Final answer:
The first ionization energy (IE₁) is the energy required to remove one electron from an atom. The arrangement in decreasing IE₁ for the given sets of elements are: K, Na, Li; Be, C, F; Cl, Na, Ar; and Cl, Br, Se.
Step-by-step explanation:
The first ionization energy (IE₁) refers to the energy required to remove one electron from an atom in its gaseous state. It is generally observed that ionization energy increases across a period and decreases down a group in the periodic table.
(a) Na, Li, K: These elements are in the same group of alkali metals. Since potassium (K) is located below sodium (Na) and lithium (Li) in the periodic table, K has the lowest ionization energy, followed by Na, and then Li. Therefore, the arrangement in decreasing IE₁ would be: K, Na, Li.
(b) Be, F, C: Beryllium (Be) has a higher ionization energy compared to carbon (C) and fluorine (F) because Be has a smaller atomic radius. Thus, the arrangement in decreasing IE₁ would be: Be, C, F.
(c) Cl, Ar, Na: Chlorine (Cl) has a higher ionization energy compared to argon (Ar) and sodium (Na) because Cl has fewer energy levels and a higher effective nuclear charge. Therefore, the arrangement in decreasing IE₁ would be: Cl, Na, Ar.
(d) Cl, Br, Se: Chlorine (Cl) has the highest ionization energy among the three elements because it requires the most energy to remove an electron. Bromine (Br) has a lower ionization energy compared to chlorine, and selenium (Se) has the lowest ionization energy of the three elements. Therefore, the arrangement in decreasing IE₁ would be: Cl, Br, Se.