What is the oxidation number of #"Mn"# in #"KMnO"_4#?

2 Answers
Jan 15, 2017

Answer:

The oxidation number of #"Mn"# is +7.

Explanation:

We assign the oxidation number per a set of rules.

For this question, the important rules are:

  1. The oxidation number of #"O"# in compounds is usually -2, but it is -1 in peroxides.

  2. The oxidation number of a Group 1 element in a compound is +1.

  3. The sum of the oxidation numbers of all of the atoms in a neutral compound is 0.

Rule 1 states that the oxidation number of #"O"# is -2.

We write the oxidation number of the element above its symbol and the total for 3 #"O"# atoms below the symbol.

This gives #"KMn"stackrelcolor(blue)("-2")("O")_4#
#color(white)(mmmmmmmll)stackrelcolor(blue)("-8"color(white)(mm))#

Rule 2 states that the oxidation number of K is +1.

This gives #stackrelcolor(blue)("+1")("K")"Mn"stackrelcolor(blue)("-2")("O")_4#
#color(white)(mmll)stackrelcolor(blue)("+1")color(white)(mmmmm)stackrelcolor(blue)("-8")color(white)(ml)#

Rule 3 states the numbers along the bottom must add up to zero.

The number under #"Mn"# must be +7.

This gives #stackrelcolor(blue)("+1")("K")"Mn"stackrelcolor(blue)("-2")("O")_4#
#color(white)(mmmml)stackrelcolor(blue)("+1")color(white)(mm)stackrelcolor(blue)("+7")(color(white)(ll))stackrelcolor(blue)("-8")color(white)(ml)#

There is only one #"Mn"# atom, so its oxidation number is +7.

This gives #stackrelcolor(blue)("+1")("K")stackrelcolor(blue)("+7")"Mn"stackrelcolor(blue)("-2")("O")_4#
#color(white)(mmmml)stackrelcolor(blue)("+1")color(white)(mm)stackrelcolor(blue)("+7")(color(white)(ll))stackrelcolor(blue)("-8")color(white)(ml)#

Jan 19, 2017

Answer:

The metal has a #+VII# oxidation state in #MnO_4^-#, #"permanganate ion"#.

Explanation:

See this link for more of the same. Remember that oxidation number is a factitious entity, however, it does help us to balance redox reactions.

And also here.