Question

Transition metals and their maximum compounds are paramagnetic. Explain it.

Hint:

Count unpaired \(d\)-electrons in the metal ion/complex to predict magnetism: any \(n>0\) \(\Rightarrow\) paramagnetic; \(n=0\) \(\Rightarrow\) diamagnetic.

Answer 1

Paramagnetism arises from the presence of unpaired electrons. Transition elements have their valence electrons in the \(\,(n-1)d\,\) and \(ns\) subshells. Because the \(d\)-subshell can accommodate ten electrons and is partially filled across the series, most transition-metal atoms and many of their ions retain one or more unpaired \(d\)-electrons.
\(\Rightarrow\) These unpaired electrons give a net magnetic moment (\(\mu \propto \sqrt{n(n+2)}\), where \(n\) is the number of unpaired electrons), hence paramagnetism.
Further, transition metals exhibit variable oxidation states; during oxidation or complex formation, electrons are removed from \(ns\) and \((n-1)d\) levels, often leaving unpaired \(d\)-electrons in the cations/complexes.
Notes/Exceptions: Closed-shell cases like \(\mathrm{Zn^{2+}(d^{10})}\), \(\mathrm{Cd^{2+}(d^{10})}\), \(\mathrm{Hg^{2+}(d^{10})}\) are diamagnetic. Some strong-field, low-spin complexes can also become diamagnetic if all electrons pair.