JEE Main 2024 — Structure Of Atom Question with Solution

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:

$$(\mathrm{C})<(\mathrm{E})<(\mathrm{D})<(\mathrm{A})<(\mathrm{B})$$