Problem 95 - Chemistry: An Atoms-Focused Approach

Question

Manganese(II) chloride was one of the first compounds to be investigated as MRI contrast agents. How many unpaired electrons does the complex have when \(\mathrm{MnCl}_{2}\) dissolves in water to form the coordination compound \(\mathrm{MnCl}_{2}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4} ?\)

Step-by-Step Solution

Answer

Answer: The coordination compound MnCl₂(H₂O)₄ has five unpaired electrons.

Step 1: Find the atomic number of the central metal ion

We first need to find the atomic number of manganese (Mn). Looking at the periodic table, the atomic number of Mn is 25.

Step 2: Determine the electron configuration of the Mn atom

To find the electron configuration of the manganese atom, we can use the following order of filling: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p. Since the atomic number of manganese is 25, the electron configuration is: \(1s^{2}\ 2s^{2}\ 2p^{6}\ 3s^{2}\ 3p^{6}\ 4s^{2}\ 3d^{5}\)

Step 3: Determine the oxidation state of Mn and find the electron configuration for Mn(II)

In MnCl\(_2\), manganese has an oxidation state of +II. Therefore, Mn(II) will have lost two electrons from its original electron configuration. The electron configuration for Mn(II) will be: \(1s^{2}\ 2s^{2}\ 2p^{6}\ 3s^{2}\ 3p^{6}\ 4s^{0}\ 3d^{5}\)

Step 4: Determine the ligands' electron donation sites

In the coordination compound \(\mathrm{MnCl}_{2}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}\), the ligands are water (H2O) and chloride (Cl-). Both water and chloride are monodentate ligands which means they donate only one electron pair each to the central metal ion.

Step 5: Determine the number of unpaired electrons

The electron configuration for Mn(II) is \(1s^{2}\ 2s^{2}\ 2p^{6}\ 3s^{2}\ 3p^{6}\ 4s^{0}\ 3d^{5}\). Thus, Mn(II) has five unpaired electrons in its 3d orbitals. In conclusion, the coordination compound \(\mathrm{MnCl}_{2}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}\) will have five unpaired electrons.