Question

The correct decreasing order of energy for the orbitals having, following set of quantum numbers:
(A) n = 3, l = 0, m = 0
(B) n = 4, l = 0, m = 0
(C) n = 3, l = 1, m = 0
(D) n = 3, l = 2, m = 1

• (D) > (B) > (C) > (A)
• (B) > (D) > (C) > (A)
• (C) > (B) > (D) > (A)
• (B) > (C) > (D) > (A)

Answer 1

The Correct Option is A

In order to determine the decreasing order of energy for the given orbitals, we need to understand the relationship between quantum numbers and energy levels of orbitals in an atom. 

The energy of an orbital in multi-electron atoms is primarily determined by the principal quantum number \(n\) and the azimuthal quantum number \(l\). The general rule is that the energy increases with increasing \(n+l\) value. If two orbitals have the same \(n+l\) value, the one with a higher \(n\) value will have higher energy.

Now, let's apply this rule to each set of quantum numbers provided:

1. (A) n = 3, l = 0, m = 0: Here \(n+l = 3+0 = 3\).
2. (B) n = 4, l = 0, m = 0: Here \(n+l = 4+0 = 4\).
3. (C) n = 3, l = 1, m = 0: Here \(n+l = 3+1 = 4\).
4. (D) n = 3, l = 2, m = 1: Here \(n+l = 3+2 = 5\).

From the above calculation:

• The orbital with the highest \((n+l)\) value is (D), so it has the highest energy.
• Orbitals (B) and (C) have the same \((n+l)\) value. Since (B) has a higher \(n\) than (C), it has slightly higher energy.
• The orbital (A) has the lowest \((n+l)\) value and thus the lowest energy.

Therefore, the correct order of decreasing energy for the orbitals is:

(D) > (B) > (C) > (A)

This analysis supports the given correct answer: (D) > (B) > (C) > (A).