Question

Acidified $K_2Cr_2O_7 $ solution turns green when $Na_2SO_3 $ is added to it. This is due to the formation of

• A. $CrO^{2-}_4 $
• B. $Cr_2(SO_3)_3 $
• C. $CrSO_4 $
• D. $Cr_2(SO_4)_3 $

Answer 1

The Correct Option is D

Chemistry often involves understanding how different substances interact and the resulting compounds. When an acidified solution of \(K_2Cr_2O_7\) (Potassium dichromate) is mixed with \(Na_2SO_3\) (Sodium sulfite), a chemical reaction occurs that causes the solution to turn green. Let us break down why this happens:

1. Initial Compounds:
   • \(K_2Cr_2O_7\): Potassium dichromate is known for its orange color in solutions. 
   • \(Na_2SO_3\): Sodium sulfite is a reducing agent.
2. Nature of Reaction:

In an acidic medium, potassium dichromate acts as an oxidizing agent. The dichromate ions \((Cr_2O_7^{2-})\) in acidic conditions get reduced, while the sulfite ions \((SO_3^{2-})\) act as reducing agents.

1. Transformation Process:

The reaction between dichromate ions and sulfite ions leads to the reduction of \(Cr_2O_7^{2-}\) to chromium sulfate \((Cr_2(SO_4)_3)\):

\(Cr_2O_7^{2-}\)

+

\(3SO_3^{2-}\)

+

8H+

→

2Cr3+

+

\(3SO_4^{2-}\)

+

4H2O

1. Results of the Reaction:

During the reaction, the dichromate ions are reduced to chromium ions, which form chromium sulfate \((Cr_2(SO_4)_3)\). The characteristic green color of the solution is due to the formation of \(Cr_2(SO_4)_3\), holding the chromium in the +3 oxidation state, which is typically green in solutions.

1. Conclusion:

The formation of \(Cr_2(SO_4)_3\) is the reason the solution turns green when \(Na_2SO_3\) is added to acidified \(K_2Cr_2O_7\). Thus, the correct answer is \(Cr_2(SO_4)_3\).