Question

name: pd: chemistry u3l12_q2 ct module day 1 hw note: annotate all questions & diagrams! cross out part of the answer choices that makes that answer incorrect! x(g)→x+(g)+e− ie₁ = 740 kj/mol x+(g)→x²+(g)+e− ie₂ = 1450 kj/mol x²+(g)→x³+(g)+e− ie₃ = 7730 kj/mol for element x represented above, which of the following is the most likely explanation for the large difference between the second and third ionization energies? (a) the effective nuclear charge decreases with successive ionizations. (b) the shielding of outer electrons increases with successive ionizations. (c) the electron removed during the third ionization is, on average, much closer to the nucleus than the first two electrons removed were. (d) the ionic radius increases with successive ionizations. 2. which of the following correctly compares periodic properties of two elements and provides an accurate explanation for that difference? (a) the first ionization energy of al is greater than that of b because al has a larger nuclear charge than b does. (b) the first ionization energy of f is greater than that of o because o has a higher electronegativity than f has. (c) the atomic radius of ca is larger than that of mg because the valence electrons in mg experience more shielding than the valence electrons in ca do. (d) the atomic radius of cl is smaller than that of s because cl has a larger nuclear charge than s does. 3. the first five ionization energies of a second - period element are listed in the table above. which of the following correctly identifies the element and best explains the data in the table? (a) b, because it has five core electrons (b) b, because it has three valence electrons (c) n, because it has five valence electrons (d) n, because it has three electrons in the p sublevel

Explanation:

Question 1

• Option A: The effective nuclear charge increases with successive ionizations as electrons are removed, so this option is incorrect.
• Option B: Shielding of outer electrons decreases with successive ionizations (fewer electrons to shield), so this option is incorrect.
• Option C: When the third electron is removed, if it is from an inner - shell (closer to the nucleus), the ionization energy will be much larger. For example, elements in Group 2 (like Mg: \(1s^{2}2s^{2}2p^{6}3s^{2}\)), after losing two \(3s\) electrons (\(IE_1\) and \(IE_2\)), the third electron is from the \(2p\) sub - shell (closer to the nucleus), so \(IE_3\) is much larger. This option is correct.
• Option D: The ionic radius decreases with successive ionizations (fewer electrons, more protons pulling the remaining electrons in), so this option is incorrect.

• Option A: \(Al\) (\(1s^{2}2s^{2}2p^{6}3s^{2}3p^{1}\)) has a first ionization energy less than \(B\) (\(1s^{2}2s^{2}2p^{1}\)) because the \(3p\) electron in \(Al\) is shielded more. So this option is incorrect.
• Option B: The first ionization energy of \(F\) (\(1s^{2}2s^{2}2p^{5}\)) is greater than \(O\) (\(1s^{2}2s^{2}2p^{4}\)) because \(F\) has a higher effective nuclear charge (more protons in the nucleus for the same shell). Also, electron - negativity of \(F\) is higher than \(O\). So this option is incorrect.
• Option C: The atomic radius of \(Ca\) (\(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}\)) is larger than \(Mg\) (\(1s^{2}2s^{2}2p^{6}3s^{2}\)) because \(Ca\) has an additional shell (\(n = 4\) for valence electrons in \(Ca\) vs \(n=3\) for \(Mg\)). The valence electrons in \(Ca\) experience more shielding. So this option is incorrect.
• Option D: For \(Cl\) (\(1s^{2}2s^{2}2p^{6}3s^{2}3p^{5}\)) and \(S\) (\(1s^{2}2s^{2}2p^{6}3s^{2}3p^{4}\)), they are in the same period. \(Cl\) has more protons (\(Z = 17\) for \(Cl\) and \(Z = 16\) for \(S\)). The increased nuclear charge in \(Cl\) pulls the valence electrons closer (higher effective nuclear charge), so the atomic radius of \(Cl\) is smaller. This option is correct.

• The large jump in ionization energy occurs between the third and fourth ionization energies. This indicates that after losing three electrons, the fourth electron is from an inner (core) shell.
• Boron (\(B:1s^{2}2s^{2}2p^{1}\)) has three valence electrons. After losing three electrons (\(IE_1\), \(IE_2\), \(IE_3\)), the fourth electron is from the \(1s\) or \(2s\) (core) shell, which requires a much higher energy (\(IE_4\)).
• Nitrogen (\(N:1s^{2}2s^{2}2p^{3}\)) has five valence electrons. The jump in ionization energy would not be between the third and fourth if it had five valence electrons.

Answer:

C. The electron removed during the third ionization is, on average, much closer to the nucleus than the first two electrons removed were.

Question 2