Question

question 41 (1 point)
which of the following factors contributes to the decrease in ionization energy within a group in the periodic table as the atomic number increases?
o a increase in number of protons
o b increase in atomic size
o c increase in size of the nucleus
o d fewer electrons in the highest occupied energy level

question 42 (1 point)
which of the following elements has the smallest first ionization energy?
o a calcium
o b magnesium
o c sodium
o d potassium

question 43 (1 point)
which of the following elements has the lowest electronegativity?
o a lithium
o b carbon
o c bromine
o d fluorine

Explanation:

Question 41

Ionization energy is the energy required to remove an electron from an atom. As atomic size increases within a group (due to more electron shells), the outermost electrons are farther from the nucleus and experience less nuclear attraction. This makes it easier to remove them (lower ionization energy). An increase in protons (a) would actually increase nuclear pull (if electron shielding didn't occur, but within a group, electron - shell increase dominates). The size of the nucleus (c) is less relevant compared to atomic size. For a group, the number of electrons in the highest - occupied energy level (d) is the same (e.g., Group 1 has 1 valence electron for all elements in the group).

Ionization energy decreases down a group and increases across a period. Sodium (\(Na\)) and potassium (\(K\)) are in Group 1 (more metallic, lower ionization energy compared to Group 2 elements like calcium (\(Ca\)) and magnesium (\(Mg\))). Potassium is below sodium in Group 1. As we go down Group 1, atomic size increases, and ionization energy decreases. So \(K\) has a lower first ionization energy than \(Na\).

Electronegativity is the ability of an atom to attract electrons in a chemical bond. It increases across a period and decreases down a group. Lithium (\(Li\)) is in Group 1 (left - most group among the options). Carbon (\(C\)) is in Group 14. Bromine (\(Br\)) is in Group 17. Fluorine (\(F\)) is in Group 17 (and is the most electronegative element). Since \(Li\) is the left - most and lower (compared to \(C\) in terms of group position for this trend) among the options, it has the lowest electronegativity.

Answer:

B. increase in atomic size

Question 42