The Ionic Model

Question 1

What does the van Arkel triangle of bonding look like? What does this tell us about bonding?

Answer 1

Bonding is not always restrictive to ionic/covalent/metallic and instead there are contributions for other types within each other

Question 2

What are the assumptions of the ionic model?

Answer 2

Solids are hard, incompressible, non-polarisable, charged spheres
Charges are integer multiples
The only interaction between ions is electrostatic

Question 3

What evidence is there supporting the ionic model?

Answer 3

Appearance: ionic typically non-volatile, brittle, transparent- share similar properties
Electrical conductivity only when molten or dissolved in a polar solvent
Spectroscopy and magnetism: similar absorption spectra/ magnetic properties as free-gas phase ions
X-ray diffraction: shows electron density with larger anions and smaller cations in a lattice

Question 4

How can you determine the size of an ion?

Answer 4

The lattice parameter contains the size of r+ + r-
If the radius of one ion is known, subtract from 1/2 lattice parameter (assumes ions touching)
Or an estimate from Cn/Zeff = R
where Cn is a constant dependent on quantum number
Given in pm, 10^-12
Consult database for ion size

Question 5

What are the trends for ionic radius in the periodic table and why? What about between cations and anions?

Answer 5

Across the period, size of the cations/anions decreasing
This is because Zeff is increasing as protons are added whilst shielding does not increase at the same rate
The outer electrons are more greatly attracted by the stronger nuclear charge

Down the group, ionic radius increases, as although Zeff increases, the increase in quantum number has the larger effect, and so overall the higher energy, more diffuse orbitals results in a larger ionic radius

Cations<Anions as cations have fewer electrons for the same nuclear charge

Question 6

How does ionic radius change with coordination number and oxidation state and why?

Answer 6

Higher coordination numbers increase ionic radius
This is because the ion is no longer fully incompressible and so larger

Higher oxidation states mean fewer electrons for the same nuclear charge
This means the valence electrons are more strongly attracted to the nucleus, and so the orbitals contract resulting in a smaller ionic radius

Question 7

Define the lattice energy of a compound?

Answer 7

The amount of energy required to disassociate 1 mol of an ionic solid into a gas of its ions at infinite separation, so from 0K
The negative of this is lattice enthalpy

Question 8

How can lattice enthalpy be calculated from a thermodynamics perspective?

Answer 8

This quantity cannot be directly measured

Question 9

How has the Madelung Constant been derived? And the Coulombic interaction component of the Born-Lande equation?

Answer 9

6 - 12 + 8
Over 1 , root 2 , root 3

Question 10

What is the Born-Lande equation? What are the two components to it?

Answer 10

The coulombic electrostatic attraction between ions
And the repulsion between same charges

Question 11

From the derivation of the coulombic interactions, derive the Born-Lande equation? What are the main steps for the whole derivation?

Answer 11

Find the coulombic attractions from Coulombs law and summing all the interactions
Derive the Madelung constant by considering the interactions next to, same plane repulsion, to centre, 6, 12, 8, add r as a parameter
Born repulsion as inversely proportional to r to the power of n
Add them, from graph, need to minimise energy, so differentiate with respect to r, equate to 0
Find B (constant for repulsion) in terms of the other
Substitute back into the equation and factorise
1- 1/n not 1-r

Question 12

How does the Born-Mayer equation differ from Born Lande?

Answer 12

Differ in terms of repulsion, with Mayer using ae^ -r/constant, which is related to the compressibility of the crystal

Question 13

What is the Kapustinskii equation and what assumptions have been used/

Answer 13

The madelung constant has been replaced with 0.87v, where v is the number of species in the unit cell
The interatomic radius r (1/2 A) has been replaced by r+ + r-
n=9 (rocksalt for all)

Question 14

Derive the Kapustinskii equation from the Born-Lande equation?

Answer 14

Make sure to change m to pm, and J to Kj

Question 15

What is the value for e (fundamental charge?

Answer 15

1.602 x 10^-19

Question 16

What is the value for epsilon 0?

Answer 16

8.85 x 10 ^ -12

Question 17

What should a Hess cycle for NaCl look like?

Answer 17

Includes Formation of NaCl(S) from standard states
Sublimation and atomisation energies for both
Ionisation energy
Electron affinity
Lattice enthalpy

Question 18

What is the synthesis and general reactions of the alkali metal hydrides?

Answer 18

Formed via direct reaction with hydrogen
M + 1/2 H2 ⟶ MH
All hydrolyse to formed MOH and H2
Used a deprotonating agents to form H2 and Na+
And to make other hydrides e.g
4LiH + AlCl3 ⟶ LiAlH4 + 3LiCl

Question 19

What happens to the thermal stability of the G1 metal hydrides down the group? Using a Hess cycle, explain why this occurs?

What type of lattice are they?

Answer 19

Thermal stability decreases, decomposes at lower temperatures
Using the Gibbs equation at change in G=0 at equilibrium, decomposition at
T=H/S
Entropy change is the same for both, so H must be smaller for formation of the hydride
This is due to the lattice enthalpy decreasing down the group as the cation ionic radius increases, resulting in smaller charge densities, meaning decreasingly energetic and thermal stability

All NaCl type structures

Question 20

What is the bonding like for the group 2 hydrides?

Answer 20

BeH2: covalent, linear molecule in gas phase, tetrahedral in solid
MgH2: Ionic, rutile, evident from XRD electron density
- reversibly, so potential for hydrogen fuel storage
CaH2, SrH2, BaH2: ionic, PbCl2 structure with 9 fold coordination

Question 21

Compare the stability of the G1 vs G2 hydrides?

Answer 21

G2 higher stability, as they have a higher lattice enthalpy, arising from higher charges and a greater number of ions in the formula unit
Use Kapustinskii for comparison (radi similar)

Question 22

What compounds can form from combustion of a metal in air?

Answer 22

Oxides, O (2-)
Peroxides (O-O) (2-)
Superoxides (O-O) (1-)

Question 23

How do formation enthalpies relate to stability?

Answer 23

The more negative the enthalpy of formation, the more stable the compound
This is because more energy is released from formation, and so a lower energy, more stable compound

Question 24

Why are the oxides more stable for smaller cations whilst peroxides/superoxides more stable for larger cations?

Answer 24

To form the oxide, an addition electron must be added to the O- ion, which is an endothermic process
This is most favourable when the energy given out as the lattice enthalpy is large enough to compensate for the additional energy requirement
For small ion, the lattice enthalpy is more negative, so does compensate, and so the oxide is more favourable as there is a better option
For larger ions, the lattice is enthalpy is not as negative, so rather only the first electron affinity for superoxides is favourable (although all theoretically are thermodynamically favourable, but superoxides the most)

So the superoxides decompose into peroxides, and peroxides into oxides when the cation is small enough as their is a more thermodynamically favourable product available

Question 25

What are the thermodynamically favourable compounds for G1 and G2? And their solid structure type?

Answer 25

Lithium Oxide- antifluorite Sodium Peroxide, K/Rb/Cs superoxide- all with peroxide/superoxide subunit BeO- covalent, Wurtzite MgO,CaO,SrO- rocksalt BaO2- peroxide, but decomposes upon heating

Question 26

What is the thermochemical radius and how can it be calculated?

Answer 26

A way of finding the radius of ions from thermochemical experiments e.g difficult to identify radius of CO3 2- so instead calculate the lattice enthalpy using Hess Cycles and experimental data Then rearrange for the anion/cation radius

Question 27

Why is lattice enthalpy not exactly the same as internal energy change? What do our equations give and why are they valid?

Answer 27

Lattice enthalpy defined from absolute 0K Equations derived to give change in internal energy change in H= change in U + change in pv However, as pv is small, almost negligible so we can assume enthalpy change is approximately internal energy change

Question 28

What happens to the thermal stability of the carbonates down a group? Why?

Answer 28

Thermal stability increases This is because cation size increases, meaning the stability of the oxide decreases, as the lattice enthalpy will fall Therefore, the difference in energy between the carbonate and oxide is smaller down the group, meaning decomposition is favourable but becomes less favourable Decomposition relies on the high lattice enthalpy of MO, but as this decreases, higher temperatures are needed for decomposition

Question 29

Does CaCl exist and why?

Answer 29

Theoretically, there is an exothermic change for formation of CaCl, however, CaCl will disproportionate into CaCl2 so it does not exist

Question 30

When will a reaction favour disproportionation and how can this be calculated?

Answer 30

When the enthalpy for disproportionation < 0, so more stable and exothermic Start with the hypothetical disproportion reaction, and use Hess Cycles to calculate the enthalpy change i.e ionisation energies, electron affinities, atomisation...

Question 31

Out of the G2 halides, why is Iodide the species which would most likely stabilise MX the most?

Answer 31

To form MI2, needs energy for additional ionisation which is very endothermic As the electron affinity for iodine is the lowest for the halides, and formation enthalpy of iodide the least exothermic, the addition of an extra iodide is the least likely to compensate for the additional energy of ionisation Additionally, the lattice enthalpy of MI2 is the smallest of the halides due to the large ionic radius of I- These factors would limit the ability to form MX2, however, the M+ cation is still too large to adequately stabilise so disproportionation still occurs

Question 32

Why does NaF2 not form?

Answer 32

The second ionisation energy is far to endothermic to be compensated by the energy released from formation of NaF2 and additional F- This is because the additional electron is being removed from a shell of a lower quantum number, experiencing a far larger Zeff

Question 33

Why does the lattice enthalpy AgCl not agree with the Kapustinskii prediction? How can this be alleviated?

Answer 33

There is a degree of covalency within AgCl, not purely/perfectly ionic This means there are additional covalent interactions to overcome which the Kapustinskii equation does not consider, so the prediction is too low Instead, use the thermochemical radius for a more accurate estimate

Question 34

Why does the addition of fluoride ions support higher oxidation states relative to the other halides?

Answer 34

The enthalpy of formation of F- is the most exothermic out of the halides, as it has the most negative electron affinity Additionally the difference in lattice enthalpy between Mxn and Mxn+1 is largest for F-, as F- has the smallest anion radius This results in F- having the most negative enthalpy of formation of a higher oxidation state of the halides

Question 35

What happens to ions dissolved in a polar solvent?

Answer 35

The ions will be surrounded by molecules of the polar solvent, and coordinate to the ion The lattice will break apart Stabilises the ion, with ion-dipole interactions the main interaction

Question 36

What happens to an ion in a structure-less medium?

Answer 36

Average ion-dipole interactions are considered Polarisable, so described by the dielectric constant which tells us about the polarisability of a given medium in response to an electric field

Question 37

What Hess Cycle should be considered for whether a substance dissolves?

Answer 37

Includes lattice enthalpy, hydration enthalpies of gaseous ions, and solution enthalpies for dissolving if sol enthalpies <0, dissolves

Question 38

What is the equation for the hydration enthalpy of a substance?

Answer 38

Question 39

What simplifications can be made to known equations to enable comparisons of solubility? How can this be interpreted?

Answer 39

From the equations, the only time when the lattice enthalpy is small but hydration is large, is when one ion is small and the other large There is the greatest solubility when there is a mismatch in ion size

Question 40

What is the chelate effect? Why does it occur?

Answer 40

Larger, the more polydentate ligands will replace less polydentate ligands already coordinated This is because there is an increase in entropy from the ligands liberated from the metal ion, replaced by fewer ligands Therefore, the gibbs energy for the reaction is more negative, a driving force The entropy increase is the dominant factor, will very low enthalpy values

Question 41

What happens when an alkali metal is placed in ammonia?

Answer 41

Alkali metals dissolve in liquid ammonia Na(s) --> Na+ (solv) + e- (solv) Free electrons localised in the cavity of the solvent Optical observation with electron excitation, colours However solution metastable (will decay over time) e.g NH3 + e- --> NH2- (sol) + 1/2 H2(g)

Question 42

What are grignard reagents? When can Mg be found in an uncommon state?

Answer 42

Mg-R compounds, formed from reactions with RX, good nucleophiles Mg(I) when coordinated by very bulky carbon ligands with nitrogen

Question 43

What are carbides? And dizaenides?

Answer 43

[C2] 2- or C 4- [N2] 2-, from thermal decomposition of azides

Question 44

How would you justify your answers from a hess cycle from a stability perspect?

Answer 44

The enthalpy of the reaction > 0, and so the reaction is not thermodynamically feasible from an enthalpy perspective The energy demand for ionisation e.g, is not compensated by the additional stabilisation of the (lattice enthalpy) so the reaction is unfavourable

Question 45

What should you include when showing Kapustinskii from Born-Lande?

Answer 45

Why the approximations have been used - 0.87v as for most solids a good replacement for their madelung constant