JEE Advanced 2023 Paper 1 Q04 Chemistry Inorganic Chemistry Qualitative Analysis Medium

JEE Advanced 2023 Paper 1 · Q04 · Qualitative Analysis

In the scheme given below, $X$ and $Y$, respectively, are: $MnCl_2 \xrightarrow{aq.\,NaOH} Mn(OH)_2 \downarrow$(P, white ppt) $\xrightarrow{PbO_2,\,H^+} HMnO_4$(violet) = X. $X + NaCl \xrightarrow{\text{conc}\,H_2SO_4/\Delta} MnO(OH)_2$(Q) + Y + ...

  1. A. $CrO_4^{2-}$ and $Br_2$
  2. B. $MnO_4^{2-}$ and $Cl_2$
  3. C. $MnO_4^-$ and $Cl_2$
  4. D. $MnSO_4$ and $HOCl$
Reveal answer + step-by-step solution

Correct answer:C

Solution

$Mn^{2+}$ oxidized by $PbO_2/H^+$ gives $HMnO_4$ (violet, permanganate ion $MnO_4^-$), so $X = MnO_4^-$. Permanganate + $NaCl$ in hot concentrated $H_2SO_4$ liberates $Cl_2$ gas. Hence $Y = Cl_2$.

Want to drill deeper? Open this question inside SolveKar AI and tap "Why?" on any step — the AI Mentor reads your full solution context and explains until it clicks. Download SolveKar AI →