Question

The quantum number of four electron are given below-

{I.\;\;\;n = 4,l = 2{\text{ }},{m_l} = - 2,{m_s} = - \dfrac{1}{2}} \\\ {II.\;\;n = 3,{\text{ }}l = 2,{\text{ }},{m_l} = 1,{\text{ }}{m_s} = + \dfrac{1}{2}} \\\ {III.\;n = 4,{\text{ }}l = 1,{\text{ }}{m_l} = 0,{\text{ }}{m_s} = + \dfrac{1}{2}} \\\ {IV.n = 3,l = 1,{m_l} = 1,{\text{ }}{m_s} = - \dfrac{1}{2}} \end{array}$$ The correct order of their increasing energy will be: a. IV < III < II < I b. IV < II < III < I c. I < II < III < IV d. I < III < II < IV

Answer 1

Apply n+l rule to check the energy of the electron. The one which have lower value are more stable than others.

Complete answer:
According to (n+l) rule: Orbital which has the least value of (n+l) will be filled first to the electrons. Means whose sum of n+l is more will have more energy and will be filled by electrons later. And if the sum of n+l is the same then the energy depends on n i.e. more the value of n than more will be its energy and later it will be filled by an electron.
Let’s check all the options first.
I.n=4, l=2
(n+l) = 4+2
=6

II.n=3, l=2
(n+l) = 3+2
=5

III.n=4, l=1,
(n+l) = 4+1
=5

IV. n=3, l=1
(n+l) = 3+1
=4
So, the correct order is IV < II < III < I

Hence, the correct option is (B) IV < II < III < I .

Note:
Here the option II and III have the same n+l value. To decide which one has the lowest energy, check the value of n, the one that has a low n value is the one of lower energy.