• # question_answer Compound X and Y are treated with dilute ${{H}_{2}}S{{O}_{4}}$ separately. The gases liberated are P and Q. P turns acidified ${{K}_{2}}C{{r}_{2}}{{O}_{7}}$ paper green while gas Q turns lead acetate paper black. The compounds X and y are A)  $N{{a}_{2}}S{{O}_{3}}$and$N{{a}_{2}}S$        B)  $N{{a}_{2}}C{{O}_{3}}$and$N{{a}_{2}}S$ C)  $NaCl$and$N{{a}_{2}}S{{O}_{3}}$ D)  $N{{a}_{2}}S{{O}_{3}}$and$N{{a}_{2}}S{{O}_{4}}$

Idea This problem is based on identification of ${{H}_{2}}S$ and $S{{O}_{2}}$ using qualitative analysis and students are advised to complete chemical reaction first and then use qualitative techniques to identify the gaseous particles. $S{{O}_{2}}$turns acidified ${{K}_{2}}C{{r}_{2}}{{O}_{7}}$ paper green $N{{a}_{2}}S{{O}_{3}}+{{H}_{2}}S{{O}_{4}}\xrightarrow[{}]{{}}N{{a}_{2}}S{{O}_{4}}+\underset{(P)}{\mathop{S{{O}_{2}}}}\,+{{H}_{2}}O$ ${{K}_{2}}C{{r}_{2}}{{O}_{7}}+{{H}_{2}}S{{O}_{4}}+3S{{O}_{2}}\xrightarrow[{}]{{}}C{{r}_{2}}\underset{Green}{\mathop{{{(S{{O}_{4}})}_{3}}}}\,$$+{{K}_{2}}S{{O}_{4}}+{{H}_{2}}O$ ${{H}_{2}}S$ turns lead acetate paper black $N{{a}_{2}}S+{{H}_{2}}S{{O}_{4}}\xrightarrow[{}]{{}}N{{a}_{2}}S{{O}_{4}}+\underset{(Q)}{\mathop{{{H}_{2}}S}}\,$ ${{H}_{2}}S+Pb{{(C{{H}_{3}}COO)}_{2}}\xrightarrow[{}]{{}}\underset{Black}{\mathop{PbS}}\,+2C{{H}_{3}}COOH$TEST Edge Similar trends related problem with unknown starting material including variation in reagent used are also asked. Students are advised to go through study of wet test for acid radicals and basic radicals which are commonly asked.