Paramagnetism is given by the relation `mu=2sqrt(s(s+1))` magnetons where 's' is the total spin. On this basis, the paramagnetism of `Cu^+` ion is

Options

A. 3.88 magnetons
B. 2.83 magnetons
C. 1.41 magnatone
D. zero

Answer

To determine the paramagnetism of the \( Cu^+ \) ion using the formula \( \mu = 2\sqrt{s(s+1)} \), we will follow these steps:

### Step 1: Identify the electronic configuration of Copper
Copper (Cu) has an atomic number of 29. Its electronic configuration is:
\[ \text{Cu: } [Ar] \, 4s^1 \, 3d^{10} \]

### Step 2: Determine the electronic configuration of \( Cu^+ \)
When copper loses one electron to form \( Cu^+ \), it loses the 4s electron, resulting in:
\[ \text{Cu}^+: [Ar] \, 3d^{10} \]
This means that the \( Cu^+ \) ion has a completely filled 3d subshell.

### Step 3: Count the number of unpaired electrons
In the \( Cu^+ \) ion, the 3d subshell has 10 electrons, which are all paired. Therefore, the number of unpaired electrons (\( n \)) is:
\[ n = 0 \]

### Step 4: Calculate the total spin (\( s \))
The total spin (\( s \)) is calculated using the formula:
\[ s = \frac{1}{2} \times \text{(number of unpaired electrons)} \]
Since there are no unpaired electrons:
\[ s = \frac{1}{2} \times 0 = 0 \]

### Step 5: Substitute \( s \) into the paramagnetism formula
Now, we can substitute \( s \) into the paramagnetism formula:
\[ \mu = 2\sqrt{s(s+1)} \]
Substituting \( s = 0 \):
\[ \mu = 2\sqrt{0(0+1)} = 2\sqrt{0} = 0 \]

### Conclusion
The paramagnetism of the \( Cu^+ \) ion is:
\[ \mu = 0 \, \text{magnetons} \]

### Final Answer
Thus, the paramagnetism of the \( Cu^+ \) ion is **zero**.

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