Question

The arrangement of orbitals on the basis of energy is based upon their (𝑛+𝑙) value. Lower the value of (𝑛+𝑙), lower is the energy. For orbitals having the same values of (𝑛+𝑙), the orbital with lower value of n will have lower energy.

Based upon the above information, arrange the following orbitals in the increasing order of energy.

Answer 1

Calculating the value of (𝒏+𝒍)

(i) 1𝑠, 2𝑠, 3𝑠, 2𝑝

orbital	n	l	(n+1) value
1s	1	0	1
2s	2	0	2
3s	3	0	3
2p	2	1	3

Increasing order of energy:$1s<2s<2p<3s$

(ii) 4𝑠, 3𝑠, 3𝑝, 4𝑑

orbital	n	l	(n+l) value
4s	4	0	4
3s	3	0	3
3p	3	1	4
4d	4	2	6

increasing order of energy:$3s<3p<4s<4d$

(iii) 5𝑝, 4𝑑, 5𝑑, 4𝑓, 6𝑠

orbital	n	l	(n+1) value
5p	5	1	6
4d	4	2	6
5d	5	2	7
4f	4	3	7
6s	6	0	6

Increasing order of energy: $4d<5p<6s<4f<5d$

(iv) 5𝑓, 6𝑑, 7𝑠, 7𝑝

oribial	n	l	(n+l) value
5f	5	3	8
6d	6	2	8
7s	7	0	7
7p	7	1	8

Increasing order of energy: $7s<5f<6d<7p$