Question

The oxidation state of $Fe$ in the brown ring complex $\left[Fe\right(H_{2}O{)}_{5}NO]S{O}_4$ is

• A. 2
• B. 1
• C. 3

Answer 1

The correct option is B 1

This is a straightforward numerical, what we need to do is figure out the complex here, which happens to be the positive ion attaached to the $S{O}_4^{2-}$ ion.
Clearly, the positive ion must have a charge of +2 to get a neutral molecule.
You also know that the nitrosonium ion has a charge of +1, and the hydrate ligand would have no charge. We can form the following equation:

Let the oxidation state of $Fe$ be $x$
$x+(5\times H_{2}O)+NO=+2$
$x+(5\times 0)+1=+2$
$\therefore x=+1$

This is quite unique since the oxidation number of 1 is usually unheard of for iron!

Some extra information:
The brown ring test is test used to test the presence of the $N{O}_3^{-}$ ion.
You dissolve the given compond in sulphuric acid and iron (II) sulphare. If you get a brown coloured ring at the top of the solution, then presence of nitrate is confirmed. The complex in the question is the compound that forms this brown ring.
The following reactions take place:

1) $2HN{O}_3+3H2S{O}_4+6FeS{O}_4\to 3F{e}_2(S{O}_4{)}_3+2NO+4H_{2}O$
2) $\left[Fe\right(H_{2}O{)}_6]S{O}_4+NO\rightleftharpoons [Fe\left(H_{2}O{)}_5\right(NO\left)\right]S{O}_4+H_{2}O$