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

Question

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

Accepted Answer

Correct answer: B. (b) (A) n = 3; 1 = 0; m = 0 ; 3s orbital (B) n = 4; 1 = 0; m = 0 ; 4s orbital (C) n = 3; 1 = 1; m = 0 ; 3p orbital (D) n = 3; 1 = 2; m = 0 ; 3d orbital As per Hunt’s rule energy is given by (n+l) value. If value of (n+l) remains same then energy is given by n only.

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

Solution

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