Arrange the following orbitals in decreasing order of energy. A. n = 3 , l = 0 , m = 0 B. n = 4 , l = 0 , m = 0 C. n = 3 , l = 1 , m = 0 D. n = 3 , l = 2 , m = 1 The correct option for the order is:

Step-by-step solution

In multi-electronic species, energy is decided on the basis of ( n + l ) rule. 3 s   =   3   +   0   +   0 =   3 4 s   =   4   +   0   + 0   =   4 3 p   =   3   +   1   + 0   =   4 3 d   =   3   +   2   +   1   =   6 For 4s and 3p, an orbital with a higher value of n has higher energy. So increasing the order of energy is 3 d > 4 s > 3 p > 3 s .