Indicate which one of the two species in each of the following pairs is smaller: (a) Cl or Cl−; (b) Na or Na+; (c) O2− or S2−; (d) Mg2+ or Al3+; (e) Au or Au3+.
(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.
To identify: The species which will have smaller size from the given species.
Answer to Problem 4.89QP
Answer
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.
To identify: The species which will have smaller size from the given species.
Answer to Problem 4.89QP
Answer
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.
To identify: The species which will have smaller size from the given species.
Answer to Problem 4.89QP
Answer
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.
To identify: The species which will have smaller size from the given species.
Answer to Problem 4.89QP
Answer
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.
To identify: The species which will have smaller size from the given species.
Answer to Problem 4.89QP
Answer
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 4 Solutions
CHEMISTRY:ATOMS FIRST (LL)>CUSTOM PKG.<
- Using reaction free energy to predict equilibrium composition Consider the following equilibrium: 2NO2 (g) = N2O4(g) AGº = -5.4 kJ Now suppose a reaction vessel is filled with 4.53 atm of dinitrogen tetroxide (N2O4) at 279. °C. Answer the following questions about this system: Under these conditions, will the pressure of N2O4 tend to rise or fall? Is it possible to reverse this tendency by adding NO2? In other words, if you said the pressure of N2O4 will tend to rise, can that be changed to a tendency to fall by adding NO2? Similarly, if you said the pressure of N2O4 will tend to fall, can that be changed to a tendency to '2' rise by adding NO2? If you said the tendency can be reversed in the second question, calculate the minimum pressure of NO 2 needed to reverse it. Round your answer to 2 significant digits. 00 rise ☐ x10 fall yes no ☐ atm G Ar 1arrow_forwardWhy do we analyse salt?arrow_forwardCurved arrows are used to illustrate the flow of electrons. Using the provided starting and product structures, draw the curved electron-pushing arrows for the following reaction or mechanistic step(s). Be sure to account for all bond-breaking and bond-making steps. H H CH3OH, H+ H Select to Add Arrows H° 0:0 'H + Q HH ■ Select to Add Arrows CH3OH, H* H. H CH3OH, H+ HH ■ Select to Add Arrows i Please select a drawing or reagent from the question areaarrow_forward
- What are examples of analytical methods that can be used to analyse salt in tomato sauce?arrow_forwardA common alkene starting material is shown below. Predict the major product for each reaction. Use a dash or wedge bond to indicate the relative stereochemistry of substituents on asymmetric centers, where applicable. Ignore any inorganic byproducts H Šali OH H OH Select to Edit Select to Draw 1. BH3-THF 1. Hg(OAc)2, H2O =U= 2. H2O2, NaOH 2. NaBH4, NaOH + Please select a drawing or reagent from the question areaarrow_forwardWhat is the MOHR titration & AOAC method? What is it and how does it work? How can it be used to quantify salt in a sample?arrow_forward
- Predict the major products of this reaction. Cl₂ hv ? Draw only the major product or products in the drawing area below. If there's more than one major product, you can draw them in any arrangement you like. Be sure you use wedge and dash bonds if necessary, for example to distinguish between major products with different stereochemistry. If there will be no products because there will be no significant reaction, just check the box under the drawing area and leave it blank. Note for advanced students: you can ignore any products of repeated addition. Explanation Check Click and drag to start drawing a structure. 80 10 m 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Center | Accessibility DII A F1 F2 F3 F4 F5 F6 F7 F8 EO F11arrow_forwardGiven a system with an anodic overpotential, the variation of η as a function of current density- at low fields is linear.- at higher fields, it follows Tafel's law.Calculate the range of current densities for which the overpotential has the same value when calculated for both cases (the maximum relative difference will be 5%, compared to the behavior for higher fields).arrow_forwardUsing reaction free energy to predict equilibrium composition Consider the following equilibrium: N2 (g) + 3H2 (g) = 2NH3 (g) AGº = -34. KJ Now suppose a reaction vessel is filled with 8.06 atm of nitrogen (N2) and 2.58 atm of ammonia (NH3) at 106. °C. Answer the following questions about this system: rise Under these conditions, will the pressure of N2 tend to rise or fall? ☐ x10 fall Is it possible to reverse this tendency by adding H₂? In other words, if you said the pressure of N2 will tend to rise, can that be changed to a tendency to fall by adding H2? Similarly, if you said the pressure of N will tend to fall, can that be changed to a tendency to rise by adding H₂? If you said the tendency can be reversed in the second question, calculate the minimum pressure of H₂ needed to reverse it. Round your answer to 2 significant digits. yes no ☐ atm Х ด ? olo 18 Ararrow_forward
- Introduction to General, Organic and BiochemistryChemistryISBN:9781285869759Author:Frederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar TorresPublisher:Cengage LearningChemistry & Chemical ReactivityChemistryISBN:9781133949640Author:John C. Kotz, Paul M. Treichel, John Townsend, David TreichelPublisher:Cengage LearningChemistry & Chemical ReactivityChemistryISBN:9781337399074Author:John C. Kotz, Paul M. Treichel, John Townsend, David TreichelPublisher:Cengage Learning
- Chemistry: Principles and ReactionsChemistryISBN:9781305079373Author:William L. Masterton, Cecile N. HurleyPublisher:Cengage LearningChemistryChemistryISBN:9781305957404Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCostePublisher:Cengage Learning





