Quick Summary
Concept Tested: Energy Order of Orbitals based on Madelung (n+ℓ) Rule
Chapter-Subtopic: Structure of Atom – Electronic Configuration
Difficulty: ★☆☆☆☆ (Easy)
Time Required: 1 min
Key Formula: $$n + \ell$$ (Madelung Constant)
Correct Answer: (A)
Reason: The orbital with the highest value of $$n + \ell$$ has the highest energy. If $$n + \ell$$ is equal, the orbital with the higher principal quantum number $$n$$ has higher energy. Calculation yields: $$5 > 4 > 4 > 3$$, resulting in the order: (D) > (B) > (C) > (A).
The Question
Determine the correct decreasing order of energy for the following orbitals:
- (A) $$n = 3, \ell = 0$$
- (B) $$n = 4, \ell = 0$$
- (C) $$n = 3, \ell = 1$$
- (D) $$n = 3, \ell = 2$$
Options:
(A) (D) > (B) > (C) > (A)
(B) (B) > (D) > (C) > (A)
(C) (C) > (B) > (D) > (A)
(D) (B) > (C) > (D) > (A)
Quick Answer
The correct answer is Option (A). The energy order is determined by calculating the sum of the principal quantum number ($$n$$) and the azimuthal quantum number ($$\ell$$) for each orbital. Orbital (D) has the highest sum ($$5$$), Orbital (B) and (C) have a sum of $$4$$, and Orbital (A) has the lowest sum ($$3$$). When (B) and (C) have the same sum, (B) is higher because it has a higher $$n$$ value ($$4 > 3$$).
Why Other Options Are Incorrect
Option (B): (B) > (D) > (C) > (A)
This option incorrectly places Orbital (B) above Orbital (D). According to the n + ℓ rule, the orbital with the higher value of $$n + \ell$$ possesses higher energy. Orbital (D) has a sum of $$5$$ ($$3 + 2$$), whereas Orbital (B) has a sum of $$4$$ ($$4 + 0$$). Since $$5 > 4$$, Orbital (D) must have higher energy than (B).
Option (C): (C) > (B) > (D) > (A)
This option contains two errors. First, it places Orbital (C) above Orbital (B). Both orbitals have an $$n + \ell$$ value of $$4$$; however, since $$n$$ is higher in (B) ($$4$$ vs $$3$$), (B) must be higher in energy than (C). Second, it places Orbital (D) below Orbital (B), ignoring the fact that (D) has a significantly higher $$n + \ell$$ sum than (B).
Option (D): (B) > (C) > (D) > (A)
This option incorrectly places Orbital (D) below Orbital (C). Orbital (D) has an $$n + \ell$$ value of $$5$$, while Orbital (C) has a value of $$4$$. The rule explicitly states that higher $$n + \ell$$ values correspond to higher energies. Therefore, (D) must be higher than (C).
Video Solution
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Understanding the Concept
In multi-electron atoms, the energy of an electron in an orbital is not determined solely by the principal quantum number ($$n$$). Instead, it is determined by a combination of the principal quantum number ($$n$$) and the azimuthal quantum number ($$\ell$$), known as the Madelung rule or the $$n + \ell$$ rule.
The logic is as follows:
- Primary Rule: Orbitals with a lower value of $$n + \ell$$ are filled first and have lower energy.
- Tiebreaker Rule: If two orbitals have the same $$n + \ell$$ value, the orbital with the lower principal quantum number ($$n$$) is filled first (has lower energy).
This rule helps predict the order in which orbitals are filled (Aufbau principle).
Detailed Step-by-Step Solution
Step 1: Recall the Energy Ordering Rule
The energy of an orbital is primarily determined by the sum of the principal quantum number ($$n$$) and the azimuthal quantum number ($$\ell$$). The formula for this sum is:
$$n + \ell$$
According to the Madelung rule:
- Orbitals with a lower $$n + \ell$$ value have lower energy.
- If two orbitals have the same $$n + \ell$$ value, the orbital with the higher principal quantum number ($$n$$) has higher energy.
Step 2: Calculate $$n + \ell$$ for Each Orbital
Let’s compute the sum for each given orbital using the formula $$n + \ell$$:
- Orbital (A): $$3 + 0 = 3$$
- Orbital (B): $$4 + 0 = 4$$
- Orbital (C): $$3 + 1 = 4$$
- Orbital (D): $$3 + 2 = 5$$
Step 3: Compare the Values to Determine Energy Order
Now, we compare the calculated sums to rank the orbitals from highest energy to lowest energy:
- Orbital (D) has the highest sum of $$5$$, so it has the highest energy.
- Next, we compare Orbital (B) and Orbital (C), both having a sum of $$4$$.
- Since their sums are equal, we look at the principal quantum number ($$n$$):
- Orbital (B) has $$n = 4$$.
- Orbital (C) has $$n = 3$$.
Because $$4 > 3$$, Orbital (B) has higher energy than Orbital (C).
- Orbital (A) has the lowest sum of $$3$$, so it has the lowest energy.
Combining these results, the decreasing order of energy is:
(D) > (B) > (C) > (A)
Final Answer
✔ Correct Option: (A)
The energy order of the given orbitals is $$n + \ell = 5 > 4 > 4 > 3$$, which corresponds to (D) > (B) > (C) > (A).
Essential Formulas for This Topic
1. Madelung Rule (n + ℓ Rule): $$n + \ell$$ determines the relative energy of orbitals in multi-electron atoms.
2. Tiebreaker Condition: If $$n + \ell$$ is constant, energy increases with increasing $$n$$.
3. Quantum Numbers: $$n$$ (Principal), $$\ell$$ (Azimuthal, s=0, p=1, d=2, f=3).
Common Mistakes to Avoid
Mistake 1: Ignoring the Tiebreaker Rule
Wrong Thinking: Assuming that if two orbitals have the same $$n + \ell$$ value, they have the same energy.
Correction: The rule states that if $$n + \ell$$ is equal, the orbital with the higher $$n$$ has higher energy. For example, 4s and 3d have the same sum ($$4$$), but 4s is filled before 3d because 4s has a higher $$n$$.
Mistake 2: Comparing Only the Principal Quantum Number (n)
Wrong Thinking: Thinking that any orbital with a higher $$n$$ is always higher in energy than one with a lower $$n$$.
Correction: The azimuthal quantum number ($$\ell$$) plays a significant role. For instance, a 4p orbital ($$n + \ell = 5$$) has higher energy than a 3d orbital ($$n + \ell = 5$$), even though 4p has a higher $$n$$, because 3d has a higher $$\ell$$.
Mistake 3: Misidentifying Quantum Numbers
Wrong Thinking: Confusing the values of $$\ell$$ (azimuthal quantum number) with subshell labels.
Correction: Remember the mapping: $$\ell = 0$$ for s, $$1$$ for p, $$2$$ for d, and $$3$$ for f. Using this mapping correctly is essential for calculating $$n + \ell$$.
Key Concept Summary
- Energy increases as the value of $$n + \ell$$ increases.
- If $$n + \ell$$ is the same, energy increases as $$n$$ increases.
- The order of increasing energy for subshells is: 1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p ...
- Orbitals are filled following the Madelung rule, also known as the Aufbau principle.
Golden Rule: Always calculate $$n + \ell$$ first. If the sums are equal, compare $$n$$.
Frequently Asked Questions
Q: Why is the 4s orbital filled before the 3d orbital?
A: This is a common point of confusion. According to the $$n + \ell$$ rule, 4s has $$n + \ell = 4$$, while 3d has $$n + \ell = 5$$. Since 4 is less than 5, 4s is filled first (has lower energy).
Q: What happens if two orbitals have the same $$n + \ell$$ and the same $$n$$?
A: This is impossible for standard orbitals. If $$n + \ell$$ is the same and $$n$$ is the same, then $$\ell$$ must be the same, meaning they are the exact same orbital.
Q: Is the $$n + \ell$$ rule valid for Hydrogen atoms?
A: No. The rule applies to multi-electron atoms. In Hydrogen, all orbitals with the same $$n$$ (e.g., 2s and 2p) have the same energy because there is no shielding or electron-electron repulsion.
Q: How does the energy of an orbital relate to its radius?
A: Generally, as energy increases (higher $$n + \ell$$), the orbital becomes larger and further from the nucleus.
Prerequisites to Solve This Question
- Knowledge of Quantum Numbers: Understanding the definitions of Principal ($$n$$) and Azimuthal ($$\ell$$) quantum numbers.
- Familiarity with Subshells: Knowing that $$\ell = 0$$ corresponds to s, $$1$$ to p, $$2$$ to d, and $$3$$ to f.
- Basic Algebra: Ability to add numbers and compare integers.
After Solving This, You Can:
- ✔ Predict the order of filling for any set of orbitals.
- ✔ Determine which orbital is higher in energy without memorizing the entire periodic table order.
- ✔ Solve questions related to electronic configuration of elements.
- ✔ Understand the concept of shielding and penetration indirectly.
Study Tips for This Topic
1. Memorize the Chart: Create a small chart showing the increasing order of $$n + \ell$$ values (1, 2, 3, 4, 5…).
2. Visualize: Draw the orbitals and write the $$n + \ell$$ value next to them to visualize the concept.
3. Practice Variations: Try changing the quantum numbers (e.g., $$n=5, \ell=3$$) and see if you can predict the energy relative to others.
4. Connect to Periodic Table: Relate this concept to the filling of s, p, d, and f blocks in the periodic table.
Difficulty Rating & Exam Frequency
Difficulty: ★☆☆☆☆ (Very Easy)
JEE Main Frequency: High. This concept is fundamental and appears in almost every Chemistry section of JEE Main papers.
JEE Advanced Frequency: Medium. Often used as a part of a larger question involving electronic configuration.
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Written by Nishant Kumar Gupta
Founder of Padho Likho JEE & Senior Chemistry Educator — 12+ Years Experience, Ex-Faculty Allen/Aakash/Narayana.
Last Updated: July 2026
Question Source: JEE Main 2026 PYQ
Topic: Structure Of Atom