In qualitative analysis, dichromate ions (Cr2O72–) are commonly used as oxidizing agents. Reduction of these ions often results in a color change, which can help identify the reducing species.
Step 1: Reaction Involved
When \(\text{SO}_3^{2-}\) reacts with dilute \(\text{H}_2\text{SO}_4\), \(\text{SO}_2\) gas is evolved. The \(\text{SO}_2\) gas reduces the dichromate ion (\(\text{Cr}_2\text{O}_7^{2-}\)) to \(\text{Cr}^{3+}\), which is green in color. The reaction is as follows:
\[\text{Cr}_2\text{O}_7^{2-} + \text{SO}_3^{2-} + \text{H}^+ \rightarrow \text{Cr}^{3+} + \text{SO}_4^{2-}.\]
Step 2: Color Change
- Dichromate ion (\(\text{Cr}_2\text{O}_7^{2-}\)) is orange in color.
- After reduction, \(\text{Cr}^{3+}\) ions form, which are green in color.
Conclusion: The solution turns green due to the formation of \(\text{Cr}^{3+}\). Therefore, the correct answer is \((3)\) Green.
In Carius method for estimation of halogens, 180 mg of an organic compound produced 143.5 mg of AgCl. The percentage composition of chlorine in the compound is ___________%. [Given: Molar mass in g mol\(^{-1}\) of Ag = 108, Cl = 35.5]


For the circuit shown above, the equivalent gate is:
Let \( f : \mathbb{R} \to \mathbb{R} \) be a twice differentiable function such that \[ (\sin x \cos y)(f(2x + 2y) - f(2x - 2y)) = (\cos x \sin y)(f(2x + 2y) + f(2x - 2y)), \] for all \( x, y \in \mathbb{R}. \)
If \( f'(0) = \frac{1}{2} \), then the value of \( 24f''\left( \frac{5\pi}{3} \right) \) is: