Presence of nitrate ions can be easily detected in a solution by adding to it a freshly prepared $FeSO_4$, solution. Subsequently few drops of concentrated sulphuric acid are added along the walls of the test-tube. This results in the formation of a brown ring at the interface (Brown Ring Test) having the formula. $[Fe(H_2O)_5(NO)]^{2+}$.

The oxidation state of Fe in this complex is _____.

(+1)

The correct answer is option 1. (+1).

The Brown Ring Test is used to detect nitrate ions \((NO_3^-)\) in a solution. In this test, nitrate is reduced to \(NO\) (nitric oxide) by ferrous sulfate \((FeSO_4)\) in the presence of concentrated sulfuric acid. The resulting \(NO\) (nitric oxide) forms a complex with \(Fe^{2+}\) ions, producing the characteristic brown ring complex \([Fe(H_2O)_5(NO)]^{2+}\).

Structure of the Complex \([Fe(H_2O)_5(NO)]^{2+}\):

The complex consists of:

\(Fe\) (iron), which is the central metal ion.

5 \(H_2O\) (water) ligands, which are neutral

\(NO\) (nitrosyl) ligand, which can exist in different forms:

As a neutral ligand (NO).

As a positively charged nitrosonium ion \((NO^+)\).

Oxidation State Calculation:

The overall charge on the complex ion is +2. Each water molecule (H₂O) is a neutral ligand and does not contribute to the charge. NO (nitrosyl) in this case is considered as NO⁺ (nitrosonium ion), contributing +1 to the overall charge.

Let the oxidation state of Fe be \(x\).

Now, use the equation for the total charge of the complex:

\(\text{Oxidation state of }Fe + \text{Charge of }NO^+ = \text{Overall charge of the complex}\)

\(x + 1 = +2\)

Solving for \)x\):

\(x = +2 - 1 = +1\)

Therefore, the oxidation state of \(Fe\) in this complex is +1

Summary:

In the brown ring complex \([Fe(H_2O)_5(NO)]^{2+}\), the iron exists in the +1 oxidation state because:

The overall charge on the complex is +2.

The nitrosyl ligand is contributing +1 as \(NO^+\).

The water ligands are neutral and do not affect the charge.