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?

Short answer

(a) Li has a larger first ionization energy than Na due to its smaller atomic radius, as it belongs to the 2nd period on the periodic table, while Na belongs to the 3rd period. Therefore, the valence electron in Li experiences a stronger attraction to the nucleus, making it harder to remove. (b) The difference between the 3rd and 4th ionization energies of scandium (Sc) is much larger than that of titanium (Ti) because removing the 4th electron from Sc involves the deeper 2p orbital in the [Ar] core, which is more tightly bound, while both 3rd and 4th removals in Ti involve the 3d orbital. (c) Li has a much larger second ionization energy than Be as removing the second electron from Li+ disrupts the stable noble gas configuration of Helium (He), whereas Be+ doesn't have a noble gas configuration, making it relatively easier to remove the second electron.

Step-by-step solution

(a) Compare Li and Na positions on the periodic table

Both Lithium(Li) and Sodium(Na) are alkali metals. Li belongs to the 2nd period, while Na belongs to the 3rd period on the periodic table. Their positions on the periodic table are crucial in determining their ionization energies.

(a) Factors affecting Ionization Energy

The ionization energy depends on the following factors: 1. Nuclear charge: Greater the nuclear charge, higher the ionization energy 2. Atomic radius: Smaller the atomic radius, higher the ionization energy 3. Electron shielding: Electron shielding decreases the effective nuclear charge on valence electrons, making it easier to remove electrons.

(a) Ionization energies of Li and Na

Lithium has a smaller atomic radius than Sodium due to its position on the periodic table (Li is in period 2 and Na is in period 3). As a result, the valence electron in Li experiences a stronger attraction to the nucleus, making it harder to remove. Therefore, Li has a larger first ionization energy than Na.

(b) Electron configurations of Sc and Ti

Scandium (Sc) has the electron configuration of [Ar] 3d1 4s2, and Titanium (Ti) has [Ar] 3d2 4s2.

(b) 3rd and 4th Ionization energies comparison

In Sc, the first 3 electrons to be removed reside in the 4s and 3d orbitals, which have a similar energy and shielding effect. When the 4th electron is removed from Sc, it is taken from the deeper 2p orbital in the [Ar] core, which is much closer to the nucleus and more tightly bound, resulting in a significant increase in ionization energy. In the case of Ti, removing the 3rd and 4th electrons both involve the 3d orbital. Therefore, the difference between removing the third and fourth electron is smaller in Ti, as both involve the same 3d orbital. That explains why 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.

(c) Second ionization energies of Li and Be

First, let's look at the electron configurations of Lithium (Li) and Beryllium (Be): Li has the electron configuration of 1s2 2s1 Be has the electron configuration of 1s2 2s2

(c) Second ionization energies comparison

After the first ionization, Li becomes Li+ with an electronic configuration of 1s2, which is similar to the configuration of noble gas Helium (He). Removing the second electron from Li+ would disrupt the stable noble gas configuration, making it extremely difficult. In the case of Be, after the first ionization, Be+ has the configuration 1s2 2s1, which is not similar to any noble gas configuration, so, it is relatively easier to remove the second electron from Be. This is why Li has a much larger second ionization energy than Be.