Introductory Chemistry: A Foundation
Introductory Chemistry: A Foundation
9th Edition
ISBN: 9781337399425
Author: Steven S. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 5, Problem 24CR
Interpretation Introduction

(a)

Interpretation:

To determine the number of proton and electron of the ionic species Mg2+.

Concept Introduction:

Ions are formed by loss or gain of an electron. If an atom gain the electron it will carry positive charge and if it lose the electron it will carry the positive charge. The electron present in the outermost shell of an atom is mostly responsible for formation of ions. If the number of proton is less than number of electron the species is anionic and if the number of proton is higher than number of electron the species if cationic.

Expert Solution
Check Mark

Answer to Problem 24CR

The number of proton and electron of Mg2 + species is 12 and 10 respectively.

Explanation of Solution

Atomic number of an ionic species represents the number of proton so, number of proton of Mg2 + is 12 and the number of electron of cationic species is the difference of number of proton and charge given. Number of electron = 12-2 = 10.

Interpretation Introduction

(b)

Interpretation:

To determine the number of proton and electron of the ionic species Fe2+.

Concept Introduction:

Ions are formed by loss or gain of an electron. If an atom gain the electron it will carry positive charge and if it lose the electron it will carry the positive charge. The electron present in the outermost shell of an atom is mostly responsible for formation of ions. If the number of proton is less than number of electron the species is anionic and if the number of proton is higher than number of electron the species if cationic.

Expert Solution
Check Mark

Answer to Problem 24CR

The number of proton and electron of Fe2 + species is 26 and 24 respectively.

Explanation of Solution

Atomic number of an ionic species represents the number of proton so, number of proton of Fe2 + is 26 and the number of electron of cationic species is the difference of number of proton and charge given. Number of electron = 26-2 = 24.

Interpretation Introduction

(c)

Interpretation:

To determine the number of proton and electron of the ionic species Fe3+.

Concept Introduction:

Ions are formed by loss or gain of an electron. If an atom gain the electron it will carry positive charge and if it lose the electron it will carry the positive charge. The electron present in the outermost shell of an atom is mostly responsible for formation of ions. If the number of proton is less than number of electron the species is anionic and if the number of proton is higher than number of electron the species if cationic.

Expert Solution
Check Mark

Answer to Problem 24CR

The number of proton and electron of Fe3 + species is 26 and 23 respectively.

Explanation of Solution

Atomic number of an ionic species represents the number of proton so, number of proton of Fe2 + is 26 and the number of electron of cationic species is the difference of number of proton and charge given. Number of electron = 26-3 = 23.

Interpretation Introduction

(d)

Interpretation:

To determine the number of proton and electron of the ionic species F-.

Concept Introduction:

Ions are formed by loss or gain of an electron. If an atom gain the electron it will carry positive charge and if it lose the electron it will carry the positive charge. The electron present in the outermost shell of an atom is mostly responsible for formation of ions. If the number of proton is less than number of electron the species is anionic and if the number of proton is higher than number of electron the species if cationic.

Expert Solution
Check Mark

Answer to Problem 24CR

The number of proton and electron of F- species is 9 and 10 respectively.

Explanation of Solution

Atomic number of an ionic species represents the number of proton so, number of proton of Mg2 + is 9 and the number of electron of cationic species is the addition of number of proton and charge given. Number of electron = 9+1 = 10.

Interpretation Introduction

(e)

Interpretation:

To determine the number of proton and electron of the ionic species Ni2+.

Concept Introduction:

Ions are formed by loss or gain of an electron. If an atom gain the electron it will carry positive charge and if it lose the electron it will carry the positive charge. The electron present in the outermost shell of an atom is mostly responsible for formation of ions. If the number of proton is less than number of electron the species is anionic and if the number of proton is higher than number of electron the species if cationic.

Expert Solution
Check Mark

Answer to Problem 24CR

The number of proton and electron of Ni2 + species is 28 and 26 respectively.

Explanation of Solution

Atomic number of an ionic species represents the number of proton so, number of proton of Ni2 + is 28 and the number of electron of cationic species is the difference of number of proton and charge given. Number of electron = 28-2 = 26.

Interpretation Introduction

(f)

Interpretation:

To determine the number of proton and electron of the ionic species Zn2+.

Concept Introduction:

Ions are formed by loss or gain of an electron. If an atom gain the electron it will carry positive charge and if it lose the electron it will carry the positive charge. The electron present in the outermost shell of an atom is mostly responsible for formation of ions. If the number of proton is less than number of electron the species is anionic and if the number of proton is higher than number of electron the species if cationic.

Expert Solution
Check Mark

Answer to Problem 24CR

The number of proton and electron of Zn2 + species is 30 and 28 respectively.

Explanation of Solution

Atomic number of an ionic species represents the number of proton so, number of proton of Zn2 + is 30 and the number of electron of cationic species is the difference of number of proton and charge given. Number of electron = 30-2 = 28.

Interpretation Introduction

(g)

Interpretation:

To determine the number of proton and electron of the ionic species Co3+.

Concept Introduction:

Ions are formed by loss or gain of an electron. If an atom gain the electron it will carry positive charge and if it lose the electron it will carry the positive charge. The electron present in the outermost shell of an atom is mostly responsible for formation of ions. If the number of proton is less than number of electron the species is anionic and if the number of proton is higher than number of electron the species if cationic.

Expert Solution
Check Mark

Answer to Problem 24CR

The number of proton and electron of Co3 + species is 27 and 24 respectively.

Explanation of Solution

Atomic number of an ionic species represents the number of proton so, number of proton of Co3 + is 27 and the number of electron of cationic species is the difference of number of proton and charge given. Number of electron = 27-3 = 24.

Interpretation Introduction

(h)

Interpretation:

To determine the number of proton and electron of the ionic species N3-.

Concept Introduction:

Ions are formed by loss or gain of an electron. If an atom gain the electron it will carry positive charge and if it lose the electron it will carry the positive charge. The electron present in the outermost shell of an atom is mostly responsible for formation of ions. If the number of proton is less than number of electron the species is anionic and if the number of proton is higher than number of electron the species if cationic.

Expert Solution
Check Mark

Answer to Problem 24CR

The number of proton and electron of N3 - species is 7 and 10 respectively.

Explanation of Solution

Atomic number of an ionic species represents the number of proton so, number of proton of N3 - is 7 and the number of electron of cationic species is the addition of number of proton and charge given. Number of electron = 7+3 = 10.

Interpretation Introduction

(i)

Interpretation:

To determine the number of proton and electron of the ionic species S2-.

Concept Introduction:

Ions are formed by loss or gain of an electron. If an atom gain the electron it will carry positive charge and if it lose the electron it will carry the positive charge. The electron present in the outermost shell of an atom is mostly responsible for formation of ions. If the number of proton is less than number of electron the species is anionic and if the number of proton is higher than number of electron the species if cationic.

Expert Solution
Check Mark

Answer to Problem 24CR

The number of proton and electron of S2 - species is 16 and 18 respectively.

Explanation of Solution

Atomic number of an ionic species represents the number of proton so, number of proton of S2 - is 16 and the number of electron of cationic species is the addition of number of proton and charge given. Number of electron = 16+2 = 18.

Interpretation Introduction

(j)

Interpretation:

To determine the number of proton and electron of the ionic species Rb+.

Concept Introduction:

Ions are formed by loss or gain of an electron. If an atom gain the electron it will carry positive charge and if it lose the electron it will carry the positive charge. The electron present in the outermost shell of an atom is mostly responsible for formation of ions. If the number of proton is less than number of electron the species is anionic and if the number of proton is higher than number of electron the species if cationic.

Expert Solution
Check Mark

Answer to Problem 24CR

The number of proton and electron of Rb+ species is 37and 36 respectively.

Explanation of Solution

Atomic number of an ionic species represents the number of proton so, number of proton of Rb+ is 37 and the number of electron of cationic species is the difference of number of proton and charge given. Number of electron = 37-1=36.

Interpretation Introduction

(k)

Interpretation:

To determine the number of proton and electron of the ionic species Se2-.

Concept Introduction:

Ions are formed by loss or gain of an electron. If an atom gain the electron it will carry positive charge and if it lose the electron it will carry the positive charge. The electron present in the outermost shell of an atom is mostly responsible for formation of ions. If the number of proton is less than number of electron the species is anionic and if the number of proton is higher than number of electron the species if cationic.

Expert Solution
Check Mark

Answer to Problem 24CR

The number of proton and electron of Se2 - species is 34 and 36 respectively.

Explanation of Solution

Atomic number of an ionic species represents the number of proton so, number of proton of Se2 - is 34 and the number of electron of cationic species is the addition of number of proton and charge given. Number of electron = 34+2 = 36.

Interpretation Introduction

(l)

Interpretation:

To determine the number of proton and electron of the ionic species K+.

Concept Introduction:

Ions are formed by loss or gain of an electron. If an atom gain the electron it will carry positive charge and if it lose the electron it will carry the positive charge. The electron present in the outermost shell of an atom is mostly responsible for formation of ions. If the number of proton is less than number of electron the species is anionic and if the number of proton is higher than number of electron the species if cationic.

Expert Solution
Check Mark

Answer to Problem 24CR

The number of proton and electron of K+ species is 19 and 18 respectively.

Explanation of Solution

Atomic number of an ionic species represents the number of proton so, number of proton of K+ is 19 and the number of electron of cationic species is the difference of number of proton and charge given. Number of electron = 19-1 = 18.

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
Consider the ions Ag+ and Cl-: How many protons and electrons does each of these ions posses? protons in Ag+=electrons in Ag+= protons in Cl-=electrons in Cl-=
Bromine monochloride is formed from the elements:Cl₂(g) +Br₂(g) ⇌2BrCl(g) ΔH°ᵣₙₓ =-1.35 kJ/mol ΔG°(f)=-0.88 kJ/molCalculate (a) ΔH°(f) and (b) S° of BrCl(g).
Are these correct? Polar chemicals Bromine trichloride (BrCl3) Bromine pentafluoride (BrF5) Sulfur dioxide (SO2) Sulfur tetrachloride (SCl4) Oxygen difluoride (OF2) Silicon tetrafluoride (SiF4) Nonpolar chemicals  Beryllium chloride (BeCl2) Nitrogen trichloride (NCl3) Arsenic pentafluoride (AsF5) Sulfur hexafluoride (SF6) Boron trihydride (BH3) Krypton difluoride (KrF2) Krypton tetrachloride (KrCl4)

Chapter 5 Solutions

Introductory Chemistry: A Foundation

Ch. 5 - In some cases the Roman numeral in a name is the...Ch. 5 - Prob. 2ALQCh. 5 - The formulas MgO and CO look very similar. What is...Ch. 5 - Explain how to use the periodic table to determine...Ch. 5 - Prob. 5ALQCh. 5 - Prob. 6ALQCh. 5 - Name each of the following compounds. SO5 P2S5Ch. 5 - Why do we callBa(NO3)2 barium nitrate hut...Ch. 5 - What is the difference between sulfuric acid and...Ch. 5 - The “Chemistry in Focus” segment Sugar of Lead...Ch. 5 - Prob. 2QAPCh. 5 - Prob. 3QAPCh. 5 - Prob. 4QAPCh. 5 - Prob. 5QAPCh. 5 - Prob. 6QAPCh. 5 - Prob. 7QAPCh. 5 - We indicate the charge of a metallic element that...Ch. 5 - Prob. 9QAPCh. 5 - Prob. 10QAPCh. 5 - Prob. 11QAPCh. 5 - Prob. 12QAPCh. 5 - Prob. 13QAPCh. 5 - Prob. 14QAPCh. 5 - Prob. 15QAPCh. 5 - Prob. 16QAPCh. 5 - Write the name of each of the following binary...Ch. 5 - Write the name for each of the following binary...Ch. 5 - Name each of the following binary compounds, using...Ch. 5 - The formulasNa2O andN2O look very similar. What is...Ch. 5 - Name each of the following binary compounds, using...Ch. 5 - Name each of the following binary compounds, using...Ch. 5 - What is apolyatomicion? Give examples of five...Ch. 5 - Prob. 24QAPCh. 5 - Prob. 25QAPCh. 5 - Prob. 26QAPCh. 5 - Prob. 27QAPCh. 5 - Prob. 28QAPCh. 5 - Prob. 29QAPCh. 5 - Prob. 30QAPCh. 5 - Prob. 31QAPCh. 5 - Prob. 32QAPCh. 5 - Give the name of each of the following polyatomic...Ch. 5 - Prob. 34QAPCh. 5 - Prob. 35QAPCh. 5 - Prob. 36QAPCh. 5 - Give a simple definition of anacid.Ch. 5 - Prob. 38QAPCh. 5 - Prob. 39QAPCh. 5 - Prob. 40QAPCh. 5 - Prob. 41QAPCh. 5 - Prob. 42QAPCh. 5 - Prob. 43QAPCh. 5 - Prob. 44QAPCh. 5 - Prob. 45QAPCh. 5 - Prob. 46QAPCh. 5 - Prob. 47QAPCh. 5 - Prob. 48QAPCh. 5 - Prob. 49QAPCh. 5 - Prob. 50QAPCh. 5 - Prob. 51APCh. 5 - Prob. 52APCh. 5 - Prob. 53APCh. 5 - Prob. 54APCh. 5 - Prob. 55APCh. 5 - Prob. 56APCh. 5 - Name the following compounds. Ca(C2H3O2)2 PCl3...Ch. 5 - Prob. 58APCh. 5 - Prob. 59APCh. 5 - Prob. 60APCh. 5 - Most metallic elements formoxides, and often the...Ch. 5 - Consider a hypothetical simple ionDetermine the...Ch. 5 - Prob. 63APCh. 5 - A metal ion with a 2+ charge has 23 electrons and...Ch. 5 - Prob. 65APCh. 5 - Prob. 66APCh. 5 - The noble metals gold, silver, and platinum are...Ch. 5 - Prob. 68APCh. 5 - The elements of Group 7 (fluorine, chlorine,...Ch. 5 - Prob. 70APCh. 5 - Prob. 71APCh. 5 - An ion with one less electron than it has protons...Ch. 5 - An atom that has lost three electrons will have a...Ch. 5 - An ion with two more electrons than it has protons...Ch. 5 - For each of the negative ions listed in column 1,...Ch. 5 - For each of the following processes that show the...Ch. 5 - For each of the following atomic numbers, use the...Ch. 5 - For the following pairs of ions, use the principle...Ch. 5 - Prob. 79APCh. 5 - Prob. 80APCh. 5 - Prob. 81APCh. 5 - Prob. 82APCh. 5 - Prob. 83APCh. 5 - Name each of the following compounds....Ch. 5 - Prob. 85APCh. 5 - Prob. 86APCh. 5 - Write the foḿu1a for each of the following...Ch. 5 - Give the name of each of the following polyatomic...Ch. 5 - Prob. 89APCh. 5 - Prob. 90APCh. 5 - Prob. 91APCh. 5 - Prob. 92APCh. 5 - Prob. 93APCh. 5 - Complete the following table to predict whether...Ch. 5 - Prob. 95CPCh. 5 - Prob. 96CPCh. 5 - Prob. 97CPCh. 5 - Prob. 98CPCh. 5 - Prob. 1CRCh. 5 - Prob. 2CRCh. 5 - Prob. 3CRCh. 5 - Without consulting your textbook or notes, state...Ch. 5 - Prob. 5CRCh. 5 - What is meant by anuclear atom? Describe the...Ch. 5 - Prob. 7CRCh. 5 - Prob. 8CRCh. 5 - Prob. 9CRCh. 5 - Are most elements found in nature in the elemental...Ch. 5 - What are bus? How are ions formed from atoms? Do...Ch. 5 - Prob. 12CRCh. 5 - Prob. 13CRCh. 5 - Prob. 14CRCh. 5 - Prob. 15CRCh. 5 - Prob. 16CRCh. 5 - Prob. 17CRCh. 5 - Prob. 18CRCh. 5 - Prob. 19CRCh. 5 - Prob. 20CRCh. 5 - Prob. 21CRCh. 5 - How many electrons, protons, and neutrons are...Ch. 5 - What simple ion does each of the following...Ch. 5 - Prob. 24CRCh. 5 - Prob. 25CRCh. 5 - Prob. 26CRCh. 5 - Prob. 27CRCh. 5 - Prob. 28CRCh. 5 - Prob. 29CRCh. 5 - Prob. 30CRCh. 5 - Prob. 31CR
Knowledge Booster
Background pattern image
Chemistry
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Similar questions
Recommended textbooks for you
Text book image
Introductory Chemistry: A Foundation
Chemistry
ISBN:9781337399425
Author:Steven S. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Text book image
Chemistry & Chemical Reactivity
Chemistry
ISBN:9781337399074
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:Cengage Learning
Text book image
Chemistry & Chemical Reactivity
Chemistry
ISBN:9781133949640
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:Cengage Learning
Text book image
Chemical Principles in the Laboratory
Chemistry
ISBN:9781305264434
Author:Emil Slowinski, Wayne C. Wolsey, Robert Rossi
Publisher:Brooks Cole
Text book image
Chemistry for Engineering Students
Chemistry
ISBN:9781337398909
Author:Lawrence S. Brown, Tom Holme
Publisher:Cengage Learning
Text book image
Chemistry
Chemistry
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Atomic Number, Atomic Mass, and the Atomic Structure | How to Pass ChemistryThe Nucleus: Crash Course Chemistry #1; Author: Crash Course;https://www.youtube.com/watch?v=FSyAehMdpyI;License: Standard YouTube License, CC-BY