Question

consider the orbitals 7p, 5d, 4f, 6s,3d .arrange them in increasing order of energies​

Answer 1

Answer:

Solution :

I (a) `(n +l)` values are `2s =1 +0 =1, 2s = 2 + 0= 2, 3s = 3 + 0 = 3, 2p = 2 + 1 = 3`

Hence, increasing order of their energy is `1s lt 2s lt 2p lt 3s`.

(b) `4s = 4 + 0 = 4, 3s = 3 + 0 = 3, 3p = 3 +1 = 4, 4d = 4 + 2 = 6` Hence, `3s lt 3p lt 4s lt 4d`

(c) `5p = 5 +1 = 6, 4d = 4 + 2 = 6, 5d= 5 + 2 = 7, 4f = 4 + 3 = 7, 6s = 6 + 0 = 6` Hence, `4d lt 5p lt 6s lt 4f lt 5d`

(d) `5f = 5 + 3 = 8, 6d = 6 + 2 = 8, 7s = 7 + 0= 7, 7p = 7 + 1 = 8`. Hence, `7s lt 5f l 6d lt 7p`

II. (a) `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.