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Solution


Q.

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:                          [2023]
1 D > B > C > A  
2 B > D > C > A  
3 A > C > B > D  
4 D > B > A > C  

Ans.

(1)

In multi-electronic species, energy is decided on the basis of (n+l) rule. So increasing order of energy is

3d>4s>3p>3s