(a)
Interpretation:
The orbital which is lower in energy in many electron atoms should be identified in the given pairs of orbitals.
Concept Introduction:
Energy of an orbital in a many electron atom depends on both the values of principle quantum number (n) and
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).
To find: Identify the orbital which is lower in energy in the given pair 2s, 2p orbitals of many electron atoms
Find the value of ‘n’
The principle quantum number (n) of the 2s, 2p orbitals is 2. Find the value of ‘l’
(b)
Interpretation:
The orbital which is lower in energy in many electron atoms should be identified in the given pairs of orbitals.
Concept Introduction:
Energy of an orbital in a many electron atom depends on both the values of principle quantum number (n) and angular momentum quantum number (l). For a given value of principle quantum number (n), the energy of orbital increases with increasing value of the angular momentum quantum number (l) in a many electron atom.
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 Bohr’s model of the hydrogen atom. If all orbitals have the same value of ‘n’, they are said to be in the same shell (level). The total number of orbitals for a given n value is n2. As the value of ‘n’ increases, the energy of the electron also increases.
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).
To find: Identify the orbital which is lower in energy in the given pair 3p, 3d orbitals of many electron atoms
Find the value of ‘n’
The principle quantum number (n) of the 3p, 3d orbitals is 3. Find the value of ‘l’
(c)
Interpretation:
The orbital which is lower in energy in many electron atoms should be identified in the given pairs of orbitals.
Concept Introduction:
Energy of an orbital in a many electron atom depends on both the values of principle quantum number (n) and angular momentum quantum number (l). For a given value of principle quantum number (n), the energy of orbital increases with increasing value of the angular momentum quantum number (l) in a many electron atom.
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 Bohr’s model of the hydrogen atom. If all orbitals have the same value of ‘n’, they are said to be in the same shell (level). The total number of orbitals for a given n value is n2. As the value of ‘n’ increases, the energy of the electron also increases.
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).
To find: Identify the orbital which is lower in energy in the given pair 3s, 4s orbitals of many electron atoms Find the value of ‘n’
The principle quantum number (n) of the 3s orbital is 3 whereas the principle quantum number (n) of the 4s orbital is 4. Hence 3s orbital is lower in energy than 4s orbital as the 3s orbital has the lower value of ‘n’ than 4s orbital. Find the value of ‘l’
(d)
Interpretation:
The orbital which is lower in energy in many electron atoms should be identified in the given pairs of orbitals.
Concept Introduction:
Energy of an orbital in a many electron atom depends on both the values of principle quantum number (n) and angular momentum quantum number (l). For a given value of principle quantum number (n), the energy of orbital increases with increasing value of the angular momentum quantum number (l) in a many electron atom.
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 Bohr’s model of the hydrogen atom. If all orbitals have the same value of ‘n’, they are said to be in the same shell (level). The total number of orbitals for a given n value is n2. As the value of ‘n’ increases, the energy of the electron also increases.
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).
To find: Identify the orbital which is lower in energy in the given pair 4d, 5f orbitals of many electron atoms Find the value of ‘n’
The principle quantum number (n) of the 4d orbital is 4 whereas the principle quantum number (n) of the 5f orbital is 5. Hence 4d orbital is lower in energy than 5f orbital as the 4d orbital has the lower value of ‘n’ than 5f orbital. Find the value of ‘l’

Want to see the full answer?
Check out a sample textbook solution
Chapter 7 Solutions
EBK GENERAL CHEMISTRY: THE ESSENTIAL CO
- C. I, II, III Consider the reaction sequence below to answer the following questions: 0 0 1. NaOEt, EtOH ΕΙΟ OEt 2 Compound X CO₂Et NaOEt, EtOH CO₂Et Br Compound Y A Compound Z A. Compound X, diethyl propanedioate, is more commonly known as a. ethyl acetoacetate acetoacetic ester b. C. oxalic ester d. malonic ester B. Write the complete stepwise mechanism for the conversion of Compound X into Compound Y. Show all electron flow with arrows and draw all intermediate structures.arrow_forwardDiethyl malonate can be prepared by the following reaction sequence. Draw the structures of each of the missing intermediates in the boxes provided EtO 0 H3C 11 C 1. Br₂ PBr OH 2 H₂O 010 0 CH3CH₂OH C CH2 OEt Ha CH3CH2OH на NaCN H₂SO4 NC H₂O, heat CH2 OCH2CH3arrow_forwardShow how you would accomplish each of the following transformations. More than one step may be quired. Show all reagents and all intermediate structures. [three only] A. 0 CH3 B. C. D. H 0 0 OCH 3 CH₂CO₂CH2CH3 H3C ➤ HN C NO₂ Clarrow_forward
- Choose the BEST reagent for carrying out each of the following conversions. A. CO₂CH3 CO₂CH3 0 CO₂H a. LiAlH4, ether C. CrO3, pyridine B. 0 H a. C. NaBH4, ethanol NaOH, H2O CO₂H OH HD b. NaBH4, ethanol d. H₂/Pd CH₂OH b. CH₂PPh3 d. All of the abovearrow_forwardWrite the complete stepwise mechanism for the acid-catalyzed hydrolysis of the following amide to yield mandelic acid. Show all electron flow with arrows and draw the structures of all intermediate species. OH H-OH₂ CnH2 :0: OH C OH + NH4 10: The purpose of the acid catalyst in the hydrolysis of an amide is: to enhance the electrophilicity of the amide carbonyl carbon a. to enhance the nucleophilicity of the water molecule b. C. to enhance the electrophilicity of the water molecule d. to shift the equilibrium of the reactionarrow_forward1.arrow_forward
- Can I please get help with this?arrow_forward. Provide IUPAC names for each of the following structures OR draw structures corresponding to each of the following names: [Three only]kk a. H₂N- 0 COCH2CH3 benzocaine b. What is the correct structure for phenylbenzoate? C a. 0 C-O O b. H3C-C-O 0 0 C-O-CH3 d. CH₂O C-CHZ c. Acetyl chloride d. 3,4,5-trimethoxybenzoyl chloridearrow_forward. Draw structures corresponding to each of the following names or Provide IUPAC names for each of the ollowing structures [for 4 ONLY]. A. 2-propylpentanoic acid. B. m-chlorobenzoic acid. C. 0 0 HOC(CH2) COH glutaricadd D. E. F. 0 OH HO OH HO INCO salicylicadd H3C CH3 C=C tgicadd H COOH CH₂C=N 4arrow_forward
- The reaction of a carboxylic acid with an alcohol in the presence of acid is termed Fischer esterification. 0 0 C .C. OH + CH3OH OCH3 + H₂O HCI A B C A. The nucleophile in this reaction is B. Compound C functions as a. a base scavenger b. a solvent C. a catalyst in this reaction. d. a neutralizer C. Fischer esterification is an example of: ........ a. nucleophilic acyl addition b. nucleophilic acyl substitution c. nucleophilic acyl elimination d. nucleophilic acyl rearrangementarrow_forwardThe Handbook of Chemistry and Physics gives solubilities of the following compounds in grams per 100 mL of water. Because these compounds are only slightly soluble, assume that the volume does not change on dissolution and calculate the solubility product for each. (a) BaSeO4, 0.0118 g/100 mLarrow_forwardCan I please get help with answering this?arrow_forward
- ChemistryChemistryISBN:9781305957404Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCostePublisher:Cengage LearningChemistryChemistryISBN:9781259911156Author:Raymond Chang Dr., Jason Overby ProfessorPublisher:McGraw-Hill EducationPrinciples of Instrumental AnalysisChemistryISBN:9781305577213Author:Douglas A. Skoog, F. James Holler, Stanley R. CrouchPublisher:Cengage Learning
- Organic ChemistryChemistryISBN:9780078021558Author:Janice Gorzynski Smith Dr.Publisher:McGraw-Hill EducationChemistry: Principles and ReactionsChemistryISBN:9781305079373Author:William L. Masterton, Cecile N. HurleyPublisher:Cengage LearningElementary Principles of Chemical Processes, Bind...ChemistryISBN:9781118431221Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. BullardPublisher:WILEY





