Completed electron configuration practice worksheet showing ground state configurations for various elements.

Let's go through the electron configuration practice worksheet and evaluate each answer, checking for correctness based on the Aufbau principle, Hund’s rule, and Pauli exclusion principle. We’ll also explain any mistakes, especially focusing on problem #6 (Ag) which is marked with a star — indicating it might be incorrect or needs clarification.

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🔍 How to Write Electron Configurations:

1. Determine the atomic number (number of electrons in neutral atom).
2. Fill orbitals in order of increasing energy using the Aufbau principle:
`1s → 2s → 2p → 3s → 3p → 4s → 3d → 4p → 5s → 4d → 5p → 6s → 4f → 5d → 6p → 7s → 5f → 6d → 7p`
3. Maximize unpaired electrons (Hund’s Rule) and follow Pauli exclusion (max 2 e⁻ per orbital).
4. Check that total electrons = atomic number.

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✔ Let's evaluate each one:

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#### 1. Na (Sodium) – Atomic Number: 11
Given: `1s² 2s² 2p⁶ 3s¹`
✔ Correct!
- 2+2+6+1 = 11 electrons ✔️
- Filled up to 3s¹ (correct order)

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#### 2. Pb (Lead) – Atomic Number: 82
Given:
`1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f¹⁴ 5d¹⁰ 6p²`

Wait — this has a typo in order. The correct order should be:
- After 6s² comes 4f¹⁴, then 5d¹⁰, then 6p²
But 4f¹⁴ is written after 6s², which is fine, but let’s count:

Let’s add up:
- Up to 5p⁶: 54 electrons
- 6s² → 56
- 4f¹⁴ → 70
- 5d¹⁰ → 80
- 6p² → 82

So total = 82 ✔️

But order matters — we must write subshells in order of increasing n and ℓ.

The correct sequence is:
> 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f¹⁴ 5d¹⁰ 6p²

This is almost correct, but note: 4f comes before 5d? Actually, yes, because 4f (n=4) has lower energy than 5d (n=5), even though 5d appears earlier in the periodic table.

But wait: In filling order, after 6s² comes 4f, then 5d, then 6p.

So yes, the order is correct.

✔ Correct — just check the sequence:
- 6s² → 4f¹⁴ → 5d¹⁰ → 6p² → YES

✔️ So Pb: [Xe] 4f¹⁴ 5d¹⁰ 6s² 6p² → Correct

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#### 3. Sr (Strontium) – Atomic Number: 38
Given: `1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s²`

Add:
2+2+6+2+6+2+10+6+2 = 38 ✔️

Order:
After 4p⁶ → 5s² → YES

✔ Correct

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#### 4. U (Uranium) – Atomic Number: 92
Given:
`1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f¹⁴ 5d¹⁰ 6p⁶ 7s² 5f³ 6d¹`

Now, let’s count:

- Up to 6p⁶: 86 electrons
- 7s² → 88
- 5f³ → 91
- 6d¹ → 92

Total: 92 ✔️

But order: After 7s², next is 5f, then 6d, then 7p

So:
- 7s² → 5f³ → 6d¹ → YES

However, Uranium (U) has 5f³ 6d¹ 7s², not 5f³ 6d¹ with 7s² already placed.

But the configuration given ends with `7s² 5f³ 6d¹` — which is acceptable if written in filling order, but we usually write in order of increasing n and ℓ.

Standard convention: write in order of principal quantum number (n), not filling order.

So better to write:

`1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f¹⁴ 5d¹⁰ 6p⁶ 7s² 5f³ 6d¹`

But since 5f is part of the 5th shell, it should come before 6d in order of n?

Actually, no: 5f has n=5, 6d has n=6 → so 5f comes before 6d.

So correct order:
... 6p⁶ → 7s² → 5f³ → 6d¹

Wait — 5f is filled after 7s²? Yes, due to energy levels.

So filling order: 7s² → 5f³ → 6d¹ → 7p⁰

So writing as `7s² 5f³ 6d¹` is acceptable, but not in n-order.

But in standard notation, we write in order of n, so:

`1s² ... 6p⁶ 7s² 5f³ 6d¹` → incorrect order

We should group by n:

- n=1: 1s²
- n=2: 2s² 2p⁶
- ...
- n=7: 7s² 7p⁰
- But 5f is n=5 → so should be grouped with n=5

So correct way:
`... 5p⁶ 4f¹⁴ 5d¹⁰ 5f³ 6s² 6p⁶ 6d¹ 7s²`

Wait — no: 5f is filled after 6s², so it belongs to n=5, but filled after 6s².

But in electron configuration notation, we write in order of increasing n, so:

- All n=1 → n=2 → n=3 → n=4 → n=5 → n=6 → n=7

So:

- n=5: 5s² 5p⁶ 5d¹⁰ 5f³
- n=6: 6s² 6p⁶ 6d¹
- n=7: 7s²

But 5f is not filled until after 6s², so it’s okay to place it after 6s², but in n=5 group.

So correct format:
`1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 4f¹⁴ 5d¹⁰ 5f³ 6s² 6p⁶ 6d¹ 7s²`

But the student wrote:
`... 6s² 4f¹⁴ 5d¹⁰ 6p⁶ 7s² 5f³ 6d¹`

→ This has 4f¹⁴ and 5d¹⁰ before 6p⁶, which is wrong.

Mistake here:

- 6s² → 4f¹⁴ → 5d¹⁰ → 6p⁶ → 7s² → 5f³ → 6d¹

But 4f¹⁴ should be after 6s², but before 5d¹⁰? No.

Actual filling order:
- 6s² → 4f¹⁴ → 5d¹⁰ → 6p⁶ → 7s² → 5f³ → 6d¹

So the student wrote:
`... 6s² 4f¹⁴ 5d¹⁰ 6p⁶ 7s² 5f³ 6d¹`

That is correct in filling order, but out of n-order.

But in standard notation, we write in order of n, so:

- Group all n=4: 4s² 3d¹⁰ 4p⁶ 4f¹⁴
- n=5: 5s² 4d¹⁰ 5p⁶ 5d¹⁰ 5f³
- n=6: 6s² 6p⁶ 6d¹
- n=7: 7s²

So correct configuration should be:

`1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 4f¹⁴ 5s² 4d¹⁰ 5p⁶ 5d¹⁰ 5f³ 6s² 6p⁶ 6d¹ 7s²`

But the student wrote:
`... 6s² 4f¹⁴ 5d¹⁰ 6p⁶ 7s² 5f³ 6d¹` → 4f¹⁴ after 6s², but before 5d¹⁰ — that's okay, but 5f³ after 7s² — is that right?

Yes, uranium has 7s² 5f³ 6d¹, so it's fine.

But the order of writing is not in increasing n.

However, most textbooks allow writing in filling order, so this may be accepted.

But strictly speaking, 5f³ should be under n=5, so written before 6s²?

No — 5f is filled after 6s², so it's okay to write it later.

So final judgment: Acceptable, but not ideal.

But the total electrons are correct, and the subshells are correct.

✔ Acceptable — minor formatting issue

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#### 5. N (Nitrogen) – Atomic Number: 7
Given: `1s² 2s² 2p³`
✔ Correct!
- 2+2+3 = 7 ✔️
- 2p³ means three unpaired electrons (Hund’s Rule)

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#### 6. Ag (Silver) – Atomic Number: 47
Given: `1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s¹ 4d¹⁰`

This is marked with a star, meaning likely incorrect.

Let’s analyze:

Atomic number of Ag = 47

Normal filling order would be:
- After Kr (36): 5s² 4d⁹ → but Ag is an exception!

Silver (Ag) has a half-filled or fully filled d-subshell preference.

Expected: `[Kr] 5s² 4d⁹`
Actual: `[Kr] 5s¹ 4d¹⁰` → more stable because 4d¹⁰ is full

So correct configuration is:
`1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s¹ 4d¹⁰`

Wait — the student wrote exactly that!

So why is it marked with a star?

Let’s check the order:

Student wrote:
`1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s¹ 4d¹⁰`

Is this in correct order?

- After 4p⁶ → 5s¹ → 4d¹⁰ → YES

But 4d¹⁰ should come after 5s¹? Yes — in filling order.

But in writing, we usually write by increasing n, so:

- n=4: 4s² 3d¹⁰ 4p⁶ 4d¹⁰
- n=5: 5s¹

So correct order should be:
`... 4p⁶ 4d¹⁰ 5s¹`

But the student wrote: `5s¹ 4d¹⁰` — which is filling order, not n-order.

But again, this is common — many accept it.

But the configuration is correct:
- Total electrons:
- Up to Kr: 36
- 5s¹ → 37
- 4d¹⁰ → 47 → YES

✔ So configuration is correct.

But perhaps the star is because of order — should be `4d¹⁰` before `5s¹`?

No — 4d¹⁰ is filled after 5s¹, so it’s okay to write it after.

Wait — actually, 5s fills before 4d, so 5s¹ is written first, then 4d¹⁰.

So `5s¹ 4d¹⁰` is correct.

✔ So Ag is correct — the star may be a mistake, or perhaps the teacher wants it in n-order.

But standard notation allows writing in filling order.

So student is correct.

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#### 7. Ti (Titanium) – Atomic Number: 22
Given: `1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d²`

✔ Correct!
- 2+2+6+2+6+2+2 = 22 ✔️
- 4s² before 3d² — correct (filling order)

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#### 8. Ce (Cerium) – Atomic Number: 58
Given: `1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f¹ 5d¹`

Total electrons:
- Up to 5p⁶: 54
- 6s² → 56
- 4f¹ → 57
- 5d¹ → 58 ✔️

But Ce is an exception — it has 4f¹ 5d¹ instead of 4f².

And the configuration is correct.

But order: 6s² → 4f¹ → 5d¹ → YES

But 4f¹ should be written after 6s², but before 5d¹? Yes.

But in n-order, 4f is n=4, so should be grouped with n=4.

So correct way:
`... 4p⁶ 4f¹ 5s² 4d¹⁰ 5p⁶ 6s² 5d¹`

But the student wrote:
`... 5p⁶ 6s² 4f¹ 5d¹` → 4f¹ after 6s² — correct

But 4f¹ should be before 5s²? No — 4f fills after 6s².

So `6s² 4f¹ 5d¹` is acceptable.

✔ Correct

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#### 9. Cl (Chlorine) – Atomic Number: 17
Given: `1s² 2s² 2p⁶ 3s² 3p⁵`
✔ Correct!
- 2+2+6+2+5 = 17 ✔️

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#### 10. Hg (Mercury) – Atomic Number: 80
Given: `1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f¹⁴ 5d¹⁰`

Count:
- Up to 5p⁶: 54
- 6s² → 56
- 4f¹⁴ → 70
- 5d¹⁰ → 80 ✔️

But Hg has a full 5d¹⁰ and 6s², and no 6p electrons.

But is 4f¹⁴ before 5d¹⁰? Yes — filling order: 6s² → 4f¹⁴ → 5d¹⁰

But in writing, we write in order of n, so:

- n=4: 4s² 3d¹⁰ 4p⁶ 4f¹⁴
- n=5: 5s² 4d¹⁰ 5p⁶ 5d¹⁰
- n=6: 6s²

So correct order:
`... 4p⁶ 4f¹⁴ 5s² 4d¹⁰ 5p⁶ 5d¹⁰ 6s²`

But the student wrote:
`... 5p⁶ 6s² 4f¹⁴ 5d¹⁰`

→ This has 6s² before 4f¹⁴ — wrong order

Because 4f¹⁴ is filled after 6s², so it can be written after, but should be grouped with n=4

But more importantly: 4f¹⁴ should come before 5d¹⁰, but after 6s²

So the sequence is okay: `6s² 4f¹⁴ 5d¹⁰` — acceptable

But missing 5d¹⁰? No — it's there.

Wait: the configuration ends at `5d¹⁰`, but no 6p or anything else — good.

But Hg has 5d¹⁰ 6s², and 4f¹⁴ is present.

But the order is not in increasing n.

But filling order: 6s² → 4f¹⁴ → 5d¹⁰ → YES

So `6s² 4f¹⁴ 5d¹⁰` is acceptable.

✔ Correct

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🟡 Summary of Errors:

| Element | Given | Correct? | Notes |
|--------|-------|---------|------|
| Na | `1s² 2s² 2p⁶ 3s¹` | ✔ | Correct |
| Pb | Long config | ✔ | Correct, minor order |
| Sr | `... 5s²` | ✔ | Correct |
| U | `... 7s² 5f³ 6d¹` | ⚠️ | Order off, but content correct |
| N | `1s² 2s² 2p³` | ✔ | Correct |
| Ag | `... 5s¹ 4d¹⁰` | ✔ | Correct — star may be error |
| Ti | `... 4s² 3d²` | ✔ | Correct |
| Ce | `... 6s² 4f¹ 5d¹` | ✔ | Correct |
| Cl | `... 3p⁵` | ✔ | Correct |
| Hg | `... 6s² 4f¹⁴ 5d¹⁰` | ⚠️ | Order not ideal, but acceptable |

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❗ Final Note on Ag (#6):

Student wrote: `1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s¹ 4d¹⁰`
✔ This is correct — Silver does have a [Kr] 5s¹ 4d¹⁰ configuration due to stability of full d-subshell.

So the star is likely misplaced — this is correct.

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✔ Final Answer:

All configurations are correct except possibly order issues, but only Ag is truly correct and the star is likely a mistake.

Recommendation:
- Ag is correct — remove the star.
- For U and Hg, suggest writing in increasing n order for clarity.
- Otherwise, all answers are accurate.

Let me know if you want the corrected versions in proper order!