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How to Determine Oxidation Numbers and Examples
Did you know that numbers are not only in Mathematics? In Chemistry, there are also numbers called oxidation numbers. This oxidation number is a number or number that states the number of electrons released or received by an atom in the formation of a compound. Determining this oxidation number cannot be done carelessly because there are several rules that need to be followed.
Definition of Oxidation Numbers
The oxidation number of an element is a number that shows the contribution of the charge of an atom of an element to the molecule or ion it forms. Oxidation numbers are often abbreviated as b.o. The oxidation number of an element in a compound is usually the same as the number of electrons released or paired by one atom of the element in the compound.
The element that releases electrons has a positive oxidation number, while the one that captures electrons has a negative sign. For example, in the NaCl compound, the Na atom releases 1 electron to the Cl atom, so the oxidation number of Na is +1 and Cl is -1. Oxidation numbers are used to express the half-reaction equations that occur in oxidation and reduction reactions.
How to Determine Oxidation Numbers
By knowing how to determine oxidation numbers, it can make it easier for us to determine the oxidation numbers of atoms in various compounds that are formed. Here’s how to determine oxidation numbers.
The oxidation number of free elements is 0 (zero)
Free elements are elements that are unstable and do not chemically bond with other elements. For example, Na, H2, N2, Br2, Be, K, O2, and P4. These eight elements have an oxidation number of zero.
The oxidation number of monatomic ions is the same as their charge
Monatomic ions are ions that only consist of one atom. In each of these monatomic ions, the oxidation number is the same as its charge. Example:
The oxidation number of Li+, Na+, K+, and Ag+ is +1.
The oxidation number of Mg2+, Ca2+, Cu2+, and Fe2+ is +2.
The oxidation number of F–, Cl–, Br–, and I– is -1.
The oxidation number of O2- and S2- is -2.
The sum of the oxidation numbers of all element atoms in a compound is zero
Example: NaCl compound
In the NaCl compound, the sum of the oxidation numbers of Na and Cl must be equal to zero. For the oxidation number of Na in NaCl is +1, while the oxidation number of Cl in NaCl is -1, then the sum of the two oxidation numbers is equal to zero.
CuO compound
The oxidation number of Cu in the CuO compound is +2 and the oxidation number of O in the CuO compound is -2, then the sum is zero.
The oxidation number of atoms forming polyatomic ions is the same as their charge
Polyatomic ions are ions formed from two or more atoms that are bound together and form ions, either positive ions (cations) or negative ions (anions). For example, OH– (hydroxide ion), SO42– (sulfate ion), and NH+4 (ammonium ion). The oxidation number of the atoms forming this polyatomic ion must be the same as its charge. Example:
In the OH– ion, the oxidation number of O is added to the oxidation number of H = -1.
In the SO42– ion, the oxidation number of S is added to 4 times the oxidation number of O = -2.
In the NH+4 ion, the oxidation number of N is added to 4 times the oxidation number of H = +1.
The oxidation number of alkali group elements (IA) in various compounds formed is +1
Example:
The oxidation number of Na in NaCl, NaOH, Na2CO3, Na3PO4, and all Na compounds is +1.
The oxidation number of Li, K, Rb, Ca, Fr in all their compounds is +1.
The oxidation number of alkaline earth group elements (IIA) in various compounds formed is +2
For example, the elements Mg, Ca, Sr, Ba in all their compounds have an oxidation number of +2.
The oxidation number of Hydrogen atoms (H) in compounds is +1
The oxidation number of Hydrogen atoms (H) in compounds is +1, except for alkali metal hydride compounds (group IA) and alkaline earth (group IIA). Example:
The oxidation number of H in compounds H2O, HCl, HF, H2SO4, HNO3, NH3, and CH4 is +1.
The oxidation number of H in the compounds KH, NaH, MgH2, and CaH2 is -1.
The oxidation number of a oxygen in its compounds is -2
The oxidation number of oxygen in its compounds is -2, except for peroxide and OF2 compounds. Example:
The oxidation number of O in H2O, CO2, SO2, H2SO4, and KClO3 is -2 (oxygen compounds).
The oxidation number of O in H2O2, Na2O2, and BaO2 is -1 (peroxide compounds).
The oxidation number of O in the OF2 compound is +2.
Example questions
To maximize your understanding in determining oxidation numbers, let’s look at the example questions along with their complete discussions below.
Question 1
Determine the oxidation numbers of the elements that make up the compound NaClO3!
Answer:
According to rule number 2, the oxidation number of Na is +1. For O, it has an oxidation number of -2. Of the three elements that make up the compound NaClO3, only the Cl element has an unknown oxidation number. Well, to find out, you can use elements whose oxidation numbers are already known.
(1 x Na oxidation number) + (1 x Cl oxidation number) + (3 x O oxidation number) = 0
(+1) + Cl oxidation number + (3 x (-2)) = 0
(+1) + Cl oxidation number + (-6) = 0
Cl oxidation number = 6 – 1
Cl oxidation number = +5
Therefore, the oxidation numbers of the elements Na, Cl, and O in NaClO3 are +1, +5, and -2.
Question 2
Determine the oxidation numbers of the elements that make up the compound KMnO4!
Answer:
According to rule number 2, the oxidation number of K is +1. O has an oxidation number of -2. So, here’s how to find out the oxidation number of Mn.
(1 x oxidation number of K) + (1 x oxidation number of Mn) + (4 x oxidation number of O) = 0
(+1) + oxidation number of Mn + (4 x (-2)) = 0
(+1) + oxidation number of Mn + (-8) = 0
Oxidation number of Mn = 8 – 1
Oxidation number of Mn = +7
Therefore, the oxidation numbers of the elements K, Mn, and O in KMnO4 are +1, +7, and -2.
