Chemical bonding II

Question 1

Hybridisation

Answer 1

• Used to explain observed shapes of molecules
• Set of atomic orbitals mixed to generate a set of equivalent hybrid orbitals with same shape and energy
• Equal bond lengths → form σ bonds

Question 2

Sp3 hybridisation

Answer 2

• C atoms bonded to 4 other atoms, e.g. CH4
• Tetrahedral (109.5°)
• One 2s orbital + three 2p orbitals → four equivalent sp3 hybrid orbitals

Question 3

Sp2 hybridisation

Answer 3

• C atoms bonded to 3 other atoms, e.g. C2H4
• Trigonal planar (120°)
• One 2s orbital + two 2p orbitals → 3 equivalent sp2 hybrid orbitals
• Unhybridised 2p orbital remains → unpaired electron → form 𝝅 bond

Question 4

Sp hybridisation

Answer 4

• C atoms bonded to 2 other atoms, e.g. C2H2
• Linear (180°)
• One 2s orbital + one 2p orbitals → 2 equivalent sp hybrid orbitals
• 2 unhybridised 2p orbitals remain (perpendicular to each other) → unpaired electron → form 2 𝝅 bonds

Question 5

Effect of hybridisation on bond length and strength

Answer 5

• s orbital spherical and electrons closer to nucleus
• Higher s-character → less diffuse hybrid orbital → more tightly the shared electrons are held by nuclei → more effective overlap → stronger bond + shorter bond length

Question 6

Resonance structures/canonical forms

Answer 6

Same placement of atoms but different arrangement of electrons → diff placement of 𝝅 bonds and non-bonding electrons

Question 7

Molecules/ions exhibit resonance when there is

Answer 7

Continuous side-on overlap of p-orbitals over at least 3 adjacent atoms, allowing for delocalisation of 𝝅 electrons

Question 8

Resonance hybrid

Answer 8

• Hybrid of resonance structures
• 𝝅 electrons delocalised over adjacent atoms via continuous side-on overlapping of p orbitals
• Delocalises electron density over larger volume → more stable
• Bond length equal, mid-way between single and double bond length (partial double bond character)

Question 9

Diamond

Answer 9

• Each C atom sp3 hybridised → covalently bonded to 4 other C atoms → tetrahedral
• Extremely strong and rigid
• Electrical insulator → no delocalised electrons

Question 10

Graphite

Answer 10

• Layer structure → planes of interconnected hexagonal rings
• Sp2 hybridised → 3 σ bonds → trigonal planar
• Unhybridised p-orbital with single electron → overlaps collaterally with p-orbitals of immediate neighbour → extended 𝝅-electron cloud above and below the plane
• 𝝅 electrons delocalised over whole layer → electrical conductor → only parallel to layers

Question 11

Why does graphite have a higher melting point than diamond? (2)

Answer 11

• Both have giant covalent structure → melting involves breaking of covalent bonds
  1. Each C atom in graphite has unhybridised p orbital containing 1 electron each → overlap collaterally with p orbitals of immediate neighbour → additional electron density between C atoms → hold nuclei more closely → strengthens C-C bond
  2. Hybrid orbitals of C atoms in graphite have higher s character → less diffuse → more effective overlap