Problem 124 - Chemistry A Molecular Approach

Question

The outermost valence electron in atom A experiences an effective nuclear charge of \(2+\) and is on average \(225 \mathrm{pm}\) from the nucleus. The outermost valence electron in atom B experiences an effective nuclear charge of \(1+\) and is on average \(175 \mathrm{pm}\) from the nucleus. Which atom (A or B) has the higher first ionization energy? Explain.

Step-by-Step Solution

Answer

Atom A has the higher first ionization energy, due to its valence electron experiencing a higher effective nuclear charge of +2, compared to a +1 charge for atom B, despite being further from the nucleus.

Step 1: Understand the Concepts of Effective Nuclear Charge and Ionization Energy

Effective nuclear charge (Z_eff) is the net positive charge experienced by an electron in a multi-electron atom. The ionization energy is the energy required to remove the outermost electron. Generally, the higher the effective nuclear charge experienced by an electron, the more strongly it is held by the nucleus, and the higher the ionization energy.

Step 2: Assess the Effective Nuclear Charge on the Outermost Electrons

According to the given information, the outermost electron of atom A experiences an effective nuclear charge of +2, whereas that of atom B experiences a charge of +1. This implies that the valence electron in atom A is more strongly attracted to the nucleus than the electron in atom B.

Step 3: Compare Distances from the Nuclei

The distance of the outermost electron from the nucleus affects the ionization energy; closer electrons are more strongly bound. The given distances are 225 pm for atom A and 175 pm for atom B. Despite being closer and therefore more strongly bound due to distance, the electron in B experiences a lesser charge and is expected to be easier to remove than the electron in A.

Step 4: Determine the Higher Ionization Energy

Considering the effective nuclear charge and the distance from the nucleus, atom A's electron, being subjected to a higher effective nuclear charge, will require more energy to be ionized than atom B's electron. Thus, atom A has the higher first ionization energy.

Key Concepts

Effective Nuclear Charge

The concept of effective nuclear charge, often represented as \(Z_{eff}\), is fundamental to understanding atomic structure and behavior. It describes the net positive charge an electron experiences within a multi-electron atom. This force is a result of the actual nuclear charge (the number of protons in the nucleus) minus the shielding effects caused by other electrons, particularly the inner-shell electrons.

For a valence electron, the more protracted the inner electron cloud, the less straightforward the nuclear charge's pull. In more straightforward terms, if atom A's valence electron feels \(2+\) effective nuclear charge and atom B's only \(1+\), it means that in atom A, the nucleus wields more influence over its electrons compared to atom B. Consequently, electrons in atom A are held more tightly to the nucleus, suggesting that it would take more energy to dislodge one—an aspect directly linked to the ionization energy.

Ionization Energy

Ionization energy is the energy needed to completely remove an electron from an atom or ion in its gaseous state. A higher ionization energy means that the electron is more strongly bound to the nucleus and requires more energy to be 'freed'.

The rule of thumb is that elements with a higher effective nuclear charge will have a higher ionization energy because the nucleus exerts a stronger pull on the electrons. For instance, atom A with a higher \(Z_{eff}\) than atom B will also have a higher first ionization energy. It is also affected by electron orbital type, with s-orbital electrons usually being more tightly held than p-orbital ones, and atomic radius, as electrons further out from the nucleus are easier to remove.

Atomic Radius

The atomic radius of an element is the distance from its nucleus to the outer boundary of its electrons. This distance can influence an atom's properties, including its ionization energy. Generally, as the atomic radius increases, the electrons are further from the nucleus and are less strongly attracted, leading to a decrease in ionization energy.

However, consider our example of atoms A and B, with radii of \(225 \text{pm}\) and \(175 \text{pm}\), respectively. Although atom B's valence electron is closer to its nucleus, which usually means a higher ionization energy, the stronger \(Z_{eff}\) in atom A actually predicts a higher ionization energy, showing that \(Z_{eff}\) can have a more dominant role than atomic radius in certain contexts.

Valence Electrons

Valence electrons are the electrons in the outermost shell of an atom and play a critical role in determining an atom’s chemical properties, including reactivity and bonding behavior. These electrons are the ones typically involved in chemical reactions as they can be lost, gained, or shared to form chemical bonds.

The number of valence electrons, along with effective nuclear charge, shapes the ionization energy. The more valence electrons there are, the more they can shield each other from the nucleus, often leading to a lower effective nuclear charge and, thus, a lower ionization energy. In our comparison of atom A and atom B, each has one valence electron, but due to A's higher \(Z_{eff}\), its valence electron is bound more tightly to the nucleus, thus reflecting a higher ionization energy.