textn and ℓ for four electrons are given as ( texti) textn = 4 , ℓ = 1 ( textii) textn = 4 , ℓ = 0 ( textiii) textn = 3 , ℓ = 2 ( textiv) textn = 3 , ℓ = 1. Arrange them in order of increasing energy from the lowest to highest:

Question

textn and ℓ for four electrons are given as ( texti) textn = 4 , ℓ = 1 ( textii) textn = 4 , ℓ = 0 ( textiii) textn = 3 , ℓ = 2 ( textiv) textn = 3 , ℓ = 1. Arrange them in order of increasing energy from the lowest to highest: (A) (ii) < (iv) < (i) < (iii) (B) (i) < (iii) < (ii) < (iv) (C) (iii) < (i) < (iv) < (ii) (D) (iv) < (ii) < (iii) < (i)

Tardigrade Admin · Accepted Answer

Lower the value of ( textn + ℓ ) , lower is the energy. When ( textn + ℓ ) value is the same for two orbitals, lower textn value implies lower energy of the electron. Thus, the order will be: (iv) < (ii) < (iii) < (i).

Q. $n$ and $ℓ$ for four electrons are given as $(i) n = 4, ℓ = 1 (ii) n = 4, ℓ = 0 (iii) n = 3, ℓ = 2 (iv) n = 3, ℓ = 1.$
Arrange them in order of increasing energy from the lowest to highest:

(ii) < (iv) < (i) < (iii)

(i) < (iii) < (ii) < (iv)

(iii) < (i) < (iv) < (ii)

(iv) < (ii) < (iii) < (i)

Lower the value of $(n + ℓ)$ , lower is the energy. When $(n + ℓ)$ value is the same for two orbitals, lower $n$ value implies lower energy of the electron. Thus, the order will be:
(iv) < (ii) < (iii) < (i).

Q. n and ℓ for four electrons are given as (i)n=4,ℓ=1(ii)n=4,ℓ=0(iii)n=3,ℓ=2(iv)n=3,ℓ=1. Arrange them in order of increasing energy from the lowest to highest: