NCERT Class XI Chemistry Structure of Atom Solutions
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Question : 62
Total: 67
The quantum numbers of six electrons are given below. Arrange them in order of increasing energies. If any of these combination (s) has/have the same energy lists :
1. n = 4, l = 2,m l m s m l
3. n = 4, l = 1,m l m s m l
5. n = 3, l = 1,m l m s m l
1. n = 4, l = 2,
3. n = 4, l = 1,
5. n = 3, l = 1,
Solution:
1. 4d (n + l = 4 + 2 = 6)
2. 3d (n + l = 3 + 2 = 5)
3. 4p (n + l = 4 + 1 = 5)
4. 3d (n + l = 3 + 2 = 5)
5. 3p (n + l = 3 + 1 = 4)
6. 4p (n + l = 4 + 1 = 5)
Greater the value of n + l, higher will be the energy of orbital. If two orbitals have same n + l value then the orbital having higher n value will possess higher energy.
Therefore, the required order is : 5 < 2 = 4 < 6 = 3 < 1
2. 3d (n + l = 3 + 2 = 5)
3. 4p (n + l = 4 + 1 = 5)
4. 3d (n + l = 3 + 2 = 5)
5. 3p (n + l = 3 + 1 = 4)
6. 4p (n + l = 4 + 1 = 5)
Greater the value of n + l, higher will be the energy of orbital. If two orbitals have same n + l value then the orbital having higher n value will possess higher energy.
Therefore, the required order is : 5 < 2 = 4 < 6 = 3 < 1
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