The magnetic moment of `K_3[Fe(CN)_6]` is found to be `1.7 B.M`. How many unpaired electron(s) is/are present per molecule?

Options

A. (a) `2`
B. (b) `3`
C. (c) `4`
D. (d) `1`

Explanation

To determine the number of unpaired electrons in the complex `K_3[Fe(CN)_6]` based on the given magnetic moment of `1.7 B.M`, we can follow these steps:

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

### Step 2: Substitute the given magnetic moment value
We are given that the magnetic moment \( \mu = 1.7 \, B.M \). We can substitute this value into the formula:
\[ 1.7 = \sqrt{n(n + 2)} \]

### Step 3: Square both sides to eliminate the square root
Squaring both sides gives us:
\[ (1.7)^2 = n(n + 2) \]
Calculating \( (1.7)^2 \):
\[ 2.89 = n(n + 2) \]

### Step 4: Rearrange the equation
Rearranging the equation gives us:
\[ n^2 + 2n - 2.89 = 0 \]

### Step 5: Solve the quadratic equation
We can solve this quadratic equation using the quadratic formula:
\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = 2 \), and \( c = -2.89 \):
\[ n = \frac{-2 \pm \sqrt{(2)^2 - 4 \cdot 1 \cdot (-2.89)}}{2 \cdot 1} \]
Calculating the discriminant:
\[ n = \frac{-2 \pm \sqrt{4 + 11.56}}{2} \]
\[ n = \frac{-2 \pm \sqrt{15.56}}{2} \]
Calculating \( \sqrt{15.56} \):
\[ n = \frac{-2 \pm 3.94}{2} \]

### Step 6: Calculate the possible values for n
Calculating the two possible values:
1. \( n = \frac{-2 + 3.94}{2} = \frac{1.94}{2} \approx 0.97 \) (not a valid solution since n must be a whole number)
2. \( n = \frac{-2 - 3.94}{2} = \frac{-5.94}{2} \) (not valid as n cannot be negative)

### Step 7: Use approximation for unpaired electrons
Since we know that the magnetic moment is approximately 1.7, we can infer that the number of unpaired electrons is likely to be 1, as the magnetic moment values typically correlate with the number of unpaired electrons.

### Conclusion
Thus, the number of unpaired electrons in `K_3[Fe(CN)_6]` is **1**.