Question

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: (i) n = 4, l = 2, $m_{l}$ = - 2, $m_{s}$ = - 1/2 (ii) n = 3, l = 2, $m_{l}$ = - 1, $m_{s}$ = + 1/2 (iii) n = 4, l = 1, $m_{l}$ = 0, $m_{s}$ = + 1/2 (iv) n = 3, l = 2, $m_{l}$ = - 2, $m_{s}$ = - 1/2 (v) n = 3, l = 1, $m_{l}$ = - 1, $m_{s}$ = + 1/2 (vi) n = 4, l = 1, $m_{l}$ = 0, $m_{s}$ = + 1/2

Answer 1

Four quantum numbers requiring for the complete explanation of electron in an atom.

Number: Principal quantum number

Symbol: n

Possible values: 1, 2, 3, 4...

Number: “Angular momentum quantum number”

Symbol: I

Possible values: 0, 1, 2, 3, 4..... (n-1)

Number: Magnetic quantum number

Symbol: m

Possible values: -1 to +1

Number: Spin quantum number

Symbol: s

Possible values: $+\frac { 1 }{ 2 } ,\quad -\frac { 1 }{ 2 }$

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 $5(3p)\quad <\quad 2(3d)\quad =\quad 4(3d)\quad <\quad 3(4p)\quad =\quad 6(4p)\quad <\quad 1(4d)$

Answer 2

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THE ORDER IS......☺☺

$5(3p)<2(3d)=4(3d)<3(4p)=6(4p)<1(4d)$