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F11SPC > Structure + Periodicity > Flashcards

Structure + Periodicity Flashcards

1
Q

State the Nernst equation.

A

∆Go (red) = -nFEo (red) n: number of electrons involved F: Faraday constant Eo (red): reduction potential

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2
Q

Since oxidation is the reverse of reduction, state the link between equations.

A

∆Go (red) = -∆Go (ox)

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3
Q

If Eo(red) is negative…

A

then ∆Go (ox)/F is negative for the OXIDATION couple.

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4
Q

If Eo(red) is positive…

A

then ∆Go (red)/F is negative for the REDUCTION couple.

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5
Q

What equation do we use to calculate relative free energy changes?

A

Rearrange the Nernst equation to give: ∆Go (red)/F = -nEo (red) Units are in volts.

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6
Q

State Slater’s rules (4).

A

Zeff = Z - S 1. No contribution to S from electrons in groups to the right of the one being considered. 2. Contribution of 0.35 added to S for each electron in the SAME GROUP as the one being considered. 3. IF electron being considered is in an ns or np orbital then the electrons in the next lowest shell (n-1) each contribute 0.85 and electrons in lower shells (n-2) and lower contribute 1.00. 4. IF electron being considered is in an nd or nf orbital, then all electrons below contribute 1.00.

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7
Q

What is the definition of a mole?

A

1 mole of specified particles is equal to the number of atoms in exactly 12g of carbon-12 (12C).

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8
Q

What are the different flame test colours?

A

Lithium: red Potassium: lilac Sodium: yellow Barium: green

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9
Q

What is the principle quantum number and what does it determine?

A

n - determines overall size of orbital - allowed values are anything but 0 - for species with just one electron it determines the energy of an allowed solution of the wave equation

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10
Q

What is the orbital angular momentum quantum number and what does it determine?

A

l - maximum value = n-1 - has no effect on the energy of 1 electron atoms - l + 1 allowed values - l = 0 s orbital - l = 1 p orbital - l = 2 d orbital - l = 3 f orbital

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11
Q

What is the magnetic quantum number and what does it determine?

A

ml - has no effect on electron energy except when in a magnetic field - allowed values are -l –> +l in integer steps, hence when l = 1 (p-orbital), ml values are -1, 0, +1 hence why there are 3 x 2p orbitals etc. - determines orbital orientation

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12
Q

What is spin quantum number and what does it determine?

A

ms - allowed values are +1/2 and -1/2

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13
Q

Define the Aufbau principle.

A

To move from one element to the next, add one proton and x neutrons to the nucleus, and one electron into the orbital of lowest energy which is available.

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14
Q

Define the Pauli exclusion principle.

A

No two electrons in any system can have identical values for all 4 quantum numbers.

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15
Q

How is the Pauli exclusion principle achieved? (3).

A
  1. All electrons in a given orbital have the same values for n, l and ml. 2. Therefore by the Pauli exclusion principle every electron in an orbital must have different values for ms. 3. Only two values for ms (+1/2 -1/2) therefore there is a maximum of TWO electrons in any orbital and they must have opposed spins.
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16
Q

How does a many electron atom compare to H1 energies? (4).

A
  1. Both only allowed certain energies. 2. Each solution for both is characterised by the 3 quantum numbers n l and ml. 3. Both have the same angular wave functions therefore shapes of s p d and f orbitals are the same. 4. The difference is that energies for many electron atoms depend on BOTH n and l.
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17
Q

Define Hund’s rule.

A

The most stable electronic state is one with the most PARALLEL spins.

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18
Q

Why is the order of energies 3s < 3p < 3d when the probability diagrams predict 3s > 3p > 3d? (4).

A
  1. Whilst the maximum probability decreases in radius from 3s to 3d, the presence of other peaks for 3s and 3p diagrams explains the order of energies. 2. The smaller peaks might not be the most probable place to find an electron but at some points in time electrons will be present at the radius of the smaller peaks. 3. These electrons will be subjected to an increase nuclear charge (Zeff) making the electron more stable. 4. Increased stability = lower energy.
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19
Q

Define the term electronegativity.

A

The ability of an atom in a molecule to attract electron density to itself.

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20
Q

Define the term resonance.

A

The blending of two or more Lewis structures, where the atoms have the same relative positions, but with different electronic arrangements.

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21
Q

Draw the structure of dinitrogen tetroxide (N2O4) and explain its properties.

A

*see notes* - very long and weak N-N bond due to repulsion between the two positive charges. - molecule readily splits into two NO2 fragments.

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22
Q

Explain the difference in O-O bond length in H2O2 compared to F2O2 with the aid of Lewis structures.

A

*see notes* - F2O2 shorter bonds due to resonance. - highly electronegative F pulls electrons towards itself.

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23
Q

With the aid of Lewis structures, explain why the NCO- ion is stable yet the CNO- ion is UNstable.

A

NCO- - one formal negative charge with two resonance forms - negative charge predominately on O due to electronegativity CNO- - three formal charges, one +ve on electronegative N atom - too many electrons on C atom

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24
Q

What is the shape and bond angle(s) for an AB2 structure?

A

Linear 180

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25
Q

What is the shape and bond angle(s) for an AB3 structure?

A

Trigonal planar 120

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26
Q

What is the shape and bond angle(s) for an AB4 structure?

A

Tetrahedral 109.5

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27
Q

What is the shape and bond angle(s) for an AB5 structure?

A

Trigonal bypyramid ax-eq bonds 90 eq-eq bonds 120

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28
Q

What is the shape and bond angle(s) for an AB6 structure?

A

Octahedral 90

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29
Q

What is the shape and bond angle(s) for an AB2L structure?

A
  • Three electron pairs but only two bonds - V-shaped 95
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30
Q

What is the shape and bond angle(s) for an AB3L structure?

A
  • Four electron pairs but only three bonds - Pyramidal 107
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31
Q

What is the shape and bond angle(s) for an AB2L2 structure?

A
  • Four electron pairs but only two bonds - V-shaped 104.5
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32
Q

What is the shape and bond angle(s) for an AB4L structure?

A
  • Five electron pairs but only four bonds - Seesaw - ax-eq bonds <90 - eq-eq bonds <120
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33
Q

What is the shape and bond angle(s) for an AB3L2 structure?

A
  • Five electron pairs but only three bonds - T-shaped 87
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34
Q

What is the shape and bond angle(s) for an AB2L3 structure?

A
  • Linear 180 - Lone pair repulsions all in eq positions cancel out
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35
Q

What is the shape and bond angle(s) for an AB5L structure?

A
  • Square-based pyramidal - ax-eq <90
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36
Q

What is the shape and bond angle(s) for an AB4L2 structure?

A

Square planar

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37
Q

How do we account for coordinate bonds in VSEPR?

A

For the coordinate bond count: - 2 electrons at acceptor atom - 0 electrons at donor atom

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38
Q

How do we account for multiple bonds in VSEPR?

A

Electron count: +1 for every single bond -1 for every double bond -2 for every triple bond

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39
Q

What are dispersion forces?

A
  1. Electrons moving around can cause uneven distribution creating an instantaneous dipole. 2. This induces a dipole in a neighbouring molecule. 3. The two molecules are then attracted together by electrostatic interactions.
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40
Q

State the features of dispersion forces (4).

A
  1. ALWAYS exist for all molecules. 2. Always attractive, independent of molecule orientation. 3. Increase with molecular weight because more electrons creates are larger dipole. 4. For molecules of a similar weight, the more compact molecule is less easily polarised + the dispersion forces are weaker.
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41
Q

Why are dipole-dipole forces weaker than dispersion forces?

A

Due to the orientation dependency of the permanent dipole.

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42
Q

What molecules are harder to polarise for dipole-induced-dipole forces?

A

Electronegative atoms due to the electrons being tightly held.

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43
Q

What are the features of ionic compounds? (5)

A
  1. Form large 3D matrices. 2. Strong electrostatic interactions between the ions. 3. Results in high melting points. 4. Lots of thermal energy required to break bonds. 5. Water can usually dissolve ionic compounds by solvating the ions via hydrogen bonding.
44
Q

What solvents dissolve polar molecules?

A

Polar solvents.

45
Q

What solvents dissolve non-polar molecules?

A

Non-polar solvents.

46
Q

Why are shape and symmetry important for molecules? (5).

A
  1. Affects how the molecules assemble and react. 2. Affect how the molecules pack together into a crystal. 3. Affects whether the molecule has a dipole moment. 4. Determines whether atoms in a molecule are equivalent. 5. Determines whether a molecule is chiral.
47
Q

What is the quantitative explanation for why oxygen exists as O2 and sulphur S8?

A

4X2 (g) –> X8 (g) - For O2 the ∆Hr value is positive, therefore the reaction won’t proceed making O2 the most stable form of oxygen. - For S8 the ∆Hr value is negative, therefore the reaction will proceed making S8 the most stable form of sulphur.

48
Q

What is the bonding explanation for why oxygen exists as O2 and sulphur S8?

A
  • 2p-pi - 2p-pi interaction gives shorter interatomic distance providing good overlap and a strong bond. - 3p-pi - 3p-pi interaction gives longer interatomic distance, poor overlap + weak bond making S8 more stable with two single bonds.
49
Q

Draw the Lewis and VSEPR structures of ozone.

A

O=O+-O- 117o bond angle, AB2L V-shaped

50
Q

Write the equations for the formation of ozone.

A

O2 –> 2O O2 + O –> O3

51
Q

Write the equations for ozone reacting with nitrogen oxide.

A

NO2 + O3 –> NO3 + O2 NO3 –> NO + O2 NO + O3 –> NO2 + O2 Overall: 2O3 –> 3O2

52
Q

Write the equations for ozone reacting with halogens.

A

Cl + O3 –> ClO + O2 ClO + O –> Cl + O2 Overall: O3 + O –> 2O2

53
Q

Describe the structure and properties of diamond (4).

A
  • tetrahedral sp3 carbon atoms - extended network of covalent bonds - electrical insulator - good conductor of heat
54
Q

Describe the structure and properties of graphite (4).

A
  • trigonal planar sp2 carbon atoms - arranged in layers - weak bonds between each layer - solid state lubricant
55
Q

Describe the structure of buckminsterfullerene (2).

A
  • C60 - truncated isocahedron made of hexagons + pentagons
56
Q

State and describe the 4 types of oxidation state.

A
  1. Minimum point - stable with respect to neighbouring oxidation state. 2. Maximum point - UNstable with respect to neighbouring oxidation state. 3. Concave point - thermodynamically stable with respect to disproportionation to neighbouring oxidation state. 4. Convex point - thermodynamically Unstable with respect to disproportionation to neighbouring oxidation state.
57
Q

Outline the features of the OFSE diagram for the first row of transition metals (6).

A
  1. Group oxidation state adopted until manganese. 2. Group oxidation state most stable for Sc (III) + Ti (IV). 3. Oxidising power increases from V (V) to Mn (VII). 4. Beyond Mn, group oxidation state disappears + a few unstable strongly oxidising O.S. above (III) exist. 5. Beyond Ti, either (II) or (III) O.S. is most stable. 6. Cu (I), Mn (III) + Mn (VI) are unstable with respect to disproportionation.
58
Q

Outline the features of the OFSE diagram for transition metal triads (3).

A
  1. Higher oxidation state more accessible for heavier transition metals. 2. Oxidising power of higher oxidation states reduces down a triad. 3. Occurrence of lower oxidation state reduces down a triad.
59
Q

Outline the features of the OFSE diagram for the main group of elements (groups 14+16) (4).

A
  1. Higher oxidation state more accessible for lighter elements. 2. Oxidising power of higher oxidation state increases down a group. 3. Reducing power of (-II) oxidation state for group 16 elements increases down the group. 4. S (III) is unstable with respect to disproportionation.
60
Q

State the three ways of achieving nobel gas configuration and for which elements they apply.

A
  1. Loss of electrons - atoms of low electronegativity. 2. Gain of electrons - atoms of high electronegativity. 3. Sharing of electrons in a covalent bond.
61
Q

How can we show the thermodynamics of ionic compound formation?

A

Born-Haber cycles.

62
Q

What is ∆Hf?

A

Enthalpy of formation, a sum of all the other processes.

63
Q

What is ∆Hsub?

A

Enthalpy of sublimation, converting a solid into a gas.

64
Q

What is ∆Hdiss?

A

Enthalpy of dissociation, obtaining a single atom of element i.e. X2 –> X

65
Q

What is Ie?

A

Ionisation energy, the removal of an electron.

66
Q

What is Ea?

A

Electron affinity, addition of an electron.

67
Q

What is Ul?

A

Lattice energy.

68
Q

What is the ionisation energy trend for the first Ie of group 1 elements?

A

Decreases down the group as size increases.

69
Q

What is the ionisation energy trend for the third Ie of group 13 elements?

A

General decrease. Increase at Al and In due to d-block contraction and f-block contraction respectively.

70
Q

How does the d/f-block contraction reduce the amount of energy needed to remove an electron?

A

d/f-block contractions cause the atom to become small with less shielding, therefore less energy is required to remove an electron.

71
Q

What is the ionisation energy trend for the first row of elements? (4)

A
  1. General upward trend due to increasing size + proton number. 2. Increases at Be due to breaking into a full 2s sub-shell. 3. Increases at N due to breaking into a half-full 2p sub-shell. 4. High value at Ne due too breaking into NGC.
72
Q

What is the ionisation energy trend for the sequential ionisation of tin?

A

Increasing values as atoms is becoming more positively charged.

73
Q

What is the electron affinity trend for the group 17 elements? (2).

A
  1. Increases from F- to Cl- 2. General decrease from Cl-, less energy being released as electron is being added to a larger atom with more shielding.
74
Q

What is the electron affinity trend for the first row of elements?

A

All negative values except for Nitrogen (+7 kJmol-1)

75
Q

Annotate Kapustinskii’s equation.

A

z+ cation charge z- anion charge v number of ions per formula unit r+ radius of cation in pm r- radius of anion in pm NOTE: in pm NOT SI units

76
Q

How do we increase lattice energy? (3).

A
  1. Increasing ion charge (z+/z-). 2. Increasing number of ions (v). 3. Decreasing ionic radius (r+/r-).
77
Q

What are the periodic trends going down the group 1 elements? (3).

A
  1. Size increases due to increasing principle quantum number n. 2. Zeff initially increases then stays constant after Na. 3. Ionisation energy decreases due to increasing size + increasing principle quantum number n.
78
Q

Which group 1 metal forms a nitride and why?

A
  • ONLY Li - large lattice energy is required to dissociate N2 triple bond and overcome the large positive Ea for nitrogen - lattice energy is inversely proportional to ionic radii therefore to smallest cations give the largest lattice energies
79
Q

Outline oxide formation and hydrolysis of group 1 compounds (2).

A
  1. Combustion of alkali metals in air to form MO2 M = Li, Na, K, Rb, Cs 2. Hydrolysis of which gives 2MO2 + 2H2O –> H2O2 + O2 + 2MOH
80
Q

What is a superoxide and what is it used for?

A

4MO2 Used in breathing apparatus to generate oxygen: 4MO2 + 2CO2 –> 2M2CO3 + 3O2

81
Q

Which 3 group 1 metals form complexes and what is their geometry?

A

Li form tetrahedral complexes Na/K form octahedral complexes

82
Q

What are the periodic trends going down the group 2 metals? (3).

A
  • size of atomic/ionic radii increases due to increasing principle quantum number - Zeff increases initially then stay constant from Mg - ionisation energy decreases due to increasing size + principle quantum number
83
Q

Outline halide formation for group 2 metals.

A
  1. Be forms covalent compound BeX2 2. Others form ionic compound MX2 M = Mg, Ca, Sr, Ba
84
Q

Outline oxide formation for group 2 metals.

A
  1. Be forms covalent compound BeO 2. Others form ionic compound MO M = Mg, Ca, Sr, Ba
85
Q

Outline nitride formation for group 2 metals.

A

Possibility of two ionic compounds: Be3N2 and Mg3N2

86
Q

Outline organometallic formation for group 2 metals.

A
  1. Be and Mg form covalent compounds [(CH3)2 M] which are fairly reactive. 2. Ca, Sr and Ba form ionic compounds (CH3)2 M which are extremely reactive.
87
Q

Outline complex formation for group 2 metals (3).

A
  1. Be forms tetrahedral complexes. 2. Mg, Ca, Sr and Ba form octahedral complexes. 3. Sr and Ba can form cubic complexes [ML8].
88
Q

What are the features of the diagonal relationship between Li and Mg? (5).

A
  1. *draw diagonal positioning ft. Na* 2. Direct formation of the nitride with dinitrogen to form Li3N/Mg3N2 whereas Na doesn’t react. 3. Direct formation of the oxide by combustion in dioxygen to form Li2O/MgO whereas Na forms Na2O2. 4. Oxysalts readily decompose to the oxides Li2CO3 –> Li2O + CO2 MgCO3 –> MgO + CO2 whereas Na2CO3 is stable to moderate heating. 5. Formation of covalent organometallics whereas Na forms ionic organometallics.
89
Q

Why do Li and Mg have a diagonal relationship?

A

Mg2+ and Li+ have similar charge densities because the higher charge of Mg2+ is offset by its larger size.

90
Q

Define the term bond dissociation energy.

A

The enthalpy required to dissociate a specific bond in a gaseous molecule (in its ground state) to form two gaseous fragments (in their ground states).

91
Q

Outline the trends in BDE down a group (4).

A
  • atoms increase in size - orbitals become more diffuse - overlap becomes poorer - bonds become weaker NOTE: same for homo and heteronuclear bonds just with higher energies for heater due to overall stronger bonds.
92
Q

Outline the trends in BDE for multiple bonds (4).

A
  • as bond order increases - bond strength increases - bond distances decrease - increase in BDE
93
Q

Outline the anomalous behaviour of F, N + O (3).

A
  • small size - increased inter electronic repulsion destabilises the bond - smaller BDE values
94
Q

Why are homonuclear bonds weaker than heteronuclear bonds? (2).

A
  • heteronuclear bonds have BOTH electrostatic interactions and covalent bond interactions - homonuclear bonds ONLY have covalent bond interactions
95
Q

State the Schomaker-Stevenson relationship and annotate.

A

d(A-B) = rA + rB - 9(XA - XB) d(A-B): interatomic distance between A-B rA, rB: covalent radii, their addition is value of the atoms joining to form a covalent bond -9(XA - XB): value of electrostatic interaction/difference in electronegativity

96
Q

State the Pauling equation and annotate.

A

BDE(A-B) - 0.5[BDE(A-A) + BDE(B-B)] = 96.49(XA-XB)^2

97
Q

Describe the structure and bonding of BeCl2.

A

Forms polynuclear [BeCl2] chains with bridging Cl atoms.

98
Q

How does Be achieve an octet?

A
  1. Two electrons from Be. 2. Two electrons from two one-electron donor chlorines. 3. Four electrons from two two-electron donor chlorines.
99
Q

Describe the structure and bonding of AlCl3.

A

Forms dinuclear [Al2Cl6] with bridging Cl atoms.

100
Q

How does Al achieve an octet?

A
  1. Three electrons from Al. 2. Three electrons from three one-electron donor chlorines. 3. Two electrons from one two-electron donor chlorine.
101
Q

Why does the number of EX5 compounds decrease with increasing atomic number?

A
  1. The heavier halides are more easily oxidised than lighter halides because E-X BDE decreases with increasing atomic number of the group 17 element. 2. O.S. of the group 15 elements becomes more strongly oxidising with increasing atomic number.
102
Q

Describe the structure of BF3 (4).

A
  • planar mononuclear molecule - short B-F interatomic distance - permits effective ppi - ppi intramolecular interaction - good Lewis acid
103
Q

Describe the structure of BX3 (4).

A

X = Cl, Br, I - planar mononuclear molecules - longer B-X interatomic distances - weaker ppi - ppi intramolecular interaction - powerful Lewis acids

104
Q

Describe the structure of AlF3 (4).

A
  • large difference in electronegativity - ionic character - high melting polymeric solid - built up from fluoride-bridged octahedra
105
Q

Describe the structure of AlCl3 (3).

A
  • polymeric solid - built from chloride-bridged octahedra - dimer in liquid and gas phase
106
Q

Describe the structure of AlX3 (4).

A

X = Br, I - dimer in solid, liquid and gas phase - limited ionic character - reduced difference in electronegativity - power Lewis acids

107
Q

What is the inert pair effect?

A
  1. Increased nuclear charge means electron(s) can be accelerated to very high speeds and interact more with nucleus penetrating shielding electrons. 2. Causes the contraction of orbital to lower energy therefore electrons less likely to become involved in bonding.

F11SPC (2 decks)

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