The correct option is D 2s<3s<3px=3py<4s<4pz<4dxy
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.
For 2s orbital:
n=2, l=0
∴(n+l)=(2+0)=2
For 3s orbital:
n=3, l=0
∴(n+l)=(3+0)=3
For 3px orbital:
n=3, l=1
∴(n+l)=(3+1)=4
For 3py orbital:
n=3, l=1
∴(n+l)=(3+1)=4
For 4s orbital:
n=4, l=0
∴(n+l)=(4+0)=4
Both 3py, 3px 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.
For 4dxy orbital:
n=4, l=2
∴(n+l)=(4+2)=6.
For 4pz orbital:
n=4, l=1
∴(n+l)=(4+1)=5
The order of increasing energies of orbitals is
2s<3s<3px=3py<4s<4pz<4dxy