Free Printable Electron Configuration Worksheets - Free Printable

Here is the complete solution to your Electron Configuration Worksheet, with clear explanations for each problem.

---

1. Name the elements that have the following electron configurations.

i. `1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁵`

- Add up all the superscripts (electrons):
2 + 2 + 6 + 2 + 6 + 2 + 5 = 25 electrons
- Atomic number 25 → Manganese (Mn)

✔ Answer: Manganese (Mn)

---

ii. `1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁴`

- Total electrons: 2+2+6+2+6+2+10+4 = 34 electrons
- Atomic number 34 → Selenium (Se)

✔ Answer: Selenium (Se)

---

2. Write the ground state electron configuration for a neutral atom of the following elements.

i. Oxygen (O) — atomic number 8

- Fill orbitals in order: 1s, 2s, 2p
- 1s² 2s² 2p⁴

✔ Answer: `1s² 2s² 2p⁴`

---

ii. Krypton (Kr) — atomic number 36

- Full configuration:
1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶

✔ Answer: `1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶`

*(Note: 4s fills before 3d, but we write in order of principal quantum number: 3d before 4p)*

---

iii. Chromium (Cr) — atomic number 24

- Expected: [Ar] 4s² 3d⁴
- But Chromium is an exception — it’s more stable with half-filled d subshell:
→ [Ar] 4s¹ 3d⁵

Full configuration:
1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d⁵

✔ Answer: `1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d⁵`

---

3. Using the long method, give the electron configuration of

*(“Long method” means writing out all orbitals without noble gas shorthand)*

i. Magnesium (Mg) — atomic number 12

- 1s² 2s² 2p⁶ 3s²

✔ Answer: `1s² 2s² 2p⁶ 3s²`

---

ii. Potassium (K) — atomic number 19

- 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹

✔ Answer: `1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹`

---

iii. Lithium (Li) — atomic number 3

- 1s² 2s¹

✔ Answer: `1s² 2s¹`

---

iv. Nickel (Ni) — atomic number 28

- Expected: [Ar] 4s² 3d⁸ → full config:
1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁸

✔ Answer: `1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁸`

*(Note: Nickel does NOT have an exception like Cr or Cu)*

---

v. Sulfur (S) — atomic number 16

- 1s² 2s² 2p⁶ 3s² 3p⁴

✔ Answer: `1s² 2s² 2p⁶ 3s² 3p⁴`

---

4. Identify the following elements.

i. `1s² 2s² 2p²`

- Total electrons: 2+2+2 = 6 → Carbon (C)

✔ Answer: Carbon

---

ii. `1s² 2s² 2p⁶`

- Total: 10 electrons → Neon (Ne)

✔ Answer: Neon

---

iii. `[Ar] 4s² 3d¹⁰ 4p⁵`

- [Ar] = 18 electrons
- Add: 2 (4s) + 10 (3d) + 5 (4p) = 17 → Total = 35
- Atomic number 35 → Bromine (Br)

✔ Answer: Bromine

---

iv. `[Kr] 5s² 4d¹`

- [Kr] = 36 electrons
- Add: 2 (5s) + 1 (4d) = 3 → Total = 39
- Atomic number 39 → Yttrium (Y)

✔ Answer: Yttrium

---

v. `[Ne] 3s² 3p³`

- [Ne] = 10 electrons
- Add: 2 + 3 = 5 → Total = 15
- Atomic number 15 → Phosphorus (P)

✔ Answer: Phosphorus

---

5. Determine the electron configuration using the short method.

*(“Short method” = noble gas core notation)*

i. Strontium (Sr) — atomic number 38

- Previous noble gas: Krypton (Kr, Z=36)
- Remaining: 2 electrons → 5s²
→ [Kr] 5s²

✔ Answer: `[Kr] 5s²`

---

ii. Bromine (Br) — atomic number 35

- Previous noble gas: Argon (Ar, Z=18)
- Remaining: 17 electrons → 4s² 3d¹⁰ 4p⁵
→ [Ar] 4s² 3d¹⁰ 4p⁵

✔ Answer: `[Ar] 4s² 3d¹⁰ 4p⁵`

*(Note: Some may write [Ar] 3d¹⁰ 4s² 4p⁵ — same thing; order doesn’t matter as long as subshells are correct)*

---

iii. Zirconium (Zr) — atomic number 40

- Previous noble gas: Krypton (Kr, Z=36)
- Remaining: 4 electrons → 5s² 4d²
→ [Kr] 5s² 4d²

✔ Answer: `[Kr] 5s² 4d²`

---

iv. Molybdenum (Mo) — atomic number 42

- Expected: [Kr] 5s² 4d⁴
- BUT Mo is an exception — it prefers half-filled 4d subshell:
→ [Kr] 5s¹ 4d⁵

✔ Answer: `[Kr] 5s¹ 4d⁵`

---

v. Silver (Ag) — atomic number 47

- Expected: [Kr] 5s² 4d⁹
- BUT Ag is an exception — full d subshell is more stable:
→ [Kr] 5s¹ 4d¹⁰

✔ Answer: `[Kr] 5s¹ 4d¹⁰`

---

## ✔ Final Answers Summary:

1.

i. Manganese (Mn)
ii. Selenium (Se)

2.

i. `1s² 2s² 2p⁴`
ii. `1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶`
iii. `1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d⁵`

3.

i. `1s² 2s² 2p⁶ 3s²`
ii. `1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹`
iii. `1s² 2s¹`
iv. `1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁸`
v. `1s² 2s² 2p⁶ 3s² 3p⁴`

4.

i. Carbon
ii. Neon
iii. Bromine
iv. Yttrium
v. Phosphorus

5.

i. `[Kr] 5s²`
ii. `[Ar] 4s² 3d¹⁰ 4p⁵`
iii. `[Kr] 5s² 4d²`
iv. `[Kr] 5s¹ 4d⁵`
v. `[Kr] 5s¹ 4d¹⁰`

---

Let me know if you’d like diagrams, orbital filling rules, or exceptions explained further! 🧪⚛️