
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.

Want to see the full answer?
Check out a sample textbook solution
Chapter 3 Solutions
EBK CHEMISTRY: ATOMS FIRST
- Provide the structure, circle or draw, of the monomeric unit found in the biological polymeric materials given below. HO OH amylose OH OH 행 3 HO cellulose OH OH OH Ho HOarrow_forwardWhat units (if any) does K have? Does K depend upon how the concentration is expressed (e.g. molarity, ppm, ppb, etc.)? in calculating the response factorarrow_forwardDon't used hand raiting and don't used Ai solutionarrow_forward
- Don't used Ai solution and don't used hand raitingarrow_forwardOA. For the structure shown, rank the bond lengths (labeled a, b and c) from shortest to longest. Place your answer in the box. Only the answer in the box will be graded. (2 points) H -CH3 THe b Нarrow_forwardDon't used hand raitingarrow_forward
- Quizzes - Gen Organic & Biological Che... ☆ myd21.lcc.edu + O G screenshot on mac - Google Search savings hulu youtube google disney+ HBO zlib Homework Hel...s | bartleby cell bio book Yuzu Reader: Chemistry G periodic table - Google Search b Home | bartleby 0:33:26 remaining CHEM 120 Chapter 5_Quiz 3 Page 1: 1 > 2 > 3 > 6 ¦ 5 > 4 > 7 ¦ 1 1 10 8 ¦ 9 a ¦ -- Quiz Information silicon-27 A doctor gives a patient 0.01 mC i of beta radiation. How many beta particles would the patient receive in I minute? (1 Ci = 3.7 x 10 10 d/s) Question 5 (1 point) Saved Listen 2.22 x 107 222 x 108 3.7 x 108 2.22 x 108 none of the above Question 6 (1 point) Listen The recommended dosage of 1-131 for a test is 4.2 μCi per kg of body mass. How many millicuries should be given to a 55 kg patient? (1 mCi = 1000 μСi)? 230 mCiarrow_forwardDon't used hand raiting and don't used Ai solutionarrow_forwardDon't used hand raiting and don't used Ai solutionarrow_forward
- Q3: Arrange each group of compounds from fastest SN2 reaction rate to slowest SN2 reaction rate. CI Cl H3C-Cl CI a) A B C D Br Br b) A B C Br H3C-Br Darrow_forwardQ4: Rank the relative nucleophilicity of halide ions in water solution and DMF solution, respectively. F CI Br | Q5: Determine which of the substrates will and will not react with NaSCH3 in an SN2 reaction to have a reasonable yield of product. NH2 Br Br Br .OH Brarrow_forwardClassify each molecule as optically active or inactive. Determine the configuration at each H соон Chirality center OH 애 He OH H3C Ноос H H COOH A K B.arrow_forward
- Chemistry: Principles and ReactionsChemistryISBN:9781305079373Author:William L. Masterton, Cecile N. HurleyPublisher:Cengage LearningChemistry: Matter and ChangeChemistryISBN:9780078746376Author:Dinah Zike, Laurel Dingrando, Nicholas Hainen, Cheryl WistromPublisher:Glencoe/McGraw-Hill School Pub CoIntroductory Chemistry: An Active Learning Approa...ChemistryISBN:9781305079250Author:Mark S. Cracolice, Ed PetersPublisher:Cengage Learning
- General Chemistry - Standalone book (MindTap Cour...ChemistryISBN:9781305580343Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; DarrellPublisher:Cengage LearningChemistry for Engineering StudentsChemistryISBN:9781337398909Author:Lawrence S. Brown, Tom HolmePublisher:Cengage Learning




