For the given complex the orbital splitting diagram has to be drawn using spectrochemial series. [ C r ( C N ) 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 . Spectrochemical series: The list of ligands arranged in an ascending order of (Δ) (the splitting of d-orbitals in presence of various ligands). → I - < Br - < SCN - < Cl - < S 2- < F - < OH - < O 2- < H 2 O < NCS - < edta 4- < NH 3 < en < NO 2- < CN - < CO weak-field increasing(Δ) strong-field ligands ligands Crystal field splitting: The energy gap between the splitting of d-orbitals of the metal ion in presence of ligands is known as the crystal field splitting (Δ) . The magnitude of (Δ) is depends on the nature of metal ions and the ligands. Splitting of five d-orbitals in an octahedral crystals field is as follows: Figure 1

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Answer 1

Question

Chapter 23, Problem 23.89P

(a)

Interpretation Introduction

Interpretation:

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

$[Cr(CN)6]3−$

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

$[noble gas]ns2(n− 1)dx.$

Spectrochemical series: The list of ligands arranged in an ascending order of $(Δ)$ (the splitting of d-orbitals in presence of various ligands).

$→I-< Br-< SCN-< Cl-< S2-< F-< OH-< O2-< H2O < NCS-< edta4-< NH3< en < NO2-< CN-< COweak-field increasing(Δ) strong-field ligands ligands$

Crystal field splitting: The energy gap between the splitting of d-orbitals of the metal ion in presence of ligands is known as the crystal field splitting $(Δ)$ . The magnitude of $(Δ)$ is depends on the nature of metal ions and the ligands.

Splitting of five d-orbitals in an octahedral crystals field is as follows:

Figure 1

(a)

Expert Solution

Explanation of Solution

Given,

$[Cr(CN)6]3−$

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.

CONNECT ACCESS CARD FOR CHEMISTRY: MOLECULAR NATURE OF MATTER AND CHANGE, Chapter 23, Problem 23.89P , additional homework tip 2

(b)

Interpretation Introduction

Interpretation:

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

$[Rh(CO)6]3+$

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

$[noble gas]ns2(n− 1)dx.$

Spectrochemical series: The list of ligands arranged in an ascending order of $(Δ)$ (the splitting of d-orbitals in presence of various ligands).

$→I-< Br-< SCN-< Cl-< S2-< F-< OH-< O2-< H2O < NCS-< edta4-< NH3< en < NO2-< CN-< COweak-field increasing(Δ) strong-field ligands ligands$

Crystal field splitting: The energy gap between the splitting of d-orbitals of the metal ion in presence of ligands is known as the crystal field splitting $(Δ)$ . The magnitude of $(Δ)$ is depends on the nature of metal ions and the ligands.

Splitting of five d-orbitals in an octahedral crystals field is as follows:

Figure 1

(b)

Expert Solution

Explanation of Solution

Given,

$[Rh(CO)6]3+$

Electron configuration of $Rh: [Ar]4s13d8$

Charge on $Rh$ : The carbonyl ligands are neutral, so $Rh$ has a +3 charge.

Electron configuration of $Rh3+: [Ar]3d6$

Four ligands and a $d6$ configuration indicate an octahedral geometry. Use Hund’s rule to fill in the nine d electrons. Thus, the carbonyl ligands is strong so the correct orbital-energy splitting diagram shows no unpaired electron.

CONNECT ACCESS CARD FOR CHEMISTRY: MOLECULAR NATURE OF MATTER AND CHANGE, Chapter 23, Problem 23.89P , additional homework tip 4

(c)

Interpretation Introduction

Interpretation:

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

$[Co(OH)6]4−$

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

$[noble gas]ns2(n− 1)dx.$

Spectrochemical series: The list of ligands arranged in an ascending order of $(Δ)$ (the splitting of d-orbitals in presence of various ligands).

$→I-< Br-< SCN-< Cl-< S2-< F-< OH-< O2-< H2O < NCS-< edta4-< NH3< en < NO2-< CN-< COweak-field increasing(Δ) strong-field ligands ligands$

Crystal field splitting: The energy gap between the splitting of d-orbitals of the metal ion in presence of ligands is known as the crystal field splitting $(Δ)$ . The magnitude of $(Δ)$ is depends on the nature of metal ions and the ligands.

Splitting of five d-orbitals in an octahedral crystals field is as follows:

Figure 1

(c)

Expert Solution

Explanation of Solution

Electron configuration of $Co: [Ar]4s23d7$

Charge on $Co$ : water is a neutral ligand so $Co$ has a +2 charge to make the overall complex charge equal to –4.

Electron configuration of $Co2+: [Ar]3d7$

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

Water – 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.

CONNECT ACCESS CARD FOR CHEMISTRY: MOLECULAR NATURE OF MATTER AND CHANGE, Chapter 23, Problem 23.89P , additional homework tip 6

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can you please draw out and list step-by-step the synthetic strategy for this rxn? thank you sm in advance

Answer 2

Question

Chapter 23, Problem 23.89P

(a)

Interpretation Introduction

Interpretation:

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

$[Cr(CN)6]3−$

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

$[noble gas]ns2(n− 1)dx.$

Spectrochemical series: The list of ligands arranged in an ascending order of $(Δ)$ (the splitting of d-orbitals in presence of various ligands).

$→I-< Br-< SCN-< Cl-< S2-< F-< OH-< O2-< H2O < NCS-< edta4-< NH3< en < NO2-< CN-< COweak-field increasing(Δ) strong-field ligands ligands$

Crystal field splitting: The energy gap between the splitting of d-orbitals of the metal ion in presence of ligands is known as the crystal field splitting $(Δ)$ . The magnitude of $(Δ)$ is depends on the nature of metal ions and the ligands.

Splitting of five d-orbitals in an octahedral crystals field is as follows:

Figure 1

(a)

Expert Solution

Explanation of Solution

Given,

$[Cr(CN)6]3−$

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.

CONNECT ACCESS CARD FOR CHEMISTRY: MOLECULAR NATURE OF MATTER AND CHANGE, Chapter 23, Problem 23.89P , additional homework tip 2

(b)

Interpretation Introduction

Interpretation:

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

$[Rh(CO)6]3+$

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

$[noble gas]ns2(n− 1)dx.$

Spectrochemical series: The list of ligands arranged in an ascending order of $(Δ)$ (the splitting of d-orbitals in presence of various ligands).

$→I-< Br-< SCN-< Cl-< S2-< F-< OH-< O2-< H2O < NCS-< edta4-< NH3< en < NO2-< CN-< COweak-field increasing(Δ) strong-field ligands ligands$

Crystal field splitting: The energy gap between the splitting of d-orbitals of the metal ion in presence of ligands is known as the crystal field splitting $(Δ)$ . The magnitude of $(Δ)$ is depends on the nature of metal ions and the ligands.

Splitting of five d-orbitals in an octahedral crystals field is as follows:

Figure 1

(b)

Expert Solution

Explanation of Solution

Given,

$[Rh(CO)6]3+$

Electron configuration of $Rh: [Ar]4s13d8$

Charge on $Rh$ : The carbonyl ligands are neutral, so $Rh$ has a +3 charge.

Electron configuration of $Rh3+: [Ar]3d6$

Four ligands and a $d6$ configuration indicate an octahedral geometry. Use Hund’s rule to fill in the nine d electrons. Thus, the carbonyl ligands is strong so the correct orbital-energy splitting diagram shows no unpaired electron.

CONNECT ACCESS CARD FOR CHEMISTRY: MOLECULAR NATURE OF MATTER AND CHANGE, Chapter 23, Problem 23.89P , additional homework tip 4

(c)

Interpretation Introduction

Interpretation:

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

$[Co(OH)6]4−$

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

$[noble gas]ns2(n− 1)dx.$

Spectrochemical series: The list of ligands arranged in an ascending order of $(Δ)$ (the splitting of d-orbitals in presence of various ligands).

$→I-< Br-< SCN-< Cl-< S2-< F-< OH-< O2-< H2O < NCS-< edta4-< NH3< en < NO2-< CN-< COweak-field increasing(Δ) strong-field ligands ligands$

Crystal field splitting: The energy gap between the splitting of d-orbitals of the metal ion in presence of ligands is known as the crystal field splitting $(Δ)$ . The magnitude of $(Δ)$ is depends on the nature of metal ions and the ligands.

Splitting of five d-orbitals in an octahedral crystals field is as follows:

Figure 1

(c)

Expert Solution

Explanation of Solution

Electron configuration of $Co: [Ar]4s23d7$

Charge on $Co$ : water is a neutral ligand so $Co$ has a +2 charge to make the overall complex charge equal to –4.

Electron configuration of $Co2+: [Ar]3d7$

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

Water – 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.

CONNECT ACCESS CARD FOR CHEMISTRY: MOLECULAR NATURE OF MATTER AND CHANGE, Chapter 23, Problem 23.89P , additional homework tip 6

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Answer 3

Question

Chapter 23, Problem 23.89P

(a)

Interpretation Introduction

Interpretation:

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

$[Cr(CN)6]3−$

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

$[noble gas]ns2(n− 1)dx.$

Spectrochemical series: The list of ligands arranged in an ascending order of $(Δ)$ (the splitting of d-orbitals in presence of various ligands).

$→I-< Br-< SCN-< Cl-< S2-< F-< OH-< O2-< H2O < NCS-< edta4-< NH3< en < NO2-< CN-< COweak-field increasing(Δ) strong-field ligands ligands$

Crystal field splitting: The energy gap between the splitting of d-orbitals of the metal ion in presence of ligands is known as the crystal field splitting $(Δ)$ . The magnitude of $(Δ)$ is depends on the nature of metal ions and the ligands.

Splitting of five d-orbitals in an octahedral crystals field is as follows:

Figure 1

(a)

Expert Solution

Explanation of Solution

Given,

$[Cr(CN)6]3−$

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.

CONNECT ACCESS CARD FOR CHEMISTRY: MOLECULAR NATURE OF MATTER AND CHANGE, Chapter 23, Problem 23.89P , additional homework tip 2

(b)

Interpretation Introduction

Interpretation:

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

$[Rh(CO)6]3+$

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

$[noble gas]ns2(n− 1)dx.$

Spectrochemical series: The list of ligands arranged in an ascending order of $(Δ)$ (the splitting of d-orbitals in presence of various ligands).

$→I-< Br-< SCN-< Cl-< S2-< F-< OH-< O2-< H2O < NCS-< edta4-< NH3< en < NO2-< CN-< COweak-field increasing(Δ) strong-field ligands ligands$

Crystal field splitting: The energy gap between the splitting of d-orbitals of the metal ion in presence of ligands is known as the crystal field splitting $(Δ)$ . The magnitude of $(Δ)$ is depends on the nature of metal ions and the ligands.

Splitting of five d-orbitals in an octahedral crystals field is as follows:

Figure 1

(b)

Expert Solution

Explanation of Solution

Given,

$[Rh(CO)6]3+$

Electron configuration of $Rh: [Ar]4s13d8$

Charge on $Rh$ : The carbonyl ligands are neutral, so $Rh$ has a +3 charge.

Electron configuration of $Rh3+: [Ar]3d6$

Four ligands and a $d6$ configuration indicate an octahedral geometry. Use Hund’s rule to fill in the nine d electrons. Thus, the carbonyl ligands is strong so the correct orbital-energy splitting diagram shows no unpaired electron.

CONNECT ACCESS CARD FOR CHEMISTRY: MOLECULAR NATURE OF MATTER AND CHANGE, Chapter 23, Problem 23.89P , additional homework tip 4

(c)

Interpretation Introduction

Interpretation:

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

$[Co(OH)6]4−$

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

$[noble gas]ns2(n− 1)dx.$

Spectrochemical series: The list of ligands arranged in an ascending order of $(Δ)$ (the splitting of d-orbitals in presence of various ligands).

$→I-< Br-< SCN-< Cl-< S2-< F-< OH-< O2-< H2O < NCS-< edta4-< NH3< en < NO2-< CN-< COweak-field increasing(Δ) strong-field ligands ligands$

Crystal field splitting: The energy gap between the splitting of d-orbitals of the metal ion in presence of ligands is known as the crystal field splitting $(Δ)$ . The magnitude of $(Δ)$ is depends on the nature of metal ions and the ligands.

Splitting of five d-orbitals in an octahedral crystals field is as follows:

Figure 1

(c)

Expert Solution

Explanation of Solution

Electron configuration of $Co: [Ar]4s23d7$

Charge on $Co$ : water is a neutral ligand so $Co$ has a +2 charge to make the overall complex charge equal to –4.

Electron configuration of $Co2+: [Ar]3d7$

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

Water – 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.

CONNECT ACCESS CARD FOR CHEMISTRY: MOLECULAR NATURE OF MATTER AND CHANGE, Chapter 23, Problem 23.89P , additional homework tip 6

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Answer 4

Question

Chapter 23, Problem 23.89P

(a)

Interpretation Introduction

Interpretation:

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

$[Cr(CN)6]3−$

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

$[noble gas]ns2(n− 1)dx.$

Spectrochemical series: The list of ligands arranged in an ascending order of $(Δ)$ (the splitting of d-orbitals in presence of various ligands).

$→I-< Br-< SCN-< Cl-< S2-< F-< OH-< O2-< H2O < NCS-< edta4-< NH3< en < NO2-< CN-< COweak-field increasing(Δ) strong-field ligands ligands$

Crystal field splitting: The energy gap between the splitting of d-orbitals of the metal ion in presence of ligands is known as the crystal field splitting $(Δ)$ . The magnitude of $(Δ)$ is depends on the nature of metal ions and the ligands.

Splitting of five d-orbitals in an octahedral crystals field is as follows:

Figure 1

(a)

Expert Solution

Explanation of Solution

Given,

$[Cr(CN)6]3−$

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.

CONNECT ACCESS CARD FOR CHEMISTRY: MOLECULAR NATURE OF MATTER AND CHANGE, Chapter 23, Problem 23.89P , additional homework tip 2

(b)

Interpretation Introduction

Interpretation:

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

$[Rh(CO)6]3+$

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

$[noble gas]ns2(n− 1)dx.$

Spectrochemical series: The list of ligands arranged in an ascending order of $(Δ)$ (the splitting of d-orbitals in presence of various ligands).

$→I-< Br-< SCN-< Cl-< S2-< F-< OH-< O2-< H2O < NCS-< edta4-< NH3< en < NO2-< CN-< COweak-field increasing(Δ) strong-field ligands ligands$

Crystal field splitting: The energy gap between the splitting of d-orbitals of the metal ion in presence of ligands is known as the crystal field splitting $(Δ)$ . The magnitude of $(Δ)$ is depends on the nature of metal ions and the ligands.

Splitting of five d-orbitals in an octahedral crystals field is as follows:

Figure 1

(b)

Expert Solution

Explanation of Solution

Given,

$[Rh(CO)6]3+$

Electron configuration of $Rh: [Ar]4s13d8$

Charge on $Rh$ : The carbonyl ligands are neutral, so $Rh$ has a +3 charge.

Electron configuration of $Rh3+: [Ar]3d6$

Four ligands and a $d6$ configuration indicate an octahedral geometry. Use Hund’s rule to fill in the nine d electrons. Thus, the carbonyl ligands is strong so the correct orbital-energy splitting diagram shows no unpaired electron.

CONNECT ACCESS CARD FOR CHEMISTRY: MOLECULAR NATURE OF MATTER AND CHANGE, Chapter 23, Problem 23.89P , additional homework tip 4

(c)

Interpretation Introduction

Interpretation:

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

$[Co(OH)6]4−$

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

$[noble gas]ns2(n− 1)dx.$

Spectrochemical series: The list of ligands arranged in an ascending order of $(Δ)$ (the splitting of d-orbitals in presence of various ligands).

$→I-< Br-< SCN-< Cl-< S2-< F-< OH-< O2-< H2O < NCS-< edta4-< NH3< en < NO2-< CN-< COweak-field increasing(Δ) strong-field ligands ligands$

Crystal field splitting: The energy gap between the splitting of d-orbitals of the metal ion in presence of ligands is known as the crystal field splitting $(Δ)$ . The magnitude of $(Δ)$ is depends on the nature of metal ions and the ligands.

Splitting of five d-orbitals in an octahedral crystals field is as follows:

Figure 1

(c)

Expert Solution

Explanation of Solution

Electron configuration of $Co: [Ar]4s23d7$

Charge on $Co$ : water is a neutral ligand so $Co$ has a +2 charge to make the overall complex charge equal to –4.

Electron configuration of $Co2+: [Ar]3d7$

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

Water – 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.

CONNECT ACCESS CARD FOR CHEMISTRY: MOLECULAR NATURE OF MATTER AND CHANGE, Chapter 23, Problem 23.89P , additional homework tip 6

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Answer 5

Chapter 23, Problem 23.89P

(a)

Interpretation Introduction

Interpretation:

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

$[Cr(CN)6]3−$

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

$[noble gas]ns2(n− 1)dx.$

Spectrochemical series: The list of ligands arranged in an ascending order of $(Δ)$ (the splitting of d-orbitals in presence of various ligands).

$→I-< Br-< SCN-< Cl-< S2-< F-< OH-< O2-< H2O < NCS-< edta4-< NH3< en < NO2-< CN-< COweak-field increasing(Δ) strong-field ligands ligands$

Crystal field splitting: The energy gap between the splitting of d-orbitals of the metal ion in presence of ligands is known as the crystal field splitting $(Δ)$ . The magnitude of $(Δ)$ is depends on the nature of metal ions and the ligands.

Splitting of five d-orbitals in an octahedral crystals field is as follows:

Figure 1

(a)

Expert Solution

Explanation of Solution

Given,

$[Cr(CN)6]3−$

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.

CONNECT ACCESS CARD FOR CHEMISTRY: MOLECULAR NATURE OF MATTER AND CHANGE, Chapter 23, Problem 23.89P , additional homework tip 2

(b)

Interpretation Introduction

Interpretation:

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

$[Rh(CO)6]3+$

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

$[noble gas]ns2(n− 1)dx.$

Spectrochemical series: The list of ligands arranged in an ascending order of $(Δ)$ (the splitting of d-orbitals in presence of various ligands).

$→I-< Br-< SCN-< Cl-< S2-< F-< OH-< O2-< H2O < NCS-< edta4-< NH3< en < NO2-< CN-< COweak-field increasing(Δ) strong-field ligands ligands$

Crystal field splitting: The energy gap between the splitting of d-orbitals of the metal ion in presence of ligands is known as the crystal field splitting $(Δ)$ . The magnitude of $(Δ)$ is depends on the nature of metal ions and the ligands.

Splitting of five d-orbitals in an octahedral crystals field is as follows:

Figure 1

(b)

Expert Solution

Explanation of Solution

Given,

$[Rh(CO)6]3+$

Electron configuration of $Rh: [Ar]4s13d8$

Charge on $Rh$ : The carbonyl ligands are neutral, so $Rh$ has a +3 charge.

Electron configuration of $Rh3+: [Ar]3d6$

Four ligands and a $d6$ configuration indicate an octahedral geometry. Use Hund’s rule to fill in the nine d electrons. Thus, the carbonyl ligands is strong so the correct orbital-energy splitting diagram shows no unpaired electron.

CONNECT ACCESS CARD FOR CHEMISTRY: MOLECULAR NATURE OF MATTER AND CHANGE, Chapter 23, Problem 23.89P , additional homework tip 4

(c)

Interpretation Introduction

Interpretation:

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

$[Co(OH)6]4−$

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

$[noble gas]ns2(n− 1)dx.$

Spectrochemical series: The list of ligands arranged in an ascending order of $(Δ)$ (the splitting of d-orbitals in presence of various ligands).

$→I-< Br-< SCN-< Cl-< S2-< F-< OH-< O2-< H2O < NCS-< edta4-< NH3< en < NO2-< CN-< COweak-field increasing(Δ) strong-field ligands ligands$

Crystal field splitting: The energy gap between the splitting of d-orbitals of the metal ion in presence of ligands is known as the crystal field splitting $(Δ)$ . The magnitude of $(Δ)$ is depends on the nature of metal ions and the ligands.

Splitting of five d-orbitals in an octahedral crystals field is as follows:

Figure 1

(c)

Expert Solution

Explanation of Solution

Electron configuration of $Co: [Ar]4s23d7$

Charge on $Co$ : water is a neutral ligand so $Co$ has a +2 charge to make the overall complex charge equal to –4.

Electron configuration of $Co2+: [Ar]3d7$

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

Water – 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.

CONNECT ACCESS CARD FOR CHEMISTRY: MOLECULAR NATURE OF MATTER AND CHANGE, Chapter 23, Problem 23.89P , additional homework tip 6

Answer 6

Chapter 23, Problem 23.89P

(a)

Interpretation Introduction

Interpretation:

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

$[Cr(CN)6]3−$

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

$[noble gas]ns2(n− 1)dx.$

Spectrochemical series: The list of ligands arranged in an ascending order of $(Δ)$ (the splitting of d-orbitals in presence of various ligands).

$→I-< Br-< SCN-< Cl-< S2-< F-< OH-< O2-< H2O < NCS-< edta4-< NH3< en < NO2-< CN-< COweak-field increasing(Δ) strong-field ligands ligands$

Crystal field splitting: The energy gap between the splitting of d-orbitals of the metal ion in presence of ligands is known as the crystal field splitting $(Δ)$ . The magnitude of $(Δ)$ is depends on the nature of metal ions and the ligands.

Splitting of five d-orbitals in an octahedral crystals field is as follows:

Figure 1

(a)

Expert Solution

Explanation of Solution

Given,

$[Cr(CN)6]3−$

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.

CONNECT ACCESS CARD FOR CHEMISTRY: MOLECULAR NATURE OF MATTER AND CHANGE, Chapter 23, Problem 23.89P , additional homework tip 2

Answer 7

Chapter 23, Problem 23.89P

(a)

Interpretation Introduction

Interpretation:

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

$[Cr(CN)6]3−$

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

$[noble gas]ns2(n− 1)dx.$

Spectrochemical series: The list of ligands arranged in an ascending order of $(Δ)$ (the splitting of d-orbitals in presence of various ligands).

$→I-< Br-< SCN-< Cl-< S2-< F-< OH-< O2-< H2O < NCS-< edta4-< NH3< en < NO2-< CN-< COweak-field increasing(Δ) strong-field ligands ligands$

Crystal field splitting: The energy gap between the splitting of d-orbitals of the metal ion in presence of ligands is known as the crystal field splitting $(Δ)$ . The magnitude of $(Δ)$ is depends on the nature of metal ions and the ligands.

Splitting of five d-orbitals in an octahedral crystals field is as follows:

Figure 1

Answer 8

(a)

Expert Solution

Explanation of Solution

Given,

$[Cr(CN)6]3−$

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.

CONNECT ACCESS CARD FOR CHEMISTRY: MOLECULAR NATURE OF MATTER AND CHANGE, Chapter 23, Problem 23.89P , additional homework tip 2

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.

$[Rh(CO)6]3+$

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

$[noble gas]ns2(n− 1)dx.$

Spectrochemical series: The list of ligands arranged in an ascending order of $(Δ)$ (the splitting of d-orbitals in presence of various ligands).

$→I-< Br-< SCN-< Cl-< S2-< F-< OH-< O2-< H2O < NCS-< edta4-< NH3< en < NO2-< CN-< COweak-field increasing(Δ) strong-field ligands ligands$

Crystal field splitting: The energy gap between the splitting of d-orbitals of the metal ion in presence of ligands is known as the crystal field splitting $(Δ)$ . The magnitude of $(Δ)$ is depends on the nature of metal ions and the ligands.

Splitting of five d-orbitals in an octahedral crystals field is as follows:

Figure 1

(b)

Expert Solution

Explanation of Solution

Given,

$[Rh(CO)6]3+$

Electron configuration of $Rh: [Ar]4s13d8$

Charge on $Rh$ : The carbonyl ligands are neutral, so $Rh$ has a +3 charge.

Electron configuration of $Rh3+: [Ar]3d6$

Four ligands and a $d6$ configuration indicate an octahedral geometry. Use Hund’s rule to fill in the nine d electrons. Thus, the carbonyl ligands is strong so the correct orbital-energy splitting diagram shows no unpaired electron.

CONNECT ACCESS CARD FOR CHEMISTRY: MOLECULAR NATURE OF MATTER AND CHANGE, Chapter 23, Problem 23.89P , additional homework tip 4

Answer 13

(b)

Interpretation Introduction

Interpretation:

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

$[Rh(CO)6]3+$

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

$[noble gas]ns2(n− 1)dx.$

Spectrochemical series: The list of ligands arranged in an ascending order of $(Δ)$ (the splitting of d-orbitals in presence of various ligands).

$→I-< Br-< SCN-< Cl-< S2-< F-< OH-< O2-< H2O < NCS-< edta4-< NH3< en < NO2-< CN-< COweak-field increasing(Δ) strong-field ligands ligands$

Crystal field splitting: The energy gap between the splitting of d-orbitals of the metal ion in presence of ligands is known as the crystal field splitting $(Δ)$ . The magnitude of $(Δ)$ is depends on the nature of metal ions and the ligands.

Splitting of five d-orbitals in an octahedral crystals field is as follows:

Figure 1

Answer 14

(b)

Expert Solution

Explanation of Solution

Given,

$[Rh(CO)6]3+$

Electron configuration of $Rh: [Ar]4s13d8$

Charge on $Rh$ : The carbonyl ligands are neutral, so $Rh$ has a +3 charge.

Electron configuration of $Rh3+: [Ar]3d6$

Four ligands and a $d6$ configuration indicate an octahedral geometry. Use Hund’s rule to fill in the nine d electrons. Thus, the carbonyl ligands is strong so the correct orbital-energy splitting diagram shows no unpaired electron.

CONNECT ACCESS CARD FOR CHEMISTRY: MOLECULAR NATURE OF MATTER AND CHANGE, Chapter 23, Problem 23.89P , additional homework tip 4

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.

$[Co(OH)6]4−$

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

$[noble gas]ns2(n− 1)dx.$

Spectrochemical series: The list of ligands arranged in an ascending order of $(Δ)$ (the splitting of d-orbitals in presence of various ligands).

$→I-< Br-< SCN-< Cl-< S2-< F-< OH-< O2-< H2O < NCS-< edta4-< NH3< en < NO2-< CN-< COweak-field increasing(Δ) strong-field ligands ligands$

Crystal field splitting: The energy gap between the splitting of d-orbitals of the metal ion in presence of ligands is known as the crystal field splitting $(Δ)$ . The magnitude of $(Δ)$ is depends on the nature of metal ions and the ligands.

Splitting of five d-orbitals in an octahedral crystals field is as follows:

Figure 1

(c)

Expert Solution

Explanation of Solution

Electron configuration of $Co: [Ar]4s23d7$

Charge on $Co$ : water is a neutral ligand so $Co$ has a +2 charge to make the overall complex charge equal to –4.

Electron configuration of $Co2+: [Ar]3d7$

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

Water – 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.

CONNECT ACCESS CARD FOR CHEMISTRY: MOLECULAR NATURE OF MATTER AND CHANGE, Chapter 23, Problem 23.89P , additional homework tip 6

Answer 19

(c)

Interpretation Introduction

Interpretation:

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

$[Co(OH)6]4−$

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

$[noble gas]ns2(n− 1)dx.$

Spectrochemical series: The list of ligands arranged in an ascending order of $(Δ)$ (the splitting of d-orbitals in presence of various ligands).

$→I-< Br-< SCN-< Cl-< S2-< F-< OH-< O2-< H2O < NCS-< edta4-< NH3< en < NO2-< CN-< COweak-field increasing(Δ) strong-field ligands ligands$

Crystal field splitting: The energy gap between the splitting of d-orbitals of the metal ion in presence of ligands is known as the crystal field splitting $(Δ)$ . The magnitude of $(Δ)$ is depends on the nature of metal ions and the ligands.

Splitting of five d-orbitals in an octahedral crystals field is as follows:

Figure 1

Answer 20

(c)

Expert Solution

Explanation of Solution

Electron configuration of $Co: [Ar]4s23d7$

Charge on $Co$ : water is a neutral ligand so $Co$ has a +2 charge to make the overall complex charge equal to –4.

Electron configuration of $Co2+: [Ar]3d7$

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

Water – 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.

CONNECT ACCESS CARD FOR CHEMISTRY: MOLECULAR NATURE OF MATTER AND CHANGE, Chapter 23, Problem 23.89P , additional homework tip 6