Concept Explanation: Reaction of NaCl with Conc. H₂SO₄ and K₂Cr₂O₇

This question involves identifying the reddish fumes (B) and the yellow solution (C) formed in a series of chemical reactions. Let's break it down step by step.

Step 1: Formation of Reddish Fumes (B)

When NaCl reacts with concentrated H₂SO₄, hydrogen chloride gas (HCl) is produced:

$\mathrm{NaCl}+H{\mathrm{SO}}_4\to \mathrm{NaHSO}{4}_{}+\mathrm{HCl}$

This HCl gas then reacts with K₂Cr₂O₇ (potassium dichromate) to form chromyl chloride (CrO₂Cl₂), which is a reddish-brown fumes:

$K{\mathrm{Cr}}_2{O}_7+4\mathrm{HCl}\to 2\mathrm{KCl}+2\mathrm{CrO}{\mathrm{Cl}}_2+3H_{2}O$

So, (B) is chromyl chloride, $\mathrm{CrO}{\mathrm{Cl}}_2$.

Step 2: Reaction of (B) with NaOH to Form Yellow Solution (C)

Chromyl chloride (CrO₂Cl₂) reacts with sodium hydroxide (NaOH) to form sodium chromate (Na₂CrO₄), which has a yellow color:

$\mathrm{CrO}{\mathrm{Cl}}_2+4\mathrm{NaOH}\to \mathrm{Na}{\mathrm{CrO}}_4+2\mathrm{NaCl}+2H_{2}O$

Thus, (C) is sodium chromate, $\mathrm{Na}{\mathrm{CrO}}_4$.

Final Answer

Therefore, (B) is CrO₂Cl₂ and (C) is Na₂CrO₄. The correct option is:

CrO₂Cl₂, Na₂CrO₄

Related Topics

• Chromyl Chloride Test: A test for chloride ions where NaCl with conc. H₂SO₄ and K₂Cr₂O₇ gives reddish fumes of CrO₂Cl₂.
• Properties of Chromium Compounds: Chromates (yellow) and dichromates (orange) are interconvertible in acidic or basic conditions.
• Acid-Base Reactions: NaOH neutralizes acidic compounds like CrO₂Cl₂ to form salts.

Important Formulae and Theory

• Formation of HCl: $\mathrm{NaCl}+H{\mathrm{SO}}_4\to \mathrm{NaHSO}{4}_{}+\mathrm{HCl}$
• Chromyl Chloride Formation: $K{\mathrm{Cr}}_2{O}_7+4\mathrm{HCl}\to 2\mathrm{KCl}+2\mathrm{CrO}{\mathrm{Cl}}_2+3H_{2}O$
• Sodium Chromate Formation: $\mathrm{CrO}{\mathrm{Cl}}_2+4\mathrm{NaOH}\to \mathrm{Na}{\mathrm{CrO}}_4+2\mathrm{NaCl}+2H_{2}O$
• Theory: Chromyl chloride (CrO₂Cl₂) is an oxychloride of chromium, and sodium chromate (Na₂CrO₄) is a yellow compound due to the chromate ion (CrO₄^2-).