Use electron configurations to explain the following observa tions: (a) The first ionization energy of phosphorus is greater than that of sulfur. (b) The electron afnity 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.

The electron configuration of phosphorus (P) is [Ne]\(3s^2 3p^3\) and sulfur (S) is [Ne]\(3s^2 3p^4\). Both elements are in the third period, and phosphorus is to the left of sulfur. The first ionization energy refers to the energy required to remove the first electron. Since both of these elements have electrons in the 3p sub-shell, we will compare their energies in the same sub-shell. For phosphorus, this would mean removing one of its three 3p electrons. For sulfur, this would mean removing one of its four 3p electrons. Phosphorus has half-filled 3p orbitals, which results in greater stability due to the lower electron-electron repulsion. Sulfur, on the other hand, has one of its 3p orbitals with two electrons, causing an extra electron-electron repulsion. Since phosphorus has a more stable electron configuration, it requires more energy to remove an electron, resulting in a larger first ionization energy.

The electron configurations of nitrogen (N), carbon (C), and oxygen (O) are [He]\(2s^2 2p^3\), [He]\(2s^2 2p^2\), and [He]\(2s^2 2p^4\), respectively. Nitrogen has a half-filled 2p sub-shell, which results in increased stability due to lower electron-electron repulsion. Adding an electron would disturb this stability and would require energy input. In the case of carbon and oxygen, both elements would gain stability upon adding an electron, thus they have more negative electron affinity values than nitrogen.

The electron configuration of oxygen (O) is [He]\(2s^2 2p^4\) and fluorine (F) is [He]\(2s^2 2p^5\). The first ionization energy for fluorine would involve the removal of a single electron from its half-filled 2p sub-shell. When one electron is removed from oxygen (second ionization energy), it would result in a configuration with a half-filled 2p sub-shell just like nitrogen, offering it stability, so more energy is required to remove the second electron from oxygen as compared to the first electron from fluorine.

The electron configurations of manganese (Mn), chromium (Cr), and iron (Fe) are [Ar]\(4s^2 3d^5\), [Ar]\(4s^1 3d^5\), and [Ar]\(4s^2 3d^6\), respectively. For manganese, after removing two electrons (4s), the third ionization energy involves removal of an electron from the half-filled and stable 3d sub-shell. For both chromium and iron, the third ionization energy refers to the removal of an electron from the same 3d sub-shell but without breaking any half-filled sub-shells. Since manganese has a stable half-filled 3d level, more energy is required to remove an electron, resulting in a greater third ionization energy compared to chromium and iron.