Question

Oxidation state of Fe in Mohr’s salt $\left[ {FeS{O_4} \cdot {{\left( {N{H_4}} \right)}_2} \cdot 6{H_2}O} \right]$
A.$ + 1$
B.$ + 2$
C.$ + 3$
D.$ + 4$

Answer 1

Hint: We can define oxidation state as oxidation degree (loss of an electron) in a chemical compound. We can calculate the oxidation state by an element in a compound by the help of the rules of oxidation numbers.

Complete step by step answer:
We know that oxidation state is the loss of an electron in a chemical compound. We can now see a few rules of oxidation numbers.
-A free element would have zero as its oxidation number.
-Monatomic ions would have an oxidation number same as the charge of the ion.
-In hydrogen, the oxidation number is ${\text{ + 1,}}$ when combined with elements showing less electronegativity, the oxidation number of hydrogen would -1.
-In compounds of oxygen, the oxidation number of oxygen would be -2 and in peroxides, the oxidation state would be -1.
-Elements present in group 1 could contain +1 oxidation number.
-Elements present in group 2 could contain +2 oxidation numbers.
-Elements in group 17 could contain -1 oxidation number.
-Total of oxidation number of all atoms seen in neutral compounds is zero.
-In polyatomic ions, the sum of the oxidation number is similar to the ionic charge.
-The given compound is $\left[ {FeS{O_4} \cdot {{\left( {N{H_4}} \right)}_2} \cdot 6{H_2}O} \right]$.
Let us keep the oxidation state of $Fe$ as x.
-The oxidation number of $N{H_4}$ is $ + 1$
-The oxidation number of $S{O_4}$ is $ - 2$
-The oxidation number of $H$ is $ + 1$
-The oxidation number of $O$ is $ - 2$
-The oxidation number of $Fe$ in $\left[ {FeS{O_4} \cdot {{\left( {N{H_4}} \right)}_2} \cdot 6{H_2}O} \right]$ is calculated as,
$x + 2\left( { + 1} \right) + 2\left( { - 2} \right) + 6\left( {2\left( 1 \right)2} \right) = 0$
$ \Rightarrow x + 2 - 4 + 6\left[ {2 - 2} \right] = 0$
On simplifying we get,
$ \Rightarrow x - 2 + 0 = 0$
$ \Rightarrow x - 2 = 0$
$ \Rightarrow x = + 2$
The oxidation state of $Fe$ in $\left[ {FeS{O_4} \cdot {{\left( {N{H_4}} \right)}_2} \cdot 6{H_2}O} \right]$ is $ + 2$.
Therefore, the option (B) is correct.

Note: We have to remember that the Mohr salt is otherwise called ammonium ferrous sulfate or ammonium iron(II)sulfate. We know that it is the inorganic compound that has two different cations $F{e^{2 + }}$ and $N{H_4}^ + $. It comes under a double salt of ferrous sulfate and ammonium sulfate.
In ferrous sulfate, a polyatomic anion which has -2 charges is sulfate. Therefore, +2 charged iron is required to neutralize the -2 charge of sulfate ion. Hence, the oxidation state of Fe in Mohr salt is +2.