Question

Select the element from each pair expected to have the lower first ionization energy. Explain. (a) \(\mathrm{N}\) and \(\mathrm{F}\) (b) \(\mathrm{Mg}\) and \(\mathrm{Ba}\) (c) \(\mathrm{N}\) and \(\mathrm{Ca}\)

Short answer

(a) Nitrogen (N) will have a lower ionization energy than Fluorine (F) because ionization energy increases from left to right across a period. (b) Barium (Ba) will have a lower ionization energy than Magnesium (Mg) because ionization energy decreases down a group. (c) Calcium (Ca) is expected to have a lower ionization energy than Nitrogen (N) since it is further to the left and lower in the periodic table.

Step-by-step solution

(a) N and F

: Nitrogen (N) is in the 2nd period and 15th Group, while Fluorine (F) is in the 2nd period and 17th Group. We are given two elements in the same period, so we should look into their positions in the groups. Since the ionization energy increases from left to right, we can conclude that Nitrogen (N) will have a lower ionization energy than Fluorine (F).

(b) Mg and Ba

: Magnesium (Mg) is in the 3rd period and 2nd Group, while Barium (Ba) is in the 6th period and 2nd Group. We are given two elements in the same group, so we should look into their positions in the periods. Since the ionization energy decreases going down a group, Barium (Ba) will have a lower ionization energy than Magnesium (Mg).

(c) N and Ca

: Nitrogen (N) is in the 2nd period and 15th Group, while Calcium (Ca) is in the 4th period and 2nd Group. Since nitrogen is further to the right and higher up in the periodic table, ionization energy should be generally higher for it than for calcium. Thus, Calcium (Ca) is expected to have a lower ionization energy than Nitrogen (N).

Key concepts

Understanding Periodic Trends

When exploring the periodic table, understanding the trends in first ionization energy is crucial for making accurate predictions about an element's chemical behavior. Ionization energy is the energy required to remove an electron from a gaseous atom or ion. As a broad trend, first ionization energy increases across a period from left to right and decreases down a group.

Why is this the case? As you move from left to right across a period, electrons are being added to the same principal energy level while the nucleus gains more protons. This results in a stronger attraction between the positively charged nucleus and the negatively charged electrons, which, in turn, requires more energy to remove an electron. Conversely, moving down a group, the additional electron shells cause electrons to be further from the nucleus and to experience more shielding from the inner electrons. This makes the outer electrons easier to remove, thus lowering the ionization energy.

This trend has important implications: it can indicate the reactivity of elements, particularly the metals which tend to lose electrons, and non-metals that tend to gain electrons. Understanding these trends, students can better predict element reactivity and bonding capabilities.

Group and Period Comparison in Ionization Energy

Comparing elements within the same group or period requires an understanding of how atomic structure impacts ionization energy. For elements in the same period, as we move to the right, electrons are added to orbitals that are similarly shielded from the atomic nucleus, which means that effective nuclear charge increases and it is harder to remove an electron, resulting in higher ionization energies.

When evaluating elements within the same group, we consider the element's period. The vertical movement in a group involves increasing the principal quantum number; that is, each successive element has an additional electron shell. This extra distance, combined with the increased electron shielding, reduces the effective nuclear charge felt by the outer electrons and, in turn, lowers the ionization energy.

For instance, when comparing Magnesium (Mg) and Barium (Ba), both from Group 2 but in different periods, Ba has more filled electron shells, leading to a larger atomic radius and more shielded outer electrons. Therefore, Ba requires less energy to ionize than Mg.

Atomic Structure and its Relation to Ionization Energy

At the heart of these periodic trends lies atomic structure. An atom's ionization energy is influenced by its atomic radius, nuclear charge, and the electron configuration. Let's dive into each factor:

• Atomic Radius: The distance between the nucleus and the valence electrons. Larger atoms have lower ionization energies because the valence electrons are further from the nucleus and feel a weaker attraction.
• Nuclear Charge: The total positive charge of the nucleus. Higher nuclear charge attracts electrons more strongly, thus increasing ionization energy.
  
• Electron Configuration: Atoms with full or half-full orbitals have a more stable configuration, resulting in a higher ionization energy necessary to disturb that stability.

Understanding these details is essential, as they explain not only the overarching trends but also the exceptions. For example, the first ionization energy of Nitrogen is lower than that of Fluorine, despite being in the same period, due to its half-filled p-orbital which provides additional electron stability. Similarly, Calcium's ionization energy is lower compared to Nitrogen even though it is to the left in the periodic table, because the added electron shells and radius size significantly decrease the nuclear attraction on the valence electron.