Question

Arrange the orbitals of H atom in the increasing order of their energy levels:

${3p}_x,2s,{4d}_{xy},s,{4p}_z,{3p}_y,4s$

• A. $2s<3s<{3p}_x={3p}_y<4s<{4p}_z<{4d}_{xy}$
• B. $2s<3s<{3p}_x={3p}_y<4s={4p}_z={4d}_{xy}$
• C. $2s<3s<{3p}_x={3p}_y<4s={4p}_z<{4d}_{xy}$
• D. $2s<3s={3p}_x={3p}_y<4s={4p}_z={4d}_{xy}$
  

Answer 1

The correct option is A $2s<3s<{3p}_x={3p}_y<4s<{4p}_z<{4d}_{xy}$

Orbitals are arranged with increasing energy on the basis of (n + l) rule.
The higher the value of (n + l) for an orbital the higher is its energy. Similarly the lower the value of (n + l) for an orbital the lower is its energy.
If two orbitals possess same (n + l) value, the orbital with lower (n + l) value will have the lower energy.
Let us find the energies of different orbitals
3s orbital:
n =3 , l= 0
(n + l) = (3 + 0) = 3
3p orbital:
n =3 , l= 1
(n + l) = (3 + 1) = 4
4s orbital:
n =4 , l= 0
(n + l) = (4+ 0) = 4
Both 3p and 4s orbitals have same (n + l) value, but the n value of 3p orbital is lower than 4s orbital hence, it will have lower energy.
3d orbital:
n =3 , l= 2
(n + l) = (3+ 2) = 5
The order of increasing energies of orbitals is $3s<3p<4s<3d$.
Option A is correct.