Question

When a manganous salt is fused with a mixture of $KN{O_3}$ and solid $NaOH$ , the oxidation number of $Mn$ changes from +2 to?
A.) +4
B.) +3
C.) +6
D.) +7

Answer 1

Hint: In order to deal with this question first we will define the term oxidation number further we will write the proper reaction of given compounds and according to it we will evaluate the charges when reaction is completed.

Complete step-by-step answer:

Oxidation number: The number of an atom's oxidation is defined as the charge an atom tends to have on the forming of ionic bonds with other heteroatoms. An atom with higher electronegativity is given a negative oxidation state (even though it forms a covalent bond).
The reaction can be given as follows when a manganese salt is fused with a mixture of $KN{O_3}$ and solid $NaOH$ :
$M{n^{ + 2}} + NO_3^ - \to MnO_4^{ - 2} + {H_2}O$
Clearly we can see that the charge of $Mn{O_4}$ is -2
Charge of ${O_4} = \left( { - 2} \right) \times 4 = - 8$
Let charge on $Mn = x$
Hence charge of $Mn{O_4}$ in terms of individual charge is $\left( x \right) + \left( { - 8} \right) = - 2$
$
   \Rightarrow x = - 2 + 8 \\
   \Rightarrow x = + 6 \\
 $
So, charge of $Mn = 6$
Hence, when a manganous salt is fused with a mixture of $KN{O_3}$ and solid $NaOH$ , the oxidation number if $Mn$ changes from +2 to 6
So, the correct answer is option C.

Note- Oxidation status or number usually lets us understand electron transfer. Students will note, however, that this is distinct from a formal charge that specifies the structure of atoms. The number / state of oxidation are also used to determine changes which occur in redox reactions. Meanwhile, valence electrons are very similar. Redox reaction is a type of reaction to oxidation and a reduction. In a redox reaction these two reactions occur concurrently.