For the given complex the orbital splitting diagram has to be drawn using spectrochemial series. [ C r ( H 2 O ) 6 ] 3 + Concept introduction: The element in the periodic table and count its position in the respective transition series. These elements are in Periods 5 and 6, so the general configuration is [ n o b l e g a s ] n s 2 ( n − 1 ) d x . 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

Question

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Answer 1

Question

Chapter 23, Problem 23.88P

(a)

Interpretation Introduction

Interpretation:

For the given complex the orbital splitting diagram has to be drawn using spectrochemial series.

[Cr(H2O)6]3+

Concept introduction:

The element in the periodic table and count its position in the respective transition series. These elements are in Periods 5 and 6, so the general configuration is

[noble gas]ns2(n− 1)dx.

The spectrochemical series is

I−<Cl−<F−<OH−<H2O<SCN−<en<NO2−<CN−<CO

(a)

Expert Solution

Explanation of Solution

Electron configuration of Cr: [Ar]4s13d5

Charge on Cr: The aqua ligands are neutral, so the charge on Cr is +3.

Electron configuration of Cr3+: [Ar]3d3

Six ligands indicate an octahedral arrangement. Using Hund’s rule, fill the lower energy

t2g orbitals first, filling empty orbitals before pairing electrons within an orbital.

MCGRAW: CHEMISTRY THE MOLECULAR NATURE, Chapter 23, Problem 23.88P , additional homework tip 1

(b)

Interpretation Introduction

Interpretation:

For the given complex the orbital splitting diagram has to be drawn using spectrochemial series.

[Cu(H2O)4]2+

Concept introduction:

The element in the periodic table and count its position in the respective transition series. These elements are in Periods 5 and 6, so the general configuration is

[noble gas]ns2(n− 1)dx.

The spectrochemical series is

I−<Cl−<F−<OH−<H2O<SCN−<en<NO2−<CN−<CO

(b)

Expert Solution

Explanation of Solution

Electron configuration of Cu: [Ar]4s13d10

Charge on Cu: The aqua ligands are neutral, so Cu has a +2 charge.

Electron configuration of Cu2+: [Ar]3d9

Four ligands and a d9 configuration indicate a square planar geometry (only filled d sublevel ions exhibit tetrahedral geometry). Use Hund’s rule to fill in the nine d electrons. Thus, the correct orbital-energy splitting diagram shows one unpaired electron.

MCGRAW: CHEMISTRY THE MOLECULAR NATURE, Chapter 23, Problem 23.88P , additional homework tip 2

(c)

Interpretation Introduction

Interpretation:

For the given complex the orbital splitting diagram has to be drawn using spectrochemial series.

[FeF6]3−

Concept introduction:

The element in the periodic table and count its position in the respective transition series. These elements are in Periods 5 and 6, so the general configuration is

[noble gas]ns2(n− 1)dx.

The spectrochemical series is

I−<Cl−<F−<OH−<H2O<SCN−<en<NO2−<CN−<CO

(c)

Expert Solution

Explanation of Solution

Electron configuration of Fe: [Ar]4s23d6

Charge on Fe: Each fluoride ligand has a –1 charge for a total charge of –6, so Fe has a +3 charge to make the overall complex charge equal to –3.

Electron configuration of Fe3+: [Ar]3d5

Six ligands indicate an octahedral arrangement. Use Hund’s rule to fill the orbitals.

F– is a weak-field ligand, so the splitting energy, Δ, is not large enough to overcome the resistance to electron pairing. The electrons remain unpaired, and the complex is called high-spin.

MCGRAW: CHEMISTRY THE MOLECULAR NATURE, Chapter 23, Problem 23.88P , additional homework tip 3

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Answer 2

Question

Chapter 23, Problem 23.88P

(a)

Interpretation Introduction

Interpretation:

For the given complex the orbital splitting diagram has to be drawn using spectrochemial series.

[Cr(H2O)6]3+

Concept introduction:

The element in the periodic table and count its position in the respective transition series. These elements are in Periods 5 and 6, so the general configuration is

[noble gas]ns2(n− 1)dx.

The spectrochemical series is

I−<Cl−<F−<OH−<H2O<SCN−<en<NO2−<CN−<CO

(a)

Expert Solution

Explanation of Solution

Electron configuration of Cr: [Ar]4s13d5

Charge on Cr: The aqua ligands are neutral, so the charge on Cr is +3.

Electron configuration of Cr3+: [Ar]3d3

Six ligands indicate an octahedral arrangement. Using Hund’s rule, fill the lower energy

t2g orbitals first, filling empty orbitals before pairing electrons within an orbital.

MCGRAW: CHEMISTRY THE MOLECULAR NATURE, Chapter 23, Problem 23.88P , additional homework tip 1

(b)

Interpretation Introduction

Interpretation:

For the given complex the orbital splitting diagram has to be drawn using spectrochemial series.

[Cu(H2O)4]2+

Concept introduction:

The element in the periodic table and count its position in the respective transition series. These elements are in Periods 5 and 6, so the general configuration is

[noble gas]ns2(n− 1)dx.

The spectrochemical series is

I−<Cl−<F−<OH−<H2O<SCN−<en<NO2−<CN−<CO

(b)

Expert Solution

Explanation of Solution

Electron configuration of Cu: [Ar]4s13d10

Charge on Cu: The aqua ligands are neutral, so Cu has a +2 charge.

Electron configuration of Cu2+: [Ar]3d9

Four ligands and a d9 configuration indicate a square planar geometry (only filled d sublevel ions exhibit tetrahedral geometry). Use Hund’s rule to fill in the nine d electrons. Thus, the correct orbital-energy splitting diagram shows one unpaired electron.

MCGRAW: CHEMISTRY THE MOLECULAR NATURE, Chapter 23, Problem 23.88P , additional homework tip 2

(c)

Interpretation Introduction

Interpretation:

For the given complex the orbital splitting diagram has to be drawn using spectrochemial series.

[FeF6]3−

Concept introduction:

The element in the periodic table and count its position in the respective transition series. These elements are in Periods 5 and 6, so the general configuration is

[noble gas]ns2(n− 1)dx.

The spectrochemical series is

I−<Cl−<F−<OH−<H2O<SCN−<en<NO2−<CN−<CO

(c)

Expert Solution

Explanation of Solution

Electron configuration of Fe: [Ar]4s23d6

Charge on Fe: Each fluoride ligand has a –1 charge for a total charge of –6, so Fe has a +3 charge to make the overall complex charge equal to –3.

Electron configuration of Fe3+: [Ar]3d5

Six ligands indicate an octahedral arrangement. Use Hund’s rule to fill the orbitals.

F– is a weak-field ligand, so the splitting energy, Δ, is not large enough to overcome the resistance to electron pairing. The electrons remain unpaired, and the complex is called high-spin.

MCGRAW: CHEMISTRY THE MOLECULAR NATURE, Chapter 23, Problem 23.88P , additional homework tip 3

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Chapter 23, Problem 23.88P

(a)

Interpretation Introduction

Interpretation:

For the given complex the orbital splitting diagram has to be drawn using spectrochemial series.

[Cr(H2O)6]3+

Concept introduction:

The element in the periodic table and count its position in the respective transition series. These elements are in Periods 5 and 6, so the general configuration is

[noble gas]ns2(n− 1)dx.

The spectrochemical series is

I−<Cl−<F−<OH−<H2O<SCN−<en<NO2−<CN−<CO

(a)

Expert Solution

Explanation of Solution

Electron configuration of Cr: [Ar]4s13d5

Charge on Cr: The aqua ligands are neutral, so the charge on Cr is +3.

Electron configuration of Cr3+: [Ar]3d3

Six ligands indicate an octahedral arrangement. Using Hund’s rule, fill the lower energy

t2g orbitals first, filling empty orbitals before pairing electrons within an orbital.

MCGRAW: CHEMISTRY THE MOLECULAR NATURE, Chapter 23, Problem 23.88P , additional homework tip 1

(b)

Interpretation Introduction

Interpretation:

For the given complex the orbital splitting diagram has to be drawn using spectrochemial series.

[Cu(H2O)4]2+

Concept introduction:

The element in the periodic table and count its position in the respective transition series. These elements are in Periods 5 and 6, so the general configuration is

[noble gas]ns2(n− 1)dx.

The spectrochemical series is

I−<Cl−<F−<OH−<H2O<SCN−<en<NO2−<CN−<CO

(b)

Expert Solution

Explanation of Solution

Electron configuration of Cu: [Ar]4s13d10

Charge on Cu: The aqua ligands are neutral, so Cu has a +2 charge.

Electron configuration of Cu2+: [Ar]3d9

Four ligands and a d9 configuration indicate a square planar geometry (only filled d sublevel ions exhibit tetrahedral geometry). Use Hund’s rule to fill in the nine d electrons. Thus, the correct orbital-energy splitting diagram shows one unpaired electron.

MCGRAW: CHEMISTRY THE MOLECULAR NATURE, Chapter 23, Problem 23.88P , additional homework tip 2

(c)

Interpretation Introduction

Interpretation:

For the given complex the orbital splitting diagram has to be drawn using spectrochemial series.

[FeF6]3−

Concept introduction:

The element in the periodic table and count its position in the respective transition series. These elements are in Periods 5 and 6, so the general configuration is

[noble gas]ns2(n− 1)dx.

The spectrochemical series is

I−<Cl−<F−<OH−<H2O<SCN−<en<NO2−<CN−<CO

(c)

Expert Solution

Explanation of Solution

Electron configuration of Fe: [Ar]4s23d6

Charge on Fe: Each fluoride ligand has a –1 charge for a total charge of –6, so Fe has a +3 charge to make the overall complex charge equal to –3.

Electron configuration of Fe3+: [Ar]3d5

Six ligands indicate an octahedral arrangement. Use Hund’s rule to fill the orbitals.

F– is a weak-field ligand, so the splitting energy, Δ, is not large enough to overcome the resistance to electron pairing. The electrons remain unpaired, and the complex is called high-spin.

MCGRAW: CHEMISTRY THE MOLECULAR NATURE, Chapter 23, Problem 23.88P , additional homework tip 3

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Answer 4

Question

Chapter 23, Problem 23.88P

(a)

Interpretation Introduction

Interpretation:

For the given complex the orbital splitting diagram has to be drawn using spectrochemial series.

[Cr(H2O)6]3+

Concept introduction:

The element in the periodic table and count its position in the respective transition series. These elements are in Periods 5 and 6, so the general configuration is

[noble gas]ns2(n− 1)dx.

The spectrochemical series is

I−<Cl−<F−<OH−<H2O<SCN−<en<NO2−<CN−<CO

(a)

Expert Solution

Explanation of Solution

Electron configuration of Cr: [Ar]4s13d5

Charge on Cr: The aqua ligands are neutral, so the charge on Cr is +3.

Electron configuration of Cr3+: [Ar]3d3

Six ligands indicate an octahedral arrangement. Using Hund’s rule, fill the lower energy

t2g orbitals first, filling empty orbitals before pairing electrons within an orbital.

MCGRAW: CHEMISTRY THE MOLECULAR NATURE, Chapter 23, Problem 23.88P , additional homework tip 1

(b)

Interpretation Introduction

Interpretation:

For the given complex the orbital splitting diagram has to be drawn using spectrochemial series.

[Cu(H2O)4]2+

Concept introduction:

The element in the periodic table and count its position in the respective transition series. These elements are in Periods 5 and 6, so the general configuration is

[noble gas]ns2(n− 1)dx.

The spectrochemical series is

I−<Cl−<F−<OH−<H2O<SCN−<en<NO2−<CN−<CO

(b)

Expert Solution

Explanation of Solution

Electron configuration of Cu: [Ar]4s13d10

Charge on Cu: The aqua ligands are neutral, so Cu has a +2 charge.

Electron configuration of Cu2+: [Ar]3d9

Four ligands and a d9 configuration indicate a square planar geometry (only filled d sublevel ions exhibit tetrahedral geometry). Use Hund’s rule to fill in the nine d electrons. Thus, the correct orbital-energy splitting diagram shows one unpaired electron.

MCGRAW: CHEMISTRY THE MOLECULAR NATURE, Chapter 23, Problem 23.88P , additional homework tip 2

(c)

Interpretation Introduction

Interpretation:

For the given complex the orbital splitting diagram has to be drawn using spectrochemial series.

[FeF6]3−

Concept introduction:

The element in the periodic table and count its position in the respective transition series. These elements are in Periods 5 and 6, so the general configuration is

[noble gas]ns2(n− 1)dx.

The spectrochemical series is

I−<Cl−<F−<OH−<H2O<SCN−<en<NO2−<CN−<CO

(c)

Expert Solution

Explanation of Solution

Electron configuration of Fe: [Ar]4s23d6

Charge on Fe: Each fluoride ligand has a –1 charge for a total charge of –6, so Fe has a +3 charge to make the overall complex charge equal to –3.

Electron configuration of Fe3+: [Ar]3d5

Six ligands indicate an octahedral arrangement. Use Hund’s rule to fill the orbitals.

F– is a weak-field ligand, so the splitting energy, Δ, is not large enough to overcome the resistance to electron pairing. The electrons remain unpaired, and the complex is called high-spin.

MCGRAW: CHEMISTRY THE MOLECULAR NATURE, Chapter 23, Problem 23.88P , additional homework tip 3

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 23, Problem 23.88P

(a)

Interpretation Introduction

Interpretation:

For the given complex the orbital splitting diagram has to be drawn using spectrochemial series.

[Cr(H2O)6]3+

Concept introduction:

The element in the periodic table and count its position in the respective transition series. These elements are in Periods 5 and 6, so the general configuration is

[noble gas]ns2(n− 1)dx.

The spectrochemical series is

I−<Cl−<F−<OH−<H2O<SCN−<en<NO2−<CN−<CO

(a)

Expert Solution

Explanation of Solution

Electron configuration of Cr: [Ar]4s13d5

Charge on Cr: The aqua ligands are neutral, so the charge on Cr is +3.

Electron configuration of Cr3+: [Ar]3d3

Six ligands indicate an octahedral arrangement. Using Hund’s rule, fill the lower energy

t2g orbitals first, filling empty orbitals before pairing electrons within an orbital.

MCGRAW: CHEMISTRY THE MOLECULAR NATURE, Chapter 23, Problem 23.88P , additional homework tip 1

(b)

Interpretation Introduction

Interpretation:

For the given complex the orbital splitting diagram has to be drawn using spectrochemial series.

[Cu(H2O)4]2+

Concept introduction:

The element in the periodic table and count its position in the respective transition series. These elements are in Periods 5 and 6, so the general configuration is

[noble gas]ns2(n− 1)dx.

The spectrochemical series is

I−<Cl−<F−<OH−<H2O<SCN−<en<NO2−<CN−<CO

(b)

Expert Solution

Explanation of Solution

Electron configuration of Cu: [Ar]4s13d10

Charge on Cu: The aqua ligands are neutral, so Cu has a +2 charge.

Electron configuration of Cu2+: [Ar]3d9

Four ligands and a d9 configuration indicate a square planar geometry (only filled d sublevel ions exhibit tetrahedral geometry). Use Hund’s rule to fill in the nine d electrons. Thus, the correct orbital-energy splitting diagram shows one unpaired electron.

MCGRAW: CHEMISTRY THE MOLECULAR NATURE, Chapter 23, Problem 23.88P , additional homework tip 2

(c)

Interpretation Introduction

Interpretation:

For the given complex the orbital splitting diagram has to be drawn using spectrochemial series.

[FeF6]3−

Concept introduction:

The element in the periodic table and count its position in the respective transition series. These elements are in Periods 5 and 6, so the general configuration is

[noble gas]ns2(n− 1)dx.

The spectrochemical series is

I−<Cl−<F−<OH−<H2O<SCN−<en<NO2−<CN−<CO

(c)

Expert Solution

Explanation of Solution

Electron configuration of Fe: [Ar]4s23d6

Charge on Fe: Each fluoride ligand has a –1 charge for a total charge of –6, so Fe has a +3 charge to make the overall complex charge equal to –3.

Electron configuration of Fe3+: [Ar]3d5

Six ligands indicate an octahedral arrangement. Use Hund’s rule to fill the orbitals.

F– is a weak-field ligand, so the splitting energy, Δ, is not large enough to overcome the resistance to electron pairing. The electrons remain unpaired, and the complex is called high-spin.

MCGRAW: CHEMISTRY THE MOLECULAR NATURE, Chapter 23, Problem 23.88P , additional homework tip 3

Answer 6

Chapter 23, Problem 23.88P

(a)

Interpretation Introduction

Interpretation:

For the given complex the orbital splitting diagram has to be drawn using spectrochemial series.

[Cr(H2O)6]3+

Concept introduction:

The element in the periodic table and count its position in the respective transition series. These elements are in Periods 5 and 6, so the general configuration is

[noble gas]ns2(n− 1)dx.

The spectrochemical series is

I−<Cl−<F−<OH−<H2O<SCN−<en<NO2−<CN−<CO

(a)

Expert Solution

Explanation of Solution

Electron configuration of Cr: [Ar]4s13d5

Charge on Cr: The aqua ligands are neutral, so the charge on Cr is +3.

Electron configuration of Cr3+: [Ar]3d3

Six ligands indicate an octahedral arrangement. Using Hund’s rule, fill the lower energy

t2g orbitals first, filling empty orbitals before pairing electrons within an orbital.

MCGRAW: CHEMISTRY THE MOLECULAR NATURE, Chapter 23, Problem 23.88P , additional homework tip 1

Answer 7

Chapter 23, Problem 23.88P

(a)

Interpretation Introduction

Interpretation:

For the given complex the orbital splitting diagram has to be drawn using spectrochemial series.

[Cr(H2O)6]3+

Concept introduction:

The element in the periodic table and count its position in the respective transition series. These elements are in Periods 5 and 6, so the general configuration is

[noble gas]ns2(n− 1)dx.

The spectrochemical series is

I−<Cl−<F−<OH−<H2O<SCN−<en<NO2−<CN−<CO

Answer 8

(a)

Expert Solution

Explanation of Solution

Electron configuration of Cr: [Ar]4s13d5

Charge on Cr: The aqua ligands are neutral, so the charge on Cr is +3.

Electron configuration of Cr3+: [Ar]3d3

Six ligands indicate an octahedral arrangement. Using Hund’s rule, fill the lower energy

t2g orbitals first, filling empty orbitals before pairing electrons within an orbital.

MCGRAW: CHEMISTRY THE MOLECULAR NATURE, Chapter 23, Problem 23.88P , additional homework tip 1

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

(b)

Interpretation Introduction

Interpretation:

For the given complex the orbital splitting diagram has to be drawn using spectrochemial series.

[Cu(H2O)4]2+

Concept introduction:

The element in the periodic table and count its position in the respective transition series. These elements are in Periods 5 and 6, so the general configuration is

[noble gas]ns2(n− 1)dx.

The spectrochemical series is

I−<Cl−<F−<OH−<H2O<SCN−<en<NO2−<CN−<CO

(b)

Expert Solution

Explanation of Solution

Electron configuration of Cu: [Ar]4s13d10

Charge on Cu: The aqua ligands are neutral, so Cu has a +2 charge.

Electron configuration of Cu2+: [Ar]3d9

Four ligands and a d9 configuration indicate a square planar geometry (only filled d sublevel ions exhibit tetrahedral geometry). Use Hund’s rule to fill in the nine d electrons. Thus, the correct orbital-energy splitting diagram shows one unpaired electron.

MCGRAW: CHEMISTRY THE MOLECULAR NATURE, Chapter 23, Problem 23.88P , additional homework tip 2

Answer 13

(b)

Interpretation Introduction

Interpretation:

For the given complex the orbital splitting diagram has to be drawn using spectrochemial series.

[Cu(H2O)4]2+

Concept introduction:

The element in the periodic table and count its position in the respective transition series. These elements are in Periods 5 and 6, so the general configuration is

[noble gas]ns2(n− 1)dx.

The spectrochemical series is

I−<Cl−<F−<OH−<H2O<SCN−<en<NO2−<CN−<CO

Answer 14

(b)

Expert Solution

Explanation of Solution

Electron configuration of Cu: [Ar]4s13d10

Charge on Cu: The aqua ligands are neutral, so Cu has a +2 charge.

Electron configuration of Cu2+: [Ar]3d9

Four ligands and a d9 configuration indicate a square planar geometry (only filled d sublevel ions exhibit tetrahedral geometry). Use Hund’s rule to fill in the nine d electrons. Thus, the correct orbital-energy splitting diagram shows one unpaired electron.

MCGRAW: CHEMISTRY THE MOLECULAR NATURE, Chapter 23, Problem 23.88P , additional homework tip 2

Answer 15

(b)

Expert Solution

Answer 16

(b)

Expert Solution

Answer 17

Expert Solution

Answer 18

(c)

Interpretation Introduction

Interpretation:

For the given complex the orbital splitting diagram has to be drawn using spectrochemial series.

[FeF6]3−

Concept introduction:

The element in the periodic table and count its position in the respective transition series. These elements are in Periods 5 and 6, so the general configuration is

[noble gas]ns2(n− 1)dx.

The spectrochemical series is

I−<Cl−<F−<OH−<H2O<SCN−<en<NO2−<CN−<CO

(c)

Expert Solution

Explanation of Solution

Electron configuration of Fe: [Ar]4s23d6

Charge on Fe: Each fluoride ligand has a –1 charge for a total charge of –6, so Fe has a +3 charge to make the overall complex charge equal to –3.

Electron configuration of Fe3+: [Ar]3d5

Six ligands indicate an octahedral arrangement. Use Hund’s rule to fill the orbitals.

F– is a weak-field ligand, so the splitting energy, Δ, is not large enough to overcome the resistance to electron pairing. The electrons remain unpaired, and the complex is called high-spin.

MCGRAW: CHEMISTRY THE MOLECULAR NATURE, Chapter 23, Problem 23.88P , additional homework tip 3

Answer 19

(c)

Interpretation Introduction

Interpretation:

For the given complex the orbital splitting diagram has to be drawn using spectrochemial series.

[FeF6]3−

Concept introduction:

The element in the periodic table and count its position in the respective transition series. These elements are in Periods 5 and 6, so the general configuration is

[noble gas]ns2(n− 1)dx.

The spectrochemical series is

I−<Cl−<F−<OH−<H2O<SCN−<en<NO2−<CN−<CO

Answer 20

(c)

Expert Solution

Explanation of Solution

Electron configuration of Fe: [Ar]4s23d6

Charge on Fe: Each fluoride ligand has a –1 charge for a total charge of –6, so Fe has a +3 charge to make the overall complex charge equal to –3.

Electron configuration of Fe3+: [Ar]3d5

Six ligands indicate an octahedral arrangement. Use Hund’s rule to fill the orbitals.

F– is a weak-field ligand, so the splitting energy, Δ, is not large enough to overcome the resistance to electron pairing. The electrons remain unpaired, and the complex is called high-spin.

MCGRAW: CHEMISTRY THE MOLECULAR NATURE, Chapter 23, Problem 23.88P , additional homework tip 3