How many electrons are present in `t_(2g)` set of d-orbitals of central metal cation in `[Fe(H_(2)O)_(5)(NO)]SO_(4)` brown ring complex?

Explanation

To determine how many electrons are present in the \( t_{2g} \) set of d-orbitals of the central metal cation in the brown ring complex \([Fe(H_2O)_5(NO)]SO_4\), we can follow these steps:

### Step 1: Identify the oxidation state of iron in the complex
The complex can be broken down into its components:
- The coordination part is \([Fe(H_2O)_5(NO)]\).
- The sulfate ion \(SO_4^{2-}\) is the counter ion.

To find the oxidation state of iron (let's denote it as \( x \)):
- The neutral ligands \( H_2O \) contribute \( 0 \) to the charge.
- The nitrosyl \( NO \) is a neutral ligand, contributing \( 0 \) as well.

Thus, the equation for the oxidation state becomes:
\[ x + 0 + 0 = +1 \]
This indicates that iron is in the \( +1 \) oxidation state.

### Step 2: Write the electron configuration of iron in the \( +1 \) oxidation state
Iron (Fe) has the atomic number 26. The ground state electron configuration of neutral iron is:
\[ [Ar] 4s^2 3d^6 \]

For \( Fe^{+1} \), one electron is removed from the \( 4s \) orbital:
\[ Fe^{+1}: [Ar] 4s^1 3d^6 \]

### Step 3: Determine the total number of d-electrons
In the \( +1 \) oxidation state, the total number of d-electrons is \( 6 \) (from \( 3d^6 \)) and \( 1 \) (from \( 4s^1 \)), giving a total of \( 7 \) valence electrons.

### Step 4: Analyze the ligand field and electron distribution
The ligands in this complex are water and nitrosyl, which are considered weak field ligands. This means that the splitting energy (\( \Delta \)) is less than the pairing energy, leading to the following filling of the d-orbitals:
- The \( 3d \) orbitals split into \( t_{2g} \) and \( e_g \) sets.
- The \( t_{2g} \) set consists of three orbitals: \( d_{xy}, d_{xz}, d_{yz} \).
- The \( e_g \) set consists of two orbitals: \( d_{x^2-y^2}, d_{z^2} \).

### Step 5: Fill the d-orbitals according to Hund's rule
Since we have \( 7 \) electrons to fill:
1. Fill each of the \( t_{2g} \) orbitals singly first (3 electrons).
2. Then fill the \( e_g \) orbitals (2 electrons).
3. Pair one electron in the \( t_{2g} \) set (1 electron).

The distribution will be:
- \( t_{2g} \): 4 electrons (3 singly + 1 paired)
- \( e_g \): 2 electrons (both singly)

### Step 6: Count the electrons in the \( t_{2g} \) set
Thus, the total number of electrons in the \( t_{2g} \) set is \( 4 \).

### Final Answer
The number of electrons present in the \( t_{2g} \) set of d-orbitals of the central metal cation in \([Fe(H_2O)_5(NO)]SO_4\) is **4**.

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