Problem 42 - Chemistry: The Central Science

Question

(a) Why does Li have a larger first ionization energy than Na? (b) The difference between the third and fourth ionization energies of scandium is much larger than the difference between the third and fourth ionization energies of titanium. Why? (c) Why does Li have a much larger second ionization energy than Be?

Step-by-Step Solution

Answer

In summary, (a) Li has a larger first ionization energy than Na due to its smaller atomic radius and slightly stronger effective nuclear charge. (b) The difference between the third and fourth ionization energies of Sc is much larger compared to Ti because Sc involves electron removal from different subshells (4s to 3d), while Ti involves electron removal from the same subshell (3d). (c) Li has a much larger second ionization energy than Be because the removed electron comes from a closer, more tightly bound 1s subshell in Li, compared to the 2s subshell in Be.

Step 1: Part (a) - Comparing first ionization energies of Li and Na

The first ionization energy is the energy required to remove an electron from a neutral atom in its gaseous state. The first ionization energy of an element generally increases as the atomic radius decreases (atoms get smaller) and as the effective nuclear charge increases (the force experienced by an electron from the protons in the nucleus). Lithium has a smaller atomic radius compared to sodium, as Li is in the second period while Na is in the third period. This means that the electrons are closer to the nucleus in Li compared to Na. Additionally, the effective nuclear charge experienced by the outermost electrons in Li is slightly stronger than in Na, as there is less electron shielding in Li. Both factors lead to a higher first ionization energy for Li compared to Na.

Step 2: Part (b) - Comparing third and fourth ionization energies of Sc and Ti

The large difference between the third and fourth ionization energies of scandium can be attributed to the removal of an electron from a different subshell or energy level. For scandium, the electron configuration is [Ar] 3d1 4s2. The third ionization energy corresponds to the removal of an electron from the 4s subshell, while the fourth ionization energy corresponds to the removal of an electron from the 3d subshell. Since the 3d subshell is closer to the nucleus and is more tightly bound, it takes significantly more energy to remove the electron from that subshell, leading to a large increase in the ionization energy from the third to the fourth ionization energy in scandium. In contrast, the electron configuration of titanium is [Ar] 3d2 4s2. The third ionization energy corresponds to the removal of an electron from the 4s subshell, and the fourth ionization energy corresponds to the removal of another electron from the 3d subshell. Since both electrons are removed from the same subshell, the difference in energy between the third and fourth ionization energies in titanium is smaller.

Step 3: Part (c) - Comparing second ionization energies of Li and Be

The second ionization energy of an element generally increases as the atomic radius decreases and the effective nuclear charge increases, just like the first ionization energy. In lithium, after the removal of the first electron, the electron configuration becomes [He] 1s², whereas in beryllium, the electron configuration is [He] 2s². Be has a larger atomic radius, and its outer electrons experience less effective nuclear charge than the electrons in Li. Additionally, the removed electron in Li comes from the 1s subshell, which is closer to the nucleus compared to the 2s subshell in Be. Since electrons in lower energy shells are more tightly bound to the nucleus, it takes considerably more energy to remove an electron from the 1s subshell in Li, leading to a much larger second ionization energy for Li compared to Be.