\(X_2\) is used in the refining of Ti metal by van Arkel method. \(Y_2\) does not liberate \(\mathrm{O}_2\)…
\(X_2\) is used in the refining of Ti metal by van Arkel method. \(Y_2\) does not liberate \(\mathrm{O}_2\) from water and does not form \(\mathrm{HY}\) and \(\mathrm{HOY}\) with water. \(X_2\) and \(Y_2\) are respectively
\(\mathrm{I}_2, \mathrm{Cl}_2\)
\(\mathrm{Cl}_2, \mathrm{I}_2\)
\(\mathrm{I}_2, \mathrm{I}_2\)
\(\mathrm{Cl}_2, \mathrm{Cl}_2\)
Solution
Given, \(X_2\) is used for refining of Ti-metal using van Arkel method. Also \(Y_2\) does not give \(\mathrm{O}_2\) from water and does not form \(\mathrm{HY}\) and \(\mathrm{HOY}\) with water. \(X_2\) and \(Y_2\) respectively.
\(\mathrm{I}_2\) is used for refining Ti metal using van Arkel method. Also when \(\mathrm{I}_2\) reacts with \(\mathrm{H}_2 \mathrm{O}\), it does not give \(\mathrm{O}_2\) and does not form \(\mathrm{HI}\) or \(\mathrm{HOI}\). Therefore, both \(X_2\) and \(Y_2\) are \(\mathrm{I}_2\). The reactions are given below:
(i) \(\underset{\text { Impure }}{\mathrm{Ti}}+2 \mathrm{I}_2 \rightarrow \mathrm{TiI}_4 \stackrel{\Delta}{\longrightarrow} \mathrm{Ti}+2 \mathrm{I}_2\)
(ii) \(\mathrm{I}_2+\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{IO}^{-}+2 \mathrm{H}^{+}+\mathrm{I}^{-}\)
Thus, option (c) is the correct answer.