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Chapter 8: Problem 21

As the \(2 p\) levels fill. from \(Z=5\) to \(10,\) the elements' first ionization energies tend to increase. But, as Figure 8.16 clearly mdicates, it is slightly easier to remove an electron from \(Z=8\) oxygen than from \(Z=7\) nitrogen. What might explain this?

Short Answer

Expert verified
It is easier to remove an electron from oxygen (Z=8) than from nitrogen (Z=7) because nitrogen has a half-filled 2p orbital, which makes it more stable than oxygen. The additional electron in oxygen, which breaks the balance and creates a less stable condition, making it easier to remove.

Step by step solution

01

Understand the concept of Ionization Energy

Ionization energy is the energy required to remove an electron from an atom in its gaseous state. It tells us about how tightly the electrons are held within an atom. The larger the ionization energy, the harder it is to remove an electron from the atom. The ionization energy tends to increase across a period from left to right on the periodic table.
02

Consider the Electron Configurations

Examine the electron configurations of nitrogen and oxygen. Nitrogen, with atomic number 7, has an electron configuration of 1s虏2s虏2p鲁, and oxygen, with atomic number 8, has an electron configuration of 1s虏2s虏2p鈦�. This indicates that nitrogen has half-filled 2p orbitals, while oxygen has one electron more than a half-filled 2p orbital.
03

Realize the Stability of Half-filled Orbitals

Half-filled and fully-filled orbitals are more stable due to the balance of electrons. The 2p orbital in nitrogen is half-filled and is in a more stable state. On the other hand, oxygen has one 2p orbital with a single electron, and the others are doubly occupied, making it less stable. Therefore, it's easier to remove an electron from oxygen than from nitrogen, despite the general trend.

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Key Concepts

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

Periodic Table
The periodic table is more than just a chart with atomic symbols; it holds the secrets of element properties and behaviors. Elements in the periodic table are arranged based on their atomic number, which is the number of protons in an atom. As you move across a period from left to right, certain patterns in properties become evident.
One key trend is the increase in ionization energy across a period. This is because as you move across a period, the nuclear charge increases, pulling the electrons closer to the nucleus and making them harder to remove. Despite this general trend, there are exceptions due to the unique electron configurations present for specific elements. Observing these anomalies can tell us a lot about the intricacies in chemistry, which you can explore further when analyzing specific cases like nitrogen and oxygen.
Electron Configuration
Electron configuration refers to the arrangement of electrons in an atom. It's like a map where each "address" corresponds to an electron's location. Understanding electron configuration is crucial because it helps explain chemical behavior and trends.
An atom's electrons fill orbitals in a defined sequence. For instance, nitrogen with atomic number 7 has an electron configuration of 1s虏 2s虏 2p鲁. This suggests three electrons in the 2p subshell, filling half of it. Oxygen, with atomic number 8, has a configuration of 1s虏 2s虏 2p鈦�, showing four electrons in the 2p subshell, surpassing the half-filled mark.
This arrangement impacts not only the stability of the atom but also its reactivity. The outermost or "valence" electrons play a pivotal role in bonding and are vital in understanding trends like ionization energy along the periodic table.
Orbital Stability
The concept of orbital stability is linked to how electrons are distributed within an atom. Orbitals hold electrons, and their stability can greatly influence an element's properties.
Half-filled and fully-filled orbitals carry a special stability due to electron balance and symmetry. This balance minimizes electron repulsion within an orbital group, making configurations such as nitrogen's 2p鲁 more stable. In contrast, oxygen's 2p鈦� configuration has more electron repulsion due to the additional electron, which disturbs this balance.
  • A half-filled 2p subshell, like in nitrogen, has a balanced distribution that adds a level of inherent stability.
  • Oxygen, with one extra electron beyond a half-filled state, experiences increased repulsion, disrupting stability.
This added repulsion in oxygen makes it slightly easier to remove an electron compared to the more stable nitrogen. Hence, orbital stability nuances explain exceptions in ionization trends across periods.

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Most popular questions from this chapter

Write the electronis configurations for phosphorus. germanium. and cesium.

The 21 cm Line: One of the most important windows to the mysteries of the cosmos is the 21 cm line. With it, astronomers map hydrogen throughout the universe. An important trait is that it involves a highly forbidden transition that is, accordingly, quite long-lived. But it is also an excellent example of the coupling of angular momenta. Hydrogen's ground state has no spin-orbit interaction - for \(\boldsymbol{\theta}=0,\) there is no orbit. However, the proton and electron magnetic moments do interact. Consider the following simple model. (a) The proton sees itself surrounded by a spherically symmetric cloud of Is electron, which has an intrinsic magnetic dipole moment/spin that, of course, has a direction. For the purpose of investigating its effect on the proton, treat this dispersed magnetic moment as behaving effectively like a single loop of current whose radius is \(a_{0}\), then find the magnetic tield at the middle of the loop in tenns of \(e, h, m_{e}, \mu_{0},\) and \(a_{0} .\) (b) The proton sits right in the middle of the electron's magnetic moment. Like the electron. the proton is a spin- \(\frac{1}{2}\) particle, with only two possible orientations in a magnetic field. Noting, however, that its spin and magnetic moment are parallel rather than opposite, would the interaction energy be lower with the proton's spin aligned or antialigned with that of the electron'? (c) For the proton, \(g_{p}\) is \(5.6 .\) Obtain a rough value for the energy difference between the two orientations. (d) What would be the wavelength of a photon that carries away this energy difference?

Assuming that the spin-orbit interaction is not overwhelmed by an extemal magnetic field, what is the minimum angle the toral angular momentum vector J may make with the \(z\) -axis in a \(3 d\) state of hydrogen?

Concisely, why is the table periodic?

The Zeeman effect occurs in sodium just as in hydrogen - sodium's lone 3 s valence electron behaves much as hydrogen's 1.5. Suppose sodium atoms are immersed in a \(0.1 \mathrm{~T}\) magnetic field. (a) Into how many levels is the \(3 p_{1 / 2}\) level split? (b) Determine the energy spacing between these states. (c) Into how many lines is the \(3 p_{1 / 2}\) to \(3 s_{1 / 2}\) spectral line split by the field? (d) Describe quantitatively the spacing of these lines. (e) The sodium doublet \((589.0 \mathrm{nm}\) and \(589.6 \mathrm{nm}\) ) is two spectral lines. \(3 p_{3 n} \rightarrow 3 s_{1 / 2}\) and \(3 p_{1 / 2} \rightarrow 3 s_{1 / 2}\) which are split according to the two different possible spin-orbit ener gies in the \(3 p\) state (see Exercise 60 ). Detemine the splitting of the sodium doublet (the energy diff erence between the two photons). How does it compare with the line splitting of part (d), and why?
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