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Interpretation:
The four quantum numbers which is applied to characterize an electron in an atom should be described.
Concept Introduction:
The electron density gives the probability of finding an electron in a particular region in an atom. An atomic orbital is the region of three-dimensional space, defined by ψ2 (the square of the wave function, ψ), where the probability of finding an electron is high. An atomic orbital can accommodate a maximum of two electrons.
A wave function for an electron in an atom is called an atomic orbital; this atomic orbital describes a region of space in which there is a high probability of finding the electron. Energy changes within an atom are the result of an electron changing from a wave pattern with one energy to a wave pattern with a different energy (usually accompanied by the absorption or emission of a photon of light).
Each electron in an atom is described by four different quantum numbers. The first three (n, l, ml) specify the particular orbital of interest, and the fourth (ms) specifies how many electrons can occupy that orbital.
Principal Quantum Number (n)
The principal quantum number (n) assigns the size of the orbital and specifies the energy of an electron. If the value of n is larger, then the average distance of an electron in the orbital from the nucleus will be greater. Therefore the size of the orbital is large. The principal quantum numbers have the integral values of 1, 2, 3 and so forth and it corresponds to the quantum number in
The angular momentum quantum number (l) explains the shape of the atomic orbital. The values of l are integers which depend on the value of the principal quantum number, n. For a given value of n, the possible values of l range are from 0 to n − 1. If n = 1, there is only one possible value of l (l=0). If n = 2, there are two values of l: 0 and 1. If n = 3, there are three values of l: 0, 1, and 2. The value of l is selected by the letters s, p, d, and f. If l = 0, we have an s orbital; if l = 1, we have a p orbital; if l = 2, we have a d orbital and finally if l = 3, we have a f orbital. A collection of orbitals with the same value of n is called a shell. One or more orbitals with the same n and l values are referred to a subshell (sublevel). The value of l also has a slight effect on the energy of the subshell; the energy of the subshell increases with l (s < p < d < f).
Magnetic Quantum Number (ml)
The magnetic quantum number (ml ) explains the orientation of the orbital in space. The value of ml depends on the value of l in a subshell. This number divides the subshell into individual orbitals which hold the electrons. For a certain value of l, there are (2l + 1) integral values of ml which is explained as follows:
ml = ‒ l, ..., 0, ..., +l
If l = 0, there is only one possible value of ml: 0.
If l = 1, then there are three values of ml: −1, 0, and +1.
If l = 2, there are five values of ml, namely, −2, −1, 0, +1, and +2.
If l = 3, there are seven values of ml, namely, −3, −2, −1, 0, +1, +2, and +3, and so on.
The number of ml values indicates the number of orbitals in a subshell with a particular l value. Therefore, each ml value refers to a different orbital.
Electron Spin Quantum Number (ms)
It specifies the orientation of the spin axis of an electron. An electron can spin in only one of two directions. There are two possible ways to represent ms values. They are +½ and ‒½. One electron spins in the clockwise direction. Another electron spins in the anticlockwise direction. But, no two electrons should have the same spin quantum number.
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Chapter 3 Solutions
Chemistry Atoms First, Second Edition
- 2. 200 LOD For an unknown compound with a molecular ion of 101 m/z: a. Use the molecular ion to propose at least two molecular formulas. (show your work) b. What is the DU for each of your possible formulas? (show your work) C. Solve the structure and assign each of the following spectra. 8 6 4 2 (ppm) 150 100 50 ō (ppm) 4000 3000 2000 1500 1000 500 HAVENUMBERI-11arrow_forwardComplete the spectroscopy with structurearrow_forwardComplete the spectroscopy with structurearrow_forward
- Given the following concentrations for a system, calculate the value for the reaction quotient: Cl2(g)+ CS2(g) ⇌ CCl4(g)+ S2Cl2(g) Cl2 = 31.1 atm CS2 = 91.2 atm CCl4 = 2.12 atm S2Cl2 = 10.4 atmarrow_forwardMatch each chemical or item with the proper disposal or cleanup mwthod, Not all disposal and cleanup methods will be labeled. Metal sheets C, calcium, choroide solutions part A, damp metal pieces Part B, volumetric flask part A. a.Return to correct lables”drying out breaker. Place used items in the drawer.: Rinse with deionized water, dry as best you can, return to instructor. Return used material to the instructor.: Pour down the sink with planty of running water.: f.Pour into aqueous waste container. g.Places used items in garbage.arrow_forwardWrite the equilibrium constant expression for the following reaction: HNO2(aq) + H2O(l) ⇌ H3O+(aq) + NO2-(aq)arrow_forward
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