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

${{\text{M}}^{0}}\to {{\text{M}}^{+}}+{{\text{e}}^{-}}$ 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.