Question

An ionic liquid consisting of a bulky amine cation and chloride anion has a melting point of \(41^{\circ} \mathrm{C}\). For a green chemistry solvent application, it is desirable to have a melting point below room temperature \(\left(25^{\circ} \mathrm{C}\right)\) so energy is not required to heat the compound. The melting point of the ionic liquid can be altered by replacing chloride with a different anion. The anion possibilities are: \(\mathrm{F}^{-}, \mathrm{Se}^{2-}, \mathrm{O}^{2-}\), and \(\mathrm{Br}^{-}\) (a) Write the electron configuration for each of these ions. (b) Which ions are isoelectronic? (c) Which ion is the best choice for replacing the chloride ion to make an ionic liquid with a lower melting temperature?

Short answer

(a) Configurations: \(\mathrm{F}^-: [\mathrm{He}] 2s^2 2p^6, \mathrm{Se}^{2-}: [\mathrm{Ar}]4s^2 3d^{10} 4p^6, \mathrm{O}^{2-}: [\mathrm{He}] 2s^2 2p^6, \mathrm{Br}^-: [\mathrm{Ar}]4s^2 3d^{10} 4p^6\); (b) Isoelectronic: \(\mathrm{F}^-\) and \(\mathrm{O}^{2-}\), \(\mathrm{Se}^{2-}\) and \(\mathrm{Br}^-\); (c) Best choice: \(\mathrm{Br}^-\).

Step-by-step solution

Write Electron Configurations

Begin by writing the electron configuration for each given anion. - **For** \(\mathrm{F}^-\): This is the fluoride ion. The neutral fluorine atom has the configuration \([\mathrm{He}] 2s^2 2p^5\). Adding one electron gives: \[ \mathrm{F}^-: [\mathrm{He}] 2s^2 2p^6 \]- **For** \(\mathrm{Se}^{2-}\): The neutral selenium atom has the configuration \([\mathrm{Ar}]4s^2 3d^{10} 4p^4\). Adding two electrons gives: \[ \mathrm{Se}^{2-}: [\mathrm{Ar}]4s^2 3d^{10} 4p^6 \]- **For** \(\mathrm{O}^{2-}\): The neutral oxygen atom has the configuration \([\mathrm{He}] 2s^2 2p^4\). Adding two electrons gives: \[ \mathrm{O}^{2-}: [\mathrm{He}] 2s^2 2p^6 \]- **For** \(\mathrm{Br}^-\): The neutral bromine atom has the configuration \([\mathrm{Ar}]4s^2 3d^{10} 4p^5\). Adding one electron gives: \[ \mathrm{Br}^-: [\mathrm{Ar}]4s^2 3d^{10} 4p^6 \]

Determine Isoelectronic Ions

Isoelectronic ions have the same number of electrons and thus, the same electron configuration. Compare the configurations from Step 1:Both \(\mathrm{F}^-\) and \(\mathrm{O}^{2-}\) have the electron configuration \([\mathrm{He}] 2s^2 2p^6\), making them isoelectronic.Similarly, \(\mathrm{Se}^{2-}\) and \(\mathrm{Br}^-\) both have the electron configuration \([\mathrm{Ar}]4s^2 3d^{10} 4p^6\), making them isoelectronic with each other.

Evaluate Anion for Lower Melting Point

The goal is to replace chloride with an anion that results in a lower melting point. Generally, larger ions and those with lower charge density reduce lattice energy, resulting in a lower melting point.- **Consider ion size**: \(\mathrm{Br}^-\) is larger than \(\mathrm{Cl}^-\), \(\mathrm{F}^-\) is smaller.- **Consider charge**: Higher charge usually increases lattice energy. Hence \(\mathrm{O}^{2-}\) and \(\mathrm{Se}^{2-}\) might increase melting point.Thus, \(\mathrm{Br}^-\) is more likely to replace \(\mathrm{Cl}^-\) to lower the melting point, due to its larger size and similar charge.

Key concepts

Electron Configuration

Electron configuration is the arrangement of electrons in an atom or ion's atomic orbitals. It helps in understanding the electronic structure of different elements and ions.
For the given anions:

• Fluoride ion (\(\mathrm{F}^-\)): The electron configuration starts with neutral fluorine \([\mathrm{He}] 2s^2 2p^5\) and gains one electron, resulting in \([\mathrm{He}] 2s^2 2p^6\).
• Selenide ion (\(\mathrm{Se}^{2-}\)): Selenium in its neutral state is \([\mathrm{Ar}]4s^2 3d^{10} 4p^4\). Adding two electrons gives \([\mathrm{Ar}]4s^2 3d^{10} 4p^6\).
• Oxide ion (\(\mathrm{O}^{2-}\)): Oxygen starts as \([\mathrm{He}] 2s^2 2p^4\); two electrons added gives \([\mathrm{He}] 2s^2 2p^6\).
• Bromide ion (\(\mathrm{Br}^-\)): Neutral bromine is \([\mathrm{Ar}]4s^2 3d^{10} 4p^5\), and adding one electron results in \([\mathrm{Ar}]4s^2 3d^{10} 4p^6\).

Understanding these configurations helps in predicting properties and behaviors of ions.

Isoelectronic Ions

Isoelectronic ions are ions that have the same number of electrons, thus sharing identical electron configurations. Identifying isoelectronic species help in understanding trends and similarities in chemical properties.
The exercise showed:

• \(\mathrm{F}^-\) and \(\mathrm{O}^{2-}\) both have the electron configuration \([\mathrm{He}] 2s^2 2p^6\). Thus, they are isoelectronic with each other.
• Similarly, \(\mathrm{Se}^{2-}\) and \(\mathrm{Br}^-\) share the configuration \([\mathrm{Ar}]4s^2 3d^{10} 4p^6\), making them isoelectronic.

Knowing which ions are isoelectronic can assist in predicting chemical behavior, stability, and reactions.

Lattice Energy

Lattice energy is the energy required to separate one mole of an ionic solid into gaseous ions. It indicates the strength of the bonds in an ionic compound.

• Lattice energy influences the melting point of an ionic compound. Higher lattice energy means stronger ionic bonds, which increases melting points.
• \(\mathrm{Br}^-\) is larger than \(\mathrm{Cl}^-\), potentially lowering lattice energy due to its larger ionic radius and resulting in a lower melting point.
• Higher-charged ions like \(\mathrm{O}^{2-}\) or \(\mathrm{Se}^{2-}\) increase lattice energy, thus usually increasing the melting point.

Understanding lattice energy allows chemists to predict how changes in ionic composition affect the physical properties of a compound.

Melting Point

Melting point is the temperature at which a solid becomes a liquid. In ionic compounds, it is affected by ion size and charge, which relate closely to lattice energy.

• In the exercise, the goal was to substitute \(\mathrm{Cl}^-\) with an anion that lowers the melting point.
• Bromide \(\mathrm{Br}^-\) is a promising candidate since its larger size reduces lattice energy, and thus, its melting point is likely to be lower.
• Conversely, \(\mathrm{O}^{2-}\) and \(\mathrm{Se}^{2-}\) have a higher negative charge, increasing melting point due to stronger ionic interactions.

By considering both lattice energy and ion characteristics, one can predict and manipulate melting points for practical applications like green chemistry.