Question

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

Answer 1

Answer:

(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

Answer 2

Answer:

Explanation:

(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.