1.

Indentify disproportionation rectionA. `CH_(4)+2O_(2)rarrCO_(2)+2H_(2)O`B. `CH_(4)+4CI_(2)rarrC CI_(4)+4HCI`C. `2F_(2)+2Oh^(-)rarr2F^(-)+OF_(2)+H_(2)O`D. `2NO_(2)+2OH^(-)rarrNO_(2)^(-)+NO_(3)^(-)+H_(2)O`

Answer» It is disproportionation reaction
`2NO_(2)+2Oh^(-)rarrNO_(3)^(-)+H_(2)O`
`NO_(2)`: Oxidation state of `N=+4`
`NO_(2)^(-)`: oxidation state of N=+3
`NO_(3)^(-)`: Oxidation state of N=+5
It is disproportion rection since there is decrease as well as increase in oxidation state of N.


Discussion

No Comment Found

Related InterviewSolutions