To determine which complex shows a magnetic moment of 1.73 B.M., we need to analyze the magnetic properties of the given complexes based on the number of unpaired electrons. A magnetic moment of 1.73 B.M. indicates the presence of one unpaired electron, as the magnetic moment (μ) can be calculated using the formula:
\[
\mu = \sqrt{n(n+2)}
\]
where \( n \) is the number of unpaired electrons. For \( \mu = 1.73 \) B.M., we can set up the equation:
\[
1.73 = \sqrt{n(n+2)}
\]
Squaring both sides gives:
\[
1.73^2 = n(n+2)
\]
Calculating \( 1.73^2 \):
\[
2.9929 \approx n(n+2)
\]
Now we can test integer values for \( n \):
1. If \( n = 1 \):
\[
1(1+2) = 3 \quad \text{(not equal to 2.9929)}
\]
2. If \( n = 2 \):
\[
2(2+2) = 8 \quad \text{(not equal to 2.9929)}
\]
3. If \( n = 3 \):
\[
3(3+2) = 15 \quad \text{(not equal to 2.9929)}
\]
Since \( n = 1 \) is the only integer that gives a value close to 1.73 B.M., we conclude that the complex must have one unpaired electron.
Now, let's analyze the given complexes to find out which one has one unpaired electron.
### Step 1: Analyze the first complex, \( \text{Cu(NH}_3\text{)}_4^{2+} \)
1. **Determine the oxidation state of Copper (Cu)**:
- Assume the oxidation state of Cu is \( x \).
- The charge of the complex is \( +2 \).
- Amine (NH3) is neutral.
- Therefore, \( x + 0 = +2 \) implies \( x = +2 \).
2. **Determine the electron configuration of \( \text{Cu}^{2+} \)**:
- The electron configuration of Cu is \( [Ar] 4s^1 3d^{10} \).
- For \( \text{Cu}^{2+} \), it loses one 4s electron and one 3d electron, resulting in \( 3d^9 \).
3. **Draw the orbital configuration for \( 3d^9 \)**:
- The \( 3d \) subshell has 9 electrons, which means there will be one unpaired electron.
4. **Hybridization**:
- With 4 ligands (NH3), the hybridization is \( sp^3 \).
5. **Conclusion**:
- Since there is one unpaired electron, \( \text{Cu(NH}_3\text{)}_4^{2+} \) is paramagnetic and has a magnetic moment of 1.73 B.M.
### Step 2: Analyze the second complex, \( \text{Ni(CN)}_4^{2-} \)
1. **Determine the oxidation state of Nickel (Ni)**:
- Assume the oxidation state of Ni is \( x \).
- The charge of the complex is \( -2 \).
- Therefore, \( x - 4 = -2 \) implies \( x = +2 \).
2. **Determine the electron configuration of \( \text{Ni}^{2+} \)**:
- The electron configuration of Ni is \( [Ar] 4s^2 3d^8 \).
- For \( \text{Ni}^{2+} \), it loses two 4s electrons, resulting in \( 3d^8 \).
3. **Draw the orbital configuration for \( 3d^8 \)**:
- The \( 3d \) subshell has 8 electrons, which means all electrons are paired.
4. **Hybridization**:
- With 4 ligands (CN), the hybridization is \( dsp^2 \).
5. **Conclusion**:
- Since there are no unpaired electrons, \( \text{Ni(CN)}_4^{2-} \) is diamagnetic.
### Final Conclusion:
The complex that shows a magnetic moment of 1.73 B.M. is \( \text{Cu(NH}_3\text{)}_4^{2+} \) (Option A).
