The complex `[Fe(H_(2)O)_(5)NO]^(2+)` is formed in Brown ring test for nitrates select correct statement for the complex ?

Options

A. Hybridisation of iron is `sp^(3)d^(2)`
B. Iron and nitrosyl has +1 oxidation state
C. It has magnetic moment of 3.87 B.M.
D. All are correct

Explanation

To solve the question regarding the complex \([Fe(H_2O)_5NO]^{2+}\) formed in the Brown ring test for nitrates, we will follow these steps:

### Step 1: Determine the Oxidation State of Iron
In the complex \([Fe(H_2O)_5NO]^{2+}\), we need to find the oxidation state of iron (Fe). The nitrosyl (NO) group has a +1 oxidation state in this complex.

Let the oxidation state of Fe be \(X\). The overall charge of the complex is +2. The equation can be set up as follows:

\[ X + 5(0) + 1 = +2 \]

This simplifies to:

\[ X + 1 = 2 \]

Thus, solving for \(X\):

\[ X = 2 - 1 = +1 \]

### Step 2: Determine the Electronic Configuration of Iron
Iron has an atomic number of 26, and its ground state electronic configuration is:

\[ [Ar] 3d^6 4s^2 \]

When iron is in the +1 oxidation state, it loses one electron, which will be removed from the 4s orbital first. Thus, the electronic configuration becomes:

\[ [Ar] 3d^6 4s^1 \]

### Step 3: Identify the Hybridization
The coordination number of the complex is 6 (5 water molecules and 1 nitrosyl group). For a coordination number of 6, the geometry is typically octahedral.

In an octahedral complex, the hybridization can be determined as follows:

- The 3d, 4s, and 4p orbitals are involved in hybridization.
- Since water is a weak field ligand, it does not cause pairing of electrons in the 3d orbitals.
- Therefore, the hybridization will involve one 4s, three 4p, and two 3d orbitals.

Thus, the hybridization is \(d^2sp^3\), indicating that it is an outer orbital complex.

### Step 4: Calculate the Magnetic Moment
The number of unpaired electrons in the complex is 3 (since the 3d subshell has 6 electrons, and there are no paired electrons due to the weak field nature of water).

The magnetic moment (\(\mu\)) can be calculated using the formula:

\[ \mu = \sqrt{n(n + 2)} \text{ Bohr magneton} \]

Where \(n\) is the number of unpaired electrons. Here, \(n = 3\):

\[ \mu = \sqrt{3(3 + 2)} = \sqrt{15} \approx 3.87 \text{ Bohr magneton} \]

### Conclusion
The correct statement regarding the complex \([Fe(H_2O)_5NO]^{2+}\) is that it has an oxidation state of +1 for iron, exhibits octahedral geometry, is an outer orbital complex, and has a magnetic moment of approximately 3.87 Bohr magneton.