Question:
K\(_2\)Cr\(_2\)O\(_7\) on heating with aqueous caustic soda gives
K\(_2\)Cr\(_2\)O\(_7\) on heating with aqueous caustic soda gives
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The equilibrium: Cr\(_2\)O\(_7^{2-}\) (orange) + H\(_2\)O \(\rightleftharpoons\) 2CrO\(_4^{2-}\) (yellow) + 2H\(^+\). In acidic medium, dichromate is favored; in basic medium, chromate is favored.
Updated On: May 2, 2026
- CrO\(_4^{2-}\)
- Cr(OH)\(_2\)
- Cr(OH)\(_3\)
- Cr\(_2\)O\(_7^{2-}\)
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept:
Dichromate ions in alkaline medium convert to chromate ions.
Step 2: Detailed Explanation:}}
Potassium dichromate (K\(_2\)Cr\(_2\)O\(_7\)) in aqueous solution exists as Cr\(_2\)O\(_7^{2-}\) ions. When heated with an alkali (caustic soda, NaOH), the dichromate ion reacts to form chromate ion (CrO\(_4^{2-}\)). \[ Cr_2O_7^{2-} + 2OH^- \rightarrow 2CrO_4^{2-} + H_2O \] The color changes from orange (dichromate) to yellow (chromate).
Step 3: Final Answer:
K\(_2\)Cr\(_2\)O\(_7\) on heating with aqueous caustic soda gives chromate ions (CrO\(_4^{2-}\)), option (A).
Dichromate ions in alkaline medium convert to chromate ions.
Step 2: Detailed Explanation:}}
Potassium dichromate (K\(_2\)Cr\(_2\)O\(_7\)) in aqueous solution exists as Cr\(_2\)O\(_7^{2-}\) ions. When heated with an alkali (caustic soda, NaOH), the dichromate ion reacts to form chromate ion (CrO\(_4^{2-}\)). \[ Cr_2O_7^{2-} + 2OH^- \rightarrow 2CrO_4^{2-} + H_2O \] The color changes from orange (dichromate) to yellow (chromate).
Step 3: Final Answer:
K\(_2\)Cr\(_2\)O\(_7\) on heating with aqueous caustic soda gives chromate ions (CrO\(_4^{2-}\)), option (A).
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