Problem 79 - Chemistry The Molecular Science

Question

Use electron configurations to explain why (a) sulfur has a lower electron affinity than chlorine. (b) boron has a lower first ionization energy than beryllium. (c) chlorine has a lower first ionization energy than fluorine. (d) oxygen has a lower tirst ionization energy than nitrogen. (e) iodine has a lower electron affinity than bromine.

Step-by-Step Solution

Answer

(a) Chlorine completes 3p-orbital; (b) Boron starts filling p-sublevel; (c) Chlorine's larger size; (d) Oxygen's electron repulsion; (e) Iodine's larger size.

Step 1: Understanding Electron Affinity and Ionization Energy

Electron affinity is the energy change that occurs when an electron is added to an atom, while ionization energy is the energy required to remove an electron from an atom. These properties are influenced by electron configurations.

Step 1: Comparing Sulfur and Chlorine Electron Affinity

Sulfur \((1s^2 2s^2 2p^6 3s^2 3p^4)\) has a lower electron affinity than chlorine \((1s^2 2s^2 2p^6 3s^2 3p^5)\) because chlorine has one more electron which completes its p-orbital, making it more energetically favorable.

Step 2: Analyzing Boron and Beryllium's Ionization Energy

Boron \(1s^2 2s^2 2p^1\) has a lower first ionization energy than beryllium \(1s^2 2s^2\) because boron starts the filling of the p-sublevel, which is easier to remove than paired electrons in the s-sublevel of beryllium.

Step 3: Comparing Chlorine and Fluorine's Ionization Energy

Chlorine \(1s^2 2s^2 2p^6 3s^2 3p^5\) has a lower first ionization energy than fluorine \(1s^2 2s^2 2p^5\) because chlorine has a larger atomic size and its outer electrons are farther from the nucleus, making them easier to remove.

Step 4: Evaluating Oxygen and Nitrogen's Ionization Energy

Oxygen \(1s^2 2s^2 2p^4\) has a lower first ionization energy than nitrogen \(1s^2 2s^2 2p^3\) because the additional electron in oxygen leads to increased electron-electron repulsion in the p-orbital making it easier to remove.

Step 5: Examining Iodine and Bromine's Electron Affinity

Iodine \(5p^5\) has a lower electron affinity than bromine \(4p^5\) because iodine's larger atomic radius reduces nuclear attraction for added electrons compared to bromine.

Key Concepts

Electron Affinity

Electron affinity is a measure of how much an atom needs or wants to gain an electron. When you add an electron to an atom, energy is released. The more energy released, the higher the electron affinity. Atoms with high electron affinity strongly attract additional electrons.

For example, chlorine has a higher electron affinity than sulfur because it only needs one more electron to fill its outer shell. This makes it very eager to accept an electron, releasing more energy in the process compared to sulfur.

• Chlorine: \(1s^2 2s^2 2p^6 3s^2 3p^5\)
• Sulfur: \(1s^2 2s^2 2p^6 3s^2 3p^4\)

Chlorine's electron affinity is thus higher, as it completes the 3p orbital with the additional electron, making it energetically favorable.

Ionization Energy

Ionization energy is the energy needed to remove an electron from an atom. The first ionization energy is the amount of energy needed to remove the outermost electron. Atoms with high ionization energy hold onto their electrons tightly.

An example of this is boron and beryllium. Boron has a lower first ionization energy than beryllium. This is because, in boron \(1s^2 2s^2 2p^1\), the p-electron is easier to remove than the paired s-electrons found in beryllium \(1s^2 2s^2\). Boron's loosely held 2p electron requires less energy to be taken away compared to the tightly held s-electron in beryllium.

• Beryllium's configuration ends with fully paired s-electrons making them more stable.
• Boron begins filling its p-orbitals, which are higher in energy and less tightly bound.

Periodic Trends

Periodic trends describe patterns across periods and groups in the periodic table. These trends arise because of similar electron configurations in different elements.

A noticeable trend is that ionization energy generally increases across a period. This is because electrons are added to the same energy level while nuclear charge increases, pulling electrons tighter. However, there are exceptions, such as chlorine having a lower ionization energy than fluorine.

• Fluorine: \(1s^2 2s^2 2p^5\)
• Chlorine: \(1s^2 2s^2 2p^6 3s^2 3p^5\)

Chlorine's outer electrons are further from the nucleus than fluorine's electrons, making them easier to remove despite the general trend.

Atomic Structure

Atomic structure is fundamental to understanding periodic trends. Atoms consist of a dense central nucleus surrounded by electrons in various shells or orbitals. These electrons are organized based on energy levels and sublevels.

In oxygen and nitrogen, their atomic structures contribute to their ionization energies. Nitrogen's half-filled \(1s^2 2s^2 2p^3\) is particularly stable due to all p-orbitals having one electron each. However, oxygen \(1s^2 2s^2 2p^4\) has an additional electron that causes repulsion, making it easier to remove.

• Nitrogen holds onto its electrons due to a half-filled subshell.
• Oxygen experiences more electron-electron repulsion in the 2p orbitals.

Correctly understanding atomic configurations helps explain why some elements deviate from expected trends.