Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Explain electron from a quantum mechanical perspective, including a discussion of atomic radii, probabilities, and orbitals.

Short Answer

Expert verified
From a quantum mechanical perspective, electrons exhibit wave-particle duality and can only be described by probabilities. Atomic radii represent the average distance between an atom's nucleus and its outermost electrons, with effective atomic radius accounting for the distribution of electrons in orbitals and the effective nuclear charge. Electron orbitals are regions defined by wave functions that describe the probability distribution of electrons around the nucleus, with each orbital shape (s, p, d, and f) corresponding to different quantum numbers. The electron probability density helps visualize the regions where electrons are most likely to be found, ultimately providing insights into the behavior of electrons in atoms.

Step by step solution

01

Quantum Mechanics and Electrons

Quantum mechanics is a fundamental theory in physics that deals with the behavior of particles on a very small scale, such as electrons in atoms. In classical mechanics, particles like electrons are considered as particles with definite positions and velocities. However, quantum mechanics proposes that particles like electrons behave both as particles and waves, meaning that their exact position and velocity cannot be precisely determined simultaneously. This is called wave-particle duality and is described by the Heisenberg Uncertainty Principle.
02

Atomic Radii and Electrons

In the context of quantum mechanics, the term atomic radius describes the average distance between the nucleus of an atom and its outermost electrons. As electrons in an atom are not in fixed positions, we can't precisely determine the actual radius of an atom. Instead, we often use the concept of an effective atomic radius. The atomic radius depends on the number of electrons in the atom, their distribution in orbitals, and the effective nuclear charge experienced by the electrons.
03

Electron Orbitals and Probabilities

Electron orbitals are the regions in an atom where electrons are most likely to be found. They are described by wave functions, which are mathematical functions that give the probability distribution of an electron around the nucleus. An electron's wave function is determined by three quantum numbers (n, l, and m), which arise from the solution of the Schrödinger equation for the hydrogen atom. These quantum numbers define the size, shape, and orientation of the orbitals. The orbitals are categorized into different shapes: s, p, d, and f orbitals. Each shape corresponds to different values of the quantum numbers: - s orbitals (l=0) are spherical - p orbitals (l=1) have a dumbbell shape - d orbitals (l=2) have more complex shapes - f orbitals (l=3) are even more complex The probability of finding an electron in an orbital is given by the square of the magnitude of the wave function at a given point. This is known as the electron probability density. The electron probability density helps visualize the regions in space where electrons are most likely to be found.
04

Putting It All Together

In conclusion, an electron from a quantum mechanical perspective exhibits wave-particle duality and can only be described by probabilities. Atomic radii provide an estimate of the average distance between an atom's nucleus and its outermost electrons. Electron orbitals are regions of space defined by wave functions, where electrons are most likely to be found. The electron probability density provides a visualization of these regions and is critical for understanding the behavior of electrons in atoms.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A carbon-oxygen double bond in a certain organic molecule absorbs radiation that has a frequency of \(6.0 \times 10^{13} \mathrm{s}^{-1}\) . a. What is the wavelength of this radiation? b. To what region of the spectrum does this radiation belong? c. What is the energy of this radiation per photon? per mole of photons? d. A carbon-oxygen bond in a different molecule absorbs radiation with frequency equal to \(5.4 \times 10^{13} \mathrm{s}^{-1} .\) Is this radiation more or less energetic?

Arrange the following groups of atoms in order of increasing size. a. Te, S, Se b. K, Br, Ni c. Ba, Si, F

It takes 208.4 kJ of energy to remove 1 mole of electrons from an atom on the surface of rubidium metal. How much energy does it take to remove a single electron from an atom on the surface of solid rubidium? What is the maximum wavelength of light capable of doing this?

Small daily doses of lithium carbonate taken orally are often given to treat manic-depressive psychoses. This dosage maintains the level of lithium ion in the blood at about \(1 \times 10^{-3} \mathrm{mol} / \mathrm{L}\) . a. What is the formula of lithium carbonate? b. What mass of lithium ion is present per liter of blood in these patients?

Neutron diffraction is used in determining the structures of molecules. a. Calculate the de Broglie wavelength of a neutron moving at 1.00\(\%\) of the speed of light. b. Calculate the velocity of a neutron with a wavelength of 75 \(\mathrm{pm}\left(1 \mathrm{pm}=10^{-12} \mathrm{m}\right)\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free