The comple `[Fe(H_(2)O)_(6)NO]^(2+)` is formed in the brown ring test fro nitrates when freshly prepared `FesO_(4)` solution is added to aqueous solution of `NO_(3)^(-)` ions followed by addition of conc. `H_(2)SO_(4)`. Select correct statement about this complex.

Options

A. Hybridisation of iron is `sp^(3)d^(2)`
B. Iron has +1 oxidation state.
C. It has magnetic moment of 3.87 B.M. confirming three unpaired electrons is Fe.
D. All the above are correct statement.

Answer

To solve the question regarding the complex \([Fe(H_2O)_5NO]^{2+}\) formed in the brown ring test for nitrates, we will analyze the given statements step by step.

### Step 1: Identify the Complex and Its Components
The complex given is \([Fe(H_2O)_5NO]^{2+}\). Here, we have:
- \(Fe\) (Iron) as the central metal atom.
- \(H_2O\) (water) as a neutral ligand.
- \(NO\) (nitrosyl) as a ligand with a +1 charge.

### Step 2: Determine the Oxidation State of Iron
Let the oxidation state of Iron be \(x\). The total charge of the complex is \(+2\). The equation can be set up as follows:

\[ x + (0 \times 5) + (+1) = +2 \]

This simplifies to:

\[ x + 1 = 2 \]

Solving for \(x\):

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

Thus, the oxidation state of Iron in the complex is +1.

### Step 3: Determine the Electronic Configuration of Iron
The atomic number of Iron (Fe) is 26. Its electronic configuration is:

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

When Iron is in the +1 oxidation state, it loses one electron, resulting in:

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

### Step 4: Determine the Hybridization
In the complex, Iron is surrounded by 6 ligands (5 water molecules and 1 nitrosyl). The hybridization can be determined based on the number of ligands:

- 6 ligands suggest \(sp^3d^2\) hybridization.

### Step 5: Determine the Number of Unpaired Electrons
The electronic configuration of \(Fe^{+1}\) is \(4s^1 3d^6\). In the \(3d\) subshell, the configuration can be represented as:

- 3d: ↑ ↑ ↑ ↑ ↑ (6 electrons, 3 unpaired)

Thus, there are 3 unpaired electrons.

### Step 6: Calculate the Magnetic Moment
The magnetic moment (\(\mu\)) can be calculated using the formula:

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

Where \(n\) is the number of unpaired electrons. For 3 unpaired electrons:

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

### Conclusion
Now, let's summarize the findings:
1. The oxidation state of Iron is +1.
2. The hybridization of Iron is \(sp^3d^2\).
3. The magnetic moment is approximately 3.87 Bohr magneton, confirming 3 unpaired electrons.

### Final Answer
All the statements provided about the complex \([Fe(H_2O)_5NO]^{2+}\) are correct.