Problem 35 - Chemistry and Chemical Reactivity

Question

5\. Explain each answer briefly. (a) Place the following elements in order of increasing ionization energy: \(\mathrm{F}, \mathrm{O},\) and \(\mathrm{S}\) (b) Which has the largest ionization energy: \(\mathrm{O}, \mathrm{S}\) or Se? (c) Which has the most negative electron attachment enthalpy: Se, Cl, or Br? (d) Which has the largest radius: \(\mathrm{O}^{2-}, \mathrm{F}^{-},\) or \(\mathrm{F} ?\)

Step-by-Step Solution

Answer

(a) S < O < F; (b) O; (c) Cl; (d) O^{2-}

Step 1: Ionization Energy Trend

Ionization energy generally increases across a period and decreases down a group. For the elements \( \mathrm{F}, \mathrm{O}, \text{ and } \mathrm{S} \), identify their positions on the periodic table. \( \mathrm{F} \) is in Period 2, Group 17, \( \mathrm{O} \) is Period 2, Group 16, and \( \mathrm{S} \) is Period 3, Group 16. Thus, the order by increasing ionization energy is \( \mathrm{S} < \mathrm{O} < \mathrm{F} \).

Step 2: Comparing Ionization Energies

Among \( \mathrm{O}, \mathrm{S}, \) and \( \mathrm{Se} \), \mathrm{O} has the highest ionization energy because it is higher up a group in the periodic table, where ionization energies increase as atomic number decreases.

Step 3: Electron Attachment Enthalpy Trend

Electron attachment enthalpy becomes more negative across a period and less negative down a group. Since \( \mathrm{Cl} \), \( \mathrm{Br} \), and \( \mathrm{Se} \) are being compared, \( \mathrm{Cl} \), which is in the same group but a smaller atom than \( \mathrm{Br} \) and \( \mathrm{Se} \), has the most negative electron attachment enthalpy.

Step 4: Ionic Radius Comparison

The ionic radius increases with more electrons and is inversely proportional to nuclear charge for ions with similar electron configurations. \( \mathrm{O}^{2-} \) has more electrons (and therefore larger size) than \( \mathrm{F}^{-} \) and \( \mathrm{F} \). Hence, \( \mathrm{O}^{2-} \) has the largest ionic radius.

Key Concepts

Ionization Energy

Ionization Energy is the amount of energy required to remove an electron from an atom. Understanding the trend of ionization energy across the periodic table is crucial to predict how easily an atom can lose an electron. Generally, ionization energy increases as you move from left to right across a period on the periodic table. This is because atoms have more protons (positive charge), attracting electrons more strongly and making them harder to remove.

Moving down a group, ionization energy decreases. This is because the electrons are further from the nucleus due to more electron shells, reducing the hold of the nucleus on these electrons.

So, if we compare elements like fluorine (F), oxygen (O), and sulfur (S), even though they are in different periods, applying this trend means that fluorine has the highest ionization energy (since it is higher up and further to the right), followed by oxygen, and then sulfur.

Electron Attachment Enthalpy

Electron Attachment Enthalpy is the enthalpy change when an electron is added to a neutral atom. It reflects how much an atom desires additional electrons. The more negative the electron attachment enthalpy, the more an atom tends to gain electrons.

In the periodic table, moving across a period from left to right makes the electron attachment enthalpy more negative. This means these atoms are more eager to gain electrons, usually due to higher nuclear charges which attract electrons stronger.

In contrast, moving down a group, the electron attachment enthalpy becomes less negative. This is because atomic size increases, placing added electrons further from the nucleus, reducing the atom's tendency to accept electrons.

Between chlorine (Cl), bromine (Br), and selenium (Se), chlorine has the most negative electron attachment enthalpy, meaning it is most likely to gain an electron compared to bromine and selenium.

Ionic Radius

The Ionic Radius of an atom refers to the size of its ion. It varies significantly between different ions. Generally, when an atom gains electrons (forming anions), its ionic radius increases because additional electrons repel each other, spreading out more within the atom. Conversely, when an atom loses electrons (forming cations), its ionic radius decreases as the remaining electrons are drawn closer to the nucleus.

When comparing similar electron configurations, a greater number of electrons results in a larger ionic radius. The attractive pull of the nucleus on an additional electron decreases when the number of electrons increases.

For example, between oxide ion \(O^{2-}\), fluoride ion \(F^{-}\), and a neutral fluorine atom \(F\), \(O^{2-}\) has the largest ionic radius owing to possessing more electrons than \(F^{-}\) and \(F\), which results in larger size due to electron-electron repulsion.

Periodic Table Positioning

The Periodic Table is thoughtfully arranged to showcase elements with similar chemical properties. Its position brings insight into periodic trends like ionization energy, electron attachment enthalpy, and ionic radius.

• Elements across a period have similar energy levels (shells) but increasing proton numbers, affecting trends like ionization energy and electron affinity.
• Elements down a group have additional electron shells, influencing their size and reactivity.

Understanding the periodic table position helps predict an element's properties and its chemical behavior in reactions. For instance, the position of oxygen, sulfur, and their neighboring elements helps determine these atoms' tendency to gain or lose electrons and their relative sizes compared to ions of related elements, such as fluorine.