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JEE Main 202621 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
21 January
Shift
Morning Shift
Subject
Chemistry
Chapter
d and f Block Elements

JEE Main 2026 · 21 January · Morning Shift · Chemistry · d and f Block Elements

Consider the following reactions: NaCl+K_ 2 Cr_ 2 O_ 7 +H_ 2 SO_ 4...

Consider the following reactions:
NaCl+K2Cr2O7+H2SO4→ A+KHSO4+NaHSO4+H2O\mathrm{NaCl}+\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}+\mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{~A}+\mathrm{KHSO}_{4}+\mathrm{NaHSO}_{4}+\mathrm{H}_{2} \mathrm{O}
A+NaOH→B+NaCl+H2O\mathrm{A}+\mathrm{NaOH} \rightarrow \mathrm{B}+\mathrm{NaCl}+\mathrm{H}_{2} \mathrm{O}
B+H2SO4+H2O2→C+Na2SO4+H2O\mathrm{B}+\mathrm{H}_{2} \mathrm{SO}_{4}+\mathrm{H}_{2} \mathrm{O}_{2} \rightarrow \mathrm{C}+\mathrm{Na}_{2} \mathrm{SO}_{4}+\mathrm{H}_{2} \mathrm{O}
In the product ' CC ', ' XX ' is the number of O22−O_{2}^{2-} units, ' YY ' is the total number oxygen atoms present and ' ZZ ' is the oxidation state of CrC r. The value of X+Y+ZX+Y+Z is ____\_\_\_\_.

Correct answer

0

Step-by-step solution

Reaction 1: NaCl+K2Cr2O7+H2SO4→CrO2Cl2NaCl + K_2Cr_2O_7 + H_2SO_4 \rightarrow CrO_2Cl_2 (A, chromyl chloride)
Reaction 2: CrO2Cl2+4NaOH→Na2CrO4+2NaCl+2H2OCrO_2Cl_2 + 4NaOH \rightarrow Na_2CrO_4 + 2NaCl + 2H_2O, B = Na2CrO4Na_2CrO_4
Reaction 3: Na2CrO4+H2SO4+H2O2→CrO5+Na2SO4+H2ONa_2CrO_4 + H_2SO_4 + H_2O_2 \rightarrow CrO_5 + Na_2SO_4 + H_2O
C = CrO5CrO_5 or CrO(O2)2CrO(O_2)_2 (chromium peroxide, blue)
Structure: one oxo group and two peroxide (O22−O_2^{2-}) units
X = number of O22−O_2^{2-} units = 2
Y = total oxygen atoms = 5
Z = oxidation state of Cr: O2−(−2)+2×O22−(−4)=−6O^{2-}(-2) + 2 \times O_2^{2-}(-4) = -6, so Cr = +6
X+Y+Z=2+5+6=13X + Y + Z = 2 + 5 + 6 = 13
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