Question

14. The allowable combinations of quantum
numbers for each of the electron in 4s, 3p, 5d
orbitals respectively
(1) n = 4,1 = 0, m, = +1; n = 3,1 = 2,
m, = 1; n = 5,1 = 3, m, = 0
(2) n = 4,1 = 0, m, = 0; n = 3, 1 = 2, m, = -1
; n = 5, 1 = 3, m, = -2
= 0;n
(3) n = 4, 1 = 0, m, = 0; n = 3, 1 = 1, m,
F5, 1 = 2, m, = -1
-
0; n = 3, 1 = 0, m,
0:
(4) n = 4,1 = 0, m
n = 5, 1 = 1, m, = 0​

Answer 1

Explanation:

Dekh Lo we'll first energy equal and in combination 21 numbers of electron 4312 there is two electron and magnesium