The magneitc moment of `._(25)Mn` in ionic state is `4.83` B.M., then Mn is in :

Options

A. `+2` state
B. `+3` state
C. `+4` state
D. `+5` state

Explanation

To determine the oxidation state of manganese (Mn) in its ionic form based on the given magnetic moment, we can follow these steps:

### Step 1: Understand the magnetic moment formula
The magnetic moment (μ) is calculated using the formula:
\[ \mu = \sqrt{n(n + 2)} \]
where \( n \) is the number of unpaired electrons.

### Step 2: Use the given magnetic moment
We are given that the magnetic moment of Mn in its ionic state is 4.83 B.M. We can set this equal to the formula:
\[ 4.83 = \sqrt{n(n + 2)} \]

### Step 3: Square both sides to eliminate the square root
Squaring both sides gives:
\[ (4.83)^2 = n(n + 2) \]
Calculating \( (4.83)^2 \):
\[ 23.3289 = n(n + 2) \]

### Step 4: Rearrange the equation
This can be rearranged to form a quadratic equation:
\[ n^2 + 2n - 23.3289 = 0 \]

### Step 5: Solve the quadratic equation
Using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = 2, c = -23.3289 \):
\[ n = \frac{-2 \pm \sqrt{(2)^2 - 4 \cdot 1 \cdot (-23.3289)}}{2 \cdot 1} \]
Calculating the discriminant:
\[ n = \frac{-2 \pm \sqrt{4 + 93.3156}}{2} \]
\[ n = \frac{-2 \pm \sqrt{97.3156}}{2} \]
Calculating \( \sqrt{97.3156} \approx 9.86 \):
\[ n = \frac{-2 \pm 9.86}{2} \]
Calculating the two possible values for \( n \):
1. \( n = \frac{7.86}{2} \approx 3.93 \) (approximately 4)
2. \( n = \frac{-11.86}{2} \) (not a valid solution since \( n \) cannot be negative)

Thus, we find \( n \approx 4 \).

### Step 6: Determine the oxidation state
Manganese has an atomic number of 25, and its electron configuration is:
\[ [Ar] 4s^2 3d^5 \]
In the ionic state, if there are 4 unpaired electrons, we can conclude that 3 electrons have been removed (2 from 4s and 1 from 3d), leading to an oxidation state of +3.

### Final Answer
Thus, the oxidation state of Mn in its ionic state is +3.

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