The four quantum numbers for the electron in the outermost orbital of potassium (atomic no. 19) are
- n = 3, l = 0, m = 1, s = + 12
- n = 4, l = 0, m = 0, s = + 12
- n = 2, l = 0, m = 0, s = + 12
- n = 4, l = 2, m = -1, s = + 12
Answer
(2) n = 4, l = 0, m = 0, s = + 12
Solution
K (Z = 19): 1s2 2s2 2p6 3s2 3p6 4s1 ⇒ outermost electron in 4s
For 4s: n = 4, l = 0 ⇒ m = 0 (only value for l = 0), s = + 12
(A): l = 0 with m = 1 is impossible; (C): n = 2 is an inner shell; (D): l = 2 is a d orbital ⇒ all wrong
Answer: (2)