Solution:

Given: NH₃ (ammonia) Find: Hybridization of N, count σ and π bonds

Step 1: Draw Lewis structure

$H - N - H$ $|$ $H$

N has 3 N-H bonds and 1 lone pair

Step 2: Identify central atom

Central atom: N (nitrogen)

Step 3: Count electron domains on N

• 3 N-H single bonds = 3 domains
• 1 lone pair = 1 domain
• Total: 4 electron domains

Step 4: Determine steric number

$\text{SN} = 4$

Step 5: Determine hybridization

SN = 4 → sp³ hybridization

Important: The lone pair occupies one of the sp³ hybrid orbitals!

Step 6: Count sigma (σ) bonds

• 3 N-H bonds = 3 σ bonds
• Each single bond = 1 σ bond

Step 7: Count pi (π) bonds

• No double or triple bonds
• 0 π bonds

Answer:

Hybridization of N: sp³

Bonds:

• σ bonds: 3
• π bonds: 0

Explanation:

Nitrogen uses sp³ hybridization because it has 4 electron domains (3 bonds + 1 lone pair). The 4 sp³ hybrid orbitals are arranged tetrahedrally:

• 3 sp³ orbitals form σ bonds with H atoms (overlap with H 1s orbitals)
• 1 sp³ orbital contains the lone pair

The molecular geometry is trigonal pyramidal (not tetrahedral) because we only "see" the 3 H atoms in the molecular shape, though the electron geometry is tetrahedral.

Bond angle: ~107° (less than ideal 109.5° due to lone pair repulsion)

Key takeaway: Hybridization is determined by TOTAL electron domains (including lone pairs), not just bonding domains.

Solution:

Given: H₂CO (formaldehyde) Find: (a) Hybridization of C and O, (b) σ and π bond counts, (c) why planar

Step 1: Draw Lewis structure

$H$ $|$ $H - C = O$

Or more clearly: C has 2 single bonds to H and 1 double bond to O O has 2 lone pairs

Part (a): Determine hybridization

For Carbon (C):

Count electron domains:

• 2 C-H single bonds = 2 domains
• 1 C=O double bond = 1 domain (multiple bonds count as 1!)
• 0 lone pairs
• Total: 3 electron domains

Steric number: SN = 3

Hybridization: SN = 3 → sp² hybridization

For Oxygen (O):

Count electron domains:

• 1 C=O double bond = 1 domain
• 2 lone pairs = 2 domains
• Total: 3 electron domains

Steric number: SN = 3

Hybridization: SN = 3 → sp² hybridization

Part (b): Count σ and π bonds

Sigma bonds:

• C-H bond (left): 1 σ (sp²-s overlap)
• C-H bond (right): 1 σ (sp²-s overlap)
• C=O bond: 1 σ (sp²-sp² overlap)
• Total σ bonds: 3

Pi bonds:

• C=O double bond: 1 π (p-p side overlap)
• Total π bonds: 1

Verification:

• Total bonds: 2 single + 1 double = 4 bonds
• 4 bonds = 3 σ + 1 π ✓

Part (c): Explain planarity

Why is H₂CO planar?

1. Carbon hybridization: C is sp² hybridized

   • sp² creates trigonal planar electron geometry
   • All 3 sp² orbitals lie in the same plane
   • Bond angles ~120°

2. Pi bond requirement: The C=O double bond requires:

   • 1 σ bond (sp² hybrid overlap)
   • 1 π bond (unhybridized p orbital overlap)

3. P orbital orientation:

   • Each sp² hybridized atom (C and O) has 1 unhybridized p orbital
   • These p orbitals must be parallel to overlap and form π bond
   • P orbitals are perpendicular to the plane of sp² orbitals
   • This forces all atoms to lie in the same plane

4. Restricted rotation:

   • π bond prevents rotation around C=O
   • If molecule tried to twist, p orbitals wouldn't overlap
   • Planarity is maintained

Answer:

(a) Hybridization:

• Carbon: sp²
• Oxygen: sp²

(b) Bonds:

• Sigma (σ) bonds: 3
• Pi (π) bonds: 1

(c) Planarity explanation:

The molecule is planar because both C and O are sp² hybridized. sp² hybridization creates a trigonal planar arrangement where all atoms lie in the same plane. Additionally, the π bond in the C=O double bond requires parallel p orbitals, which forces the molecule to remain flat. Any deviation from planarity would disrupt the p orbital overlap needed for the π bond.

Molecular geometry:

• Around C: Trigonal planar (3 groups)
• Around O: Bent (1 bond + 2 lone pairs, but sp² geometry)
• All 4 atoms (H, C, H, O) in same plane

Bond angles:

• H-C-H: ~120°
• H-C=O: ~120°
• Total around C: ~360° (planar)

Solution:

Given: C₂H₂ (acetylene), structure H-C≡C-H Find: (a) Hybridization, (b) σ and π counts, (c) orbital overlaps, (d) geometry explanation

Step 1: Draw Lewis structure

$H - C ≡ C - H$

Each C has:

• 1 single bond to H
• 1 triple bond to other C

Part (a): Determine hybridization of each C

For left carbon (C₁):

Count electron domains:

• 1 C-H single bond = 1 domain
• 1 C≡C triple bond = 1 domain (triple bond counts as 1!)
• 0 lone pairs
• Total: 2 electron domains

Steric number: SN = 2

Hybridization: SN = 2 → sp hybridization

For right carbon (C₂):

Same analysis as C₁:

• SN = 2 → sp hybridization

Orbital composition for sp hybridization:

• Mix 1 s + 1 p orbital → 2 sp hybrid orbitals
• Remaining: 2 unhybridized p orbitals (py and pz)

Part (b): Count σ and π bonds

Sigma bonds:

• H-C bond (left): 1 σ
• C≡C bond: 1 σ (sp-sp overlap)
• C-H bond (right): 1 σ
• Total σ bonds: 3

Pi bonds:

• C≡C triple bond: 2 π bonds (two sets of p-p overlaps)
• Total π bonds: 2

Total bonds: 3 σ + 2 π = 5 bonds

• 2 single bonds (H-C, C-H) = 2 bonds
• 1 triple bond (C≡C) = 3 bonds
• Total: 5 bonds ✓

Part (c): Describe orbital overlap for each bond

Left H-C bond:

• Type: σ bond
• Orbitals: H (1s) overlaps with C (sp hybrid)
• Overlap: Head-on (along internuclear axis)

C≡C bond (triple bond):

σ bond (1):

• Orbitals: C (sp hybrid) overlaps with C (sp hybrid)
• Overlap: Head-on (sp-sp)
• Orientation: Along C-C axis

π bond (1):

• Orbitals: C (unhybridized py) overlaps with C (unhybridized py)
• Overlap: Side-by-side (parallel p orbitals)
• Orientation: Above and below C-C axis

π bond (2):

• Orbitals: C (unhybridized pz) overlaps with C (unhybridized pz)
• Overlap: Side-by-side (parallel p orbitals)
• Orientation: Front and back of C-C axis

Right C-H bond:

• Type: σ bond
• Orbitals: C (sp hybrid) overlaps with H (1s)
• Overlap: Head-on

Visualization of p orbitals:

• py orbitals: create π bond in vertical plane
• pz orbitals: create π bond in horizontal plane
• Together: cylindrical electron cloud around C-C axis

Part (d): Explain geometry and collinearity

Why are all atoms collinear (in a straight line)?

1. sp Hybridization geometry:

• sp hybridization → linear electron geometry
• 2 sp hybrid orbitals point in opposite directions (180° apart)
• This minimizes repulsion between 2 electron domains

2. Orbital arrangement:

Left carbon:

• One sp orbital points toward H (forms H-C σ bond)
• One sp orbital points toward other C (forms C-C σ bond)
• 180° angle between H and C

Right carbon:

• One sp orbital points toward other C (forms C-C σ bond)
• One sp orbital points toward H (forms C-H σ bond)
• 180° angle between C and H

3. Pi bond requirement:

• Two sets of p orbitals must be perpendicular to each other
• Both sets must be perpendicular to the sp orbital axis
• This arrangement requires linear geometry
• Any bending would destroy p orbital overlap

4. Maximum separation:

• SN = 2 means 2 electron domains
• Maximum separation = 180° (linear)
• Minimizes electron-electron repulsion

Answer:

(a) Hybridization:

• Both carbon atoms: sp

(b) Bonds:

• Sigma (σ) bonds: 3 (2 C-H + 1 C-C)
• Pi (π) bonds: 2 (both in C≡C)

(c) Orbital overlaps:

• H-C bonds: H(1s)-C(sp) σ bonds
• C≡C bond:
  • 1 σ bond from C(sp)-C(sp) overlap
  • 2 π bonds from C(py)-C(py) and C(pz)-C(pz) overlaps

(d) Geometry:

All atoms are collinear (linear) because:

1. sp hybridization creates 2 hybrid orbitals at 180°
2. Steric number = 2 requires linear geometry (maximum separation)
3. π bonds require unhybridized p orbitals to be perpendicular to internuclear axis
4. This forces H-C-C-H into straight line

Bond angle: H-C-C = C-C-H = 180°

Overall molecular geometry: Linear (all 4 atoms in straight line)

Key insight: The triple bond (1 σ + 2 π) creates a very strong, rigid, linear structure. The two π bonds form a cylindrical electron cloud around the C-C σ bond, preventing any rotation or bending.