In defining the sizes of orbitals, why must we use an arbitrary value, such as \(90 \%\) of the probability of finding an electron in that region?

Short Answer

Expert verified
In short, the use of an arbitrary value such as 90% probability when defining the size of orbitals stems from the inability to precisely determine electron positions. This value provides a reasonable balance between coverage and simplicity, allowing us to visualize orbitals while representing most of the electron's probability distribution. The choice of 90% is arbitrary but widely accepted in the field due to its balance between accuracy and ease of understanding.

Step by step solution

01

Understanding orbitals

An orbital is a region in an atom where an electron with a certain energy level is most likely to be found. They are usually represented by shapes and symbols (such as s, p, d, f) and are determined by the wave function of electrons in quantum mechanics. However, it is important to note that electrons in orbitals do not follow fixed paths, and their exact position cannot be precisely determined.
02

Electron probability distribution

Since it is impossible to determine the exact location of an electron due to the uncertainty principle in quantum mechanics, we rely on the probability distribution of electrons. The probability distribution represents where an electron is most likely to be found by considering the wave function. In other words, electron probability describes the likelihood of finding an electron in a particular region of space.
03

The need for an arbitrary value

When we attempt to represent orbitals graphically or define their sizes, we need to set a boundary to display them. The problem lies in the nature of electron probability distributions, which diminish gradually and have no sharp borders. Therefore, an arbitrary value, such as 90% probability, is needed to represent orbitals' form or size while still having most of the electron's probability density covered.
04

Significance of 90% probability

The choice of 90% probability is arbitrary, but it is widely accepted because it provides a reasonable balance between coverage and simplicity. By considering 90% of the electron density probability, orbitals can be visualized and understood relatively easily while still representing most of the electron's probability distribution. Other values could be chosen, but they might make the orbital representation too small or overly complicated to visualize and understand. In summary, the use of an arbitrary value like 90% probability when defining the size of orbitals is necessary because it allows us to represent and visualize an orbital's shape while taking into account the probability distribution of electrons to a considerable extent. It strikes a balance between simplicity and accuracy.

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Most popular questions from this chapter

Cesium was discovered in natural mineral waters in 1860 by R. W. Bunsen and G. R. Kirchhoff using the spectroscope they invented in \(1859 .\) The name came from the Latin caesius ("sky blue") because of the prominent blue line observed for this element at \(455.5 \mathrm{~nm} .\) Calculate the frequency and energy of a photon of this light.

One type of electromagnetic radiation has a frequency of \(107.1\) MHz, another type has a wavelength of \(2.12 \times 10^{-10} \mathrm{~m}\), and another type of electromagnetic radiation has photons with energy equal to \(3.97 \times 10^{-19} \mathrm{~J} /\) photon. Identify each type of electromagnetic radiation and place them in order of increasing photon energy and increasing frequency.

Write the expected electron configurations for each of the following atoms: \(\mathrm{Cl}, \mathrm{Sb}, \mathrm{Sr}, \mathrm{W}, \mathrm{Pb}, \mathrm{Cf}\).

An electron is excited from the \(n=1\) ground state to the \(n=3\) state in a hydrogen atom. Which of the following statements are true? Correct the false statements to make them true. a. It takes more energy to ionize (completely remove) the electron from \(n=3\) than from the ground state. b. The electron is farther from the nucleus on average in the \(n=3\) state than in the \(n=1\) state. c. The wavelength of light emitted if the electron drops from \(n=3\) to \(n=2\) will be shorter than the wavelength of light emitted if the electron falls from \(n=3\) to \(n=1\). d. The wavelength of light emitted when the electron returns to the ground state from \(n=3\) will be the same as the wavelength of light absorbed to go from \(n=1\) to \(n=3\). e. For \(n=3\), the electron is in the first excited state.

Answer the following questions based on the given electron configurations and identify the elements. a. Arrange these atoms in order of increasing size: \([\mathrm{Kr}] 5 s^{2} 4 d^{10} 5 p^{6} ;[\mathrm{Kr}] 5 s^{2} 4 d^{10} 5 p^{1} ;[\mathrm{Kr}] 5 s^{2} 4 d^{10} 5 p^{3}\) b. Arrange these atoms in order of decreasing first ionization energy: \([\mathrm{Ne}] 3 s^{2} 3 p^{5} ;[\mathrm{Ar}] 4 s^{2} 3 d^{10} 4 p^{3} ;[\mathrm{Ar}] 4 s^{2} 3 d^{10} 4 p^{5}\)

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