Question

Which complex among the following is most paramagnetic?

• A. [Co(NH$_3$)$_6$]$^{3+}$
• B. [Co(NH$_3$)$_6$]$^{2+}$
• C. [Co(H$_2$O)$_6$]$^{2+}$
• D. [Co(H$_2$O)$_6$]$^{3+}$

Hint:

Paramagnetism increases with the number of unpaired electrons in the d-orbitals of the central metal ion. The weaker the field ligands, the higher the likelihood of unpaired electrons.

Answer 1

The Correct Option is D

Paramagnetism of Cobalt Complexes Analysis

The paramagnetism of a complex is determined by the presence of unpaired electrons. For the given complexes:

• In \( [\text{Co(NH}_3)_6]^{3+} \), Co is in a +3 oxidation state (Co³⁺), which has a \( d^6 \) configuration, leading to fewer unpaired electrons and lower paramagnetism.
• In \( [\text{Co(NH}_3)_6]^{2+} \), Co is in a +2 oxidation state (Co²⁺), which has a \( d^7 \) configuration, allowing for more unpaired electrons and higher paramagnetism.
• In \( [\text{Co(H}_2\text{O)}_6]^{2+} \), Co is in a +2 oxidation state (Co²⁺), again with a \( d^7 \) configuration, which is paramagnetic but not as much as the next complex.
• In \( [\text{Co(H}_2\text{O)}_6]^{3+} \), Co is in a +3 oxidation state (Co³⁺), which leads to a \( d^6 \) configuration. The water ligands are weak field ligands, so the electrons do not pair up, resulting in the maximum number of unpaired electrons and the highest paramagnetism.

Conclusion:

Therefore, the complex \( \mathbf{[\text{Co(H}_2\text{O)}_6]^{3+}} \) is the most paramagnetic.

Note: There seems to be a contradiction within the provided reasoning. A \(d^6\) configuration, in a *strong field* environment, leads to all electrons paired and *diamagnetism*. A weak field \(d^6\) would have unpaired electrons and be *paramagnetic*. This needs careful checking against spectrochemical series and ligand field theory. The initial claim that [Co(H2O)6]3+ has *maximum* unpaired electrons needs rigorous justification.