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 same values of (n+l). The orbital with lower value of n will have lower energy.
I. Based upon the above information arrange the following orbitals in the increasing order of energy.
(a) 1s,2s,3s,2p (b) 4s,3s,3p,4d
(c) 2p,4d,5d,4f,6s (d) 5f,6d,7s,7p
II. Based upon the above information Solve the question. give below.
(a) which of the following orbitals has the lowest energy
`4d,4f,5s,5p`
(b) which of the following orbitals has the highest energy?
`5p,5d,5f,6s,6p`

Explanation

(a) (n+l) values of 1s=1+0=1,2s=2+0=2,3s=3+0-3.2p=2xx1=2
Hence, increasing order of their energy is
`1slt2slt2p lt3s`
(b) `4s=4+0=4,3s=3+0=3,3p=3+1=4 4d=4+2= 6`
Hence `3slt3plt4slt4d`
(c) ` 5p=5+1=6,4d=4+2= 6,5d=5+2 =7 ,4f=4+3=7, 6s= 6+0= 6`
Hence, `4dlt5dlt6slt4flt5d`
(d) ` 5f=5+3= 8,7s=7+0=7,7p=7+1=8`.
Hence, `7slt5flt6dlt7p`
II. (a) (n+l) valeus of 4d `=4+2=6,4f=4+3=7,5s=5+0=5,7p= 7+1=8`
Hence, 5s has the lowest energy.
(b) `5p=5+1 = 6,5d=5+2=7,5f=5+3=8,6s=6+0=6 6p=6+1=7`
Hence, 5f has highest energy.