Solution:

Given: CO₂ Find: Molecular geometry and bond angle

Step 1: Draw Lewis structure

Total valence electrons: 4 (C) + 2(6) (O) = 16

$O=C=O$

Each O has 2 double bonds and 2 lone pairs C has 2 double bonds (no lone pairs)

Step 2: Identify central atom

Central atom: C (carbon)

Step 3: Count electron domains on C

• 1 double bond to left O = 1 electron domain
• 1 double bond to right O = 1 electron domain
• 0 lone pairs on C

Total electron domains: 2

Remember: Multiple bonds count as ONE domain!

Step 4: Determine steric number

$\text{SN} = 2$

Step 5: Identify electron geometry

SN = 2 → Linear electron geometry

Step 6: Count lone pairs on central atom

C has 0 lone pairs

Step 7: Determine molecular geometry

SN = 2, 0 lone pairs → Linear molecular geometry

Step 8: Predict bond angle

Linear geometry → 180°

Answer:

• Molecular geometry: Linear
• Bond angle: 180°
• Structure: O=C=O (all atoms in straight line)

Verification:

• 2 electron domains ✓
• 0 lone pairs ✓
• Electron geometry = Molecular geometry (no lone pairs) ✓
• Maximum separation of 2 domains = 180° ✓

Note: CO₂ is also nonpolar because the two C=O dipoles point in opposite directions and cancel out due to the linear, symmetrical geometry.

Solution:

(a) Lewis structure:

• Xe: 8 valence electrons (noble gas)
• 4 F: 4 × 7 = 28 valence electrons
• Total: 36 electrons

Xe in center with 4 F atoms bonded, plus 2 lone pairs on Xe

(b) Geometries:

• Electron pairs around Xe: 4 bonding + 2 lone = 6 electron pairs
• Electron geometry: Octahedral
• The 2 lone pairs occupy opposite positions (180° apart to minimize repulsion)
• Molecular geometry: Square planar

(c) Bond angles:

• F-Xe-F (adjacent): 90°
• F-Xe-F (opposite): 180°

(d) Polarity: Nonpolar - Despite having polar Xe-F bonds, the square planar geometry is perfectly symmetric. The bond dipoles cancel out (4 F atoms arranged symmetrically around Xe in a plane), resulting in no net dipole moment.

Solution:

Given: H₂O and CH₄ both have SN = 4 Find: Explain different geometries and bond angles

Step 1: Draw Lewis structures

H₂O (Water):

$H - O - H$

O has 2 bonding pairs and 2 lone pairs

CH₄ (Methane):

$H - C - H$ (with 2 more H atoms)

C has 4 bonding pairs and 0 lone pairs

Step 2: Calculate steric numbers

H₂O:

• 2 O-H bonds = 2 domains
• 2 lone pairs = 2 domains
• SN = 4

CH₄:

• 4 C-H bonds = 4 domains
• 0 lone pairs = 0 domains
• SN = 4

Both have SN = 4 ✓

Step 3: Determine electron geometry

Both have SN = 4 → Tetrahedral electron geometry

Step 4: Determine molecular geometry

H₂O:

• SN = 4, 2 lone pairs
• Molecular geometry: Bent
• Only "see" the 2 H atoms (ignore lone pairs in molecular shape)

CH₄:

• SN = 4, 0 lone pairs
• Molecular geometry: Tetrahedral
• "See" all 4 H atoms

Step 5: Explain bond angles

CH₄:

• Ideal tetrahedral: 109.5°
• No lone pairs → no compression
• H-C-H angle = 109.5°

H₂O:

• Ideal tetrahedral would be 109.5°
• BUT: 2 lone pairs present
• Lone pairs repel more than bonding pairs
• Lone pairs push H atoms closer together
• H-O-H angle = 104.5° (compressed)

Step 6: Visualize repulsions

Repulsion strength: $\text{LP-LP} > \text{LP-BP} > \text{BP-BP}$

H₂O has:

• 1 LP-LP repulsion (strongest)
• 4 LP-BP repulsions
• 1 BP-BP repulsion (weakest)

The strong LP-LP and LP-BP repulsions compress the bond angle.

CH₄ has:

• 0 LP-LP repulsions
• 0 LP-BP repulsions
• 6 BP-BP repulsions (all equal)

Equal repulsions maintain ideal 109.5° angle.

Answer:

Why different despite same SN?

1. Electron geometry is the same (tetrahedral for both)

2. Molecular geometry differs due to lone pairs:

   • H₂O: Bent (2 LP compress bonding pairs)
   • CH₄: Tetrahedral (0 LP, all positions equivalent)

3. Bond angles differ due to lone pair repulsion:

   • H₂O: 104.5° (compressed by 2 lone pairs)
   • CH₄: 109.5° (ideal tetrahedral, no lone pairs)

Key principle: Molecular geometry depends on BOTH steric number AND number of lone pairs. Lone pairs are "invisible" in the molecular shape but strongly affect bond angles through greater repulsion.

Summary:

Solution:

All three molecules have tetrahedral electron geometry (4 electron pairs), but different molecular geometries:

(a) CH₄ - Tetrahedral, 109.5°

• 4 bonding pairs, 0 lone pairs
• All electron pairs are equivalent
• Perfect tetrahedral angle: 109.5°

(b) NH₃ - Trigonal pyramidal, 107°

• 3 bonding pairs, 1 lone pair
• Lone pairs occupy more space than bonding pairs (electrons closer to nucleus)
• Lone pair-bonding pair repulsion > bonding pair-bonding pair repulsion
• Lone pair pushes bonding pairs together: 107° (smaller than 109.5°)

(c) H₂O - Bent, 104.5°

• 2 bonding pairs, 2 lone pairs
• Two lone pairs create even more repulsion
• Both lone pairs push bonding pairs together
• Even smaller angle: 104.5°

VSEPR Repulsion Order: Lone pair-Lone pair > Lone pair-Bond pair > Bond pair-Bond pair

Summary: As lone pairs replace bonding pairs, the bond angle decreases because lone pairs exert greater repulsive forces.

Angle trend: CH₄ (109.5°) > NH₃ (107°) > H₂O (104.5°)

Solution:

Given: XeF₄ (xenon tetrafluoride) Find: Molecular geometry, bond angles, polarity

Step 1: Draw Lewis structure

Count valence electrons:

• Xe: 8 valence electrons (Group 18, noble gas)
• F: 7 valence electrons (×4)
• Total: 8 + 28 = 36 electrons

Skeletal structure: Xe is central

$F - Xe - F$ (with 2 more F atoms)

Connect with single bonds: 4 Xe-F bonds = 8 electrons used Remaining: 36 - 8 = 28 electrons

Complete octets of F atoms: Each F needs 6 more electrons (3 lone pairs) 4 F × 6 = 24 electrons Remaining: 28 - 24 = 4 electrons

Place remaining electrons on Xe: 4 electrons = 2 lone pairs on Xe

Lewis structure:

$F$ $|$ $F - Xe - F$ $|$ $F$

Xe has 4 bonds and 2 lone pairs

Step 2: Count electron domains on Xe

• 4 Xe-F single bonds = 4 domains
• 2 lone pairs = 2 domains
• Total: 6 electron domains

Step 3: Determine steric number

$\text{SN} = 6$

Step 4: Identify electron geometry

SN = 6 → Octahedral electron geometry

Step 5: Count lone pairs

Xe has 2 lone pairs

Step 6: Determine molecular geometry

SN = 6, 2 lone pairs → Square planar

Why square planar?

• In octahedral geometry, all 6 positions are equivalent
• Lone pairs occupy positions opposite each other (minimizes repulsion)
• Remaining 4 F atoms form a square in a plane
• Xe at center, lone pairs above and below

Step 7: Predict bond angles

F-Xe-F angles:

• Adjacent F atoms: 90° (corners of square)
• Opposite F atoms: 180° (across square)

Ideal octahedral angles: 90°

With 2 lone pairs:

• LP-LP repulsion is strongest
• LP-BP repulsion compresses slightly
• But symmetry maintains approximately 90° and 180°

Step 8: Determine polarity

Check bond polarity:

• Xe-F bonds are polar (ΔEN ≈ 2.0)
• Each Xe-F bond has dipole toward F (F is more electronegative)

Check molecular symmetry:

$F$ $↑$ $F ← Xe → F$ $↓$ $F$

Dipole analysis:

• Top F dipole ↑ cancels with bottom F dipole ↓
• Left F dipole ← cancels with right F dipole →
• Perfect symmetry in square planar arrangement
• Net dipole = 0

Answer:

Molecular geometry: Square planar

Bond angles:

• F-Xe-F (adjacent): 90°
• F-Xe-F (opposite): 180°

Polarity: Nonpolar

Explanation of polarity: Even though Xe-F bonds are polar, the square planar geometry is perfectly symmetrical. The four bond dipoles point toward the corners of a square and cancel out vectorially. The two lone pairs are positioned opposite each other (above and below the plane), maintaining symmetry.

Additional notes:

1. Expanded octet: Xe (Period 5) can accommodate 12 electrons (6 electron domains) using d orbitals

2. Lone pair placement: In octahedral geometry, lone pairs prefer to be 180° apart (trans position) to minimize LP-LP repulsion

3. Why not other geometries?

   • If lone pairs were adjacent (90°), LP-LP repulsion would be much stronger
   • Square planar with trans lone pairs is most stable

4. Other examples of square planar: ICl₄⁻, BrF₄⁻

Summary: