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NCERT Class XI Chemistry Structure of Atom Solutions
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Question : 62 of 67
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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, = –2, = –1/2 2. n = 3, l = 2, = 1, ms = +1/2 3. n = 4, l = 1, = 0, = +1/2 4. n = 3, l = 2, = –2, ms = –1/2 5. n = 3, l = 1, = –1, = +1/2 6. n = 4, l = 1, = 0, ms = +1/2
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
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