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, $$m_l$$ = –2 , ms = $$-\dfrac{1}{2}$$

2. n = 3, l = 2, $$m_l$$= 1 , ms = $$+\dfrac{1}{2}$$

3. n = 4, l = 1, $$m_l$$ = 0 , ms = $$+\dfrac{1}{2}$$

4. n = 3, l = 2, $$m_l$$= –2 , ms = $$-\dfrac{1}{2}$$

5. n = 3, l = 1, $$m_l$$ = –1 , ms= $$+\dfrac{1}{2}$$

6. n = 4, l = 1, $$m_l$$ = 0 , ms = $$+\dfrac{1}{2}$$

Asked by Abhisek | 1 year ago |  161

##### Solution :-

Electrons  1, 2, 3, 4, 5, and 6 reside in the 4d, 3d, 4p, 3d, 3p, and 4p orbitals (respectively). Ranking these orbitals in the increasing order of energies: (3p) < (3d) < (4p)  < (4d).

Answered by Abhisek | 1 year ago

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