Electron Configuration Practice worksheet for learning atomic structure and electron arrangement.

Let's solve each part of this Electron Configuration Practice worksheet step by step.

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A. For each element: identify the number of electrons, fill in the electron diagram with up-spin & down-spin electrons, and write the electron configuration.

We will use:
- `^` for up-spin (↑)
- `v` for down-spin (↓)
- Use `^` to represent superscript (e.g., 2s^2 means 2s²)

#### 1. Nitrogen (N)
- Atomic number = 7 → 7 electrons
- Electron configuration: 1s² 2s² 2p³
- Orbital filling:
- 1s: ↑↓
- 2s: ↑↓
- 2p: ↑ ↑ ↑ (Hund’s rule: parallel spins first)

```
Number of electrons: 7
Electron configuration: 1s^2 2s^2 2p^3

Orbital Diagram:
[↑↓] [↑↓] [↑] [↑] [↑] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ]
1s 2s 2p_x 2p_y 2p_z 3s 3p 4s 3d 4p 5s 4d 5p
```

> Note: Only fill up to 2p.

---

#### 2. Silicon (Si)
- Atomic number = 14 → 14 electrons
- Electron configuration: 1s² 2s² 2p⁶ 3s² 3p²
- Orbital filling:
- 1s: ↑↓
- 2s: ↑↓
- 2p: ↑↓ ↑↓ ↑↓
- 3s: ↑↓
- 3p: ↑ ↑

```
Number of electrons: 14
Electron configuration: 1s^2 2s^2 2p^6 3s^2 3p^2

Orbital Diagram:
[↑↓] [↑↓] [↑↓] [↑↓] [↑↓] [↑↓] [↑↓] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ]
1s 2s 2p_x 2p_y 2p_z 3s 3p_x 3p_y 3p_z 4s 3d 4p 5s 4d 5p
```

> Fill 3p orbitals with two electrons (parallel spins).

---

#### 3. Neon (Ne)
- Atomic number = 10 → 10 electrons
- Electron configuration: 1s² 2s² 2p⁶
- All orbitals filled.

```
Number of electrons: 10
Electron configuration: 1s^2 2s^2 2p^6

Orbital Diagram:
[↑↓] [↑↓] [↑↓] [↑↓] [↑↓] [↑↓] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ]
1s 2s 2p_x 2p_y 2p_z 3s 3p_x 3p_y 3p_z 4s 3d 4p 5s 4d 5p
```

---

#### 4. Copper (Cu)
- Atomic number = 29 → 29 electrons
- Exception: Cu has [Ar] 3d¹⁰ 4s¹ instead of 3d⁹ 4s² (for stability)
- Full config: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d¹⁰
- But written as: 1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^1

Orbital filling:
- 3d is completely filled (10 electrons), 4s has 1.

```
Number of electrons: 29
Electron configuration: 1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^1

Orbital Diagram:
[↑↓] [↑↓] [↑↓] [↑↓] [↑↓] [↑↓] [↑↓] [↑↓] [↑↓] [↑↓] [↑↓] [↑↓] [↑↓] [↑↓] [↑↓]
1s 2s 2p_x 2p_y 2p_z 3s 3p_x 3p_y 3p_z 4s 3d_1 3d_2 3d_3 3d_4 3d_5
[↑↓] [↑↓] [↑↓] [↑↓] [↑↓] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ]
3d_6 3d_7 3d_8 3d_9 3d_10 4p_x 4p_y 4p_z 5s 4d 5p
```

> 3d has 10 electrons (all paired), 4s has one ↑

---

#### 5. Zirconium (Zr)
- Atomic number = 40 → 40 electrons
- Electron configuration: [Kr] 5s² 4d², but actually it's [Kr] 5s² 4d² — no exception here.
- Full config: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d²

```
Number of electrons: 40
Electron configuration: 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^2

Orbital Diagram:
[↑↓] [↑↓] [↑↓] [↑↓] [↑↓] [↑↓] [↑↓] [↑↓] [↑↓] [↑↓] [↑↓] [↑↓] [↑↓] [↑↓] [↑↓]
1s 2s 2p_x 2p_y 2p_z 3s 3p_x 3p_y 3p_z 4s 3d_1 3d_2 3d_3 3d_4 3d_5
[↑↓] [↑↓] [↑↓] [↑↓] [↑↓] [↑↓] [↑↓] [↑↓] [↑↓] [↑↓] [↑↓] [↑↓] [↑↓] [↑↓] [↑↓]
3d_6 3d_7 3d_8 3d_9 3d_10 4p_x 4p_y 4p_z 5s 4d_1 4d_2 4d_3 4d_4 4d_5 4d_6
[ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ]
4d_7 4d_8 4d_9 4d_10 5p_x 5p_y 5p_z 6s 5d 6p
```

> 4d has two electrons: ↑ ↑ (same spin, Hund’s rule)

---

B. Write the symbol of the element with the following ground state electron configurations:

#### 6. 1s² 2s²
- Total electrons = 4 → Beryllium (Be)

#### 7. 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d²
- Count: 2+2+6+2+6+2+2 = 22 electrons → Titanium (Ti)

#### 8. 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁴
- Count: 2+2+6+2+6+2+10+4 = 34 electrons → Selenium (Se)

#### 9. 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s¹
- Count: 2+2+6+2+6+2+10+6+1 = 37 electrons → Rubidium (Rb)

> Wait! 5s¹ after 4p⁶ → Rb is [Kr] 5s¹, Kr is 36, so 37 → yes.

#### 10. 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 5d⁹
- Count: Add all:
- Up to 5p⁶: 36 (Kr) + 5s²=2 → 38, +4d¹⁰=10 → 48, +5p⁶=6 → 54, +6s²=2 → 56, +5d⁹=9 → 65
- 65 electrons → Terbium (Tb)? Wait: atomic number 65 is Tb, but let’s check:

Wait: After 5p⁶ (Xe, 54), then 6s² → 56 (Ba), 4f¹⁴ → 70 (Yb), but here we have 5d⁹.

Actually: Xe (54) + 6s² (56) + 4f¹⁴ (70) + 5d⁹ → wait, this sequence is wrong.

But the order is:
- 1s²...5p⁶ = Xe (54)
- Then 6s² = 56 (Ba)
- Then 4f¹⁴? No, here it's 5d⁹ — that’s not correct.

Wait: The configuration given is:
1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 5d⁹

So:
- Up to 5p⁶ = 54 (Xe)
- 6s² = 56
- 5d⁹ = 9 → total 56 + 9 = 65

So atomic number = 65 → Terbium (Tb)

But Tb has configuration: [Xe] 4f¹⁴ 5d¹ 6s² → not 5d⁹

Wait — something's off.

Wait: This configuration skips 4f!

But after 6s², 4f comes before 5d.

So this configuration is invalid unless it's an excited state.

But it says "ground state".

So let's count:

- 1s² (2)
- 2s² (4)
- 2p⁶ (10)
- 3s² (12)
- 3p⁶ (18)
- 4s² (20)
- 3d¹⁰ (30)
- 4p⁶ (36)
- 5s² (38)
- 4d¹⁰ (48)
- 5p⁶ (54)
- 6s² (56)
- 5d⁹ (65)

Total: 65 electrons → Element 65 is Terbium (Tb)

But Tb’s actual ground state is: [Xe] 4f¹⁴ 5d¹ 6s² → not 5d⁹

So this configuration is not a valid ground state.

Wait — perhaps it's Gold (Au)?

Au is 79: [Xe] 4f¹⁴ 5d¹⁰ 6s¹

Not matching.

Wait — maybe it's Lanthanum (La)? No.

Wait — let's look again.

Is there an element with 65 electrons and 5d⁹?

No — because after 6s², 4f fills first.

So this configuration must be incorrect or excited.

But if we assume it's correct as written, then total electrons = 65 → Tb

But Tb does not have 5d⁹.

Wait — what about Cadmium (Cd)? 48 electrons.

No.

Wait — perhaps the configuration is misordered.

But the notation is written in order of increasing n.

Wait — the last part: 6s² 5d⁹ — so 5d before 6p, etc.

But 5d should come after 6s, but before 4f?

No — 4f comes before 5d in filling order.

So this configuration is incorrect.

But if we ignore that and just sum the electrons, it's 65 → Terbium (Tb)

But Tb’s ground state is [Xe] 4f¹⁴ 5d¹ 6s²

So this configuration is not ground state.

Perhaps it's a typo?

Wait — could it be Iridium (Ir)? Ir is 47 → too small.

Wait — let's recheck:

Wait — the configuration ends with 5d⁹ — only one element has 5d⁹ in ground state?

No — Copper-like exception?

But no.

Wait — perhaps it's Platinum (Pt)? Pt is 78: [Xe] 4f¹⁴ 5d⁹ 6s¹ → YES!

But Pt has 78 electrons.

Our total is 65 → no.

Wait — our total is 65 → Tb.

But Tb is [Xe] 4f¹⁴ 5d¹ 6s² → not 5d⁹

So this configuration is invalid.

But perhaps the question meant to write:

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f¹⁴ 5d⁹

Then total: 54 + 2 + 14 + 9 = 79 → Gold (Au)

And Au is [Xe] 4f¹⁴ 5d¹⁰ 6s¹ — not 5d⁹

Still not matching.

Wait — Silver (Ag): [Kr] 5s¹ 4d¹⁰ → 47 electrons

No.

Wait — Copper: [Ar] 4s¹ 3d¹⁰ → 29

No.

So perhaps the configuration is wrong.

But let's accept the sum: 65 electrons → Tb

But Tb doesn’t have that config.

Wait — maybe it's Hafnium (Hf)? Hf is 72.

No.

Wait — perhaps the configuration is:

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 5d⁹

That’s 54 (Xe) + 2 (6s²) + 9 (5d⁹) = 65

So element 65 → Terbium (Tb)

Even though it's not the ground state, the problem asks for the element with that configuration — perhaps it's hypothetical.

But in reality, no element has 5d⁹ in ground state.

Wait — Rhodium (Rh) is [Kr] 5s¹ 4d⁸ → 45

No.

Wait — Cobalt?

No.

Perhaps it's Palladium (Pd)? Pd is 46: [Kr] 4d¹⁰ → no 5d.

I think there might be a mistake.

Wait — what if it's Lutetium (Lu)? Lu is 71: [Xe] 4f¹⁴ 5d¹ 6s²

Still not 5d⁹.

Wait — maybe it's a typo and it's supposed to be 5d¹⁰?

But let's go back.

Wait — element with 65 electrons is Terbium (Tb) — so answer is Tb

Even if the configuration is not ground state, the problem says “with the following ground state” — so it must be a valid ground state.

So perhaps the configuration is wrong.

Wait — let me double-check:

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 5d⁹

Sum:
- 1s² = 2
- 2s² = 4
- 2p⁶ = 10
- 3s² = 12
- 3p⁶ = 18
- 4s² = 20
- 3d¹⁰ = 30
- 4p⁶ = 36
- 5s² = 38
- 4d¹⁰ = 48
- 5p⁶ = 54
- 6s² = 56
- 5d⁹ = 65

Yes, 65 electrons.

But Tb (atomic number 65) has configuration: [Xe] 4f¹⁴ 5d¹ 6s²

So this configuration is not the ground state.

Therefore, there is no element with this ground state configuration.

But perhaps the problem has a typo.

Alternatively, maybe it's Cesium (Cs)? No.

Wait — perhaps the configuration is for Gold (Au)?

Au: 79 → [Xe] 4f¹⁴ 5d¹⁰ 6s¹

Not matching.

Wait — what about Platinum (Pt)? 78: [Xe] 4f¹⁴ 5d⁹ 6s¹ → YES!

So configuration: 1s²…5p⁶ = 54 (Xe) + 4f¹⁴ = 68 + 5d⁹ = 77 + 6s¹ = 78

So if the configuration were:

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 4f¹⁴ 5d⁹ 6s¹

Then it would be Pt.

But here, it's missing 4f¹⁴ and has 6s² 5d⁹ — so not correct.

So likely, the configuration is invalid.

But since the sum is 65, and the problem expects an answer, we'll go with Tb.

But I suspect a typo.

Wait — perhaps it's 5d¹⁰? Then 54 + 2 + 10 = 66 → Dy

No.

Wait — let's move on.

For now, based on electron count:

Answer: Terbium (Tb)

But note: this configuration is not the ground state.

---

C. Write the ground state electron configuration for the following:

#### 11. Iron (Fe)
- Atomic number 26
- Config: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁶
- Or: 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^6

#### 12. Sulfur (S)
- Atomic number 16
- Config: 1s² 2s² 2p⁶ 3s² 3p⁴
- Or: 1s^2 2s^2 2p^6 3s^2 3p^4

#### 13. Krypton (Kr)
- Atomic number 36
- Config: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶
- Or: 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6

---

D. Use shorthand (noble gas) notation

#### 14. Chlorine (Cl)
- Atomic number 17
- Noble gas before: Neon (Ne, 10)
- Cl: [Ne] 3s² 3p⁵
- So: [Ne] 3s^2 3p^5

#### 15. Antimony (Sb)
- Atomic number 51
- Noble gas before: Kr (36)
- After Kr: 5s² 4d¹⁰ 5p³
- So: [Kr] 5s^2 4d^10 5p^3

#### 16. Lanthanum (La)
- Atomic number 57
- Noble gas before: Xe (54)
- After Xe: 6s² 5d¹
- So: [Xe] 6s^2 5d^1

> Note: La is the first lanthanide, and its configuration is [Xe] 5d¹ 6s²

---

✔ Final Answers:

---

A.

| Element | Electrons | Configuration | Orbital Diagram (partial) |
|--------|----------|---------------|---------------------------|
| 1. Nitrogen | 7 | 1s^2 2s^2 2p^3 | 1s: ↑↓, 2s: ↑↓, 2p: ↑ ↑ ↑ |
| 2. Silicon | 14 | 1s^2 2s^2 2p^6 3s^2 3p^2 | 1s-3p filled, 3p: ↑ ↑ |
| 3. Neon | 10 | 1s^2 2s^2 2p^6 | All filled |
| 4. Copper | 29 | 1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^1 | 3d full, 4s: ↑ |
| 5. Zirconium | 40 | 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^2 | 4d: ↑ ↑ |

---

B.

6. Be
7. Ti
8. Se
9. Rb
10. Tb (though configuration is not ground state; possibly error)

> Note: #10 may be intended to be Pt or Au, but based on count, it's Tb.

---

C.

11. iron: 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^6
12. sulfur: 1s^2 2s^2 2p^6 3s^2 3p^4
13. krypton: 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6

---

D.

14. chlorine: [Ne] 3s^2 3p^5
15. antimony: [Kr] 5s^2 4d^10 5p^3
16. lanthanum: [Xe] 6s^2 5d^1

---

✔ Final Answer Summary:

A.
1. 7, 1s^2 2s^2 2p^3
2. 14, 1s^2 2s^2 2p^6 3s^2 3p^2
3. 10, 1s^2 2s^2 2p^6
4. 29, 1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^1
5. 40, 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^2

B.
6. Be
7. Ti
8. Se
9. Rb
10. Tb (note: configuration is not standard ground state)

C.
11. 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^6
12. 1s^2 2s^2 2p^6 3s^2 3p^4
13. 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6

D.
14. [Ne] 3s^2 3p^5
15. [Kr] 5s^2 4d^10 5p^3
16. [Xe] 6s^2 5d^1

---

Let me know if you'd like the orbital diagrams drawn out visually!