Question

Among the following complex ions, the one which is EPR active is:

• A. \( \text{Ni(CO)}_4 \)
• B. \( [\text{Co(NH}_3)_3\text{Cl}]^{2+} \)
• C. \( [\text{Cu(C}_2\text{O}_4)_2]^{2-} \)
• D. \( [\text{Mo(CO)}_6] \)

Hint:

For a complex to be EPR active, it must have unpaired electrons, usually due to the metal being in a lower oxidation state.

Answer 1

The Correct Option is C

Step 1: Understanding EPR activity.
Electron Paramagnetic Resonance (EPR) activity depends on the presence of unpaired electrons in a molecule or complex. A complex is EPR active if it has one or more unpaired electrons.
Step 2: Analyzing the options.
- (1) \( \text{Ni(CO)}_4 \): Nickel in this complex is in the 0 oxidation state, and the complex has no unpaired electrons, making it non-EPR active. - (2) \( [\text{Co(NH}_3)_3\text{Cl}]^{2+} \): Cobalt in this complex is in the +2 oxidation state and has no unpaired electrons, making it non-EPR active. - (3) \( [\text{Cu(C}_2\text{O}_4)_2]^{2-} \): Copper in this complex is in the +2 oxidation state, and the complex has unpaired electrons, making it EPR active. - (4) \( [\text{Mo(CO)}_6] \): Molybdenum in this complex is in the 0 oxidation state, and the complex has no unpaired electrons, making it non-EPR active.
Step 3: Conclusion.
The correct answer is (1) \( \text{Ni(CO)}_4 \), as it is the only complex with unpaired electrons, making it EPR active.

Answer 2

A species is EPR (Electron Paramagnetic Resonance) active only if it has at least one unpaired electron, since EPR detects transitions between electron spin states. Let's find the d-electron count and pairing for the metal centre in each complex.

1. \( \text{Ni(CO)}_4 \): Here nickel is in the zero oxidation state, so it has a \( d^{10} \) configuration. All ten d-electrons are paired in this configuration, and CO is a strong-field ligand that does not change the electron count, so there are no unpaired electrons. This complex is not EPR active.
2. \( [\text{Co(NH}_3)_3\text{Cl}]^{2+} \): Cobalt here is in the +3 oxidation state, giving a \( d^{6} \) configuration. With strong-field ligands like ammonia, this typically gives a low-spin \( t_{2g}^{6}e_g^{0} \) arrangement, in which all electrons are paired. This complex is not EPR active.
3. \( [\text{Cu(C}_2\text{O}_4)_2]^{2-} \): Copper here is in the +2 oxidation state, giving a \( d^{9} \) configuration. Nine electrons can never be fully paired in five d-orbitals, since pairing ten electrons would need all five orbitals doubly filled; one orbital is always left with a single, unpaired electron. This unpaired electron is present regardless of the ligand field, making this complex paramagnetic and EPR active.
4. \( [\text{Mo(CO)}_6] \): Molybdenum here is in the zero oxidation state, giving a \( d^{6} \) configuration, and CO being a strong-field ligand forces a low-spin \( t_{2g}^{6} \) arrangement with all electrons paired. This complex is not EPR active.

Only the copper complex has an odd number of d-electrons, which guarantees an unpaired electron no matter how the ligand field splits the orbitals.

Therefore, the correct answer is \( [\text{Cu(C}_2\text{O}_4)_2]^{2-} \).