Four electrons in a multi-electron atom are characterized by the following sets of quantum numbers: (A) n = 4 , l = 1 (B) n = 4 , l = 0 (C) n = 3 , l = 2 (D) n = 3 , l = 1 Which of these electrons possesses the highest energy?

Step-by-step solution

The energy of an electron in a multi-electron atom is determined by the (n+l) rule. Let us calculate the (n+l) value for each electron: (A) n = 4 , l = 1 (n+l) = 4 + 1 = 5 (B) n = 4 , l = 0 (n+l) = 4 + 0 = 4 (C) n = 3 , l = 2 (n+l) = 3 + 2 = 5 (D) n = 3 , l = 1 (n+l) = 3 + 1 = 4 Electrons with a higher (n+l) value have higher energy. Here, electrons (A) and (C) both have the highest (n+l) value of 5 . According to the rule, if two orbitals have the same (n+l) value, the one with the higher principal quantum number