Free Printable Electron Configuration Worksheets - Free Printable

Let's solve each part of this Electron Configuration worksheet step by step, with explanations.

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1. Use noble gases to write the abbreviated electron configurations

We use the noble gas that comes before the element in the periodic table as a shorthand for the core electrons.

#### i. Li (Lithium)
- Atomic number: 3
- Full configuration: 1s² 2s¹
- Noble gas before Li is He (1s²)
- Abbreviated: [He] 2s¹

#### ii. B (Boron)
- Atomic number: 5
- Full: 1s² 2s² 2p¹
- Noble gas: He
- Abbreviated: [He] 2s² 2p¹

#### iii. O (Oxygen)
- Atomic number: 8
- Full: 1s² 2s² 2p⁴
- Noble gas: He
- Abbreviated: [He] 2s² 2p⁴

#### iv. Mg (Magnesium)
- Atomic number: 12
- Full: 1s² 2s² 2p⁶ 3s²
- Noble gas: Ne (1s² 2s² 2p⁶)
- Abbreviated: [Ne] 3s²

#### v. Cl (Chlorine)
- Atomic number: 17
- Full: 1s² 2s² 2p⁶ 3s² 3p⁵
- Noble gas: Ne
- Abbreviated: [Ne] 3s² 3p⁵

#### vi. V (Vanadium)
- Atomic number: 23
- Full: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d³
- Noble gas: Ar (1s² 2s² 2p⁶ 3s² 3p⁶)
- Abbreviated: [Ar] 4s² 3d³

#### vii. Se²⁻ (Selenide ion)
- Se atomic number = 34 → Se²⁻ has 36 electrons
- Full configuration of Se²⁻: same as Kr (krypton), which is 36 electrons
- So, [Kr]
- Abbreviated: [Kr]

> Note: Se²⁻ gains 2 electrons and achieves noble gas configuration of Kr.

#### viii. Cr²⁺ (Chromium ion)
- Cr atomic number = 24
- Neutral Cr: [Ar] 4s¹ 3d⁵ (exception due to half-filled stability)
- Cr²⁺ loses 2 electrons: first from 4s, then from 3d
- So: [Ar] 3d⁴
- Abbreviated: [Ar] 3d⁴

> Note: Cr loses 4s electrons before 3d.

#### ix. I⁻ (Iodide ion)
- I atomic number = 53 → I⁻ has 54 electrons
- 54 electrons = Xe (Xenon)
- Abbreviated: [Xe]

#### x. Ag (Silver)
- Atomic number: 47
- Normal order would be [Kr] 5s² 4d⁹, but Ag is an exception
- Actual configuration: [Kr] 5s¹ 4d¹⁰ (to have full d-subshell)
- Abbreviated: [Kr] 5s¹ 4d¹⁰

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✔ Answers for #1:

| Element | Abbreviated Configuration |
|--------|----------------------------|
| i. Li | [He] 2s¹ |
| ii. B | [He] 2s² 2p¹ |
| iii. O | [He] 2s² 2p⁴ |
| iv. Mg | [Ne] 3s² |
| v. Cl | [Ne] 3s² 3p⁵ |
| vi. V | [Ar] 4s² 3d³ |
| vii. Se²⁻ | [Kr] |
| viii. Cr²⁺ | [Ar] 3d⁴ |
| ix. I⁻ | [Xe] |
| x. Ag | [Kr] 5s¹ 4d¹⁰ |

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2. Identify the element with the following electron configuration

We add up the electrons and find the element with that atomic number.

#### i. 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p¹
- Total electrons: 2+2+6+2+6+2+10+1 = 31
- Element: Ga (Gallium)

#### ii. 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s²
- Total: 2+2+6+2+6+2+10+6+2 = 38
- Element: Sr (Strontium)

#### iii. 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d⁷
- Total: 2+2+6+2+6+2+10+6+2+7 = 45
- Element: Rh (Rhodium)

#### iv. 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f¹⁴ 5d¹⁰
- Count:
- Up to 6s²: 2+2+6+2+6+2+10+6+2+10+6+2 = 58
- Add 4f¹⁴ = 58 + 14 = 72
- Add 5d¹⁰ = 72 + 10 = 82
- Element: Pb (Lead)

> Wait: Let’s check carefully — this configuration ends at 5d¹⁰, but it's missing 6p⁶? No, we have:

Wait — let's re-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
- 4f¹⁴ → 70
- 5d¹⁰ → 80

So total: 80 electrons → Element: Hg (Mercury)

> Correction: The configuration is:
> 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f¹⁴ 5d¹⁰
> That’s:
> - Core up to Xe (54)
> - Then 6s² → 56
> - 4f¹⁴ → 70
> - 5d¹⁰ → 80
> → Hg (Mercury)

#### v. 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f¹⁴ 5d¹⁰ 6p⁶ 7s¹
- Up to 6p⁶: 86 (Rn)
- Then 7s¹ → 87
- So 87 electrons → Element: Fr (Francium)

But wait: Rn is 86, so 7s¹ → 87 → Fr

✔ Answers for #2:

| Configuration | Element |
|--------------|---------|
| i. ...4p¹ | Ga |
| ii. ...5s² | Sr |
| iii. ...4d⁷ | Rh |
| iv. ...5d¹⁰ | Hg |
| v. ...7s¹ | Fr |

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3. Determine if the following electron configurations are correct

#### i. 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 4d¹⁰ 4p⁵
- Order is wrong: 4d should come after 5s, not before
- Correct order: 4s before 3d, but 4d after 5s
- Here, 4d¹⁰ appears before 5s — incorrect
- Also, 4p⁵ suggests element like Br, but 4d¹⁰ implies transition metals
- This configuration skips 5s and goes to 4d — Incorrect

> No, because 4d fills after 5s.

#### ii. 1s² 2s² 2p⁶ 3s¹
- Total electrons: 2+2+6+1 = 11 → Na (Sodium)
- But sodium is 1s² 2s² 2p⁶ 3s¹ → ✔ Correct

#### iii. [Rn] 7s² 5f⁶ 6d²
- Rn = 86 electrons
- 7s² = 2 → 88
- 5f⁶ = 6 → 94
- 6d² = 2 → 96
- Element with 96 electrons: Cm (Curium), but Cm is [Rn] 5f⁷ 6d¹ 7s²
- This one has 5f⁶ 6d² 7s² → different order
- But order matters — 5f fills before 6d, but 7s before both
- However, the configuration is possible for some actinides, but this is not the ground state of any element
- Actually, no element has 5f⁶ 6d² 7s² as ground state — incorrect

> More precisely: No, because the expected configuration for element 96 (Cm) is [Rn] 5f⁷ 6d¹ 7s², not this.

So Incorrect

#### iv. [Ar] 5s² 4d¹⁰ 5p⁵
- [Ar] = 18 electrons
- 5s² = 2 → 20
- 4d¹⁰ = 10 → 30
- 5p⁵ = 5 → 35
- Total: 35 electrons → Element: Br (Bromine)
- But bromine is [Ar] 4s² 3d¹⁰ 4p⁵ → not 5s² 4d¹⁰ 5p⁵
- This configuration is for a higher period — 5s and 5p suggest Period 5
- But 4d¹⁰ is fine, but the order is wrong — 5s should be filled before 4d?
- No: 4d fills after 5s
- But here: [Ar] 5s² 4d¹⁰ 5p⁵ → seems okay in order
- But atomic number 35 is Br, which is in Period 4 → cannot have 5s or 5p
- So this configuration is for element 35 but with wrong orbitals → Incorrect

> No, because Br doesn’t have 5s or 5p electrons.

#### v. [Xe] 6s² 4f¹⁰
- [Xe] = 54
- 6s² = 56
- 4f¹⁰ = 66 electrons
- Element 66: Dy (Dysprosium) → Ground state: [Xe] 6s² 4f¹⁰ → ✔ Correct

> Yes, Dy is [Xe] 6s² 4f¹⁰

✔ Answers for #3:

| Configuration | Correct? |
|-------------|----------|
| i. ...4d¹⁰ 4p⁵ | ✘ Incorrect (wrong order: 4d before 5s) |
| ii. 1s² 2s² 2p⁶ 3s¹ | ✔ Correct |
| iii. [Rn] 7s² 5f⁶ 6d² | ✘ Incorrect (not a valid ground state) |
| iv. [Ar] 5s² 4d¹⁰ 5p⁵ | ✘ Incorrect (wrong element/period) |
| v. [Xe] 6s² 4f¹⁰ | ✔ Correct |

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4. Identify the following elements

#### i. 1s² 2s² 2p²
- Total electrons: 2+2+2 = 6 → C (Carbon)

#### ii. 1s² 2s² 2p⁶
- Total: 10 electrons → Ne (Neon)

#### iii. [Ar] 4s² 3d¹⁰ 4p⁵
- [Ar] = 18
- 4s² = 20
- 3d¹⁰ = 30
- 4p⁵ = 35 → Br (Bromine)

#### iv. [Kr] 5s² 4d¹
- [Kr] = 36
- 5s² = 38
- 4d¹ = 39 → Y (Yttrium)

#### v. [Ne] 3s² 3p³
- [Ne] = 10
- 3s² = 12
- 3p³ = 15 → P (Phosphorus)

✔ Answers for #4:

| Configuration | Element |
|-------------|---------|
| i. 1s² 2s² 2p² | C |
| ii. 1s² 2s² 2p⁶ | Ne |
| iii. [Ar] 4s² 3d¹⁰ 4p⁵ | Br |
| iv. [Kr] 5s² 4d¹ | Y |
| v. [Ne] 3s² 3p³ | P |

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✔ Final Answers Summary:

#### 1. Abbreviated Configurations
i. [He] 2s¹
ii. [He] 2s² 2p¹
iii. [He] 2s² 2p⁴
iv. [Ne] 3s²
v. [Ne] 3s² 3p⁵
vi. [Ar] 4s² 3d³
vii. [Kr]
viii. [Ar] 3d⁴
ix. [Xe]
x. [Kr] 5s¹ 4d¹⁰

#### 2. Identify Elements
i. Ga
ii. Sr
iii. Rh
iv. Hg
v. Fr

#### 3. Are they correct?
i. ✘ No
ii. ✔ Yes
iii. ✘ No
iv. ✘ No
v. ✔ Yes

#### 4. Identify Elements
i. C
ii. Ne
iii. Br
iv. Y
v. P

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