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

Copper in \( [\text{Cu(C}_2\text{O}_4)_2]^{2-} \) is present as \( \text{Cu}^{2+} \), which has a \( d^{9} \) configuration.
A \( d^{9} \) ion can never have all its electrons paired, since nine electrons spread over five orbitals must leave exactly one orbital singly occupied, regardless of whether the field is strong or weak.
This unpaired electron gives the ion a net electron spin, so it interacts with an applied magnetic field and gives an EPR signal.
The other complexes here, \( \text{Ni(CO)}_4 \) (\( d^{10} \)), \( [\text{Co(NH}_3)_3\text{Cl}]^{2+} \) (low-spin \( d^{6} \)), and \( [\text{Mo(CO)}_6] \) (low-spin \( d^{6} \)), all have fully paired electrons and give no EPR signal.
\[ \boxed{[\text{Cu(C}_2\text{O}_4)_2]^{2-}\ \text{is EPR active}} \]

Answer 2

Another way to approach this is to look at the oxidation state of the metal and the field strength of its ligands, since together these determine whether the d-electrons end up paired or unpaired.

1. \( \text{Ni(CO)}_4 \): CO is a very strong-field, \( \pi \)-accepting ligand, and nickel is reduced all the way to the 0 oxidation state here. With ten d-electrons on a zero-valent metal, the shell is completely filled and there is no room for an unpaired spin. Not EPR active.
2. \( [\text{Co(NH}_3)_3\text{Cl}]^{2+} \): Ammonia is a moderately strong-field ligand, and cobalt is in the +3 state here. The strong field forces the six d-electrons into the lower-energy orbitals in pairs, leaving no unpaired electron. Not EPR active.
3. \( [\text{Cu(C}_2\text{O}_4)_2]^{2-} \): Oxalate is only a moderate-field ligand, and copper is in the +2 state here, that is, \( d^{9} \). Unlike \( d^{6} \), \( d^{8} \), or \( d^{10} \) counts, a \( d^{9} \) count is odd, so no arrangement of the ligand field, weak or strong, can pair up all nine electrons. One electron is always left unpaired. EPR active.
4. \( [\text{Mo(CO)}_6] \): CO again is a strong-field ligand, and molybdenum is in the 0 oxidation state, giving \( d^{6} \). The strong field pairs all six electrons in the lower orbitals. Not EPR active.

The oxidation state and ligand field only change whether an even number of d-electrons pair up or not; an odd d-electron count like \( d^{9} \) can never be fully paired.

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