Question

name: pd: chemistry u3l14_q2 ct module day 3 hw
4.

the decrease in the second ionization energy of alkali metals going down the group, as shown in the table above, can be best
attributed to a decrease in the coulombic force of attraction due to
(a) the removal of core electrons
(b) an increase in effective nuclear charge
(c) an increase in the average distance of the outermost electron from the nucleus
(d) a decrease in electron - electron repulsion within the outermost shell
**5. which of the following correctly compares the atomic radius of f to that of o and provides the best explanation?
(a) the atomic radius of f is larger than that of o because f has more protons in the nucleus than o does.
(b) the atomic radius of f is larger than that of o because f has more occupied subshells than o does.
(c) the atomic radius of f is smaller than that of o because f has a greater effective nuclear charge than o does.
(d) the atomic radius of f is smaller than that of o because valence electrons in f experience more shielding than those in
o.
*6.

the photoelectron spectra of the 1s electrons of two isoelectronic species, ca²⁺ and ar, are shown above. which of the
following correctly identifies the species associated with peak x and provides a valid justification?
(a) ar, because it has completely filled energy levels
(b) ar, because its radius is smaller than the radius of ca²⁺
(c) ca²⁺, because its nuclear mass is greater than that of ar
(d) ca²⁺, because its nucleus has two more protons than the nucleus of ar has

Explanation:

Question 4

To determine the reason for the decrease in second ionization energy of alkali metals down the group, we analyze each option:

• Option A: Removing core electrons is not the reason for the trend in second ionization energy down the group. The second ionization energy involves removing an electron from a cation, but the trend down the group is related to atomic size and distance from the nucleus.
• Option B: Effective nuclear charge generally increases across a period, not down a group. Down a group, the number of electron shells increases, so effective nuclear charge does not increase.
• Option C: As we go down the group (Li, Na, K, Rb), the number of electron shells increases, so the average distance of the outermost electron from the nucleus increases. This decreases the coulombic force of attraction between the nucleus and the outermost electron, making it easier to remove (lower ionization energy). This matches the trend.
• Option D: Electron - electron repulsion within the outermost shell does not play a major role in the trend of ionization energy down a group of alkali metals.

To compare the atomic radius of F and O:

• Option A: F and O are in the same period (period 2), so they have the same number of occupied subshells. So this option is incorrect.
• Option B: F has a smaller atomic radius than O. F is to the right of O in the same period. As we move from left to right across a period, atomic radius decreases. So this option is incorrect.
• Option C: F has one more proton in the nucleus than O (F: 9 protons, O: 8 protons). The effective nuclear charge is the net positive charge experienced by an electron in an atom. Since F has more protons, the effective nuclear charge on the valence electrons of F is greater than that of O. This greater effective nuclear charge pulls the valence electrons closer to the nucleus, resulting in a smaller atomic radius for F compared to O. This option is correct.
• Option D: Valence electrons in F and O experience similar shielding because they are in the same period and have the same number of inner - shell electrons. Also, F has a smaller atomic radius than O. So this option is incorrect.

For isoelectronic species \(Ca^{2 +}\) and Ar:

• Isoelectronic species have the same number of electrons. \(Ca^{2+}\) has 18 electrons (same as Ar), but \(Ca^{2+}\) has a nucleus with 20 protons and Ar has a nucleus with 18 protons.
• Binding energy is related to the attractive force between the nucleus and the electrons. The greater the number of protons in the nucleus, the greater the attractive force on the electrons, and thus the higher the binding energy.
• Option A: Having completely filled energy levels does not explain the difference in binding energy between \(Ca^{2+}\) and Ar. Both have filled energy levels (Ar is a noble gas, \(Ca^{2+}\) has a noble - gas electron configuration).
• Option B: The radius of \(Ca^{2+}\) is smaller than that of Ar, but this is a result of the greater nuclear charge in \(Ca^{2+}\), not the reason for the binding energy difference. Also, the option says Ar has a smaller radius, which is incorrect.
• Option C: Nuclear mass is not the main factor affecting binding energy in this case. The number of protons (nuclear charge) is the key factor.
• Option D: \(Ca^{2+}\) has 2 more protons than Ar. The greater nuclear charge in \(Ca^{2+}\) means a stronger attraction for the 1s electrons, resulting in a higher binding energy. So peak X (higher binding energy) corresponds to \(Ca^{2+}\) because its nucleus has two more protons than Ar's nucleus.

Answer:

C

Question 5