PRACTICE PROBLEM 21.1 For each of the following complexes, determine the oxidation state of the metal and the total number of valence electrons it possesses. (a) (b) (c)

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Answer 1

Textbook Question

Chapter 21, Problem 1PP

PRACTICE PROBLEM 21.1

For each of the following complexes, determine the oxidation state of the metal and the total number of valence electrons it possesses.

(a) Chapter 21, Problem 1PP, PRACTICE PROBLEM 21.1
For each of the following complexes, determine the oxidation state of the , example 1

(b) Chapter 21, Problem 1PP, PRACTICE PROBLEM 21.1
For each of the following complexes, determine the oxidation state of the , example 2

(c) Chapter 21, Problem 1PP, PRACTICE PROBLEM 21.1
For each of the following complexes, determine the oxidation state of the , example 3

Expert Solution & Answer

Interpretation Introduction

Interpretation:

The oxidation state of a metal and the total number of valence electrons in metals in the complexes is to be determined.

Concept introduction:

The oxidation state of a metal in a complex is the charge on the metal that would be there even if all the anionic ligands and counter ions were removed.

The total number of valence electrons of a metal in a complex is obtained by the following formula:

Total number of valence electrons of metal in complex = $dn$ $+$ electrons donated by ligands

Here, $dn$ is the d-electrons in the metal.

Answer to Problem 1PP

Solution:

(a)

Oxidation state of Rh is $+1$ and the total number of valence electrons in Rh is $15$ .

(b)

Oxidation state of Hg is $+2$ and the number of total valence electrons of Hg is $16$ .

(c)

Oxidation state of Ni is $0$ and the number of total valence electrons of Ni is $16$ .

Explanation of Solution

Given information:

Organic Chemistry, 12e Study Guide/Student Solutions Manual, Chapter 21, Problem 1PP

The oxidation state of rhodium is $+1$ in the complex. This is because Cl is the only ligand with a charge in the complex, phosphine triphenyl is a neutral ligand. The charge on the Cl ligand is $−1$ . To satisfy this charge, rhodium carries a $+1$ charge.

Charge on rhodium is as follows:

$x+(−1)=0$

Here, $x$ is the charge on rhodium and its value is as follows:

$x=1$

Now, d-electrons in rhodium are $8$ and the number of electrons donated by ligands is $7$ . Each phosphine triphenyl donate two electrons and the chlorine atom donates one electron. So, according to the equation, the total number of valence electrons of the metal, in the complex are:

Total number of valence electrons of metal in complex = $dn$ $+$ electrons donated by ligands

Here, $dn$ is the number of d-electrons in the metal.

For rhodium, the total number of valence electrons in the complex $=8+7=15$

Thus, the number of valence electrons in rhodium is $15$ .

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Answer 2

Textbook Question

Chapter 21, Problem 1PP

PRACTICE PROBLEM 21.1

For each of the following complexes, determine the oxidation state of the metal and the total number of valence electrons it possesses.

(a) Chapter 21, Problem 1PP, PRACTICE PROBLEM 21.1
For each of the following complexes, determine the oxidation state of the , example 1

(b) Chapter 21, Problem 1PP, PRACTICE PROBLEM 21.1
For each of the following complexes, determine the oxidation state of the , example 2

(c) Chapter 21, Problem 1PP, PRACTICE PROBLEM 21.1
For each of the following complexes, determine the oxidation state of the , example 3

Expert Solution & Answer

Interpretation Introduction

Interpretation:

The oxidation state of a metal and the total number of valence electrons in metals in the complexes is to be determined.

Concept introduction:

The oxidation state of a metal in a complex is the charge on the metal that would be there even if all the anionic ligands and counter ions were removed.

The total number of valence electrons of a metal in a complex is obtained by the following formula:

Total number of valence electrons of metal in complex = $dn$ $+$ electrons donated by ligands

Here, $dn$ is the d-electrons in the metal.

Answer to Problem 1PP

Solution:

(a)

Oxidation state of Rh is $+1$ and the total number of valence electrons in Rh is $15$ .

(b)

Oxidation state of Hg is $+2$ and the number of total valence electrons of Hg is $16$ .

(c)

Oxidation state of Ni is $0$ and the number of total valence electrons of Ni is $16$ .

Explanation of Solution

Given information:

Organic Chemistry, 12e Study Guide/Student Solutions Manual, Chapter 21, Problem 1PP

The oxidation state of rhodium is $+1$ in the complex. This is because Cl is the only ligand with a charge in the complex, phosphine triphenyl is a neutral ligand. The charge on the Cl ligand is $−1$ . To satisfy this charge, rhodium carries a $+1$ charge.

Charge on rhodium is as follows:

$x+(−1)=0$

Here, $x$ is the charge on rhodium and its value is as follows:

$x=1$

Now, d-electrons in rhodium are $8$ and the number of electrons donated by ligands is $7$ . Each phosphine triphenyl donate two electrons and the chlorine atom donates one electron. So, according to the equation, the total number of valence electrons of the metal, in the complex are:

Total number of valence electrons of metal in complex = $dn$ $+$ electrons donated by ligands

Here, $dn$ is the number of d-electrons in the metal.

For rhodium, the total number of valence electrons in the complex $=8+7=15$

Thus, the number of valence electrons in rhodium is $15$ .

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Answer 3

Textbook Question

Chapter 21, Problem 1PP

PRACTICE PROBLEM 21.1

For each of the following complexes, determine the oxidation state of the metal and the total number of valence electrons it possesses.

(a) Chapter 21, Problem 1PP, PRACTICE PROBLEM 21.1
For each of the following complexes, determine the oxidation state of the , example 1

(b) Chapter 21, Problem 1PP, PRACTICE PROBLEM 21.1
For each of the following complexes, determine the oxidation state of the , example 2

(c) Chapter 21, Problem 1PP, PRACTICE PROBLEM 21.1
For each of the following complexes, determine the oxidation state of the , example 3

Expert Solution & Answer

Interpretation Introduction

Interpretation:

The oxidation state of a metal and the total number of valence electrons in metals in the complexes is to be determined.

Concept introduction:

The oxidation state of a metal in a complex is the charge on the metal that would be there even if all the anionic ligands and counter ions were removed.

The total number of valence electrons of a metal in a complex is obtained by the following formula:

Total number of valence electrons of metal in complex = $dn$ $+$ electrons donated by ligands

Here, $dn$ is the d-electrons in the metal.

Answer to Problem 1PP

Solution:

(a)

Oxidation state of Rh is $+1$ and the total number of valence electrons in Rh is $15$ .

(b)

Oxidation state of Hg is $+2$ and the number of total valence electrons of Hg is $16$ .

(c)

Oxidation state of Ni is $0$ and the number of total valence electrons of Ni is $16$ .

Explanation of Solution

Given information:

Organic Chemistry, 12e Study Guide/Student Solutions Manual, Chapter 21, Problem 1PP

The oxidation state of rhodium is $+1$ in the complex. This is because Cl is the only ligand with a charge in the complex, phosphine triphenyl is a neutral ligand. The charge on the Cl ligand is $−1$ . To satisfy this charge, rhodium carries a $+1$ charge.

Charge on rhodium is as follows:

$x+(−1)=0$

Here, $x$ is the charge on rhodium and its value is as follows:

$x=1$

Now, d-electrons in rhodium are $8$ and the number of electrons donated by ligands is $7$ . Each phosphine triphenyl donate two electrons and the chlorine atom donates one electron. So, according to the equation, the total number of valence electrons of the metal, in the complex are:

Total number of valence electrons of metal in complex = $dn$ $+$ electrons donated by ligands

Here, $dn$ is the number of d-electrons in the metal.

For rhodium, the total number of valence electrons in the complex $=8+7=15$

Thus, the number of valence electrons in rhodium is $15$ .

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Answer 4

Textbook Question

Chapter 21, Problem 1PP

PRACTICE PROBLEM 21.1

For each of the following complexes, determine the oxidation state of the metal and the total number of valence electrons it possesses.

(a) Chapter 21, Problem 1PP, PRACTICE PROBLEM 21.1
For each of the following complexes, determine the oxidation state of the , example 1

(b) Chapter 21, Problem 1PP, PRACTICE PROBLEM 21.1
For each of the following complexes, determine the oxidation state of the , example 2

(c) Chapter 21, Problem 1PP, PRACTICE PROBLEM 21.1
For each of the following complexes, determine the oxidation state of the , example 3

Expert Solution & Answer

Interpretation Introduction

Interpretation:

The oxidation state of a metal and the total number of valence electrons in metals in the complexes is to be determined.

Concept introduction:

The oxidation state of a metal in a complex is the charge on the metal that would be there even if all the anionic ligands and counter ions were removed.

The total number of valence electrons of a metal in a complex is obtained by the following formula:

Total number of valence electrons of metal in complex = $dn$ $+$ electrons donated by ligands

Here, $dn$ is the d-electrons in the metal.

Answer to Problem 1PP

Solution:

(a)

Oxidation state of Rh is $+1$ and the total number of valence electrons in Rh is $15$ .

(b)

Oxidation state of Hg is $+2$ and the number of total valence electrons of Hg is $16$ .

(c)

Oxidation state of Ni is $0$ and the number of total valence electrons of Ni is $16$ .

Explanation of Solution

Given information:

Organic Chemistry, 12e Study Guide/Student Solutions Manual, Chapter 21, Problem 1PP

The oxidation state of rhodium is $+1$ in the complex. This is because Cl is the only ligand with a charge in the complex, phosphine triphenyl is a neutral ligand. The charge on the Cl ligand is $−1$ . To satisfy this charge, rhodium carries a $+1$ charge.

Charge on rhodium is as follows:

$x+(−1)=0$

Here, $x$ is the charge on rhodium and its value is as follows:

$x=1$

Now, d-electrons in rhodium are $8$ and the number of electrons donated by ligands is $7$ . Each phosphine triphenyl donate two electrons and the chlorine atom donates one electron. So, according to the equation, the total number of valence electrons of the metal, in the complex are:

Total number of valence electrons of metal in complex = $dn$ $+$ electrons donated by ligands

Here, $dn$ is the number of d-electrons in the metal.

For rhodium, the total number of valence electrons in the complex $=8+7=15$

Thus, the number of valence electrons in rhodium is $15$ .

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution & Answer

Interpretation Introduction

Interpretation:

The oxidation state of a metal and the total number of valence electrons in metals in the complexes is to be determined.

Concept introduction:

The oxidation state of a metal in a complex is the charge on the metal that would be there even if all the anionic ligands and counter ions were removed.

The total number of valence electrons of a metal in a complex is obtained by the following formula:

Total number of valence electrons of metal in complex = $dn$ $+$ electrons donated by ligands

Here, $dn$ is the d-electrons in the metal.

Answer to Problem 1PP

Solution:

(a)

Oxidation state of Rh is $+1$ and the total number of valence electrons in Rh is $15$ .

(b)

Oxidation state of Hg is $+2$ and the number of total valence electrons of Hg is $16$ .

(c)

Oxidation state of Ni is $0$ and the number of total valence electrons of Ni is $16$ .

Explanation of Solution

Given information:

Organic Chemistry, 12e Study Guide/Student Solutions Manual, Chapter 21, Problem 1PP

The oxidation state of rhodium is $+1$ in the complex. This is because Cl is the only ligand with a charge in the complex, phosphine triphenyl is a neutral ligand. The charge on the Cl ligand is $−1$ . To satisfy this charge, rhodium carries a $+1$ charge.

Charge on rhodium is as follows:

$x+(−1)=0$

Here, $x$ is the charge on rhodium and its value is as follows:

$x=1$

Now, d-electrons in rhodium are $8$ and the number of electrons donated by ligands is $7$ . Each phosphine triphenyl donate two electrons and the chlorine atom donates one electron. So, according to the equation, the total number of valence electrons of the metal, in the complex are:

Total number of valence electrons of metal in complex = $dn$ $+$ electrons donated by ligands

Here, $dn$ is the number of d-electrons in the metal.

For rhodium, the total number of valence electrons in the complex $=8+7=15$

Thus, the number of valence electrons in rhodium is $15$ .

Answer 8

Expert Solution & Answer

Interpretation Introduction

Interpretation:

The oxidation state of a metal and the total number of valence electrons in metals in the complexes is to be determined.

Concept introduction:

The oxidation state of a metal in a complex is the charge on the metal that would be there even if all the anionic ligands and counter ions were removed.

The total number of valence electrons of a metal in a complex is obtained by the following formula:

Total number of valence electrons of metal in complex = $dn$ $+$ electrons donated by ligands

Here, $dn$ is the d-electrons in the metal.

Answer to Problem 1PP

Solution:

(a)

Oxidation state of Rh is $+1$ and the total number of valence electrons in Rh is $15$ .

(b)

Oxidation state of Hg is $+2$ and the number of total valence electrons of Hg is $16$ .

(c)

Oxidation state of Ni is $0$ and the number of total valence electrons of Ni is $16$ .

Explanation of Solution

Given information:

Organic Chemistry, 12e Study Guide/Student Solutions Manual, Chapter 21, Problem 1PP

The oxidation state of rhodium is $+1$ in the complex. This is because Cl is the only ligand with a charge in the complex, phosphine triphenyl is a neutral ligand. The charge on the Cl ligand is $−1$ . To satisfy this charge, rhodium carries a $+1$ charge.

Charge on rhodium is as follows:

$x+(−1)=0$

Here, $x$ is the charge on rhodium and its value is as follows:

$x=1$

Now, d-electrons in rhodium are $8$ and the number of electrons donated by ligands is $7$ . Each phosphine triphenyl donate two electrons and the chlorine atom donates one electron. So, according to the equation, the total number of valence electrons of the metal, in the complex are:

Total number of valence electrons of metal in complex = $dn$ $+$ electrons donated by ligands

Here, $dn$ is the number of d-electrons in the metal.

For rhodium, the total number of valence electrons in the complex $=8+7=15$

Thus, the number of valence electrons in rhodium is $15$ .

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Interpretation Introduction

Interpretation:

The oxidation state of a metal and the total number of valence electrons in metals in the complexes is to be determined.

Concept introduction:

The oxidation state of a metal in a complex is the charge on the metal that would be there even if all the anionic ligands and counter ions were removed.

The total number of valence electrons of a metal in a complex is obtained by the following formula:

Total number of valence electrons of metal in complex = $dn$ $+$ electrons donated by ligands

Here, $dn$ is the d-electrons in the metal.

Answer 12

Answer to Problem 1PP

Solution:

(a)

Oxidation state of Rh is $+1$ and the total number of valence electrons in Rh is $15$ .

(b)

Oxidation state of Hg is $+2$ and the number of total valence electrons of Hg is $16$ .

(c)

Oxidation state of Ni is $0$ and the number of total valence electrons of Ni is $16$ .

Answer 13

Explanation of Solution

Given information:

Organic Chemistry, 12e Study Guide/Student Solutions Manual, Chapter 21, Problem 1PP

The oxidation state of rhodium is $+1$ in the complex. This is because Cl is the only ligand with a charge in the complex, phosphine triphenyl is a neutral ligand. The charge on the Cl ligand is $−1$ . To satisfy this charge, rhodium carries a $+1$ charge.

Charge on rhodium is as follows:

$x+(−1)=0$

Here, $x$ is the charge on rhodium and its value is as follows:

$x=1$

Now, d-electrons in rhodium are $8$ and the number of electrons donated by ligands is $7$ . Each phosphine triphenyl donate two electrons and the chlorine atom donates one electron. So, according to the equation, the total number of valence electrons of the metal, in the complex are:

Total number of valence electrons of metal in complex = $dn$ $+$ electrons donated by ligands

Here, $dn$ is the number of d-electrons in the metal.

For rhodium, the total number of valence electrons in the complex $=8+7=15$

Thus, the number of valence electrons in rhodium is $15$ .

Answer 14

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PRACTICE PROBLEM 21.1 For each of the following complexes, determine the oxidation state of the metal and the total number of valence electrons it possesses. (a) (b) (c)

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Answer to Problem 1PP

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