Question

Two compounds of iron are given below:
$FeS{O_4}F{e_2}{(S{O_4})_3}$
(The oxidation state of sulfate radical is $- 2$ )
Which of these compounds show $+ 2$ oxidation state for $Fe$ ?

Answer 1

Oxidation number, also called oxidation state, the total number of electrons that an atom either gains or losses in order to form a chemical bond with another atom. Oxidation state indicates the degree of oxidation for an atom in a chemical compound; it is the hypothetical charge that an atom would have if all bonds to atoms of different elements were completely ionic. Oxidation states are typically represented by integers, which can be positive, negative, or zero.

Complete Step By Step Answer:
The oxidation state, sometimes referred to as oxidation number, describes the degree of oxidation (loss of electrons) of an atom in a chemical compound. Conceptually, the oxidation state, which may be positive, negative or zero, is the hypothetical charge that an atom would have if all bonds to atoms of different elements were $100\%$ ionic, with no covalent component. This is never exactly true for real bonds.
$FeS{O_4}$ The oxidation number calculated should be $0$ , as it is a neutral compound. and $FeS{O_4}$ show $+ 2$ oxidation state of $Fe$
For $FeS{O_4}$
$X + ( - 2) = 0$
$X = + 2$
For $F{e_2}{(S{O_4})_3}$
$2 \times X + (3 \times - 2) = 0$
$X = + 3$ .

Note:
A loss of negatively-charged electrons corresponds to an increase in oxidation number, while a gain of electrons corresponds to a decrease in oxidation number. Therefore, the element or ion that is oxidized undergoes an increase in oxidation number. The oxidation states of atoms in a given compound may vary depending on the choice of electronegativity scale used in their calculation.