Solid K2Cr2O7 is heated with solid…

Solid K2Cr2O7 is heated with solid NaCl and concentrated H2SO4 and the orange-red vapours X evolved are passed into aqueous NaOH to give a yellow…

Single correct option

Solid

\mathrm{K_2Cr_2O_7}

is heated with solid

\mathrm{NaCl}

and concentrated

\mathrm{H_2SO_4}

and the orange-red vapours

\mathbf{X}

evolved are passed into aqueous

\mathrm{NaOH}

to give a yellow solution

\mathbf{Y}

. Acidifying

\mathbf{Y}

with acetic acid and adding lead acetate solution gives a yellow precipitate

\mathbf{Z}

. Which of the following is correct?

A. $\mathbf{X}$ is $\mathrm{CrO_2Cl_2}$ , in which chromium is in the $+4$ state, and $\mathbf{Z}$ is $\mathrm{PbCrO_4}$
B. $\mathbf{X}$ is $\mathrm{CrCl_3}$ , and acidifying $\mathbf{Y}$ with dilute $\mathrm{H_2SO_4}$ gives orange $\mathrm{Cr_2O_7^{2-}}$ by a redox change
C. $\mathbf{X}$ is tetrahedral $\mathrm{CrO_2Cl_2}$ , and the yellow colour of $\mathbf{Y}$ arises from d-d transitions of chromium
D. $\mathbf{X}$ is tetrahedral $\mathrm{CrO_2Cl_2}$ and $\mathbf{Y}$ contains $\mathrm{CrO_4^{2-}}$ ; chromium is in the $+6$ state in $\mathbf{X}$ , $\mathbf{Y}$ and $\mathbf{Z}$ , and the yellow colour of $\mathbf{Z}$ arises from ligand-to-metal charge transfer

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Answer: (D) $\mathbf{X}$ is tetrahedral $\mathrm{CrO_2Cl_2}$ and $\mathbf{Y}$ contains $\mathrm{CrO_4^{2-}}$ ; chromium is in the $+6$ state in $\mathbf{X}$ , $\mathbf{Y}$ and $\mathbf{Z}$ , and the yellow colour of $\mathbf{Z}$ arises from ligand-to-metal charge transfer

The chromyl chloride test:

\mathrm{K_2Cr_2O_7+4NaCl+6H_2SO_4\rightarrow2KHSO_4+4NaHSO_4+2CrO_2Cl_2+3H_2O}

\mathbf{X}=\mathrm{CrO_2Cl_2}

(orange-red, tetrahedral, Cr in

+6

state since

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

). With alkali,

\mathrm{CrO_2Cl_2+4OH^-\rightarrow CrO_4^{2-}+2Cl^-+2H_2O}

, so

\mathbf{Y}

contains yellow

\mathrm{CrO_4^{2-}}

(

+6

). With

\mathrm{Pb^{2+}}

in acetic acid,

\mathbf{Z}=\mathrm{PbCrO_4}

(yellow, Cr

+6

). Cr(VI) is

d^0

, so no d-d transition is possible and the colour is due to ligand-to-metal charge transfer. Option (A) wrongly gives

+4

; (B)

\mathrm{CrCl_3}

is not volatile and

\mathrm{CrO_4^{2-}\rightarrow Cr_2O_7^{2-}}

is a non-redox, pH-driven condensation; (C) invokes d-d transitions, impossible for

d^0

. Hence (D).

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Original question written in the JEE Advanced 2023 Paper 2 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake