Question

For each of the following pairs, indicate which species is larger. Explain your reasoning in terms of the electron structure of each species. a. \(\mathrm{Mg}^{2+}\) or \(\mathrm{Mg}\) b. \(\mathrm{Ca}^{2+}\) or \(\mathrm{K}^{+}\) c. \(\mathrm{Rb}^{+}\) or \(\mathrm{Br}^{-}\) d. \(\mathrm{Se}^{2-}\) or \(\mathrm{Se}\)

Short answer

a. \(\mathrm{Mg}\) is larger than \(\mathrm{Mg}^{2+}\) because after losing two electrons, the effective nuclear charge increases and electron-electron repulsion decreases, causing \(\mathrm{Mg}^{2+}\) to be smaller. b. \(\mathrm{K}^{+}\) is larger than \(\mathrm{Ca}^{2+}\) since they have the same electron configuration, but \(\mathrm{Ca}^{2+}\) has a higher effective nuclear charge, resulting in a stronger attraction between electrons and nucleus. c. \(\mathrm{Br}^{-}\) is larger than \(\mathrm{Rb}^{+}\) due to \(\mathrm{Rb}^{+}\) having a higher effective nuclear charge, which causes the electrons to be more strongly attracted to the nucleus. d. \(\mathrm{Se}^{2-}\) is larger than \(\mathrm{Se}\) because the gain of two electrons increases electron-electron repulsion, making \(\mathrm{Se}^{2-}\) larger.

Step-by-step solution

Identify electron configurations for each species

Firstly, we need to determine the electron configurations of each species in order to understand their electron structures and valence electrons. a. \(\mathrm{Mg}^{2+}\) has the electron configuration: \([Ne]\) \(\mathrm{Mg}\) has the electron configuration: \([Ne] 3s^2\) b. \(\mathrm{Ca}^{2+}\) has the electron configuration: \([Ar]\) \(\mathrm{K}^{+}\) has the electron configuration: \([Ar]\) c. \(\mathrm{Rb}^{+}\) has the electron configuration: \([Kr]\) \(\mathrm{Br}^{-}\) has the electron configuration: \([Kr] 4d^{10} 5s^2 5p^6\) d. \(\mathrm{Se}^{2-}\) has the electron configuration: \([Kr] 4d^{10} 5s^2 5p^6\) \(\mathrm{Se}\) has the electron configuration: \([Ar] 3d^{10} 4s^2 4p^4\)

Analyze the shielding effect, electron repulsion, and effective nuclear charge

In this step, we will use the electron configurations obtained earlier to analyze how each species is affected by shielding effects, electron repulsions, and effective nuclear charges: a. \(\mathrm{Mg}^{2+}\) has lost two electrons from its valence shell, resulting in an increase in effective nuclear charge and decrease in electron-electron repulsion. This causes the remaining electrons to be drawn closer to the nucleus, making the ion smaller than the neutral atom \(\mathrm{Mg}\). b. In the case of \(\mathrm{Ca}^{2+}\) and \(\mathrm{K}^{+}\), both species have the same electron configuration. However, \(\mathrm{Ca}^{2+}\) has a higher effective nuclear charge (due to having more protons) than \(\mathrm{K}^{+}\), resulting in stronger attraction between electrons and the nucleus for \(\mathrm{Ca}^{2+}\). Therefore, \(\mathrm{Ca}^{2+}\) will be smaller than \(\mathrm{K}^{+}\). c. For \(\mathrm{Rb}^{+}\) and \(\mathrm{Br}^{-}\), although both have a similar number of electrons, \(\mathrm{Rb}^{+}\) has a higher effective nuclear charge, which causes the electrons to be more strongly attracted to the nucleus. Therefore, \(\mathrm{Rb}^{+}\) will be smaller than \(\mathrm{Br}^{-}\). d. Just like in the first example, \(\mathrm{Se}^{2-}\) has gained two extra electrons, increasing electron-electron repulsion. This causes the electrons to be repelled further from the nucleus, making \(\mathrm{Se}^{2-}\) larger than the neutral atom \(\mathrm{Se}\).

Determine the larger species in each pair

Based on the analysis in Step 2, we can now determine the larger species in each pair: a. \(\mathrm{Mg}\) is larger than \(\mathrm{Mg}^{2+}\). b. \(\mathrm{K}^{+}\) is larger than \(\mathrm{Ca}^{2+}\). c. \(\mathrm{Br}^{-}\) is larger than \(\mathrm{Rb}^{+}\). d. \(\mathrm{Se}^{2-}\) is larger than \(\mathrm{Se}\).

Key concepts

Effective Nuclear Charge

When talking about atoms and ions, the term 'effective nuclear charge' is key to understanding how electrons are held in place. This concept refers to the net positive charge experienced by an electron in an atom. Although the nucleus contains many protons, which attract electrons, these effects are partially offset by the electrons themselves, specifically the inner ones that shield their outer counterparts.
To calculate the effective nuclear charge (Z_{eff}), we often use the formula: Z_{eff} = Z - S, where \( Z \) is the atomic number (total number of protons), and \( S \) is the number of shielding electrons.

• For ions, losing electrons usually increases the effective nuclear charge because there are fewer electrons to shield the nucleus' pull.
• This increased charge means the remaining electrons are held more tightly, often resulting in a smaller ion.

Shielding Effect

The shielding effect describes how inner electrons hinder the pull of the nucleus on outer electrons. Simply put, the more inner electrons there are, the more interference there is with the nucleus. This makes it easier for outer electrons to stay further away.

• In atoms with more electron shells, outer electrons experience significant shielding, resulting in reduced attraction to the nucleus.
• This reduced attraction allows the outer electrons to be held less tightly, affecting both the size and reactivity of an atom or ion.

What's interesting is that in species gaining electrons (like Se^{2-} ), the increase in electron count enhances the shielding effect. This increase causes those new electrons to experience less pull from the nucleus, leading to the atom expanding in size.

Electron Repulsion

Particles with the same charge repel each other, and this concept is vital in understanding electron configuration. Electrons, as negatively charged particles, naturally repel one another. When atoms gain additional electrons, as in the case of anions, the electron cloud becomes more crowded.

• This repulsion between electrons causes the atom or ion to increase in size because the electrons push each other apart.
• In Se^{2-} , for example, adding more electrons increases the repulsive forces among them, causing the ion to be larger than its neutral form, Se .

Ion Size Comparison

When comparing the sizes of ions, we must consider several factors: effective nuclear charge, shielding, and electron repulsion. These contribute to how compact or expanded an ion may be.

• Cations like Mg^{2+} and Ca^{2+} lose electrons, which increases the effective nuclear charge per electron and generally results in a smaller ionic size compared to their neutral atoms.
• Anions such as Br^{-} and Se^{2-} gain electrons, experiencing increased electron repulsion and shielding. This often leads to a larger ion compared to the neutral atom due to the extra electrons demanding more space.

Understanding these factors can help explain why some ions are larger or smaller than others. For example, although Ca^{2+} and K^{+} have the same electron configuration, Ca^{2+} is smaller due to a higher effective nuclear charge minimizing the ionic radius.