Question

Among the following ions which one has the highest paramagnetism
A \[{{\left[ Cr{{\left( {{H}_{2}}O \right)}_{6}} \right]}^{3+}}\]
B \[{{\left[ Fe{{\left( {{H}_{2}}O \right)}_{6}} \right]}^{2+}}\]
C \[{{\left[ Cu{{\left( {{H}_{2}}O \right)}_{6}} \right]}^{2+}}\]
D \[{{\left[ Zn{{\left( {{H}_{2}}O \right)}_{6}} \right]}^{2+}}\]

Answer 1

Hint: The magnetic character (paramagnetic or diamagnetic) is determined by the number of electrons present in the outermost shell of the central metal atom and the paring of the outermost electron. If the number of electron present are all paired so we say that compound is diamagnetic otherwise paramagnetic.

Complete answer:To determine the number of electrons present in the outermost shell we need to know the oxidation state of the central metal atom so that we can exactly find the electronic configuration of metal and thus the number of electrons present in the outermost shell.

In all the options, ligands (\[{{H}_{2}}O\]) are neutral then the oxidation state of complex cation (+ sign indicates cation) is the same oxidation state of a central metal atom. In option second, the central metal is iron (Fe) and its oxidation state is +2 so the electronic configuration of \[F{{e}^{2+}}\]is given as
\[Fe\text{ }=\text{ }3{{d}^{6}},\text{ }4{{s}^{2}}\]
\[F{{e}^{2+}}\text{ }=\text{ }3{{d}^{6}}\]
As the d subshell has 5 orbitals so as per Hund's rule all electrons are filled one by one in all orbitals first before we fill another electron (double electrons) by starting with the opposite spin (as per Pauli exclusion principle). So, five electrons will be filled first then the sixth electron with the opposite spin filled in the first orbital. So we were left with four unpaired numbers of electrons in the outermost subshell of iron metal of \[{{\left[ Fe{{\left( {{H}_{2}}O \right)}_{6}} \right]}^{2+}}\]complex.
The coordinate compound in options 1, 3, and 4 has less number of unpaired electrons.

Note: The electronic configuration of chromium, Cr in +3 oxidation state (option 1) is \[3{{d}^{3}}\] (3 unpaired electrons), the electronic configuration of copper, Cu in +2 oxidation state is \[3{{d}^{7}}\] (3 unpaired electrons) and the electronic configuration of Zinc, Zn in the oxidation state of +2 is \[3{{d}^{10}}\] (no unpaired electron diamagnetic in nature).