
Interpretation:
The possible values of l and ml for n = 3 quantum level should be identified using the concept of quantum numbers.
Concept Introduction:
Quantum numbers are explained for the distribution of electron density in an atom. They are derived from the mathematical solution of Schrodinger’s equation for the hydrogen atom. The types of quantum numbers are the principal quantum number (n), the
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
Angular Momentum Quantum Number (l)
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.
To find: Get the possible values of l and ml for n = 3 quantum level
Find the value of ‘l’ for n = 3

Want to see the full answer?
Check out a sample textbook solution
Chapter 3 Solutions
Chemistry: Atoms First
- a) Write out 6 completely different reactions of acetophenone (reagent, product). b) Write out 3 preparations of 1-methylcyclohexanol, using a different starting material for each one. You may use preps where you just change the functional group, and/or preps where you construct the carbon chain. c) Write out 3 preparations of 2-ethoxybenzoic acid, a different starting material for each one. You may use preps where you just change the functional group, and/or preps where you construct the carbon chain.arrow_forward12. CH3 OH OH H&C CH3 H₂C N OH H₂C CH3 H&C CH3 H₂C' CH3 H.C CH3OH H.C CH2CH3OH CH3CEN Which one of these 17 compounds is represented by this IR and this 'H NMR spectrum? IR Spectrum 3000 4000 3000 NMR Spectrum 2000 £500 RAVENUMBER 2000 1500 9 8 6 5 10 HP-00-290 ppm m 1000 500 1000 4 °arrow_forwardDraw the structure of (E,6R) 6-methoxy-4-hepten-2-one. Give the IUPAC name of this compound, including stereochemistry. Draw the most stable chair conformation of (cis) 1,3-isobutylcyclohexane. H HC=CCH₂ CH2CH3 EN(CH3)2 -CN(CH3)2arrow_forward
- 10. Write out the mechanism (intermediate/transition state) for this reaction; indicate stereochemistry in product. H3C CH₂OH CH3 SN1 Harrow_forwardWrite "most" under the member of each trio which is most stable. Write "least under the member of each trio which is least stable. b) Draw a Fischer projection of a pair of enantiomers with three chiral carbons. Which of these two would you expect to be more soluble in water? Why? 1-butanol 1-heptanol Which of these two would you expect to have the higher boiling point? Why? hexyl methyl ether 1-heptanolarrow_forwardWrite "most" under the most acidic compound. Write "least" under the least acidic compound. OH NO₂ OCH3 Br 9. Compound X, C50H84F2, reacts with excess H2/Pd to give a C50H88F2 compound. How many rings are in X? How many double bonds are in X? Show your work.arrow_forward
- 4. State whether these two are: a) the same molecule b) c) d) different compounds that are not isomers constitutional isomers diastereomers e) enantiomers CH3 CH₁₂ H OH HO H H OH HO H CH, CH₂ 5. a) How many stereocenters does this compound have? b) How many stereoisomers are possible for this compound? CH₂ OH CHCHarrow_forwardCalculating the pH at equivalence of a titration A chemist titrates 210.0 mL of a 0.1003 M hydrobromic acid (HBr) solution with 0.7550M KOH solution at 25 °C. Calculate the pH at equivalence. Round your answer to 2 decimal places. Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of KOH solution added. pH = ] ☑ o0o 18 Ararrow_forwardDo you do chemistry assignmentsarrow_forward
- Using the conditions of spontaneity to deduce the signs of AH and AS Use the observations about each chemical reaction in the table below to decide the sign (positive or negative) of the reaction enthalpy AH and reaction entropy AS. Note: if you have not been given enough information to decide a sign, select the "unknown" option. reaction observations conclusions A This reaction is always spontaneous, but proceeds slower at temperatures above 120. °C. ΔΗ is (pick one) AS is (pick one) ΔΗ is (pick one) B This reaction is spontaneous except above 117. °C. AS is (pick one) ΔΗ is (pick one) This reaction is slower below 20. °C than C above. AS is |(pick one) ? 18 Ar 1arrow_forwardCalculating the pH at equivalence of a titration Try Again Your answer is incorrect. 0/5 a A chemist titrates 70.0 mL of a 0.7089 M hydrocyanic acid (HCN) solution with 0.4574M KOH solution at 25 °C. Calculate the pH at equivalence. The pK of hydrocyanic acid is 9.21. Round your answer to 2 decimal places. Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of KOH solution added. pH = 11.43] G 00. 18 Ar B•arrow_forwardBiological Macromolecules Naming and drawing the products of aldose oxidation and reduction aw a Fischer projection of the molecule that would produce L-ribonic acid if it were subjected to mildly oxidizing reaction conditions. Click and drag to start drawing a structure. X AP ‡ 1/5 Naor Explanation Check McGraw Hill LLC. All Rights Reserved. Terms of Use Privacy Center Accessibilarrow_forward
- 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 & Chemical ReactivityChemistryISBN:9781133949640Author:John C. Kotz, Paul M. Treichel, John Townsend, David TreichelPublisher:Cengage LearningChemistry: Principles and PracticeChemistryISBN:9780534420123Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward MercerPublisher:Cengage Learning
- Chemistry: An Atoms First ApproachChemistryISBN:9781305079243Author:Steven S. Zumdahl, Susan A. ZumdahlPublisher:Cengage LearningChemistry: The Molecular ScienceChemistryISBN:9781285199047Author:John W. Moore, Conrad L. StanitskiPublisher:Cengage LearningChemistry for Engineering StudentsChemistryISBN:9781337398909Author:Lawrence S. Brown, Tom HolmePublisher:Cengage Learning





