Lewis Dot Structure Worksheet for drawing electron dot diagrams of various molecules and ions.
Lewis Dot Structure Worksheet with 12 molecular formulas to draw Lewis structures for, including NI₃, H₂Se, I⁻, HCN, BrF₅, CO, SiF₄, OF₂, IBr, NH₂Cl, BF₃, CH₂F₂, CH₃Br, PO₃³⁻, and NH₄⁺.
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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 molecules and ions, we follow a standard process:
1. Count Valence Electrons: Add up the valence electrons for all atoms. Add electrons for negative charges and subtract for positive charges.
2. Determine the Central Atom: Usually the least electronegative atom (except Hydrogen, which is never central).
3. Draw Skeleton Structure: Connect atoms with single bonds.
4. Fill Octets: Distribute remaining electrons to outer atoms first to give them 8 electrons (duet for Hydrogen).
5. Check Central Atom: If the central atom doesn't have 8 electrons, move lone pairs from outer atoms to form double or triple bonds.
Here are the solutions for each box in your worksheet:
1. $NI_3$ (Nitrogen triiodide)
* Valence e⁻: N (5) + 3 × I (7) = 26 electrons.
* Structure: Nitrogen is in the center bonded to three Iodine atoms.
* Lone Pairs: Each Iodine gets 3 lone pairs (6 dots). Nitrogen gets 1 lone pair (2 dots).
* Result: N in center, single bonds to 3 I's. All atoms have full octets.
2. $H_2Se$ (Hydrogen selenide)
* Valence e⁻: Se (6) + 2 × H (1) = 8 electrons.
* Structure: Selenium is central bonded to two Hydrogens.
* Lone Pairs: The remaining 4 electrons go on Selenium as 2 lone pairs.
* Result: Se in center, single bonds to 2 H's. Se has two pairs of dots on top/bottom.
3. $I_3^-$ (Triiodide ion)
* Valence e⁻: 3 × I (7) + 1 (charge) = 22 electrons.
* Structure: Three Iodines in a row ($I-I-I$). This is an exception to the octet rule; the central atom expands its octet.
* Lone Pairs: Fill outer Iodines with 3 lone pairs each (uses 12 e⁻). Put remaining 6 electrons (3 lone pairs) on the central Iodine.
* Result: Linear shape. Central I has 3 lone pairs and 2 bonds (10 electrons total). Enclose in brackets with a minus sign.
4. $HCN$ (Hydrogen cyanide)
* Valence e⁻: H (1) + C (4) + N (5) = 10 electrons.
* Structure: Carbon is central ($H-C-N$).
* Bonding: Single bond H-C uses 2 e⁻. Remaining 8 go to N? No, C needs 8. We form a triple bond between C and N.
* Result: $H-C \equiv N$. Nitrogen has one lone pair. Carbon has 4 bonds (8 e⁻). Hydrogen has 1 bond (2 e⁻).
5. $BrF_5$ (Bromine pentafluoride)
* Valence e⁻: Br (7) + 5 × F (7) = 42 electrons.
* Structure: Bromine central bonded to 5 Fluorines.
* Lone Pairs: Fill all F's with 3 lone pairs each (30 e⁻). Remaining 2 electrons go on Br as 1 lone pair.
* Result: Br in center, 5 single bonds to F. Br has one lone pair.
6. $CO$ (Carbon monoxide)
* Valence e⁻: C (4) + O (6) = 10 electrons.
* Structure: C bonded to O.
* Bonding: To satisfy the octet for both with only 10 electrons, they must share 3 pairs.
* Result: $C \equiv O$ (Triple bond). Both Carbon and Oxygen have 1 lone pair each.
7. $SiF_4$ (Silicon tetrafluoride)
* Valence e⁻: Si (4) + 4 × F (7) = 32 electrons.
* Structure: Silicon central bonded to 4 Fluorines.
* Lone Pairs: Fill all F's with 3 lone pairs each. Silicon has no lone pairs (4 bonds = 8 e⁻).
* Result: Si in center, 4 single bonds to F. Tetrahedral shape.
8. $OF_2$ (Oxygen difluoride)
* Valence e⁻: O (6) + 2 × F (7) = 20 electrons.
* Structure: Oxygen is more electropositive than Fluorine, so O is central. Bonded to 2 F's.
* Lone Pairs: Fill F's with 3 lone pairs each (12 e⁻). Remaining 4 e⁻ go on Oxygen as 2 lone pairs.
* Result: O in center, single bonds to 2 F's. Oxygen has two lone pairs. Bent shape.
9. $IBr$ (Iodine monobromide)
* Valence e⁻: I (7) + Br (7) = 14 electrons.
* Structure: Single bond between I and Br.
* Lone Pairs: Remaining 12 electrons are split evenly. 3 lone pairs on I, 3 lone pairs on Br.
* Result: $I-Br$ with 3 pairs of dots around each atom.
10. $NH_2Cl$ (Chloramine)
* Valence e⁻: N (5) + 2 × H (1) + Cl (7) = 14 electrons.
* Structure: Nitrogen is central. Bonded to 2 H's and 1 Cl.
* Lone Pairs: Bonds use 6 e⁻. Fill Cl with 3 lone pairs (6 e⁻). Remaining 2 e⁻ go on N as 1 lone pair.
* Result: N in center, bonded to H, H, and Cl. N has one lone pair. Cl has three lone pairs.
11. $BF_3$ (Boron trifluoride)
* Valence e⁻: B (3) + 3 × F (7) = 24 electrons.
* Structure: Boron central bonded to 3 Fluorines.
* Note: Boron is an exception; it is stable with only 6 valence electrons (incomplete octet). Do not force a double bond unless necessary for formal charge (here single bonds are standard).
* Lone Pairs: Fill F's with 3 lone pairs each. Boron has no lone pairs.
* Result: B in center, 3 single bonds to F. Trigonal planar.
12. $CH_2F_2$ (Difluoromethane)
* Valence e⁻: C (4) + 2 × H (1) + 2 × F (7) = 20 electrons.
* Structure: Carbon central bonded to 2 H's and 2 F's.
* Lone Pairs: Fill F's with 3 lone pairs each. Carbon has no lone pairs.
* Result: C in center, single bonds to two H's and two F's.
13. $CH_3Br$ (Bromomethane)
* Valence e⁻: C (4) + 3 × H (1) + Br (7) = 14 electrons.
* Structure: Carbon central bonded to 3 H's and 1 Br.
* Lone Pairs: Fill Br with 3 lone pairs. Carbon has no lone pairs.
* Result: C in center, single bonds to three H's and one Br.
14. $PO_3^{3-}$ (Phosphite ion)
* Valence e⁻: P (5) + 3 × O (6) + 3 (charge) = 26 electrons.
* Structure: Phosphorus central bonded to 3 Oxygens.
* Lone Pairs: Give each Oxygen 3 lone pairs (octets). P has 1 lone pair.
* Note: While P can expand its octet to form a double bond (reducing formal charge), the structure satisfying the basic octet rule with a -3 charge distributed is often accepted in introductory chemistry. However, the most stable resonance structure usually has one $P=O$ double bond and two $P-O^-$ single bonds, with a lone pair on P. Let's provide the standard octet-compliant version often taught first: P single bonded to 3 O's, each O has 3 lone pairs and a -1 formal charge, P has 1 lone pair.
* Result: P in center, single bonds to 3 O's. P has one lone pair. Each O has 3 lone pairs. Enclose in brackets with 3- charge.
15. $NH_4^+$ (Ammonium ion)
* Valence e⁻: N (5) + 4 × H (1) - 1 (charge) = 8 electrons.
* Structure: Nitrogen central bonded to 4 Hydrogens.
* Lone Pairs: All 8 electrons are used in the 4 bonds. No lone pairs remain.
* Result: N in center, single bonds to 4 H's. Enclose in brackets with + charge.
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Final Answer:
1. $NI_3$: Nitrogen central, single bonded to 3 Iodines. N has 1 lone pair. Each I has 3 lone pairs.
2. $H_2Se$: Selenium central, single bonded to 2 Hydrogens. Se has 2 lone pairs.
3. $I_3^-$: Linear $[I-I-I]^-$. Central I has 3 lone pairs. Outer I's have 3 lone pairs each.
4. $HCN$: $H-C \equiv N$. Triple bond between C and N. N has 1 lone pair.
5. $BrF_5$: Bromine central, single bonded to 5 Fluorines. Br has 1 lone pair. Each F has 3 lone pairs.
6. $CO$: $C \equiv O$. Triple bond. C has 1 lone pair, O has 1 lone pair.
7. $SiF_4$: Silicon central, single bonded to 4 Fluorines. No lone pairs on Si. Each F has 3 lone pairs.
8. $OF_2$: Oxygen central, single bonded to 2 Fluorines. O has 2 lone pairs. Each F has 3 lone pairs.
9. $IBr$: $I-Br$. Single bond. Both I and Br have 3 lone pairs.
10. $NH_2Cl$: Nitrogen central, bonded to 2 H's and 1 Cl. N has 1 lone pair. Cl has 3 lone pairs.
11. $BF_3$: Boron central, single bonded to 3 Fluorines. No lone pairs on B. Each F has 3 lone pairs.
12. $CH_2F_2$: Carbon central, bonded to 2 H's and 2 F's. No lone pairs on C. Each F has 3 lone pairs.
13. $CH_3Br$: Carbon central, bonded to 3 H's and 1 Br. No lone pairs on C. Br has 3 lone pairs.
14. $PO_3^{3-}$: Phosphorus central, single bonded to 3 Oxygens. P has 1 lone pair. Each O has 3 lone pairs. (Bracket with 3- charge).
15. $NH_4^+$: Nitrogen central, bonded to 4 Hydrogens. No lone pairs. (Bracket with + charge).
1. Count Valence Electrons: Add up the valence electrons for all atoms. Add electrons for negative charges and subtract for positive charges.
2. Determine the Central Atom: Usually the least electronegative atom (except Hydrogen, which is never central).
3. Draw Skeleton Structure: Connect atoms with single bonds.
4. Fill Octets: Distribute remaining electrons to outer atoms first to give them 8 electrons (duet for Hydrogen).
5. Check Central Atom: If the central atom doesn't have 8 electrons, move lone pairs from outer atoms to form double or triple bonds.
Here are the solutions for each box in your worksheet:
Row 1
1. $NI_3$ (Nitrogen triiodide)
* Valence e⁻: N (5) + 3 × I (7) = 26 electrons.
* Structure: Nitrogen is in the center bonded to three Iodine atoms.
* Lone Pairs: Each Iodine gets 3 lone pairs (6 dots). Nitrogen gets 1 lone pair (2 dots).
* Result: N in center, single bonds to 3 I's. All atoms have full octets.
2. $H_2Se$ (Hydrogen selenide)
* Valence e⁻: Se (6) + 2 × H (1) = 8 electrons.
* Structure: Selenium is central bonded to two Hydrogens.
* Lone Pairs: The remaining 4 electrons go on Selenium as 2 lone pairs.
* Result: Se in center, single bonds to 2 H's. Se has two pairs of dots on top/bottom.
3. $I_3^-$ (Triiodide ion)
* Valence e⁻: 3 × I (7) + 1 (charge) = 22 electrons.
* Structure: Three Iodines in a row ($I-I-I$). This is an exception to the octet rule; the central atom expands its octet.
* Lone Pairs: Fill outer Iodines with 3 lone pairs each (uses 12 e⁻). Put remaining 6 electrons (3 lone pairs) on the central Iodine.
* Result: Linear shape. Central I has 3 lone pairs and 2 bonds (10 electrons total). Enclose in brackets with a minus sign.
Row 2
4. $HCN$ (Hydrogen cyanide)
* Valence e⁻: H (1) + C (4) + N (5) = 10 electrons.
* Structure: Carbon is central ($H-C-N$).
* Bonding: Single bond H-C uses 2 e⁻. Remaining 8 go to N? No, C needs 8. We form a triple bond between C and N.
* Result: $H-C \equiv N$. Nitrogen has one lone pair. Carbon has 4 bonds (8 e⁻). Hydrogen has 1 bond (2 e⁻).
5. $BrF_5$ (Bromine pentafluoride)
* Valence e⁻: Br (7) + 5 × F (7) = 42 electrons.
* Structure: Bromine central bonded to 5 Fluorines.
* Lone Pairs: Fill all F's with 3 lone pairs each (30 e⁻). Remaining 2 electrons go on Br as 1 lone pair.
* Result: Br in center, 5 single bonds to F. Br has one lone pair.
6. $CO$ (Carbon monoxide)
* Valence e⁻: C (4) + O (6) = 10 electrons.
* Structure: C bonded to O.
* Bonding: To satisfy the octet for both with only 10 electrons, they must share 3 pairs.
* Result: $C \equiv O$ (Triple bond). Both Carbon and Oxygen have 1 lone pair each.
Row 3
7. $SiF_4$ (Silicon tetrafluoride)
* Valence e⁻: Si (4) + 4 × F (7) = 32 electrons.
* Structure: Silicon central bonded to 4 Fluorines.
* Lone Pairs: Fill all F's with 3 lone pairs each. Silicon has no lone pairs (4 bonds = 8 e⁻).
* Result: Si in center, 4 single bonds to F. Tetrahedral shape.
8. $OF_2$ (Oxygen difluoride)
* Valence e⁻: O (6) + 2 × F (7) = 20 electrons.
* Structure: Oxygen is more electropositive than Fluorine, so O is central. Bonded to 2 F's.
* Lone Pairs: Fill F's with 3 lone pairs each (12 e⁻). Remaining 4 e⁻ go on Oxygen as 2 lone pairs.
* Result: O in center, single bonds to 2 F's. Oxygen has two lone pairs. Bent shape.
9. $IBr$ (Iodine monobromide)
* Valence e⁻: I (7) + Br (7) = 14 electrons.
* Structure: Single bond between I and Br.
* Lone Pairs: Remaining 12 electrons are split evenly. 3 lone pairs on I, 3 lone pairs on Br.
* Result: $I-Br$ with 3 pairs of dots around each atom.
Row 4
10. $NH_2Cl$ (Chloramine)
* Valence e⁻: N (5) + 2 × H (1) + Cl (7) = 14 electrons.
* Structure: Nitrogen is central. Bonded to 2 H's and 1 Cl.
* Lone Pairs: Bonds use 6 e⁻. Fill Cl with 3 lone pairs (6 e⁻). Remaining 2 e⁻ go on N as 1 lone pair.
* Result: N in center, bonded to H, H, and Cl. N has one lone pair. Cl has three lone pairs.
11. $BF_3$ (Boron trifluoride)
* Valence e⁻: B (3) + 3 × F (7) = 24 electrons.
* Structure: Boron central bonded to 3 Fluorines.
* Note: Boron is an exception; it is stable with only 6 valence electrons (incomplete octet). Do not force a double bond unless necessary for formal charge (here single bonds are standard).
* Lone Pairs: Fill F's with 3 lone pairs each. Boron has no lone pairs.
* Result: B in center, 3 single bonds to F. Trigonal planar.
12. $CH_2F_2$ (Difluoromethane)
* Valence e⁻: C (4) + 2 × H (1) + 2 × F (7) = 20 electrons.
* Structure: Carbon central bonded to 2 H's and 2 F's.
* Lone Pairs: Fill F's with 3 lone pairs each. Carbon has no lone pairs.
* Result: C in center, single bonds to two H's and two F's.
Row 5
13. $CH_3Br$ (Bromomethane)
* Valence e⁻: C (4) + 3 × H (1) + Br (7) = 14 electrons.
* Structure: Carbon central bonded to 3 H's and 1 Br.
* Lone Pairs: Fill Br with 3 lone pairs. Carbon has no lone pairs.
* Result: C in center, single bonds to three H's and one Br.
14. $PO_3^{3-}$ (Phosphite ion)
* Valence e⁻: P (5) + 3 × O (6) + 3 (charge) = 26 electrons.
* Structure: Phosphorus central bonded to 3 Oxygens.
* Lone Pairs: Give each Oxygen 3 lone pairs (octets). P has 1 lone pair.
* Note: While P can expand its octet to form a double bond (reducing formal charge), the structure satisfying the basic octet rule with a -3 charge distributed is often accepted in introductory chemistry. However, the most stable resonance structure usually has one $P=O$ double bond and two $P-O^-$ single bonds, with a lone pair on P. Let's provide the standard octet-compliant version often taught first: P single bonded to 3 O's, each O has 3 lone pairs and a -1 formal charge, P has 1 lone pair.
* Result: P in center, single bonds to 3 O's. P has one lone pair. Each O has 3 lone pairs. Enclose in brackets with 3- charge.
15. $NH_4^+$ (Ammonium ion)
* Valence e⁻: N (5) + 4 × H (1) - 1 (charge) = 8 electrons.
* Structure: Nitrogen central bonded to 4 Hydrogens.
* Lone Pairs: All 8 electrons are used in the 4 bonds. No lone pairs remain.
* Result: N in center, single bonds to 4 H's. Enclose in brackets with + charge.
──────────────────────────────────────
Final Answer:
1. $NI_3$: Nitrogen central, single bonded to 3 Iodines. N has 1 lone pair. Each I has 3 lone pairs.
2. $H_2Se$: Selenium central, single bonded to 2 Hydrogens. Se has 2 lone pairs.
3. $I_3^-$: Linear $[I-I-I]^-$. Central I has 3 lone pairs. Outer I's have 3 lone pairs each.
4. $HCN$: $H-C \equiv N$. Triple bond between C and N. N has 1 lone pair.
5. $BrF_5$: Bromine central, single bonded to 5 Fluorines. Br has 1 lone pair. Each F has 3 lone pairs.
6. $CO$: $C \equiv O$. Triple bond. C has 1 lone pair, O has 1 lone pair.
7. $SiF_4$: Silicon central, single bonded to 4 Fluorines. No lone pairs on Si. Each F has 3 lone pairs.
8. $OF_2$: Oxygen central, single bonded to 2 Fluorines. O has 2 lone pairs. Each F has 3 lone pairs.
9. $IBr$: $I-Br$. Single bond. Both I and Br have 3 lone pairs.
10. $NH_2Cl$: Nitrogen central, bonded to 2 H's and 1 Cl. N has 1 lone pair. Cl has 3 lone pairs.
11. $BF_3$: Boron central, single bonded to 3 Fluorines. No lone pairs on B. Each F has 3 lone pairs.
12. $CH_2F_2$: Carbon central, bonded to 2 H's and 2 F's. No lone pairs on C. Each F has 3 lone pairs.
13. $CH_3Br$: Carbon central, bonded to 3 H's and 1 Br. No lone pairs on C. Br has 3 lone pairs.
14. $PO_3^{3-}$: Phosphorus central, single bonded to 3 Oxygens. P has 1 lone pair. Each O has 3 lone pairs. (Bracket with 3- charge).
15. $NH_4^+$: Nitrogen central, bonded to 4 Hydrogens. No lone pairs. (Bracket with + charge).
Parent Tip: Review the logic above to help your child master the concept of lewis dot structures worksheet with answers.
