Question

Compare the energies of following sets of quantum numbers for multielectron system.
(A) n = 4, 1 = 1
(B) n= 4, l = 2
(C) n = 3, l = 1
(D) n = 3, l = 2
(E) n = 4, 1 = 0
Choose the correct answer from the options given below :

• A. $(B)>(A)>(C)>(E)>(D)$
• B. $(E)>(C)<(D)<(A)<(B)$
• C. $(E)>(C)>(A)>(D)>(B)$
• D. $(C)<(E)<(D)<(A)<(B)$

Answer 1

Answer: (D)

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

Answer 2

In multielectron systems, the energy of an electron in an orbital depends on both the principal quantum number ($n$) and the azimuthal quantum number ($l$). The energy increases as the value of $n + l$ increases. For orbitals with the same $n + l$, the one with the lower $n$ has lower energy.

• (A) $n + l = 4 + 1 = 5$
• (B) $n + l = 4 + 2 = 6$
• (C) $n + l = 3 + 1 = 4$
• (D) $n + l = 3 + 2 = 5$
• (E) $n + l = 4 + 0 = 4$

Order by $n + l$:

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

For orbitals with the same $n + l$, compare $n$:

$(C) < (E)$, as $n = 3$ for (C) and $n = 4$ for (E).

Final Answer: $(C) < (E) < (D) < (A) < (B)$.