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

The Correct Option is D

To determine the order of energy levels for the given sets of quantum numbers in a multi-electron atom, we use the concept of the n + l rule. According to this rule, the energy of an electron is determined first by the sum of the principal quantum number \( n \) and the azimuthal (angular momentum) quantum number \( l \). The higher the sum of \( n + l \), the higher the energy. If two orbitals have the same \( n + l \) value, the one with the lower \( n \) value is lower in energy.

Let's evaluate each given set:

1. (A) \( n = 4, l = 1 \) → \( n + l = 4 + 1 = 5 \)
2. (B) \( n = 4, l = 2 \) → \( n + l = 4 + 2 = 6 \)
3. (C) \( n = 3, l = 1 \) → \( n + l = 3 + 1 = 4 \)
4. (D) \( n = 3, l = 2 \) → \( n + l = 3 + 2 = 5 \)
5. (E) \( n = 4, l = 0 \) → \( n + l = 4 + 0 = 4 \)

Now, let's order the orbitals by energy:

• Sets (C) and (E) both have the smallest \( n + l \) values (4), but set (C) has a lower \( n \) (\(n = 3\) compared to \(n = 4\)), so set (C) is lower in energy than set (E).
• Sets (A) and (D) have the same \( n + l \) values (5), but set (D) has a lower \( n \), making it lower in energy than set (A).
• Set (B) has the highest \( n + l \) value (6), so it is the highest in energy.

Thus, the order of energy levels is: (C) < (E) < (D) < (A) < (B)

The correct answer is: \((C)<(E)<(D)<(A)<(B)\)

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: \((C) < (E) < (D) < (A) < (B)\).