Topic 2 - Chemical Bonds

Question 1

What is the wave function?

Answer 1

Wavefunction: The probability of finding the electron in a particular area.

• The reason why the wave function is used is because electrons show wave-like properties. This means they don’t always have a fixed position (Wave is not a single particle). Hence, the wave function informs us of the probable locations of the electrons.

Note –> diagram shows different shades –> representative of different probabilities.

Question 2

How does one represent the wave function?

Answer 2

Mathematical description –> Ψ (x,y,z) –>Three axis as we are dealing with three dimensions.

Question 3

How does the graph of the wave function look like?

Answer 3

Question 4

What is the electron density?

Answer 4

Electron density (Ψ²)

• Obtained by squaring the wave function
• Gives the number of electrons in a given area.

Question 5

What is the graphical representation of the electron density?

Answer 5

Question 6

What is the radial electron distribution?

Answer 6

Radial electron distribution (4πr²)(Ψ²)

• Multiply the volume of a sphere with electron density gives the e^- distribution in a given volume. Basically, gives the probability of finding an electron in a given ‘layer’ relative to the nucleus.
• Equation provides us with the most likely radius where e^- are found –> represented by maximum on graph.

Question 7

What does the graphical representation of the radial wave function look like?

Answer 7

Question 8

What is the wave function + electron distribution of the 2s and 2p orbitals?

Answer 8

Note –> shading/ ± is used to represent wave function direction, not charge.

Question 9

What is the equation for electrostatic potential energy?

Answer 9

Electrostatic potential energy –> Coulobmb’s Law

E α (q₁ x q₂)/(r)

which is equivalent to….

E = K (q₁ x q₂)/(r)

Where…

q = charge

r = seperation distance

k = constant

Question 10

What can coulomb’s law (Electrostatic potential energy) be used for?

Answer 10

When two objects with charge q₁ and q₂ respectively are located r distance away from each other we can calculate the electrostatic potential energy using the constant K.

• Greater charges/smaller distance –> greater E (visa versa)

Results:

• Positive E –> forces are repulsive
• Negative E –> forces are attracting
• Neutral object –> 0 interaction.

Question 11

What is the Aufbau principal?

Answer 11

Electrons are placed in orbitals starting with the lowest energy, working up.

Question 12

What is the Pauli exclusion principle?

Answer 12

It states that there are a maximum of 2 electrons per orbital and they must spin in opposite directions.

Question 13

What is Hund’s rule?

Answer 13

It states that when multiple orbitals of the same energy are available, electrons are distributed among them and spin parallel (before pairing electrons).

Question 14

What are molecular orbitals?

Answer 14

Molecular orbitals explain what happens to the orbitals when a molecule is formed.

They are helpful in explaining a specific type of bonding (delocalization) and properties (magnetism) of molecules which the Lewis structure doesn’t explain.

• Atomic orbitals are associated with atoms
• Molecular orbitals are associated with molecules –> are spread across the entire molecule.

Question 15

In what situations are molecular orbitals formed?

Answer 15

Atoms only bond to form a molecule when there are more favourable interactions than unfavourable interactions.

For example: H₂ –> 4 favourable and 2 unfavourable –> H₂ molecule is formed.

Question 16

What determines whether a successful bond is formed?

Answer 16

When talking about the molecular orbitals - a successful bond depends on whether the wave functions of the two atoms are in-phase or out of phase.

In-Phase –> Succesful bond

Out of phase –> No successful bond

Note:

In-Phase –> Waves combine together –> amplitude of waves is added.

Out of phase –> Waves cancel each other out –> destruction of the wave.

Question 17

What do the A.O wave function graph, M.O wave function graph, radial distribution and electron density look like for H₂.

A.O –> Atomic orbital (before combined)

M.O –> After orbitals are combined.

Answer 17

In phase

• Bond is formed –> cylindrical symmetry
• Bond is always at a fixed distance - energetically favourable.
• Known as bonding molecular orbitals.

Out of phase

• No bond formed (also cylindrical symmetry).
• Known as an anti-bonding orbital

Question 18

Do the number of atomic orbitals and molecular orbitals need to equal each other?

Answer 18

Yes, the number of atom orbitals and molecular orbitals must equal each other.

Question 19

What do the energy level diagrams of H₂ look like? What can they tell us about the bonding taking place?

Answer 19

The molecular bonding orbital is at a lower energy than the individual atomic orbitals –> hence, the formation of the M.O is energetically favourable –> provides stability.

Question 20

What do the energy level diagrams of He look like? What can they tell us about the bonding taking place?

Answer 20

In He’s case both of the bonding and anti-bonding orbitals get filled.

Stabilisation gained by the bonding orbitals is counter-balanced by the destabilisation of the anti-bonding orbital. No energetic reason to He₂ –> hence it is not found in nature in this form.

Question 21

Can the molecular orbital theory be extended to π orbitals?

Answer 21

Yes, the M.O theory can also be extended to π bonds.

Question 22

Draw the bonding and anti-bonding orbitals for the sigma bonds formed by pi bond head-on overlap.

Answer 22

Question 23

Draw the bonding and anti-bonding orbitals for the formation of π bonds.

Answer 23

Important to realize that the Pi bond isn’t as strong as the sigma bond. This results in destabilisation of the of π^* orbital having a lower energy than σ^* .

Question 24

Draw the three different bonds that can be formed in an O₂ molecule.

Hint - Think about different Pi overlaps.

Answer 24

The geometry of molecular orbitals in O₂.

1. Head on collision of 2p orbitals –> sigma bond
2. Vertical side-ways overlap of 2p orbitals –> Pi bond
3. Horizontal side-ways overlap of 2p orbitals –> Pi bond.

Question 25

What does the orbitals size correspond to?

Answer 25

The orbital size correspond to the point where there is the highest probability of finding electrons.

Question 26

When 2p orbitals are involved in bonding what happens to the 2s and 1s Orbitals?

Answer 26

When 2p orbitals are bonding the 2s and 1s orbital can’t participate in bonding as they lack the sufficient size –> they can’t physically reach.

Question 27

Draw the energy level diagram for O₂ .

Answer 27

Question 28

The formula for calculating bond order?

Answer 28

Bond order = (e^- in bonding orbitals - e^- in anti-bonding orbitals)/2

Bond order –> gives the net number of shared pairs of electrons. i.e. A double bond corresponds to a bond order of 2.

Question 29

Explain the bonding in methane (reference hybridisation)

Answer 29

According to the atomic structure of carbon, carbon wouldn’t be able to form the tetrahedral shape of methane.

This is overcome by the hybridisation of orbitals to form Sp³ hybrid orbitals.

1. One electron in 2s orbital promoted to 2p
2. 2s and 2p orbitals hybridize to form Sp³ orbitals.

Each orbital has one electron so it can now form 4 sigma bonds with hydrogen –> this has a tetrahedral arrangement (109.5^o).

Question 30

What is a Fisher projection?

Answer 30

A diagram that shows 3D nature of the molecule.

• Atoms coming out from the side –> go out of the plane of the paper.
• Atoms at the top and bottom go into the plane of the paper.

Question 31

Explain the bonding in ethene (reference hybridisation).

Answer 31

The carbons in ethene need to form 3 sigma bonds and one double bond. Orbitals need to be hybridized to Sp² hybrid orbitals in order to make this possible.

1. The electron is promoted to P orbital
2. Orbitals hybridize to form Sp²
   - Sp² form sigma bonds and the extra P orbital forms Pi bond.
   - Note we would expect the electron in the P orbital to move down to the Sp² orbital but the orbitals are close in terms of energy that pairing electrons is more energetically unfavourable.

Question 32

Explain the energy level diagram for the bonds in methane.

Answer 32

Question 33

Explain the energy level diagram for the double bond in ethene.

Answer 33

Question 34

What is the average bond energy + bond length of a single and double bond?

Answer 34

Single bond (σ) –> 350 KJ/Mole –> 1.54 Å (0.154nm)

Double bond (σ + π) –> 600 KJ/mole –> 1.34 Å (0.134nm)

As we can see from this a Pi bond is roughly 50 kJ/mole weaker than a sigma bond.

Question 35

Definition of an isomer?

Answer 35

Isomer –> same molecular formula but a different structural formula (different arrangement of atoms)

Question 36

Structural isomer definition? Characteristics?

Answer 36

Same molecular formula but a completely different arrangement of atoms.

1. Chemically different
2. Same number of bonds
3. Composition is the same –> different arrangement

Question 37

What is geometric isomerism?

Answer 37

This type of isomerism involves the restricted rotation of a bond due to the presence of a double bond or ring structure. Results in the formation of Cis (same) and trans (opposite) isomers.

The only way to interconvert would be to break the bond and rotate the bond by 180 degrees –> This process requires energy as electrons need to be returned to their atomic orbitals (higher energy level).

Question 38

What are different types of stereoisomers?

Answer 38

Stereoisomerism: Molecules that have the same molecular formula but have a different arrangement of the atoms in space.

Types

1. Configurational –> interconverted by breaking bonds
   a) Optical Isomerism
   b) Geometric isomers

C) Diastereomers

2. Conformational –> no bonds need to break

Question 39

When are two chiral molecules enantiomers?

Answer 39

They are enantiomers (isomers of each other) if they are mirror images and non-superimposable.

Note - Chiral centre must have 4 different groups attached.

Question 40

What are diastereomers?

Answer 40

These are stereoisomers that are not mirror images and non-superimposable.

Both molecules have different chemical and physical properties.

Question 41

What’s the difference between absolute and relative stereochemistry?

Answer 41

Absolute stereochemistry: is the precise arrangement of atoms in space –> use the stereochemical description of R and S.

Relative stereochemistry: Compares the arrangement of atoms in space of one compound with those of another. Use the stereochemical description of L and D.

Basically comparing and matching similar functional groups to figure out whether something is L or D.

Question 42

Explain the stereoisomerism of glucose.

Answer 42

All carbons on the glucose molecules are diastereomers apart from one which is an enantiomer.

• Change in the position of the -OH only takes place on one carbon which is carbon 4 –> involves breaking bonds to interconvert.

CHECK SLIDE!!!

Question 43

What is conformational isomerism?

Answer 43

Conformational isomers (or conformers or rotational isomers or rotamers) are stereoisomers produced by rotation (twisting) about σ bonds.

Question 44

What are the different conformations?

Answer 44

There are a total of 4 different conformations that one must consider.

1. One must examine the molecule in order to see whether it is eclipsed or staggered.
2. If it is staggered one must examine the bond angles between the two groups (usually examine the atom with the most protons –> repel the most –> further away –> more stable).

Furthest angle –> Called Anti

Anything else –> Gauche

Question 45

Differences in in eclipsed and staggered on an energy level diagram?

Answer 45

Question 46

Differences between anti and gauche on an energy level diagram?

Answer 46

Question 47

What is a Newman projection?

Answer 47

Diagram used to show different conformations.

Note the angle between the groups on the different carbons –> known as a torsion angle.

Question 48

Explain the dissociation of H₂O and how the value of Kw is derived.

Answer 48

Dissociation of H₂O

H₂O <—> H⁺ + OH^-

Equilibrium expression:

Kw = Keq = ([H⁺][OH^-])/(1) = 10^-14

Note that H₂O is ingored –> concentration is so large that change have minimal impact on Equilibrium constant.

Hence, in a neutral solution….

[H⁺][OH^-] = 10^-7

Question 49

What’s the definition of pH?

Answer 49

pH = -log[H⁺]

Question 50

Write out the equation for the dissociation of a proton from acetic acid and create an expression for Ka.

Answer 50

CH₃COOH <—–> CH₃COO^- + H⁺

Acid <—–> Conjugate base (acetate) + proton

Ka = ([CH₃COO^-][H⁺])/[CH₃COOH]

Question 51

From a general expression of Ka, derive the Henderson Hassel bach equation.

Answer 51

Question 52

When will the pH = Pka in the Henderson Hasselbach equation?

Answer 52

When there are equal concentrations of Acid and conjugate base.

Log[A^-]/[HA] = 0

Which means that…

pH = Pka

Question 53

Using the Henderson-Hasselbach equation figure out whether the protonated or deprotonated form of aspartic acid is preferred at physiological pH.

Answer 53

Physiological pH = 7.2

pKa of aspartic acid = 4.2

7.2 = 4.2 + log [A^-]/[HA]

3 = log [A^-​]/[HA]

10³ = [A^-​]/[HA]

Ratio is 1000:1

For every 1000 A^- there is 1 HA.

Hence, the most common form of the acid is the aspartate ion.

Question 54

Explain using orbitals why aspartate is preferred instead of aspartic acid.

Answer 54

The carboxylate ion forms a resonance structure –> Charged is distributed over a larger area –> dispersed charge –> increase stability –> explains why aspartate is preferred.

Question 55

Using the Henderson-Hasselbach equation figure out whether the protonated or deprotonated form of lysine side chain is preferred at physiological pH.

Answer 55

pH = 7.2

pKa = 10.2

7. 2 = 10.2 + log[B]/[BH⁺]
   - 3 = log[B]/[BH⁺]

10^-3 = [B]/[BH⁺]

10³ = [BH⁺]/ [B]

For every 1 unprotonated lysine there are 1000 protonated lysine molecules.

Amino acid side chain will be positively charged at physiological pH.

Question 56

Explain the charge distribution in the arginine side chain.

Answer 56

Basically what happens is that the C=NH group bonds to H⁺ to form C=NH₂⁺ –> subsequently the electrons from a lone pair from another nitrogen move towards positive charge –> breaks double bond –> but nitrogen that has now lost a lone pair needs to form a double bond –> Double bond moves between amine groups.

Question 57

Explain why arginine is more basic than lysine.

Answer 57

It is more basic as it more readily accepts the proton because it can form a resonance structure –> distributes the charge –> more stable.

Question 58

Explain how phosphate is able to form 5 bonds in phosphoric acid.

Answer 58

1. Because the orbitals in the third energy level are similar in terms of energy + electrons paired in 3s orbital experience repulsion –> the electron from the 3s promotes to the 3d orbital (as shown in picture).
2. The 3s and 3p orbitals hybridize –> sp³ hybrid orbital

Now the phosphate can form 4 sigma bonds and 1 pi bond.

Question 59

Explain the resonance system in phosphate ions.

Answer 59

Once phosphoric acid has become completely deprotonated it can form resonance structure –> the double bond moves between all the P — O bonds –> disperses the negative charge –> increases stability.

Hence, all oxygens have a partial negative charge.

IMPORTANT –> favours the hydrolysis of ATP to ADP and Pi can form stable resonance.

Question 60

What does a titration graph between a strong acid and base look like?

Answer 60

Question 61

What does a titration graph of a weak acid and strong base look like?

Answer 61

Question 62

Where is the buffer region of a weak acid?

Answer 62

The buffer is represented by the plateau on the graph before the sharp increase.

It occurs when the pH = pKa –> good buffer region (plus minus 1 pH).

Question 63

Why is a weak acid a good buffer when the pH = pKa?

Answer 63

This is because the concentration of both acid and conjugate base are equal. Hence, these molecules will be able to absorb any H⁺ or OH^- added to the solution.

Question 64

What are two important points to remember about a weak acid buffer?

Answer 64

2 main points

1. Weak acid can act as a goof buffer between plus minus 1 pH of pKa.
2. pH of the buffer is independent of dilution –> this is because when diluted the ratio of protonated to unprotonated remains the same –> no change in pH.

However, there is a decrease in buffering capacity.

Question 65

What does the titration graph of phosphoric acid look like?

Answer 65

Question 66

How does the fraction of the different protonated/deprotonated phosphoric acid molecules change when pH changes?

Answer 66

1. pH = pKa = 2.1
   50: 50 —> H₃PO₄ and H₂PO₄^-
2. pH = pKa = 6.9
   50: 50 —> H₂PO₄^- and HPO₄^2-
3. pH = pKa = 12.4
   50: 50 —> HPO₄^2- and PO₄^3-

Question 67

What is the dominant species of Phosphoric acid when pH is equal to 7.2.

Answer 67

Examining the ratio of

H₂PO4^- and HPO₄^2- at a pH = 7.2

7.2 = 6.9 + log [HPO₄^2-]/[H₂PO4^-]

10^0.3 = [HPO₄^2-]/[H₂PO4^-]

2 = [HPO₄^2-]/[H₂PO4^-]

ratio is 2:1

For every H₂PO4^- you will have two HPO₄^2-

Question 68

What are the pKa values for the oxygens on phosphate?

Answer 68

Pka for the bottom -OH –> 1.0 –> 100% of the time deprotonated.

Pka for side -OH –> 6.1 –> 90% of the time will be found deprotonated

Question 69

How can amino acids act as a buffer?

Answer 69

Amino acids are Zwitterions –> NH₂ forms NH₃⁺ and COOH forms COO^-

Hence, amino acids can absorb/neutralize both H⁺ ions and OH^- ions –> important for amino acids floating freely.

Question 70

How can a weak acid buffer be created?

Answer 70

Weak acid buffer can be created by mixing equal weak acid and its salt –> weak acid high concentrated –> salt fully dissociates –> conjugate base highly concentrated.

Weak acid + C.B –> Good buffer.

Question 71

What is the relationship between frequency and wavelength?

Answer 71

λ = C / ν

Wavelength = Speed of light / Frequency

This shows us that the frequency and wavelength are inversely proportionate.

Question 72

What the relationship between energy and wavelength?

Answer 72

Because we know now that Energy = h/ν

We can derive the following equation:

E = (hc)/λ –> energy is inversely proportionate to the wavelength but proportionate to the frequency.

Question 73

What can happen to electrons in bonding orbitals when provided with energy?

Answer 73

The electrons can promote and move into the anti-bonding orbitals.

Question 74

How can one calculate the absorbance? What are the different results?

Answer 74

Absorbance = log (I_Input/I_transmitance)

This is the log of the ratio of light that entered versus the light that actually passed through.

The log compresses the scale to small positive numbers. Thus…

Transmittance = 100% = 1.0 –> Absorbance = 0

Transmittance = 10% = 0.1 –> Absorbance = 1

Transmittance = 1% = 0.01 –> Absorbance = 2

Absorbance of 2 is usually the limit as the spectrophotometer can’t detect smaller quantities.

Question 75

In reality, do orbitals have many different energy levels?

Answer 75

Yes, this is because the orbitals have different vibrational energy levels.

Consequently, when electrons are promoted, a whole range of wavelengths are absorbed not just a specific wavelength –> results in a curved shaped graph.

However, there is a peak absorbance –> show wavelength that corresponds to the most frequent energy transition.

Question 76

What is the relationship between double bonds and absorption?

Answer 76

The more double bonds present –> the more absorption takes place in the light spectrum.

Question 77

How can absorbance be used to identify specific ring structures?

Answer 77

Ring structures absorb a particular wavelength which can be identified.

Question 78

What is the Beer-Lambert law state?

Answer 78

It states that….

A = E x c x I

A = Abosrbance

E = Molar extinction coefficient (value derived from graph –> as it changes depending on wavelength).

(M^-1 cm^-1)

c = concentration (Molar)

I = Path length (cm)

Question 79

How does the property of fluorescence work?

Answer 79

Question 80

Examples of heterocyclic aromatic systems?

Answer 80

Hetero - meaning containing something else than carbon (N/O).

1. Pyridine
2. Pyrimidine
3. Pyrole

Question 81

Explain the ring structure in pyridine?

Answer 81

Each atom in the ring provides 1 electron to the delocalised system –> results in aromatic ring formation.

Question 82

Explain the ring structure in Pyrimidine?

Answer 82

Each atom provides one electron to ring structure –> allows for aromatic ring formation –> Nitrogen still have two lone pairs.

Note that all the atoms are Sp² hybridised which matches the 120 bond angle of the ring.

Thymine + uracil both have this ring structure –> can be detected using UV.

Question 83

Explain the ring structure in Pyrole?

Answer 83

Each atom provides one electron to ring structure except for nitrogen which provides 2 electrons (gets it from the lone pair) –> allows for aromatic ring formation.

Note that all the atoms are Sp2 hybridised which doesn’t match the 108 bond angle of the ring –> makes the ring structure less stable.

Question 84

What is the ratio of protonated to unprotonated histidine at physiological pH?

Answer 84

pH of 7.2 and pKa of 6.6.

80% unprotonated

20% protonated

Question 85

Draw the different resonance and tautomers of the histidine ring.

Answer 85

Question 86

Why do atoms bond together in the first place?

Answer 86

Atoms bond with one another in order to gain stability –> electrons distribute themselves within the atomic orbitals in arrangements that have the lowest possible energy –> to give the atom the greatest stability.

When chemical bonds form between atoms to generate compounds –> valence electrons redistribute themselves so that the atoms gain full valence shells.

Question 87

Do you consider all orbitals when talking about full valency>

Answer 87

No –> refer to the S and P orbitals –> only concerned with those 8 electrons –> this is where the Octet rule was derived from (caution –> not an absolute rule –> applies to atoms up to period 2.

Question 88

Definition of electronegativity?

Answer 88

It indicates how strongly an atom of that element can attract an electron.

Question 89

The link between electronegativity and bonding?

Answer 89

High electronegativity –> complete transfer of electrons from one atom to another –> one atom has the ‘strength’ to completely tear an electron away from the neighbouring atom –> Ionic bonding

Lower electronegativity –> atoms exert a similar pull –> complete transfer is unlikely –> neither atom has the strength to tear of electrons from the other –> covalent bonding.

The difference in electronegativity greater or equal to 1.7 –> ionic bonding –> difference less –> Covalent.

Question 90

How do bonding and anti-bonding orbitals differ when it comes to covalent bond formation?

Answer 90

Bonding and anti-bonding orbitals have opposing effects on the formation of a covalent bond.

A pair of electrons occupying the bonding orbitals facilitates covalent bonding.

A pair of electrons occupying anti-bonding orbitals inhibits covalent bond formation.

Question 91

What is a hypervalent atom? What determines whether an atom can expand its octet?

Answer 91

An atom with more than eight valence electrons

The answer may lie in the presence or absence of d orbitals in the valence shell of the element being considered.

Good biological example –> Phosphorus –> uses a vacant 3d orbital to hold some of its valence electrons.

Question 92

Do atoms of the same element show different valencies?

Answer 92

Yes, some elements can exhibit different valencies and therefore contribute to different numbers of covalent bonds.

Question 93

What is a conjugate system?

Answer 93

A conjugate system is one which contains electrons that are delocalised.

It arises from the overlap of adjacent P orbitals in any molecule featuring a sequence of alternating single and double bonds.

Question 94

Can delocalisation occur in non-conjugate system?

Answer 94

Yes, delocalization of electrons in compounds that do not possess a conjugate system of bond does occur. It happens when there are two or more different arrangements of bond –> ozone.

Question 95

What are the 6 key types of non-covalent interactions?

Answer 95

1. Dispersion forces
2. Permanent dipolar interactions
3. Steric repulsion
4. Hydrogen bonds
5. Ionic interactions
6. Hydrophobic forces.

Question 96

Are non-covalent interactions significant?

Answer 96

Not significant when examining the interactions as single events/in isolation but overall the net effect of many non-covalent interactions become noticeable and significant.

Question 97

Difference between inter and intramolecular forces

Answer 97

1. Intramolecular interactions operate between separate parts of the same molecule –> within the molecule.
2. Intermolecular interactions –> operate between different molecules.

Question 98

What is a permanent dipole?

Answer 98

Permanent dipole –> a molecule that has a permanently partially positively charged region and a permanently partially negatively charged region.

Note –> polar bonds can cancel each other out if the molecule is symmetrical

Question 99

What are dispersion forces?

Answer 99

A dispersion force is a force of attraction that exists between the two areas of opposite charge, which form the induced dipole.

Occurs due to the random movement of electrons –> creates temporary dipole –> induces dipoles in the surrounding molecule –> attraction.

Question 100

What two factors influence the prevalence of dispersion forces?

Answer 100

1. Shape –> Planar flat molecules are able to associate more than molecules with irregular shapes –> as dispersion forces are stronger in close proximity –> when the molecules are able to closely associate –> stronger interactions.
2. Size –> Larger molecule –> more atoms –> more electrons –> more/larger induced dipoles may exist.

Question 101

What are permanent dipolar interactions?

Answer 101

Permanent dipolar interactions are the attractive forces that exist between opposite partial charges on polar molecules.

• They are permanent/last longer

Question 102

What are induced dipolar interactions?

Answer 102

Electrostatic interactions that arises when a permanent dipole on a polar molecule induces a temporary dipole on a non-polar molecule –> creates an electrostatic force of attraction between both molecules

Question 103

What is steric repulsion?

Answer 103

Steric repulsion –> When two molecules approach each other –> the molecules on their surface first interact –> both groups of electrons will both carry a negative charge –> thus these groups repel.

Question 104

Definition of van der Waals forces?

Answer 104

A term used to describe the overall interaction between two molecular species once both the attractive dispersion forces and dipolar interactions and the repulsive forces of steric repulsion are taken into account.

Question 105

What are hydrogen bonds?

Answer 105

Hydrogen bonds represent a special class of dipolar interactions –> strong interaction between the hydrogen on one molecule to an electronegative atom either on another molecule or a spatially distinct part of the same molecule.

Question 106

What are two critical factors needed for hydrogen bond formation?

Answer 106

1. Hydrogen atom must be bonded to an electronegative atom (Oxy, Nirto, fluorine) –> compound containing this arrangement is known as the hydrogen bond donor.
2. The hydrogen atom can only form a hydrogen bond with a non-bonding pair of electrons on another electronegative atom –> compound containing this arrangement is known as the hydrogen bond acceptor.

Question 107

What are ionic forces?

Answer 107

Ionic forces are the attractive forces that exist between ionic species –> species carrying a fully positive and negative charge.

This is not limited to ionic compounds –> also important for covalent molecules with a full positive or negative charge.

Question 108

What is a salt-bridge?

Answer 108

Ionic forces that operate between two oppositely charged amino acid side chains in a protein is called a salt bridge.

An important force that stabilizes 3D structures.

Question 109

What are hydrophobic forces?

Answer 109

Hydrophobic interaction occurs when structural rearrangements move the hydrophobic portions of a molecule away from their aqueous surroundings.

Once clustered together the hydrophobic portions are stabilized through a network of dispersion forces that operate between them.

Basically –> hydrophobic forces drive the shielding of hydrophobic molecules from their aqueous surroundings.

Question 110

What three key factors determine the shape and structure of a molecule?

Answer 110

1. Bond Lengths
2. Bond Angles
3. Bond rotation

Question 111

Definition of bond length?

Answer 111

The distance between the nuclei of two covalently bonded atoms.

Dictated by:

1. Atomic radii of the atoms joined together
2. Type of covalent bond

Question 112

Definition of the atomic radius?

Answer 112

Usually defined as half the distance between two covalently bonded atoms when the two atoms are identical.

Question 113

What two main factors influence atomic radii?

Answer 113

1. Valence shell number increases –> atomic radius increases.
2. An increasing number of valence electrons enter the same valence shell –> atomic radius decreases slightly –> more valence electrons –> greater attraction between nucleus and valence electrons.

Question 114

What does the VSEPR theory state?

Answer 114

The valence pairs surrounding a central atom arrange themselves in such a way that they minimize repulsive forces that operate between them –> basically they position themselves as far away as possible –> decreases interaction –> increased stability.

Applies to non-bonding and bonding pairs.

Question 115

Why do non-bonding electrons distort the expected geometry?

Answer 115

The distortion in geometry arises because non-bonding pairs occupy more space than bonding pairs –> so surrounding electrons pairs have to move slightly further away in order to evade repulsive forces from the non-bonding pair –> this relatively larger push exerted by non-bonding pairs results in a slight distortion of the bond angles.

Question 116

What is the hybridisation of atomic orbitals?

Answer 116

The hybridisation of atomic orbitals –> a change in shape of the orbitals which allow the electrons they contain to adopt new geometries.

Helps the explain the geometries seen using the VSEPR theory.

Question 117

Difference between configuration and conformation?

Answer 117

Configuration –> describes how its composite atoms are joined together in three-dimensional space –> the configuration is fixed and can only be changed by breaking bonds.

Conformation –> Describes the overall three-dimensional structure at a particular moment in time –> this is variable –> applies to single bonds –> double/triple bonds can’t rotate.

Question 118

What is steric hindrance?

Answer 118

There are circumstances where bond rotation around a single sigma bond is not possible –> rotation is blocked because different groups of atoms simply get in the way of each other.

Question 119

Can delocalised systems of electrons rotate?

Answer 119

When delocalization occurs, the electrons are Pi-bonding electrons that are shared between more than two atoms –> shares both single and double bond character.

But they are UNABLE to rotate.

Question 120

Can bonds in cyclic molecules rotate?

Answer 120

No, they are unable to rotate –> they lack the flexibility to allow it to move into and adopt different conformations.

Question 121

What do the values of Ka and Kb tell you about acid/base strength?

Answer 121

Large Ka –> Strong acid

Small Ka –> Weak acid

Large Kb –> Strong Base

Small Kb –> Weak base

Note –> there is an inverse relationship between the strength of a base and the value of Ka

• Weak base –> large Ka and small Kb
• Strong base –> small Ka and large Kb

Question 122

The important relationship between Ka, Kb and Kw?

Answer 122

Multiplying Ka and Kb together will always equal Kw (1x10^-14).

This relationship tell us that if we know the value for Kb of a weak base, we can calculate Ka, the acid dissociation constant for the conjugate acid (Visa Versa).

This is true for all weak acids and bases.

Question 123

Impact of changing pH on the dissociation of a weak acid?

Answer 123

• A decrease in pH leads to a decrease in the extent of dissociation of a weak acid.
• An increase in pH leads to an increase in the extent of dissociation of a weak acid.

Due to Le Chatelier’s principle –>, for example, a decrease in pH means there are more H⁺ in solution –> drives the reaction away from the dissociation of the weak acid.

Question 124

An important point to remember when pH=pka

Answer 124

When pH equals the Pka –> acid and conjugate base are found in equal concentrations in solution.

pH > Pka –> there will be more dissociated acid in solution.

pH < pKa –> there will be more undissociated acid in solution.

Question 125

Important points to remember when making a buffer.

Answer 125

Two types of buffer

1. Weak acid and the salt of its conjugate base
2. Weak base and the salt of its conjugate acid

Requirements

• We must use an acid-base conjugate pair
• Must use a weak acid or base
• Both compounds must be present at sufficiently high concentrations.

Question 126

How can we calculate the pH of the buffer solution?

Answer 126

Using the Henderson-Hasselbach equation.