1S\(^2\)2S\(^2\)2PX\(^2\)2py\(^2\)2pz\(^0\)3s\(^0\)
1S\(^2\)2S\(^2\)2PX\(^2\)2py\(^1\)2pz\(^1\)3s\(^0\)
1S\(^2\)2S\(^2\)2PX\(^1\)2py\(^1\)2pz\(^1\)3s\(^1\)
1S\(^2\)2S\(^2\)2PX\(^2\)2py\(^2\)2pz\(^0\)3s\(^1\)
Explanation
Video Explanation
Post your Contribution
Discussions (19)
Your first statement is correct: electrons fill subshells in increasing energy order, meaning the 2p subshell is filled before the 3s subshell. However, your second statement needs clarification:
Hund’s Rule states that electrons fill degenerate orbitals (same energy level) singly before pairing.
In the 2p subshell (which contains 2px, 2py, and 2pz), electrons do not fill in a specific order (x before y before z). Instead, each orbital gets one electron first before pairing starts.
Thus, the correct electron filling order follows 1s → 2s → 2px, 2py, 2pz (each gets one before pairing) → 3s.
So the correct answer remains:
C. 1S² 2S² 2PX¹ 2py¹ 2pz¹ 3s¹
A. 1s² 2s² 2px² 2py² 2pz⁰ 3s⁰
B. 1s² 2s² 2px² 2py¹ 2pz¹ 3s⁰
C. 1s² 2s² 2px¹ 2py¹ 2pz¹ 3s¹
D. 1s² 2s² 2px² 2py² 2pz⁰ 3s¹
look closely in option C let me tell you why it's Wrong.
Normally, before filling 3S orbitals, you have to fill those that have 1 & 2 first.
2px 2py and 2pz needs to have ( ² )on top of them before you can consider 3S.
But since the electron isn't enough, that's why option B is correct 
.
The correct option is "C"
option B indicate that pairing started too early
2px², 2py¹, 2pz¹ → This is not correct
• 1s²
• 2s²
• 2px¹, 2py¹, 2pz¹ —Each 2p orbital gets 1 electron before any pairing (Hund’s rule)
• 3s¹ — Higher energy orbital gets 1 electron after 2p
Electrons are supposed to fill in the orbit singly before pairing, so C should be the answer
But the question refers to aufbau principle, but you follow Hunds rules, if that is the correct answer 🤔
Electrons enter orbitals in order of increasing energy?
u wan fill 3s u neva finish 2p 
This question is based on the Aufbau Principle, which states that:
Electrons fill orbitals starting from the lowest energy to the highest energy.
Also remember:
The three p orbitals (px, py, pz) have equal energy
According to Hund’s Rule, electrons fill them singly first before pairing
Check the options:
Option A:
2px², 2py², 2pz⁰ 
Electrons are pairing before all orbitals are singly occupied → violates Hund’s rule
Option B:
2px², 2py¹, 2pz¹
Electrons are distributed properly:
First one goes into each orbital (px, py, pz)
Then pairing starts → correct order
Option C:
3s¹ appears before 2p is fully filled
This violates the Aufbau principle
Option D:
Similar issue as A (uneven filling of p orbitals)
Final Answer:
B.
Option B is incorrect because it violates Hund’s rule. Electrons must occupy degenerate p orbitals singly before pairing occurs. Option C correctly shows electrons filling 2p orbitals singly before pairing and then proceeding to 3s, which follows the order of increasing energy.
Myschool is correct
The question is testing the Aufbau principle (electrons fill orbitals in order of increasing energy) and Hund’s rule (degenerate orbitals are singly occupied before pairing).
Energy order:
1s < 2s < 2p (px, py, pz) < 3s
Now check the options:
A:
2pₓ² 2pᵧ² 2p_z⁰ → ❌ pairs electrons before occupying all 2p orbitals (violates Hund’s rule)
B:
2pₓ² 2pᵧ¹ 2p_z¹ → ✅
1s filled
2s filled
2p orbitals singly occupied before pairing
3s not filled yet
✔ Correct application of Aufbau + Hund’s rule
C:
3s¹ while 2p is not full → ❌ violates Aufbau principle
D:
3s¹ before completing 2p → ❌ violates Aufbau principle
✅ Correct answer: B
Electrons fill orbitals in order of increasing energy according to Aufbau's principle. This principle states that electrons will occupy the lowest available energy levels first.
In the case of the p-orbitals (which are degenerate, meaning they have the same energy), the Hund's rule comes into play: electrons will fill degenerate orbitals (like 2px, 2py, 2pz) singly and with parallel spins before pairing up.
Looking at the options:
Option C : 1S² 2S² 2PX¹ 2PY¹ 2PZ¹ 3S¹ is the correct option
