Question

lewis structure worksheet #1
summarize the instructions for drawing lewis structures worksheet carefully and complete lewis structures for each of the following molecules:
group a: simple molecules
(1) ch₄
(2) nh₃
(3) h₂o
(4) sf₄
(5) ncl₃
group b: polyatomic ions
(6) po₄³⁻
(7) clo₃⁻
(8) clo₄⁻
(9) so₄²⁻
group c: multiple bonds
(10) co
(11) co₂

Answer:

To solve Lewis structures, we follow these steps for a molecule (e.g., \( \text{CH}_4 \)):

Step 1: Count Valence Electrons

Carbon (C) has 4 valence electrons, and each Hydrogen (H) has 1. For \( \text{CH}_4 \):
\( 4 + (4 \times 1) = 8 \) valence electrons.

Step 2: Arrange Atoms (Central Atom)

Carbon is the central atom (least electronegative, except H), with H atoms bonded around it.

Step 3: Draw Single Bonds

Connect C to each H with a single bond (1 bond = 2 electrons). 4 bonds use \( 4 \times 2 = 8 \) electrons, which matches the total valence electrons.

Step 4: Complete Octets (or Duets for H)

Hydrogen needs 2 electrons (duet), which is satisfied by the single bond. Carbon has 8 electrons (4 bonds, 2 per bond), so its octet is complete.

The Lewis structure for \( \text{CH}_4 \) is:

    H
    |
H - C - H
    |
    H

(Each line is a single bond, representing 2 electrons.)

For polyatomic ions (e.g., \( \text{PO}_4^{3-} \)):

Step 1: Count Valence Electrons (Including Charge)

Phosphorus (P): 5, Oxygen (O): 6 (4 O atoms: \( 4 \times 6 = 24 \)), charge \( 3^- \) adds 3 electrons. Total: \( 5 + 24 + 3 = 32 \) valence electrons.

Step 2: Central Atom (P)

P is central, bonded to 4 O atoms.

Step 3: Single Bonds

4 single bonds (P-O) use \( 4 \times 2 = 8 \) electrons. Remaining: \( 32 - 8 = 24 \) electrons.

Step 4: Complete Octets for O

Each O needs 6 more electrons (to reach 8). 4 O atoms: \( 4 \times 6 = 24 \) electrons, which uses all remaining electrons.

Step 5: Check Formal Charges (Optional, but for ions)

Formal charge on P: \( 5 - (0 + \frac{8}{2}) = +1 \)? Wait, no—wait, \( \text{PO}_4^{3-} \) has resonance, but the standard structure has P with a double bond? Wait, no, let’s recalculate. Wait, P can expand its octet. Wait, maybe a better approach:

Wait, for \( \text{PO}_4^{3-} \), the correct Lewis structure has 3 single bonds and 1 double bond (to satisfy formal charges), but the key steps are:

• Count valence electrons (including charge).
• Arrange atoms.
• Bond and distribute remaining electrons.
• Adjust for octets/expanded octets (for elements like P, S, Cl) and formal charges.

Example for \( \text{NH}_3 \) (Ammonia):

Step 1: Valence Electrons

Nitrogen (N): 5, Hydrogen (H): 1 (3 H atoms: \( 3 \times 1 = 3 \)). Total: \( 5 + 3 = 8 \) valence electrons.

Step 2: Central Atom (N)

N is central, bonded to 3 H atoms.

Step 3: Single Bonds

3 N-H bonds use \( 3 \times 2 = 6 \) electrons. Remaining: \( 8 - 6 = 2 \) electrons (1 lone pair on N).

Step 4: Complete Octets

N has 8 electrons (3 bonds + 1 lone pair), H has 2 each. The Lewis structure is:

    H
    |
H - N - H
    |
    (Lone pair on N: two dots)

Key Rules:

• Octet Rule: Atoms (except H, He, Li, Be) aim for 8 valence electrons.
• Formal Charge: \( \text{Formal Charge} = \text{Valence Electrons} - (\text{Nonbonding Electrons} + \frac{\text{Bonding Electrons}}{2}) \).
• Resonance: For ions/molecules with multiple valid structures (e.g., \( \text{NO}_3^- \)), draw all resonance forms.

Example for \( \text{H}_2\text{O} \):

Step 1: Valence Electrons

Oxygen (O): 6, Hydrogen (H): 1 (2 H atoms: \( 2 \times 1 = 2 \)). Total: \( 6 + 2 = 8 \) valence electrons.

Step 2: Central Atom (O)

O is central, bonded to 2 H atoms.

Step 3: Single Bonds

2 O-H bonds use \( 2 \times 2 = 4 \) electrons. Remaining: \( 8 - 4 = 4 \) electrons (2 lone pairs on O).

Step 4: Complete Octets

O has 8 electrons (2 bonds + 2 lone pairs), H has 2 each. The Lewis structure is:

H - O - H
   ..
   (Two lone pairs on O)

For \( \text{SF}_4 \) (Sulfur Tetrafluoride):

Step 1: Valence Electrons

Sulfur (S): 6, Fluorine (F): 7 (4 F atoms: \( 4 \times 7 = 28 \)). Total: \( 6 + 28 = 34 \) valence electrons.

Step 2: Central Atom (S)

S is central, bonded to 4 F atoms.

Step 3: Single Bonds

4 S-F bonds use \( 4 \times 2 = 8 \) electrons. Remaining: \( 34 - 8 = 26 \) electrons.

Step 4: Complete Octets for F

Each F needs 6 more electrons (to reach 8). 4 F atoms: \( 4 \times 6 = 24 \) electrons. Remaining: \( 26 - 24 = 2 \) electrons (1 lone pair on S).

Step 5: Expanded Octet for S

Sulfur (a 3rd-period element) can expand its octet. It has \( 4 \) bonds (\( 8 \) electrons) + \( 1 \) lone pair (\( 2 \) electrons) = \( 10 \) electrons (allowed for 3rd-period and below).

The Lewis structure has S in the center, 4 single bonds to F, and 1 lone pair on S.

Final Answer (Example for \( \text{CH}_4 \)):

The Lewis structure for \( \text{CH}_4 \) is a central C atom bonded to 4 H atoms with single bonds, with no lone pairs on C or H.

(For other molecules/ions, follow the same steps: count valence electrons, arrange atoms, bond, distribute remaining electrons, and adjust for octets/charges.)