The magnetic moment of `MnSO_(4).4H_(2)O` is:

Options

A. 1.73 BM
B. 3.87 BM
C. 4.09 BM
D. 5.92 BM

Answer

To find the magnetic moment of `MnSO4.4H2O`, we will follow these steps:

### Step 1: Determine the oxidation state of manganese (Mn) in `MnSO4`.
- The sulfate ion (SO4) has a charge of -2.
- The compound is neutral, so the sum of the oxidation states must equal 0.
- Let the oxidation state of Mn be \( x \).
- The equation is:
\[ x + (-2) = 0 \implies x = +2 \]

### Step 2: Write the electronic configuration of Mn and Mn²⁺.
- The atomic number of manganese (Mn) is 25.
- The electronic configuration of Mn is:
\[ [Ar] 3d^5 4s^2 \]
- For Mn²⁺, we remove 2 electrons (from 4s):
\[ [Ar] 3d^5 4s^0 \]

### Step 3: Analyze the electron configuration of Mn²⁺.
- The configuration of Mn²⁺ is \( 3d^5 \).
- Since water (H2O) is a weak field ligand, it does not cause pairing of electrons.
- Therefore, all 5 electrons in the 3d orbitals will remain unpaired.

### Step 4: Count the number of unpaired electrons.
- The number of unpaired electrons (N) in Mn²⁺ is 5.

### Step 5: Calculate the magnetic moment using the formula.
- The formula for magnetic moment (μ) is:
\[ \mu = \sqrt{N(N + 2)} \]
- Substituting the value of N:
\[ \mu = \sqrt{5(5 + 2)} = \sqrt{5 \times 7} = \sqrt{35} \]
- Calculating this gives:
\[ \mu \approx 5.92 \text{ Bohr Magneton} \]

### Conclusion
The magnetic moment of `MnSO4.4H2O` is **5.92 Bohr Magneton**.

---