Chemistry · 2026
JEE Main · 21 January 2026, Shift 1 · Q72
Consider the following reactions: NaCl + K_2 Cr_2 O_7+ H_2 SO_4 → A + KHSO_4+ NaHSO_4+ H_2 O; A + NaOH → B + NaCl + H_2 O; B + H_2 SO_4+ H_2 O_2 → C…
Consider the following reactions :
$$\begin{aligned}
& \mathrm{NaCl}+\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{~A}+\mathrm{KHSO}_4+\mathrm{NaHSO}_4+\mathrm{H}_2 \mathrm{O} \\
& \mathrm{~A}+\mathrm{NaOH} \rightarrow \mathrm{~B}+\mathrm{NaCl}+\mathrm{H}_2 \mathrm{O} \\
& \mathrm{~B}+\mathrm{H}_2 \mathrm{SO}_4+\mathrm{H}_2 \mathrm{O}_2 \rightarrow \mathrm{C}+\mathrm{Na}_2 \mathrm{SO}_4+\mathrm{H}_2 \mathrm{O}
\end{aligned}
$$
In the product ' C ', ' X ' is the number of $\displaystyle \mathrm{O}_2^{2-}$ units, ' Y ' is the total number oxygen atoms present and 'Z' is the oxidation state of Cr. The value of X+Y+Z is $\displaystyle \_\_\_\_$.
Official answer
From NTA’s final answer key for this paper.
13
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JEE Main 2026 Chemistry question, with the answer from NTA’s final answer key. Where our answers come from.