Which of the following species is diamagnetic?

The correct answer is option 4. A low spin octahedral \(Co^{3+}\) complex.

To determine which species is diamagnetic, let us analyze each option based on their electron configurations and the magnetic properties associated with them:

1. An isolated, gas-phase \(V^{3+}\) ion:

Vanadium in the \(+3\) oxidation state (\(V^{3+}\)) has the electron configuration \([Ar] 3d^2\).

Configuration: \(3d^2\). This configuration has 2 unpaired electrons, making it paramagnetic, not diamagnetic.

2. A high spin octahedral \(Fe^{2+}\) complex:

Iron in the \(+2\) oxidation state (\(Fe^{2+}\)) typically has the electron configuration \([Ar] 3d^6\). High spin octahedral complexes typically have a \(t_{2g}^4 e_g^2\) configuration in an octahedral crystal field.

Configuration: \(3d^6\).

This configuration has 4 unpaired electrons, making it paramagnetic, not diamagnetic.

3. An isolated, gas phase \(Cu^{2+}\) ion:

Copper in the +2 oxidation state (\(Cu^{2+}\)) has the electron configuration \([Ar] 3d^9\).

Configuration: \(3d^9\).

This configuration has 1 unpaired electron, making it paramagnetic, not diamagnetic.

4. A low spin octahedral Co3+ complex:

Cobalt in the \(+3\) oxidation state (\(Co^{3+}\)) typically has the electron configuration \([Ar] 3d^6\). Low spin octahedral complexes typically have a \(t_{2g}^6\) configuration in an octahedral crystal field.

Configuration: \(3d^6\).

This configuration has 0 unpaired electrons, making it diamagnetic.

The species that is diamagnetic among the given options is: \(\text{A low spin octahedral }Co^{3+} \text{ complex}\)

This complex has a \(3d^6\) electron configuration in a low spin octahedral field, resulting in all electrons being paired, hence exhibiting diamagnetic behavior.