Brown coloured ring formed in nitrate test has formula `[Fe(H_(2)O)_(5)NO]SO_(4)`. Unpaired electron in `Fe` is

Explanation

To determine the number of unpaired electrons in the iron (Fe) present in the compound `[Fe(H₂O)₅NO]SO₄`, we will follow these steps:

### Step 1: Identify the oxidation state of iron (Fe)
The formula given is `[Fe(H₂O)₅NO]SO₄`. In this complex:
- H₂O is a neutral ligand, contributing 0 to the charge.
- The nitrosonium ion (NO) has a charge of +1.
- The sulfate ion (SO₄) has a charge of -2.

Let the oxidation state of iron be \( x \). The overall charge of the complex is +2 (since it is balanced by the sulfate ion). Therefore, we can set up the equation:

\[ x + 0 \cdot 5 + 1 = +2 \]

From this, we can solve for \( x \):

\[ x + 1 = 2 \implies x = 2 - 1 = +1 \]

### Step 2: Determine the electron configuration of iron in the +1 oxidation state
Iron (Fe) has an atomic number of 26, with the electron configuration of:

\[ \text{Fe: } [Ar] 3d^6 4s^2 \]

In the +1 oxidation state, iron loses one electron. The electron is removed from the 4s orbital first, leading to the configuration:

\[ \text{Fe}^{+1}: [Ar] 3d^6 \]

### Step 3: Analyze the 3d subshell for unpaired electrons
The 3d subshell can hold a maximum of 10 electrons. The 3d electrons are filled according to Hund's rule, which states that every orbital in a subshell is singly occupied before any orbital is doubly occupied.

For 3d^6, the filling will look like this:

- 3d: ↑ ↑ ↑ ↑ ↑ (5 electrons in separate orbitals) and then one pair in one of the orbitals.

This means that there will be 4 unpaired electrons (the first five are singly occupied and one additional electron pairs with one of them).

### Conclusion
Thus, the number of unpaired electrons in the iron (Fe) in the compound `[Fe(H₂O)₅NO]SO₄` is **4**.