Formula for the octahedral complex ion has to be written that has Fe 3+ with NCS − ligands and with NO 2 − ligands.

Question

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Answer 1

Question

Chapter 20, Problem 73QRT

(a)

Interpretation Introduction

Interpretation:

Formula for the octahedral complex ion has to be written that has $Fe3+$ with $NCS−$ ligands and with $NO2−$ ligands.

(a)

Expert Solution

Explanation of Solution

Octahedral complex contains six ligands inside the coordination sphere. Central metal atom is given as $Fe3+$ . The two ligands are $NCS−$ and $NO2−$ . Octahedral complex ion that is formed from $Fe3+$ and $NCS−$ is given as $[Fe(NCS)6]3−$ . Octahedral complex ion that is formed from $Fe3+$ and $NO2−$ is given as $[Fe(NO2)6]3−$ .

(b)

Interpretation Introduction

Interpretation:

Complex that is formed from $Fe3+$ ion, $NCS−$ ligand and with $NO2−$ ligand are high-spin or low-spin has to be predicted.

(b)

Expert Solution

Explanation of Solution

Octahedral complex ion that is formed from $Fe3+$ and $NCS−$ is given as $[Fe(NCS)6]3−$ . $NCS−$ ion is a weak field ligand. Hence, the complex ion formed will be high spin.

Octahedral complex ion that is formed from $Fe3+$ and $NO2−$ is given as $[Fe(NO2)6]3−$ . $NO2−$ is a strong field ligand. Hence, the complex ion formed will be low spin.

(c)

Interpretation Introduction

Interpretation:

Complex that has more paramagnetism has to be chosen.

(c)

Expert Solution

Explanation of Solution

Octahedral complex ion that is formed from $Fe3+$ and $NCS−$ is given as $[Fe(NCS)6]3−$ . $NCS−$ ion is a weak field ligand. Hence, the complex ion formed will be high spin.

Octahedral complex ion that is formed from $Fe3+$ and $NO2−$ is given as $[Fe(NO2)6]3−$ . $NO2−$ is a strong field ligand. Hence, the complex ion formed will be low spin.

The crystal field model for high-spin and low-spin complex of $Fe3+$ can be given as,

OWLv2 for Moore/Stanitski's Chemistry: The Molecular Science, 5th Edition, [Instant Access], 1 term (6 months), Chapter 20, Problem 73QRT , additional homework tip 1

Number of unpaired electrons:

The ligand $NCS−$ is a weak field ligand and therefore the complex $[Fe(NCS)6]3−$ forms high spin complex. There are five unpaired electrons. The ligand $NO2−$ is a strong field ligand and therefore the complex $[Fe(NO2)6]3−$ forms low spin complex. There is one unpaired electron.

Paramagnetism depends upon the number of unpaired electrons. Therefore, the complex ion $[Fe(NCS)6]3−$ will be more paramagnetic.

(c)

Interpretation Introduction

Interpretation:

Crystal field diagram for the $d$ electrons in complex $[Fe(NCS)6]3−$ and $[Fe(NO2)6]3−$ has to be drawn.

(c)

Expert Solution

Explanation of Solution

Octahedral complex ion that is formed from $Fe3+$ and $NCS−$ is given as $[Fe(NCS)6]3−$ . $NCS−$ ion is a weak field ligand. Hence, the complex ion formed will be high spin.

Octahedral complex ion that is formed from $Fe3+$ and $NO2−$ is given as $[Fe(NO2)6]3−$ . $NO2−$ is a strong field ligand. Hence, the complex ion formed will be low spin.

The central metal ion is $Fe3+$ . $Fe3+$ ion is said to contain five $3d$ electrons. For high-spin complex, the number of unpaired electrons will be more and for low-spin complex, the number of unpaired electrons will be less. The splitting of orbitals depends upon the ligand. If the ligand is weak field ligand, then the crystal field splitting energy ( $Δo$ ) will be smaller resulting in unpaired electrons. If the ligand is a strong field ligand, then the crystal field splitting energy will be larger resulting in pairing of electrons. The crystal field model for high-spin and low-spin complex of $Fe3+$ can be given as,

OWLv2 for Moore/Stanitski's Chemistry: The Molecular Science, 5th Edition, [Instant Access], 1 term (6 months), Chapter 20, Problem 73QRT , additional homework tip 2

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Answer 2

Question

Chapter 20, Problem 73QRT

(a)

Interpretation Introduction

Interpretation:

Formula for the octahedral complex ion has to be written that has $Fe3+$ with $NCS−$ ligands and with $NO2−$ ligands.

(a)

Expert Solution

Explanation of Solution

Octahedral complex contains six ligands inside the coordination sphere. Central metal atom is given as $Fe3+$ . The two ligands are $NCS−$ and $NO2−$ . Octahedral complex ion that is formed from $Fe3+$ and $NCS−$ is given as $[Fe(NCS)6]3−$ . Octahedral complex ion that is formed from $Fe3+$ and $NO2−$ is given as $[Fe(NO2)6]3−$ .

(b)

Interpretation Introduction

Interpretation:

Complex that is formed from $Fe3+$ ion, $NCS−$ ligand and with $NO2−$ ligand are high-spin or low-spin has to be predicted.

(b)

Expert Solution

Explanation of Solution

Octahedral complex ion that is formed from $Fe3+$ and $NCS−$ is given as $[Fe(NCS)6]3−$ . $NCS−$ ion is a weak field ligand. Hence, the complex ion formed will be high spin.

Octahedral complex ion that is formed from $Fe3+$ and $NO2−$ is given as $[Fe(NO2)6]3−$ . $NO2−$ is a strong field ligand. Hence, the complex ion formed will be low spin.

(c)

Interpretation Introduction

Interpretation:

Complex that has more paramagnetism has to be chosen.

(c)

Expert Solution

Explanation of Solution

Octahedral complex ion that is formed from $Fe3+$ and $NCS−$ is given as $[Fe(NCS)6]3−$ . $NCS−$ ion is a weak field ligand. Hence, the complex ion formed will be high spin.

Octahedral complex ion that is formed from $Fe3+$ and $NO2−$ is given as $[Fe(NO2)6]3−$ . $NO2−$ is a strong field ligand. Hence, the complex ion formed will be low spin.

The crystal field model for high-spin and low-spin complex of $Fe3+$ can be given as,

OWLv2 for Moore/Stanitski's Chemistry: The Molecular Science, 5th Edition, [Instant Access], 1 term (6 months), Chapter 20, Problem 73QRT , additional homework tip 1

Number of unpaired electrons:

The ligand $NCS−$ is a weak field ligand and therefore the complex $[Fe(NCS)6]3−$ forms high spin complex. There are five unpaired electrons. The ligand $NO2−$ is a strong field ligand and therefore the complex $[Fe(NO2)6]3−$ forms low spin complex. There is one unpaired electron.

Paramagnetism depends upon the number of unpaired electrons. Therefore, the complex ion $[Fe(NCS)6]3−$ will be more paramagnetic.

(c)

Interpretation Introduction

Interpretation:

Crystal field diagram for the $d$ electrons in complex $[Fe(NCS)6]3−$ and $[Fe(NO2)6]3−$ has to be drawn.

(c)

Expert Solution

Explanation of Solution

Octahedral complex ion that is formed from $Fe3+$ and $NCS−$ is given as $[Fe(NCS)6]3−$ . $NCS−$ ion is a weak field ligand. Hence, the complex ion formed will be high spin.

Octahedral complex ion that is formed from $Fe3+$ and $NO2−$ is given as $[Fe(NO2)6]3−$ . $NO2−$ is a strong field ligand. Hence, the complex ion formed will be low spin.

The central metal ion is $Fe3+$ . $Fe3+$ ion is said to contain five $3d$ electrons. For high-spin complex, the number of unpaired electrons will be more and for low-spin complex, the number of unpaired electrons will be less. The splitting of orbitals depends upon the ligand. If the ligand is weak field ligand, then the crystal field splitting energy ( $Δo$ ) will be smaller resulting in unpaired electrons. If the ligand is a strong field ligand, then the crystal field splitting energy will be larger resulting in pairing of electrons. The crystal field model for high-spin and low-spin complex of $Fe3+$ can be given as,

OWLv2 for Moore/Stanitski's Chemistry: The Molecular Science, 5th Edition, [Instant Access], 1 term (6 months), Chapter 20, Problem 73QRT , additional homework tip 2

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Answer 3

Question

Chapter 20, Problem 73QRT

(a)

Interpretation Introduction

Interpretation:

Formula for the octahedral complex ion has to be written that has $Fe3+$ with $NCS−$ ligands and with $NO2−$ ligands.

(a)

Expert Solution

Explanation of Solution

Octahedral complex contains six ligands inside the coordination sphere. Central metal atom is given as $Fe3+$ . The two ligands are $NCS−$ and $NO2−$ . Octahedral complex ion that is formed from $Fe3+$ and $NCS−$ is given as $[Fe(NCS)6]3−$ . Octahedral complex ion that is formed from $Fe3+$ and $NO2−$ is given as $[Fe(NO2)6]3−$ .

(b)

Interpretation Introduction

Interpretation:

Complex that is formed from $Fe3+$ ion, $NCS−$ ligand and with $NO2−$ ligand are high-spin or low-spin has to be predicted.

(b)

Expert Solution

Explanation of Solution

Octahedral complex ion that is formed from $Fe3+$ and $NCS−$ is given as $[Fe(NCS)6]3−$ . $NCS−$ ion is a weak field ligand. Hence, the complex ion formed will be high spin.

Octahedral complex ion that is formed from $Fe3+$ and $NO2−$ is given as $[Fe(NO2)6]3−$ . $NO2−$ is a strong field ligand. Hence, the complex ion formed will be low spin.

(c)

Interpretation Introduction

Interpretation:

Complex that has more paramagnetism has to be chosen.

(c)

Expert Solution

Explanation of Solution

Octahedral complex ion that is formed from $Fe3+$ and $NCS−$ is given as $[Fe(NCS)6]3−$ . $NCS−$ ion is a weak field ligand. Hence, the complex ion formed will be high spin.

Octahedral complex ion that is formed from $Fe3+$ and $NO2−$ is given as $[Fe(NO2)6]3−$ . $NO2−$ is a strong field ligand. Hence, the complex ion formed will be low spin.

The crystal field model for high-spin and low-spin complex of $Fe3+$ can be given as,

OWLv2 for Moore/Stanitski's Chemistry: The Molecular Science, 5th Edition, [Instant Access], 1 term (6 months), Chapter 20, Problem 73QRT , additional homework tip 1

Number of unpaired electrons:

The ligand $NCS−$ is a weak field ligand and therefore the complex $[Fe(NCS)6]3−$ forms high spin complex. There are five unpaired electrons. The ligand $NO2−$ is a strong field ligand and therefore the complex $[Fe(NO2)6]3−$ forms low spin complex. There is one unpaired electron.

Paramagnetism depends upon the number of unpaired electrons. Therefore, the complex ion $[Fe(NCS)6]3−$ will be more paramagnetic.

(c)

Interpretation Introduction

Interpretation:

Crystal field diagram for the $d$ electrons in complex $[Fe(NCS)6]3−$ and $[Fe(NO2)6]3−$ has to be drawn.

(c)

Expert Solution

Explanation of Solution

Octahedral complex ion that is formed from $Fe3+$ and $NCS−$ is given as $[Fe(NCS)6]3−$ . $NCS−$ ion is a weak field ligand. Hence, the complex ion formed will be high spin.

Octahedral complex ion that is formed from $Fe3+$ and $NO2−$ is given as $[Fe(NO2)6]3−$ . $NO2−$ is a strong field ligand. Hence, the complex ion formed will be low spin.

The central metal ion is $Fe3+$ . $Fe3+$ ion is said to contain five $3d$ electrons. For high-spin complex, the number of unpaired electrons will be more and for low-spin complex, the number of unpaired electrons will be less. The splitting of orbitals depends upon the ligand. If the ligand is weak field ligand, then the crystal field splitting energy ( $Δo$ ) will be smaller resulting in unpaired electrons. If the ligand is a strong field ligand, then the crystal field splitting energy will be larger resulting in pairing of electrons. The crystal field model for high-spin and low-spin complex of $Fe3+$ can be given as,

OWLv2 for Moore/Stanitski's Chemistry: The Molecular Science, 5th Edition, [Instant Access], 1 term (6 months), Chapter 20, Problem 73QRT , additional homework tip 2

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Answer 4

Question

Chapter 20, Problem 73QRT

(a)

Interpretation Introduction

Interpretation:

Formula for the octahedral complex ion has to be written that has $Fe3+$ with $NCS−$ ligands and with $NO2−$ ligands.

(a)

Expert Solution

Explanation of Solution

Octahedral complex contains six ligands inside the coordination sphere. Central metal atom is given as $Fe3+$ . The two ligands are $NCS−$ and $NO2−$ . Octahedral complex ion that is formed from $Fe3+$ and $NCS−$ is given as $[Fe(NCS)6]3−$ . Octahedral complex ion that is formed from $Fe3+$ and $NO2−$ is given as $[Fe(NO2)6]3−$ .

(b)

Interpretation Introduction

Interpretation:

Complex that is formed from $Fe3+$ ion, $NCS−$ ligand and with $NO2−$ ligand are high-spin or low-spin has to be predicted.

(b)

Expert Solution

Explanation of Solution

Octahedral complex ion that is formed from $Fe3+$ and $NCS−$ is given as $[Fe(NCS)6]3−$ . $NCS−$ ion is a weak field ligand. Hence, the complex ion formed will be high spin.

Octahedral complex ion that is formed from $Fe3+$ and $NO2−$ is given as $[Fe(NO2)6]3−$ . $NO2−$ is a strong field ligand. Hence, the complex ion formed will be low spin.

(c)

Interpretation Introduction

Interpretation:

Complex that has more paramagnetism has to be chosen.

(c)

Expert Solution

Explanation of Solution

Octahedral complex ion that is formed from $Fe3+$ and $NCS−$ is given as $[Fe(NCS)6]3−$ . $NCS−$ ion is a weak field ligand. Hence, the complex ion formed will be high spin.

Octahedral complex ion that is formed from $Fe3+$ and $NO2−$ is given as $[Fe(NO2)6]3−$ . $NO2−$ is a strong field ligand. Hence, the complex ion formed will be low spin.

The crystal field model for high-spin and low-spin complex of $Fe3+$ can be given as,

OWLv2 for Moore/Stanitski's Chemistry: The Molecular Science, 5th Edition, [Instant Access], 1 term (6 months), Chapter 20, Problem 73QRT , additional homework tip 1

Number of unpaired electrons:

The ligand $NCS−$ is a weak field ligand and therefore the complex $[Fe(NCS)6]3−$ forms high spin complex. There are five unpaired electrons. The ligand $NO2−$ is a strong field ligand and therefore the complex $[Fe(NO2)6]3−$ forms low spin complex. There is one unpaired electron.

Paramagnetism depends upon the number of unpaired electrons. Therefore, the complex ion $[Fe(NCS)6]3−$ will be more paramagnetic.

(c)

Interpretation Introduction

Interpretation:

Crystal field diagram for the $d$ electrons in complex $[Fe(NCS)6]3−$ and $[Fe(NO2)6]3−$ has to be drawn.

(c)

Expert Solution

Explanation of Solution

Octahedral complex ion that is formed from $Fe3+$ and $NCS−$ is given as $[Fe(NCS)6]3−$ . $NCS−$ ion is a weak field ligand. Hence, the complex ion formed will be high spin.

Octahedral complex ion that is formed from $Fe3+$ and $NO2−$ is given as $[Fe(NO2)6]3−$ . $NO2−$ is a strong field ligand. Hence, the complex ion formed will be low spin.

The central metal ion is $Fe3+$ . $Fe3+$ ion is said to contain five $3d$ electrons. For high-spin complex, the number of unpaired electrons will be more and for low-spin complex, the number of unpaired electrons will be less. The splitting of orbitals depends upon the ligand. If the ligand is weak field ligand, then the crystal field splitting energy ( $Δo$ ) will be smaller resulting in unpaired electrons. If the ligand is a strong field ligand, then the crystal field splitting energy will be larger resulting in pairing of electrons. The crystal field model for high-spin and low-spin complex of $Fe3+$ can be given as,

OWLv2 for Moore/Stanitski's Chemistry: The Molecular Science, 5th Edition, [Instant Access], 1 term (6 months), Chapter 20, Problem 73QRT , additional homework tip 2

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Answer 5

Chapter 20, Problem 73QRT

(a)

Interpretation Introduction

Interpretation:

Formula for the octahedral complex ion has to be written that has $Fe3+$ with $NCS−$ ligands and with $NO2−$ ligands.

(a)

Expert Solution

Explanation of Solution

Octahedral complex contains six ligands inside the coordination sphere. Central metal atom is given as $Fe3+$ . The two ligands are $NCS−$ and $NO2−$ . Octahedral complex ion that is formed from $Fe3+$ and $NCS−$ is given as $[Fe(NCS)6]3−$ . Octahedral complex ion that is formed from $Fe3+$ and $NO2−$ is given as $[Fe(NO2)6]3−$ .

(b)

Interpretation Introduction

Interpretation:

Complex that is formed from $Fe3+$ ion, $NCS−$ ligand and with $NO2−$ ligand are high-spin or low-spin has to be predicted.

(b)

Expert Solution

Explanation of Solution

Octahedral complex ion that is formed from $Fe3+$ and $NCS−$ is given as $[Fe(NCS)6]3−$ . $NCS−$ ion is a weak field ligand. Hence, the complex ion formed will be high spin.

Octahedral complex ion that is formed from $Fe3+$ and $NO2−$ is given as $[Fe(NO2)6]3−$ . $NO2−$ is a strong field ligand. Hence, the complex ion formed will be low spin.

(c)

Interpretation Introduction

Interpretation:

Complex that has more paramagnetism has to be chosen.

(c)

Expert Solution

Explanation of Solution

Octahedral complex ion that is formed from $Fe3+$ and $NCS−$ is given as $[Fe(NCS)6]3−$ . $NCS−$ ion is a weak field ligand. Hence, the complex ion formed will be high spin.

Octahedral complex ion that is formed from $Fe3+$ and $NO2−$ is given as $[Fe(NO2)6]3−$ . $NO2−$ is a strong field ligand. Hence, the complex ion formed will be low spin.

The crystal field model for high-spin and low-spin complex of $Fe3+$ can be given as,

OWLv2 for Moore/Stanitski's Chemistry: The Molecular Science, 5th Edition, [Instant Access], 1 term (6 months), Chapter 20, Problem 73QRT , additional homework tip 1

Number of unpaired electrons:

The ligand $NCS−$ is a weak field ligand and therefore the complex $[Fe(NCS)6]3−$ forms high spin complex. There are five unpaired electrons. The ligand $NO2−$ is a strong field ligand and therefore the complex $[Fe(NO2)6]3−$ forms low spin complex. There is one unpaired electron.

Paramagnetism depends upon the number of unpaired electrons. Therefore, the complex ion $[Fe(NCS)6]3−$ will be more paramagnetic.

(c)

Interpretation Introduction

Interpretation:

Crystal field diagram for the $d$ electrons in complex $[Fe(NCS)6]3−$ and $[Fe(NO2)6]3−$ has to be drawn.

(c)

Expert Solution

Explanation of Solution

Octahedral complex ion that is formed from $Fe3+$ and $NCS−$ is given as $[Fe(NCS)6]3−$ . $NCS−$ ion is a weak field ligand. Hence, the complex ion formed will be high spin.

Octahedral complex ion that is formed from $Fe3+$ and $NO2−$ is given as $[Fe(NO2)6]3−$ . $NO2−$ is a strong field ligand. Hence, the complex ion formed will be low spin.

The central metal ion is $Fe3+$ . $Fe3+$ ion is said to contain five $3d$ electrons. For high-spin complex, the number of unpaired electrons will be more and for low-spin complex, the number of unpaired electrons will be less. The splitting of orbitals depends upon the ligand. If the ligand is weak field ligand, then the crystal field splitting energy ( $Δo$ ) will be smaller resulting in unpaired electrons. If the ligand is a strong field ligand, then the crystal field splitting energy will be larger resulting in pairing of electrons. The crystal field model for high-spin and low-spin complex of $Fe3+$ can be given as,

OWLv2 for Moore/Stanitski's Chemistry: The Molecular Science, 5th Edition, [Instant Access], 1 term (6 months), Chapter 20, Problem 73QRT , additional homework tip 2

Answer 6

Chapter 20, Problem 73QRT

(a)

Interpretation Introduction

Interpretation:

Formula for the octahedral complex ion has to be written that has $Fe3+$ with $NCS−$ ligands and with $NO2−$ ligands.

(a)

Expert Solution

Explanation of Solution

Octahedral complex contains six ligands inside the coordination sphere. Central metal atom is given as $Fe3+$ . The two ligands are $NCS−$ and $NO2−$ . Octahedral complex ion that is formed from $Fe3+$ and $NCS−$ is given as $[Fe(NCS)6]3−$ . Octahedral complex ion that is formed from $Fe3+$ and $NO2−$ is given as $[Fe(NO2)6]3−$ .

Answer 7

Chapter 20, Problem 73QRT

(a)

Interpretation Introduction

Interpretation:

Formula for the octahedral complex ion has to be written that has $Fe3+$ with $NCS−$ ligands and with $NO2−$ ligands.

Answer 8

(a)

Expert Solution

Explanation of Solution

Octahedral complex contains six ligands inside the coordination sphere. Central metal atom is given as $Fe3+$ . The two ligands are $NCS−$ and $NO2−$ . Octahedral complex ion that is formed from $Fe3+$ and $NCS−$ is given as $[Fe(NCS)6]3−$ . Octahedral complex ion that is formed from $Fe3+$ and $NO2−$ is given as $[Fe(NO2)6]3−$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

(b)

Interpretation Introduction

Interpretation:

Complex that is formed from $Fe3+$ ion, $NCS−$ ligand and with $NO2−$ ligand are high-spin or low-spin has to be predicted.

(b)

Expert Solution

Explanation of Solution

Octahedral complex ion that is formed from $Fe3+$ and $NCS−$ is given as $[Fe(NCS)6]3−$ . $NCS−$ ion is a weak field ligand. Hence, the complex ion formed will be high spin.

Octahedral complex ion that is formed from $Fe3+$ and $NO2−$ is given as $[Fe(NO2)6]3−$ . $NO2−$ is a strong field ligand. Hence, the complex ion formed will be low spin.

Answer 13

(b)

Interpretation Introduction

Interpretation:

Complex that is formed from $Fe3+$ ion, $NCS−$ ligand and with $NO2−$ ligand are high-spin or low-spin has to be predicted.

Answer 14

(b)

Expert Solution

Explanation of Solution

Octahedral complex ion that is formed from $Fe3+$ and $NCS−$ is given as $[Fe(NCS)6]3−$ . $NCS−$ ion is a weak field ligand. Hence, the complex ion formed will be high spin.

Octahedral complex ion that is formed from $Fe3+$ and $NO2−$ is given as $[Fe(NO2)6]3−$ . $NO2−$ is a strong field ligand. Hence, the complex ion formed will be low spin.

Answer 15

(b)

Expert Solution

Answer 16

(b)

Expert Solution

Answer 17

Expert Solution

Answer 18

(c)

Interpretation Introduction

Interpretation:

Complex that has more paramagnetism has to be chosen.

(c)

Expert Solution

Explanation of Solution

Octahedral complex ion that is formed from $Fe3+$ and $NCS−$ is given as $[Fe(NCS)6]3−$ . $NCS−$ ion is a weak field ligand. Hence, the complex ion formed will be high spin.

Octahedral complex ion that is formed from $Fe3+$ and $NO2−$ is given as $[Fe(NO2)6]3−$ . $NO2−$ is a strong field ligand. Hence, the complex ion formed will be low spin.

The crystal field model for high-spin and low-spin complex of $Fe3+$ can be given as,

OWLv2 for Moore/Stanitski's Chemistry: The Molecular Science, 5th Edition, [Instant Access], 1 term (6 months), Chapter 20, Problem 73QRT , additional homework tip 1

Number of unpaired electrons:

The ligand $NCS−$ is a weak field ligand and therefore the complex $[Fe(NCS)6]3−$ forms high spin complex. There are five unpaired electrons. The ligand $NO2−$ is a strong field ligand and therefore the complex $[Fe(NO2)6]3−$ forms low spin complex. There is one unpaired electron.

Paramagnetism depends upon the number of unpaired electrons. Therefore, the complex ion $[Fe(NCS)6]3−$ will be more paramagnetic.

Answer 19

(c)

Interpretation Introduction

Interpretation:

Complex that has more paramagnetism has to be chosen.

Answer 20

(c)

Expert Solution

Explanation of Solution

Octahedral complex ion that is formed from $Fe3+$ and $NCS−$ is given as $[Fe(NCS)6]3−$ . $NCS−$ ion is a weak field ligand. Hence, the complex ion formed will be high spin.

Octahedral complex ion that is formed from $Fe3+$ and $NO2−$ is given as $[Fe(NO2)6]3−$ . $NO2−$ is a strong field ligand. Hence, the complex ion formed will be low spin.

The crystal field model for high-spin and low-spin complex of $Fe3+$ can be given as,

OWLv2 for Moore/Stanitski's Chemistry: The Molecular Science, 5th Edition, [Instant Access], 1 term (6 months), Chapter 20, Problem 73QRT , additional homework tip 1

Number of unpaired electrons:

The ligand $NCS−$ is a weak field ligand and therefore the complex $[Fe(NCS)6]3−$ forms high spin complex. There are five unpaired electrons. The ligand $NO2−$ is a strong field ligand and therefore the complex $[Fe(NO2)6]3−$ forms low spin complex. There is one unpaired electron.

Paramagnetism depends upon the number of unpaired electrons. Therefore, the complex ion $[Fe(NCS)6]3−$ will be more paramagnetic.

Answer 21

(c)

Expert Solution

Answer 22

(c)

Expert Solution

Answer 23

Expert Solution

Answer 24

(c)

Interpretation Introduction

Interpretation:

Crystal field diagram for the $d$ electrons in complex $[Fe(NCS)6]3−$ and $[Fe(NO2)6]3−$ has to be drawn.

(c)

Expert Solution

Explanation of Solution

Octahedral complex ion that is formed from $Fe3+$ and $NCS−$ is given as $[Fe(NCS)6]3−$ . $NCS−$ ion is a weak field ligand. Hence, the complex ion formed will be high spin.

Octahedral complex ion that is formed from $Fe3+$ and $NO2−$ is given as $[Fe(NO2)6]3−$ . $NO2−$ is a strong field ligand. Hence, the complex ion formed will be low spin.

The central metal ion is $Fe3+$ . $Fe3+$ ion is said to contain five $3d$ electrons. For high-spin complex, the number of unpaired electrons will be more and for low-spin complex, the number of unpaired electrons will be less. The splitting of orbitals depends upon the ligand. If the ligand is weak field ligand, then the crystal field splitting energy ( $Δo$ ) will be smaller resulting in unpaired electrons. If the ligand is a strong field ligand, then the crystal field splitting energy will be larger resulting in pairing of electrons. The crystal field model for high-spin and low-spin complex of $Fe3+$ can be given as,

OWLv2 for Moore/Stanitski's Chemistry: The Molecular Science, 5th Edition, [Instant Access], 1 term (6 months), Chapter 20, Problem 73QRT , additional homework tip 2

Answer 25

(c)

Interpretation Introduction

Interpretation:

Crystal field diagram for the $d$ electrons in complex $[Fe(NCS)6]3−$ and $[Fe(NO2)6]3−$ has to be drawn.

Answer 26

(c)

Expert Solution

Explanation of Solution

Octahedral complex ion that is formed from $Fe3+$ and $NCS−$ is given as $[Fe(NCS)6]3−$ . $NCS−$ ion is a weak field ligand. Hence, the complex ion formed will be high spin.

Octahedral complex ion that is formed from $Fe3+$ and $NO2−$ is given as $[Fe(NO2)6]3−$ . $NO2−$ is a strong field ligand. Hence, the complex ion formed will be low spin.

The central metal ion is $Fe3+$ . $Fe3+$ ion is said to contain five $3d$ electrons. For high-spin complex, the number of unpaired electrons will be more and for low-spin complex, the number of unpaired electrons will be less. The splitting of orbitals depends upon the ligand. If the ligand is weak field ligand, then the crystal field splitting energy ( $Δo$ ) will be smaller resulting in unpaired electrons. If the ligand is a strong field ligand, then the crystal field splitting energy will be larger resulting in pairing of electrons. The crystal field model for high-spin and low-spin complex of $Fe3+$ can be given as,

OWLv2 for Moore/Stanitski's Chemistry: The Molecular Science, 5th Edition, [Instant Access], 1 term (6 months), Chapter 20, Problem 73QRT , additional homework tip 2