The arrangement of orbitals based on energy is based upon their (n+l ) value. Lower the value of (n+l ), 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.
I. Based upon the above information, arrange the following orbitals in the increasing order of energy
(a) 1s, 2s, 3s, 2p
(b) 4s, 3s, 3p, 4d
(c) 5p, 4d, 5d, 4f, 6s
(d) 5f, 6d, 7s, 7p
II. Based upon the above information, solve the questions given below :
(a) Which of the following orbitals has the lowest energy?
4d, 4f, 5s, 5p
(b) Which of the following orbitals has the highest energy?
5p, 5d, 5f, 6s, 6p
(i) (a) the increasing order of energy of the given orbital is : 1s >2s >2p> 3s
(b) the increasing order of energy of the given orbital is : 3s<3p<4s<4d
(c) the increasing order of energy of the given orbital is : 4d<5p<6s<4f<5d
(d)the increasing order of energy of the given orbital is: 7s<5f<6d<7p
(ii) (a) among the orbitals, 5s has the lowest energy.
the (n+l) value for 5s is the lowest = 5 + 0 = 5. Other orbitals have (n+l )value more than 5 –
5p= 5 + 1 = 6 , 4f = 4 + 3 = 7 , 4d = 4 + 2 = 6.
(b) among the orbitals , 5f has the highest energy because the (n +l ) value – 5 + 3 = 8 is highest.
5d = 5 + 2 = 7 , 5p = 5 + 1= 6 , 6s =6 + 0 = 6 , 6p =6 + 1 = 7.