Orange solution A turns yellow (B) with NaOH. B gives a yellow precipitate C with Pb(NO3)2. Acidified A with H2O2 and ether gives a blue ether-soluble compound D. The correct identification and the oxidation state of Cr in D are

C is PbCrO4; D is CrO5 with Cr at +6

C is PbCrO4; D is CrO5 with Cr at +10

C is PbCr2O7; D is CrO5 with Cr at +6

C is PbCrO4; D is Cr2O3 with Cr at +3

JEE Advanced levelSingle correct optionDifficulty 3 of 3df102

Orange solution A turns yellow (B) with NaOH. B gives a yellow precipitate C with $Pb (NO X_{3}) X_{2}$ . Acidified A with $H X_{2} O X_{2}$ and ether gives a blue ether-soluble compound D. The correct identification and the oxidation state of Cr in D are

A is

Cr X_{2} O X_{7} X^{2 -}

and B is

CrO X_{4} X^{2 -}

;

CrO X_{4} X^{2 -} + Pb X^{2 +} \to PbCrO X_{4}

(yellow). With

H X_{2} O X_{2}

in acid,

CrO X_{5}

(blue peroxide, one oxo and two peroxo groups) forms. Oxidation state:

x + (- 2) + 4 (- 1) = 0

, so

x = + 6

. No change from A, so it is not a redox reaction of Cr.