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
b increase in atomic size
oc fewer electrons in the highest occupied energy level
d increase in size of the nucleus

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

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

Explanation:

• Question 41: Ionization energy is the energy required to remove an electron from an atom. As atomic size increases (down a group), the outermost electrons are farther from the nucleus. The electrostatic attraction between the nucleus and these electrons decreases, making it easier to remove an electron (lower ionization energy). An increase in protons (a) would actually increase the attraction (if electron - shell distance is constant). For a group, the number of electrons in the highest - occupied energy level is the same (c is wrong). The size of the nucleus (d) is not the direct factor affecting ionization energy in the context of group trends; it's the electron - nucleus distance that matters more.
• Question 42: Ionization energy decreases down a group and increases across a period. Potassium (\(K\)) is in period 4, while sodium (\(Na\)) is in period 3, magnesium (\(Mg\)) is in period 3 (group 2, more protons than \(Na\) in the same period), and calcium (\(Ca\)) is in period 4 (group 2). Since \(K\) is in group 1 (alkali metals) and period 4, it has a larger atomic size compared to the other elements (except \(Ca\) in the same period, but \(K\) has fewer protons than \(Ca\) in period 4). Larger atomic size (due to more electron shells) leads to lower ionization energy.
• Question 43: Electronegativity is the ability of an atom to attract electrons in a chemical bond. It increases across a period and decreases down a group. Fluorine (\(F\)) is the most electronegative element (period 2, group 17). Carbon (\(C\)) is in period 2, group 14. Bromine (\(Br\)) is in period 4, group 17. Lithium (\(Li\)) is in period 2, group 1. Among these, \(Li\) has the lowest electronegativity because it is in group 1 (metals have lower electronegativity compared to non - metals in the same period, and it is less electronegative than \(Br\) which is in a lower group but higher period effect is less dominant here as the group trend (metals vs non - metals) is more significant for \(Li\) vs \(Br\) comparison).

Answer:

• Question 41: B. increase in atomic size
• Question 42: B. potassium
• Question 43: D. lithium