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. \(L i^{+}\) or \(F^{-}\) b. \(\mathrm{Na}^{+}\) or \(\mathrm{Cl}^{-}\) c. \(\mathrm{Ca}^{2+}\) or \(\mathrm{Ca}\) d. \(\mathrm{Cs}^{+}\) or \(\mathrm{I}^{-}\)

Short answer

a. \(Li^+\) is smaller due to its higher effective nuclear charge (\(Z_\text{eff}\)). b. \(Na^+\) is smaller due to its higher effective nuclear charge (\(Z_\text{eff}\)). c. \(Ca^{2+}\) is smaller due to its higher effective nuclear charge (\(Z_\text{eff}\)). d. \(I^-\) is smaller due to its higher effective nuclear charge (\(Z_\text{eff}\)).

Step-by-step solution

a. Comparing \(Li^+\) and \(F^-\)

The electron configurations for \(Li^+\) and \(F^-\) are as follows: \(Li^+: \ 1s^2\) \(F^-: \ 1s^2 2s^2 2p^6\) Both \(Li^+\) and \(F^-\) are isoelectronic, meaning they have the same number of electrons. However, the effective nuclear charge (\(Z_\text{eff}\)) for \(Li^+\) (with \(Z=3\)) is higher than that for \(F^-\) (with \(Z=9\)). Therefore, the electrons in \(Li^+\) will be drawn closer to the nucleus and \(Li^+\) will be smaller than \(F^-\).

b. Comparing \(Na^+\) and \(Cl^-\)

The electron configurations for \(Na^+\) and \(Cl^-\) are as follows: \(Na^+: \ 1s^2 2s^2 2p^6\) \(Cl^-: \ 1s^2 2s^2 2p^6\) Both \(Na^+\) and \(Cl^-\) are isoelectronic with the same electron configuration. However, the effective nuclear charge (\(Z_\text{eff}\)) for \(Na^+\) (with \(Z=11\)) is higher than that for \(Cl^-\) (with \(Z=17\)). Therefore, the electrons in \(Na^+\) will be drawn closer to the nucleus and \(Na^+\) will be smaller than \(Cl^-\).

c. Comparing \(Ca^{2+}\) and \(Ca\)

The electron configurations for \(Ca^{2+}\) and \(Ca\) are as follows: \(Ca^{2+}: \ 1s^2 2s^2 2p^6 3s^2 3p^6\) \(Ca: \ 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2\) As \(Ca^{2+}\) has lost two electrons compared to the neutral \(Ca\) atom, the effective nuclear charge (\(Z_\text{eff}\)) is higher in \(Ca^{2+}\). The increased \(Z_\text{eff}\) pulls the electrons closer to the nucleus, making \(Ca^{2+}\) smaller than \(Ca\).

d. Comparing \(Cs^+\) and \(I^-\)

The electron configurations for \(Cs^+\) and \(I^-\) are as follows: \(Cs^+: \ 1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^2 4p^6 4d^{10} 5s^2 5p^6\) \(I^-: \ 1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^2 4p^6 4d^{10} 5s^2 5p^6\) Both \(Cs^+\) and \(I^-\) are isoelectronic with the same electron configuration. However, the effective nuclear charge (\(Z_\text{eff}\)) for \(Cs^+\) (with \(Z=55\)) is lower than that for \(I^-\) (with \(Z=53\)). The lower \(Z_\text{eff}\) in \(Cs^+\) allows the electrons to be further from the nucleus, making \(Cs^+\) larger than \(I^-\) in size.

Key concepts

Isoelectronic Species

When we talk about isoelectronic species, we are referring to atoms or ions that share the same number of electrons, or have identical electron configurations. Despite having the same number of electrons, isoelectronic species can have different atomic sizes. This is because their nuclei have different numbers of protons, which affects the nuclear charge.

Consider two ions, such as \(Li^+\) and \(F^-\). Both ions have the electron configuration \(1s^2\), yet \(Li^+\) is smaller than \(F^-\). This is due to the difference in their effective nuclear charge. Since \(Li^+\) has fewer protons (3) than \(F^-\) (9), the electrons in \(Li^+\) are more strongly attracted to the nucleus, resulting in a smaller radii.

When comparing isoelectronic species, keep in mind:

• Same number of electrons does not mean same size.
• The species with more protons will likely be smaller.
• Effective nuclear charge plays a major role.

This understanding helps students see why species with the same number of electrons can be different in size.

Electron Configuration

Electron configuration refers to the distribution of electrons in an atom or ion. It's a way to map out where each electron resides, typically listed in order of increasing energy levels. This information gives insight into which electrons are involved in bonding and reactivity.

For example, consider \(Ca^{2+}\) and neutral \(Ca\). A neutral calcium atom has the configuration \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2\). When it loses two electrons to form \(Ca^{2+}\), the new configuration becomes \(1s^2 2s^2 2p^6 3s^2 3p^6\). Losing these electrons, especially from an outer shell, affects the size of the ion.

Key aspects of electron configuration include:

• Determining valence electrons, which determine chemical properties.
• Predicting ionic charges based on electron loss or gain.
• Understanding trends across the periodic table, such as ionization energy and atomic size trends.

The electron configuration serves as a foundation for predicting how atoms will behave and interact with each other.

Atomic Size

Atomic size or atomic radii is the distance from the nucleus of an atom to the outer boundary of the surrounding cloud of electrons. It gives a visual as to how 'big' an atom or ion appears in space. This property depends on several factors, such as the number of electron shells and the effective nuclear charge.

In terms of electron shells, more shells generally mean a larger atomic size. However, the effective nuclear charge can counteract this. For example, although \(Na^+\) and \(Cl^-\) are isoelectronic, the stronger pull from a higher effective nuclear charge in \(Na^+\) makes it smaller than \(Cl^-\).

Some notable points about atomic size include:

• Atomic size decreases across a period due to increased nuclear charge without a significant addition of electron shells.
• Atomic size generally increases down a group as more electron shells are added.

Understanding atomic size is crucial as it influences properties like ionization energy and how atoms pack together in solids.