Question

Oxidation state of Fe in Mohr’s salt $\left[FeS{O}_4\cdot {\left(N{H}_4\right)}_2\cdot 6H_{2}O\right]$
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 $\left[FeS{O}_4\cdot {\left(N{H}_4\right)}_2\cdot 6H_{2}O\right]$.
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 $\left[FeS{O}_4\cdot {\left(N{H}_4\right)}_2\cdot 6H_{2}O\right]$ 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 $\left[FeS{O}_4\cdot {\left(N{H}_4\right)}_2\cdot 6H_{2}O\right]$ 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.