Question

Prove mathematically that it is more energetically favorable for a fluorine atom to take an electron from a sodium atom than for a fluorine atom to take an electron from another fluorine atom.

Short answer

The ionization energy for sodium is \(I_1(Na) = 495.8\, kJ/mol\), and the second ionization energy for fluorine is \(I_2(F) = 3,401\, kJ/mol\). Comparing these values, we find that the energy difference \(\Delta E = 2,905.2\, kJ/mol\), which is positive and significant. Therefore, it is more energetically favorable for a fluorine atom to take an electron from a sodium atom than from another fluorine atom.

Step-by-step solution

Understand the terms of the problem

In this exercise, we are dealing with ionization energy. Ionization energy is the amount of energy required to remove an electron from an atom or ion. In this case, we will compare the ionization energy of sodium and the second ionization energy of fluorine.

Ionization energy for sodium

For sodium (Na), it has one electron in its outermost shell. When sodium loses this outermost electron, it forms a sodium ion (Na⁺). The amount of energy required to remove this electron from sodium atom is called its ionization energy, which is denoted as \(I_1(Na)\). The value of ionization energy for sodium is approximately \(495.8\, kJ/mol\).

Ionization energy for fluorine

For fluorine (F), it has seven electrons in its outermost shell. When fluorine gains one electron to form a stable fluoride ion (F⁻), a certain amount of energy is released, which is known as its electron affinity. However, we are asked to consider a situation where fluorine atom loses an electron to form a positive ion (F⁺). This process will require additional energy to overcome the electron-electron repulsion, and this energy is called the second ionization energy of fluorine, denoted as \(I_2(F)\). The value of second ionization energy for fluorine is approximately \(3,401\, kJ/mol\).

Compare the ionization energies

Now that we have the ionization energy for sodium and the second ionization energy for fluorine, we can compare these values to determine which is more energetically favorable. We know that \(I_1(Na) = 495.8\, kJ/mol\) and \(I_2(F) = 3,401\, kJ/mol\). The difference in energy can be represented by the following expression: \(\Delta E = I_2(F) - I_1(Na)\) By comparing these values, we can show that taking an electron from sodium is energetically more favorable than taking an electron from another fluorine atom.

Calculate the energy difference and draw conclusion

Now let's calculate the energy difference: \(\Delta E = 3,401\, kJ/mol - 495.8\, kJ/mol = 2,905.2\, kJ/mol\) Since \(\Delta E > 0\), we can conclude that taking an electron from a sodium atom requires significantly less energy (\(495.8\, kJ/mol\)) as compared to taking an electron from another fluorine atom (\(3,401\, kJ/mol\)). Thus, it is more energetically favorable for a fluorine atom to take an electron from a sodium atom than for a fluorine atom to take an electron from another fluorine atom.

Key concepts

Electron Affinity

Electron affinity refers to the amount of energy released when an atom in the gas phase acquires an electron to form a negative ion. This property is particularly significant when considering the formation of ions in reactions. For instance, halogens like fluorine have a high electron affinity since they require only one more electron to complete their valence shell.

When a fluorine atom gains an electron to become a fluoride ion (F⁻), energy is liberated. This energy release occurs because the electron adds to an already negatively charged area, creating a more stable electron configuration. To envision this, think of a magnet: bringing the opposite poles together releases energy, just as adding an electron to an atom with a high electron affinity does.

Energy Difference Calculation

The energy difference calculation involves comparing the energy required to remove or add an electron. It is a straightforward subtraction that yields the energy advantage or disadvantage of a specific electron transfer. In chemical reactions, this calculation is crucial for predicting which atom is more likely to gain or lose an electron.

In our exercise example, the energy difference (\(\Delta E\))) between the ionization energy of sodium (\(I_1(Na)\)) and the second ionization energy of fluorine (\(I_2(F)\)) revealed which process required less energy. Such calculations are pivotal in understanding chemical reactivity and stability of the resulting ions.

Electron-Electron Repulsion

Electron-electron repulsion is a fundamental principle in chemistry that explains the force exerted between electrons due to their like charges. This repulsion affects how electrons are arranged in an atom and plays a key role in processes such as ionization.

Considering our exercise, the ionization energy for fluorine to lose an electron (thus forming F⁺) is notably high, because removing an electron from a crowded electron shell must overcome significant electron-electron repulsion. Envisioning a crowded room, asking one person to leave requires navigating through the packed space, just as an electron must overcome the repulsion of the remaining electrons.

Formation of Ions

The formation of ions involves the addition or removal of electrons from an atom, resulting in a charged entity. Ions are crucial for the conception of salts and ionic compounds, as seen in the reaction of sodium with fluorine to form sodium fluoride (NaF).

Understanding the formation of ions ties together concepts about ionization energy and electron affinity. Elements with low ionization energies, such as sodium, tend to lose electrons and form positive ions easily, while elements with high electron affinities, like fluorine, readily gain electrons to form negative ions. When these processes happen with less energy consumption, the formation of ions is considered energetically favorable—as highlighted in our comparison between sodium and fluorine.