Question

In the preparation of $KMn{{O}_{4}}$, pyrolusite $\left( Mn{{O}_{2}} \right)$ is first converted to potassium manganate $\left( {{K}_{2}}Mn{{O}_{4}} \right)$. In this conversion, the oxidation state of manganese changes from:
(A) +1 to +3
(B) +2 to +4
(C) +3 to +5
(D) +4 to +6

Answer 1

In a neutral molecule, the algebraic sum of oxidation number of all the atoms in a compound must be zero. The oxidation number of oxygen is generally considered as -2 when it is bonded with the metal elements due to its higher electronegativity value and its ability to gain 2 electrons to attain the stable configuration of neon.

Complete step by step solution:
-Oxidation number is a number assigned to an element that represents the number of electrons lost or gained by an atom of that element of the compound which is neutral.
-Any chemical reaction involves either the sharing of electrons or the transfer of electrons. Generally in inorganic chemistry, we deal with the transfer of electrons so that the reactions occur. Oxidation number tells us about the number of electrons lost or gained by that particular atom.
-The importance of oxidation number was recognized and so some basic rules were developed such that we could find the exact number of electrons being transferred by several atoms. This led to the development of some basic rules of finding the oxidation number of the atoms present in the compounds undergoing the reactions. These rules are:
-Always a cation is written first in a chemical formula. After that only we can write the anions. This makes it easy for us to understand which ion will be cation and which will be anion for many compounds, whether they are made of atoms with similar electronegativities or different electronegativities. Eg, in the compound NaCl, sodium forms cation and chlorine forms anion. In the compound $C{{O}_{2}}$, carbon forms cation and oxygen forms anion though both is non-metals.
-In elements, each atom bears an oxidation number of zero when it occurs in the free or uncombined state. So the molecules formed by the same type of atoms generally have 0 oxidation number like all the noble gases in their free state, oxygen molecule ${{O}_{2}}$, nitrogen molecule ${{N}_{2}}$, etc.
-Oxidation number of oxygen is mostly -2 but it shows the deviation in case of peroxides and superoxides and also when oxygen is bonded to fluorine because fluorine is the most electronegative atom and it always has an oxidation number of -1. Oxygen has oxidation numbers like +1, -1, +2 when bonded with such atoms.
-Hydrogen atoms usually have an oxidation number of +1. Its oxidation number becomes -1 only when it is bonded to alkali metals and alkali earth metals which always have the oxidation number of +1 and +2 respectively.
-In a neutral molecule, the algebraic sum of oxidation number of all atoms in a compound must be zero and in polyatomic ions, the algebraic sum of oxidation number of the atoms of ion is equal to charge present on that ion.
-In the question, we see that manganese and potassium are metals and will have positive oxidation numbers. Also, the sum of the oxidation numbers of all the atoms of the compound will be zero. So, in $Mn{{O}_{2}}$, the sum of oxidation numbers of Mn and O will be zero. Oxygen will have its usual oxidation number of -2 as it is bonded with metal.
So, we can write, x+(-2x2)=0 where x=oxidation number of Mn.
Thus, Mn has an oxidation number of +4 in this compound.
-Similarly, if we see the compound ${{K}_{2}}Mn{{O}_{4}}$, we see that both K and Mn are metals and they will have positive oxidation numbers and oxygen will have -2 oxidation number. K belongs to group 1 elements and so it will have a fixed oxidation number of +1. Writing the equation for this compound, we get
(+1x2) + x + (-2x4) = 0
So we get x=8-2=+6
So, the oxidation number of Mn in this compound is +6.

Therefore, in this conversion, the oxidation state of manganese changes from +4 to +6 and so the correct option is D.

Note: The oxidation number of group 17 elements is generally considered as -1 only. But if the elements are combined with more electronegative atoms, then their oxidation number becomes positive. Except for F, if any other halogen is combined with O or N, then they have a positive oxidation number only. N is practically more electronegative than Cl though we say it is less electronegative than Cl in theory.