IndietroStudy Guide: Periodic Trends, Atomic Structure, and Ionic Compounds
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Q1. Which element, Li or Be, has a smaller atomic radius? Justify your answer in terms of atomic structure and Coulomb’s law.
Background
Topic: Periodic Trends – Atomic Radius
This question tests your understanding of how atomic radius changes across a period and how atomic structure and Coulomb’s law explain these trends.
Key Terms and Formulas
Atomic radius: The distance from the nucleus to the outermost electron shell.
Coulomb’s law: , where is the force between charges, and are the magnitudes of the charges, and is the distance between them.
Effective nuclear charge (Zeff): The net positive charge experienced by valence electrons.
Step-by-Step Guidance
Identify the period and group for Li and Be. Both are in period 2, but Li is in group 1 and Be is in group 2.
Recall that as you move across a period from left to right, the number of protons increases, increasing the nuclear charge.
Consider that the added proton in Be increases the attraction between the nucleus and the electrons, pulling the electrons closer to the nucleus.
Use Coulomb’s law to explain that a greater nuclear charge (with similar shielding) results in a stronger attraction, leading to a smaller atomic radius.
Try solving on your own before revealing the answer!
Final Answer:
Be has a smaller atomic radius than Li. This is because both elements have their valence electrons in the same energy level (n = 2), but Be has more protons (higher nuclear charge), resulting in a stronger attraction between the nucleus and the electrons. According to Coulomb’s law, this stronger attraction pulls the electrons closer, decreasing the atomic radius.
Q2. Which element, Li or Na, has a smaller atomic radius? Justify your answer in terms of atomic structure and Coulomb’s law.
Background
Topic: Periodic Trends – Atomic Radius (Group Trends)
This question tests your understanding of how atomic radius changes down a group and how atomic structure and Coulomb’s law explain these trends.
Key Terms and Formulas
Atomic radius
Energy levels (shells): Higher principal quantum number (n) means electrons are farther from the nucleus.
Coulomb’s law:
Step-by-Step Guidance
Identify the group and period for Li and Na. Both are in group 1, but Li is in period 2 and Na is in period 3.
Recall that as you move down a group, each element has an additional energy level, so the valence electrons are farther from the nucleus.
According to Coulomb’s law, as the distance (r) increases, the attractive force between the nucleus and valence electrons decreases.
Consider the effect of increased shielding from inner electrons as you move down a group.
Try solving on your own before revealing the answer!
Final Answer:
Li has a smaller atomic radius than Na. This is because Li’s valence electrons are in the second energy level, while Na’s are in the third, which is farther from the nucleus. The increased distance and additional shielding in Na result in a larger atomic radius, as explained by Coulomb’s law.
Q3. Based on your answers to Questions #1 and #2, arrange the atoms Li, Be, and Na in order of increasing atomic radius.
Background
Topic: Periodic Trends – Atomic Radius
This question asks you to synthesize your understanding of atomic radius trends across periods and down groups.
Key Terms
Atomic radius
Step-by-Step Guidance
Recall from Q1 and Q2 which element has the smallest and largest atomic radius among Li, Be, and Na.
Arrange the elements in order from smallest to largest atomic radius based on your previous reasoning.
Write the sequence using arrows to indicate increasing atomic radius.
Try solving on your own before revealing the answer!
Final Answer:
Order of increasing atomic radius: Be < Li < Na. Be has the smallest atomic radius, followed by Li, and Na has the largest.
Q4. The atomic radius of the Na atom is different than the ionic radius of the Na+ ion.
(a) Write the complete ground state electron configuration for Na and for Na+.
Background
Topic: Electron Configuration and Ionic Radius
This question tests your ability to write electron configurations for atoms and ions and understand how losing electrons affects the configuration.
Key Terms
Ground state electron configuration
Na atom: 11 electrons
Na+ ion: 10 electrons (one less than neutral atom)
Step-by-Step Guidance
Write the electron configuration for a neutral Na atom (Z = 11).
Remove one electron (from the outermost shell) to write the configuration for Na+.
Try solving on your own before revealing the answer!
Final Answer:
Na: 1s2 2s2 2p6 3s1 Na+: 1s2 2s2 2p6
Na+ has lost its single 3s electron, resulting in a configuration identical to neon.
(b) Which particle, Na or Na+, has a larger radius? Justify your answer in terms of atomic structure.
Background
Topic: Atomic vs. Ionic Radius
This question tests your understanding of how losing an electron affects the size of an atom.
Key Terms
Atomic radius
Ionic radius
Step-by-Step Guidance
Compare the number of electrons and the electron configuration for Na and Na+.
Consider the effect of losing the outermost electron shell on the size of the ion.
Think about the increased effective nuclear charge per electron in Na+ compared to Na.
Try solving on your own before revealing the answer!
Final Answer:
Na has a larger radius than Na+. When Na loses an electron to become Na+, it loses its entire 3s shell, resulting in a smaller radius. Additionally, the remaining electrons experience a greater effective nuclear charge, pulling them closer to the nucleus.
Q5. The ionic radii of two different ions are shown in the table above.
(a) Write the ground state electron configuration for Fe2+ and for Fe3+.
Background
Topic: Electron Configuration of Transition Metal Ions
This question tests your ability to write electron configurations for transition metal ions, considering the order in which electrons are lost.
Key Terms
Fe atom: Z = 26
Fe2+: 24 electrons
Fe3+: 23 electrons
Step-by-Step Guidance
Write the electron configuration for a neutral Fe atom.
Remove electrons first from the 4s orbital, then from the 3d orbital as needed for Fe2+ and Fe3+.
Try solving on your own before revealing the answer!
Final Answer:
Fe: 1s2 2s2 2p6 3s2 3p6 4s2 3d6 Fe2+: 1s2 2s2 2p6 3s2 3p6 3d6 Fe3+: 1s2 2s2 2p6 3s2 3p6 3d5
(b) In terms of atomic structure, explain why the ionic radius of Fe2+ is larger than that of Fe3+.
Background
Topic: Ionic Radius and Effective Nuclear Charge
This question tests your understanding of how the loss of additional electrons affects ionic size.
Key Terms
Ionic radius
Effective nuclear charge
Step-by-Step Guidance
Compare the number of electrons in Fe2+ and Fe3+ for the same number of protons.
Consider how removing an additional electron increases the effective nuclear charge per electron.
Think about how this increased attraction pulls the remaining electrons closer to the nucleus, reducing the ionic radius.
Try solving on your own before revealing the answer!
Final Answer:
Fe2+ has a larger ionic radius than Fe3+ because Fe3+ has one fewer electron but the same number of protons, resulting in a greater effective nuclear charge per electron. This stronger attraction pulls the electrons closer, decreasing the ionic radius.
Q6. The atomic radius of the F atom is different than the ionic radius of the F– ion.
(a) Write the complete ground state electron configuration for F and for F–.
Background
Topic: Electron Configuration of Anions
This question tests your ability to write electron configurations for atoms and their anions.
Key Terms
F atom: Z = 9
F– ion: 10 electrons
Step-by-Step Guidance
Write the electron configuration for a neutral F atom.
Add one electron to write the configuration for F–.
Try solving on your own before revealing the answer!
Final Answer:
F: 1s2 2s2 2p5 F–: 1s2 2s2 2p6
(b) Which particle, F or F–, has a larger radius? Justify your answer in terms of atomic structure.
Background
Topic: Atomic vs. Ionic Radius (Anions)
This question tests your understanding of how gaining an electron affects the size of an atom.
Key Terms
Atomic radius
Ionic radius
Step-by-Step Guidance
Compare the number of electrons and the electron configuration for F and F–.
Consider the effect of adding an electron to the same shell, increasing electron-electron repulsion.
Think about how this repulsion causes the electron cloud to expand, increasing the ionic radius.
Try solving on your own before revealing the answer!
Final Answer:
F– has a larger radius than F. Adding an extra electron increases electron-electron repulsion in the same shell, causing the electron cloud to expand and the ionic radius to increase.
Q7. Each of the ions shown in the table (K+, Ca2+, S2–, Cl–) are members of an isoelectronic series. (a) Arrange these ions in order of increasing ionic radius.
Background
Topic: Isoelectronic Series and Ionic Radius
This question tests your understanding of how ionic radius changes for ions with the same number of electrons but different nuclear charges.
Key Terms
Isoelectronic series: Ions with the same number of electrons.
Ionic radius
Step-by-Step Guidance
Determine the number of electrons for each ion (all have 18 electrons).
Compare the number of protons (nuclear charge) for each ion.
Recall that for isoelectronic ions, a higher nuclear charge pulls electrons closer, resulting in a smaller radius.
Arrange the ions from smallest to largest radius based on increasing nuclear charge.
Try solving on your own before revealing the answer!
Final Answer:
Order of increasing ionic radius: Ca2+ < K+ < Cl– < S2–. The ion with the highest nuclear charge (Ca2+) has the smallest radius, and the one with the lowest nuclear charge (S2–) has the largest.
(b) Justify your answer.
Background
Topic: Isoelectronic Series and Effective Nuclear Charge
This question asks you to explain the trend in ionic radius for isoelectronic ions.
Key Terms
Effective nuclear charge
Step-by-Step Guidance
Explain that all ions have the same number of electrons but different numbers of protons.
Describe how a greater number of protons increases the effective nuclear charge, pulling electrons closer.
Relate this to the observed order of ionic radii.
Try solving on your own before revealing the answer!
Final Answer:
As the number of protons increases for isoelectronic ions, the effective nuclear charge increases, pulling the electrons closer and decreasing the ionic radius. Therefore, Ca2+ (20 protons) is smallest, and S2– (16 protons) is largest.
Q8. As you move from left to right across a horizontal row (period) on the periodic table, atomic radius values tend to ______ from left to right, and first ionization energy values tend to ______ from left to right.
Background
Topic: Periodic Trends – Atomic Radius and Ionization Energy
This question tests your knowledge of how atomic radius and ionization energy change across a period.
Key Terms
Atomic radius
Ionization energy
Step-by-Step Guidance
Recall the trend for atomic radius across a period (left to right).
Recall the trend for first ionization energy across a period.
Fill in the blanks with the correct trend directions.
Try solving on your own before revealing the answer!
Final Answer:
Atomic radius values tend to decrease from left to right, and first ionization energy values tend to increase from left to right.
Q9. As you move from top to bottom down a vertical column (group) on the periodic table, atomic radius values tend to ______ from top to bottom, and first ionization energy values tend to ______ from top to bottom.
Background
Topic: Periodic Trends – Group Trends
This question tests your knowledge of how atomic radius and ionization energy change down a group.
Key Terms
Atomic radius
Ionization energy
Step-by-Step Guidance
Recall the trend for atomic radius down a group.
Recall the trend for first ionization energy down a group.
Fill in the blanks with the correct trend directions.
Try solving on your own before revealing the answer!
Final Answer:
Atomic radius values tend to increase from top to bottom, and first ionization energy values tend to decrease from top to bottom.
Q10. (a) In terms of atomic structure and Coulomb’s law, explain why the ionization energy values increase as successive electrons are removed from an atom.
Background
Topic: Successive Ionization Energies
This question tests your understanding of why it becomes harder to remove each additional electron from an atom.
Key Terms
Ionization energy
Coulomb’s law
Step-by-Step Guidance
Consider what happens to the number of electrons and protons as each electron is removed.
Think about how the effective nuclear charge per remaining electron changes after each removal.
Use Coulomb’s law to explain why the attraction between the nucleus and the remaining electrons increases.
Try solving on your own before revealing the answer!
Final Answer:
As each electron is removed, the remaining electrons experience a greater effective nuclear charge, so the attraction between the nucleus and the remaining electrons increases. According to Coulomb’s law, this stronger attraction means more energy is required to remove each successive electron, so ionization energy increases.
(b) In terms of atomic structure and Coulomb’s law, explain why the 2nd IE for Na is much higher than the 2nd IE for Mg.
Background
Topic: Successive Ionization Energies and Electron Configuration
This question tests your understanding of why there is a large jump in ionization energy for certain electrons, based on their location in the atom.
Key Terms
Electron configuration
Core electrons
Step-by-Step Guidance
Write the electron configurations for Na and Mg after the first electron is removed.
Identify which electron is being removed for the 2nd ionization energy in each case (valence vs. core).
Consider the effective nuclear charge and the energy level of the electron being removed.
Try solving on your own before revealing the answer!
Final Answer:
The 2nd ionization energy for Na is much higher because it involves removing a core electron from a stable, closed shell, which is much closer to the nucleus and more strongly attracted. For Mg, the 2nd electron removed is still a valence electron, which is less tightly held. According to Coulomb’s law, the closer the electron is to the nucleus, the greater the force, so more energy is required.
Q11. Based on the information in the table above, how many valence electrons does element X have? Justify your answer.
Background
Topic: Successive Ionization Energies and Valence Electrons
This question tests your ability to interpret ionization energy data to determine the number of valence electrons.
Key Terms
Valence electrons
Ionization energy jumps
Step-by-Step Guidance
Examine the ionization energy values for element X and look for a large jump between two successive values.
Recall that a large jump indicates the removal of a core electron after all valence electrons have been removed.
Count the number of electrons removed before the large jump to determine the number of valence electrons.
Try solving on your own before revealing the answer!
Final Answer:
Element X has 4 valence electrons. The large jump in ionization energy occurs after the 4th electron is removed, indicating that the first 4 electrons are valence electrons and the 5th is a core electron.
Q12. As you move from left to right across a horizontal row (period) on the periodic table, electronegativity values tend to ______ from left to right. As you move from top to bottom down a vertical column (group) on the periodic table, electronegativity values tend to ______ from top to bottom.
Background
Topic: Periodic Trends – Electronegativity
This question tests your knowledge of how electronegativity changes across periods and down groups.
Key Terms
Electronegativity
Step-by-Step Guidance
Recall the trend for electronegativity across a period.
Recall the trend for electronegativity down a group.
Fill in the blanks with the correct trend directions.
Try solving on your own before revealing the answer!
Final Answer:
Electronegativity values tend to increase from left to right across a period and decrease from top to bottom down a group.
Q13. The smaller the atomic radius is, the ______ the electronegativity value is. The larger the atomic radius is, the ______ the electronegativity value is. The most electronegative element on the periodic table is ______.
Background
Topic: Relationship Between Atomic Radius and Electronegativity
This question tests your understanding of how atomic size affects an atom’s ability to attract electrons in a bond.
Key Terms
Atomic radius
Electronegativity
Step-by-Step Guidance
Recall the relationship between atomic radius and electronegativity.
Identify the most electronegative element using the periodic table or data provided.
Fill in the blanks with the correct terms.
Try solving on your own before revealing the answer!
Final Answer:
The smaller the atomic radius, the higher the electronegativity value. The larger the atomic radius, the lower the electronegativity value. The most electronegative element is fluorine (F).
Q14. Write the correct number of valence electrons for each of the following elements: Li, Be, B, C, N, O, F, Ne.
Background
Topic: Valence Electrons
This question tests your ability to determine the number of valence electrons for main group elements.
Key Terms
Valence electrons: Electrons in the outermost shell of an atom.
Step-by-Step Guidance
Locate each element in the periodic table and determine its group number.
Recall that for main group elements, the group number corresponds to the number of valence electrons.
Write the number of valence electrons for each element.
Try solving on your own before revealing the answer!
Final Answer:
Li: 1, Be: 2, B: 3, C: 4, N: 5, O: 6, F: 7, Ne: 8
Q15. Write the correct charge (e.g., 1+, 2+, 1–, 2–, etc.) that each of the following elements has when it forms a stable monoatomic ion: Li, Be, B, C, N, O, F, Ne.
Background
Topic: Ionic Charges of Main Group Elements
This question tests your ability to predict the charge of ions formed by main group elements based on their position in the periodic table.
Key Terms
Monoatomic ion
Step-by-Step Guidance
Determine how many electrons each element needs to gain or lose to achieve a noble gas configuration.
Assign the correct charge based on whether the element loses (positive) or gains (negative) electrons.
Write the charge for each element; note that noble gases typically do not form ions.
Try solving on your own before revealing the answer!
Final Answer:
Li: 1+, Be: 2+, B: 3+, C: 4+/4– (rarely forms monoatomic ions), N: 3–, O: 2–, F: 1–, Ne: N/A
Q16. Write the correct chemical formula for the binary ionic compound that is formed from the combination of each of the following pairs of elements: Li and F, Na and S, Mg and Cl, Al and O, Ca and P.
Background
Topic: Writing Formulas for Binary Ionic Compounds
This question tests your ability to write empirical formulas for ionic compounds based on the charges of the ions involved.
Key Terms
Binary ionic compound
Empirical formula
Step-by-Step Guidance
Determine the charge of each ion formed by the elements in the pair.
Balance the charges to ensure the compound is neutral overall.
Write the empirical formula for each compound.
Try solving on your own before revealing the answer!
Final Answer:
Li and F: LiF Na and S: Na2S Mg and Cl: MgCl2 Al and O: Al2O3 Ca and P: Ca3P2
