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inorganic y2 / s1 > lecture 6 > Flashcards

lecture 6 Flashcards

1
Q

valence e- and hybridisation seen in b

A

3
sp2

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2
Q

valence e- and hybridisation seen in N

A

5 e-
sp2

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3
Q

explain where the valence e- are in both N and B if theyre sp2 hybrids

A

sp2 = 3 hybrid orbitals
both n and b use 3 e- here.

b thereofre has an empty p orbital

n therefore has a filled p orbital which is its lone pair

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4
Q

what is the p orbital used for

A

its used for pi bonding

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5
Q

explain the sigma framework of borazine

A

sigma = sp2 bonds

think of borazine from above so u can only see the sp2 hybrids (3 on each atom)

each atom uses 2 orbitals to overlap with its neighbours
and uses the other sp2 orbital to bond with s which has a circle orbital

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6
Q

explain the pi framework in borazine

A

think of it from above and ignore the sigma framework made up of sp2.

the b’s have an empty p orbital and the n’s have a full p orbital with a lone pair in it.

in this example theyre all in phase. aka the bottom of the p orbital is coloured in while the top bit isnt

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7
Q

how many pi electrons does borazine have

A

it has 6 pi electrons which all come from the N

3 N’s and they all bring 2e- from their lone pair i nthe p orbital.

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8
Q

B-N bond both ways explained

A

resonance: (-)B-N(+)
bc the N gave B its lone pair.

partial charges: (+)B-N(-)
bc N is more electronegative than B.

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9
Q

explain the pi framework for benzene

A

each C has a p orbital with 1e- in it,, they all contribute 1e- to the pi system.

1e- in the upper p orbital

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10
Q

in borazine,, what contributes more to the bonding MO

A

the N
bc it has a larger Zeff,, more close e- density

also more electronegative

therefore more stable and lower in energy

remember bonding mos are lower in energy than antibonding

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11
Q

how many MO are in borazine and benzene seperately

A

they each have6 MO

bc 6AO

3/6 = bonding
3/6= antibonding

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12
Q

the filled mo in borazine have what character

A

the filled mo in borazine have more N character bc the lone pair came from N

N sp2 is also more electronegative than Np and B – so the sigma bond is more poalrised to the N

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13
Q

does benzene or borazine have more aromatic stabilisation energy and why

A

benzene has a greater degree of aromatic stabilisaion energy

this is bc in borazine,, the e- in the pi system come from N,, N also has a larger Zeff so it has a large e- density surrounding it.

this means the e- are less delocalised in the pi system and are found closer to the N. due to the pi bond polarisation

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14
Q

reules for aromaticity

A

huckels rule of pi e-: 4n+2
flat
conjugated
cyclic

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15
Q

rules for antiaromaticity

A

when there is 4n pi e-

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16
Q

if theres 6 p atomic orbitals thennnn

A

theres gona be 6 molecular orbitals

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17
Q

what are the 6 molecular orb itals for both benzene and borazine

A

3 bonding
- no nodes - all in phase
- one node x2

3 antibonding
- 2 nodes x2
- 3 nodes - all out of phase

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18
Q

what do we need to thibk about when drawing the borazine MO

A

we need to remmeber that N has a larger coefficient aka more e- density around it which means the N contributes for to the bonding MO’s.

this means that when we draw it,, we need a bigger circle around the N when draeing the bonding MO. and a smaller circle than B when drawing the antibonding orbitals.

all in phase
half half
2 nodes
3 nodes and all out of phase

in benzene all the coefficients are the same as they all have the same Zeff and the same contribution to the pi system.

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19
Q

why does N get a larger circle in the bonding MO diagrams

A

N has a greater Z effective
meaning its orbitals are lower in energy
so they have a larger coefficient in the BMO

this gives a lumpy pi system as the p orbitals are different sizes.

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20
Q

are all the bonding MOs filled for benzene and borazine

A

yesss
all the bonding mos are filled.

the sigma and pi’s are filled with a total of 6e-. thats why theyre both so stable + aromatic.

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21
Q

whats the difference between the bonding and antibonding mos of borazine

A

larger coefficient on N for the bonding MOs

larger coefficient for B on the antibonding MOs

shown by the larger circle // orbital size.

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22
Q

when theres a nodal plane // node in a moelcule shown by like a ninja fruit slice type of line,, what happens

A

every time u pass across the node,, the phase of the orbitsals change.

bc theres a change in the wave function.

23
Q

borazine molecular formula

A

B3N3H6

24
Q

physical properties of borazine are similar to benzene bc theyreee

A

isoelectronic

same amount of valence e- in both of them

25
Q

physical similarities between borazine and benzene

A

both colourless liquids
similar densities
similar enthalpies of evaporation

26
Q

whats more reactive,, borazine or benzene and why

A

borazine is more reactive

lower pi stabilisation energy due to less e- delocalisation (due to N having a greater Zeff)

and bc the bonds are polar: 2 different polarity possibilities depending on if we think of resonance or partial charges.

the pi stabilisation energy is lower for borazine

27
Q

N and B being lewis acids or bases

A

lewis acid accepts so B is the lewis acid

N is the lewis base bc it donates its lone pair.

28
Q

if theres an EWG attached to the B,, what happens

A

B is a lewis acid meaning it accepts e-

by having an EWG next to it,, the complexes formed will no longer be 1:1 .

this is bc the B already accepts e- from N,, making it,, the B,, a weaker acid. and the N is a weaker base bc its already donated its lone pair to the B.

29
Q

can borazine undergo addition reactions

A

yessss

with 3Br2,, then reflux to remove 3HBr,, leaving 1H on each N.

30
Q

name a simple borazine derivative that can react with a Grignard reagent (RMgX, RLi, R2Hg) to form b substituted derivative

A

B3N3H3Cl3

or (HBNCl)3

31
Q

what does the borazine derivative (HBNCl)3 forming b substituted derivatives mean

A

it means the things it react with,, normally grignard reagents,, get added onto the B

the driving force of the reaction is salt formation,, LiCl // XCl

32
Q

the borazine derivative (HBNCl)3 and 3PhLi reaction

A

b substituted derivative

Ph goes onto the B’s,, Cl is removed from the B’s

3LiCl is made,, salt formation which is the driving force of the reaction.

N’s still have a H attached.

33
Q

descrive the borazine derivative (HBNCl)3

A

borazine but with 1 H on every N

and 1 Cl on every B

33
Q

thermolysis and self condensation of borazine gives what products

A

loss of H2 // HX

polycyclic borazines (2 borazines bonded together)

or further condensation to give boron nitride (isoelectronic to graphite but the layers are eclipsed) eg hexagonal layers eclipsed

34
Q

hexagonal boron nitride is isoelectronic to what

A

isoelectronic to graphite,, but the graphite layers are staggered and the hexagonal boron nitride layers are eclipsed

35
Q

cubic boron nitride is isoelectronic to what

A

its isoelectronic to diamond
theyre both hard.

36
Q

what is a cyclophosphazene

A

inorganic,, cyclic pi delocalised ring made up of N and P and a halogen

(NPX2)n
n = number
X = halogen

37
Q

in a cyclophosphazene,, what coordinate geometry does N and P have

A

N has 2 coordinate geometry as its bonded to 2 diff things

P has 4 coordinate geometry as its bonded to the N’s and 2 X’s

X’s being halogens

38
Q

in cyclophosphazenes,, how many e- does each element provide

A

the N and the P both provide 1 e- each!! for pi bonding.

39
Q

the starting materials for most cyclophosphazene reactions areeee

A

the Cl derivatives of cyclophosphazenes.

40
Q

how can cyclophosphazenes be made

A

condensation route
azide route

41
Q

cyclophosphazene condensation route

A

high boiling solvents
150*C heat for 5 hours

cyclic oligomers and polymers are produced and then these are normally separated by fractionation.

4n NH4Cl + nPCl5 –> (NPCl2)n + 4nHCl

42
Q

azide route for cyclophosphazenes

A

sometimes used to form disubstituted cyclophosphazenes!!1

R2PCl is dissolved in MeCN
LiN3 or NaN3 (salt formation) is added to form the azide (N3)
azide is heated to generate intermediate R2PN –> oligomerises.

R2PCl + LiN3 – loss of LiCl —> [R2PN3] —loss of N2–> 1/n (NPR2)n

43
Q

trimeric meaning

A

polymer made up of 3 monomer units

44
Q

most cyclic trimeric cyclophosphazenes areeee

A

planarrrrr

n = 3 so its trimeric!! bc theres 3x PN

(NPCl2)n

45
Q

is (NPF2)3 perfectly planar?? and are the other trimeric rings also planar // with other halogens

A

the (NPF2)3 is perfectly planar!!!

but the other trimeric rings are more//less planar.

46
Q

when n = 4 for a cyclophosphazene,, (NPC2)4,, whats it called

A

tetramer!! tetrameric instead of trimeric

47
Q

are tetrameric cyclophosphazenes planar,, if not,, what shape do they have

A

non planar – they have a large variety of structures including chair, boat, saddle and crown derivatives.

chair looks like a chair,, just keep doing P-N

boat is also like a boat,, just keep writing P-N down

48
Q

bonding in cyclophosphazes must take what into account

A
  • very strong + stable P-N bonds
  • P-N bond lengths are all the same + short
  • N-P-N has 120* angle
  • N is weakly basic,, it can be protonated// involved in pi bonding.
  • hard to add e- to the structure,, its therefore difficult to electrochemically reduce the phosphazene skeleteon

(r = gain of h // e- and loss of o)

49
Q

describe the tetramic rings

A

boat and chair
single PN bonds only
just keep writing p - n

flip one side to get boat from chair.

50
Q

P and N hybrids and other exp for cyclophosphazes

A

N:
sp2 hybrid
so 120* angle
1e- for pi bonding
1 lone pair in the ring plane

P:
sp3 hybrid
120* but due to bond pair bond pair repulsions
1e- for pi bonding

6 pi electrons in total: 1 from each N and P…

51
Q

explain N in cyclophosphazes

A

N has 5 valence e-!!
and is sp2 hybridised

lone pair on the hybrid
1 e- in the other hybrids
1e- in the p orbital!! for pi bonding. this is different to borazine where its lone pair is in the p orbital.

this is bc P has 1e- in the p orbital whilst B had an empty p orbital.

52
Q
A

inorganic y2 / s1 (10 decks)

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