Problem 8 - Chemistry An Atoms First Approach

Question

A certain first-row transition metal ion forms many different colored solutions. When four coordination compounds of this metal, each having the same coordination number, are dissolved in water, the colors of the solutions are red, yellow, green, and blue. Further experiments reveal that two of the complex ions are paramagnetic with four unpaired electrons and the other two are diamagnetic. What can be deduced from this information about the four coordination compounds?

Step-by-Step Solution

Answer

From the given information, we can deduce that the transition metal ion in the four coordination compounds is either Cr or Mn. The paramagnetic compounds have different oxidation states, either Cr(III) or Mn(II), with four unpaired 3d electrons. Meanwhile, the diamagnetic compounds have their 3d electrons paired up due to strong-field ligands, with oxidation states of Cr(IV) or Mn(III). The different colors of the solutions are a result of the electronic configurations and the ligands present in each coordination compound.

Step 1: Identify possible transition metal ions and their oxidation states based on unpaired electrons

Since we have four unpaired electrons in the paramagnetic species, we should look for possible oxidation states that would result in such an electronic configuration. The first-row transition metals and their d-electron counts are: - Sc: \(3d^1\) - Ti: \(3d^2\) - V: \(3d^3\) - Cr: \(3d^5\) - Mn: \(3d^5\) - Fe: \(3d^6\) - Co: \(3d^7\) - Ni: \(3d^8\) - Cu: \(3d^{10}\) - Zn: \(3d^{10}\) We are looking for an oxidation state that would result in \(3d^4\). Based on the possible oxidation states and electron configurations of first-row transition metals, Cr(III) and Mn(II) can have this configuration.

Step 2: Compare diamagnetic and paramagnetic properties

We know that two of the coordination compounds are paramagnetic with four unpaired electrons, and the other two are diamagnetic. For both Cr(III) and Mn(II), we have a \(3d^4\) electronic configuration, making them paramagnetic with four unpaired electrons. The diamagnetic nature of the other two coordination compounds indicates no unpaired electrons. Comparing the oxidation states of first-row transition metals, Cr(IV) and Mn(III) have a \(3d^2\) electronic configuration, resulting in two unpaired electrons. However, when considering the ligand-field stabilization energy, strong-field ligands can cause the electrons in the \(3d\) orbitals to pair up, resulting in a diamagnetic compound.

Step 3: Conclusions about the coordination compounds

Based on the information provided, we can deduce that the four coordination compounds have the same transition metal ion, either Cr or Mn, with two oxidation states (III and IV for Cr or II and III for Mn). The paramagnetic compounds are due to the four unpaired \(3d\) electrons, while the diamagnetic ones have their \(3d\) electrons paired up due to strong-field ligands. The different colors of the solutions are a result of the electronic configurations and the ligands present in each coordination compound.