QUESTION IMAGE
Question
- write lewis structures for the following molecules:
a. nh₃ b. co c. co₂ d. n₂ e. o₂
f. c₂h₂ g. c₂h₄ h. h₂o i. cl₂ j. h₂
k. hcn
- match electron configurations.
- write ionic formulas.
Step1: Analyze $NH_3$ (Ammonia)
- Valence Electrons: N has 5, each H has 1. Total: $5 + 3\times1 = 8$.
- Central Atom: N (least electronegative, not H).
- Bonding: N forms 3 single bonds with H (3 pairs shared), leaving 1 lone pair on N.
- Lewis Structure: $\ce{H - \overset{..}{N} - H}$ (with a H on top too, represented as $\ce{H_{3}N:}$ or $\ce{H - N(-H)-H}$ with two dots on N).
Step2: Analyze $CO$ (Carbon Monoxide)
- Valence Electrons: C has 4, O has 6. Total: $4 + 6 = 10$.
- Central Atom: C (or O, but C is less electronegative here).
- Bonding: Triple bond (6 electrons shared) between C and O, with 1 lone pair on C and 1 on O.
- Lewis Structure: $\ce{:C\equiv O:}$ (or $\ce{:C#O:}$).
Step3: Analyze $CO_2$ (Carbon Dioxide)
- Valence Electrons: C has 4, each O has 6. Total: $4 + 2\times6 = 16$.
- Central Atom: C.
- Bonding: Double bonds (4 electrons each) between C and each O. No lone pairs on C, 2 lone pairs on each O.
- Lewis Structure: $\ce{O::C::O}$ (or $\ce{O=C=O}$ with two lone pairs on each O).
Step4: Analyze $N_2$ (Nitrogen Gas)
- Valence Electrons: Each N has 5. Total: $5 + 5 = 10$.
- Bonding: Triple bond (6 electrons shared) between N atoms, 1 lone pair on each N.
- Lewis Structure: $\ce{:N\equiv N:}$ (or $\ce{:N#N:}$).
Step5: Analyze $O_2$ (Oxygen Gas)
- Valence Electrons: Each O has 6. Total: $6 + 6 = 12$.
- Bonding: Double bond (4 electrons shared) between O atoms, 2 lone pairs on each O. (Note: In reality, it has a double bond with unpaired electrons, but Lewis structure here is $\ce{O=O}$ with two lone pairs on each O, or $\ce{:O=O:}$ with two lone pairs on each, but actual has a double bond and two unpaired electrons, but Lewis structure for simplicity: $\ce{:O - O:}$ with a double bond? Wait, no: 12 electrons. Double bond (4) + 2 lone pairs (4) on each O: $4 + 4 + 4 = 12$. So $\ce{:O=O:}$ (two lone pairs on each O).
Step6: Analyze $C_2H_2$ (Acetylene)
- Valence Electrons: Each C has 4, each H has 1. Total: $2\times4 + 2\times1 = 10$.
- Central Atoms: Two C atoms.
- Bonding: Triple bond between C atoms, each C bonded to 1 H (single bond).
- Lewis Structure: $\ce{H - C\equiv C - H}$ (or $\ce{H:C:::C:H}$).
Step7: Analyze $C_2H_4$ (Ethylene)
- Valence Electrons: Each C has 4, each H has 1. Total: $2\times4 + 4\times1 = 12$.
- Central Atoms: Two C atoms.
- Bonding: Double bond between C atoms, each C bonded to 2 H (single bonds).
- Lewis Structure: $\ce{H_{2}C=CH_{2}}$ (or $\ce{H - C(=C)-H}$ with H on each C, represented as $\ce{H_{2}C=CH_{2}}$ with two H on each C, and a double bond between C's).
Step8: Analyze $H_2O$ (Water)
- Valence Electrons: O has 6, each H has 1. Total: $6 + 2\times1 = 8$.
- Central Atom: O.
- Bonding: 2 single bonds with H, 2 lone pairs on O.
- Lewis Structure: $\ce{H - \overset{..}{O} - H}$ (with two dots on top and bottom of O).
Step9: Analyze $Cl_2$ (Chlorine Gas)
- Valence Electrons: Each Cl has 7. Total: $7 + 7 = 14$.
- Bonding: Single bond (2 electrons shared) between Cl atoms, 3 lone pairs on each Cl.
- Lewis Structure: $\ce{:Cl - Cl:}$ (with three lone pairs on each Cl).
Step10: Analyze $H_2$ (Hydrogen Gas)
- Valence Electrons: Each H has 1. Total: $1 + 1 = 2$.
- Bonding: Single bond (2 electrons shared) between H atoms.
- Lewis Structure: $\ce{H - H}$ (or $\ce{H:H}$).
Step11: Analyze $HCN$ (Hydrogen Cyanide)
- Valence Electrons: H has 1, C has 4, N has 5. Total: $1 + 4 + 5 = 10$.
- Central Atom: C.
-…
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- a. $\boldsymbol{\ce{H_{3}N:}}$ (or $\ce{H - \overset{..}{N} - H}$ with H on top)
- b. $\boldsymbol{\ce{:C\equiv O:}}$
- c. $\boldsymbol{\ce{O=C=O}}$ (with two lone pairs on each O)
- d. $\boldsymbol{\ce{:N\equiv N:}}$
- e. $\boldsymbol{\ce{:O=O:}}$ (with two lone pairs on each O)
- f. $\boldsymbol{\ce{H - C\equiv C - H}}$
- g. $\boldsymbol{\ce{H_{2}C=CH_{2}}}$
- h. $\boldsymbol{\ce{H - \overset{..}{O} - H}}$ (with two lone pairs on O)
- i. $\boldsymbol{\ce{:Cl - Cl:}}$ (with three lone pairs on each Cl)
- j. $\boldsymbol{\ce{H - H}}$
- k. $\boldsymbol{\ce{H - C\equiv N:}}$