Which of following species has the magnetic moment value of 3.87 BM? 

Options

A. `Fe^(3+)`
B. `Cr^(2+)`
C. `Co^(2+)`
D. `Cu^(+)`

Answer

To determine which species has a magnetic moment value of 3.87 BM, we will use the formula for magnetic moment:

\[ \mu = \sqrt{n(n + 2)} \]

where \( n \) is the number of unpaired electrons.

### Step 1: Calculate the value of \( n \)
Given that the magnetic moment \( \mu \) is 3.87 BM, we can set up the equation:

\[ 3.87 = \sqrt{n(n + 2)} \]

Squaring both sides gives:

\[ (3.87)^2 = n(n + 2) \]

Calculating \( (3.87)^2 \):

\[ 14.9769 = n(n + 2) \]

### Step 2: Solve the quadratic equation
Rearranging the equation:

\[ n^2 + 2n - 14.9769 = 0 \]

Using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

Here, \( a = 1, b = 2, c = -14.9769 \).

Calculating the discriminant:

\[ b^2 - 4ac = 2^2 - 4 \times 1 \times (-14.9769) = 4 + 59.9076 = 63.9076 \]

Now, calculating \( n \):

\[ n = \frac{-2 \pm \sqrt{63.9076}}{2} \]

Calculating \( \sqrt{63.9076} \):

\[ \sqrt{63.9076} \approx 7.99 \]

So,

\[ n = \frac{-2 \pm 7.99}{2} \]

Calculating the two possible values for \( n \):

1. \( n = \frac{5.99}{2} \approx 2.995 \) (approximately 3)
2. \( n = \frac{-9.99}{2} \) (not valid as \( n \) cannot be negative)

Thus, \( n \approx 3 \).

### Step 3: Identify the correct species
Now we need to check which of the given species has 3 unpaired electrons:

1. **Fe³⁺**: Electron configuration is \( [Ar] 3d^5 \). All 5 electrons are unpaired. \( n = 5 \) (not correct).
2. **Cr²⁺**: Electron configuration is \( [Ar] 3d^4 \). \( n = 4 \) (not correct).
3. **Co²⁺**: Electron configuration is \( [Ar] 3d^7 \). This has 3 unpaired electrons. \( n = 3 \) (correct).
4. **Cu⁺**: Electron configuration is \( [Ar] 3d^{10} \). All 10 electrons are paired. \( n = 0 \) (not correct).

### Conclusion
The species with a magnetic moment value of 3.87 BM is **Co²⁺**.

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