Q19 / 75 · JEE Main 2026 (21 January Shift 2) · Chemistry

On heating a mixture of common salt and $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ in equal amount along with concentrated $\mathrm{H}_{2} \mathrm{SO}_{4}$ in a test tube, a gas is evolved. Formula of the gas evolved and oxidation state of the central metal atom in the gas respectively are:

  • A
    $\mathrm{Cr}_{2} \mathrm{O}_{2} \mathrm{Cl}_{2} \text { and }+3$
  • B
    $\mathrm{CrO}_{2} \mathrm{Cl}_{2} \text { and }+5$
  • Correct answer
    C
    $\mathrm{CrO}_{2} \mathrm{Cl}_{2} \text { and }+6$
  • D
    $\mathrm{Cr}_{2} \mathrm{O}_{2} \mathrm{Cl}_{2} \text { and }+6$

Solution

When a mixture of NaCl and K2Cr2O7 is heated with concentrated H2SO4, the dichromate ion oxidizes chloride ions. The reaction produces chromyl chloride (CrO2Cl), which is a volatile deep-red colored gas. The dichromate ion contains chromium in the +6 oxidation state. In the reaction, the dichromate is reduced and chloride is oxidized, but the chromium in the product remains in the +6 oxidation state as CrO2Cl (or as CrO2⁺ in ionic form). The gas evolved is CrO2Cl with chromium in the +6 oxidation state

Hence, the correct answer is option: 'C'.

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