The quantum numbers of four electrons are given below. [2024]

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Solution

Q.
The quantum numbers of four electrons are given below.

I. $n = 4; l = 2; m_{l} = - 2; s = - \frac{1}{2}$

II. $n = 3; l = 2; m_{l} = 1; s = + \frac{1}{2}$

III. $n = 4; l = 1; m_{l} = 0; s = + \frac{1}{2}$

IV. $n = 3; l = 1; m_{l} = - 1; s = + \frac{1}{2}$

The correct decreasing order of energy of these electrons is [2024]

1 IV > II > III > I

2 I > III > II > IV

3 III > I > II > IV

4 I > II > III > IV

Ans.
(2)

The increasing order of energy is $1 s, 2 s, 2 p, 3 s, 3 p, 4 s, 3 d, 4 p, 5 s, 4 d, 5 p, 4 f, 5 d, 6 p, 7 s, \dots$

$n$ $l$ Orbital

4 2 $4 d$

3 2 $3 d$

4 1 $4 p$

3 1 $3 p$

Thus, $I (4 d) > III (4 p) > II (3 d) > IV (3 p) .$

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