Step 1: Understanding unpaired electrons.
To determine which species has the most unpaired electrons, we need to examine the electron configuration of the metal center and consider its oxidation state. The more unpaired electrons in the d-orbitals, the more paramagnetic the species.
Step 2: Analyzing the options.
- (A) VCl\(_4\): Vanadium (V) in the +4 oxidation state has the electron configuration [Ar] 3d\(^2\), resulting in 2 unpaired electrons.
- (B) Ni(CO)\(_4\): Nickel in the +0 oxidation state has the electron configuration [Ar] 3d\(^8\), with no unpaired electrons.
- (C) [AuCl\(_4\)]\(^-\): Gold in the +1 oxidation state has the electron configuration [Xe] 4f\(^1\) 5d\(^9\), with no unpaired electrons.
- (D) [CdBr\(_4\)]\(^2-\): Cadmium in the +2 oxidation state has the electron configuration [Kr] 4d\(^1\), with no unpaired electrons.
Step 3: Conclusion.
The correct answer is (A) VCl\(_4\), as it has 2 unpaired electrons in the 3d orbitals.