The magnetic moments of `[Cu(NH_(3))_(4)]^(2+)` was found to be 1.73 B.M. The number of unpaired electrons in the complex is:

Options

A. 0
B. 1
C. 2
D. 3

Explanation

To find the number of unpaired electrons in the complex \([Cu(NH_3)_4]^{2+}\) given that its magnetic moment is 1.73 B.M., we can follow these steps:

### Step 1: Understand the Formula for Magnetic Moment
The magnetic moment (\(\mu\)) is given by the formula:
\[ \mu = \sqrt{n(n + 2)} \]
where \(n\) is the number of unpaired electrons.

### Step 2: Substitute the Given Magnetic Moment
We know from the question that the magnetic moment is 1.73 B.M. Thus, we can set up the equation:
\[ 1.73 = \sqrt{n(n + 2)} \]

### Step 3: Square Both Sides
To eliminate the square root, we square both sides of the equation:
\[ (1.73)^2 = n(n + 2) \]
Calculating \(1.73^2\):
\[ 1.73^2 = 2.9929 \approx 3 \]
So, we have:
\[ 3 = n(n + 2) \]

### Step 4: Rearrange the Equation
Rearranging gives us:
\[ n^2 + 2n - 3 = 0 \]

### Step 5: Factor the Quadratic Equation
Now, we will factor the quadratic equation:
\[ n^2 + 3n - n - 3 = 0 \]
This can be factored as:
\[ (n - 1)(n + 3) = 0 \]

### Step 6: Solve for \(n\)
Setting each factor to zero gives us:
1. \(n - 1 = 0 \Rightarrow n = 1\)
2. \(n + 3 = 0 \Rightarrow n = -3\)

Since the number of unpaired electrons cannot be negative, we discard \(n = -3\).

### Conclusion
Thus, the number of unpaired electrons in the complex \([Cu(NH_3)_4]^{2+}\) is:
\[ \boxed{1} \]

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