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14 October, 16:21

The following quantum number combinations are not allowed. Assuming the n and ml values are correct, change the l value to create an allowable combination:

(a) n = 3; l = 0; ml = - 1

(b) n = 3; l = 3; ml = + 1

(c) n = 7; l = 2; ml = + 3

(d) n = 4; l 0 1; ml = - 2

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Answers (2)
  1. 14 October, 18:32
    0
    (a) n = 3; l = 1; ml = - 1

    (b) n = 3; l = 1; ml = + 1

    (c) n = 7; l = 3; ml = + 3

    (d) n = 4; l = 2; ml = - 2

    This is assuming the ml are largest possible + / - values.
  2. 14 October, 18:39
    0
    (a) n = 3; l = 1; ml = - 1

    (b) n = 3; l = 1; ml = ±1

    (c) n = 7; l = 3; ml = + 3

    (d) n = 4; l = 2; ml = - 2

    Explanation:

    The rules for electron quantum numbers are:

    1. Shell number, 1 ≤ n

    2. Subshell number, 0 ≤ l ≤ n - 1

    3. Orbital energy shift, - l ≤ ml ≤ l

    4. Spin, either - 1/2 or + 1/2

    So apply in our cases,

    (a). ml = - 1, so l must be at least 1

    ⇒ n = 3; l = 1; ml = - 1

    (b). n = 3, so l must be less than 3, 2 or 1 is fine because ml = ±1

    ⇒ n = 3; l = 1; ml = ±1

    (c) ml = ± 3, so l must be at least 3, n = 7 so l can be 3, 4, 5 or 6

    ⇒ n = 7; l = 3; ml = + 3

    (d) ml = - 2, so l must be at least 2, n = 4 so l can be 2 or 3

    ⇒ n = 4; l = 2; ml = - 2
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