Consider the following reactions: \(\mathrm{NaCl}+\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}+\underset{\text { (Conc.) }}{\mathrm{H}_{2} \mathrm{SO}_{4}} \longrightarrow(\mathrm{~A})+\) side products \((\mathrm{A})+\mathrm{NaOH} \longrightarrow(\mathrm{B})+\) Side products \((\mathrm{B})+\underset{\text { (dilute) }}{\mathrm{H}_{2} \mathrm{SO}_{4}}+\mathrm{H}_{2} \mathrm{O}_{2} \longrightarrow(\mathrm{C})+\) Side products The sum of the total number of atoms in one molecule each of (A), (B) and (C) is \_\_\_\_

JEE Main · 2020 · 07 Jan 2020 · Shift 2 · d and f Block Elements

Consider the following reactions:

NaCl + K_{2} Cr_{2} O_{7} + (Conc.) H_{2} SO_{4} ⟶ (A) +

side products

(A) + NaOH ⟶ (B) +

Side products

(B) + (dilute) H_{2} SO_{4} + H_{2} O_{2} ⟶ (C) +

Side products
The sum of the total number of atoms in one molecule each of (A), (B) and (C) is ____

Answer: 18

The sequence runs

NaCl + K X_{2} Cr X_{2} O X_{7} + H X_{2} SO X_{4} CrO X_{2} Cl X_{2}

, so A is chromyl chloride; then

CrO X_{2} Cl X_{2} + NaOH Na X_{2} CrO X_{4}

, so B is sodium chromate; then

Na X_{2} CrO X_{4} + H X_{2} SO X_{4} + H X_{2} O X_{2} CrO X_{5}

, so C is chromium peroxide. Counting atoms, A (

CrO X_{2} Cl X_{2}

) has

5

, B (

Na X_{2} CrO X_{4}

) has

7

, and C (

CrO X_{5}

) has

6

, for a total of

5 + 7 + 6 = 18

JEE Main · 2020 · 07 Jan 2020 · Shift 2 · Integer answer