Arrange the electrons represented by the following set of quantum numbers in the decreasing order of energy.
(i) $n = 4, l = 0, m = 0, m_{s} = + 1/2$ (ii) $n = 3, l = 1, m = 1, m_{s} = - 1/2$
(iii) $n = 3, l = 2, m = 0, m_{s} = + 1/2$ (iv) $n = 3, l = 0, m = 0, m_{s} = - 1/2$

Question

Arrange the electrons represented by the following set of quantum numbers in the decreasing order of energy.

(i) $n = 4, l = 0, m = 0, m_{s} = + 1/2$ (ii) $n = 3, l = 1, m = 1, m_{s} = - 1/2$

(iii) $n = 3, l = 2, m = 0, m_{s} = + 1/2$ (iv) $n = 3, l = 0, m = 0, m_{s} = - 1/2$

TestHub · Accepted Answer

The energy of electrons is determined by the $(n + l)$ rule. Higher $(n + l)$ means higher energy. If $(n + l)$ is same, higher $n$ means higher energy.

(i) $n = 4, l = 0 ⟹ n + l = 4 + 0 = 4$

(ii) $n = 3, l = 1 ⟹ n + l = 3 + 1 = 4$

(iii) $n = 3, l = 2 ⟹ n + l = 3 + 2 = 5$

(iv) $n = 3, l = 0 ⟹ n + l = 3 + 0 = 3$

Comparing (i) and (ii), both have $n + l = 4$ . Since $n$ for (i) is 4 and for (ii) is 3, (i) has higher energy than (ii).

So, the order of energy is (iii) > (i) > (ii) > (iv).

Final Answer: C

Chemistry - Atomic Structure Question with Solution | TestHub

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