Question

What is the oxidation state of Mn in the following compound?
$KMn{{O}_{3}}$
A.3
B.4
C.5
D.7

Answer 1

Hint: Oxidation state of an element or an atom refers to a number which can be positive, negative or zero. It can be any integer.

Complete answer:
Oxidation state indicates the degree of oxidation for an atom in a very chemical compound; it's a hypothetical charge that an atom would have if all bonds to atoms of various elements were completely ionic. Oxidation states are typically represented by integers, which might be positive, negative, or zero.
We need to know about oxidation no of some elements first
The oxidation number of a free element has to be always 0.
The oxidation number of a monatomic ion is equal to the charge of the ion.
The oxidation number of H is generally +1, but it is -1 in when it is combined with some less electronegative elements.
The oxidation number of O in compounds is generally -2, but it is -1 in case of peroxides.
The oxidation number of Group 1 elements in a compound is always +1.
The oxidation number of Group 2 elements in a compound is always +2.
The oxidation number of a Group 17 elements is -1.
The sum of the oxidation numbers of all of the atoms present in a neutral compound is 0.
The sum of oxidation numbers in a polyatomic ion is equal to the charge of the ion present.
Now let's come to the solution.
So the oxidation number of potassium is +1 and that of oxygen is -2.
Now in case of $KMn{{O}_{3}}$,
Let oxidation number of Manganese be x
Now ,
$+1+x+(-2\times 3)=0$
Therefore x=5

So the correct option is c that is 5.

Note:
Oxidation results in the increase of oxidation number where reduction results in the decrease of oxidation number. More the oxidation number the more deficient the element is in electrons.