Question

For each of the following pairs, indicate which species is smaller. Explain your reasoning in terms of the electron structure of each species. a. H or \(\mathrm{H}^{-}\) b. \(\mathrm{N}\) or \(\mathrm{N}^{3-}\) c. Al or \(\mathrm{Al}^{3+}\) d. F or Cl

Short answer

In summary: a. H^- is larger than H because of the additional electron, electron shielding, and repulsion. b. N^3- is larger than N due to the increased electron-electron repulsion and shielding from the extra electrons. c. Al^3+ is smaller than Al since the removal of three electrons leads to increased effective nuclear charge and less shielding. d. Cl is larger than F as the additional energy level and shielding from extra electrons outweigh the increase in protons.

Step-by-step solution

a. Compare H and H- by analyzing the electron structure:

In this comparison, we have Hydrogen (H) with an electron configuration of 1s^1, while Hydrogen anion (H^-) has an electron configuration of 1s^2. Since Hydrogen anion has an extra electron, the electron shielding and repulsion in the H^- species make it larger than H.

Answer for a.

H^- is larger than H.

b. Compare N and N^3- by analyzing the electron structure:

Nitrogen (N) has an electron configuration of 1s^2 2s^2 2p^3, and Nitrogen anion (N^3-) has 1s^2 2s^2 2p^6. Adding three more electrons to Nitrogen will increase electron-electron repulsion and shielding in the N^3- species, making it larger than N.

Answer for b.

N^3- is larger than N.

c. Compare Al and Al^3+ by analyzing the electron structure:

Aluminum (Al) has an electron configuration of 1s^2 2s^2 2p^6 3s^2 3p^1, while Aluminum cation (Al^3+) has 1s^2 2s^2 2p^6. By removing three electrons from Aluminum, we have less electron-electron repulsion and shielding in the Al^3+ species. The effective nuclear charge increases, and the remaining electrons are drawn closer to the nucleus, making the Al^3+ species smaller than Al.

Answer for c.

Al^3+ is smaller than Al.

d. Compare F and Cl by analyzing the electron structure:

Fluorine (F) has an electron configuration of 1s^2 2s^2 2p^5, while Chlorine (Cl) has 1s^2 2s^2 2p^6 3s^2 3p^5. Chlorine has an additional energy level compared to Fluorine, thus having a larger electron cloud. This contributes to the size of Cl being larger than the size of F, even though the effective nuclear charge of Chlorine is greater than that of Fluorine. Increased shielding effect in Chlorine due to the presence of an additional quantum level outweighs the increase in protons, making Cl larger than F.

Answer for d.

Cl is larger than F.

Key concepts

Electron Configuration

When comparing the size of atomic species like ions and atoms, it's crucial to understand their electron configuration. The electron configuration of an atom describes the distribution of electrons among the different orbitals in an atom. Electrons fill these orbitals according to the Aufbau principle, which states that electrons occupy the lowest energy orbital available. For example, Hydrogen (H) has a simple configuration of 1s¹, with a single electron in the 1s orbital. On the other hand, a Hydrogen anion (H^-) adds an electron, resulting in a configuration of 1s². This additional electron increases repulsion between electrons, thereby slightly expanding the electron cloud and resulting in a larger atomic size for H^- compared to H.

Similarly, as we see in the exercise, Nitrogen (N) has the configuration of 1s² 2s² 2p³, while its anion N^3- achieves a full octet with the configuration 1s² 2s² 2p⁶. The additional electrons increase electron-electron repulsion and reduce the overall pull that the nucleus has on each electron, resulting in a larger atomic radius for the anion.

Effective Nuclear Charge

The effective nuclear charge (Z_eff) is the net positive charge experienced by an electron in a multi-electron atom. The more protons in the nucleus, the higher the nuclear charge; however, not all electrons feel this full charge due to shielding by other electrons. Consider Aluminium (Al) with a neutral electron configuration of 1s² 2s² 2p⁶ 3s² 3p¹. When it becomes a cation (Al³⁺), it loses three electrons, resulting in the configuration 1s² 2s² 2p⁶. The removal of these electrons increases the Z_eff on remaining electrons as there is less electron shielding. This stronger nuclear pull makes the Al³⁺ ion smaller than its neutral atom.

The comparison between Fluorine (F) and Chlorine (Cl) also hinges on Z_eff. Despite Chlorine having more protons and consequently a higher nuclear charge, the increase in shielding from the extra energy level in Cl compared to F means that the outer electrons are less firmly held by the nucleus, resulting in a larger atomic radius for Chlorine.

Electron-Electron Repulsion

Electron-electron repulsion within an atom can significantly affect atomic size. As electrons are added to an atom, they repel each other due to their negative charge. This repulsion can cause the electrons to occupy more space, leading to an increase in atomic size. In the cases of anions such as H^- and N^3-, the additional electrons increase repulsion, which moves the electrons slightly further apart, contributing to a larger atomic or ionic radius.

In contrast, when electrons are removed as is the case with cations like Al³⁺, electron repulsion decreases. This allows the electron cloud to contract as the effective nuclear charge on the remaining electrons is greater, leading to a smaller atomic or ionic radius. Understanding electron-electron repulsion is pivotal when predicting changes in atomic size due to ion formation. In essence, more electrons mean increased repulsion and a larger size, while fewer electrons mean less repulsion and a smaller size.