question_answer

    When KBr is treated with concentrated \[\text{Ba}{{\text{O}}_{\text{2}}}\]redish brown gas evolved, gas is :

A)  mixture of bromine and HBr

B)  \[{{[NiC{{l}_{4}}]}^{2-}}\]

C)  bromine

D)  none of these

Correct Answer: A

Solution :

                \[2KBr+3{{H}_{2}}S{{O}_{4}}+Mn{{O}_{2}}\xrightarrow{\Delta }\] \[2KHS{{O}_{4}}+MnS{{O}_{4}}+2{{H}_{2}}O+B{{r}_{2}}\] \[[N{{i}^{2+}}in\,persence\,of\,CN-=[Ar]\] i.e.,        \[\frac{3-4{{\sin }^{2}}\theta }{1+4{{\sin }^{2}}\theta }=0\] \[\Rightarrow \]               \[3-4{{\sin }^{2}}\theta =0\] \[\Rightarrow \]               \[\sin \theta =\pm \frac{\sqrt{3}}{2}\] \[=\sin \left( \pm \frac{\pi }{3} \right)\] \[\Rightarrow \]               \[\theta =n\pi +{{(-1)}^{n}}\left( \pm \frac{\pi }{3} \right)=n\pi \pm \frac{\pi }{3}\]