Question

The quantum number of four electrons are given below:

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 = 1, m_l = 1, m_s = -1/2

The correct order of their increasing energies will be:
1. IV < III < II < I
2. I < II < III < IV
3. IV < II < III < I
4. I < III < II < IV

Answer 1

Correct Answer - Option 3 : IV < II < III < I

Concept:

The value of (n + l) is directly proportional to the energy of the orbital.

If the value of (n + l) is same then the value of n is considered. The higher the value of n, the higher the energy of the orbital.

Calculation:

I. n + l = 6 while n = 4

II. n + l = 5 while n = 3

III. n + l = 5 while n = 4

IV. n + l = 4 while n = 3

Thus, the increasing order of the energy of the orbital is given as:

IV < II < III < I