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2. One among the following set of quantum numbers defines the highest energy electron in scandium (I) ion
(A) n = 3, l = 1, m = 0, s = –1/2
(B) n = 3, l = 0, m = 0, s = –1/2
(C) n = 4, l = 0, m = 0, s = +1/2
(D) n = 3, l = 2, m = 2, s = +1/2.
21Se+ ion has conf. = 1s2 2s2 2p6 3s2 3d1 4s1.
Highest energy electron is 4s.
During ionization such electron is removed fist from 4s orbital only. Although 3d is filled after 4s.
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