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Definitions Flashcards

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1
Q

Relative atomic mass (Ar)

A

Ratio of the mass of the average mass of one atom of an element to 1/12 the mass of an atom of 12C isotope.

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2
Q

Relative molecular mass (Mr)

A

Ratio of the average mass of one molecule of the substance to 1/12 the mass of an atom of 12C isotope

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3
Q

Relative formula mass (Mr)

A

‘Of an ionic compound’

Ratio of the average mass of one formula unit of a compound to 1/12 the mass of an atom of 12C isotope

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4
Q

Relative isotopic mass (Ar)

A

Ratio of the mass of one atom of an isotope of an element to 1/12 the mass of an atom of 12C isotope

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5
Q

Isotopes

A

Same number of protons, different number of neutrons.
Similar chemical properties (since same no. electrons) but different physical properties (since different mass)

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6
Q

Mole

A

Amount of substance containing no. of particles that is equal to avogadro’s constant (6.02 x 10^23)

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7
Q

Molar mass (M)

A

Mass of one mole of a substance, units g/mol

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8
Q

Molar gas volume

A

Volume that 1 mole of gas occupies at a particular set of temperature and pressure

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9
Q

Concentration

A

Amount of solute dissolved per unit volume of a solution, g/dm^3

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10
Q

Empirical vs molecular formula

A

Empirical: Simplest ratio of the number of atoms of each element of a compound
Molecular: Actual number of atoms of each element of a compound

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11
Q

Avogadro’s hypothesis

A

Same volume of 2 gases under same temp and pressure contain the same number of molecules

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12
Q

Equivilance point

A

Reactants have just reacted according to stiochiometric ratio given by balanced equation of the reaction

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13
Q

End point

A

Indicator in titration has just changed colour

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14
Q

Sampling vs dilution

A

Sampling: collection of a portion of a given solution
Dilution: addition of more solvent to a given solution

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15
Q

Mass number

A

Number of nucleons in a nucleus

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16
Q

Atomic number

A

Number of protons in a nucleus

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17
Q

Orbitals

A

Region of space in which there is a high probability of locating electrons

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18
Q

Degenerate orbitals

A

Orbitals of the same energy level

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19
Q

Aufbau principle

A

Electrons will always occupy the orbitals of lower energy levels first before occupying orbitals of higher energy levels

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20
Q

Pauli exclusion principal

A

The electrons in the same orbital must have opposite spins so that they can generate sufficient magnetic force to overcome force of repulsion due to like charge

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21
Q

Hund’s rule

A

When degenerate orbitals are available, electrons will always occupy orbitals singly first before any pairing occurs to minimise interelectrostatic repulsion.

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22
Q

Ionisation energy

A

Energy required to remove 1 mole of electrons from 1 mole of gaseous atoms to form 1 mole of singly positive gaseous ions.

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23
Q

Spontaneous

A

Thermodynamically favourable

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24
Q

How to choose indicator?

A

End point should coincide with equivalence point

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25
Q

Why solutions can be stable for long time despite E° value of reaction being spontaneous?

A

Actual reaction conditions could vary greatly from the standard conditions. (Temp, pressure, concentration)

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26
Q

Catalyst vs enzymes

A

Increases rate of reaction by providing an alternative pathway with lower activation energy for reaction to occur, and is chemically unchanged (and regenerated) by the end of the reaction.

Enzymes just add ‘biological’ and ‘protein in nature’

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27
Q

Enzymes characteristics

A

Highly selective
Only work over a narrow pH range and temperature range (easily denatured)
Specific in nature (can only be used for 1 task)
highly efficient
Neither homogenous nor heterogeneous, colloidal in nature

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28
Q

Homogenous catalyst

A

Catalyst is in the same physical state as the reactants

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29
Q

Heterogenous catalyst

A

Catalyst is in a different physical state from the reactants

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30
Q

Homogenous mixture

A

Reactants and products are in the same physical state

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31
Q

Heterogenous mixture

A

Reactants and products are in different physical states

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32
Q

What to exclude in Kp/Kc calculations?

A

Conc of solids, conc of more liquids in heterogenous, conc of solvents

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33
Q

Why TM can act as heterogenous and homogenous catalyst?

A

Heterogenous: Reactants form weak interactions with surface of TM, covalent bonds in reactant molecules weakened. Ea falls, and surface conc of reactant molecules increases (allows more to come into close contact with each other in correct orientation) —> rate of reaction increases —> products desorp and catalyst can be used for other reactants

Homogenous: variable oxidation states allow TM to oxidise/reduce compounds and be regenerated by transitioning btw oxidation states —> lower activation energy needed for reaction to occur (especially btw like charge ions with very high repulsion btw them)

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34
Q

Complex or complex ion

A

Central metal atom or ions surrounded by ligands that are bonded to it through a dative bond

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35
Q

Ligand

A

Ligands are neutral molecules or anions that have at least 1 lone pair of electrons that can be used to form a dative bond with a transition metal atom or ion.

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36
Q

Coordination number of transition metal complex

A

Number of dative bonds formed between ligands and transition metal atom or ion in a complex

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37
Q

How is a complex formed?

A

TM high charge density, highly polarising, attract LP of electrons from surrounding particles towards itself.

Can accommodate more electrons also because energetically accessible vacant 3d, 4s, 4p, 4d electron subshells. Overall of empty orbital in TM atom or ion and fully filled orbital of ligand —> forms a dative bond

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38
Q

How stability of TM ions can be affected by dif ligands?

A

2024 prelim p2 qn
- haemoglobin + 4O2 ⇌ oxyhaemoglobin + 4H2O
- at high conc of O2 (eg lungs), ligand exchange happens, water ligands replaced with O2 ligands, POE shift right, H2O and oxyhaemoglobin released

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39
Q

Colour of TM complexes

A

Cu:
Cu(H2O)6 2+ blue
Cu(NH3)4(H2O)2 2+ dark blue
CuCl4 2- yellow
CuI white ppt
Cu2O brick red ppt
Cu(OH)2 blue ppt

Cr:
Cr2O7 2- orange
CrO4 2- yellow
Cr(H2O)6 3+ green
Cr(OH)3 grey green ppt
Cr(OH)6 3- dark green

Fe:
Fe(H2O)6 2+ pale green
Fe(H2O)6 3+ yellow
Fe(H2O)5SCN 2+ blood red
Fe(OH)2 green ppt
Fe(OH)3 reddish-brown ppt

Mn:
MnO4 - purple
Mn(H2O)6 2+ colourless/pale pink
Mn(OH)2 off-white ppt
Mn(OH)3 brown
MnO2 brown ppt

Al:
Al(OH)3 white
Al(OH)4 - colourless

Ag:
Ag(NH3)2 + colourless
AgCl white ppt
AgBr off-white ppt
AgI yellow ppt

Zn:
Zn(OH)2 white
Zn(OH)4 2- colourless
Zn(H2O)6 2+ colourless
Zn(NH3)4 2+ colourless

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40
Q

How colour arises in complexes

A

Assuming shape of complex as octahedral, due to ligand approach, orbitals are split into 2 dif energy levels (d-d splitting) An electron from the lower energy level absorbs a photon from the electromagnetic spectrum and is promoted to a higher energy level. Wavelength of photon is determined by degree of splitting. The colour observed is complement to those absorbed in visible region of spectrum.

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41
Q

Ionic bond

A

eFOA btw oppositely charged ions

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42
Q

Metallic bond

A

eFOA between sea of delocalised electrons and lattice of positively charged ions

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43
Q

Covalent bond

A

eFOA between shared pair of electrons and positively charged surrounding nuclei

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44
Q

How to determine strength of ionic bond?

A

Ionic bond strength: lattice energy (energy released when 1 mole of ionic solid is formed from its isolated gaseous ions an infinite distance apart)

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45
Q

Factors affecting metallic bond strength

A

Charge density of cation
No of delocalised electrons

—> more extensive/stronger eFOA

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46
Q

Factors affecting covalent bond strength and explain why each factor affects

A

Multiplicity of bond: more shared electron pairs = stronger eFOA between shared pair of electrons and nuclei of atoms = stronger covalent bond
Size of atom: smaller = less diffuse orbitals = more effective overlap = stronger covalent bond
Proximity of lone pairs: closer together = more (excessive) repulsion between LP = weaker covalent bond
Polarity: comparable size, induction of partial charges due to electronegativity differences = more polar bond, eFOA between partial charges strengthens covalent bond

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47
Q

Formation of dative bonds

A

One compound has a lone pair of electrons available for donation and the other has energetically accessible orbitals that it can use to accept electrons

NOT THE SAME as H bond! Don’t mix up!
Neither need to be highly electronegative for this

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48
Q

Sigma bond

A

Head on overlap, electron density concentrated between the nuclei of the 2 bonding atoms

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49
Q

Pi bond

A

Side on overlap, less effective, region of overlap is above and below nuclear axis, node present

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50
Q

Node

A

Region of space in which there is a zero probability of locating electrons

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51
Q

Ionic bonds with covalent character

A

Ionic bonding, assumed to be perfectly symmetrical spheres. But when oppositely charged ions are close to each other, partial sharing of electrons because cation attracts valence electrons of anion towards itself and polarises electron cloud of anion. Partial covalent character

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52
Q

Factors for degree of covalent character in ionic bond

A

Charge density of cation
Polarisibility of anion

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53
Q

Ionic character in covalent bond

A

Perfect covalent bond—> non polar only exists when electronegativities are exactly the same
Most molecules, not. Partial positive and partial negative charges induced, ionic character

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54
Q

Factors affecting ionic character in covalent bond

A

Net dipole moment (dipole moment = charge x distance between the nuclei of the 2 atoms)

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55
Q

Id-id, pd-pd interactions description

A

eFOA between polar ends of the molecule / between opposite partial charges

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56
Q

H bond description

A

eFOA between highly electron deficient H (that is covalently bonded to a highly electronegative atoms) and lone pair of electrons on highly electronegative atom

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57
Q

Drawing of dot and cross diagram: is it central atom or side atoms that must be satisfied?

A

Always satisfy Side atoms first then consider whether centre one has energetically accessible centre

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58
Q

Nature or bonding in AlCl3, BeCl3 and BCl3

A

All covalent, because high charge density of cations causes electron cloud of anion to be polarised to a very large extent (GMS)

But for AlCl3 if they ask structure, it is actually GICLS as a solid, SMS as molten/gas (period 3 notes) so this one specifically needa see context

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59
Q

Why mostly bonds will be hybridised orbitals?

A

Much more effective overlaps than s-s or p-p so much stronger bond

(s orbitals spend more time close to their respective nuclei, and less in the binding region. p orbitals spend more time in the binding region; that is close both the nuclei; thus s-p overlap forming a stronger bond than s-s overlap - https://www.physicsforums.com/threads/why-s-p-orbital-overlap-is-stronger-than-s-s-overlap.498941/#)

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60
Q

Why there is a difference between experimental and theoretical values of LE?
(Emphasis on ‘difference’ not larger experimental than theoretical; you can’t predict that)

A

Explain the covalent character thing
LE calculated on assumption that ions are 2 perfectly spherical point charges but there’s a disagreement in values for experimental (consider polarising power of cation and polarisability of anion —> degree of electron cloud distortion, electron sharing and covalent character). Higher covalent character = larger disagreement.

FYI
Hey apparently they use coloumb’s law to calculate; link btw phys electric fields and like chem ionic charge
Larger theoretical than experiential suggests covalent bonds —> ionic weakened (When ionic bonds exhibit covalent character, there is a degree of sharing of electron density between the ions. This sharing reduces the effective charge that each ion experiences, as the electron cloud is distorted so strength of ionic bonds reduced.)
vs larger experimental than theoretical suggests more energy is needed to overcome the covalent bonds between the lattice ions to break up the lattice. (Since additional bonds and interactions created ig)

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61
Q

Structures and descriptions

A

Giant metallic crystal lattice structure: orderly arrangement of metal cations and sea of delocalised electrons held together by strong eFOA between …

Giant molecular structure: each c atom is covalent bonded to .. other atoms with eFOA between shared pair of electron and nuclei of surrounding atoms … (hexagonal + layers w id-id due to delocalisation of remaining electron / extensive tetrahedral network)

Simple molecular structure: weak interactions (pd-pd/id-id/h bond) between … molecules

Giant ionic crystal lattice structure: regular arrangement of cations and anions held together by strong eFOA between opp charged ions

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62
Q

Electrical conductivity for each structure (what is the key thing to say?)

A

Ions/electrons can act as mobile charge carriers to carry charges around —> can conduct electricity

Non-conductor: no mobile charge carrier since covalent molecules in SMS are neutral/since ions are held in fixed position by strong ionic bonds (GIS)

GMS- graphite structure, good conductors in direction parallel to layers as delocalised electrons act as mobile charge carrier under applied PD but poor conductor in direction perpendicular to plane as delocalised electrons cannot jump across layers

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63
Q

Physical properties (for each structure)

A

GMCLS: ductile and malleable (mobility of delocalised electrons allows layers of cations to slide over each other without shattering structure)
GICLS: brittle (slight displacement brings together ions of like charge, shattering lattice structure), hard (strong eFOA)
GMS: diamond: hard (strong eFOA), graphite: flaky and slippery (weak idid between layers)
SMS: soft (weak idid)

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64
Q

Bp/mp format

A

Structure
Bonding
Energy

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65
Q

Why lone pair more repulsion that bond pair?

A

Lone pair held closer to nucleus than bond pair since bp shared w another atom

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66
Q

VSEPR theory

A

Electron pairs arrange themselves ard central atom to reduce repulsion, because LP-LP repulsion>LP-BP>BP-BP

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67
Q

Coordination number and factors

A

Number of closest neighbours surrounding an ion in a lattice structure

Ionic radius

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68
Q

VSEPR shapes, name and angle

A

2BP Linear, 180
3BP Trigonal planar, 120
2BP, 1LP Bent, 117.5
4BP Tetrahedral 109.5
3BP, 1 LP Trigonal pyramidal, 107
2BP, 2 LP Bent, 104.5
5BP Trigonal bipyramidal, 120, 90
4BP, 1 LP See-saw shape, 117.5, <90
3BP, 2 LP T-shaped 115, <90
2BP, 3 LP Linear, 180
6BP Octahedral, 90
5BP, 1 LP square pyramidal, <90
4BP, 2 LP square planar, <90

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69
Q

Electrolytic vs Electrochemical cell

A

Electrolytic cell: electric energy to chemical energy
Electrochemical cell: chemical energy to electrical energy

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70
Q

Cathode vs anode; polarity in electrochemical and electrolytic cell

A

Cathode: Place where reduction takes place
Anode: Place where oxidation takes place
Red cat and ox-an

Electrochemical: Cathode +, anode -
Electrolytic: Cathode -, anode +

71
Q

Types of half cells (electrochemical) and respective electrodes

A

Ion-ion (Pt electrode)
Ion-metal (metal electrode, usually reactive)
gas-ion (pt electrode)

72
Q

Function of salt bridge/membrane

A

To maintain electrical neutrality. Charges flow from salt bridge to 2 half cells to maintain charge balance.
OR
To act as electrical connector. Ions can flow through salt bridge to complete the circuit between the 2 half cells.

73
Q

Where do electrons flow?

A

External circuit via connecting wires NOT through salt bridge

74
Q

Type of solution that can be used for salt bridge

A

concentrated electrolyte

75
Q

Standard hydrogen electrode

A

Platinum electrode in 1.0 mol/dm^-3 solution with H2 gas at 1 bar bubbled in at 298K

76
Q

Standard conditions

A

298K, partial pressure of each gas is 1 bar, concentration of all aqueous ions is 1.0 mol dm^-3

NB: pressure of whole system doesn’t have to be 1 bar

77
Q

Standard electrode potential

A

Potential difference between (M2+|M electrode) / (M3+/M2+| pt electrode) / (X/X-| Pt electrode) and standard hydrogen electrode at standard conditions of partial pressure of each gas at 1 bar, concentration of aq ions at 1.0 mol dm^-3 and temperature at 298K.

78
Q

Explain what standard electrode potential measures

A

Tendency of half-cell to undergo reduction relative to the standard hydrogen electrode

79
Q

Factors affecting electrode potential

A

change concentration of ions
POE to shift left/right
Reduction/oxidation pre-dominates
Reduction potential (more/less positive/negative)

80
Q

How to calculate standard electrode potential?

A

See which ions go cathode vs anode. Most + at cathode —> reduced; most - at anode —> oxidation
E(cell) = E(red)- E(oxi)

81
Q

Predict feasibility with E0 cell value

A

<0 thermodynamically infeasible as written
=0 reaction at eqm, not spontaneous in either direction
>0 thermodynamically feasible as written

82
Q

Limitations of using E to determine feasibility

A

Might be thermodynamically feasible but not kinetically feasible
Possible reasons: high Ea (strong covalent bonds require a lot of energy to break/similar charge ions require a lot of energy to overcome repulsion first) or very slow rate (unobservable in the short term)

OR

conditions non-standard. POE shift —> E and E cell shifts —> thermodynamically feasible under these conditions

83
Q

Calculate standard Gibbs free energy from standard cell potential

A

ΔG = -nFE

NB: Stoichiometric coefficient affects G (by affecting no. of moles of e transferred) but NOT E

84
Q

T/F? More than one ion can be discharged at cathode/anode simultaneously.

A

T. As long as E +, thermodynamically spontaneous.
Seen in electrolytic purification of copper

85
Q

Factors affecting discharge of ions

A

Standard electrode potential
State of electrolyte (molten/aq)
Nature of electrode (active/inactive)
concentration of reactants (shift POE, E cell positive —> reaction spontaneous) [this one is the brine eg]

86
Q

Electrolysis of brine, equations

A

2H2O + 2e- —> H2 + 2OH- [R]
2Cl- —> Cl2 + 2e- [O]
(Because of conc difference for cathode side, POE shift, E cell of Cl- can rise above that of H2O —> preferentially oxidised)

Cathode: steel
Anode: titanium

Cats like to ‘steal’; can ‘titan’ ‘anot’

87
Q

Purpose of electrolysing Brine
Purpose of anodising aluminium
Purpose of electroysing copper

A

Brine: Manufacture chlorine from salt
Anodising aluminium: Create an insoluble Al2O3 layer/thicken it around the aluminium object to resist corrosion and act as an electrical insulator
Copper: To purify impure copper

88
Q

Equations for anodising aluminium

A

2H2O —> O2+ 4H+ + 4e- [O]
4Al + 3O2 —> 2Al2O3

2H+ + 2e- —> H2 [R]

Cathode: pt electrode
Anode: aluminium object

89
Q

Electrolysis of Brine, anodising of aluminium and electrolytic purification of copper (what solutions are used?)

A

Electrolysis of Brine: concentrated NaCl (aq)
Anodising Aluminium: H2SO4 (aq)
Electrolytic purification of copper: CuSO4 (aq)

90
Q

Electrolysis of impure copper (equations)

A

Cu —> Cu2+ + 2e- [O]
Zn —> Zn2+ + 2e- [O]
Sn —> Sn2+ + 2e- [O]
Fe —> Fe2+ + 2e- [O]

Cu2+ + 2e- —> Cu [R]

NB all the reactions above will happen and dissolve as ion. But those (metal ion|metal) with E value more than +0.34V will fall to the bottom (anode sludge).

Cathode (pure)
Anode (impure)

91
Q

Free radical definition

A

A particle (atom/molecule/ion) that has at least 1 unpaired electron.

92
Q

Faraday’s Second law

A

Number of moles of electrons transferred to discharge one mole of ion at an electrode is equal to the number of charges on an ion.

93
Q

Faraday’s first law

A

Mass of a substance produced at an electrode during electrolysis is directly proportional to the amount of electricity passed.

94
Q

Faraday equations

A

Q = It
Q = F x no. of moles of electrons
One faraday = avogadro’s constant x charge of 1 electron (F=Le)

95
Q

How to find avogadro’s constant?

A

Set up electrolysis cell with CuSO4 (aq) as electrolyte and 2 Cu electrodes **of known mass*.
Pass current of known I for a fixed period of time (t seconds)
Remove electrodes and find gain in mass of cathode (m grams) (gain in mass at cathode should be same as loss in mass at anode) —> wash, dry and re-weigh

Use m/63.5 to find number of moles of Cu
Then use 2 x It/F to find number of moles of Cu (2 times the number of moles of electrons transferred)
Solve for F
Use equation L=F/e to find avogadro’s constant (L)

96
Q

Transition metal

A

d-block element that is able to form one or more stable ions with partially filled d subshell.

97
Q

Which elements are not considered TM?

A

Sc and Zn

98
Q

Explain unique electric config of Cr and Cu

A

Half and fully filled shells associated with higher stability as inter-electronic repulsion between 4s electrons are minimised

99
Q

Explain trend in 1st IE (TM)

A

Roughly invariant
NC increase (since protons increase) offset by increase in shielding effect since electrons added to penultimate 3d electron shell —> ENC roughly constant —> 1st IE roughly constant

100
Q

Explain anomalies in 2nd IE of TM

A

Roughly invariant
NC increase (since protons increase) offset by increase in shielding effect since electrons added to penultimate 3d electron shell —> ENC roughly constant —> 2nd IE roughly constant

Except for Cu and Cr, because 2nd electron removed from inner 3d shell —> shielding effect much lower than other TM, ENC higher, eFOA stronger, more energy needed to remove 2nd valence e of Cr and Cu —> higher than expected 2nd IE

101
Q

Explain trend in 3rd IE of TM and anomaly

A

NC increase (since protons increase) but roughly constant shielding effect since all have some number of inner shell electrons —> ENC increases across the group —> 3rd IE increases across the group

Anomaly: Fe; removal of 3rd electron is from 3d orbital with paired electrons (which experience repulsion) and would form a half-filled 3d subshell associated with higher stability —> eFOA between valence e and positive nucleus lower, much less energy needed to remove 3rd valence electron —> lower than 3rd IE of other TM

102
Q

Melting point of TM vs s block metals

A

TM Higher
More electrons can be delocalised per atom (4s and unpaired 3d can all be delocalised compared to only 4s electrons for s block metals so eFOA between sea of delocalised e and positive cations increased) —> more energy needed to overcome —> higher mp

103
Q

Density of TM vs s block

A

TM higher
Proton number higher —> NC higher and shielding effect lower (d orbitals more diffuse, shielding effect less effective for TM); ENC higher, eFOA between positively charged cations and valence electrons higher —> v e held closer to nucleus —> atomic radii and volume smaller
Mass higher because atomic number is higher for TM

Density = mass/volume; larger mass, smaller volume —> density higher

104
Q

Explain why TM have multiple oxidation’s states compared to s block elements, and how to determine how many oxidation states it has.

A

For TM, 4s and unpaired 3d can all be delocalised because 3d and 4s electrons are close in energy compared to only 4s electrons being delocalised for s block metals because 4s and 3p electrons vary greatly in energy

Maximum oxidation state is sum of unpaired 3d electrons and 4s electrons.

105
Q

Why Zn does not show +3 oxidation state?

A

Max os of Zn is +2 (has 0 unpaired 3d electrons, and 2 4s electrons)

106
Q

Explain different OS of TM in covalent and ionic compounds, and how it affects nature of oxides

A

Low OS: Ionic, basic
High OS: Covalent, acidic

High OS = higher polarising power (since charge density = charge/radius is higher) —> highly able to polarise electron cloud of anion and form high degree of covalent bonding

107
Q

Why steel harder than iron but more brittle?
(Sec 4 concept)

A
  • carbon and iron atoms of different sizes —> lattice structure less orderly —> more difficult to displace without breaking metallic bonds/shattering lattice structure

But also more brittle than iron
Harder to displace but once lattice structure is no longer orderly they can’t easily slide over one another anymore —> more easily broken/shattered

108
Q

Why chromium in stainless steel?

A

Formation of chromium oxide so water can’t reach iron inside so can’t form rust

109
Q

Zwitterionic bonds in water

A

Ion-dipole! NOT just H bond!

110
Q

Why might bond angles not align with VSEPR theory?

A

Size of neighbouring atoms —> repulsion between electron cloud of neighbouring atoms (more significant for larger molecules) —> larger bond angle
Electronegativity of atoms: central atom more electronegative —> pull lone pair of electrons towards the nucleus of the central atom —> more repulsion between electron pairs —> larger bond angle

111
Q

Why might bond angles not align with VSEPR theory?

A

Size of neighbouring atoms —> repulsion between electron cloud of neighbouring atoms (more significant for larger molecules) —> larger bond angle
Electronegativity of atoms: central atom more electronegative —> pull lone pair of electrons towards the nucleus of the central atom —> more repulsion between electron pairs —> larger bond angle

112
Q

Factors affecting id-id

A

Number of electrons/electron cloud size (affects polarisability and stuff)
Degree of branching (affect extent of id-id formed)

113
Q

Factors affecting pd-pd

A

Net dipole moment, affects strength of pd-pd

114
Q

Factors affecting H bond

A

Extent of H bonds (no. Of H bonds per molecule)
Dipole moment of H-X bond (where X=F/O/N) because affects the magnitude of the partial positive charge on H

115
Q

Explain trend of melting points for groups 15, 16 and 17 and explain why NH3, H2O and HF do not follow the trend.

A

All structures are SMS (covalent bonds within molecule with weak __ between molecules)
In same period, idid gets stronger since size of electron cloud increases and higher polarisability of e cloud = higher extent of id-id = more energy needed to overcome = rising melting point

BUT FON are highly electronegative, making the 3 molecules highly polar and held by H bonds —> stronger and require more energy to overcome —> higher than expected boiling points when compared to hydrides formed by other elements in their respective periods

116
Q

Explain differences in density of ice and liquid water, in terms of structure and bonding

A

Liquid, water interact in clusters, can form H bonds with 2 neighbouring water molecules
H bonds constantly break and reform —>

117
Q

Explain differences in density of ice and liquid water, in terms of structure and bonding

A

Liquid, water interact in clusters, can form H bonds with 2 neighbouring water molecules
H bonds constantly break and reform —> water molecules slide past each other easily
vs solid, form H bonds w 4 others, tetrahedral arrangement —> rigid, highly ordered, open structure, water molecules more spaced out
When ice melts, water molecules can move into the cavities of the ice structure —> increases number of water molecules per unit volume —> ice is denser

118
Q

Explain phenomenon of dimerisation, and conditions

A

Only in non-polar solvent / gaseous phase

2 carboxylic acids with similar masses form intermolecular hydrogen bonds and bond together, hence exist as dimers
Polar solvent, preferentially forms H bonds with much higher proportion of polar solvent molecules

119
Q

How intramolecular H bonding affects mp/bp?

A

Close proximity of ___ groups allows the molecules to experience intramolecular H bonds between them —> fewer sites available for formation of intermolecular H bonds —> extent of intermolecular H bonds less, less energy needed to overcome —> lower mp/bp

120
Q

Soluble/insoluble? (Explanation)

A

Energy released from the formation of ___ (in)sufficient to overcome energy that needs to be absorbed to overcome the ____ bonds.

121
Q

Solubility of the different structures in polar and non-polar solvents?

A

GMS: insoluble in both
GIS: Soluble in polar, insoluble in non-polar
SMS: like dissolves like

122
Q

Explain arrangement and movement of particles in solid, liquid and gaseous states

A

Solid: particles are very close together, held in orderly arrangement, vibrate about their fixed positions
Liquid: particles are close together and arranged in fairly orderly manner, but can vibrate, rotate and move through the liquid
Gas: particles are far apart, mainly empty space between particles, move freely through the container with continuous random motion.

123
Q

State basic assumptions of kinetic theory applied to ideal gas and which one applies to real gases as well?

A

Perfectly elastic collisions (no loss of kinetic energy upon collision)
Size and Volume of gas particles with respect to volume of container is negligible
Intermolecular forces of attraction between gaseous atoms is negligible
Ave KE of molecules proportional to temp (K)

Applies to both: in continuous, rapid and random linear motion

124
Q

Conditions for real gases to behave most ideally

A

High temp, low pressure

125
Q

Maxwell-Boltzmann distribution curve axes and describe how the curve changes when temp is reduced

A

Number of molecules with given energy against energy ; area under curve must remain the same

Peak shifts left, fraction of particles with lower energy increases, and fraction with higher energy decreases
More particles with most probable energy and lower proportion at lower energy

126
Q

Charles’ Law

A

At constant pressure, volume is directly proportional to temperature in Kelvin.

127
Q

Boyle’s law

A

At constant temp, volume is inversely proportional to pressure of the gas.

128
Q

Gay-lussac’s law

A

At constant volume, pressure of the gas is directly proportional to temperature of the gas in kelvin.

129
Q

Avogadro’s law

A

Same volume of gases at same temp and pressure have the same number of particles.

130
Q

Dalton’s law of partial pressure

A

Total pressure is the sum of partial pressures of all the gases in the system, provided that the gases do not react chemically.

131
Q

Deviation graph, key points to note and explain why there are negative and positive deviations

A

Axes: PV/RT against external pressure
Negative deviation: low external pressure, molecules about to strike the walls experience significant IMFOA to neighbouring molecules, reducing impact with which they hit the walls of the container. Observed pressure hence lower than expected, and PV/RT is less than expected —> negative deviation
Positive deviation: at high external pressure, volume of molecules with respect to container no longer negligible, because they are pushed very close together and less compressible. Volume of observed gas is larger than ideal volume —> PV/RT larger than expected —> positive deviation.

132
Q

Explain factors affecting extent of deviation from ideal gas behaviour in real gases

A

Type of IMFOA (for negative deviation)
Large/heavy/big gas particles (can use Mr to estimate)

133
Q

Understand van der waals equation and explain its difference from ideal gas eqn

A

Addition to pressure to correct for IMFOA and subtraction from volume to correct for molecular size

134
Q

What do sign conventions for enthalpy reactions tell you? (Heat transfer, temperature change, stability?)

A

Negative: Exothermic, temperature rises, heat released, stability increases (energy level decreases)
Positive: Endothermic, temperature falls, heat absorbed, stability decreases (energy level increases)

135
Q

Characteristics of ∆H (always must be accompanied by? Is it affected by stoichiometric ratio and physical states? Do you flip the sign if you flip equation?)

A

Always accompanied by balanced chemical eqn with state symbols
yeah affected by ratio
Yeah flip sign

136
Q

Apply equation to calculate heat evolved or absorbed in aq solution and dissolving a solid in water, and hence ∆H of reaction

A

Q = mc∆T =pvc∆T, (v here is the volume of the TOTAL liquid)
∆H = Q/amount of limiting reagent
But then you needa use this value that you find to find like the different ∆H you need (eg neut, sol, etc.)

137
Q

Describe calorimetric method, precautions and understand limitations

A

Add 2 solutions together, find max/min temp, use that to calculate the ∆H

Precautions: cups must be good thermal insulators (eg styrofoam) as good thermal insulation and high specific heat capacity

Limitations: does not take into consideration heat loss/heat gain from surroundings

138
Q

Thermometric titration procedure (imp for planning also)

A

Procedure
1. Using a 100cm^3 measuring cylinder, measure v cm^3 of solution 1 and transfer into a 250cm^3 beaker placed in a styrofoam cup to minimise heat loss to surroundings.
2. Using an electronic weighing balance, weigh the mass of the weighing bottle (m1), weighing bottle and about 5.0g of the solid (m2).
3. Place thermometer in beaker with substance 1 and start stopwatch. Record temperature at 0 minute. At regular intervals of 1 minute, record the temperature of the solution.
4. At 4 minute mark, pour solution 2/solid into the same beaker and stir continuously with a glass rod. Do not attempt to take the temperature at this point (temp changes too rapidly to get a steady/accurate reading)
5. Continue to record temperature at regular intervals of every minute from 5-12 minute mark.
6. Reweigh empty weighing bottle (m3) and take m3-m1 to get mass of solid reacted.

Graph plotting:
1. Plot a graph of temperature against time
2. Draw a best fit line for all the points before solid was added (4 min mark) and another best fit line for all the points after solid was added (4 min mark). Extrapolate both lines (in DOTTED lines) so they cross at t=4 min.
3. Determine ∆T by finding temperature difference between 2 graphs at 4 minute mark.

Calc:
1. Q = vcp∆T
2. n= (m3-m1)/molar mass = x
3. ∆H = Q/x

139
Q

Standard enthalpy change of formation

A

Energy absorbed when one mole of pure solid is formed from its constituent elements in standard state under standard conditions of 298K and 1 bar.

140
Q

What is the sign convention of standard enthalpy change of formation and what does it indicate wrt stability?

A

Can be positive or negative. Positive indicates that stability of system decreased (reactants more stable than products) but negative indicates that stability of system increased (products more stable than reactants).

141
Q

Standard enthalpy change of combustion

A

Energy released when 1 mole of substance is completely burnt in excess oxygen under standard conditions of 298K and 1 bar.

142
Q

Explain why standard enthalpy change of combustion has a fixed sign convention

A

Always negative. More energy released when strong covalent bonds within CO2 and H2O molecules are formed than energy absorbed from breaking of covalent bonds in reactant molecules because CO2 and H2O are very small molecules with high degree of orbital overlap (very effective) and hence very strong covalent bonds.

143
Q

Mg (s) + 1/2O2 (g) —> MgO (s)
Identify the type (s) of enthalpy change.

A

Enthalpy change of formation of MgO (s)
Enthalpy change of combustion of Mg (s)

144
Q

1/2 F2 (g) —> F (g)

A

1/2 bond energy of F-F bond
Enthalpy change of atomisation of F (g)

145
Q

Hess’ law of Constant Heat Summation

A

Enthalpy change of reaction is independent of pathway, provided that initial and final state of reaction is the same.

146
Q

Shortcut calculation using enthalpy change of formation, bond energy and combustion, and a brief description of how to derive

A

Combustion: reactant-product
Bond energy: reactant-product
Formation: product-reactant

Derivation: write out the base equation for the larger compound (eg for formation the standard state stuff and for combustion the number of CO2 and H2O after etc…) then draw the born haber cycle and add in the missing components (eg enthalpy values and balancing of products/reactants)

147
Q

Standard enthalpy change of neutralisation

A

Energy released when 1 mole of water is formed when an acid neutralises a base in an infinitely dilute aqueous solution under standard conditions of 298K and 1 bar.

148
Q

Standard enthalpy change of neutralisation

A

Energy released when 1 mole of water is formed when an acid neutralises a base in an infinitely dilute aqueous solution under standard conditions of 298K and 1 bar.

149
Q

Explain why standard enthalpy change of neutralisation has a fixed sign convention and a fixed value for strong acids and base neutralisation

A

Heat is always evolved due to formation of bonds between H+ and OH- to form H2O molecules.
Always a fixed value as strong acids and bases undergo full dissociation hence all their neutralisation reactions can be simplified to the same equation: H+ + OH- —> H2O

150
Q

Why standard enthalpy change of neutralisation of weak acids and bases is less exothermic than strong acids and bases?

A

Strong acids and bases, full dissociation.
Weak acids and bases, only partial dissociation. Hence additional energy needed to be absorbed to ionise the weak acid/base before complete neutralisation can take place, hence less exothermic ∆H(Neut).

151
Q

Define first and subsequent ionisation energies of an element

A

Energy absorbed when one mole of electrons is removed from one mole of gaseous atoms to form one mole of singly positive gaseous ions.
…
Energy absorbed when one mole of electrons is removed from one mole singly positive gaseous ions to form one mole of doubly positive gaseous ions.

152
Q

Lattice energy

A

Energy released when 1 mole of ionic solid is formed from isolated gaseous ions from an infinite distance apart.

153
Q

Why lattice energy has a fixed sign convention

A

Always negative
Energy released from the formation of strong ionic bonds between oppositely charged ions.

154
Q

Factors affecting ionic bond strength

A

Ie factors affecting LE
Quote formula of LE
Charge and radius; since it affects magnitude of LE which affects bond strength

And another one is crystal structure (In enthalpy notes); arrangement of ions affects attraction and repulsion between ion in the compound

155
Q

Understand what dif magnitudes of LE represent (think abt mp/bp)

A

higher magnitude of LE = stronger ionic bonds = more energy needed to overcome stronger ionic bonds = higher mp/bp

But also, Larger LE = products more stable

156
Q

Bond energy

A

Energy absorbed to break 1 mole of covalent bonds between 2 atoms in a gaseous molecule.

157
Q

Why bond energy must be wrt a fixed state

A

If in solid state, energy also absorbed for sublimination of the object, so it’s not reflective of the true energy to just break the bond.

158
Q

Why bond energy has a fixed sign convention

A

Energy absorbed to break bonds

159
Q

Limitations of bond energy

A
  1. Average values, not specific to the bond for the specific molecule
  2. Only for molecules in gaseous state
160
Q

Standard enthalpy change of atomisation

A

Energy absorbed to form 1 mole of isolated gaseous atoms from constituent elements in standard state under standard conditions of 298K and 1 bar.

161
Q

Explain why enthalpy change of atomisation has a fixed sign convention

A

Energy has to be absorbed so that atoms can break away from rigid structures in other physical states/break covalent bonds to liberate atoms

162
Q

Electron affinity

A

Enthalpy change when one mole of atoms or anions gains one mole of electrons.

163
Q

Electron affinity

A

Enthalpy change when one mole of atoms or anions gains one mole of electrons.

164
Q

Sign convention (Electron affinity)

A

1st: negative; eFOA formed when highly positive nucleus attracts the incoming electrons, releasing energy
2nd onwards: positive; energy needs to be absorbed to overcome repulsion between increasingly negatively charged anion and incoming negatively charged electrons.

165
Q

Energy profile diagram vs energy level diagram axes

A

Energy profile: y - energy/KJ mol^-1; x - progress of reaction
Energy level: y - energy/KJ mol^-1

166
Q

Energy profile diagram vs energy level diagram axes

A

Energy profile: y - energy/KJ mol^-1; x - progress of reaction
Energy level: y - energy/KJ mol^-1

167
Q

Standard enthalpy change of hydration

A

Energy released when 1 mole of gaseous ions is dissolved in a large amount of water under standard conditions of 298K and 1 bar.

NO isolated! (isolated is ALe, infinite is LeNS)

168
Q

Why enthalpy change of hydration has a fixed sign convention

A

Formation of ion-dipole interactions releases a lot of energy (sufficient to overcome energy absorbed from breaking of H bonds between water molecules)

Same type of ions NO interactions between them k!

169
Q

Why enthalpy change of hydration has a fixed sign convention

A

Formation of ion-dipole interactions releases a lot of energy (sufficient to overcome energy absorbed from breaking of H bonds between water molecules and weak IMFOA between ions)

170
Q

Ion-dipole interactions description

A

eFOA between ions and opposite partial charges. (Like btw cation and partial negative charge and anion and partial positive charge)

171
Q

Factors affecting enthalpy change of hydration

A

Charge density, higher charge density = strong ion-dipole interactions = more exorthermic ∆H(Hyd)

172
Q

Standard enthalpy change of solution

A

Enthalpy change when one mole of solute is completely dissolved in excess solvent to form an infinitely dilute solution under standard conditions of 298K and 1 bar.

173
Q

∆H (sol) derivation/relationship

A

∆H (sol) = ∆H (Hyd) - ∆H (LE) ; solid ionic compound —> isolated aq ions
Energy absorbed when ionic bonds in solid lattice structure have to be broken (ions separated) (-LE) and interaction between solvent molecules and solute molecules are broken
Energy released when ion-dipole interactions between water molecules and liberated ions are formed (H)

174
Q

Explain change in solubility of ionic salts of metals (hydroxides and sulfates) down the group

A

Hydroxides: solubility increases
- H by hydroxide no change, but by cation less exothermic since radius increasing.
- LE becomes less exothermic because radius increasing significantly since radius of OH- is so small.
- Fall in magnitude of H much more significant than fall in magnitude of LE since r- is so small —> S becomes more exothermic —> solubility increases

Sulfates: solubility decreases.
- H directly proportional to charge density. (H by sulfate no change, but H by cation less exothermic because radius increasing; charge density directly proportional to charge/radius)
- LE roughly constant because sulfate so much bigger than metal ion
- S = H-LE —> S becomes more positive —> solubility decreases)

Chemistry (8 decks)

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