Q55 / 180 · COMEDK 2025 (10 May Shift 3) · Chemistry
Find the correct matches.
$\begin{array}{|l|l|l|l|} \hline \text{A} & \mathrm{Cr}_2 \mathrm{O}_7^{2-} & \mathrm{P} & \text{is predominantly a basic compound.} \\ \hline \text{B} & \mathrm{FeCr}_2 \mathrm{O}_4 & \mathrm{Q} & \text{is predominantly an acidic compound.} \\ \hline \text{C} & \mathrm{CrO}_3 & \mathrm{R} & \text{forms yellow chromate on reaction with } \mathrm{Na}_2 \mathrm{CO}_3 \text{ in air.} \\ \hline \text{D} & \mathrm{CrO} & \mathrm{S} & \text{is interconvertible with } \mathrm{MO}_4^{2-} \text{ depending on the pH of the medium.} \\ \hline \end{array}$
- A$\mathrm{A}=\mathrm{R} \quad \mathrm{B}=\mathrm{S} \quad \mathrm{C}=\mathrm{P} \quad \mathrm{D}=\mathrm{Q}$
- B$\mathrm{A}=\mathrm{R} \quad \mathrm{B}=\mathrm{Q} \quad \mathrm{C}=\mathrm{S} \quad \mathrm{D}=\mathrm{P}$
- Correct answerC$\mathrm{A}=\mathrm{S} \quad \mathrm{B}=\mathrm{R} \quad \mathrm{C}=\mathrm{Q} \quad \mathrm{D}=\mathrm{P}$
- D$\mathrm{A}=\mathrm{S} \quad \mathrm{B}=\mathrm{R} \quad \mathrm{C}=\mathrm{P} \quad \mathrm{D}=\mathrm{Q}$
Solution
The dichromate ion $\mathrm{Cr}_2 \mathrm{O}_7^{2-}$ exists in equilibrium with the chromate ion $\mathrm{CrO}_4^{2-}$ depending on the pH of the medium, where $\mathrm{Cr}_2 \mathrm{O}_7^{2-} + \mathrm{H}_2 \mathrm{O} \rightleftharpoons 2 \mathrm{CrO}_4^{2-} + 2 \mathrm{H}^{+}$. Thus, A matches S.
Chromite ore $\mathrm{FeCr}_2 \mathrm{O}_4$ reacts with $\mathrm{Na}_2 \mathrm{CO}_3$ in the presence of air to form sodium chromate, which is yellow. The reaction is $4 \mathrm{FeCr}_2 \mathrm{O}_4 + 8 \mathrm{Na}_2 \mathrm{CO}_3 + 7 \mathrm{O}_2 \rightarrow 8 \mathrm{Na}_2 \mathrm{CrO}_4 + 2 \mathrm{Fe}_2 \mathrm{O}_3 + 8 \mathrm{CO}_2$. Thus, B matches R.
Chromium trioxide $\mathrm{CrO}_3$ is an acidic oxide because it dissolves in water to form chromic acid $\mathrm{H}_2 \mathrm{CrO}_4$. Thus, C matches Q.
Chromium monoxide $\mathrm{CrO}$ is a basic oxide, as it is a metal oxide in a low oxidation state. Thus, D matches P.
Answer: $\mathrm{A}=\mathrm{S} \quad \mathrm{B}=\mathrm{R} \quad \mathrm{C}=\mathrm{Q} \quad \mathrm{D}=\mathrm{P}$
Chromite ore $\mathrm{FeCr}_2 \mathrm{O}_4$ reacts with $\mathrm{Na}_2 \mathrm{CO}_3$ in the presence of air to form sodium chromate, which is yellow. The reaction is $4 \mathrm{FeCr}_2 \mathrm{O}_4 + 8 \mathrm{Na}_2 \mathrm{CO}_3 + 7 \mathrm{O}_2 \rightarrow 8 \mathrm{Na}_2 \mathrm{CrO}_4 + 2 \mathrm{Fe}_2 \mathrm{O}_3 + 8 \mathrm{CO}_2$. Thus, B matches R.
Chromium trioxide $\mathrm{CrO}_3$ is an acidic oxide because it dissolves in water to form chromic acid $\mathrm{H}_2 \mathrm{CrO}_4$. Thus, C matches Q.
Chromium monoxide $\mathrm{CrO}$ is a basic oxide, as it is a metal oxide in a low oxidation state. Thus, D matches P.
Answer: $\mathrm{A}=\mathrm{S} \quad \mathrm{B}=\mathrm{R} \quad \mathrm{C}=\mathrm{Q} \quad \mathrm{D}=\mathrm{P}$
Hence, the correct answer is option: 'C'.