What is the magnetic moment ( spin only) and hybridisation of the brown ring complex `[Fe(H_(2)O)_(5)NO]SO_(4)` ?

Options

A. `sqrt(3)BM,sp^(3)d^(2)`
B. `sqrt(3)BM,d^(2)sp^(3)`
C. `sqrt(15)BM,sp^(3)d^(2)`
D. `sqrt(15)BM,d^(2)sp^(3)`

Explanation

To solve the question regarding the magnetic moment (spin-only) and hybridization of the brown ring complex \([Fe(H_2O)_5NO]SO_4\), we will follow these steps:

### Step 1: Determine the Oxidation State of Iron
1. The complex is \([Fe(H_2O)_5NO]^{2+}\) with a sulfate ion \(SO_4^{2-}\) outside the coordination sphere.
2. Water (\(H_2O\)) is a neutral ligand, and nitric oxide (\(NO\)) is a cationic ligand with a charge of +1.
3. Let the oxidation state of iron be \(x\).
4. The equation for the charge balance is:
\[ x + 0 + 1 = +2 \]
Thus, \(x = +1\).

### Step 2: Determine the Electron Configuration of Iron
1. The atomic number of iron (Fe) is 26.
2. The electron configuration of neutral iron is \( [Ar] 3d^6 4s^2 \).
3. In the +1 oxidation state, iron loses one electron from the 4s orbital:
\[ \text{Fe}^{+1}: [Ar] 3d^6 4s^1 \]

### Step 3: Determine the Number of Unpaired Electrons
1. In the \(3d\) subshell, we have 6 electrons.
2. The distribution of electrons in the \(3d\) orbitals will be:
- 3d: ↑↑↑↑↑ (one pairing occurs due to the presence of the strong field ligand \(NO\)).
3. The presence of the weak field ligands (5 \(H_2O\)) will not cause further pairing.
4. Thus, the number of unpaired electrons is 4 (3 in \(3d\) and 1 in \(4s\)).

### Step 4: Calculate the Magnetic Moment
1. The formula for the spin-only magnetic moment is:
\[ \mu = \sqrt{n(n+2)} \]
where \(n\) is the number of unpaired electrons.
2. Here, \(n = 4\):
\[ \mu = \sqrt{4(4+2)} = \sqrt{4 \times 6} = \sqrt{24} = 2\sqrt{6} \text{ Bohr magnetons} \]

### Step 5: Determine the Hybridization
1. The complex has 6 ligands (5 \(H_2O\) and 1 \(NO\)).
2. The hybridization can be determined by the number of ligands:
- 6 ligands suggest \(sp^3d^2\) hybridization.
3. The hybridization is \(sp^3d^2\) due to the involvement of 3 \(d\) orbitals, 1 \(s\) orbital, and 2 \(p\) orbitals.

### Final Answer
- The magnetic moment (spin-only) of the complex \([Fe(H_2O)_5NO]SO_4\) is \(2\sqrt{6}\) Bohr magnetons.
- The hybridization of the complex is \(sp^3d^2\).