Question

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 :

• A. $D > B > C > A$
• B. $B > D > C > A$
• C. $A > C > B > D$
• D. $D > B > A > C$

Answer 1

Answer: (A)

$D > B > C > A$

Answer 2

(A) n=3;l=0;m=0;3s orbital
(B) n=4;l=0;m=0;4s orbital
(C) n=3;l=1;m=0;3p orbital
(D) n=3;l=2;m=0;3d orbital
As per Hund's rule energy is given by (n+l) value. If value of (n+l) remains same then energy is given by n only.
So, the correct option is (A).