their penultimate dd-subshell and exhibit various oxidation state.
(ii) Zr\mathrm{Zr} and Hf\mathrm{Hf} have almost same radii because of lanthanoid contraction. Filling up of electrons in 4f4f series shows poor shielding effect.
(iii) 4d4d series and other transition metals act as catalyst because they participate in various type of chemical reactions i.e.,
N2+3H2⇌Fe High temp. High pressureMo2NH3\mathrm{N}_2+3\mathrm{H}_2 \underset{\frac{\mathrm{Fe}\,\text{High temp. High pressure}}{\mathrm{Mo}}}{\rightleftharpoons} 2\mathrm{NH}_3
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