A) fluorine from \[NaF\]
B) iodine from \[NaI\]
C) bromine from \[NaBr\]
D) None of the above
Correct Answer: A
Solution :
A non-metal placed below in electrochemical series displaces the non-metal placed above it from that's salt solution (ion). A non-metal which lies below in electrochemical series is more reactive. Greater the value of \[{{E}^{o}},\] greater is the reactivity of the metal. \[\therefore \]\[C{{l}_{2}}\] displaces \[B{{r}_{2}}\] from \[NaBr\] and \[{{I}_{2}}\] from \[NaI\] but does not displace \[{{F}_{2}}\] from \[NaF\]. \[B{{r}_{2}}\]displaces \[{{I}_{2}}\] from \[NaI\] but not \[C{{l}_{2}}\] from \[NaCl\]. \[{{I}_{2}}\] cannot displace \[C{{l}_{2}}\] or \[B{{r}_{2}}\] from chlorides or bromides.
You need to login to perform this action.
You will be redirected in
3 sec
