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Explain why the electron affinity of nitrogen is approximately zero, while the elements on either side, carbon and oxygen, have substantial positive electron affinities

Short Answer

Expert verified
Nitrogen has a stable half-filled p subshell; adding an electron disrupts this stability, unlike for carbon and oxygen.

Step by step solution

01

Understanding Electron Affinity

Electron affinity is the energy change that occurs when an electron is added to a neutral atom in the gaseous state to form a negative ion. High electron affinity indicates that the atom releases more energy when accepting an electron, meaning it readily gains electrons.
02

Examining Nitrogen's Electron Configuration

Nitrogen has the electron configuration 1s² 2s² 2p³. Its valence shell has three unpaired electrons in the 2p subshell, making it half-filled, which is relatively stable due to symmetrical electron distribution and exchange energy.
03

Identifying Stability in Nitrogen

Nitrogen's half-filled p subshell provides extra stability due to equal energy distribution among the p orbitals and exchange energy. Adding an extra electron would disrupt this symmetry and result in a less stable, lower exchange energy state.
04

Contrasting with Carbon and Oxygen

Carbon, with configuration 1s² 2s² 2p², gains an electron to achieve a half-filled p subshell, making it more stable. Oxygen, with configuration 1s² 2s² 2p⁴, benefits from a decrease in electron-electron repulsion when an electron is added, moving towards a more stable 2p⁶ configuration.

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Key Concepts

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

Nitrogen Electron Configuration
To understand nitrogen's unique behavior regarding electron affinity, we must first look at its electron configuration. Nitrogen's configuration is 1s² 2s² 2p³. This means the 2p subshell contains three electrons. Visualize it: the 1s and 2s subshells are filled to capacity, while the 2p subshell has three unpaired electrons. This "half-filled" state of the 2p subshell is crucial. It's stable due to the symmetrical arrangement of electrons, allowing for maximum exchange energy. Exchange energy refers to the stability afforded by symmetrically distributed electrons in degenerate orbitals. This stability is one reason why nitrogen does not easily gain an extra electron.
Half-Filled Subshell Stability
A half-filled subshell, like nitrogen's 2p subshell, possesses unique stability. This stability arose due to several reasons:
  • Symmetrical distribution: Electrons are spread evenly across available orbitals, minimizing repulsion.
  • Exchange energy: This is the energy stabilization resultant from permuting electrons among orbitals of the same energy (degenerate orbitals).
In simple terms, the electrons in nitrogen's 2p subshell achieve a balance that makes further electron addition energetically unfavorable. Adding another electron would disrupt this balance, leading to a less energetically favorable configuration.
Energy Change in Atom
The concept of electron affinity is intrinsically linked to energy change in an atom. When an atom captures an electron, energy is either absorbed or released, depending on the stability changes involved. In nitrogen, the addition of an extra electron results in an energy state that disturbs its half-filled stability, leading to little to no energy release. Unlike nitrogen, elements like carbon and oxygen have electron configurations that become more stable when an electron is added, thereby releasing significant energy. For nitrogen, the energy change is closer to zero, reflecting minimal benefit in electron capture.
Electron-Electron Repulsion
Electron-electron repulsion plays a significant role in determining whether an atom will release or absorb energy when gaining an electron. In nitrogen, adding an extra electron leads to increased repulsion among the electrons in the 2p subshell. This increased repulsion makes the entire negative ion formation process energetically unfavorable. Meanwhile, oxygen benefits from added electron reducing repulsion within its 2p⁴ subshell, as it approaches a noble gas-like configuration (2p⁶). On the other hand, nitrogen's stable half-filled configuration naturally results in high repulsion deterring further electron addition, thus explaining its near-zero electron affinity.

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Most popular questions from this chapter

Indicate which one of the two species in each of the following pairs is smaller: (a) \(\mathrm{Cl}\) or \(\mathrm{Cl}^{-},\) (b) Na or \(\mathrm{Na}^{+}\), (c) \(\mathrm{O}^{2-}\) or \(\mathrm{S}^{2-}\). (d) \(\mathrm{Mg}^{2+}\) or \(\mathrm{Al}^{3+}\) (e) \(\mathrm{Au}^{+}\) or \(\mathrm{Au}^{3+}\)

Use the second period of the periodic table as an example to show that the size of atoms decreases as we move from left to right. Explain the trend.

The first four ionization energies of an element are approximately \(738,1450,7.7 \times 10^{3},\) and \(1.1 \times 10^{4} \mathrm{~kJ} / \mathrm{mol}\). To which periodic group does this element belong? Explain your answer.

In halogen displacement reactions a halogen element can be generated by oxidizing its anions with a halogen element that lies above it in the periodic table. This means that there is no way to prepare elemental fluorine, because it is the first member of Group \(7 \mathrm{~A} .\) Indeed, for years the only way to prepare elemental fluorine was to oxidize \(\mathrm{F}^{-}\) ions by electrolytic means. Then, in 1986 , a chemist reported that by combining potassium hexafluoromanganate(IV) \(\left(\mathrm{K}_{2} \mathrm{MnF}_{6}\right)\) with antimony pentafluoride \(\left(\mathrm{SbF}_{5}\right)\) at \(150^{\circ} \mathrm{C}\), he had generated elemental fluorine. Balance the following equation representing the reaction: $$ \mathrm{K}_{2} \mathrm{MnF}_{6}+\mathrm{SbF}_{5} \longrightarrow \mathrm{KSbF}_{6}+\mathrm{MnF}_{3}+\mathrm{F}_{2} $$

Based on knowledge of the electronic configuration of titanium, state which of the following compounds of titanium is unlikely to exist: \(\mathrm{K}_{3} \mathrm{TiF}_{6}, \mathrm{~K}_{2} \mathrm{Ti}_{2} \mathrm{O}_{5}, \mathrm{TiCl}_{3},\) \(\mathrm{K}_{2} \mathrm{TiO}_{4}, \mathrm{~K}_{2} \mathrm{TiF}_{6}\)

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