Use LEFT and RIGHT arrow keys to navigate between flashcards;
Use UP and DOWN arrow keys to flip the card;
H to show hint;
A reads text to speech;
58 Cards in this Set
- Front
- Back
|
What is the contribution to sigma by an electron in the same energy level: ns / np / nd/ nf |
0.35 |
|
|
What is the contribution to sigma by an electron in the energy level with quantum number n-1:
For a) ns / np
For b) nd/ nf |
a) 0.85 b)1 |
|
|
What is the contribution to sigma by an electron in the energy level with quantum number
For a) ns / np
For b) nd/ nf |
a) 1 b) 1 |
|
|
Formula for Slater's rules |
Z* = Z - sigma |
|
|
Are elements with low ionisation energies metals or non-metals? |
Metals |
|
|
Why does the 2s - 2p energy gap increase across a period? |
A small part of the 2s orbital is close to the nucleus inside the orbit of the 2p and 1s orbitals. Increasing the positive charge of the nucleus (z) therefore stabilises (lowers the energy of) the 2s orbital more than the 2p orbital |
|
|
What does a small 2s - 2p energy gap allow? E.g. in B2 |
2s(σ*) and 2p(σ) molecular orbitals can mix and move apart. The 2 electrons move from 2p(σ) into 2(pi) to become paramagnetic causing it to fall below the 2p(σ) on the energy level diagram |
|
|
When are covalent bonds strongest? |
When atoms are nearly the same size as orbitals overlap best when they are the same size |
|
|
Why are pi bonds weaker for larger molecules? |
Pi bonds made from the overlap of p orbitals. Larger atoms hold p orbitals further apart; hence making the sideways overlap harder and pi boding weaker |
|
|
Why is SH4 more reactive in air than CH4? |
Silicon is larger so more easily accessible for attacking O2. No space around C for 5th group |
|
|
Define electronegativity and state what it depends on |
The power of an atom to attract electrons; depends on Z* |
|
|
What does the highest oxidation state for p-block compounds require and which two elements does this usually occur with? |
requires the removal of all valence electrons; usually occurs with O and F |
|
|
Oxidation states usually change in units of two to give diamagnetic compounds. Intermediate oxidation states do occur but usually as dimers. Give an example: |
BF2 technically does not exist bu the dimer B2F3 does |
|
|
TlCl has an oxidation state of +1. Explain the existence of 2TlCl2 |
TlCl2 does not exist, oxidation states vary by +2. TlCl2 is actually Tl+[TlCl4]- |
|
|
Why can CCl4 exist but PbCl4 decomposes to PbCl2 + Cl2? |
The inert pair effect: For carbon the energy gap between 2 s and 2p is small so an electron can be promoted from 2s to 2p allowing 4 orbitals available for bonding. The 6 s - 6p energy gap for Pb is much larger, meaning the electron is not promoted from the 6s leaving it doubly occupied and existing as a lone pair, and only 2 6p orbitals available for bonding |
|
|
Why does the lone pair on Sb in SbCl6 have no effect on the geometry? |
It exists in the s orbital which is spherically symmetrical |
|
|
Explain the alternation of XCl5 stability down group 5 |
N: is small and cannot fit 5 groups around it P: is more stable As: is less stable than P as it is more electronegtaive than expected (comes after the 3d-block so atoms are small and Z* high) and is difficult to oxidise to +5. Sb: more stable than As Bi: less stable than Sb due to inert pair effect |
|
|
Give three reasons SF2 can react with 2F2 to form SF6, but SCl2 cannot react wth 2Cl2 to form Cl6 |
1) F-F bond weak as very short so electrons repel each other 2) Cl is larger and bulkier so harder to fit around S 3) F and S more similar in size so better orbital overlap |
|
|
Distinguish between Lewis acid and Lewis base |
Lewis base has a lone pair of electrons Lewis acid has a vacant orbital |
|
|
Explain the formation of a BeCl2 polymer |
Be in BeCl2 has 4 valence electrons so can accept 2 lone pairs from adjacent BeCl2 molecules; each Cl donates a lone pair |
|
|
How does the compound BF3 exist when B only has 6 electrons? |
B is sp2 hybridized with a vacant pz orbital. The lone pairs on F lying in their p orbitals form a pi bonding overlap |
|
|
Why is BBr3 a stronger Lewis acid than BF3? |
BF3 is a more stable compound as there is good pi overlap between the 2p and 2s as they are similar size. Br 4s orbitals are much larger and hence will not overlap as well, meaning B is more electron defficient |
|
|
Explain the trend in Lewis acid strength down group 3 for XY3 molecules |
Lewis acid strength is essentially how electron deficient the central atom is, i.e. easy to attack. BX3 < AlX3 > GaX3 > InX3 > TlX3
B is small so hard to fit four groups around Ga, In and Tl are large and their empty 4p,5p and 6p orbitals do not overlap effectively with the 2s/2p/3s/3p orbitals over lewis bases Al most electropositive atom in the group so has most affinity for lone pairs
|
|
|
Why is (CH3)2NPH more stable than N(CH3)3? |
The lone pair in the vacant pz orbital on the N can overlap with the pi system of the benzene ring |
|
|
Explain why the H-B bonds are different lengths in the B2H6 dimer |
The centre bonds are longer as they only contain 1 electron. This is due to 2-centre 3-electron bonding. The two B's each share one of the others B-H bonds |
|
|
Why would adding more electrons not help stabilise a B2H6 2-centre 3-electron bond? |
Added electrons would go into non-bonding or antibonding orbitals |
|
|
Why when BCl3 and BBr3 are mixed is some BCl2Br and BBr2Cl are produced? |
They form a dimer due to the lone pairs being attracted to the central atom, this then breaks off giving the monomers |
|
|
How can X=X heavy atom bonds be made more stable? |
Large groups, highly branched groups protect the double bond. This is kinetic stabilisation, the double bond is still unstable but less accessible. Also prevent polymerisation by making the products less stable, thermodynamic effect |
|
|
Synthesising a double bond for heavy atoms is unfavourable. Name two ways to make it favourable: |
1) Precipitation of a stable salt e.g. LiCl 2) Elimination of a stable molecule e.g. HCl |
|
|
What happens to the nature of of the double bond as you descend group 14 with R2E=ER2 compounds? |
Bond moves from planar to bent Free rotation around the double bond becomes possible Dissociation into 2 ER2 in solution |
|
|
Explain using steric arguments why heavier group 14 elements have bent double bonds between them |
Bulky groups are needed to make the double bond stable, these need to be as far apart as possible which distorts the molecule |
|
|
Explain using the molecular orbital argument why R2Sn=SnR2 is bent and R2C=CR2 is planar |
Carbon has a small 2s/2p energy gap allowing the pi bond to form, but Sn has too high an energy gap meaning that the sp2 bonding orbital is doubly occupied. This bonds with the vacant p orbital causing there to be an angle |
|
|
Why do groups have to be bulkier to stabilise group 15 E=E than 14? |
Only one group can attach so has to be very bulky to protect E=E |
|
|
Why does the E-E-R bond angle gradually change from 120 degrees to 90 as you go down group 15? |
Larger s/p gap, lone pair in s orbital and 3 p orbitals make the bonds. Angle for p orbital bonding is 90 degrees angle for sp2 hybrids is 120 |
|
|
Explain the existence of R3P=O |
Back-bonding; the lone pair on the oxygen has a pi overlap with the vacant σ* from the P-R bond. This allows P to have five orbital interactions, but weakens the R-P bond |
|
|
Why is O=P-R backbonding stronger when R is mroe electronegative? |
They will draw electron density away from P so the lone pair will be able to attack it easier |
|
|
How do you prepare a phosphorus ylide? |
1) alkylation of a phosphine 2) deprotonation with strong base |
|
|
Where does the lone pair go on SF4? |
Equitorial, only interacts with two bonding electron pairs |
|
|
Why is promoting an electron from the s orbital to a 3d orbital to allow sp3d hybridization for SF5 not an option? |
d orbital too high in energy Too large and diffuse to make a bond with smaller atoms Spectroscopy shows use of 3d is minimal Axial bonds are longer than equitorial bonds; this implies that there are not 5 equivalent hybrid orbitals on P |
|
|
How can hypervalent SF6 exist when only 4 valence orbitals exist? (s+p) |
Two electron pairs are in a non-bonding orbital. This effectively leaves it in an ionic state such as SF42+ + 2F-. They form 3-centre 2 electron bonds using the bonding electrons.
|
|
|
Why are axial bonds longer for SF6 than equitorial? |
Lone pairs go on axial F, gives them bond order of 0.5 |
|
|
Why do more electronegative atoms prefer to be axial? |
Can better stabilise the lone pair |
|
|
Why does the NMR spectra for PF4(CH3) vary depending on temperature? |
At a high temperature berry pesudo rotation occurs. The trigonal bipyramid is constantly rotating giving an average of each position. At low temperatures this is slowed down and the axial and equitorial positions have an effect |
|
|
How can you make reactive and unstable organometallic compounds such as R-MgX and R-Li and keep them stable? |
Use bulky groups to protect reactive metal-carbon bond Make sure the process involves the formation of a very stable molecule e.g. LiCl or H2 Make sure equilibrium keeps moving in the right direction, e.g. remove product as it forms |
|
|
Explain what happens to the oxidation state of Mg in the following oxidative addition reaction:
R-X + Mg --> RMgBr |
Grignard reagent, metal has to have two stable oxidation states 2 units apart. Mg goes from 0 to +2 |
|
|
What do you do if metal can only do one electron oxidation in an oxidative addition reaction?
E.g. Li/Li+ |
Use 2 equivalents and get a metal halide salt by product:
H3C-Br + 2Li --> H3C-Li + LiBr |
|
|
What is metallation? |
Oxidation of metal by an acidic H on a hydrocarbon. Redox reaction |
|
|
What is transmetallation? Give 5 types of transmetallation reactions |
Conversion of one preformed organometallic compound to another
organometallic + metal organometallic + organometallic organometallic + organic halide organometallic + metal halide organometallic + acidic organic hydrocarbon |
|
|
What is hydrometallation? |
Addition of M-H across double bond E.g. BH3 + H2C=CH2 --> H3C-CH2(BH2) |
|
|
What is carbometallation? |
Addition of M-R across a double bond |
|
|
How can you prepare an organo-lithium compound? |
Reaction of R-Br with Li metal
Ch3-Br + 2 Li --> Ch3-Li + LiBr |
|
|
Why are organo-lithium compounds such as LiCH3 soluble in organic solvents? |
Li-C has a lot of ionic character, forms a tetramer (CH3Li)4 in a cubic lattice structure made of Ch3- and Li+ groups. 3 Li orbitals and 1 CH3- orbital combine to give a molecular orbital delocalised over 4 atoms. The organic groups are on the outside of the molecule making it soluble |
|
|
Explain Li-CH3 bonding in terms of molecular orbitals |
3 Li orbitals one CH3. Four possible combinations All 4 in phase 2 Non-bonding 1 anti-bonding
Leads to 4-centre 2 electron bonds, any extra electrons would not help stabilise
Can also form a hexamer with 6 x 4-centre 2-electron bonds |
|
|
Why is (MeLi)4 not soluble in alkane? Why is (BuLi)4? |
One of the sp3 orbitals on the Li will interact with the C-H bonds of the alkane. Bu is bulky preventing this |
|
|
Why is R-Li a more reactive nucleophile in THF than a hydrocarbon solvent? |
Non-coordinating solvents give larger aggregates THF and ethers are coordinating solvents and tend to give monomers and dimers and smaller aggregates are more reactive |
|
|
Why are heavier group 1 metal alkyl compounds such as Na-Me not so useful compared to Me-Li? Give an example of an R group that would be useful R-Na |
Metal is more electropositive so there is increased ionic bonding, but decreased covalent bonding due to it being larger and having worse orbital overlap. C5H6 as C5H5- anion stable meaning Na+(C5H5)- not so reactive |
|
|
Why are Grignard reagents (R-Mg-X) most reactive in THF or Et2O? |
Smaller aggregates and can only react in dry solvents as very sensitive to air and water |
|
|
Why can the presence of MgX2 increase the reactivity of a substrate? |
Can act as a Lewis acid catalyst. E.g. lone pairs on oxygen attack Mg and one of the X leaves as X- |
