18. the atomic radii of the elements in the nitrogen group in the periodic table are given in the table above. which of the following best helps explain the trend of increasing atomic radius from n to bi? (a) the number of particles in the nucleus of the atom increases. (b) the number of electrons in the outermost shell of the atom increases. (c) the attractive force between the valence electrons and the nuclei of the atoms decreases. (d) the repulsive force between the valence electrons and the electrons in the inner shells decreases. 19. which of the following best helps explain why the first ionization energy of k is less than that of ca? (a) the electronegativity of k is greater than that of ca. (b) the atomic radius of the k atom is less than that of the ca atom. (c) the valence electron of k experiences a lower effective nuclear charge than the valence electrons of ca. (d) the nucleus of the k atom has fewer neutrons, on average, than the nucleus of the ca atom has. 20. which of the following best accounts for the decrease in the first - ionization energy of the alkaline earth metals going down the group? (a) the number of protons in the nucleus increases, resulting in increased attraction between the nucleus and the electron being removed. (b) the number of protons in the nucleus increases, resulting in a greater effective nuclear charge attracting the electron being removed. (c) the number of electron shells increases, resulting in a greater average distance between the nucleus and the electron being removed. (d) the number of electron shells increases, resulting in less shielding between the nucleus and the electron being removed. 21. which of the following elements has the largest first ionization energy? (a) be (b) li (c) na (d) mg 22. which of the following best helps to explain why the atomic radius of k is greater than that of br? (a) the first ionization energy of k is higher than that of br. (b) the valence electrons in k are in a higher principal energy level than those of br. (c) in the ground state, an atom of k has fewer unpaired electrons than an atom of br has. (d) the effective nuclear charge experienced by valence electrons is smaller for k than for br. 23. for parts of the free response question that require calculations, clearly show the method used and the steps involved in arriving at your answers. you must show your work to receive credit for your answer. examples and equations may be included in your answers where appropriate.

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### Question 18
# Brief Explanations:
To determine the trend of increasing atomic radius from N to Bi in the nitrogen group, we analyze each option:
- Option A: The number of particles in the nucleus (protons + neutrons) increasing doesn't directly explain atomic radius increase. Atomic radius is more about electron - nucleus attraction and electron shielding.
- Option B: The number of electrons in the outermost shell for elements in the same group (nitrogen group) remains the same (group valence electrons are the same). So this is incorrect.
- Option C: As we go down a group (from N to Bi), the number of electron shells increases. The inner - shell electrons shield the valence electrons from the nucleus. So the attractive force between the valence electrons and the nucleus decreases. This allows the valence electrons to be further from the nucleus, increasing the atomic radius. This option is correct.
- Option D: As we go down the group, the number of inner - shell electrons increases, so the repulsive force between valence electrons and inner - shell electrons should increase, not decrease. So this option is incorrect.
# Answer: C

### Question 19
# Brief Explanations:
First ionization energy is the energy required to remove the outermost (valence) electron from an atom in the gaseous state.
- Option A: Electronegativity and first ionization energy are related but not in a way that explains the difference between K and Ca here. Also, the electronegativity of K is less than that of Ca. So this is incorrect.
- Option B: The atomic radius of K is larger than that of Ca. So this option is incorrect.
- Option C: For K (electronic configuration: $[Ar]4s^1$) and Ca (electronic configuration: $[Ar]4s^2$), the valence electrons of K experience a lower effective nuclear charge. Effective nuclear charge ($Z_{eff}$) is the net positive charge experienced by an electron in an atom. The additional electron in Ca (compared to K in the same valence shell) increases the $Z_{eff}$ for Ca's valence electrons. A lower $Z_{eff}$ in K means it is easier to remove the valence electron, so the first ionization energy of K is less than that of Ca. This option is correct.
- Option D: The number of neutrons does not directly affect the first ionization energy. First ionization energy is related to electron - nucleus interactions, not the number of neutrons. So this option is incorrect.
# Answer: C

### Question 20
# Brief Explanations:
Alkaline earth metals are in Group 2 of the periodic table. As we go down the group:
- Option A: An increase in the number of protons (nuclear charge) would increase the attraction between the nucleus and the electron, which would increase the first ionization energy, not decrease it. So this is incorrect.
- Option B: An increase in effective nuclear charge (due to more protons) would also increase the first ionization energy, which is the opposite of what we observe (ionization energy decreases down the group). So this is incorrect.
- Option C: As we go down the group, the number of electron shells increases. The average distance between the nucleus and the electron being removed (valence electron) increases. According to Coulomb's law, the force of attraction between the nucleus (positive charge) and the electron (negative charge) is inversely proportional to the square of the distance between them ($F=\frac{kq_1q_2}{r^2}$). A larger distance means a weaker attraction, so less energy is required to remove the electron, resulting in a decrease in first ionization energy. This option is correct.
- Option D: As the number of electron shells increases, the shielding between the nucleus and the valence electron increases (more inner - shell electrons), not decreases. So this is incorrect.
# Answer: C

### Question 21
# Brief Explanations:
First ionization energy generally increases across a period (from left to right) and decreases down a group.
- Be: Electronic configuration $[He]2s^2$
- Li: Electronic configuration $[He]2s^1$
- Na: Electronic configuration $[Ne]3s^1$
- Mg: Electronic configuration $[Ne]3s^2$

Li and Na are in Group 1, Be and Mg are in Group 2. Down a group (Group 1: Li to Na; Group 2: Be to Mg), ionization energy decreases. So Na has lower ionization energy than Li, and Mg has lower ionization energy than Be. Across a period (from Li to Be, from Na to Mg), ionization energy increases. Be has a higher first ionization energy than Li, and Mg has a higher first ionization energy than Na. Also, Be (in period 2) has a higher first ionization energy than Mg (in period 3) because as we go down a group, ionization energy decreases. So among Be, Li, Na, and Mg, Be has the largest first ionization energy.
# Answer: A

### Question 22
# Brief Explanations:
Atomic radius is the distance from the nucleus to the outermost electron.
- Option A: The first ionization energy of K is lower than that of Br. So this option is incorrect.
- Option B: K (electronic configuration: $[Ar]4s^1$) and Br (electronic configuration: $[Ar]3d^{10}4s^24p^5$) - the valence electrons of K are in the 4s orbital (principal energy level $n = 4$), and the valence electrons of Br are in the 4s and 4p orbitals, but the outermost electrons of Br are still in the $n = 4$ level. So the principal energy level is not the main reason. So this option is incorrect.
- Option C: The number of unpaired electrons does not determine the atomic radius. Atomic radius is related to electron - nucleus interactions and electron shielding. So this option is incorrect.
- Option D: The effective nuclear charge ($Z_{eff}$) experienced by valence electrons: K has one valence electron ($4s^1$) and Br has seven valence electrons ($4s^24p^5$). The inner - shell electrons shield the valence electrons from the nucleus. For K, the shielding is more effective in reducing the $Z_{eff}$ for the valence electron compared to Br. A smaller $Z_{eff}$ means the valence electrons are held less tightly, so the atomic radius of K is larger than that of Br. This option is correct.
# Answer: D

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