Question

What is the oxidation state of Mn in $ \text{KMn}{{\text{O}}_{4}} $ ?
(A) $ -7 $
(B) $ -3 $
(C) 0
(D) $ +3 $
(E) $ +7 $

Answer 1

Hint: $ {{\text{M}}^{0}}\to {{\text{M}}^{+}}+{{\text{e}}^{-}} $ oxidation reaction $ +1 $
 $ {{\text{M}}^{+}}+{{e}^{-}}\to {{\text{M}}^{0}} $ Reduction 0
Oxidation state is the hypothetical charge an atom would have if all bonds to atoms of different elements were treated as ionic.

Complete step by step Solution
 $ \text{KMn}{{\text{O}}_{4}} $ dissociated as $ {{\text{K}}^{+}} $ and $ \text{MnO}_{4}^{-} $ .
Thus we will now calculate the oxidation number of Mn l. let us suppose it has an oxidation state of X. Thus the equation is –
 $ 1+\text{X}-8=0 $
Thus the value of X is $ +7 $
The equation has been written in that manner because $ \text{KMn}{{\text{O}}_{4}} $ has an overall oxidation number of 0.K has an oxidation number of $ +1 $ and o has an oxidation number on both the sides should be balanced and hence we get the value to be $ +7 $
So option E correct answer

Additional information
The oxidation number of a free element is always 0.
The oxidation number of a monatomic ion equals the charge of the ion.
The oxidation number of H is $ +1 $ , but it is $ -1 $ in when combined with less electronegative elements.
The oxidation number of O in compounds is usually $ -2 $ , but it is $ -1 $ in peroxides.
Rules of Assigning Oxidation Number to Elements
Rule 1: the oxidation number of an element in its free state is zero
Rule 2: the oxidation number of monatomic ions is the same as the charge on the ion.
Rule 3: the sum of all oxidation numbers in a neutral compound is zero.

Note
Oxidation number also called oxidation state, the total number of electrons that an atom either gains or loses in order to form a chemical bond with another atom.The oxidation number of hydrogen is almost always $ +1 $ when it is in a compound.