| Consider the following reactions (unbalanced) |
| \[\text{Zn}\,+\,\text{hot}\,\text{conc}.\,{{\text{H}}_{2}}\text{S}{{\text{O}}_{4}}\to \,\text{G}\,+\,\text{R}\,+\,\text{X}\] |
| \[\text{Zn}\,+\,\text{conc}.\,\text{NaOH}\,\to \,\text{T}\,+\,\text{Q}\] |
| \[\text{G}\,+\,{{\text{H}}_{2}}\text{S}\,\,+\,\text{NaOH}\,\to \,\text{Z}\,(a\,\text{precipitate})\,+\,\text{X}\,+\text{Y}\] |
| Choose the correct option(s): |
A) R is a V-shaped molecule
B) The oxidation state of Zn in T is +1
C) Bond order of Q is 1 in its ground state
D) Z is dirty white in colour
Correct Answer: A , C , D
Solution :
(i) \[Zn+\underset{\text{hot}\,\,\text{conc}.}{\mathop{2{{H}_{2}}}}\,S{{O}_{4}}\to \underset{'\text{ }\!\!'\!\!\text{ G }\!\!'\!\!\text{ }\!\!'\!\!\text{ }}{\mathop{ZnS{{O}_{4}}}}\,+S{{O}_{2}}+2{{H}_{2}}O\] Hot conc. ??G?? (ii) \[Zn+\text{Conc}\text{.}NaOH\to N{{a}_{2}}\underset{''T''}{\mathop{Zn{{O}_{2}}}}\,+\underset{\text{Q}}{\mathop{{{H}_{2}}}}\,\] (iii) \[ZnS{{O}_{4}}+\underset{(\text{G}\text{.R}\text{.of}\,\text{IVgroup})}{\mathop{{{H}_{2}}S+N{{H}_{4}}OH}}\,\to \underset{''\text{Z}''\,(\text{dirty}\,\,\text{white}\,\,\text{ppt})}{\mathop{Zns+}}\,\]Solution :
Same as aboveSolution :
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