Problem 95 - Chemistry: The Central Science

Question

Use electron configurations to explain the following observations: (a) The first ionization energy of phosphorus is greater than that of sulfur. (b) The electron affinity of nitrogen is lower (less negative) than those of both carbon and oxygen. (c) The second ionization energy of oxygen is greater than the first ionization energy of fluorine. (d) The third ionization energy of manganese is greater than those of both chromium and iron.

Step-by-Step Solution

Answer

The first ionization energy of phosphorus (P) is greater than sulfur (S) due to its half-filled 3p subshell, providing extra stability, while sulfur's additional electron makes it easier to remove. Nitrogen (N) has a lower electron affinity than carbon (C) and oxygen (O) since its half-filled 2p subshell is relatively stable, while adding an electron to C and O increases stability. Oxygen's (O) second ionization energy is greater than fluorine's (F) first ionization energy because removing a second electron from O's full 2s subshell requires more energy than removing F's first electron from its not full 2p subshell. Finally, the third ionization energy of manganese (Mn) is greater than chromium (Cr) and iron (Fe) due to Mn losing its half-filled 4s and 3d subshells and becoming less stable after the second ionization.

Step 1: Write electron configurations for phosphorus and sulfur

First, let's find the electron configurations for phosphorus (P) and sulfur (S). Phosphorus (P, atomic number 15): \( 1s^{2}2s^{2}2p^{6}3s^{2}3p^{3} \) Sulfur (S, atomic number 16): \( 1s^{2}2s^{2}2p^{6}3s^{2}3p^{4} \)

Step 2: Explain the difference in ionization energies

Phosphorus has a half-filled 3p subshell, which provides extra stability. Removing an electron from the stable half-filled subshell requires more energy. In contrast, sulfur has one additional electron in its 3p subshell, making it easier to remove an electron as it doesn't disrupt the stability as much. Hence, the first ionization energy of P is greater than that of S. #b) The electron affinity of nitrogen is lower (less negative) than those of both carbon and oxygen.#

Step 1: Write electron configurations for nitrogen, carbon, and oxygen

Next, let's find the electron configurations for nitrogen (N), carbon (C), and oxygen (O). Nitrogen (N, atomic number 7): \( 1s^{2}2s^{2}2p^{3} \) Carbon (C, atomic number 6): \( 1s^{2}2s^{2}2p^{2} \) Oxygen (O, atomic number 8): \( 1s^{2}2s^{2}2p^{4} \)

Step 2: Explain the difference in electron affinities

Nitrogen has a half-filled 2p subshell, which is a relatively stable configuration. Adding an electron to this subshell would destabilize it, so nitrogen's electron affinity is low (less negative). Conversely, adding an electron to carbon's and oxygen's 2p subshells increases their stability, resulting in more negative electron affinities. #c) The second ionization energy of oxygen is greater than the first ionization energy of fluorine.#

Step 1: Write electron configurations for oxygen and fluorine

Now, let's find the electron configurations for oxygen (O) and fluorine (F). Oxygen (O, atomic number 8): \( 1s^{2}2s^{2}2p^{4} \) Fluorine (F, atomic number 9): \( 1s^{2}2s^{2}2p^{5} \)

Step 2: Explain the difference in ionization energies

When removing a second electron from oxygen, we are removing it from the stable, full 2s subshell. This requires more energy than removing the first electron from fluorine's 2p subshell, which is not full and therefore less stable. Thus, the second ionization energy of oxygen is greater than the first ionization energy of fluorine. #d) The third ionization energy of manganese is greater than those of both chromium and iron.#

Step 1: Write electron configurations for manganese, chromium, and iron

Finally, let's find the electron configurations for manganese (Mn), chromium (Cr), and iron (Fe). Manganese (Mn, atomic number 25): \( 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{5} \) Chromium (Cr, atomic number 24): \( 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{1}3d^{5} \) Iron (Fe, atomic number 26): \( 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{6} \)

Step 2: Explain the difference in ionization energies

After the second ionization, Mn loses its half-filled 4s and 3d subshells, becoming less stable. Removing a third electron from Mn would require more energy due to this destabilization. In contrast, removing a third electron from Cr doesn't change the half-filled 3d subshell, while removing it from Fe results in a half-filled 3d subshell, increasing stability. Hence, the third ionization energy of Mn is greater than those of both Cr and Fe.