Problem 121 - Chemistry The Molecular Science

Question

Use electron configurations to explain why (a) the electron affinity of selenium is lower than that of bromine. (b) the first ionization energy of aluminum is lower than that of magnesium. (c) sulfur has a lower first ionization energy than phosphorus. (d) bromine has a lower first ionization energy than chlorine.

Step-by-Step Solution

Answer

(a) Br is more stable with filled subshells than Se. (b) Aluminum's electron is easier to remove than magnesium's due to partial subshell. (c) Phosphorus has a more stable half-filled subshell than sulfur. (d) Bromine's electrons feel less attraction and more shielding than chlorine's.

Step 1: Identify Electron Configuration of Elements

First, identify the electron configurations for each of the elements: Selenium (Se), Bromine (Br), Aluminum (Al), Magnesium (Mg), Sulfur (S), Phosphorus (P), and Chlorine (Cl). - Se: \([Ar] \, 3d^{10} \, 4s^{2} \, 4p^{4}\) - Br: \([Ar] \, 3d^{10} \, 4s^{2} \, 4p^{5}\) - Al: \([Ne] \, 3s^{2} \, 3p^{1}\) - Mg: \([Ne] \, 3s^{2}\) - S: \([Ne] \, 3s^{2} \, 3p^{4}\) - P: \([Ne] \, 3s^{2} \, 3p^{3}\) - Cl: \([Ne] \, 3s^{2} \, 3p^{5}\)

Step 2: Electron Affinity Comparison

Electron affinity increases as we move across a period. Bromine has a nearly filled p-subshell (4p^5), requiring only one more electron to fill, whereas selenium is slightly less filled (4p^4). When an atom's p-subshell is close to being filled, it has a higher electron affinity, as in the case of bromine compared to selenium.

Step 3: Ionization Energy for Al vs. Mg

Ionization energy increases across a period. Magnesium has a filled subshell (3s^2), leading to greater stability compared to aluminum, which only starts filling the p-subshell (3p^1) after. Therefore, it requires more energy to remove an electron from magnesium than aluminum, explaining the lower ionization energy for aluminum.

Step 4: Ionization Energy for S vs. P

For sulfur, the additional electron in the 3p sublevel causes increased electron-electron repulsion. Phosphorus has a half-filled p-subshell (3p^3), which is more stable and thus requires more energy to ionize compared to sulfur (3p^4), resulting in sulfur having a lower ionization energy.

Step 5: Ionization Energy for Br vs. Cl

Ionization energy decreases down a group due to increased atomic radius and electron shielding. Even though bromine has a similar configuration to chlorine, the outer electrons in bromine are further from the nucleus and experience more shielding, leading to a lower ionization energy compared to chlorine.

Key Concepts

Electron Affinity

Electron affinity refers to the energy change that occurs when an electron is added to a neutral atom to form a negative ion. It's a measure of an atom's ability to attract electrons. Generally, electron affinity increases across a period in the periodic table as atoms move towards a complete valence shell. This is because atoms are more ready to accept electrons to achieve a stable electron configuration similar to that of noble gases.
Bromine ( Br) tends to have a higher electron affinity than selenium ( Se) because its electron configuration is anti 4p^{5} para close to a filled p-subshell. Selenium’s electron configuration is anti 4p^{4} para, meaning it is less ready than bromine to accept an additional electron as it does not bring it as close to a noble gas configuration.

• Bromine: Br -> [Ar] anti 3d^{10} 4s^{2} 4p^{5} para
• Selenium: Se -> [Ar] anti 3d^{10} 4s^{2} 4p^{4} para

Thus, atoms with nearly filled valence shells, like bromine, show higher electron affinities compared to those like selenium.

Ionization Energy

Ionization energy is the energy required to remove an electron from a neutral atom. Generally, as you move across a period on the periodic table, the ionization energy increases. This is because electrons are held more tightly by the nucleus as the effective nuclear charge increases, requiring more energy for removal.
For the comparison between aluminum ( Al) and magnesium ( Mg), magnesium has a higher ionization energy because it has a filled 3s subshell, making it more stable, whereas aluminum has its electron configuration beginning in the next subshell (3p).

• Magnesium: Mg -> [Ne] 3s^{2}
• Aluminum: Al -> [Ne] 3s^{2} 3p^{1}

Additionally, sulfur ( S) and phosphorus ( P) are compared based on their valence electron configurations. Sulfur has greater electron repulsion due to the additional 3p electron, making it easier to ionize than phosphorus, which has a half-filled, more stable 3p subshell.

• Sulfur: S -> [Ne] 3s^{2} 3p^{4}
• Phosphorus: P -> [Ne] 3s^{2} 3p^{3}

These examples show how electron configurations influence ionization energies.

Periodic Trends

Periodic trends in the periodic table include variations in properties such as electron affinity, ionization energy, atomic radius, and electronegativity across periods and groups. Understanding these trends helps explain differences in chemical properties among elements.
Ionization energy typically decreases as you move down a group because the outer electrons are farther from the nucleus and more shielded by inner electrons. This makes them easier to remove. Comparatively, bromine ( Br) has a lower ionization energy than chlorine ( Cl) because its electrons are further from the nucleus due to a larger atomic radius. The increased distance and additional electron shielding create a lower effective nuclear charge.

• Chlorine: Cl -> [Ne] 3s^{2} 3p^{5}
• Bromine: Br -> [Ar] 3d^{10} 4s^{2} 4p^{5}

As we can see, these trends are crucial for predicting and understanding the behavior of elements in chemical reactions and bonding scenarios.