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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:

- n = 4, l = 2, m
_{l}= –2 , m_{s}= - 1/2 - n = 3, l = 2, m
_{l}= 1 , m_{s}= +1/2 - n = 4, l = 1, m
_{l}= 0 , m_{s}= +1/2 - n = 3, l = 2, m
_{l}= –2 , m_{s}= –1/2 - n = 3, l = 1, m
_{l}= –1 , m_{s}= +1/2 - n = 4, l = 1, m
_{l}= 0 , m_{s}= +1/2

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#### Solution 1

For n = 4 and l = 2, the orbital occupied is 4d.

For n = 3 and l = 2, the orbital occupied is 3d.

For n = 4 and l = 1, the orbital occupied is 4p.

Hence, the six electrons i.e., 1, 2, 3, 4, 5, and 6 are present in the 4d, 3d, 4p, 3d, 3p, and 4p orbitals respectively.

Therefore, the increasing order of energies is 5(3p) < 2(3d) = 4(3d) < 3(4p) = 6(4p) < 1 (4d).

#### Solution 2

The orbitals occupied by the electrons are:

(1) 4d

(2) 3d

(3) 4p

(4) 3d

(5) 3p

(6) 4p

Same orbitals will have same energy and higher the value of (n+l) higher is the energy,

Their energies will be in order: (5) < (2) = (4)< (6) = (3) < (1).

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