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Chapter 6: Problem 70

An experiment called the Stern-Gerlach experiment helped establish the existence of electron spin. In this experiment, a beam of silver atoms is passed through a magnetic field, which deflects half of the silver atoms in one direction and half in the opposite direction. The separation between the two beams increases as the strength of the magnetic field increases. (a) What is the electron configuration for a silver atom? (b) Would this experiment work for a beam of cadmium (Cd) atoms? (c) Would this experiment work for a beam of fluorine (F) atoms?

Short Answer

Expert verified
(a) The electron configuration for a silver (Ag) atom is [Kr] 4d^10 5s^1. (b) The Stern-Gerlach experiment would not work for a beam of cadmium (Cd) atoms because all electrons are paired in the 6s^2 orbital. (c) The experiment would work for a beam of fluorine (F) atoms because it has one unpaired electron in the 2p^5 orbital.

Step by step solution

01

(a) Electron configuration for a silver atom

To find the electron configuration for a silver (Ag) atom, we can use the periodic table. Silver has an atomic number of 47, which means it has 47 electrons. The electron configuration can be obtained by filling the sublevels in order of increasing energy levels: 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^10 5p^6 6s^1 So, the electron configuration for a silver atom is [Kr] 4d^10 5s^1.
02

(b) Experiment with cadmium atoms

To answer this question, we need to find the electron configuration for a cadmium (Cd) atom. Cadmium has an atomic number of 48, which means it has 48 electrons. Its electron configuration can be obtained in the same way as for silver: 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^10 5p^6 6s^2 For a Stern-Gerlach experiment to work, the last electron must be unpaired, meaning that the electron has a spin. Silver atoms have one unpaired electron in the 6s^1 orbital, which is why the experiment works for silver. In contrast, cadmium atoms have all electrons paired in the 6s^2 orbital. Therefore, the Stern-Gerlach experiment would not work for a beam of cadmium atoms.
03

(c) Experiment with fluorine atoms

Similarly, we need to find the electron configuration for fluorine (F) to answer this question. Fluorine has an atomic number of 9, which means it has 9 electrons. Its electron configuration can be obtained as follows: 1s^2 2s^2 2p^5 Fluorine has one unpaired electron in the 2p orbital. Therefore, the Stern-Gerlach experiment would work for a beam of fluorine atoms. In summary, the Stern-Gerlach experiment would work for both silver and fluorine atoms, but it would not work for cadmium atoms.

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Key Concepts

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

Electron Configuration
Electron configuration describes the distribution of electrons in an atom's electron shells and subshells. Understanding this setup helps us determine an atom's chemical properties and behaviors, such as bonding and magnetism.
For example, silver (Ag) has the electron configuration of [Kr] 4d^{10} 5s^{1}. Here, 47 electrons fill the shells up to krypton (Kr), then fill the 4d subshell completely and place the remaining electron in the 5s orbital.
Cadmium (Cd), with an atomic number of 48, follows [Kr] 4d^{10} 5s^{2}. This means all its outer-shell electrons are paired.
In contrast, fluorine (F), having an atomic number of 9, has the electron configuration of 1s^{2} 2s^{2} 2p^{5}. This means it has one unpaired electron, making magnetism measurements possible.
  • To determine electron configuration, follow Aufbau principle: fill lower energy levels first.
  • Pauli exclusion principle ensures no two electrons in the same orbital have identical spins.
  • Hund's rule indicates that electrons spread over orbitals in the same subshell before doubling up.
Understanding electron configuration is crucial for predicting how atoms behave in various fields like chemistry and physics.
Electron Spin
Electron spin is an intrinsic property of electrons, akin to angular momentum, that is fundamental to quantum mechanics. It is responsible for many magnetic behaviors in materials. Each electron has a property called 'spin,' which can take one of two values often described as "up" (+1/2) or "down" (-1/2).

Electrons will prefer to be unpaired because opposite spins will allow them to occupy the same orbitals without repulsion, per the Pauli exclusion principle. This property explains why elements like silver, with its lone unpaired 5s electron, display magnetic properties detectable in the Stern-Gerlach experiment.
  • Electrons increase atomic stability when occupying separate orbitals with the same spin state.
  • 'Spin' is a quantum number, arising from solutions to the Schr枚dinger equation.
  • Spin plays a critical role in technologies like MRI and quantum computing.
Understanding electron spin sheds light on the magnetic characteristics and overall behavior of atoms, which is essential in both modern chemistry and physics.
Magnetic Field Deflection
In the Stern-Gerlach experiment, electron spin causes deflection when an atom passes through a magnetic field. This deflection demonstrates the quantized nature of angular momentum or "spin" of electrons.
The entire setup originated to make visible the inherent properties of quantum mechanics, particularly how electrons behave like tiny magnets because of their spin. When silver atoms with unpaired electrons, like those in the 5s^1 orbital, navigate through a magnetic field, their magnetic moments cause them to split into discrete beams.
This finding laid the groundwork for spintronics and magnetic storage technologies. Atoms like cadmium fail in such experiments because all their electrons are paired, leading to zero net magnetic moment.
  • Unpaired electrons exhibit magnetic fields that respond to external fields through deflection.
  • Deflection patterns allow for insights into atomic structure.
  • This principle is fundamental in explaining the magnetic behavior of materials.
Consequently, understanding how electron configuration and spin interact to cause magnetic field deflection facilitates advancements in material science and technology.

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Most popular questions from this chapter

Is energy emitted or absorbed when the following electronic transitions occur in hydrogen? (a) from \(n=3\) to \(n=2\), (b) from an orbit of radius \(0.846 \mathrm{nm}\) to one of radius 0.212 \(\mathrm{nm},(\mathbf{c})\) an electron adds to the \(\mathrm{H}^{+}\) ion and ends up in the \(n=2\) shell?

One type of sunburn occurs on exposure to UV light of wavelength in the vicinity of \(325 \mathrm{nm}\) (a) What is the energy of a photon of this wavelength? (b) What is the energy of a mole of these photons? (c) How many photons are in a \(1.00 \mathrm{~mJ}\) burst of this radiation? \((\mathbf{d})\) These \(\mathrm{UV}\) photons can break chemical bonds in your skin to cause sunburn-a form of radiation damage. If the \(325-\mathrm{nm}\) radiation provides exactly the energy to break an average chemical bond in the skin, estimate the average energy of these bonds in \(\mathrm{kJ} / \mathrm{mol}\).

The wavenumber \(\bar{\lambda}\) is the number of waves that exist over a specified distance, very often \(1 \mathrm{~cm}\). The wavenumber can easily be calculated by taking the reciprocal of the wavelength. Give typical wavenumbers for (a) X-rays \((\lambda=1 \mathrm{nm})\) (b) visible light \((\lambda=500 \mathrm{nm})\) (c) microwaves \((\lambda=1 \mathrm{~mm})\).

Bohr's model can be used for hydrogen-like ions-ions that have only one electron, such as \(\mathrm{He}^{+}\) and \(\mathrm{Li}^{2+} .\) (a) Why is the Bohr model applicable to \(\mathrm{Li}^{2+}\) ions but not to neutral Li atoms? (b) The ground-state energies of \(\mathrm{B}^{4+}, \mathrm{C}^{5+},\) and \(\mathrm{N}^{6+}\) are tabulated as follows: By examining these numbers, propose a relationship between the ground-state energy of hydrogen-like systems and the nuclear charge, \(Z\). (Hint: Divide by the ground-state energy of hydrogen \(\left.-2.18 \times 10^{-18} \mathrm{~J}\right)\) (c) Use the relationship you derive in part (b) to predict the ground-state energy of the \(\mathrm{Be}^{3+}\) ion.

(a) A green laser pointer emits light with a wavelength of \(532 \mathrm{nm}\). What is the frequency of this light? (b) What is the energy of one of these photons? (c) The laser pointer emits light because electrons in the material are excited (by a battery) from their ground state to an upper excited state. When the electrons return to the ground state, they lose the excess energy in the form of \(532-\mathrm{nm}\) photons. What is the energy gap between the ground state and excited state in the laser material?
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