Question

The decreasing order of energy for the electron represented by the following sets of quantum number is
(1) $n=4,l=0,m=0,s=+\frac{1}{2}$
(2) $n=3,l=2,m=+1,s=-\frac{1}{2}$
(3) $n=3,l=2,m=-1,s=+\frac{1}{2}$
(4) $n=3,l=1,m=0,s=+\frac{1}{2}$
(5) $n=5,l=2,m=+1,s=-\frac{1}{2}$

• A. 5 > 3 > 2 > 4 > 1
• B. 5 > 3 = 2 > 1 > 4
• C. 5 > 2 > 3 > 4 > 1
• D. 5 > 3 = 2 > 4 > 1

Answer 1

The correct option is B 5 > 3 = 2 > 1 > 4

The relative order of energies of various orbitals in multi electron atom can be predicted with the help of (n + l) rule
(1) n + l = 4 + 0 = 4 = 4s
(2) n + l = 3 + 2 = 5 = 3d
(3) n + l = 3 + 2 = 5 = 3d
(4) n + l = 3 + 1 = 4 = 3p
(5) n + l = 5 + 2 = 7 = 3d
For an orbital, the lower the value of (n + l), the lower the energy. If (n + l) is same, the orbital with lower value of n will have lower energy.