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 determine the magnetic moment (spin-only) and hybridization of the brown ring complex \([Fe(H_2O)_5NO]SO_4\), we can follow these steps:

### Step 1: Identify the oxidation state of Iron (Fe)
The complex can be broken down as follows:
- The sulfate ion \((SO_4^{2-})\) has a charge of -2.
- The ligand \((NO)\) is neutral and contributes 0.
- The five water molecules \((H_2O)\) are also neutral and contribute 0.

Let \(x\) be the oxidation state of iron. The overall charge of the complex is +2, so we can set up the equation:
\[ x + 0 + 0 - 2 = +2 \]
Solving for \(x\):
\[ x = +2 \]
Thus, the oxidation state of iron in this complex is +2.

### Step 2: Determine the electronic configuration of \(Fe^{2+}\)
The atomic number of iron (Fe) is 26. The electronic configuration of neutral iron is:
\[ [Ar] 4s^2 3d^6 \]
When iron loses two electrons to form \(Fe^{2+}\), the configuration becomes:
\[ [Ar] 3d^6 \]

### Step 3: Determine the number of unpaired electrons
In the \(3d^6\) configuration, the distribution of electrons in the \(d\) orbitals can be represented as follows:
- The \(3d\) orbitals will fill according to Hund's rule and the Pauli exclusion principle. Since water is a weak field ligand, it will not cause pairing of electrons.
- The configuration will be:
\[ \text{3d: } \uparrow \uparrow \uparrow \uparrow \downarrow \quad (\text{4 unpaired electrons}) \]
Thus, there are 4 unpaired electrons.

### Step 4: Calculate the magnetic moment
The formula for the spin-only magnetic moment \(\mu\) is given by:
\[ \mu = \sqrt{n(n + 2)} \]
where \(n\) is the number of unpaired electrons. Here, \(n = 4\):
\[ \mu = \sqrt{4(4 + 2)} = \sqrt{4 \times 6} = \sqrt{24} = 2\sqrt{6} \approx 4.9 \, \text{BM} \]

### Step 5: Determine the hybridization
In this complex, the central metal ion \(Fe^{2+}\) is surrounded by 5 water molecules and one nitrosyl ligand. The coordination number is 6, which suggests an octahedral geometry. The hybridization corresponding to an octahedral geometry is \(sp^3d^2\).

### Final Answer
- **Magnetic Moment**: \(2\sqrt{6} \, \text{BM} \) (approximately 4.9 BM)
- **Hybridization**: \(sp^3d^2\)