Q6. The arrangement of orbitals on the basis of energy is based upon their (n+l) value. Lower the value of (n+1), the lower is the energy. For orbitals having the same values of (n+l), the orbital with a lower value of n will have lower energy.
Answer. I. 1s < 2s < 3s < 3p 3s < 3p < 4s < 4d 4d < 5p < 6s < 4f < 5d 7s < 5f < 6d < 7p II. 5s has the lowest energy. 5f has the highest energy.
