Question

Designate the electrons having following sets of quantum numbers:
a n = 2, 1 = 0, m = 0, s = +7
b. n = 3,1 = 1, m =+1, s = +12
C. n = 1, 1 = 0, m =0, S = -2​

Answer 1

Explanation:

The energy order of the respective orbital can be compared by their n+l values. In case of equal n+l values of orbital higher energy corresponds to the orbital whose n value is greater.

In (a), the electron is present in 4s-orbital ⇒n+l=4

In (b), the electron is present in 3p-orbital ⇒n+l=4

In (c), the electron is present in 3d-orbital ⇒n+l=5

In (d), the electron is present in 3s-orbital ⇒n+l=3

Therefore, the decreasing order of energies of orbitals is: 3d>4s>3p>3s.