Question

16. An element X burns with a brilliant blue flame in the presence of oxygen to form Y. Y turns acidified potassium dichromate paper to green. Identify X and Y.

• A. nitrogen, nitrogen dioxide
• B. sulphur, sulphur dioxide
• C. phosphorus, phosphorus pentoxide
• D. sodium, sodium oxide

Answer 1

The correct option is B

sulphur, sulphur dioxide

Explanation:

• It is given that element X burns with a brilliant blue flame in the presence of Oxygen to form Y. So, X is Sulphur $\left(\mathrm{S}\right)$ a non-metal which burns with a blue coloured flame in the presence of oxygen.
• The reaction involved is:

$\underset{\mathrm{Sulphur}\left(\mathrm{X}\right)}{\mathrm{S}}+\underset{\mathrm{Oxygen}}{{\mathrm{O}}_2}\to \underset{\mathrm{Sulphur}\mathrm{dioxide}\left(\mathrm{Y}\right)}{{\mathrm{SO}}_2}$

• Sulphur dioxide turns orange coloured acidified Potassium dichromate $\left({\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7\right)$ to green due to the formation of $\mathrm{Cr}{\left({\mathrm{SO}}_4\right)}_3$.
• The chemical equation for the reaction is:

${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7+{\mathrm{H}}_2{\mathrm{SO}}_4+3{\mathrm{SO}}_2\to {\mathrm{K}}_2{\mathrm{SO}}_4+{\mathrm{Cr}}_2{\left({\mathrm{SO}}_4\right)}_3+{\mathrm{H}}_2\mathrm{O}$

Hence the correct answer is (B). Therefore, X is sulphur and Y is Sulphur dioxide.