Question

The paramagnetic complex ion which has no unpaired electrons in \( t_{2g} \) orbitals is

• A. \( [Fe(CN)_6]^{4-} \)
• B. \( [Fe(CN)_6]^{3-} \)
• C. \( [Zn(NH_3)_6]^{2+} \)
• D. \( [Ni(NH_3)_6]^{2+} \)

Hint:

When analyzing paramagnetic behavior, check the oxidation state of the central metal and the ligand strength. Strong field ligands cause electron pairing, while weak field ligands often lead to unpaired electrons in higher energy orbitals.

Answer 1

The Correct Option is D

Step 1: Identifying oxidation states and electronic configurations
- \( [Fe(CN)_6]^{4-} \)
- Fe is in the \( +2 \) oxidation state (\( 3d^6 \)).
- \( CN^- \) is a strong field ligand, causing pairing of electrons.
- The complex is low spin but still contains unpaired electrons in the \( t_{2g} \) orbitals.
- \( [Fe(CN)_6]^{3-} \)
- Fe is in the \( +3 \) oxidation state (\( 3d^5 \)).
- \( CN^- \) is a strong field ligand.
- The complex is low spin and has one unpaired electron in the \( t_{2g} \) orbitals.
- \( [Zn(NH_3)_6]^{2+} \)
- Zn is in the \( +2 \) oxidation state (\( 3d^{10} \)).
- All orbitals are completely filled, so it is diamagnetic.
- However, it does not belong to the category where paramagnetic behavior is considered.
- \( [Ni(NH_3)_6]^{2+} \)
- Ni is in the \( +2 \) oxidation state (\( 3d^8 \)).
- \( NH_3 \) is a weak field ligand, leading to high spin configuration.
- The \( t_{2g} \) orbitals are completely filled with paired electrons, making it the correct choice.
Step 2: Conclusion
Since the question asks for the paramagnetic complex with no unpaired electrons in \( t_{2g} \) orbitals, the correct answer is \( [Ni(NH_3)_6]^{2+} \).