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JEE Main 2017 · Chemistry · d- and f-Block Elements and Metallurgy · medium

An orange solution of K2Cr2O7 is acidified, and then treated with H2O2 in the presence of ether, giving a blue colour in the ether layer. The…

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An orange solution of K2Cr2O7\mathrm{K_2Cr_2O_7} is acidified, and then treated with H2O2\mathrm{H_2O_2} in the presence of ether, giving a blue colour in the ether layer. The oxidation state of chromium in the blue species and the number of O−O\mathrm{O{-}O} bonds in it are respectively
  1. A. +5+5 and 22
  2. B. +10+10 and 22
  3. C. +6+6 and 11
  4. D. +6+6 and 22
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Answer: (D) +6+6 and 22

The blue compound is CrO5\mathrm{CrO_5} (CrO(O2)2\mathrm{CrO(O_2)_2}), a butterfly-shaped peroxide: one Cr=O\mathrm{Cr=O} and two O22−\mathrm{O_2^{2-}} (peroxo) groups. Oxidation state: x+(−2)+2(−2)=0⇒x=+6x+(-2)+2(-2)=0\Rightarrow x=+6 (the peroxo oxygens are −1-1, not −2-2, so assigning −2-2 to all five oxygens would wrongly give +10+10). Two peroxo groups mean two O−O\mathrm{O{-}O} bonds. The reaction is not a redox change for Cr.
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