ATOMS AND BONDING

Question 1

What are sigma bonds?

Answer 1

End to end overlap of atomic orbitals
Always first bond formed between 2 atoms

Question 2

What are pi bonds?

Answer 2

Side to side overlap of atomic p orbitals (ONLY FORMS BETWEEN P ORBITALS)
Always found in addition to a sigma bond

Question 3

What is hybridisation?

Answer 3

Process of blending 2 atomic orbitals to create hybridised orbitals

Question 4

Explain a sp3 hybridised carbon

Answer 4

• x1 s orbital and x3 p orbitals
• carbon makes 4 sigma bonds
• hybridises one 2s orbital and three 2p orbitals
• forms four sp3 hybridised orbitals
• tetrahedron, 109.5

Question 5

Explain a sp2 hybridised carbon

Answer 5

• x1 s orbital and x2 p orbitals
  -leaves x1 p orbital unhybridized
• carbon makes 3 sigma bonds and one pi bond (resulting in formation of double bonds or delocalised pi systems)
  -trigonal planar, 120

Question 6

Explain a sp hybridised carbon

Answer 6

• x1 s orbital and x1 p orbital
  -leaves x2 p orbitals unhybridized
• carbon makes 2 sigma bonds and 2 pi bonds (resulting in formation of triple bonds - 1 sigma bond, 2 pi bonds- or delocalised pi systems)

Question 7

How does covalent bonding in oxygen and nitrogen work?

Answer 7

Nitrogen - makes 3 covalent bonds, leaves a lone pair
Oxygen - makes 2 covalent bonds, leaves 2 lone pairs

Question 8

Explain sp3 hybridised nitrogen and oxygen

Answer 8

N:
-makes 3 sigma bonds, forms three sp3 orbitals, with lone pair in a sp3 orbital
-lone pair can make sigma bond with proton, but N has to hybridise one 2s orbital and three 2p orbitals to form four sp3 orbitals (to form 4 sigma bonds)
O:
-makes 2 sigma bonds, forms four sp3 orbitals with two lone pairs of electrons in sp3 orbitals

Question 9

Explain sp2 hybridised Nitrogen and oxygen:

Answer 9

N:
-makes 2 sigma bonds and 1 pi bond, forms three sp orbitals and leaves one pi orbital unhybridized, with lone pair in an sp2 orbital
O:
-makes 1 sigma bond and one pi bond, forms three sp orbitals and one p orbital unhybridized, with two lone pairs in the p orbital

Question 10

Explain sp hybridised Nitrogen

Answer 10

N:
-makes one sigma bond and 2 pi bonds, forming two sp orbitals and two p orbitals unhybridized, leaving lone pair of electrons in p orbital

Question 11

What is the special case with sp2 hybridised Nitrogens and oxygens?

Answer 11

N’s or O’s with free pair of electrons in can switch between hybridisation states
Only forms with electrons in a hybridised orbital can form sigma bonds with a proton
Two types of sp2 hybridised N:
- one making a pi bond
- one making 3 sigma bonds (but attached to a sp2 hybridised carbon - COPYCAT RULE)

Question 12

Explain the ‘copycat’ rule

Answer 12

N will have same hybridisation state as the carbon adjacent to it
N ‘copies’ hybridisation state and bonding pattern of C but with one more valence electron

Question 13

What are the implications of orbital hybridisation on pka?

Answer 13

Acidity partly controlled by stability/ energy level or anion formed by deprotonation
Different hybridised orbitals have different stabilities
sp orbital most stable due to high degree of s character
means sp orbital closer to nucleus = electrons more tightly bound = weaker tendency to release proton = less acidic = higher pka

Question 14

What is resonance?

Answer 14

Movement of electrons between atoms
Describes delocalisation of electrons within molecules
Involves multiple lewis structures

Question 15

How do you determine if a structure can resonate?

Answer 15

1. If there’s a charge/lone (delocalised) pair in position 1
2. If there’s a double bond between position 2 and 3

Question 16

How do you draw this resonance structure?

Answer 16

3. Move charge to make double bond between 1 and 2
4. Move electrons from between 2 and 3 to make charge at position 3

Question 17

Can i resonate when a positive charge is present?

Answer 17

Yes, but do not move the positive charge, only move the electrons

Question 18

What is aromaticity?

Answer 18

Chemical property in cycloalkanes
stabilisation enhanced because electrons in p orbitals can delocalise

Question 19

State Huckels rules for aromaticity:

Answer 19

1. Molecule must be cyclic
2. Molecule must be able to be planar (all molecules lie in same plane)
3. Must be conjugation around whole ring/ must be a conjugated system (every atom must be able to participate in pi bonding- by having a p orbital)
4. must follow 4n + 2 rule

Question 20

What can affect electron density in aromatic systems?

Answer 20

Resonance effects in/out:
- due to pi electrons delocalised throughout ring stabilises distribution of electron density, increasing e.d
Inductive effects in/out:
- due to differences in electronegativity between atoms
Substituents in/out: electron withdrawing/electron donating groups

Question 21

How does electron density have implications on chemical reactivity?

Answer 21

bond polarisation: more electronegative substituent draws cloud of electrons towards itself and polarises bond

pKa: stability of conjugate base/acid determined by resonance and inductive effects
- if a conjugate base is more stable than it’s acid, then the molecule is more acidic (lower pka)
-if a acid is more stable than it’s conjugate base, it’s less acidic (higher pka)