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
The complex is \([Fe(H_2O)_5NO]SO_4\). The sulfate ion \((SO_4)^{2-}\) has a charge of -2. The water \((H_2O)\) is a neutral ligand, contributing 0 to the charge.

Let the oxidation state of iron be \(x\). The nitrogen oxide \((NO)\) has a charge of +1.

Setting up the equation based on the overall charge:
\[ x + 5(0) + 1 = +2 \]
This simplifies to:
\[ x + 1 = 2 \]
Thus, solving for \(x\):
\[ x = 2 - 1 = +1 \]
So, the oxidation state of iron is +1.

### Step 2: Determine the Electron Configuration of Iron
In the +1 oxidation state, iron has the electron configuration:
\[ [Ar] 3d^6 4s^1 \]
However, since \((NO)\) is a strong field ligand, it will cause pairing of electrons. The configuration will change to:
\[ 3d^7 4s^0 \]
This means there are 7 electrons in the 3d orbitals.

### Step 3: Determine the Number of Unpaired Electrons
In the \(3d^7\) configuration, the distribution of electrons in the d-orbitals will be:
- 3 unpaired electrons (as per Hund's rule).

Thus, \(n = 3\) (the number of unpaired electrons).

### Step 4: Calculate the Spin-Only Magnetic Moment
The formula for calculating the spin-only magnetic moment (\(\mu\)) is:
\[ \mu = \sqrt{n(n + 2)} \]
Substituting \(n = 3\):
\[ \mu = \sqrt{3(3 + 2)} = \sqrt{3 \times 5} = \sqrt{15} \text{ Bohr magneton} \]

### Step 5: Determine the Hybridization
The complex has 6 ligands (5 water molecules and 1 nitrogen oxide). The hybridization can be determined by the number of ligands:
- 6 ligands suggest \(d^2sp^3\) hybridization.

Thus, the hybridization of the complex is \(d^2sp^3\) or \(sp^3d^2\).

### Final Answers
- **Magnetic Moment**: \(\sqrt{15} \text{ Bohr magneton}\)
- **Hybridization**: \(d^2sp^3\)