Which ligand suitable for given complex C r L 6 (green colour) has to be find out. Concept introduction: The fact that colour depends on the ligand and allows us to create a spectrochemical series, which ranks the ability of a ligand to split d-orbital energies. The relative magnitude of Δ for a series of octahedral complexes of a given metal ion. As Δ increases, higher energies (shorter wavelengths) of light must be absorbed to excite electrons. The spectrochemical series is I − < C l − < F − < O H − < H 2 O < S C N − < e n < N O 2 − < C N − < C O Relationships on an artist’s colour wheel in which complementary colors are wedges opposite each other. Figure 1

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Answer 1

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Chapter 23, Problem 23.95P

Interpretation Introduction

Interpretation:

Which ligand suitable for given complex CrL6 (green colour) has to be find out.

Concept introduction:

The fact that colour depends on the ligand and allows us to create a spectrochemical series, which ranks the ability of a ligand to split d-orbital energies.

The relative magnitude of Δ for a series of octahedral complexes of a given metal ion. As Δ increases, higher energies (shorter wavelengths) of light must be absorbed to excite electrons.

The spectrochemical series is

I−<Cl−<F−<OH−<H2O<SCN−<en<NO2−<CN−<CO

Relationships on an artist’s colour wheel in which complementary colors are wedges opposite each other.

Figure 1

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Answer 2

Question

Chapter 23, Problem 23.95P

Interpretation Introduction

Interpretation:

Which ligand suitable for given complex CrL6 (green colour) has to be find out.

Concept introduction:

The fact that colour depends on the ligand and allows us to create a spectrochemical series, which ranks the ability of a ligand to split d-orbital energies.

The relative magnitude of Δ for a series of octahedral complexes of a given metal ion. As Δ increases, higher energies (shorter wavelengths) of light must be absorbed to excite electrons.

The spectrochemical series is

I−<Cl−<F−<OH−<H2O<SCN−<en<NO2−<CN−<CO

Relationships on an artist’s colour wheel in which complementary colors are wedges opposite each other.

Figure 1

Expert Solution & Answer

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Answer 3

Question

Chapter 23, Problem 23.95P

Interpretation Introduction

Interpretation:

Which ligand suitable for given complex CrL6 (green colour) has to be find out.

Concept introduction:

The fact that colour depends on the ligand and allows us to create a spectrochemical series, which ranks the ability of a ligand to split d-orbital energies.

The relative magnitude of Δ for a series of octahedral complexes of a given metal ion. As Δ increases, higher energies (shorter wavelengths) of light must be absorbed to excite electrons.

The spectrochemical series is

I−<Cl−<F−<OH−<H2O<SCN−<en<NO2−<CN−<CO

Relationships on an artist’s colour wheel in which complementary colors are wedges opposite each other.

Figure 1

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Answer 4

Question

Chapter 23, Problem 23.95P

Interpretation Introduction

Interpretation:

Which ligand suitable for given complex CrL6 (green colour) has to be find out.

Concept introduction:

The fact that colour depends on the ligand and allows us to create a spectrochemical series, which ranks the ability of a ligand to split d-orbital energies.

The relative magnitude of Δ for a series of octahedral complexes of a given metal ion. As Δ increases, higher energies (shorter wavelengths) of light must be absorbed to excite electrons.

The spectrochemical series is

I−<Cl−<F−<OH−<H2O<SCN−<en<NO2−<CN−<CO

Relationships on an artist’s colour wheel in which complementary colors are wedges opposite each other.

Figure 1

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Answer 5

Chapter 23, Problem 23.95P

Interpretation Introduction

Interpretation:

Which ligand suitable for given complex CrL6 (green colour) has to be find out.

Concept introduction:

The fact that colour depends on the ligand and allows us to create a spectrochemical series, which ranks the ability of a ligand to split d-orbital energies.

The relative magnitude of Δ for a series of octahedral complexes of a given metal ion. As Δ increases, higher energies (shorter wavelengths) of light must be absorbed to excite electrons.

The spectrochemical series is

I−<Cl−<F−<OH−<H2O<SCN−<en<NO2−<CN−<CO

Relationships on an artist’s colour wheel in which complementary colors are wedges opposite each other.

Figure 1

Answer 6

Chapter 23, Problem 23.95P

Interpretation Introduction

Interpretation:

Which ligand suitable for given complex CrL6 (green colour) has to be find out.

Concept introduction:

The fact that colour depends on the ligand and allows us to create a spectrochemical series, which ranks the ability of a ligand to split d-orbital energies.

The relative magnitude of Δ for a series of octahedral complexes of a given metal ion. As Δ increases, higher energies (shorter wavelengths) of light must be absorbed to excite electrons.

The spectrochemical series is

I−<Cl−<F−<OH−<H2O<SCN−<en<NO2−<CN−<CO

Relationships on an artist’s colour wheel in which complementary colors are wedges opposite each other.

Figure 1

Answer 7

Chapter 23, Problem 23.95P

Interpretation Introduction

Interpretation:

Which ligand suitable for given complex CrL6 (green colour) has to be find out.

Concept introduction:

The fact that colour depends on the ligand and allows us to create a spectrochemical series, which ranks the ability of a ligand to split d-orbital energies.

The relative magnitude of Δ for a series of octahedral complexes of a given metal ion. As Δ increases, higher energies (shorter wavelengths) of light must be absorbed to excite electrons.

The spectrochemical series is

I−<Cl−<F−<OH−<H2O<SCN−<en<NO2−<CN−<CO

Relationships on an artist’s colour wheel in which complementary colors are wedges opposite each other.

Figure 1

Answer 8

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Answer 9

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Answer 10

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Answer 11

Ch. 23.1 - Write partial electron configurations (no noble...Ch. 23.1 - Find the charge of each of the following ions,...Ch. 23.2 - Prob. 23.2AFP Ch. 23.2 - Prob. 23.2BFP Ch. 23.3 - Give the coordination number and the charge of the...Ch. 23.3 - Write the formula of the coordination compound...Ch. 23.3 - Prob. 23.4AFP Ch. 23.3 - Prob. 23.4BFP Ch. 23.3 - Prob. 23.5AFP Ch. 23.3 - Prob. 23.5BFP

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Chapter 23 Solutions