The quantum numbers of four electrons are given below : I. n = 4; l = 2; ml = -2; s = - 12 II. n = 3; l = 2; ml = 1; s = + 12 III. n = 4; l = 1; ml = 0; s = + 12 IV. n = 3; l = 1; ml = -1; s = + 12 The correct decreasing order of energy of these electrons is
- IV > II > III > I
- I > III > II > IV
- III > I > II > IV
- I > II > III > IV
Answer
(2) I > III > II > IV
Solution
Energy of an orbital in a multi-electron atom increases with (n + l); for equal (n + l), the one with higher n has higher energy.
I: 4d, n + l = 6; II: 3d, n + l = 5; III: 4p, n + l = 5; IV: 3p, n + l = 4.
II and III both have n + l = 5; III (n = 4) > II (n = 3).
I > III > II > IV.
Answer: (2)