### solve for the oxidation state of manganese in kmno4?

solve for the oxidation state of manganese in kmno4?

To get notifications when anyone posts a new answer to this question,

Please don't post or ask to join a "Group" or "Whatsapp Group" as a comment. It will be deleted. To join or start a group, please click here

The oxidation state of Mn in KMnO4 is +7 .
This can be obtained by taking into
account the valencies of potassium and
oxygen atoms. The valency of K is 1 but it
loses electron so to be specific it is +1.
Similarly oxygen in specific has a valency
of -2.
Follow
Solution
In KMnO4, K and O have fixed oxidation number +1 and-2 respectively.
As it is a stable and uncharged compound the sum total of all oxidation numbers should be zero.
Let oxidation number of Mn be x,So,. +1 +x+(-2)×4 = 0
1+x-8=0
x-7=0
x=7
So oxidation number of Mn is +7
Follow
In KMnO4, K and O have fixed
oxidation number +1 and -2
respectively. As it is a stable and
uncharged compound the sum total
of all oxidation numbers should be
zero.
Let oxidation number of Mn be x,
So,. +1 +x +(-2)×4 = 0
1+x-8=0
x-7=0
x=7 So oxidation number of Mn is +7
Follow
Or
Follow
Or

k(+1) +Mn(x) + 4*O(-2) =0
Mnx=+7;
Follow
KMNO4:
+1+mn+(4×-2)=0
+1+mn-8=0

mn+1=+8
mn=8-1
mn=7
potassiumtetraoxomanganate(vii)
Follow