Electron Configuration worksheet for chemistry students to practice identifying elements, writing configurations, and understanding orbital properties.
Electron Configuration worksheet with questions on determining elements from configurations, writing full and abbreviated electron configurations, orbital capacity, and orbital shapes.
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Step-by-step solution for: Free Printable Electron Configuration Worksheets
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Step-by-step solution for: Free Printable Electron Configuration Worksheets
Here’s the complete solution to the Electron Configuration worksheet, with explanations for each part.
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## 1. Determine what elements are denoted by the following configurations.
To identify the element, add up all the superscripts (which represent the number of electrons). The total equals the atomic number, which identifies the element.
- Total electrons: 2 + 2 + 6 + 2 + 4 = 16
- Element with atomic number 16: Sulfur (S)
✔ Answer: Sulfur (S)
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- Total electrons: 2+2+6+2+6+2+5 = 25
- Element with atomic number 25: Manganese (Mn)
✔ Answer: Manganese (Mn)
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- Total electrons: 2+2+6+2+6+2+10+6 = 36
- Element with atomic number 36: Krypton (Kr)
✔ Answer: Krypton (Kr)
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- [Kr] = 36 electrons
- Add: 2 (5s²) + 10 (4d¹⁰) + 3 (5p³) = 15
- Total: 36 + 15 = 51
- Element with atomic number 51: Antimony (Sb)
✔ Answer: Antimony (Sb)
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- [Ne] = 10 electrons
- Add: 2 (3s²) + 1 (3p¹) = 3
- Total: 10 + 3 = 13
- Element with atomic number 13: Aluminum (Al)
✔ Answer: Aluminum (Al)
---
## 2. Write the full electron configuration of the following elements
Use the Aufbau principle (fill orbitals in order of increasing energy).
- Order: 1s → 2s → 2p
- 1s² 2s² 2p³
✔ Answer: 1s² 2s² 2p³
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- 1s² 2s² 2p⁶ 3s² 3p²
✔ Answer: 1s² 2s² 2p⁶ 3s² 3p²
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- 1s² 2s² 2p⁶ 3s² 3p⁶ 4s²
✔ Answer: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s²
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- After Kr (36), next is 5s²
- So: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s²
✔ Answer: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s²
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- Full config:
1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f¹⁴ 5d¹⁰ 6p³
✔ Answer: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f¹⁴ 5d¹⁰ 6p³
*(Note: This follows the Aufbau order — after 5p⁶ comes 6s, then 4f, then 5d, then 6p)*
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## 3. Write the abbreviated electron configuration
Use the noble gas that comes before the element in the periodic table.
- Noble gas before F: Helium (He, 2)
- Remaining: 2s² 2p⁵
- Abbreviated: [He] 2s² 2p⁵
✔ Answer: [He] 2s² 2p⁵
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- Noble gas before Mg: Neon (Ne, 10)
- Remaining: 3s²
- Abbreviated: [Ne] 3s²
✔ Answer: [Ne] 3s²
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- Noble gas before Se: Argon (Ar, 18)
- Remaining: 4s² 3d¹⁰ 4p⁴
- Abbreviated: [Ar] 4s² 3d¹⁰ 4p⁴
✔ Answer: [Ar] 4s² 3d¹⁰ 4p⁴
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- Noble gas before Rb: Krypton (Kr, 36)
- Remaining: 5s¹
- Abbreviated: [Kr] 5s¹
✔ Answer: [Kr] 5s¹
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- Noble gas before Pb: Xenon (Xe, 54)
- Remaining: 6s² 4f¹⁴ 5d¹⁰ 6p²
- Abbreviated: [Xe] 6s² 4f¹⁴ 5d¹⁰ 6p²
✔ Answer: [Xe] 6s² 4f¹⁴ 5d¹⁰ 6p²
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## 4. If each orbital can hold a maximum of two electrons, how many can each of the following hold?
Each subshell has a fixed number of orbitals:
- s → 1 orbital → holds 2 electrons
- p → 3 orbitals → holds 6 electrons
- d → 5 orbitals → holds 10 electrons
- f → 7 orbitals → holds 14 electrons
So:
✔ Answers:
i. 2
ii. 6
iii. 14
iv. 10
v. 10
---
## 5. What is the shape of an s orbital?
- Spherical — symmetric in all directions around the nucleus.
✔ Answer: Spherical
---
## 6. What is the shape of a p orbital?
- Dumbbell-shaped — two lobes with a node at the nucleus.
✔ Answer: Dumbbell
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## ✔ Final Answers Summary:
i. Sulfur (S)
ii. Manganese (Mn)
iii. Krypton (Kr)
iv. Antimony (Sb)
v. Aluminum (Al)
i. 1s² 2s² 2p³
ii. 1s² 2s² 2p⁶ 3s² 3p²
iii. 1s² 2s² 2p⁶ 3s² 3p⁶ 4s²
iv. 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s²
v. 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f¹⁴ 5d¹⁰ 6p³
i. [He] 2s² 2p⁵
ii. [Ne] 3s²
iii. [Ar] 4s² 3d¹⁰ 4p⁴
iv. [Kr] 5s¹
v. [Xe] 6s² 4f¹⁴ 5d¹⁰ 6p²
i. 2
ii. 6
iii. 14
iv. 10
v. 10
---
Let me know if you’d like diagrams or further explanation on any part!
---
## 1. Determine what elements are denoted by the following configurations.
To identify the element, add up all the superscripts (which represent the number of electrons). The total equals the atomic number, which identifies the element.
i. `1s² 2s² 2p⁶ 3s² 3p⁴`
- Total electrons: 2 + 2 + 6 + 2 + 4 = 16
- Element with atomic number 16: Sulfur (S)
✔ Answer: Sulfur (S)
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ii. `1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁵`
- Total electrons: 2+2+6+2+6+2+5 = 25
- Element with atomic number 25: Manganese (Mn)
✔ Answer: Manganese (Mn)
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iii. `1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶`
- Total electrons: 2+2+6+2+6+2+10+6 = 36
- Element with atomic number 36: Krypton (Kr)
✔ Answer: Krypton (Kr)
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iv. `[Kr] 5s² 4d¹⁰ 5p³`
- [Kr] = 36 electrons
- Add: 2 (5s²) + 10 (4d¹⁰) + 3 (5p³) = 15
- Total: 36 + 15 = 51
- Element with atomic number 51: Antimony (Sb)
✔ Answer: Antimony (Sb)
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v. `[Ne] 3s² 3p¹`
- [Ne] = 10 electrons
- Add: 2 (3s²) + 1 (3p¹) = 3
- Total: 10 + 3 = 13
- Element with atomic number 13: Aluminum (Al)
✔ Answer: Aluminum (Al)
---
## 2. Write the full electron configuration of the following elements
Use the Aufbau principle (fill orbitals in order of increasing energy).
i. Nitrogen (N, atomic number 7)
- Order: 1s → 2s → 2p
- 1s² 2s² 2p³
✔ Answer: 1s² 2s² 2p³
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ii. Silicon (Si, atomic number 14)
- 1s² 2s² 2p⁶ 3s² 3p²
✔ Answer: 1s² 2s² 2p⁶ 3s² 3p²
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iii. Calcium (Ca, atomic number 20)
- 1s² 2s² 2p⁶ 3s² 3p⁶ 4s²
✔ Answer: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s²
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iv. Strontium (Sr, atomic number 38)
- After Kr (36), next is 5s²
- So: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s²
✔ Answer: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s²
---
v. Bismuth (Bi, atomic number 83)
- Full config:
1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f¹⁴ 5d¹⁰ 6p³
✔ Answer: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f¹⁴ 5d¹⁰ 6p³
*(Note: This follows the Aufbau order — after 5p⁶ comes 6s, then 4f, then 5d, then 6p)*
---
## 3. Write the abbreviated electron configuration
Use the noble gas that comes before the element in the periodic table.
i. Fluorine (F, 9)
- Noble gas before F: Helium (He, 2)
- Remaining: 2s² 2p⁵
- Abbreviated: [He] 2s² 2p⁵
✔ Answer: [He] 2s² 2p⁵
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ii. Magnesium (Mg, 12)
- Noble gas before Mg: Neon (Ne, 10)
- Remaining: 3s²
- Abbreviated: [Ne] 3s²
✔ Answer: [Ne] 3s²
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iii. Selenium (Se, 34)
- Noble gas before Se: Argon (Ar, 18)
- Remaining: 4s² 3d¹⁰ 4p⁴
- Abbreviated: [Ar] 4s² 3d¹⁰ 4p⁴
✔ Answer: [Ar] 4s² 3d¹⁰ 4p⁴
---
iv. Rubidium (Rb, 37)
- Noble gas before Rb: Krypton (Kr, 36)
- Remaining: 5s¹
- Abbreviated: [Kr] 5s¹
✔ Answer: [Kr] 5s¹
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v. Lead (Pb, 82)
- Noble gas before Pb: Xenon (Xe, 54)
- Remaining: 6s² 4f¹⁴ 5d¹⁰ 6p²
- Abbreviated: [Xe] 6s² 4f¹⁴ 5d¹⁰ 6p²
✔ Answer: [Xe] 6s² 4f¹⁴ 5d¹⁰ 6p²
---
## 4. If each orbital can hold a maximum of two electrons, how many can each of the following hold?
Each subshell has a fixed number of orbitals:
- s → 1 orbital → holds 2 electrons
- p → 3 orbitals → holds 6 electrons
- d → 5 orbitals → holds 10 electrons
- f → 7 orbitals → holds 14 electrons
So:
i. 2s → s subshell → 2 electrons
ii. 5p → p subshell → 6 electrons
iii. 4f → f subshell → 14 electrons
iv. 3d → d subshell → 10 electrons
v. 4d → d subshell → 10 electrons
✔ Answers:
i. 2
ii. 6
iii. 14
iv. 10
v. 10
---
## 5. What is the shape of an s orbital?
- Spherical — symmetric in all directions around the nucleus.
✔ Answer: Spherical
---
## 6. What is the shape of a p orbital?
- Dumbbell-shaped — two lobes with a node at the nucleus.
✔ Answer: Dumbbell
---
## ✔ Final Answers Summary:
1.
i. Sulfur (S)
ii. Manganese (Mn)
iii. Krypton (Kr)
iv. Antimony (Sb)
v. Aluminum (Al)
2.
i. 1s² 2s² 2p³
ii. 1s² 2s² 2p⁶ 3s² 3p²
iii. 1s² 2s² 2p⁶ 3s² 3p⁶ 4s²
iv. 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s²
v. 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f¹⁴ 5d¹⁰ 6p³
3.
i. [He] 2s² 2p⁵
ii. [Ne] 3s²
iii. [Ar] 4s² 3d¹⁰ 4p⁴
iv. [Kr] 5s¹
v. [Xe] 6s² 4f¹⁴ 5d¹⁰ 6p²
4.
i. 2
ii. 6
iii. 14
iv. 10
v. 10
5. Spherical
6. Dumbbell
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Let me know if you’d like diagrams or further explanation on any part!
Parent Tip: Review the logic above to help your child master the concept of chemfiesta electron configuration worksheet.