Chemistry · d and f Block Elements

JEE Main 2024 — 27 January, Shift 1 — Question 70

NaCl reacts with conc. $\mathrm{H}_{2} \mathrm{SO}_{4}$ and $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ to give reddish fumes (B), which react with NaOH to give yellow solution (C). (B) and (C) respectively are ;

Option A:
$\mathrm{CrO}_{2} \mathrm{Cl}_{2}, \mathrm{Na}_{2} \mathrm{CrO}_{4}$
Correct
Option B:
$\mathrm{Na}_{2} \mathrm{CrO}_{4}, \mathrm{CrO}_{2} \mathrm{Cl}_{2}$
Option C:
$\mathrm{CrO}_{2} \mathrm{Cl}_{2}, \mathrm{KHSO}_{4}$
Option D:
$\mathrm{CrO}_{2} \mathrm{Cl}_{2}, \mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$

Answer: A

Step-by-step solution

$\mathrm{NaCl}+$ conc. $\mathrm{H}_{2} \mathrm{SO}_{4}+\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ $\rightarrow \mathrm{CrO}_{2} \mathrm{Cl}_{2}+\mathrm{KHSO}_{4}+\mathrm{NaHSO}_{4}+\mathrm{H}_{2} \mathrm{O}$

                                                                                                   (B)

                                                                                           Reddish brown

\mathrm{CrO}_{2} \mathrm{Cl}_{2}+\mathrm{NaOH} \rightarrow \mathrm{Na}_{2} \mathrm{CrO}_{4}+\mathrm{NaCl}+\mathrm{H}_{2} \mathrm{O}

                                                         (C)

                                              Yellow colour

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Exam: JEE Main 2024
Paper: 27 January, Shift 1
Subject: Chemistry
Chapter: d and f Block Elements
Topic: Compounds of Chromium

The electronic configuration for Neodymium is: [Atomic Number for Neodymium 60]

Answer: A

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