
(a)
Interpretation:
In the given set of species which will be smaller in size has to be identified.
Concept Introduction:
- The distance between the nucleus and the valence shell of a cation or an anion is known as ionic radius. An ion is formed by either loss or gain of electrons from its valence shell.
- As we move down the group in periodic table the ionic radius increases as the electrons are added to new subshell. As we move across the period the ionic radius increases as the electrons are added to the same subshell.
- An anion is formed when an electron is added to the valence shell of an atom. The anion has a net negative charge in it. In anion the extra electron added occupies more space and maximizes the shielding.
- Anions will have larger size compared to cations.
- The reduction in the effective nuclear charge on the electron cloud, due to a difference in the attraction forces of the electrons in the nucleus is known as shielding effect
- When the proton number is greater than the electron, the size of the ion will be smaller due to less shielding. When the proton number is lesser than the electron, the size of the ion will be larger due to more shielding.
- A cation is formed when an electron is lost by an atom from its valence shell. The cation has a net positive charge. In cation the shielding decreases as the electron is removed from the valence shell.
- If the total number of electrons is less than the total number of protons in the ion, then the protons present can effectively attract the valence shell decreasing the size of the ion and vice-versa.
- The trend in periodic table can be described as well. As we move down the group the ionic radius decreases as the electrons are added to a new shell. But as we move across a period in periodic table the ionic radius increases as the electrons are added to the same subshell.
(a)

Answer to Problem 8.43QP
In (a)
Explanation of Solution
The number of electrons and protons in the given set of species (a) is,
Species | Total number of Electrons | Total number of Protons |
17 | 17 | |
18 | 17 |
The total number of electrons and protons present for the given species are found out and presented in the above table. From this we can see that total number of protons in all the given species is same, but the total numbers of electrons are different.
By comparing the total number of protons and electrons in the table given in the previous step and as comparing to the size of atom size of anion is larger. And also here the proton number is lesser than the electron in
(b)
Interpretation:
In the given set of species which will be smaller in size has to be identified.
Concept Introduction:
- The distance between the nucleus and the valence shell of a cation or an anion is known as ionic radius. An ion is formed by either loss or gain of electrons from its valence shell.
- As we move down the group in periodic table the ionic radius increases as the electrons are added to new subshell. As we move across the period the ionic radius increases as the electrons are added to the same subshell.
- An anion is formed when an electron is added to the valence shell of an atom. The anion has a net negative charge in it. In anion the extra electron added occupies more space and maximizes the shielding.
- Anions will have larger size compared to cations.
- The reduction in the effective nuclear charge on the electron cloud, due to a difference in the attraction forces of the electrons in the nucleus is known as shielding effect
- When the proton number is greater than the electron, the size of the ion will be smaller due to less shielding. When the proton number is lesser than the electron, the size of the ion will be larger due to more shielding.
- A cation is formed when an electron is lost by an atom from its valence shell. The cation has a net positive charge. In cation the shielding decreases as the electron is removed from the valence shell.
- If the total number of electrons is less than the total number of protons in the ion, then the protons present can effectively attract the valence shell decreasing the size of the ion and vice-versa.
- The trend in periodic table can be described as well. As we move down the group the ionic radius decreases as the electrons are added to a new shell. But as we move across a period in periodic table the ionic radius increases as the electrons are added to the same subshell.
(b)

Answer to Problem 8.43QP
In (b)
Explanation of Solution
The number of electrons and protons in the given set of species (b)
Species | Total number of Electrons | Total number of Protons |
11 | 11 | |
10 | 11 |
The total number of electrons and protons present for the given species are found out and presented in the above table. From this we can see that total numbers of protons in all the given species are same, but the total numbers of electrons are different.
By comparing the total number of protons and electrons in the table given in the previous step it is clear that the number of proton is greater than the electron in
(c)
Interpretation:
In the given set of species which will be smaller in size has to be identified.
Concept Introduction:
- The distance between the nucleus and the valence shell of a cation or an anion is known as ionic radius. An ion is formed by either loss or gain of electrons from its valence shell.
- As we move down the group in periodic table the ionic radius increases as the electrons are added to new subshell. As we move across the period the ionic radius increases as the electrons are added to the same subshell.
- An anion is formed when an electron is added to the valence shell of an atom. The anion has a net negative charge in it. In anion the extra electron added occupies more space and maximizes the shielding.
- Anions will have larger size compared to cations.
- The reduction in the effective nuclear charge on the electron cloud, due to a difference in the attraction forces of the electrons in the nucleus is known as shielding effect
- When the proton number is greater than the electron, the size of the ion will be smaller due to less shielding. When the proton number is lesser than the electron, the size of the ion will be larger due to more shielding.
- A cation is formed when an electron is lost by an atom from its valence shell. The cation has a net positive charge. In cation the shielding decreases as the electron is removed from the valence shell.
- If the total number of electrons is less than the total number of protons in the ion, then the protons present can effectively attract the valence shell decreasing the size of the ion and vice-versa.
- The trend in periodic table can be described as well. As we move down the group the ionic radius decreases as the electrons are added to a new shell. But as we move across a period in periodic table the ionic radius increases as the electrons are added to the same subshell.
(c)

Answer to Problem 8.43QP
In (c)
Explanation of Solution
The number of electrons and protons in the given set of species (c)
Species | Total number of Electrons | Total number of Protons |
8 | 10 | |
16 | 18 |
The total number of electrons and protons present for the given species are found out and presented in the above table.
The given two species belong to group “6A” of periodic table. The oxygen atom comes before the sulphur atom when we move down the periodic table. As discussed above, when we move down the group the ionic radius increases because the electrons are added to a new subshell. Hence,
(d)
Interpretation:
In the given set of species which will be smaller in size has to be identified.
Concept Introduction:
- The distance between the nucleus and the valence shell of a cation or an anion is known as ionic radius. An ion is formed by either loss or gain of electrons from its valence shell.
- As we move down the group in periodic table the ionic radius increases as the electrons are added to new subshell. As we move across the period the ionic radius increases as the electrons are added to the same subshell.
- An anion is formed when an electron is added to the valence shell of an atom. The anion has a net negative charge in it. In anion the extra electron added occupies more space and maximizes the shielding.
- Anions will have larger size compared to cations.
- The reduction in the effective nuclear charge on the electron cloud, due to a difference in the attraction forces of the electrons in the nucleus is known as shielding effect
- When the proton number is greater than the electron, the size of the ion will be smaller due to less shielding. When the proton number is lesser than the electron, the size of the ion will be larger due to more shielding.
- A cation is formed when an electron is lost by an atom from its valence shell. The cation has a net positive charge. In cation the shielding decreases as the electron is removed from the valence shell.
- If the total number of electrons is less than the total number of protons in the ion, then the protons present can effectively attract the valence shell decreasing the size of the ion and vice-versa.
- The trend in periodic table can be described as well. As we move down the group the ionic radius decreases as the electrons are added to a new shell. But as we move across a period in periodic table the ionic radius increases as the electrons are added to the same subshell.
(d)

Answer to Problem 8.43QP
In (d)
Explanation of Solution
The number of electrons and protons in the given set of species (d)
Species | Total number of Electrons | Total number of Protons |
10 | 12 | |
10 | 13 |
The total number of electrons and protons present for the given species are found out and presented in the above table. From this we can see that total number of electrons in all the given ions is same, but the total numbers of protons are different.
The number of proton in
(e)
Interpretation:
In the given set of species which will be smaller in size has to be identified.
Concept Introduction:
- The distance between the nucleus and the valence shell of a cation or an anion is known as ionic radius. An ion is formed by either loss or gain of electrons from its valence shell.
- As we move down the group in periodic table the ionic radius increases as the electrons are added to new subshell. As we move across the period the ionic radius increases as the electrons are added to the same subshell.
- An anion is formed when an electron is added to the valence shell of an atom. The anion has a net negative charge in it. In anion the extra electron added occupies more space and maximizes the shielding.
- Anions will have larger size compared to cations.
- The reduction in the effective nuclear charge on the electron cloud, due to a difference in the attraction forces of the electrons in the nucleus is known as shielding effect
- When the proton number is greater than the electron, the size of the ion will be smaller due to less shielding. When the proton number is lesser than the electron, the size of the ion will be larger due to more shielding.
- A cation is formed when an electron is lost by an atom from its valence shell. The cation has a net positive charge. In cation the shielding decreases as the electron is removed from the valence shell.
- If the total number of electrons is less than the total number of protons in the ion, then the protons present can effectively attract the valence shell decreasing the size of the ion and vice-versa.
- The trend in periodic table can be described as well. As we move down the group the ionic radius decreases as the electrons are added to a new shell. But as we move across a period in periodic table the ionic radius increases as the electrons are added to the same subshell.
(e)

Answer to Problem 8.43QP
In (e)
Explanation of Solution
The number of electrons and protons in the given set of species (e)
Species | Total number of Electrons | Total number of Protons |
79 | 79 | |
76 | 79 |
The total number of electrons and protons present for the given species are found out and presented in the above table. From this we can see that total number of protons in the given species is same, but the total numbers of electrons are different.
By comparing the total number of protons and electrons in the table given in the previous step it is clear that the number of proton is greater than the electron in
Want to see more full solutions like this?
Chapter 8 Solutions
General Chemistry
- 10. The most important reason why Br- is a better nucleophile than Cl-is ___. A. polarizability; B. size; C. solvation; D. basicity; E. polarity. Please include all steps. Thanks!arrow_forwardPredicting the qualitative acid-base properties of salts Consider the following data on some weak acids and weak bases: base acid Ка K₁₁ name formula name formula nitrous acid HNO2 4.5×10 4 pyridine CHEN 1.7 × 10 9 4 hydrofluoric acid HF 6.8 × 10 methylamine CH3NH2 | 4.4 × 10¯ Use this data to rank the following solutions in order of increasing pH. In other words, select a '1' next to the solution that will have the lowest pH, a '2' next to the solution that will have the next lowest pH, and so on. solution 0.1 M NaNO2 0.1 M KF pH choose one v choose one v 0.1 M C5H5NHBr 0.1 M CH3NH3CI choose one v ✓ choose one 1 (lowest) 2 ☑ 3 4 (highest) 000 18 Ararrow_forward4. The major product from treatment of 2-propanol with the Jonesreagent is ___.A. acetone; B. none of the other answers is correct C. propene; D.propanoic acid; E carbon dioxide. Please include all steps! Thank you!arrow_forward
- 7. All of the following compounds that are at the same oxidation levelare ___.u. methyl epoxide, v. propyne, w. propanal, x. propene,y. 2,2-dihydroxypropane, z. isopropanol?A. u,v,w,y; B. u,v,w; C. v,w,y,z; D. v, z; E. x,y,z Please include all steps. Thank you!arrow_forward9. Which one of the following substituents is the worst leaving group inan SN2 reaction? A. -NH2; B. -OH; C. –F; D. NH3; E. H2O Please include all steps. Thanks!arrow_forwardUsing the general properties of equilibrium constants At a certain temperature, the equilibrium constant K for the following reaction is 2.5 × 105: CO(g) + H2O(g) CO2(g) + H2(g) Use this information to complete the following table. Suppose a 7.0 L reaction vessel is filled with 1.7 mol of CO and 1.7 mol of H2O. What can you say about the composition of the mixture in the vessel at equilibrium? What is the equilibrium constant for the following reaction? Be sure your answer has the correct number of significant digits. CO2(9)+H2(g) CO(g)+H₂O(g) What is the equilibrium constant for the following reaction? Be sure your answer has the correct number of significant digits. 3 CO(g)+3H2O(g) = 3 CO2(g)+3H2(g) There will be very little CO and H2O. x10 There will be very little CO2 and H2. 000 Neither of the above is true. K = ☐ K = ☐ 18 Ararrow_forward
- 8. When ethane thiol is treated with hydrogen peroxide the product is___.A. ethane disulfide; B. diethyl sulfide; C. ethane sulfoxide; D. ethanesulfate; E. ethyl mercaptan. Please include all steps. Thanks!arrow_forward5. The major product of the three step reaction that takes place when 1-propanol is treated with strong acid is?A. dipropyl ether; B. propene; C. propanal; D. isopropyl propyl ether;E. 1-hexanol Please include all steps. Thank you!arrow_forward6. The formula of the product of the addition of HCN to benzaldehydeis ___.A. C8H7NO; B. C8H6NO; C. C14H11NO; D. C9H9NO; E. C9H8NO Please include all steps. Thank you!arrow_forward
- Predicting the qualitative acid-base properties of salts Consider the following data on some weak acids and weak bases: base acid K K a name formula name formula nitrous acid HNO2 4.5×10 hydroxylamine HONH2 1.1 × 10 8 hypochlorous acid HCIO 8 3.0 × 10 methylamine CH3NH2 | 4.4 × 10¯ 4 Use this data to rank the following solutions in order of increasing pH. In other words, select a '1' next to the solution that will have the lowest pH, a '2' next to the solution that will have the next lowest pH, and so on. 0.1 M KCIO solution PH choose one 0.1 M NaNO2 0.1 M CH3NH3Br 0.1 M NaBr choose one ✓ choose one v ✓ choose one 1 (lowest) ☑ 2 3 4 (highest)arrow_forwardFor this Orgo problem, don't worry about question 3 below it. Please explain your thought process, all your steps, and also include how you would tackle a similar problem. Thank you!arrow_forwardUsing the general properties of equilibrium constants At a certain temperature, the equilibrium constant K for the following reaction is 0.84: H2(g) + 2(g) 2 HI(g) = Use this information to complete the following table. Suppose a 34. L reaction vessel is filled with 0.79 mol of HI. What can you say about the composition of the mixture in the vessel at equilibrium? There will be very little H2 and 12. ☐ x10 There will be very little HI. Neither of the above is true. What is the equilibrium constant for the following reaction? Be sure your answer has the correct number of significant digits. 2 HI(g) H₂(9)+12(9) K = What is the equilibrium constant for the following reaction? Be sure your answer has the correct number of significant digits. 2 H2(g)+212(9) 4 HI(g) K = ☐ ☑arrow_forward
- ChemistryChemistryISBN:9781305957404Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCostePublisher:Cengage LearningChemistryChemistryISBN:9781259911156Author:Raymond Chang Dr., Jason Overby ProfessorPublisher:McGraw-Hill EducationPrinciples of Instrumental AnalysisChemistryISBN:9781305577213Author:Douglas A. Skoog, F. James Holler, Stanley R. CrouchPublisher:Cengage Learning
- Organic ChemistryChemistryISBN:9780078021558Author:Janice Gorzynski Smith Dr.Publisher:McGraw-Hill EducationChemistry: Principles and ReactionsChemistryISBN:9781305079373Author:William L. Masterton, Cecile N. HurleyPublisher:Cengage LearningElementary Principles of Chemical Processes, Bind...ChemistryISBN:9781118431221Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. BullardPublisher:WILEY





