Free Printable Lewis Dot Structure Worksheets - Free Printable
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Step-by-step solution for: Free Printable Lewis Dot Structure Worksheets
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Show Answer Key & Explanations
Step-by-step solution for: Free Printable Lewis Dot Structure Worksheets
To draw the Lewis dot structures for these compounds, we need to follow a few simple steps:
1. Count the total number of valence electrons for all atoms in the molecule.
2. Determine the central atom (usually the least electronegative).
3. Draw single bonds between the central atom and the surrounding atoms.
4. Distribute the remaining electrons as lone pairs on the outer atoms.
5. Check if all atoms have a full octet (or duet for hydrogen). If not, form double or triple bonds as needed.
Let's go through each compound:
- Valence electrons: P (5) + 5 × F (7) = 40
- Central atom: P
- Structure: P is bonded to 5 F atoms with single bonds. Each F has 3 lone pairs.
- Valence electrons: N (5) + O (6) = 11
- Central atom: N
- Structure: N is bonded to O with a double bond. N has one lone pair, O has two lone pairs.
- Valence electrons: 3 × I (7) + 1 (charge) = 22
- Central atom: I (middle one)
- Structure: Central I is bonded to two other I atoms with single bonds. Each I has 3 lone pairs, and there are 2 extra electrons on the central I.
- Valence electrons: H (1) + C (4) + N (5) = 10
- Central atom: C
- Structure: H is bonded to C, and C is triple-bonded to N. N has one lone pair.
- Valence electrons: Br (7) + 5 × F (7) = 42
- Central atom: Br
- Structure: Br is bonded to 5 F atoms with single bonds. Each F has 3 lone pairs, and Br has one lone pair.
- Valence electrons: C (4) + O (6) = 10
- Central atom: C
- Structure: C is triple-bonded to O. Both C and O have no lone pairs.
- Valence electrons: Br (7) + 3 × F (7) = 28
- Central atom: Br
- Structure: Br is bonded to 3 F atoms with single bonds. Each F has 3 lone pairs, and Br has 2 lone pairs.
- Valence electrons: C (4) + 2 × O (6) + 1 (charge) = 17
- Central atom: C
- Structure: C is double-bonded to one O and single-bonded to the other O. The single-bonded O has 3 lone pairs, and the double-bonded O has 2 lone pairs. There is one extra electron on the single-bonded O.
- Valence electrons: S (6) + 4 × Cl (7) = 34
- Central atom: S
- Structure: S is bonded to 4 Cl atoms with single bonds. Each Cl has 3 lone pairs, and S has one lone pair.
- Valence electrons: Al (3) + 3 × H (1) = 6
- Central atom: Al
- Structure: Al is bonded to 3 H atoms with single bonds. Al has no lone pairs.
- Valence electrons: B (3) + 3 × F (7) = 24
- Central atom: B
- Structure: B is bonded to 3 F atoms with single bonds. Each F has 3 lone pairs, and B has no lone pairs.
- Valence electrons: 2 × C (4) + 4 × H (1) = 12
- Central atom: C (both are central)
- Structure: Each C is bonded to 2 H atoms and double-bonded to the other C. Each C has no lone pairs.
- Valence electrons: P (5) + 3 × H (1) = 8
- Central atom: P
- Structure: P is bonded to 3 H atoms with single bonds. P has one lone pair.
- Valence electrons: 2 × C (4) + 6 × H (1) = 14
- Central atom: C (both are central)
- Structure: Each C is bonded to 3 H atoms and single-bonded to the other C. Each C has no lone pairs.
- Valence electrons: N (5) + 4 × H (1) - 1 (charge) = 8
- Central atom: N
- Structure: N is bonded to 4 H atoms with single bonds. N has no lone pairs.
Final Answer:
The Lewis dot structures for the given compounds are as described above.
1. Count the total number of valence electrons for all atoms in the molecule.
2. Determine the central atom (usually the least electronegative).
3. Draw single bonds between the central atom and the surrounding atoms.
4. Distribute the remaining electrons as lone pairs on the outer atoms.
5. Check if all atoms have a full octet (or duet for hydrogen). If not, form double or triple bonds as needed.
Let's go through each compound:
1. PF₅
- Valence electrons: P (5) + 5 × F (7) = 40
- Central atom: P
- Structure: P is bonded to 5 F atoms with single bonds. Each F has 3 lone pairs.
2. NO
- Valence electrons: N (5) + O (6) = 11
- Central atom: N
- Structure: N is bonded to O with a double bond. N has one lone pair, O has two lone pairs.
3. I₃⁻
- Valence electrons: 3 × I (7) + 1 (charge) = 22
- Central atom: I (middle one)
- Structure: Central I is bonded to two other I atoms with single bonds. Each I has 3 lone pairs, and there are 2 extra electrons on the central I.
4. HCN
- Valence electrons: H (1) + C (4) + N (5) = 10
- Central atom: C
- Structure: H is bonded to C, and C is triple-bonded to N. N has one lone pair.
5. BrF₅
- Valence electrons: Br (7) + 5 × F (7) = 42
- Central atom: Br
- Structure: Br is bonded to 5 F atoms with single bonds. Each F has 3 lone pairs, and Br has one lone pair.
6. CO
- Valence electrons: C (4) + O (6) = 10
- Central atom: C
- Structure: C is triple-bonded to O. Both C and O have no lone pairs.
7. BrF₃
- Valence electrons: Br (7) + 3 × F (7) = 28
- Central atom: Br
- Structure: Br is bonded to 3 F atoms with single bonds. Each F has 3 lone pairs, and Br has 2 lone pairs.
8. CO₂⁻
- Valence electrons: C (4) + 2 × O (6) + 1 (charge) = 17
- Central atom: C
- Structure: C is double-bonded to one O and single-bonded to the other O. The single-bonded O has 3 lone pairs, and the double-bonded O has 2 lone pairs. There is one extra electron on the single-bonded O.
9. SCl₄
- Valence electrons: S (6) + 4 × Cl (7) = 34
- Central atom: S
- Structure: S is bonded to 4 Cl atoms with single bonds. Each Cl has 3 lone pairs, and S has one lone pair.
10. AlH₃
- Valence electrons: Al (3) + 3 × H (1) = 6
- Central atom: Al
- Structure: Al is bonded to 3 H atoms with single bonds. Al has no lone pairs.
11. BF₃
- Valence electrons: B (3) + 3 × F (7) = 24
- Central atom: B
- Structure: B is bonded to 3 F atoms with single bonds. Each F has 3 lone pairs, and B has no lone pairs.
12. C₂H₄
- Valence electrons: 2 × C (4) + 4 × H (1) = 12
- Central atom: C (both are central)
- Structure: Each C is bonded to 2 H atoms and double-bonded to the other C. Each C has no lone pairs.
13. PH₃
- Valence electrons: P (5) + 3 × H (1) = 8
- Central atom: P
- Structure: P is bonded to 3 H atoms with single bonds. P has one lone pair.
14. C₂H₆
- Valence electrons: 2 × C (4) + 6 × H (1) = 14
- Central atom: C (both are central)
- Structure: Each C is bonded to 3 H atoms and single-bonded to the other C. Each C has no lone pairs.
15. NH₄⁺
- Valence electrons: N (5) + 4 × H (1) - 1 (charge) = 8
- Central atom: N
- Structure: N is bonded to 4 H atoms with single bonds. N has no lone pairs.
Final Answer:
The Lewis dot structures for the given compounds are as described above.
Parent Tip: Review the logic above to help your child master the concept of electron dot diagrams worksheet.
