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)

Question

Want to see more full solutions like this?

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, 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.

Want to see more full solutions like this?

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

Students have asked these similar questions

Draw the major product of the following reaction. Do not draw inorganic byproducts. H3PO4 OH

Predict the major products of this organic reaction: HBr (1 equiv) Δ ? Some important notes: • Draw the major product, or products, of this reaction in the drawing area below. • You can draw the products in any arrangement you like. • Pay careful attention to the reaction conditions, and only include the major products. • Be sure to use wedge and dash bonds when necessary, for example to distinguish between major products that are enantiomers. • Note that there is only 1 equivalent of HBr reactant, so you need not consider the case of multiple additions. Explanation Check X ©2025 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy

For the structure below, draw the resonance structure that is indicated by the curved arrow(s). Be sure to include formal charges. :ÖH Modify the second structure given to draw the new resonance structure. Include lone pairs and charges in your structure. Use the + and - tools to add/remove charges to an atom, and use the single bond tool to add/remove double bonds.

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, 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.

Want to see more full solutions like this?

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

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, 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.

Want to see more full solutions like this?

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

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, 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.

Want to see more full solutions like this?

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

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, 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, 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, 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

Ch. 21 - PRACTICE PROBLEM 21.1 For each of the following...Ch. 21 - Prob. 2PP Ch. 21 - PRACTICE PROBLEM 21.3 What product would you...Ch. 21 - Prob. 4PP Ch. 21 - PRACTICE PROBLEM 21.5 What is the product of the...Ch. 21 - Prob. 6PP Ch. 21 - Prob. 7PP Ch. 21 - Prob. 8PP Ch. 21 - Prob. 9PP Ch. 21 - Prob. 10PP

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)

Videos

Answer to Problem 1PP

Explanation of Solution

Want to see more full solutions like this?

Chapter 21 Solutions

Additional Science Textbook Solutions