Question
Few electrons have following quantum numbers,
$$\eqalign{
& \left( {\text{i}} \right)n = 4,l = 1 \cr
& \left( {{\text{ii}}} \right)n = 4,l = 0 \cr
& \left( {{\text{iii}}} \right)n = 3,l = 2 \cr
& \left( {{\text{iv}}} \right)n = 3,l = 1 \cr} $$
Arrange them in the order of increasing energy from lowest to highest.
A.
(iv) < (ii) < (iii) < (i)
B.
(ii) < (iv) < (i) < (iii)
C.
(i) < (iii) < (ii) < (iv)
D.
(iii) < (i) < (iv) < (ii)
Answer :
(iv) < (ii) < (iii) < (i)
Solution :
(i) 4$$p,$$ (ii) 4$$s,$$ (iii) 3$$d,$$ (iv) 3$$p,$$
The order of increasing energy
(iv) < (ii) < (iii) < (i)