Question

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:

• A. IV > II > III > I
• B. I > III > II > IV
• C. III > I > II > IV
• D. I > II > III > IV

Answer 1

The Correct Option is B

To determine the correct decreasing order of energy of the given electrons based on their quantum numbers, we need to understand the rules for the energy levels of electrons.

The energy of an orbital is generally determined by the principal quantum number (\(n\)) and the azimuthal quantum number (\(l\)). A higher value of \(n\) usually signifies a higher energy level. However, to determine which electron has a higher energy state, we use the (n + l) rule, also known as Madelung's rule: 

• If two electrons have the same (n + l) value, the one with the higher \(n\) will have higher energy.
• If two electrons have different (n + l) values, the electron with the higher (n + l) will have higher energy.

Let's analyze each electron:

1. Electron I: \(n = 4, \, l = 2\) → \(n + l = 4 + 2 = 6\)
2. Electron II: \(n = 3, \, l = 2\) → \(n + l = 3 + 2 = 5\)
3. Electron III: \(n = 4, \, l = 1\) → \(n + l = 4 + 1 = 5\)
4. Electron IV: \(n = 3, \, l = 1\) → \(n + l = 3 + 1 = 4\)

Now, order the electrons according to \(n + l\) values from highest to lowest:

• I: \(n + l = 6\)
• II and III: \(n + l = 5\) (for these, compare \(n\) values; III has a higher \(n\) = 4 than II, \(n\) = 3)
• IV: \(n + l = 4\)

Thus, the correct decreasing order of energy of these electrons is:

I > III > II > IV