Answer
Verified
448.5k+ views
Hint: We should know that oxidation state is a number that is assigned to an element in a chemical combination. This number represents the number of electrons that an atom can gain, lose, or share when chemically bonding with an atom of another element.
Step by step answer:
We should know that the transfer of electrons is described by the oxidation state of the molecule. We use oxidation state to determine the changes in redox reactions and it is numerically similar to valence electrons.
To find the oxidation state of a compound we should know some rules. These rules are as follows:
RULE 1: Any individual atom un-combined with other elements has the oxidation state of 0 (zero). For example: If we take Ag only, its oxidation state will be 0. The oxidation state for O (oxygen) or $O_2$ is 0 as long as it is un-combined with any other element.
RULE 2: We should know that the sum of the oxidation state of all atoms in any given species is equal to the net charge on that species.
a.) In neutral species, the total sum of the oxidation state of all atoms is 0. Example, sum for NaCl is zero because Na=+1 and Cl= -1
b.) In ions, the total sum of the oxidation state is the charge of the ion. Example: \[\,C{{a}^{2+}}\] (Calcium) is = +2. In \[CrO_{4}^{2-}\,\] (Chromate ion), the total sum is -2.
Now, we will find the oxidation state of Mn in \[{{K}_{2}}Mn{{O}_{4}}\]. So, we should let the oxidation state of Mn be x. So, oxidation state will be:
As we know, the oxidation state of O is -2 and that of H is +1.
\[\begin{align}
& {{K}_{2}}Mn{{O}_{4}}:2\times (+1)+x+(-2)\times 4=0 \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,:2+x-8=0 \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,:x=+6 \\
& KMn{{O}_{4}}\,\,:1+x+(-2)\times 4=0 \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,:1+x-8=0 \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,:x=+7 \\
\end{align}\]
So, from the above calculation we find out the oxidation state of \[{{K}_{2}}Mn{{O}_{4}}\] and \[KMn{{O}_{4}}\].
Oxidation state of \[{{K}_{2}}Mn{{O}_{4}}\] is +6. And the oxidation state of \[KMn{{O}_{4}}\]is +7. So, option C is correct.
Note: We should not confuse formal charge with oxidation states. We should know that oxidation state is commonly used to determine the changes in redox reactions and is numerically similar to valence electrons, but different from formal charge. Formal charge determines the arrangement of atoms and the likelihood of the molecule existing.
Step by step answer:
We should know that the transfer of electrons is described by the oxidation state of the molecule. We use oxidation state to determine the changes in redox reactions and it is numerically similar to valence electrons.
To find the oxidation state of a compound we should know some rules. These rules are as follows:
RULE 1: Any individual atom un-combined with other elements has the oxidation state of 0 (zero). For example: If we take Ag only, its oxidation state will be 0. The oxidation state for O (oxygen) or $O_2$ is 0 as long as it is un-combined with any other element.
RULE 2: We should know that the sum of the oxidation state of all atoms in any given species is equal to the net charge on that species.
a.) In neutral species, the total sum of the oxidation state of all atoms is 0. Example, sum for NaCl is zero because Na=+1 and Cl= -1
b.) In ions, the total sum of the oxidation state is the charge of the ion. Example: \[\,C{{a}^{2+}}\] (Calcium) is = +2. In \[CrO_{4}^{2-}\,\] (Chromate ion), the total sum is -2.
Now, we will find the oxidation state of Mn in \[{{K}_{2}}Mn{{O}_{4}}\]. So, we should let the oxidation state of Mn be x. So, oxidation state will be:
As we know, the oxidation state of O is -2 and that of H is +1.
\[\begin{align}
& {{K}_{2}}Mn{{O}_{4}}:2\times (+1)+x+(-2)\times 4=0 \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,:2+x-8=0 \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,:x=+6 \\
& KMn{{O}_{4}}\,\,:1+x+(-2)\times 4=0 \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,:1+x-8=0 \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,:x=+7 \\
\end{align}\]
So, from the above calculation we find out the oxidation state of \[{{K}_{2}}Mn{{O}_{4}}\] and \[KMn{{O}_{4}}\].
Oxidation state of \[{{K}_{2}}Mn{{O}_{4}}\] is +6. And the oxidation state of \[KMn{{O}_{4}}\]is +7. So, option C is correct.
Note: We should not confuse formal charge with oxidation states. We should know that oxidation state is commonly used to determine the changes in redox reactions and is numerically similar to valence electrons, but different from formal charge. Formal charge determines the arrangement of atoms and the likelihood of the molecule existing.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Which are the Top 10 Largest Countries of the World?
Give 10 examples for herbs , shrubs , climbers , creepers
10 examples of evaporation in daily life with explanations
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Why is there a time difference of about 5 hours between class 10 social science CBSE