Question

(a) Why does Li have a larger first ionization energy than \(\mathrm{Na}\) ? (b) The difference between the third and fourth ionization energies of scandium is much larger than that 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 because it is higher in the periodic table, leading to a stronger effective nuclear charge and smaller atomic size, making it harder to remove an electron. (b) The difference between the third and fourth ionization energies of Sc is much larger than that of Ti as the fourth ionization in Sc involves removing a core electron from a tightly bound 2p orbital, while in Ti, all four ionization energies involve removing electrons from the loosely held 4s and 3d orbitals. (c) Li has a much larger second ionization energy than Be because Li has to remove an electron from the stable, filled 1s orbital, while Be only has to remove an electron from the 2s orbital, which is comparatively easier.

Step-by-step solution

a) Why Li has a larger first ionization energy than Na

It is important to note that ionization energy increases as we go up a group and to the right across a period in the periodic table. This is because of the increasing effective nuclear charge the electrons experience due to the addition of protons in the nucleus and the decreasing atomic size. In the case of Li and Na, they are in the same group (alkali metals), with Li above Na. Therefore, Li experiences a stronger effective nuclear charge and has a smaller atomic size, making it more difficult to remove its outermost electron. Consequently, this leads to Li having a larger first ionization energy than Na.

b) Difference between the third and fourth ionization energies of Sc and Ti

First, we need to understand the electronic configurations of scandium (Sc) and titanium (Ti). Sc has the electron configuration of [Ar] 3d1 4s2, while Ti has the configuration of [Ar] 3d2 4s2. For Sc, the first three ionization energies involve removing electrons from the 4s and 3d orbitals, which are relatively loosely held. However, the fourth ionization energy requires removing a core electron from a more tightly bound 2p orbital, resulting in a larger increase in ionization energy. On the other hand, for Ti, the first four ionization energies involve removing electrons from the 4s and 3d orbitals, and since these orbitals are all relatively loosely held and similar in energy, the difference between the third and fourth ionization energies is not as significant as in Sc. This difference in electron configurations leads to a much larger difference between the third and fourth ionization energies of Sc compared to Ti.

c) Why Li has a much larger second ionization energy than Be

To understand this difference, we need to consider the electronic configurations of lithium (Li) and beryllium (Be). Li has the electron configuration of 1s2, 2s1, while Be has the electron configuration of 1s2, 2s2. When Li loses an electron, it becomes Li+, and the configuration becomes 1s2. However, to lose another electron (for a second ionization event), Li has to remove a core electron from the stable, filled 1s orbital, which is much more difficult. In the case of Be, the first ionization removes an electron from the 2s orbital, resulting in Be+ with a configuration of 1s2, 2s1. Performing the second ionization results in Be2+ and the configuration 1s2. Since the ionization process requires removing electrons from the same energy level in both cases (2s), the second ionization of Be is not as difficult as the second ionization of Li. So, Li has a much larger second ionization energy as it has to remove an electron from the stable, filled 1s orbital, whereas Be only has to remove an electron from the 2s orbital, which is comparatively easier.

Key concepts

Effective Nuclear Charge

The effective nuclear charge is a concept that explains the attractive force experienced by an electron in an atom. It is crucial in understanding why certain elements have higher ionization energies than others. Essentially, while the total number of protons determines the actual nuclear charge, not all of these protons' influence reaches the outer electrons effectively due to the presence of inner shell electrons.

These inner shell electrons act as a shield, reducing the positive charge that any outer electron "feels". Thus, the effective nuclear charge is the net positive charge experienced by an electron and is calculated by subtracting the shielding or screening effect of the core electrons from the total nuclear charge.

For elements like lithium (Li) compared to sodium (Na), Li is higher up the periodic table in the same group. Therefore, it has less shielding since it has fewer internal electrons compared to Na. This results in a higher effective nuclear charge for Li, making it more difficult to remove an electron, and hence, its ionization energy is larger.

Periodic Table Trends

The periodic table's arrangement allows us to recognize trends in elemental properties, one of which is ionization energy. As you move across a period from left to right, ionization energies typically increase. This is because the effective nuclear charge increases as more protons are added to the nucleus without adding a new energy level for electrons, leading to a stronger attraction to the nucleus.

Additionally, as you move up a group in the periodic table, ionization energies also increase. This is because the atomic size decreases, and electrons are closer to the nucleus, thus held more tightly.

These trends help explain why lithium (Li), which is above sodium (Na) in the same group, has a larger ionization energy. Li has a smaller atomic size and greater attraction between its electrons and the nucleus, compared to Na.

Electron Configuration

Electron configuration is the arrangement of electrons in an atom, which provides insight into chemical behavior and characteristics, including ionization energy. The electron configuration of an atom is determined by the order in which electron energy levels or orbitals are filled.

For lithium (Li), the electron configuration is 1s² 2s¹. Upon losing one electron (first ionization), it becomes Li⁺ with a configuration of 1s². In contrast, the electron configuration of sodium (Na) is 1s² 2s² 2p⁶ 3s¹. Comparing Li and Na explains their differing ionization energies.

The trend towards higher ionization energy can further be visualized by examining Scandium (Sc) and Titanium (Ti). Sc has an electron configuration of [Ar] 3d¹ 4s², whereas Ti has [Ar] 3d² 4s². Removing core electrons requires much more energy, hence the abrupt increase in Sc's ionization energy after the removal of the loosely bound electrons.

Core vs Valence Electrons

The differentiation between core and valence electrons is fundamental in understanding various atomic behaviors, such as ionization energy differences between elements. Valence electrons are the outermost electrons of an atom and are involved in chemical bonding and ionization processes. Core electrons are the inner electrons that shield the valence electrons from the full power of the nucleus' positive charge.

In elements like lithium (Li), after the first ionization, the electron left is in the core, making the second ionization significantly more challenging and energy-intensive. Therefore, the second ionization of Li involves removing a core electron, leading to a much larger ionization energy.

In contrast, beryllium (Be) has two valence electrons in the 2s orbital, allowing both to be removed without reaching the core, making its subsequent ionization energies less dramatically spaced. Understanding the roles of core and valence electrons helps to rationalize the stepwise nature and gradients in ionization energies observed in different elements.