QJEEM24Chae31: JEE Main 2024 Chemistry - Class 11 JEEM24Chae31 Solution: Compare the energies of following sets of quantum numbers for…

To compare the energies of electrons in a multielectron system, we must consider both the principal quantum number ( $\mathrm{n}$ ) and the azimuthal quantum number ( $\mathrm{l}$ ). In multielectron atoms, the energy levels are influenced by electron-electron repulsions, which modify the energy ordering compared to the hydrogen atom.

The general rule, called the "n+l" rule or Madelung rule, states that the energy of an electron in a multielectron atom is primarily determined by the sum of the principal quantum number ( $\mathrm{n}$ ) and the azimuthal quantum number ( $\mathrm{l}$ ). An orbital with a lower $(\mathrm{n} + \mathrm{l})$ value has lower energy. If two orbitals have the same $(\mathrm{n} + \mathrm{l})$ value, the orbital with the lower $\mathrm{n}$ value has lower energy.

Now, let's analyze each set of quantum numbers:

(A) $\mathrm{n}=4, \mathrm{l}=1$

$\mathrm{n} + \mathrm{l} = 4 + 1 = 5$

(B) $\mathrm{n}=4, \mathrm{l}=2$

$\mathrm{n} + \mathrm{l} = 4 + 2 = 6$

(C) $\mathrm{n}=3, \mathrm{l}=1$

$\mathrm{n} + \mathrm{l} = 3 + 1 = 4$

(D) $\mathrm{n}=3, \mathrm{l}=2$

$\mathrm{n} + \mathrm{l} = 3 + 2 = 5$

(E) $\mathrm{n}=4, \mathrm{l}=0$

$\mathrm{n} + \mathrm{l} = 4 + 0 = 4$

Based on the $\mathrm{n} + \mathrm{l}$ values, the order of increasing energy is:

(C) = (E) < (A) = (D) < (B)

Among (C) and (E), (C) has lower $\mathrm{n}$ value, so it is lower in energy:

(C) < (E) < (A) = (D) < (B)

Among (A) and (D), (D) has lower $\mathrm{n}$ value, so it is lower in energy:

Thus, the correct order is: (C) < (E) < (D) < (A) < (B)

The correct answer is Option C:

(\mathrm{C}) &lt; (\mathrm{E}) &lt; (\mathrm{D}) &lt; (\mathrm{A}) &lt; (\mathrm{B})