pi bonding in phosphazines and onwards // lecture 7

Question 1

how many pi electrons are there in cyclophosphazine

Answer 1

there’s 6 pi electrons

one from each n and p

Question 2

describe nitrogen in cyclophosphazines

Answer 2

sp2 hybridised

there’s 1e- in the p orbital as in cyclophosphazine,, each atom gives 1 e-!!

the lone pair is in a sp2 hybrid

Question 3

describe P in cyclophosphazine

Answer 3

it’s sp3 hybridised
forms 4 bonds and there’s an e- in each hybrid bond. (in the sigma bonding framework)

there’s one more e- tho!! and we want to figure out where this one e- is located

Question 4

what are the 2 types of pi bonding in cyclophosphazine

Answer 4

there’s in plane pi bonding

out of plane pi bonding

Question 5

what type of pi bonding was borazine and benzene

Answer 5

out of plane pi bonding!!!

Question 6

what’s the diff between in plane and out of plane pi bonding

Answer 6

deciding whether to use the P 3d orbitals
or the sigma* orbitals

ON THE P

Question 7

what do we need to think of when deciding whether to use the 3d or sigma* orbitals on P

Answer 7

the energy

the 3d orbitals are probs too high in energy to be involved in bonding.

sigma* orbitals may be lower in energy and used in any pi bonding used in the ring system.

Question 8

PF5 hybridisation used and explanation - think about how many orbitals each hybridisation gives and how many bonds the molecule has

Answer 8

so sp3 forms 4 bonds,, and this is one less than PF5 with 5 bonds

so it has to be sp3d1 hybrid
bc we need one more orbital and this must come from d as all of the p and s are used already

Question 9

PF5,, how many F environments

Answer 9

theres 2 diff environments
axial and equitorial

Question 10

spin of P

Answer 10

i. = 1/2

Question 11

spin of f

Answer 11

i = 1/2

Question 12

if P couples to the axial F’s in PF5 u gettttt

Answer 12

a triplet

bc 2x2x0.5 + 1 = 3

Question 13

if P couples to the equitorial F’s u get ,,, in PF5

Answer 13

2x3x0.5+1 = 4
a quartet

coupling is stronger and the bonds are thereofre shorter.

Question 14

so PF5 in nmr givesss

Answer 14

a quartet of triplets

the equitorial split it 3 times (quartet)

then the axisl (diff env) split each peak twice : triplet

Question 15

what happens in cold temps to PF5

Answer 15

at cold temps u can tell a difference between the axial and equitorial fluorines

in room temp they look like 5 equivalent fluorines.

Question 16

if the d orbitals in PF5 were too high in energy, how would we explain the bonding w/o them

Answer 16

we say P is sp2 hybridised

3 orbitals bond to the equitorial F’s

left over p orbital to bond to the axial F’s.

the F’s use their s orbital to bond; its a circle like in H

Question 17

3 MO for FPF,, THE AXIAL ATOMSSSSSSS

Answer 17

bonding
shaded o, shaded p, unshaded p, unshaded o (in phase with each side of the p orbital so theyre in the same phase so theyre bonding)

non bonding: shaded o,,, node at p bc any way u draw it, 1 will be inphase and 1 will be out of phase,, shaded o.

antibonding: shaded o,, unshaded p, unshaded p, shaded o. nodes between the p and the o’s

each bond has 2e-,, so u get 4e- from the FPF so u fill the bonding and non bonding AO’s

by o’s we mean the s orbital but thats the shape of the s orbital so im gonna say o.

Question 18

PFP is what type of bond

Answer 18

3c-4e

3 atoms but 4e- !!!

normally its 2c-4e-

Question 19

in PFP,, in PF5,, theres 2e- in bonding and 2e- in nonbonding,, so how many e- are holding the PFP together

Answer 19

only 2e-!!!

bc 2e- are non bonding
bond order = 1/2

so axial bonds are weaker and longer. thats why the coupling constant is smallerrrrrr!! they make the triplet after the quartet.

Question 20

phosphines are what type of ligands

Answer 20

theyre acceptor ligands

so where do they accept into

Question 21

phosphines are pi acceptor ligands,, where can the e- go

Answer 21

either the d orbitals of the metal from the P:,, then back into the P’s d orbitals via back bonding

or the sigma* orbitals of the P molecule from the metals d orbital

Question 22

pi acceptor ability of P with d orbital explanation

Answer 22

RRR P:

P has a lone pair

this can be donated to the transition metal’s d orbitals (dx^2-y^2)

this is the bonding interaction!!

same thing as amines,, sigma donation of lone pair to the metal.

Question 23

what do pi accepting mean when it comes to P being a pi acceptor ligand

Answer 23

it means that P can donate its lone pair to the metals d orbital,, but it can also accept e- into the Ps d orbitals

Question 24

wait wait wait!! explain pi acceptor ability of phosphines WITH d orbitals

Answer 24

phosphines = P RRR

p: –> metal d orbital (dx2-y2)
this is a bonding interaction

then meta d orbital –> P d orbitals
back bonding

d orbitals are available for P bc its in the 3rd period. it can accept e- here.

draw the clover and RRR bonds.

Question 25

pi acceptor ability of P without d orbitals

Answer 25

so the bonding interaction is the same bc we use the p: … the p’s lone pair to donate the lone pair to the metals d orbitals.

so bonding interaction : RRRP: –> metal (dx2-dy2)

backbonding:
accept e- density fromthe metals d orbital into the P-R sigma* orbital ( not shaded and shaded)

Question 26

when we think about P’s pi acceptor abilities we need to think offff

Answer 26

bonding interaction ,, same for both with // without d orbitals

the back bonding interactionnn

different depending on if d or sigma* is lower in energy.

Question 27

when explaining the pi acceptor ability of P,, what is the accepted method of backbonding

Answer 27

the use of their sigma* // antibonding orbitals

usually due to them having a lower energy

Question 28

when u donate e- into the sigma* of PR what happens

Answer 28

u weaken the bond between that R and P

but u strengthen the bond between P and the metal

Question 29

PF5 and phosphines is diff to

Answer 29

phosphazines

Question 30

N is phsophazines

Answer 30

sp2

Question 31

P in phosphazines is

Answer 31

sp3

Question 32

in phosphazine,, the N is sp2 but where is the lone pair

Answer 32

the lone pair is exooooo

aka its not in the p orbital its in the sp2 hybrid orbitals

Question 33

d orbitals on P can lead tooo

Answer 33

in plane // out of plane pi bonding

out of plane being similar to borazine and benzene

Question 34

using the P d orbitals in cyclophosphines

Answer 34

we say the axis coming out of the page is the z axis

we get 6 MO frpm 6 AO.

3/6 = antibonding
2/3 = strongly antibonding
1/3 = weakly antibonding

3/6 = bonding
1/3 = weakly bonding
2/3 = strongly antibonding.

for pi bonding we use dxz and dyz orbitals on P and p orbitals on the N.

bonding interaction: phase on top of d orbitals matches that of p orbitals!!!

U GET A NODE AT P BC NOTHING WILL MATCH ITS PHASE: MOST BONDING MO STILL HAS A NODE // either a N or P doesnt rlly matter.

node makes smt less stable + unstable

this isnt fully conjugated bc we have a node!!! not truley aromatic bc we have a node at the most bonding MO. less stabe than borazine// benzene

Question 35

why is cyclophosphazine less stable then borazine and benzene

Answer 35

bc even the most bonding MO have a node in them

if u draw the d orbitals that look like p’s // wings from birds eye view (dzx and dzy) and then the p orbital on N which looks like a circle,, u cant make all of them be in phase!! there will be a node on either N or P!!

and a node will weaken an interaction and leads to a degree of unstability.

Question 36

the more unstable MO of cyclophosphazine uses what d orbital

Answer 36

it uses the dyz orbital
which means the d orbital wings arent going along with the bonds but are perpendicular to them

this gives antibonding interactions outside the system but bonding interactions inside the ring.
better orbital overlap on the inside + bonding MOs so its majority bonding. still weaker tho bc u have antibonding on the outside.

Question 37

in plane pi bonding of cyclophosphazines

Answer 37

N: is donated to the empty dx2-y2 orbital on P.

aka they back donate to P, into its d orbital

in plane pi bonding bc u can see the p orbitals and d orbitals for what they rlly are.

dumbbell and clover shaped,, in plane bc the orbitals are flat in the plane

so the N lone pair is pointing out the system and is donated into one of the d orbital petals (dx2-y2) orbitalllll. STRENGTHENS THE N-P BOND IN THE RING: SHORTENS THE P-N BONDS!!!

Question 38

the strength of the NP bond in cyclophosphazines in plane bonding aka where u can see every orbital for what it is,,, the NP bond strength depends on what

Answer 38

the R grouyp u have attached onto P

bc this affects on hpow much the N lone pair can backbond ]

EWG = more back donation as P is more (+)

perfectly planar if F is used as u get more back donation as P will be more delta (+)

PN bond length decreases

which enhances out of plane pi bonding?

Question 39

advantages of in plane pi bonding and using d orbitals

Answer 39

accurately describes geometries: plananr pn rings with short NP bonds

d orbitals are too high in energy to be used in bonding

Question 40

explaining bonding in cyclophosphazines without using d orbitals,, explanation of the ring and advantages//disadvantages

Answer 40

we use the ionic model
N is more electronegative so P donates 1e- to N.

each P will be (+)
5-1e- = 4 so sp3 hybrid

N will be (-)
2 lone pairs available now!!
so u draw N with 2 lone pair orbitals
sp3 hybridised

advantage: no d orbitals
disadvantage: N is sp3 hybridised and 2 lone pairs,, so the ring should look like a cyclohexane ring! but the ring is acc planar!!

Question 41

negative hyper conjugation:

Answer 41

donation of a pi orbital // filled orbital // lone pair into a sigma* orbital

eg in an ester,, one of the oxygens lone pairs that are in plane,, they can donate lone pair to the carbonyls// C-O sigma* orbital: we want this as it stabilises the energy of the lone pair.

Question 42

whats the anomeric effect - think of the O in anOmeric

Answer 42

when the cyclohexane with an O instead of a C and an OH as a substituent is preferred to have the OH as axial.

when the o is adjacent as it allows the lone pair to bc donated to the C-OH sigma* which stabilises the energy of the lone pair. and none bonding lone pair becomes bonding by being donated into the sigma*.

Question 43

what is needed for the anOmeric effect to occur

Answer 43

the lone pair needs to be donated to the sigma* bond but the sigma* bond between at least one electronegative atom

sigma* is more stable near electronegtative groups!! non electronegative groups are higher in energy. O or F to lower the sigma* energy so it matches the lone pair better.

Question 44

explain the bonding in cyclophosphazenes using negative hyperconjugation - out of plane so p orbitals going up and looking in an angle

Answer 44

• start with ionic model
• lone pairs on N in the p orbital
• N: donated into sigma* on P,, the bond between P and X (substituent)
• if X is more EWG, the sigma* will be lower in energy,,

Question 45

is nonbonding or bonding higer in energy

Answer 45

nonbonding is higher in energy

whe nthe lone pair (nonbonding) is donated to the sigma*,, its energy is lowered as it becomes bonding.

Question 46

negative hyperconjugation in cyclophosphazine in in plane bonding

Answer 46

• N: is sticking out and pointing outside the cyclic molecule
• the N: is in the sp2 hybrid orbital
• can donate into P-N sigma* orbitals in the ring
  aka the bonds that make up the actual ring.

Question 47

in in plane bonding,, the N: is in what orbital

Answer 47

its in the pz orbital

Question 48

in out of plane bonding: where is the N:

Answer 48

its in the sp2 hybrid

Question 49

out of plane bonding in cyclophosphazines,, where is the N: and where is it donated to

Answer 49

N: is in the sp2 hybrid

its donated into the NP sigma*

Question 50

in plane bonding,, where is the N: and where is it donated

Answer 50

its in the pz orbital

its donated into the P-X sigma* orbital.

X being the substituent on P.

if X is a EWG,, we like this bc it reduces the energy of the sigma* bond.