p-block - compounds

Question 1

why do p-block elements form compounds if they can already form elemental structures that provide them with an octet?

Answer 1

stability - structures of pure elements that allow the octet rule to be obeyed are more stable than unbonded atoms, but compounds formed when combined with other elements are usually even more stable

Question 2

why are compounds more stable?

Answer 2

heteronuclear bonds tend to be stronger than homonuclear bonds, this is because 2 different elements have an electronegativity difference, so the bond is polar, there is additional electrostatic attraction between the atoms
stronger bonds = lower energy = more stable

Question 3

give 1 piece of evidence that heteronuclear bonds are stronger than homonuclear bonds?

Answer 3

comparing homonuclear bond strength to X-F bond strengths for p-block compounds shows that the X-F compounds have stronger bonds
- 2 exceptions are Cl and Br, their X-F bonds are weaker than their homonuclear bonds as F has a very small atomic radius, so X-F bonds are significantly shorter, meaning lone pairs are closer and there is greater repulsion, weakening the bond

Question 4

how does the bond strength of X-F heteronuclear bonds change across a period?

Answer 4

this is opposite to the trend in homonuclear bonds, as F is the most electronegative element and electronegativity increases across periods, the electronegativity difference between the two elements in X-F bond decreases across a period, making the bonds less polar and therefore not as strong

Question 5

how does the bond strength of X-F change down a group?

Answer 5

there is no consistent trend - this is a battle between 2 factors
- bond strength generally weakens down groups as orbital overlap gets poorer
- electronegativity difference increases going down groups as electronegativity decreases down groups, which should strengthen the bond
between groups either factor can dominate, so there is no real trend

Question 6

how are oxidation states assigned for p-block elements?

Answer 6

electronegativity is used to assign oxidation states

Question 7

what is the oxidation state of N in NH3 vs in NF3?

Answer 7

in NH3, N is more electronegative than H, so there is a δ- charge on N, so O.S = -3
in NF3, F is more electronegative than N, so there is a δ+ charge on N, so O.S = +3
- remember, oxidation state doesn’t = actual formal charge, bonds are polar but not so much that N completely transfers its electrons to F

Question 8

for each group in p-block (not 18), what is the maximum O.S?

Answer 8

maximum O.S is when the atom ‘loses’ all its valence electrons to a more electronegative element, so:
group 13, 3 valence electrons, max O.S = +3
group 14, 4 valence electrons, max O.S = +4
group 15, 5 valence electrons, max O.S = +5
group 16, 6 valence electrons, max O.S = +6
group 17, 7 valence electrons, max O.S = +7

Question 9

for each group in p-block (not 18), what is the -ve O.S that gives it an octet?

Answer 9

-ve O.S that gives an octet are electrons ‘gained’ from a less electronegative element:
group 13, needs (8-3=) 5 electrons, -ve O.S = -5
group 14, needs (8-4=) 4 electrons, -ve O.S = -4
group 15, needs (8-5=) 3 electrons, -ve O.S = -3
group 16, needs (8-6=) 2 electrons, -ve O.S = -3
group 17, needs (8-7=) 1 electrons, -veO.S = -1

Question 10

for each group in p-block (not 18), what is the inert pair O.S?

Answer 10

this is the no. valence electrons - 2, all are +ve:
group 13, 3 valence electrons, inert pair O.S = +1
group 14, 4 valence electrons, inert pair O.S = +2
group 15, 5 valence electrons, inert pair O.S = +3
group 16, 6 valence electrons, inert pair O.S = +4
group 17, 7 valence electrons, inert pair O.S = +5

Question 11

why is it not possible for a group 13 element to achieve its -ve O.S that gives it an octet?

Answer 11

the necessary oxidation state is -5, as it already has 3 valence electrons and 8-3 = 5
however group 13 elements only has 3 valence electrons, so it doesn’t have enough electrons to form 5 bonds

Question 12

what is the inert pair effect?

Answer 12

going down each group, the O.S that = maximum O.S. - 2 becomes increasingly stable, tends to be the most stable for row 6 elements
in this inert pair O.S the valence s-orbital remains filled giving ns2 valence electron configuration, these electrons are unreactive, they are an inert pair, do not form bonds
this means the maximum amount of bonds that can be formed are just from the electrons in p-orbitals, which is maximum O.S/valence electrons - 2

Question 13

how does the inert pair effect change the structures of compounds?

Answer 13

the inert pair effect changes the structures of the most stable fluorides for each element going down, e.g. in group 13, BF3, AlF3 … TiF

Question 14

what bonding models best describes bonding of p-block compounds?

Answer 14

bonding is adequately described by valence bond theory, with bonding occurring mostly through sp3 or sp2 hybrid orbitals - this requires the electrons in valence s-orbitals to be promoted to the p-orbital

Question 15

why does the inert pair O.S become more stable down the group?

Answer 15

bond strength decreases down the group, and at the bottom it is so low that the energetic pay off for forming 2 extra bonds doesn’t compensate for promotion energy required to form them

Question 16

how do reduction potentials provide proof of the inert pair effect?

Answer 16

going down group 13, for the reduction half equation X^3+ + 2e- <–> X^+ which brings an element from its maximum O.S to its inert pair O.S, the reduction potential becomes increasingly more +ve, as reduction becomes more favourable as the inert pair O.S becomes more stable

Question 17

give evidence using reduction potentials for the trend in oxidising power of halogens

Answer 17

for reduction half equation X2 + 2e- <–> 2X^- reduction potential becomes more -ve going down the group, as reduction is disfavoured, meaning oxidising power decreases down the group, they become less strong oxidants
- this is not because of electronegativity as that refers to attraction of electrons in a covalent bond, not free floating ions
- this is because of enthalpy cycles, based on atomisation, electron affinity and hydration enthalpies

Question 18

disproportionation definition

Answer 18

a reaction in which a species is both oxidised and reduced, its O.S increases and decreases

Question 19

comproportionation definition

Answer 19

a reaction in which a species with 2 different O.S reacts to form a compound with an intermediate O.S - opposite of disproportionation

Question 20

what is the purpose of latimer diagrams?

Answer 20

they are used for elements that exist in multiple O.S and can be used to show if disproportionation reactions are favourable, by allowing half equations to be made + Ecell to be determined, which needs to be +ve for the reaction to be favourable

Question 21

what are frost diagrams + their purpose?

Answer 21

frost diagrams are a graph of -nFE° against O.S
- recall ΔG = -nFE°, where n = no. electrons transferred in reaction, F = faradays constant
these diagrams provide good visual representation of relative stabilities of O.S and whether or not species are prone to disproportionation/comproportionation

the varies O.S of a element are plotted and lines are drawn between any 2 species/states, if a third species lies above the line, that species tends to disproportionate into the neighbouring species on the line
if the third species lies below the line, neighbouring species on the line tend to comproportionate into that third species

Question 22

coordination number definition

Answer 22

the number of atoms/ligands bonded to a central atom

Question 23

why do row 2 elements have a maximum coordination number of 4?

Answer 23

this is because row 2 elements are very small

Question 24

what are the typical coordination numbers + geometries of row 2 p-block elements?

Answer 24

group 13 = B, forms BF4^-, tetrahedral
group 14 = C, forms CF4, tetrahedral
group 15 = N, forms NF3, trigonal pyramidal = tetrahedral derivative
group 16 = O, forms OF2, angled = square planar derivative
group 17 = F, forms F2, linear

Question 25

what are the typical coordination numbers + geometries of row 3 p-block elements?

Answer 25

group 13 = Al, forms AlF6^3-, octahedral group 14 = Si, forms SiF6^2-, octahedral group 15 = P, forms PF6^-, octahedral group 16 = S, forms SF6, octahedral group 17 = Cl, forms ClF5, square pyramidal = octahedral derivative

Question 26

how does ligand impact coordination number?

Answer 26

size - larger ligands means less fit around the central atom, this idea makes sense with halogens as ligands, but falls apart with other atoms a better explanation is redox: e.g. for octahedral complex to form 6 bonds with a single ligand type, the O.S of the central atom is either +6 or -6 depending on which is more electronegative a large O.S like +6 requires a strong oxidant to stabilise it, the oxidising power of halides decreases down the group hence decreasing coordination number a -ve O.S may or may not be possible, as the central atom cannot exceed an octet, so its most stable -ve O.S will be used

Question 27

how do p-block elements achieve a coordination number >4 with only 4 valence orbitals and following the octet rule?

Answer 27

this is explained by the partial MO approach: e.g. PF5, P appears to have exceeded an octet as its formed 5 bonds with F, VSEPR suggets trigonal bipyramidal geometry, so has axial and equatorial bonds equatorial P-F bonds have trigonal planar geometry, so P is sp2 hybridised to form 3xsp2 hybrid orbitals + 3pz these hybrid orbitals form P-F equatorial bonds the leftover 3pz orbital is perpendicular to the hybrid orbitals and points at the axial F atoms, overlapping with their valence 2pz orbitals to form a bonding MO - only 2 electrons can be in any MO, so 2 bonding electrons are spread across the 3 atoms, this is a multicentred bond the 3-centres antibonding orbital can be formed by destructive phase overlap of the same AOs, the final MO is formed by the pair of 2p orbitals on F, this is non-bonding 2 of the 10 electrons are in non bonding MO located on Fs so P only has 8 electrons, octet rule not broken

Question 28

hypervalent definition

Answer 28

a compound in which an element appears to have exceeded an octet

Question 29

give 1 piece of evidence for the partial MO theory

Answer 29

in PF5 the axial P-F bonds are longer than the equatorial P-F bonds - this supports the presence of multicentred bonds to the axial atoms, 2 bonding electrons spread over 2 bonds, so bond order is 1/2, these bonds are weaker, therefore longer than typical bonds with bond order 1

Question 30

what are the 3 structures of p-block compounds?

Answer 30

discrete molecules, polymeric structures, ionic lattices

Question 31

discrete molecule definition

Answer 31

known number of 2 or more atoms linked by bonds

Question 32

give 2 examples of compounds with discrete molecular structures

Answer 32

lots of p-block compounds, e.g. CO2, Al2Br6 ...

Question 33

polymeric structure definition

Answer 33

very large, non-specific number of atoms bonded to one another forming extended solid structures, chains, layers or extended 3D networks

Question 34

give 2 example of a p-block compound with a polymeric structure

Answer 34

SnCl2 which is a polymeric chain structure SiO2, which has a structure similar to diamond but with O worked into it

Question 35

ionic lattice definition

Answer 35

lattice/packed structures of non-specific numbers of atoms, - like metal lattices, best considered as packed spheres representing anions + cations

Question 36

give 3 tips to predicting p-block compound types

Answer 36

1- try to draw a sensible monomeric lewis structure for the compound, if it has 8 valence electrons it will likely remain as a monomer, if not it will oligomerise/polymerise/form an ionic lattice 2- use electronegativity to tell between oligomeric, polymeric and ionic structures: small ΔX (<1.5) = oligomeric intermediate ΔX (1.5-2.1) = polymeric large ΔX (>2.1) = ionic - this works because greater ΔX means bonds are more polar, so stronger partial charges and electrostatic attractions between atoms, these allow them to aggregate into oligomers/polymers/ionic lattices 3- tip 1 can incorrectly predict monomeric structures for some compounds, where 8 valence electrons are achieved there may not be incentive to aggregate from octet rule standpoint but if ΔX is large it will happen anyway, consider 8 valence electrons and ΔX<1.9 = monomeric, if ΔX>1.9 = bigger structures

Question 37

give 1 limitation of these tips

Answer 37

tip 1 works best if the outer atoms are monovalent, but not if they have a higher valency - need some chemical intuition, what prefers single/double bonds,

Question 38

explain the structure of boron trihalides

Answer 38

BF3 --> BI3 all adopt trigonal planar monomeric structures, even though B lacks an octet - this is because these molecules can be stabilised by π-interactions B = sp2 hybridised to form 3 sp2 hybrid orbitals + 2pz orbital B-X bonds formed with hybrid orbtials and remaining empty p orbital is used for stabilisation as it filled X orbitals can overlap onto it and donate electron density B-X bonds therefore have partial π-character making them unexpectedly strong

Question 39

how does the strength of π-interactions in boron trihalides change down the group?

Answer 39

down the group the strength decreases as atomic radii of halides increases so B-X bond length increases, meaning the distance between filled halide p-orbital and empty B p-orbital is greater, so strength of π-interactions decreases

Question 40

can the rest of group 13 form similar trihalide compounds + why?

Answer 40

no, this is because they are much larger than B, which is in row 2, increasing the distance between p-orbitals and weakening π-interactions, they tend to oligomerise/polymerise to achieve an octet instead

Question 41

explain the structure of group 13 hydrides

Answer 41

BH3 isn't stable as a monomer, as H has no filled p-orbital so there can be no stabilising π-interactions to achieve an octet, BH3 dimerises to B2H6, using 3-centre-2-electron bonds formed by bridging hydrogens, which allow a kind of shared octet between the 2 B atoms other group 13 elements require this kind of multicentred bonding to achieve an octet, and tend to polymerise at standard conditions and dimerises when heated into a gas

Question 42

how does the structure of group 13 hydrides impact its properties?

Answer 42

group 13 hydrides are very reactive due to weak H-X bonds, stability decreases down the group with decreasing X-H bond strength

Question 43

give 1 function of group 13 hydrides

Answer 43

LiAlH4 and NaBH4 are widely used reducing agents in organic chemistry, these hydride salts are a source of nucleophilic H LiAlH4 is more reactive as Al-H bond strength is lower

Question 44

what structures are group 14-17 hydrides?

Answer 44

these hydrides adopt monomeric structures, as all these atoms are able to achieve an octer without needing to dimerise stability of hydrides decreases down the group due to decreasing H-X bond strength

Question 45

catenation definition

Answer 45

formation of molecules comprising of chain X-X bonds terminated by other elements e.g. hydrocarbons, specifically alkanes

Question 46

what elements are the best at catenation

Answer 46

carbon is undoubtedly best, can form alkane chains with over 100,000 carbons next best is Si which is significantly worse, maximum length ever recorded is 10

Question 47

why are other groups worse at catenation?

Answer 47

groups 15-17 have weaker homonuclear single bonds due to lone pair repulsion

Question 48

why is carbon so good at catenation?

Answer 48

this is because its C-C single bonds are decently strong, relative to the bonds it forms with other elements chain termination is favoured if X-H bond is stronger than X-X bond for C, C-C bonds are stronger than C-H bonds by 16%, this is more like 30-55% for the rest of group 14

Question 49

what are the structures of p-block oxides?

Answer 49

metals tend to adopt polymeric or ionic structures due to their large ΔX with O and preference for single bonds, cannot terminate structure with X=O nonmetals tend to adopt discrete molecular structures as they prefer to form multiple bonds and so are able to terminate via X=O, as well as their smaller ΔX e.g. CO2, NO2, P4O10, SO2, F2O, SO3, etc metalloids are inbetween, some oxides are polymeric and some are discrete molecuels

Question 50

describe the structure of silicones

Answer 50

silicones feature chains of alternating Si-O bonds, this works as these bonds are strong and highly polar 2 x R groups are attached to the Si in these polymers, these determine the properties of the compound and the size/length of the polymer

Question 51

silicate definition

Answer 51

an ion consisting of Si and O, -ve

Question 52

aluminate definition

Answer 52

an ion consisting of Al and O, -ve

Question 53

aluminosilicate definition

Answer 53

mixed silicate/aluminate compounds

Question 54

do group 18 elements form compounds?

Answer 54

normally these elements are inert, however heavier group 18 elements can react with F and O, as their high electronegativities make it easier to distort noble electron clouds, pulling electron density towards themselves and forming bonds - however extreme conditions are needed

Question 55

why is Xe the most studied noble gas?

Answer 55

reactivity increases down the group, but Rn is radioactive

Question 56

give 1 use of noble gas oxide/fluoride compounds

Answer 56

very powerful oxidants/fluorination agents, noble gas really wants to be reduced back to monatomic form, which is a very strong driving force