A) \[C{{l}_{2}}\]
B) \[B{{r}_{2}}\]
C) \[{{F}_{2}}\]
D) \[{{I}_{2}}\]
Correct Answer: C
Solution :
Disproportionation reactions give both oxidise and reduced products from the same species. \[{{F}_{2}}\]reacts with \[NaOH\]to give only the reduced product \[{{F}^{-}}\]. \[2{{F}_{2}}+4NaOH\xrightarrow{Cold}2NaF+O{{F}_{2}}+{{H}_{2}}O\] \[2{{F}_{2}}+4NaOH\xrightarrow{Hot}4NaF+2{{H}_{2}}O+{{O}_{2}}\]
You need to login to perform this action.
You will be redirected in
3 sec
