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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.

Short Answer

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In summary, (a) phosphorus has a greater first ionization energy than sulfur due to its half-filled 3p subshell being more stable; (b) nitrogen has a lower electron affinity than carbon and oxygen because adding an electron results in a stable, half-filled 2p subshell; (c) the second ionization energy of oxygen is greater than the first ionization energy of fluorine as removing an electron from oxygen results in a more stable half-filled 2p subshell compared to fluorine; (d) the third ionization energy of manganese is greater than those of chromium and iron because removing an electron from manganese's half-filled 3d^5 subshell is more energetically costly compared to removing electrons from chromium and iron.

Step by step solution

01

(a) Comparing Ionization Energies of Phosphorus and Sulfur

First, let's find the electron configurations of phosphorus (P) and sulfur (S). Phosphorus has 15 electrons and its configuration is: \(1s^2 2s^2 2p^6 3s^2 3p^3\). Sulfur has 16 electrons and its configuration is: \(1s^2 2s^2 2p^6 3s^2 3p^4\). Phosphorus has a half-filled \(3p\) subshell, while sulfur has a more than half-filled \(3p\) subshell. Removing an electron from a half-filled subshell in phosphorus requires more energy than removing one from sulfur because half-filled and fully-filled subshells are more stable. Thus, the first ionization energy of phosphorus is greater than that of sulfur.
02

(b) Comparing Electron Affinities of Nitrogen, Carbon, and Oxygen

Let's find the electron configurations of nitrogen (N), carbon (C), and oxygen (O). Nitrogen has 7 electrons and its configuration is: \(1s^2 2s^2 2p^3\). Carbon has 6 electrons and its configuration is: \(1s^2 2s^2 2p^2\). Oxygen has 8 electrons and its configuration is: \(1s^2 2s^2 2p^4\). Adding an electron to nitrogen will result in a half-filled \(2p\) subshell. The half-filled subshells have extra stability, so nitrogen is less inclined to accept an additional electron compared to carbon and oxygen. This leads to nitrogen having a lower (less negative) electron affinity than both carbon and oxygen.
03

(c) Comparing Ionization Energies of Oxygen and Fluorine

Now, let's find the electron configurations of oxygen (O) and fluorine (F). Oxygen has 8 electrons and its configuration is: \(1s^2 2s^2 2p^4\). Fluorine has 9 electrons and its configuration is: \(1s^2 2s^2 2p^5\). For oxygen, the second ionization energy corresponds to removing one of the electrons in the \(2p\) subshell, leaving a half-filled \(2p\) subshell. For fluorine, the first ionization energy corresponds to removing one of the electrons in the \(2p\) subshell, leaving a fully-filled \(2s\) subshell. Since fully-filled and half-filled subshells are more stable, removing an electron from oxygen's half-filled \(2p\) requires more energy compared to removing one from fluorine's almost fully-filled \(2p\). Therefore, the second ionization energy of oxygen is greater than the first ionization energy of fluorine.
04

(d) Comparing Third Ionization Energies of Manganese, Chromium, and Iron

Finally, let's find the electron configurations of manganese (Mn), chromium (Cr), and iron (Fe). Manganese has 25 electrons and its configuration is: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^5\). Chromium has 24 electrons and its configuration is: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 3d^5\). Iron has 26 electrons and its configuration is: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^6\). The third ionization energy of manganese involves removing an electron from the half-filled \(3d^5\) subshell, which is more stable compared to chromium and iron. Removing an electron from chromium requires removing it from the \(4s^1\) subshell, while removing an electron from iron requires removing it from the \(3d^6\) subshell, which is less stable compared to the half-filled \(3d^5\) of manganese. Thus, the third ionization energy of manganese is greater than those of both chromium and iron.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ionization Energy
Ionization energy refers to the energy needed to remove an electron from an atom or ion in its gaseous state. It plays a crucial role in determining an element's chemical properties, including reactivity and bonding behavior. A higher ionization energy means that it is harder to remove an electron because it is more tightly bound to the atom.

Factors Affecting Ionization Energy

Several factors influence ionization energy: The number of protons in the nucleus, the distance of the electron from the nucleus, and the electron shielding effect. Additionally, the stability of electron configurations plays a significant role. Half-filled and fully-filled subshells are notably stable, which means atoms will require more energy to ionize when removing electrons from such configurations. This concept is beautifully illustrated in the case of phosphorus having a higher first ionization energy than sulfur due to the stability of its half-filled subshell.
Electron Affinity
Electron affinity measures the change in energy when an electron is added to an atom or ion. If the energy change is negative, the atom releases energy when gaining an electron, indicating a higher affinity for the electron. An element's electron affinity is influenced by its electron configuration and the effective nuclear charge felt by the added electron.

Stability of Electron Configurations

The stability provided by certain electron arrangements, such as half-filled subshells, can lead to variations in electron affinities across the periodic table. For instance, nitrogen, which achieves a half-filled subshell upon adding an electron, exhibits a lower electron affinity compared to carbon and oxygen. This highlights the importance of the specific disposition of electrons in influencing an atom's desire to gain additional electrons.
Subshell Stability
Subshell stability refers to the natural tendency of electrons within an atom to occupy configurations that offer the most stability. There are certain arrangements of electrons within orbitals, particularly half-filled or fully-filled subshells, that are energetically favorable.

Significance in Ionization and Electron Gain

Atoms with stable configurations tend to resist losing or gaining electrons; thus, higher ionization energies and lower electron affinities are observed for elements with half-filled or fully-filled subshells. For example, manganese demonstrates a high third ionization energy due to the stability of its half-filled d-subshell. These variations among different elements can be largely attributed to the quest for achieving a stable electron configuration.
Periodic Trends
Understanding the periodic trends allows us to predict various chemical and physical properties of the elements. Ionization energy, electron affinity, atomic radius, and electronegativity are all influenced by the position of an element in the periodic table.

Trends Across and Down the Periodic Table

Ionization energy typically increases across a period and decreases down a group. This is due to effective nuclear charge and electron shielding. Similarly, electron affinity generally becomes more negative across a period, indicating an increase in the tendency to gain electrons. However, there are exceptions to these trends, often owing to the stabilization provided by certain electron configurations, as seen with nitrogen's electron affinity and oxygen's second ionization energy in comparison to fluorine. Acknowledging these trends is fundamental for understanding the reactivity and behavior of different elements.

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Most popular questions from this chapter

Potassium superoxide, \(\mathrm{KO}_{2}\), is often used in oxygen masks (such as those used by firefighters) because \(\mathrm{KO}_{2}\) reacts with \(\mathrm{CO}_{2}\) to release molecular oxygen. Experiments indicate that \(2 \mathrm{~mol}\) of \(\mathrm{KO}_{2}(\mathrm{~s})\) react with each mole of \(\mathrm{CO}_{2}(g) .\) (a) The products of the reaction are \(\mathrm{K}_{2} \mathrm{CO}_{3}(s)\) and \(\mathrm{O}_{2}(g) .\) Write a balanced equation for the reaction between \(\mathrm{KO}_{2}(\mathrm{~s})\) and \(\mathrm{CO}_{2}(\mathrm{~g}) .\) (b) Indicate the oxidation number for each atom involved in the reaction in part (a). What elements are being oxidized and reduced? (c) What mass of \(\mathrm{KO}_{2}(s)\) is needed to consume \(18.0 \mathrm{~g}\) \(\mathrm{CO}_{2}(g) ?\) What mass of \(\mathrm{O}_{2}(g)\) is produced during this reaction?

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