What is the only possible value of ml for an electron in an s orbital?

Question

Part A

What is the only possible value of ml for an electron in an s orbital?

What are the possible values of ml for an electron in a d orbital?

Part B

Which of the following set of quantum numbers (ordered n, l, ml, ms) are possible for an electron in an atom?

Check all that apply.

a. 5, 3, 0, 1/2

b. 3, 3, 1, -1/2

c. 3, 2, -3, 1/2

d. 4, 2, 3, -1/2

e. 3, 2, 1, -1

f. 4, 2, -1, -1/2

g. -2, 1, 0, -1/2

h. 3, 1, 0, -1/2

1

Answer ( 1 )

  1. Quantum numbers:
    1) Principal Quantum number – It is indicated by “n”. It demonstrates the quantity of shells.
    2) Azimuthal Quantum number – It is indicated by “l”. It demonstrates the orbital rakish energy and the state of the orbitals. Its range is from 0 to n-1
    3) Magnetic Quantum number – It is signified by “mlml “. It demonstrates the direction in space. Its range is – l to + l
    4) Spin Quantum number – It is meant by “msms “. It shows the turn of the electron.
    Answer and Explanation:
    Section (a) :
    The estimations of various azimuthal quantum numbers ( l )are :
    s = 0
    p = 1
    d = 2
    f = 3
    The estimation of “mlml is – l to + l.
    Since the estimation of l for s orbital is 0, subsequently the main conceivable estimation of “mlml is zero.
    The estimation of “mlml is – l to + l.
    Since the estimation of l for d orbital is 2, subsequently the potential estimations of “mlml are – 2, – 1 , 0 , 1 and 2 .
    Part (b):
    (a) n = 5, l can be 0 to (5-1) i.e 0,1,2,3 and 4. The given estimation of l is 3 which is right.
    The potential estimations of ml are – l to +l i.e – 3 to +3. The given estimation of ml is 0 which is right.
    The estimation of ms can be – 1/2 and +1/2.
    Along these lines, every one of the qualities are right, henceforth the given arrangement of Quantum numbers is legitimate.
    (b) n = 3, l can be 0 to (3-1) i.e 0,1 and 2. The given estimation of l is 3 which is absurd.
    Consequently, the given arrangement of Quantum numbers is legitimate.
    (c) n = 3, l can be 0 to (3-1) i.e 0,1 and 2. The given estimation of l is 2 which is right.
    The potential estimations of ml are – l to +l i.e – 2 to +2. The given estimation of ml is – 3 which is mistaken.
    In this manner, the given arrangement of Quantum numbers is invalid.
    (d) (a) n = 4, l can be 0 to (4-1) i.e 0,1,2 and 3. The given estimation of l is 2 which is right.
    The potential estimations of ml are – l to +l i.e – 2 to +2. The given estimation of ml is 3 which is mistaken.
    In this manner, the given arrangement of Quantum numbers is invalid.
    (e) n = 3, l can be 0 to (3-1) i.e 0,1 and 2. The given estimation of l is 2 which is right.
    The potential estimations of ml are – l to +l i.e – 2 to +2. The given estimation of ml is 1 which is right.
    The estimation of ms can be – 1/2 and +1/2. The given estimation of ms is – 1, which is off base.
    Subsequently, the given arrangement of Quantum numbers is invalid.
    (f) n = 4, l can be 0 to (4-1) i.e 0,1,2 and 3. The given estimation of l is 2 which is right.
    The potential estimations of ml are – l to +l i.e – 2 to +2. The given estimation of ml is – 1 which is right.
    The estimation of ms can be – 1/2 and +1/2. Subsequently, every one of the qualities are right, thus the given arrangement of Quantum numbers is legitimate.
    (g) n = – 2, l can be sure just. The given estimation of l is mistaken. Hence, the given arrangement of Quantum numbers is invalid.
    (h) n = 3, l can be 0 to (3-1) i.e 0,1 and 2 . The given estimation of l is 1 which is off base. The potential estimations of ml are – l to +l i.e – 1 to +1. The given estimation of ml is 0 which is right.
    The estimation of ms can be – 1/2 and +1/2.
    In this way, every one of the qualities are right, subsequently the given arrangement of Quantum numbers is substantial.

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