Question

What is the oxidation number of $Cr$ in $C{r_2}{\left( {S{O_4}} \right)_3}$ ?

Answer 1

The oxidation state or sometimes known as oxidation number, defines the degree of oxidation that an atom possesses in a chemical compound. In other words, oxidation number refers to the charge left on the central atom especially when all the bonding pairs of electrons get broken, with the charge allocated to the most electronegative atom.

Complete step by step answer:
There are few important rules to estimate the oxidation number of every atom in the compounds or ions which are as follows:
1. The oxidation number of oxygen in a compound is generally $- 2$ , except to that in peroxides, where it is $- 1$ . The oxidation number of sulphur in a compound is generally ranging from $- 2,0, + 2, + 4, + 6$ . Sulphur possesses oxidation numbers of $- 2$ in the sulphide ion. Since in the given compound $C{r_2}{\left( {S{O_4}} \right)_3}$ sulphur and oxygen atoms form covalent bond with each other (as they are enclosed in parenthesis), and are thus "sharing" electrons. We know oxygen is negatively charged so it indicates that the sulphur will be positively charged. Sulphur in sulphate$SO_4^{2 - }$has an oxidation state of $+ 6$ .
2. The summation of all the oxidation numbers in a polyatomic ion is generally equal to the charge exhibited by the ion.
Now, let us calculate the oxidation number of $Cr$ for $C{r_2}{\left( {S{O_4}} \right)_3}$ :
According to Rule 1, the oxidation number of $O$ is $- 2$ and the oxidation number of $S$ is $+ 6$
So, there are twelve $O$ atoms and three $S$atoms having a total oxidation number of $- 24$ and$+ 18$ , respectively.
According to rule 2, the summation of all the oxidation numbers is equal to the charge on the ion. In our case, $C{r_2}{\left( {S{O_4}} \right)_3}$ is neutral having a charge $0$ .
Let x be the oxidation number for $Cr$ in $C{r_2}{\left( {S{O_4}} \right)_3}$ .
Thus,
$2x + (3 \times ( + 6)) + (12 \times ( - 2)) = 0 \\\ x = \dfrac{{ + 6}}{2} = + 3 \\\$
Hence, $+ 3$ is the oxidation number of $Cr$ in $C{r_2}{\left( {S{O_4}} \right)_3}$ .

Note: The more electronegative element in a substance is always allotted a negative oxidation state whereas, the less electronegative element is allotted a positive oxidation state. Always remember the fact that electronegativity is greatest at the top-right in the periodic table which declines toward the bottom-left.