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Chapter 24: Problem 45

Draw the crystal-field energy-level diagrams and show the placement of \(d\) electrons for each of the following: (a) \(\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) (four unpaired electrons), (b) \(\left[\mathrm{Mn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) (high spin), (c) \(\left[\mathrm{Ru}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{H}_{2} \mathrm{O}\right]^{2+}\) (low spin), (d) \(\left[\mathrm{IrCl}_{6}\right]^{2-}\) (low spin), (e) \(\left[\mathrm{Cr}(\mathrm{en})_{3}\right]^{3+}\), (f) \(\left[\mathrm{NiF}_{6}\right]^{4-}\)

Short Answer

Expert verified
A short version of the answer is: (a) \(\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\): Cr虏鈦 ion with 3d鈦 configuration; high spin; energy-level diagram: 鈫戔啌 鈫 鈫 鈫. (b) \(\left[\mathrm{Mn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\): Mn虏鈦 ion with 3d鈦 configuration; high spin; energy-level diagram: 鈫戔啈鈫戔啈鈫. (c) \(\left[\mathrm{Ru}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{H}_{2} \mathrm{O}\right]^{2+}\): Ru虏鈦 ion with 4d鈦 configuration; low spin; energy-level diagram: 鈫戔啌 鈫戔啌 鈫. (d) \(\left[\mathrm{IrCl}_{6}\right]^{2-}\): Ir鈦粹伜 ion with 5d鈦 configuration; low spin; energy-level diagram: 鈫戔啌 鈫戔啌 鈫戔啌. (e) \(\left[\mathrm{Cr}(\mathrm{en})_{3}\right]^{3+}\): Cr鲁鈦 ion with 3d鲁 configuration; low spin (assumed); energy-level diagram: 鈫戔啌 鈫. (f) \(\left[\mathrm{NiF}_{6}\right]^{4-}\): Ni虏鈦 ion with 3d鈦 configuration; high spin (assumed); energy-level diagram: 鈫戔啌 鈫戔啌 鈫戔啈鈫.

Step by step solution

01

Identify the oxidation state

Cr has an oxidation state of +2, as shown by the 2+ charge on the complex.
02

Calculate the number of d-electrons

Cr has an atomic number of 24 (1s虏 2s虏 2p鈦 3s虏 3p鈦 3d鈦 4s鹿). In the +2 oxidation state, Cr loses 2 electrons, leaving it with 3d鈦 configuration.
03

Draw the energy-level diagram and place electrons

Given there are four unpaired electrons, this complex is high spin. So, in an octahedral complex, the d-electrons fill as follows: 鈫戔啌 鈫 鈫 鈫. The diagram will have three lower energy levels filled and one higher energy level filled, giving four unpaired electrons. (b) \(\left[\mathrm{Mn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) (high spin)
04

Identify the oxidation state

Mn has an oxidation state of +2, as indicated by the 2+ charge on the complex.
05

Calculate the number of d-electrons

Mn has an atomic number of 25 (1s虏 2s虏 2p鈦 3s虏 3p鈦 3d鈦 4s虏). In the +2 oxidation state, Mn loses 2 electrons from the 4s orbital, leaving it with 3d鈦 configuration.
06

Draw the energy-level diagram and place electrons

Because this complex is high spin, we place the d-electrons in an octahedral complex as follows: 鈫戔啈鈫戔啈鈫. The diagram will have three lower energy levels and two higher energy levels filled. All electrons are unpaired. (c) \(\left[\mathrm{Ru}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{H}_{2} \mathrm{O}\right]^{2+}\) (low spin)
07

Identify the oxidation state

Ru has an oxidation state of +2, as indicated by the 2+ charge on the complex.
08

Calculate the number of d-electrons

Ru has an atomic number of 44 (1s虏 2s虏 2p鈦 3s虏 3p鈦 4s虏 3d鹿鈦 4p鈦 5s虏 4d鈦). In the +2 oxidation state, Ru loses 2 electrons from the 5s orbital, leaving it with 4d鈦 configuration.
09

Draw the energy-level diagram and place electrons

Since this complex is low spin, we place the d-electrons in an octahedral complex as follows: 鈫戔啌 鈫戔啌 鈫. The diagram will have three lower energy levels completely filled and no electrons in the higher energy levels. (d) \(\left[\mathrm{IrCl}_{6}\right]^{2-}\) (low spin)
10

Identify the oxidation state

Since each chloride contributes -1 and there are six of them, the oxidation state of Ir must be +4 to balance the 2- charge on the complex.
11

Calculate the number of d-electrons

Ir has an atomic number of 77 (1s虏 2s虏 2p鈦 3s虏 3p鈦 4s虏 3d鹿鈦 4p鈦 5s虏 4d鹿鈦 5p鈦 6s虏 5d鈦). In the +4 oxidation state, Ir loses 2 electrons from the 6s and 2 electrons from the 5d orbital, leaving it with 5d鈦 configuration.
12

Draw the energy-level diagram and place electrons

As this complex is low spin, the d-electrons in the octahedral complex fill as follows: 鈫戔啌 鈫戔啌 鈫戔啌. The diagram will have three lower energy levels completely filled and two electrons paired in one higher energy level. (e) \(\left[\mathrm{Cr}(\mathrm{en})_{3}\right]^{3+}\)
13

Identify the oxidation state

Cr has an oxidation state of +3, as shown by the 3+ charge on the complex.
14

Calculate the number of d-electrons

Cr has an atomic number of 24 (1s虏 2s虏 2p鈦 3s虏 3p鈦 3d鈦 4s鹿). In the +3 oxidation state, Cr loses 3 electrons, leaving it with 3d鲁 configuration.
15

Draw the energy-level diagram and place electrons

As no specific spin configuration is given, assume low spin considering the en ligand is strong field. In an octahedral complex, the d-electrons fill as follows: 鈫戔啌 鈫. The diagram will have two lower energy levels filled and no electrons in the higher energy levels. (f) \(\left[\mathrm{NiF}_{6}\right]^{4-}\)
16

Identify the oxidation state

Since each fluoride contributes -1 and there are six of them, the oxidation state of Ni must be +2 to balance the 4- charge on the complex.
17

Calculate the number of d-electrons

Ni has an atomic number of 28 (1s虏 2s虏 2p鈦 3s虏 3p鈦 3d鈦 4s虏). In the +2 oxidation state, Ni loses 2 electrons from the 4s orbital, leaving it with 3d鈦 configuration.
18

Draw the energy-level diagram and place electrons

As no specific spin configuration is given, assume high spin considering the F ligand is weak field. In an octahedral complex, the d-electrons fill as follows: 鈫戔啌 鈫戔啌 鈫戔啈鈫. The diagram will have three lower energy levels and two higher energy levels filled, giving two unpaired electrons.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

d-Electron Configuration
Understanding the d-electron configuration is crucial when exploring transition metal complexes. The d-orbitals can hold up to ten electrons, and their arrangement plays a pivotal role in the properties and color of the complexes.

When transition metals form complexes, they do so in specific oxidation states, impacting the remaining number of electrons in their d-orbitals. For example, a chromium(II) ion, noted as Cr2+, would have a d4 electron configuration since chromium normally has six d-electrons (3d6) in its neutral state, and two are removed due to the +2 charge.

Ligands, the molecules or ions attached to the metal in a complex, influence electron distribution among d-orbitals. Strong field ligands, such as water or ammonia, can cause electrons to pair up in lower energy d-orbitals, while weak field ligands, like halides, typically result in high spin configurations with more unpaired electrons.
Octahedral Complexes
In octahedral complexes, where a central metal ion is surrounded by six ligands positioned at the corners of an octahedron, the d-orbitals split into two sets of differing energy levels due to the electric field of the surrounding ligands.

The crystal-field splitting in an octahedral complex divides the five degenerate d-orbitals into two groups鈥攖hree lower-energy t2g and two higher-energy eg orbitals. The pattern in which the d-electrons fill these orbitals gives insight into the magnetic and spectroscopic properties of the complex. A key to mastering complex formation is visualizing the energy-level diagram and accurately distributing the d-electrons according to the Hund's rule and the Pauli exclusion principle.

For instance, a high spin [Mn(H2O)6]2+ complex features five unpaired electrons arranged as 鈫戔啈鈫戔啈鈫, occupying both the t2g and eg orbitals with the maximum number of unpaired electrons.
High Spin and Low Spin Complexes
High spin and low spin complexes refer to the two possible arrangements of d-electrons based on the strength of the ligand field due to the surrounding ligands. High spin complexes occur with weak field ligands and maximize the number of unpaired electrons, leading to magnetic complexes.

For example, [NiF6]4-, a high spin complex, has electrons in both t2g and eg orbitals since fluorides are considered to produce a weaker crystal field, resulting in a configuration of 鈫戔啌 鈫戔啌 鈫戔啈鈫, with two unpaired electrons.

Conversely, low spin complexes form with strong field ligands, which have a larger splitting energy that favors pairing of electrons in the lower t2g orbitals, thus resulting in complexes with fewer unpaired electrons. Such low spin configurations are often non-magnetic. [IrCl6]2- illustrates this with its 5d5 electron configuration, creating a low spin arrangement with all d-electrons paired: 鈫戔啌 鈫戔啌 鈫戔啌.

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Most popular questions from this chapter

The molecule dimethylphosphinoethane \(\left[\left(\mathrm{CH}_{3}\right)_{2} \mathrm{PCH}_{2}-\mathrm{CH}_{2} \mathrm{P}\left(\mathrm{CH}_{3}\right)_{2}\right.\), which is abbreviated \(\left.\mathrm{dmpe}\right]\) is used as a ligand for some complexes that serve as catalysts. A complex that contains this ligand is \(\mathrm{Mo}(\mathrm{CO})_{4}(\) dmpe \()\). (a) Draw the Lewis structure for dmpe, and compare it with ethylenediammine as a coordinating ligand. (b) What is the oxidation state of Mo in \(\mathrm{Na}_{2}\left[\mathrm{Mo}(\mathrm{CN})_{2}(\mathrm{CO})_{2}(\mathrm{dmpe})\right] ?\) (c) Sketch the structure of the \(\left[\mathrm{Mo}(\mathrm{CN})_{2}(\mathrm{CO})_{2}(\mathrm{dmpe})\right]^{2-}\) ion, including all the possible isomers.

Generally speaking, for a given metal and ligand, the stability of a coordination compound is greater for the metal in the \(3+\) rather than in the \(2+\) oxidation state. Furthermore, for a given ligand the complexes of the bivalent metal ions of the first transition series tend to increase in stability in the order \(\mathrm{Mn}(\mathrm{II})<\mathrm{Fe}(\mathrm{II})<\mathrm{Co}(\mathrm{II})<\) \(\mathrm{Ni}(\mathrm{II})<\mathrm{Cu}(\mathrm{II})\). Explain how these two observations are consistent with one another and also consistent with a crystal-field picture of coordination compounds.

Sketch all the possible stereoisomers of (a) tetrahedral \(\left[\mathrm{Cd}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2} \mathrm{Cl}_{2}\right]\), (b) square-planar \(\left[\mathrm{IrCl}_{2}\left(\mathrm{PH}_{3}\right)_{2}\right]^{-}\), (c) octa- hedral \(\left[\mathrm{Fe}(0 \text { -phen })_{2} \mathrm{Cl}_{2}\right]^{+}\)

Metallic elements are essential components of many important enzymes operating within our bodies. Carbonic anhydrase, which contains \(\mathrm{Zn}^{2+}\), is responsible for rapidly interconverting dissolved \(\mathrm{CO}_{2}\) and bicarbonate ion, \(\mathrm{HCO}_{3}^{-} .\) The zinc in carbonic anhydrase is coordinated by three nitrogen-containing groups and a water molecule. The enzyme's action depends on the fact that the coordinated water molecule is more acidic than the bulk solvent molecules. Explain this fact in terms of Lewis acid-base theory (Section 16.11).

The \(\left[\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) ion has an absorption maximum at about \(725 \mathrm{~nm}\), whereas the \(\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}\) ion absorbs at about \(570 \mathrm{~nm} .\) Predict the color of each ion. (b) The \(\left[\mathrm{Ni}(\mathrm{en})_{3}\right]^{2+}\) ion absorption maximum occurs at about \(545 \mathrm{~nm}\), and that of the [Ni(bipy) \(\left._{3}\right]^{2+}\) ion occurs at about \(520 \mathrm{~nm}\). From these data, indicate the relative strengths of the ligand fields created by the four ligands involved.
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