JEE Advanced levelSingle correct optionDifficulty 3 of 3qc196
A green solution of a chromium(III) salt is treated with excess NaOH and HX2OX2 and warmed; the yellow solution is acidified with acetic acid and lead acetate is added. In a separate test, acidified HX2OX2 with the yellow ion gives a blue peroxo compound CrOX5 (one oxo and two peroxo groups). The yellow ion, the yellow precipitate and the oxidation state of Cr in CrOX5 are
ACrOX4X2−, PbCrOX4, +10
BCrOX2X−, Pb(CrOX2)X2, +3
CCrX2OX7X2−, PbCrOX4, +3
DCrOX4X2−, PbCrOX4, +6
Answer: D
Alkaline HX2OX2 oxidises Cr(III) to yellow chromate; PbX2+ then gives yellow PbCrOX4. In CrOX5 the two peroxo groups are OX2X2− (4 O at −1 each) and one oxide: x+4(−1)+(−2)=0, so x=+6 (+10 results from treating all five O as −2).