JEE Main 2026 — 21 January, Morning Shift — Question 67

Consider the following reactions: $\mathrm{NaCl}+\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}+\mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{~A}+\mathrm{KHSO}_{4}+\mathrm{NaHSO}_{4}+\mathrm{H}_{2} \mathrm{O}$ $\mathrm{A}+\mathrm{NaOH} \rightarrow \mathrm{B}+\mathrm{NaCl}+\mathrm{H}_{2} \mathrm{O}$ $\mathrm{B}+\mathrm{H}_{2} \mathrm{SO}_{4}+\mathrm{H}_{2} \mathrm{O}_{2} \rightarrow \mathrm{C}+\mathrm{Na}_{2} \mathrm{SO}_{4}+\mathrm{H}_{2} \mathrm{O}$ In the product ' C ', ' X ' is the number of $\mathrm{O}_{2}^{2-}$ units, ' Y ' is the total number oxygen atoms present and ' $Z$ ' is the oxidation state of Cr . The value of $\mathrm{X}+\mathrm{Y}+\mathrm{Z}$ is $\_\_\_\_$。