Question

What is the oxidation numbers of $Mn$in its compounds $MnC{l_2}$,$Mn{(OH)_3}$, $Mn{O_2}$ and $KMn{O_4}$ respectively?
(A) $ + 2, + 4, + 7, + 3$
(B) $ + 2, + 3, + 4, + 7$
(C) $ + 7, + 3, + 2, + 4$
(D) $ + 7, + 4, + 3, + 2$

Answer 1

Hint: As we know that oxidation state or oxidation number is the degree of oxidation of an atom present in a chemical compound. We can easily calculate the oxidation number of atoms as the oxidation number is also considered as the number of electrons that atom can gain, lose or share.

Complete Step by step answer: To find the oxidation number of atoms in a compound we should know that a free element has zero oxidation number and a monatomic ion has the oxidation number equal to its charge on ion.
Now Manganese has a stable oxidation state of $ + 2$ and when it combines with other elements to form a compound its oxidation state or oxidation number changes. We can calculate this oxidation state of manganese in each of the given compounds. Let us first recall that hydrogen and potassium have a $ + 1$ oxidation state, chlorine possess $ - 1$ oxidation state and oxygen have a $ - 2$ oxidation state according to the charge on their ions.
Now let us consider the oxidation state of manganese=$x$.
In first compound which is $MnC{l_2}$, the oxidation state of manganese will be:
$x + ( - 2) = 0$ because the compound has no overall charge.
\Rightarrow $x = + 2$, thus the oxidation number of manganese is $ + 2$.
Next is $Mn{(OH)_3}$,
$
  x + ( - 6) + ( + 3) = 0 \\
\Rightarrow x - 3 = 0 \\
 \Rightarrow x = + 3 \\
 $
Hence, in the second compound oxidation number is $ + 3$.
The third compound is $Mn{O_2}$,
$
  x + ( - 4) = 0 \\
\Rightarrow x = + 4 \\
 $
So the oxidation number of manganese is $ + 4$.
Lastly the compound is $KMn{O_4}$,
$
   + 1 + x + ( - 8) = 0 \\
\Rightarrow x = + 7 \\
 $
Hence the oxidation number of manganese is $ + 7$.
So the sequence comes out be: $ + 2, + 3, + 4, + 7$

Therefore the correct answer is (B).

Note: Manganese shows a total of oxidation states between $ + 2$ to $ + 7$. But always remember that while finding out the oxidation state the charge on ion is the oxidation number of that atom and thus the oxidation state of unknown compounds can be easily calculated.