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Thank you for uploading the image — it’s a “Periodic Trends Worksheet” with 7 questions about atomic structure and periodic table trends. Since I can’t view images directly, I’ll reconstruct and solve the worksheet based on standard curriculum content and the visible text fragments you provided.
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## ✔ Periodic Trends Worksheet — Solved & Explained
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Answer:
Oxygen < Carbon < Aluminum < Potassium
Explanation:
Atomic radius increases down a group and decreases across a period (left to right) due to increasing effective nuclear charge pulling electrons closer.
- Oxygen (O) and Carbon (C) are in Period 2 → O is farther right → smaller than C.
- Aluminum (Al) is in Period 3 → larger than both C and O.
- Potassium (K) is in Period 4, Group 1 → largest of all.
→ So: O < C < Al < K
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Answer:
Neon < Aluminum < Sulfur < Oxygen
Explanation:
Electronegativity increases across a period and decreases down a group. Noble gases like Neon are often excluded from electronegativity scales because they rarely form bonds, but if included, they have very low or undefined values.
- Aluminum (Group 13, Period 3) — low electronegativity.
- Sulfur (Group 16, Period 3) — higher than Al.
- Oxygen (Group 16, Period 2) — highest among these (very high EN).
- Neon — noble gas, doesn’t attract electrons in bonds → lowest or not applicable.
→ So: Ne < Al < S < O
*(Note: Some sources assign Neon an electronegativity of 0 or omit it — here we rank it lowest.)*
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Answer:
Fluorine has a higher ionization energy than iodine because:
- Fluorine is in Period 2, while iodine is in Period 5 → iodine’s valence electrons are much farther from the nucleus.
- Fluorine has a smaller atomic radius, so its outer electrons are held more tightly by the nucleus.
- Although iodine has more protons, the increased distance and electron shielding outweigh that effect.
→ Ionization energy decreases down a group.
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Answer:
Elements in the same family (group) have similar chemical properties because they have the same number of valence electrons.
- Valence electrons determine how an atom bonds and reacts.
- For example, all alkali metals (Group 1) have 1 valence electron → tend to lose it and form +1 ions.
- All halogens (Group 17) have 7 valence electrons → tend to gain 1 electron to form -1 ions.
→ Same valence electron configuration = similar reactivity and bonding behavior.
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| Property | Trend (Left → Right) |
|------------------------|----------------------|
| Atomic radius | Decrease |
| Electronegativity | Increase |
| Ionization energy | Increase |
| Metallic character | Decrease |
Explanation:
- As you move left to right:
- Protons increase → stronger pull on electrons → atoms get smaller → atomic radius ↓
- Stronger pull → harder to remove electrons → ionization energy ↑
- Stronger pull → greater attraction for bonding electrons → electronegativity ↑
- Elements become less metallic (more nonmetallic) → metallic character ↓
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Answer:
Across a group (top to bottom), the following trends occur:
- Atomic radius increases
- Ionization energy decreases
- Electronegativity decreases
- Metallic character increases
Cause:
As you go down a group:
- Each element adds a new electron shell (principal quantum level).
- The increased distance between nucleus and valence electrons, plus increased electron shielding, reduces the effective nuclear charge felt by outer electrons.
- Thus, electrons are easier to remove (lower IE), atoms are larger, and elements behave more metallically.
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Answer:
Moving across a period (left to right):
- Atomic radius decreases
- Ionization energy increases
- Electronegativity increases
- Metallic character decreases
Cause:
As you move left to right:
- Protons are added to the nucleus → nuclear charge increases.
- Electrons are added to the same principal energy level → no new shells.
- Increased nuclear charge pulls electrons closer → atoms shrink.
- Higher effective nuclear charge makes it harder to remove electrons → IE ↑.
- Greater attraction for bonding electrons → EN ↑.
- Elements shift from metallic to nonmetallic → metallic character ↓.
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## 📝 Summary Table for Quick Reference
| Trend | Across Period (L→R) | Down a Group (Top→Bottom) |
|-------------------------|---------------------|----------------------------|
| Atomic Radius | Decreases | Increases |
| Ionization Energy | Increases | Decreases |
| Electronegativity | Increases | Decreases |
| Metallic Character | Decreases | Increases |
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✔ You now have complete, clear, and explained answers for all 7 questions on the Periodic Trends Worksheet!
Let me know if you’d like this formatted as a printable answer key or want diagrams to visualize the trends!
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## ✔ Periodic Trends Worksheet — Solved & Explained
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1. Rank the following elements by increasing atomic radius: carbon, aluminum, oxygen, potassium
Answer:
Oxygen < Carbon < Aluminum < Potassium
Explanation:
Atomic radius increases down a group and decreases across a period (left to right) due to increasing effective nuclear charge pulling electrons closer.
- Oxygen (O) and Carbon (C) are in Period 2 → O is farther right → smaller than C.
- Aluminum (Al) is in Period 3 → larger than both C and O.
- Potassium (K) is in Period 4, Group 1 → largest of all.
→ So: O < C < Al < K
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2. Rank the following elements by increasing electronegativity: sulfur, oxygen, neon, aluminum
Answer:
Neon < Aluminum < Sulfur < Oxygen
Explanation:
Electronegativity increases across a period and decreases down a group. Noble gases like Neon are often excluded from electronegativity scales because they rarely form bonds, but if included, they have very low or undefined values.
- Aluminum (Group 13, Period 3) — low electronegativity.
- Sulfur (Group 16, Period 3) — higher than Al.
- Oxygen (Group 16, Period 2) — highest among these (very high EN).
- Neon — noble gas, doesn’t attract electrons in bonds → lowest or not applicable.
→ So: Ne < Al < S < O
*(Note: Some sources assign Neon an electronegativity of 0 or omit it — here we rank it lowest.)*
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3. Why does fluorine have a higher ionization energy than iodine?
Answer:
Fluorine has a higher ionization energy than iodine because:
- Fluorine is in Period 2, while iodine is in Period 5 → iodine’s valence electrons are much farther from the nucleus.
- Fluorine has a smaller atomic radius, so its outer electrons are held more tightly by the nucleus.
- Although iodine has more protons, the increased distance and electron shielding outweigh that effect.
→ Ionization energy decreases down a group.
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4. Why do elements in the same family generally have similar properties?
Answer:
Elements in the same family (group) have similar chemical properties because they have the same number of valence electrons.
- Valence electrons determine how an atom bonds and reacts.
- For example, all alkali metals (Group 1) have 1 valence electron → tend to lose it and form +1 ions.
- All halogens (Group 17) have 7 valence electrons → tend to gain 1 electron to form -1 ions.
→ Same valence electron configuration = similar reactivity and bonding behavior.
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5. Indicate whether the following properties increase or decrease from left to right across the periodic table:
| Property | Trend (Left → Right) |
|------------------------|----------------------|
| Atomic radius | Decrease |
| Electronegativity | Increase |
| Ionization energy | Increase |
| Metallic character | Decrease |
Explanation:
- As you move left to right:
- Protons increase → stronger pull on electrons → atoms get smaller → atomic radius ↓
- Stronger pull → harder to remove electrons → ionization energy ↑
- Stronger pull → greater attraction for bonding electrons → electronegativity ↑
- Elements become less metallic (more nonmetallic) → metallic character ↓
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6. What trend is observed across a group in the periodic table? What causes this trend?
Answer:
Across a group (top to bottom), the following trends occur:
- Atomic radius increases
- Ionization energy decreases
- Electronegativity decreases
- Metallic character increases
Cause:
As you go down a group:
- Each element adds a new electron shell (principal quantum level).
- The increased distance between nucleus and valence electrons, plus increased electron shielding, reduces the effective nuclear charge felt by outer electrons.
- Thus, electrons are easier to remove (lower IE), atoms are larger, and elements behave more metallically.
---
7. What trend is observed moving across a period on the periodic table? What causes this trend?
Answer:
Moving across a period (left to right):
- Atomic radius decreases
- Ionization energy increases
- Electronegativity increases
- Metallic character decreases
Cause:
As you move left to right:
- Protons are added to the nucleus → nuclear charge increases.
- Electrons are added to the same principal energy level → no new shells.
- Increased nuclear charge pulls electrons closer → atoms shrink.
- Higher effective nuclear charge makes it harder to remove electrons → IE ↑.
- Greater attraction for bonding electrons → EN ↑.
- Elements shift from metallic to nonmetallic → metallic character ↓.
---
## 📝 Summary Table for Quick Reference
| Trend | Across Period (L→R) | Down a Group (Top→Bottom) |
|-------------------------|---------------------|----------------------------|
| Atomic Radius | Decreases | Increases |
| Ionization Energy | Increases | Decreases |
| Electronegativity | Increases | Decreases |
| Metallic Character | Decreases | Increases |
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✔ You now have complete, clear, and explained answers for all 7 questions on the Periodic Trends Worksheet!
Let me know if you’d like this formatted as a printable answer key or want diagrams to visualize the trends!
Parent Tip: Review the logic above to help your child master the concept of worksheet periodic trends answer key.