If magnetic moment is zero the substance is diamagnetic.
If an ion of `25(Mn)` has a magnetic moment of 3.873 B.M. Then oxidation state of Mn in ion is :

Options

A. 2
B. 3
C. 4
D. 5

Answer

To determine the oxidation state of manganese (Mn) in the ion with a magnetic moment of 3.873 B.M., we can follow these steps:

### Step 1: Understand the relationship between magnetic moment and unpaired electrons
The magnetic moment (μ) is related to the number of unpaired electrons (n) in an atom or ion by the formula:
\[ \mu = \sqrt{n(n + 2)} \]
where μ is the magnetic moment in Bohr Magnetons (B.M.).

### Step 2: Set up the equation with the given magnetic moment
Given that the magnetic moment is 3.873 B.M., we can set up the equation:
\[ 3.873 = \sqrt{n(n + 2)} \]

### Step 3: Square both sides to eliminate the square root
Squaring both sides gives:
\[ (3.873)^2 = n(n + 2) \]
Calculating \( (3.873)^2 \):
\[ 15.036129 = n(n + 2) \]

### Step 4: Rearrange the equation
This can be rearranged to:
\[ n^2 + 2n - 15.036129 = 0 \]

### Step 5: Solve the quadratic equation
Using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = 2, c = -15.036129 \):
\[ n = \frac{-2 \pm \sqrt{(2)^2 - 4(1)(-15.036129)}}{2(1)} \]
Calculating the discriminant:
\[ n = \frac{-2 \pm \sqrt{4 + 60.144516}}{2} \]
\[ n = \frac{-2 \pm \sqrt{64.144516}}{2} \]
\[ n = \frac{-2 \pm 8.008}{2} \]
Calculating the two possible values for n:
1. \( n = \frac{6.008}{2} = 3.004 \) (approximately 3)
2. \( n = \frac{-10.008}{2} \) (not a valid solution since n cannot be negative)

Thus, \( n = 3 \).

### Step 6: Write the electronic configuration of Mn
The atomic number of manganese (Mn) is 25. The electronic configuration is:
\[ \text{Mn: } [Ar] 3d^5 4s^2 \]

### Step 7: Determine the oxidation state
Since we found that there are 3 unpaired electrons, we need to find how many electrons have been removed to achieve this configuration. The original configuration has 5 electrons in the 3d subshell and 2 in the 4s subshell.

To have 3 unpaired electrons, we can assume:
- 2 electrons are removed from the 4s subshell (4s^0).
- 2 electrons are removed from the 3d subshell, leaving 3 unpaired electrons (3d^3).

### Step 8: Calculate the oxidation state
In total, 4 electrons have been removed:
- 2 from 4s and 2 from 3d.

Thus, the oxidation state of Mn in this ion is:
\[ \text{Oxidation state} = +4 \]

### Final Answer
The oxidation state of Mn in the ion is **+4**.

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