Question

Compare the energies of following sets of quantum numbers for multielectron system. 

(A) n = 4, 1 = 1 

(B) n = 4, l = 2 

(C) n = 3, l = 1 

(D) n = 3, l = 2 

(E) n = 4, 1 = 0 

Choose the correct answer from the options given below :

(1) (B) > (A) > (C) > (E) > (D) 

(2) (E) > (C) < (D) < (A) < (B) 

(3) (E) > (C) > (A) > (D) > (B) 

(4) (C) < (E) < (D) < (A) < (B)

Answer 1

Correct option is (4) (C) < (E) < (D) < (A) < (B)

Energy level can be determined by comparing (n + l) values

(A) n = 4, l = 1 ⇒ (n + l) = 5 

(B) n = 4, l = 2 ⇒ (n + l) = 6 

(C) n = 3, l = 1 ⇒ (n + l) = 4 

(D) n = 3, l = 2 ⇒ (n + l) = 5 

(E) n = 4, l = 0 ⇒ (n + l) = 4 

For same value of (n + l), orbital having higher value of n, will have more energy. 

(B) > (A) > (D) > (E) > (C)