
Interpretation: The possible orbitals in an atom should be determined.
Concept Introduction: The distribution of electrons in an atom is described by the use of quantum numbers used by the
The forth quantum number is spin quantum number which is denoted by symbol s and determined the spin of the electron if it is clockwise or anticlockwise.
The all four quantum numbers are explained in detail as follows:
Principal quantum number (n) determines the size of the orbital. The orbital with principal quantum number 2 is larger than the value 1. This also determines the energy of the orbital, as the value of n increases the energy of the orbital increases as its distance from the nucleus increases.
Angular quantum number (l) determined the shape of the orbital. For
The rules for the allowed quantum numbers combinations are as follows:
- All the three quantum numbers ( n, l and m ) describes the orbital of an atom and they are integers.
- The principal quantum number, n value cannot be zero. Thus, the values allowed for the principal quantum number are 1, 2, 3, 4, and so on.
- The value of angular quantum number, l can be between 0 to n-1. Thus, if value of n is equal to 3 the value of l can be 0, 1 or 2.
- The value of magnetic quantum number, m can be between − l to +l . Thus, if value of l is equal to 2, m can be wither -2, -1, 0, +1, or +2
lFor same number of principal quantum number, an orbital form a shell. The first character denotes the shell and the second identifies the sub-shell.
Here, for s orbital value of

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Chapter 5 Solutions
EBK BASIC CHEMISTRY
- H-Br Energy 1) Draw the step-by-step mechanism by which 3-methylbut-1-ene is converted into 2-bromo-2-methylbutane. 2) Sketch a reaction coordinate diagram that shows how the internal energy (Y- axis) of the reacting species change from reactants to intermediate(s) to product. Brarrow_forward2. Draw the missing structure(s) in each of the following reactions. The missing structure(s) can be a starting material or the major reaction product(s). C5H10 H-CI CH2Cl2 CIarrow_forwardDraw the products of the stronger acid protonating the other reactant. དའི་སྐད”“ H3C OH H3C CH CH3 KEq Product acid Product basearrow_forward
- Draw the products of the stronger acid protonating the other reactant. H3C NH2 NH2 KEq H3C-CH₂ 1. Product acid Product basearrow_forwardWhat alkene or alkyne yields the following products after oxidative cleavage with ozone? Click the "draw structure" button to launch the drawing utility. draw structure ... andarrow_forwardDraw the products of the stronger acid protonating the other reactant. H3C-C=C-4 NH2 KEq CH H3C `CH3 Product acid Product basearrow_forward
- 2. Draw the missing structure(s) in each of the following reactions. The missing structure(s) can be a starting material or the major reaction product(s). C5H10 Br H-Br CH2Cl2 + enant.arrow_forwardDraw the products of the stronger acid protonating the other reactant. KEq H₂C-O-H H3C OH Product acid Product basearrow_forwardDraw the products of the stronger acid protonating the other reactant. OH KEq CH H3C H3C `CH3 Product acid Product basearrow_forward
- 2. Draw the missing structure(s) in each of the following reactions. The missing structure(s) can be a starting material or the major reaction product(s). Ph H-I CH2Cl2arrow_forward3 attempts left Check my work Draw the products formed in the following oxidative cleavage. [1] 03 [2] H₂O draw structure ... lower mass product draw structure ... higher mass productarrow_forward2. Draw the missing structure(s) in each of the following reactions. The missing structure(s) can be a starting material or the major reaction product(s). H-Br CH2Cl2arrow_forward
- Chemistry & Chemical ReactivityChemistryISBN:9781337399074Author:John C. Kotz, Paul M. Treichel, John Townsend, David TreichelPublisher:Cengage LearningChemistry & Chemical ReactivityChemistryISBN:9781133949640Author:John C. Kotz, Paul M. Treichel, John Townsend, David TreichelPublisher:Cengage Learning

