Problem 23 - Chemistry and Chemical Reactivity

Question

The following are low-spin complexes. Use the ligand field model to find the electron configuration of the central metal ion in each ion. Determine which are diamagnetic. Give the number of unpaired electrons for the paramagnetic complexes. (a) \(\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{4-}\) (c) \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}\) (b) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{3}\) (d) \(\left[\mathrm{Cr}(\mathrm{en})_{3}\right] \mathrm{SO}_{4}\)

Step-by-Step Solution

Answer

(a) and (b) are diamagnetic, (c) has 1 unpaired electron, (d) has 3 unpaired electrons.

Step 1: Determine the oxidation state

For each complex, calculate the oxidation state of the metal by considering the charge of the ligands and the overall complex charge. For (a) \([\mathrm{Mn}(\mathrm{CN})_{6}]^{4-}\), \(\text{CN}^-\) is a \(-1\) ligand and the overall charge is \(-4\), giving Mn an oxidation state of +2. For (b) \([\mathrm{Co(NH}_{3})_{6}] \mathrm{Cl}_{3}\), \(\text{NH}_3\) is a neutral ligand and \(3\text{Cl}^-\) gives an oxidation state of +3 for Co. For (c) \([\mathrm{Fe(H}_{2} \mathrm{O})_{6}]^{3+}\), \(\text{H}_2\text{O}\) is a neutral ligand, making the oxidation state of Fe +3. For (d) \([\mathrm{Cr(en)}_{3}] \mathrm{SO}_{4}\), \text{en} is a neutral ligand and \(\text{SO}_4\) is \(-2\), giving Cr an oxidation state of +3.

Step 2: Electron configuration with oxidation state

For each metal ion, subtract the oxidation state from the atomic number to find the d-electron count: - Mn: 3d^5 (7 - 2 = 5) - Co: 3d^6 (9 - 3 = 6) - Fe: 3d^5 (8 - 3 = 5) - Cr: 3d^3 (6 - 3 = 3)

Step 3: Apply ligand field theory (Low-spin configuration)

Determine the low-spin d-electron configuration for each complex using the ligand field theory, considering strong field ligands:- (a) For \([\mathrm{Mn}(\mathrm{CN})_{6}]^{4-})\), \(\text{CN}^-\) is a strong ligand, so d^5 fills as \(t_{2g}^5\), no unpaired electrons, diamagnetic.- (b) For \([\mathrm{Co(NH}_{3})_{6}]^{3+}\), \(\text{NH}_3\) is a moderate ligand, so it establishes a low-spin d^6 configuration as \(t_{2g}^6\), no unpaired electrons, diamagnetic.- (c) For \([\mathrm{Fe(H}_{2} \mathrm{O})_{6}]^{3+}\), \(\text{H}_2\text{O}\) is a weak ligand, usually high spin, but given low-spin it is \(t_{2g}^5\) in low-spin, one unpaired electron, paramagnetic.- (d) For \([\mathrm{Cr(en)}_{3}]^{3+}\), \(en\) is a strong-field ligand, placing d^3 as \(t_{2g}^3\), 3 unpaired electrons, paramagnetic.

Step 4: Identify diamagnetic complexes

Review the electron pairings determined in Step 3.- (a) Both \([\mathrm{Mn}(\mathrm{CN})_{6}]^{4-})\) and (b) \([\mathrm{Co(NH}_{3})_{6}]^{3+}\) are diamagnetic because all electrons are paired.

Step 5: Determine number of unpaired electrons in paramagnetic complexes

- In (c) \([\mathrm{Fe(H}_{2} \mathrm{O})_{6}]^{3+}\), there is 1 unpaired electron.- In (d) \([\mathrm{Cr(en)}_{3}]^{3+}\), there are 3 unpaired electrons.

Key Concepts

Diamagnetic Complexes

Diamagnetic complexes are those in which all the electrons are paired. This occurs when the energy levels of the d-orbitals in the central metal ion are fully occupied with paired electrons, generally resulting in no net magnetic moment.

In ligand field theory, the nature of the ligands interacting with the metal central ion plays a crucial role. Strong field ligands, like cyanide (\( ext{CN}^{-}\)), result in a large splitting of the d-orbitals, causing electrons to pair up in the lower energy orbitals if possible.
- In the complex \([ ext{Mn}( ext{CN})_{6}]^{4-}\), manganese achieves a \(d^5\) configuration in a low-spin state, filling the \(t_{2g}\) orbitals with paired electrons. Consequently, it becomes diamagnetic.
- Similarly, the complex \([ ext{Co(NH}_3)_{6}]^{3+}\), with cobalt in a d\(^6\) low-spin configuration due to ammonia’s moderate ligand field strength, results in a fully paired electron arrangement, leading to its diamagnetic nature.

Paramagnetic Complexes

Paramagnetic complexes are characterized by the presence of one or more unpaired electrons. These unpaired electrons generate a magnetic moment that aligns with external magnetic fields, making the complex paramagnetic.

In these cases, the strength of the ligand field affects whether the d-orbitals are partially filled without all the electrons pairing.
- In \([ ext{Fe(H}_2 ext{O})_{6}]^{3+}\), iron has a \(d^5\) configuration. Although water is a weak field ligand and typically results in high-spin complexes, the problem specifies a low-spin arrangement. This leads to five electrons occupying the t\(_{2g}\) and e\(_g\) orbitals with one unpaired electron remaining, making it paramagnetic.
- The complex \([ ext{Cr(en)}_{3}]^{3+}\), which involves chromium with a \(d^3\) configuration, has three unpaired electrons as ethylenediamine (en) is a strong field ligand, leading to the complex being paramagnetic.

Electron Configuration

Understanding electron configuration is crucial in predicting the magnetic properties of complexes. For transition metals, the valence d-orbitals are the focus.

- The d-orbitals can hold a maximum of ten electrons.
- The ligand field theory helps determine how these d-electrons are distributed among the split energy levels caused by interactions with the ligands.

For each complex:

• Calculate the metal's d-electron count by subtracting the oxidation state from the atomic number.
• Determine the arrangement of these electrons in the d-orbitals based on the strength of the ligand field and whether the configuration is high-spin or low-spin.

An example calculation has manganese in \([ ext{Mn}( ext{CN})_{6}]^{4-}\) with an oxidation state of +2. For manganese, the atomic number is 25, giving a electron count of 5 in the d-orbitals after oxidation state calculation.

Oxidation State Determination

Determining the oxidation state of the central metal ion is an essential part of analyzing any complex.

The oxidation state indicates the apparent charge on the metal when accounting for the charges on the ligands and the overall charge of the complex.
To find the oxidation state:
- Use the known charges of the ligands interacting with the metal.- Consider the total charge of the complex.

For example, in \([ ext{Co(NH}_3)_{6}]^{3+}\), ammonia is a neutral ligand, meaning it doesn't contribute to any change in charge, while the overall charge of the complex is +3. Therefore, the cobalt has to be in the +3 oxidation state to balance this out.