Question

Determine the oxidation number of the following elements or underlined elements in compounds and polyatomic ions. Enter the sign and the number, for example '+1' or '-1'.

CO2
H3O1+
MnO41-
S2O82-

Answer 1

In any compound or polyatomic ion the sum of oxidation numbers of all the elements is equal to zero therefore, the oxidation number of the central element in the given compounds or polyatomic ions can be calculated as follows:

In CO₂ molecule the oxidation number of oxygen atom is -2,. Let us assume the oxidation number of carbon atom be x. so the oxidation number of carbon atom will be

Hence, the oxidation number of C in CO₂ is +4.

In H₃O⁺ ion the oxidation number of hydrogen atom is +1. Let us assume the oxidation number of oxygen atom be x. Since thereis a +1 charge so the sum of oxidation numbers of hydrogen and oxygen atoms will be equal to +1 as shown below:

Hence, the oxidation number of O in H₃O⁺ is -2.

In MnO₄^1- ion the oxidation number of oxygen atom is -2. Let us assume the oxidation number of manganese atom be x. Since thereis a -1 charge so the sum of oxidation numbers of Mn and oxygen atoms will be equal to -1 as shown below:

Hence, the oxidation number of Mn in MnO₄^1- is +7.

In S₂O₈^2- ion the oxidation number of six oxygen atom is -2 and two oxygen atom is -1. Let us assume the oxidation number of sulfur atom be x. Since thereis a -2 charge so the sum of oxidation numbers of S and oxygen atoms will be equal to -2 as shown below:

Hence, the oxidation number of S in S₂O₈^2- is +6.