An ion `(Mn^(a+))` has the magnetic moment equal to `4.9` B.M` What is the value of (a) :

Answer

To solve the problem of determining the value of \( a \) in the ion \( Mn^{(a+)} \) with a magnetic moment of \( 4.9 \, B.M. \), we can follow these steps:

### Step 1: Use the formula for magnetic moment
The magnetic moment (\( \mu \)) of a transition metal ion can be calculated using the formula:
\[ \mu = \sqrt{n(n + 2)} \, B.M. \]
where \( n \) is the number of unpaired electrons.

### Step 2: Set up the equation
Given that the magnetic moment is \( 4.9 \, B.M. \), we can set up the equation:
\[ \sqrt{n(n + 2)} = 4.9 \]

### Step 3: Square both sides
To eliminate the square root, we square both sides of the equation:
\[ n(n + 2) = (4.9)^2 \]
Calculating \( (4.9)^2 \):
\[ (4.9)^2 = 24.01 \]
Thus, we have:
\[ n(n + 2) = 24.01 \]

### Step 4: Rearrange the equation
Rearranging the equation gives us a quadratic equation:
\[ n^2 + 2n - 24.01 = 0 \]

### Step 5: Solve the quadratic equation
We can use the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = 2, c = -24.01 \):
\[ b^2 - 4ac = 2^2 - 4(1)(-24.01) = 4 + 96.04 = 100.04 \]
Now, applying the quadratic formula:
\[ n = \frac{-2 \pm \sqrt{100.04}}{2} = \frac{-2 \pm 10}{2} \]
Calculating the two possible values for \( n \):
1. \( n = \frac{8}{2} = 4 \)
2. \( n = \frac{-12}{2} = -6 \) (not a valid solution)

Thus, \( n = 4 \).

### Step 6: Determine the oxidation state of manganese
Manganese (Mn) has an atomic number of 25 and its ground state electronic configuration is:
\[ [Ar] 3d^5 4s^2 \]
To have 4 unpaired electrons, we need to remove 3 electrons. The removal of 2 electrons from the 4s subshell and 1 electron from the 3d subshell gives us:
\[ Mn^{3+} \rightarrow [Ar] 3d^4 \]
This means the oxidation state \( a \) is 3.

### Final Answer
The value of \( a \) is \( 3 \).

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