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Suppose that the electron had no spin and that the Pauli exclusion principle still held. Which, if any, of the present noble gases would remain in that category?

Short Answer

Expert verified
None of the present noble gases would remain noble if electrons had no spin.

Step by step solution

01

Understanding the Exercise

To determine which noble gases remain noble if electrons had no spin but still followed the Pauli exclusion principle, we need to evaluate the electron configurations of noble gases assuming only one electron can occupy an orbital (because no spin means no distinguishability within the same orbital).
02

Electron Configuration with No Spin

For each electron shell and subshell, calculate the maximum number of electrons they can hold, assuming there's no spin. Normally, each orbital can hold 2 electrons (one with spin up and one with spin down). Without spin, each orbital will hold only 1 electron.
03

Determine Original Noble Gases

Recall that noble gases have a full outer shell. Under the standard model, the noble gases are He, Ne, Ar, Kr, Xe, and Rn. They have electronic configurations ending in full s and p subshells for their respective periods.
04

Evaluate Shell Capacities

1. The s subshell has 1 orbital; therefore, with no spin, it can hold 1 electron (original max was 2). 2. The p subshell has 3 orbitals; it can hold 3 electrons (original max was 6). 3. The d subshell has 5 orbitals; it can hold 5 electrons (original max was 10). 4. The f subshell has 7 orbitals; it can hold 7 electrons (original max was 14).
05

Recalculate Noble Gases Electron Configuration

1. Helium (He): Normally 1s虏, now only 1s鹿, which is not full. 2. Neon (Ne): Normally 1s虏 2s虏 2p鈦, now 1s鹿 2s鹿 2p鲁 鈥 not full. 3. Argon (Ar): Normally 1s虏 2s虏 2p鈦 3s虏 3p鈦, now 1s鹿 2s鹿 2p鲁 3s鹿 3p鲁 鈥 not full. 4. Krypton (Kr), Xenon (Xe), and Radon (Rn) are similarly incomplete since any element beyond helium follows the "3s鹿 2p鲁" pattern.
06

Conclusion

In the absence of electron spin, none of the current noble gases can have a completely filled outer shell under the Pauli exclusion principle because their typical filled configurations exceed what is possible without spin.

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Key Concepts

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

Electron Configuration
Electron configuration is a way to describe the distribution of electrons in an atom's atomic orbitals. It's like an address system for electrons to be noted around the nucleus. Normally, electrons fill atomic orbitals in a consistent order based on increasing energy levels. This is known as the Aufbau principle. Electrons live in regions called orbitals, and as energy levels increase, various types of orbitals (s, p, d, f) are filled.

Normally:
  • An s orbital can hold 2 electrons
  • A p orbital can hold 6 electrons
  • A d orbital can contain 10 electrons
  • An f orbital can hold 14 electrons
The Pauli exclusion principle, however, mandates that no two electrons can have the same set of quantum numbers within an atom. This means within a single orbital, electrons distinguish themselves by their spin, one spinning "up" and the other "down," allowing each to inhabit the same orbital.

If electrons did not have spin, as in the hypothetical scenario given, each orbital could only host one electron. This drastically changes what electron configurations look like, as seen when recalculating noble gases鈥 configurations. Instead of an s orbital holding 2 electrons, it now holds just one, limiting the number of filled orbitals and drastically altering elements鈥 chemistry.
Noble Gases
Noble gases are a family of elements known for their full outer electron shells, making them chemically inert under normal conditions. In their natural state, they goal is stability, which they achieve by having a completely filled valence shell. This makes noble gases stable and unlikely to react with other elements.

However, when considering the scenario where electrons have no spin, noble gases lose their charm. With half-occupied shells due to the lack of spin, they can no longer maintain a fully filled outer shell configuration:
  • Helium becomes 1s鹿, losing its characteristic full shell stability.
  • Neon becomes 1s鹿 2s鹿 2p鲁 instead of its usual 1s虏 2s虏 2p鈦.
  • Argon, krypton, xenon, and radon undergo similar changes.
Each of these would not be able to keep their stable, noble 'status' because they鈥檇 no longer have complete electron shells as required for being a noble gas by definition.

Thus, none of the current noble gases would remain noble due to incomplete valence shells, meaning they might be prone to chemical interactions they wouldn鈥檛 normally undergo.
Atomic Orbitals
Atomic orbitals are regions of space around an atom's nucleus where electrons are likely to be found. Each type of orbital (s, p, d, f) has a different shape and energy level associated with it, and they fill according to specific rules.

- **s orbitals** are spherical, holding up to 2 electrons. - **p orbitals** are dumbbell-shaped, encompassing up to 6 electrons across 3 orientations (px, py, pz). - **d orbitals** look like a clover and can hold up to 10 electrons, with 5 orientations. - **f orbitals** have more complex shapes, accommodating up to 14 electrons. The ability of orbitals to hold multiple electrons depends on electron spin. Spin creates a unique quantum state, allowing more than one electron in an orbital. In our hypothetical scenario, without spin, each orbital can hold only one electron, disrupting the configurations across every element.

Atomic orbitals are integral to how atoms form bonds and how elements interact. Losing the ability to fully populate these with electrons 鈥 because of the lack of electron spin 鈥 dramatically shifts an element's properties, as shown by the inability of noble gases to sustain their inert status.

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Most popular questions from this chapter

An electron in a multielectron atom is known to have the quantum number \(\ell=3 .\) What are its possible \(n, m_{\ell},\) and \(m_{s}\) quantum numbers?

A laser emits at \(424 \mathrm{nm}\) in a single pulse that lasts \(0.500 \mu \mathrm{s}\). The power of the pulse is \(2.80 \mathrm{MW}\). If we assume that the atoms contributing to the pulse underwent stimulated emission only once during the \(0.500 \mu \mathrm{s}\), how many atoms contributed?

The active volume of a laser constructed of the semiconductor GaAlAs is only \(200 \mu \mathrm{m}^{3}\) (smaller than a grain of sand), and yet the laser can continuously deliver \(5.0 \mathrm{~mW}\) of power at a wavelength of \(0.80 \mu \mathrm{m}\). At what rate does it generate photons?

In the subshell \(\ell=3,\) (a) what is the greatest (most positive) \(m_{f}\) value, (b) how many states are available with the greatest \(m_{\ell}\) value, and (c) what is the total number of states available in the subshell?

Ruby lases at a wavelength of \(694 \mathrm{nm}\). A certain ruby crystal has \(4.00 \times 10^{19} \mathrm{Cr}\) ions (which are the atoms that lase). The lasing transition is between the first excited state and the ground state, and the output is a light pulse lasting \(2.00 \mu \mathrm{s}\). As the pulse begins, \(60.0 \%\) of the \(\mathrm{Cr}\) ions are in the first excited state and the rest are in the ground state. What is the average power emitted during the pulse? (Hint: Don't just ignore the ground-state ions.)

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