Problem 57 - Chemistry: The Central Science in SI Units, Global Edition

Question

Give the number of (valence) \(d\) electrons associated with the central metal ion in each of the following complexes: (a) \(\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\right] \mathrm{Cl}_{2},\), (b) \(\mathrm{K}_{2}\left[\mathrm{Cu}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{2}\right]\), (c) \(\left[\mathrm{Os}(\mathrm{en})_{3}\right] \mathrm{Cl}_{3}\), (d) \([\mathrm{Cr}(\mathrm{EDTA})] \mathrm{SO}_{4},(\mathbf{e})\left[\mathrm{Cd}\left(\mathrm{H}_{2} ,\mathrm{O}\right)_{6}\right] \mathrm{Cl}_{2}\).

Step-by-Step Solution

Answer

The number of valence d-electrons in the central metal ion for each complex are: (a) Pt(II): 8 (b) Cu(II): 9 (c) Os(III): 5 (d) Cr(VI): 0 (e) Cd(II): 10

Step 1: (a) Determine oxidation state of Pt in the complex

In the complex \(\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\right] \mathrm{Cl}_{2}\), we have two ammonia ligands (NH3) with a neutral charge each, and two chloride ligands (Cl) with a charge of -1 each. Since the overall complex has a neutral charge, the oxidation state of Pt can be calculated as follows: Let the oxidation state of Pt be x. x + 0 * (NH3) + 2 * (-1) = 0 x = +2

Step 2: (a) Determine the number of d-electrons of Pt

Platinum (Pt) has an atomic number of 78 and the ground state electronic configuration is \([\mathrm{Xe}] \ 4f^{14} \, 5d^9 \, 6s^1\). When Pt is in the +2 oxidation state, its electronic configuration becomes: \([\mathrm{Xe}] \ 4f^{14} \, 5d^8\). Therefore, the number of valence d-electrons in Pt(II) in \(\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\right] \mathrm{Cl}_{2}\) is 8.

Step 3: (b) Determine oxidation state of Cu in the complex

In the complex \(\mathrm{K}_{2}\left[\mathrm{Cu}\left(\mathrm{C}_{2}\mathrm{O}_{4}\right)_{2}\right]\), the overall charge of the complex is -2 (since there are two potassium ions, each with a +1 charge). The oxalate ligand (C2O4) has a charge of -2. There are two oxalate ligands, thus: Let the oxidation state of Cu be x. x + 2 * (-2) = -2 x = +2

Step 4: (b) Determine the number of d-electrons of Cu

Copper (Cu) has an atomic number of 29 and the ground state electronic configuration is \([\mathrm{Ar}] \ 3d^{10} \, 4s^1\). When Cu is in the +2 oxidation state, its electronic configuration becomes: \([\mathrm{Ar}] \ 3d^9\). Therefore, the number of valence d-electrons in Cu(II) in \(\mathrm{K}_{2}\left[\mathrm{Cu}\left(\mathrm{C}_{2}\mathrm{O}_{4}\right)_{2}\right]\) is 9.

Step 5: (c) Determine oxidation state of Os in the complex

In the complex \(\left[\mathrm{Os}(\mathrm{en})_{3}\right] \mathrm{Cl}_{3}\), each ethylenediamine (en) ligand has a neutral charge. The complex has a charge of +3 due to three chloride ions outside the coordination sphere. Therefore, the oxidation state of Os can be calculated as follows: Let the oxidation state of Os be x. x + 0 * (en) = +3 x = +3

Step 6: (c) Determine the number of d-electrons of Os

Osmium (Os) has an atomic number of 76 and the ground state electronic configuration is \([\mathrm{Xe}] \ 4f^{14} \, 5d^6 \, 6s^2\). When Os is in the +3 oxidation state, its electronic configuration becomes: \([\mathrm{Xe}] \ 4f^{14} \, 5d^5\). Therefore, the number of valence d-electrons in Os(III) in \(\left[\mathrm{Os}(\mathrm{en})_{3}\right] \mathrm{Cl}_{3}\) is 5.

Step 7: (d) Determine oxidation state of Cr in the complex

In the complex \([\mathrm{Cr}(\mathrm{EDTA})] \mathrm{SO}_{4}\), ethylenediaminetetraacetic acid (EDTA) acts as a hexadentate ligand and has an overall charge of -4. The sulfate ion (SO4) outside the coordination sphere has a -2 charge. The complex must have a +2 overall charge to balance the sulfate ion. Therefore, the oxidation state of Cr can be calculated as follows: Let the oxidation state of Cr be x. x + (-4) = +2 x = +6

Step 8: (d) Determine the number of d-electrons of Cr

Chromium (Cr) has an atomic number of 24 and the ground state electronic configuration is \([\mathrm{Ar}] \, 3d^5 \, 4s^1\). When Cr is in the +6 oxidation state, its electronic configuration becomes: \([\mathrm{Ar}] \, 3d^0\). Therefore, the number of valence d-electrons in Cr(VI) in \([\mathrm{Cr}(\mathrm{EDTA})] \mathrm{SO}_{4}\) is 0.

Step 9: (e) Determine oxidation state of Cd in the complex

In the complex \(\left[\mathrm{Cd}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{2}\), there are six water molecules (H2O) with a neutral charge each and two chloride ions outside the coordination sphere, each with a -1 charge. Therefore, the oxidation state of Cd can be calculated as follows: Let the oxidation state of Cd be x. x + 0 * (H2O) = +2 x = +2

Step 10: (e) Determine the number of d-electrons of Cd

Cadmium (Cd) has an atomic number of 48 and the ground state electronic configuration is \([\mathrm{Kr}] \, 4d^{10} \, 5s^2\). When Cd is in the +2 oxidation state, its electronic configuration becomes: \([\mathrm{Kr}] \, 4d^{10}\). Therefore, the number of valence d-electrons in Cd(II) in \(\left[\mathrm{Cd}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{2}\) is 10.

Key Concepts

Oxidation State

Transition metal complexes often require us to determine the oxidation state of the central metal atom. Understanding this concept is crucial since it influences the chemical behavior and electron configuration of the metal.
To find the oxidation state, consider the charges of all components in the complex. Let's use Pt in \(\left[\text{Pt}\left(\text{NH}_3\right)_2 \text{Cl}_2\right] \text{Cl}_2\) as an example:

• Ammonia (NHa53) is a neutral ligand, contributing 0 to the overall charge.
• Each chloride (Cl) ligand has a charge of -1.
• Since the complex is neutral overall, the sum of the oxidation state of Pt and the charges of ligands must be zero.

By solving the equation:\(\text{x} + 2\cdot(-1) = 0\)we find the oxidation state of Pt to be +2.
Determining oxidation states helps us predict chemical reactivity and the coordination environment in complexes.

d-electrons

The number of d-electrons in a transition metal plays a significant role in its chemistry. By knowing the oxidation state, we can determine the number of d-electrons. This requires an understanding of the electron configuration of the neutral atom.
Take copper (Cu) as an example in the complex \(\text{K}_2\left[\text{Cu}\left(\text{C}_2\text{O}_4\right)_2\right]\).

• Copper's atomic number is 29, with an electronic configuration \([ ext{Ar}] \, 3d^{10} \, 4s^1\).
• In a +2 oxidation state, the configuration changes to:\([ ext{Ar}] \, 3d^9\).

This means Cu has 9 d-electrons.
Understanding the number of d-electrons helps explain certain properties of the metal, such as its magnetic behavior and color in complexes.

Coordination Chemistry

Coordination chemistry involves the study of compounds formed between a central metal and surrounding ligands. The metal-ligand interactions in these complexes are key to their stability and reactivity.In \([ ext{Os}( ext{en})_3] ext{Cl}_3\), osmium (Os) is coordinated by three molecules of ethylenediamine (en), a neutral bidentate ligand.

• The coordination number, which is the number of ligand "bonds" to the central metal, is 6 in this example (3 en molecules x 2 sites each).
• The coordination sphere influences the geometry of the complex, with common geometries including octahedral, square planar, and tetrahedral forms.

Coordination chemistry not only dictates geometrical arrangements but also affects the metal's electronic properties, and hence its chemical and physical behavior.

Electron Configuration

The electron configuration provides a blueprint of how electrons are distributed among the orbitals of an atom. For transition metal ions in complexes, the electron configuration is fundamentally tied to their oxidation state and their chemical identity.For chromium (Cr) in \([\text{Cr}(\text{EDTA})] \text{SO}_4\), the +6 oxidation state alters its typical \([ ext{Ar}] \, 3d^{5} \, 4s^{1}\) configuration to \([ ext{Ar}] \, 3d^{0}\).

• All d-electrons are removed when Cr is in the +6 oxidation state.
• This lack of d-electrons in Cr(VI) results in unique properties, distinguishing it from lower oxidation states.

Understanding electron configurations is critical in predicting how transition metals will react and interact with other components to form complexes.