Question

Calculate the number of unpaired electrons in the following gaseous ions: ${\text{M}}{{\text{n}}^{{\text{3 + }}}}$ , ${\text{C}}{{\text{r}}^{{\text{3 + }}}}$ , ${{\text{V}}^{{\text{3 + }}}}$ and ${\text{T}}{{\text{i}}^{{\text{3 + }}}}$ . Which one of these is the most stable in aqueous solutions?

Answer 1

Unpaired electrons are those that are not paired up in the orbitals. The ions or atoms having unpaired electrons are called paramagnetic because the unpaired electrons generate magnetic fields.

Complete step by step solution:
The atomic number of manganese is 25, hence the electronic configuration is : $\text{1}{{\text{s}}^{\text{2}}}\text{2}{{\text{s}}^{\text{2}}}\text{2}{{\text{p}}^{\text{6}}}\text{3}{{\text{s}}^{\text{2}}}\text{3}{{\text{p}}^{\text{6}}}\text{4}{{\text{s}}^{\text{2}}}\text{3}{{\text{d}}^{\text{5}}}$ . When the manganese atom loses three electrons to form $\text{M}{{\text{n}}^{\text{3+}}}$ , then the configuration becomes, $\text{1}{{\text{s}}^{\text{2}}}\text{2}{{\text{s}}^{\text{2}}}\text{2}{{\text{p}}^{\text{6}}}\text{3}{{\text{s}}^{\text{2}}}\text{3}{{\text{p}}^{\text{6}}}\text{3}{{\text{d}}^{4}}$ . Therefore there are 4 electrons in the 3d orbital that are arranged in four orbitals and hence there are 4 unpaired electrons.
The atomic number of chromium is 24, hence the electronic configuration is : $\text{1}{{\text{s}}^{\text{2}}}\text{2}{{\text{s}}^{\text{2}}}\text{2}{{\text{p}}^{\text{6}}}\text{3}{{\text{s}}^{\text{2}}}\text{3}{{\text{p}}^{\text{6}}}\text{4}{{\text{s}}^{1}}\text{3}{{\text{d}}^{\text{5}}}$ . When the manganese atom loses three electrons to form $\text{C}{{\text{r}}^{\text{3+}}}$ , then the configuration becomes, $\text{1}{{\text{s}}^{\text{2}}}\text{2}{{\text{s}}^{\text{2}}}\text{2}{{\text{p}}^{\text{6}}}\text{3}{{\text{s}}^{\text{2}}}\text{3}{{\text{p}}^{\text{6}}}\text{3}{{\text{d}}^{3}}$ , hence there are 3 unpaired electrons.
The atomic number of vanadium is 23, hence the electronic configuration is : $\text{1}{{\text{s}}^{\text{2}}}\text{2}{{\text{s}}^{\text{2}}}\text{2}{{\text{p}}^{\text{6}}}\text{3}{{\text{s}}^{\text{2}}}\text{3}{{\text{p}}^{\text{6}}}\text{4}{{\text{s}}^{2}}\text{3}{{\text{d}}^{3}}$ . When the manganese atom loses three electrons to form ${{\text{V}}^{\text{3+}}}$ , then the configuration becomes, $\text{1}{{\text{s}}^{\text{2}}}\text{2}{{\text{s}}^{\text{2}}}\text{2}{{\text{p}}^{\text{6}}}\text{3}{{\text{s}}^{\text{2}}}\text{3}{{\text{p}}^{\text{6}}}\text{3}{{\text{d}}^{2}}$ , hence there are 2 unpaired electrons.
The atomic number of titanium is 22, hence the electronic configuration is : $\text{1}{{\text{s}}^{\text{2}}}\text{2}{{\text{s}}^{\text{2}}}\text{2}{{\text{p}}^{\text{6}}}\text{3}{{\text{s}}^{\text{2}}}\text{3}{{\text{p}}^{\text{6}}}\text{4}{{\text{s}}^{2}}\text{3}{{\text{d}}^{2}}$ . When the manganese atom loses three electrons to form $\text{T}{{\text{i}}^{\text{3+}}}$ , then the configuration becomes, $\text{1}{{\text{s}}^{\text{2}}}\text{2}{{\text{s}}^{\text{2}}}\text{2}{{\text{p}}^{\text{6}}}\text{3}{{\text{s}}^{\text{2}}}\text{3}{{\text{p}}^{\text{6}}}\text{3}{{\text{d}}^{1}}$ , hence there is 1 unpaired electron.
The most stable among these cations will be the one which is most polarising so that more number of water molecules are attracted towards the cation and higher hydration energy is released. In this case chromium is the most stable one due to higher hydration energy of the cation.

Note:
The hydration energy is defined as the amount of energy released when the water molecules are arranged in the coordination sphere of the metal ions due to van der Waals attraction.