Question

The brown ring complex compound is formulated as $\left[\mathrm{Fe}{\left({\mathrm{H}}_2\mathrm{O}\right)}_5\mathrm{NO}\right]{\mathrm{SO}}_4$. The oxidation state of Iron is ?

• A. $+1$
• B. $+2$
• C. $+3$
• D. $+6$

Answer 1

The correct option is A

$+1$

Explanation for the correct option:

Option(A):$+1$

• The chemical test called a Nitrate test is conducted to find out the presence of Nitrate ions in any solution, in which the solutions form a brown ring in the test tube.
• The above test is also called the brown ring test, in which the brown ring complex compound is formulated as $\left[\mathrm{Fe}{\left({\mathrm{H}}_2\mathrm{O}\right)}_5\mathrm{NO}\right]{\mathrm{SO}}_4$.
• Here the oxidation state of Iron in the complex $\left[\mathrm{Fe}{\left({\mathrm{H}}_2\mathrm{O}\right)}_5\mathrm{NO}\right]{\mathrm{SO}}_4$, can be found as:
• We can assume that the oxidation state of Iron metal is x.
• Here the oxidation state of the Sulfate ion is $-2$, the oxidation state of water is zero and the oxidation state of Nitroso $\text{NO}$ is $+1$.
• Hence, the oxidation state of Iron can be calculated as:
  $$\begin{gathered} \Rightarrow \mathrm{X}+(+1)+5\left(0\right)=+2 \\ \Rightarrow \mathrm{X}=+1 \end{gathered}$$

Thus, the oxidation state of Iron in the brown ring complex compound is $+1$.