Question

Amongst \( \text{Ni(CO)}_4 \), \( [\text{Ni(CN)}_4]^{2-} \), and \( [\text{NiCl}_4]^{2-} \), which is paramagnetic?

• A. \( \text{Ni(CO)}_4 \) and \( \text{NiCl}_4^{2-} \) are diamagnetic but \( [\text{Ni(CN)}_4]^{2-} \) is paramagnetic
• B. \( \text{Ni(CO)}_4 \) and \( [\text{Ni(CN)}_4]^{2-} \) are diamagnetic but \( \text{NiCl}_4^{2-} \) is paramagnetic
• C. \( \text{NiCl}_4^{2-} \) and \( [\text{Ni(CN)}_4]^{2-} \) are diamagnetic but \( \text{Ni(CO)}_4 \) is paramagnetic
• D. \( \text{Ni(CO)}_4 \) is diamagnetic but \( [\text{NiCl}_4]^{2-} \) and \( [\text{Ni(CN)}_4]^{2-} \) are paramagnetic

Hint:

Paramagnetic complexes have unpaired electrons, while diamagnetic complexes have all electrons paired.

Answer 1

The Correct Option is B

Step 1: Understanding the magnetic properties of the complexes.
The magnetic properties of these complexes depend on the presence of unpaired electrons. In \( \text{NiCl}_4^{2-} \), the nickel ion has unpaired electrons, making it paramagnetic. The other two complexes are diamagnetic due to paired electrons.
Step 2: Conclusion.
Thus, \( \text{NiCl}_4^{2-} \) is paramagnetic while the other two are diamagnetic.
Final Answer: \[ \boxed{B} \]