Problem 95 - Chemistry The Central Science

Question

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.

Step-by-Step Solution

Answer

(a) Phosphorus has a half-filled 3p sub-shell, which is more stable due to lower electron-electron repulsion, resulting in a greater first ionization energy than sulfur. (b) Nitrogen's half-filled 2p sub-shell provides stability, meaning adding an electron would require energy input, resulting in a lower electron affinity compared to carbon and oxygen. (c) Oxygen requires more energy to remove its second electron, as it results in a configuration with a stable, half-filled 2p sub-shell, making its second ionization energy greater than the first ionization energy of fluorine. (d) Manganese's third ionization energy is greater because it involves removing an electron from its stable, half-filled 3d sub-shell, unlike chromium and iron.

Step 1: (a) Comparing the first ionization energy of phosphorus and sulfur

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.

Step 2: (b) Comparing the electron affinity of nitrogen, carbon, and oxygen

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.

Step 3: (c) Comparing the second ionization energy of oxygen and the first ionization energy of fluorine

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.

Step 4: (d) Comparing the third ionization energy of manganese, chromium, and iron

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.

Key Concepts

Ionization Energy

Ionization energy is the energy required to remove an electron from an atom in its gaseous state. The first ionization energy specifically refers to the energy needed to remove the first electron. Let's explore why some elements have higher ionization energies than others.

Consider phosphorus and sulfur; both are in the third period of the periodic table. Their electron configurations are as follows:

• Phosphorus: \[ [Ne]3s^2 3p^3 \]
• Sulfur: \[ [Ne]3s^2 3p^4 \]

Phosphorus has a half-filled 3p sub-shell which creates a stable configuration, reducing electron-electron repulsion. This stability means that removing an electron requires more energy. In contrast, sulfur's 3p sub-shell contains one orbital with two electrons, leading to increased repulsion between these electrons. Thus, the first ionization energy of phosphorus is greater than that of sulfur, because a more stable configuration in phosphorus makes it energetically more demanding to remove an electron.

Electron Affinity

Electron affinity measures the energy change when an electron is added to an atom, with a larger negative value indicating a greater tendency to gain an electron. Why might some elements have less negative electron affinities?

Let's compare nitrogen, carbon, and oxygen:

• Nitrogen: \[ [He]2s^2 2p^3 \]
• Carbon: \[ [He]2s^2 2p^2 \]
• Oxygen: \[ [He]2s^2 2p^4 \]

Nitrogen's half-filled 2p sub-shell provides a particularly stable electron arrangement, minimizing electron-electron repulsion. Adding another electron to nitrogen would disrupt this stability; thus, it requires energy rather than releasing it, resulting in a less negative electron affinity compared to carbon and oxygen, which would actually stabilize upon gaining an electron. Hence, nitrogen's electron affinity is lower than that of both carbon and oxygen.

Sub-shell Stability

Sub-shell stability refers to how certain electron configurations lead to increased stable states due to the distribution and pairing of electrons in sub-shells. This concept is vital to understanding the ionization energy and electron affinity of elements.

Consider the configurations when comparing ionization energies of elements like oxygen and fluorine. Oxygen's electron configuration is \[ [He]2s^2 2p^4 \], and when an electron is removed (making this the second ionization energy), it results in a half-filled \(2p\) sub-shell similar to nitrogen's highly stable arrangement. This boosted stability requires increased energy to remove this second electron.

Fluorine, with a configuration of \[ [He]2s^2 2p^5 \], sees its first electron removed from a less stable configuration in comparison to the situation where oxygen achieves a half-filled sub-shell after losing an electron. The enhanced stability of half-filled or completely filled sub-shells significantly influences the energies involved in ionization processes.

Half-filled Orbitals

Half-filled orbitals create a situation where an atom reaches a relatively stable energy state. This helps to understand why certain ionization energies are higher for some atoms compared to others.

A great illustration of this is seen in the electron configurations of manganese, chromium, and iron when considering their third ionization energies:

• Manganese: \[ [Ar]4s^2 3d^5 \]
• Chromium: \[ [Ar]4s^1 3d^5 \]
• Iron: \[ [Ar]4s^2 3d^6 \]

When manganese loses two electrons, it leaves its \(3d\) sub-shell half-filled. This is a notably stable configuration. As a consequence, removing the third electron is energetically demanding, leading to a higher third ionization energy compared to chromium and iron, where losing a third electron doesn’t disrupt a half-filled or fully filled stable arrangement. These half-filled orbitals thus have profound separations in energy requirements for electron removal, directly impacting atomic behaviors in chemical reactions.