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Question: How do you find the oxidation number of the elements in \(MnO_4^{2 - }\) ?...

How do you find the oxidation number of the elements in MnO42−MnO_4^{2 - } ?

Explanation

Solution

In above type of questions where we have to find the oxidation state, take xx for element for which you have to find the oxidation state. Oxidation state is the deficiency (to take electrons) or excess (they give up electrons) of electrons which is present for an atom to complete its oxidation state. Like for halogens, as they have deficiency of one electron to complete their oxidation state written as X−1{X^{ - 1}}

Complete step-by-step answer:
In the above question, an oxide is given of manganese it means it is a compound formed of manganese and oxygen. Here the oxidation state of MnMn is to find out, so let it be xx and other is oxygen for which we know it has (−2)( - 2) oxidation state. Now calculate by adding the oxidation state of all atoms and make the sum equals to −2 - 2 as it is given as power.

Number of charges on cationic parts is always equal to the charge on anionic parts. When cationic part in any species is absent its anionic part gets charged as negative. While when anionic part is missing its cationic part gets charged as positive.
MnO4−2MnO_4^{ - 2}
x+4(−2)=−2x + 4( - 2) = - 2

As you see, we put the oxidation state of the oxygen atom as (−2)( - 2) and 44 is multiplied because four atoms of oxygen are present. This sum is take equal to −2 - 2 , solving further we get
⇒x+4(−2)=−2\Rightarrow x + 4( - 2) = - 2
⇒x−8=−2\Rightarrow x - 8 = - 2
⇒x=8−2\Rightarrow x = 8 - 2
⇒x=+6\Rightarrow x = + 6

On solving further we get the oxidation state of manganese as +6 + 6 . To check that your calculation is correct or incorrect, put oxidation states for all atoms in the sum and if you are getting −2 - 2 on solving, it means the calculation is correct.

Note: We have taken here the oxidation state of oxygen as (−2)( - 2) oxygen is present in its simple oxide form. But there are three types of oxides which are simple oxide that is given in the question, peroxide where oxidation state is (−1)( - 1) and superoxide. If the oxidation state is written as negative then it means atom gains electrons and if it is written as positive then it means that atom gives up electrons.