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Year 3 Chem > Covalent bonding between the elements > Flashcards

Covalent bonding between the elements Flashcards

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1
Q

What is a main group cluster?
(Page 4)

A

Species with 3 dimensional shapes and direct element-element bonds

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2
Q

What is a naked cluster?
(Page 4)

A

A cluster that has only core atoms and no substituent

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3
Q

What do the lines represent in a cluster?
(Page 4)

A

Connectivity, they don’t necessarily represent a 2 centre 2e- bond

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4
Q

How are borane clusters produced?
(Page 5)

A

Different conditions and reagents will produce different borane clusters. Specifics can be found on page 5

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5
Q

What types of reactions can borane clusters undergo?
(Page 6)

A

Combustion
Hydrolysis
Electrophilic substitution
Base-induced degradation
Deprotonation reactions
Details of these reactions can be found on page 6

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6
Q

How do you calculate the total valence electron count (TVEC) of a cluster?
(Page 7)

A
  • Add up the number of valence electrons on the core atoms, eg. if there are 4 phosphorous atoms then the valence electrons on the core atoms = 4 x 5 = 20
  • Substituents count as one electron, tBu and H both count as one electron
  • Add the charge, one electron for each negative charge, remove one electron for each positive charge
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7
Q

How do you calculate the skeletal electron count (SEC) (the number of electrons that contribute towards the bonding of the cluster core)? (Page 8)

A

For clusters where each atom is bonded to three others, SEC = TVEC - 2n (n = number of cluster vertices)
eg. for [B6H6]2- SEC = 26 - (2 x 6) = 14

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8
Q

How is the number of skeletal electron pairs (SEP) calculated?
(Page 8)

A

SEP = SEC / 2

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9
Q

What is the TVEC, SEC and SEP of an electron precise cluster? (Page 8)

A

TVEC = 5n, SEC = 3n, SEP = 3n/2
Each vertex donates 2 electrons to each atom it’s connected to in the cluster (3) and 2 to the substituent its connected to (if all bonds in the cluster are 2c 2e). This explains why the TVEC is 5n if the cluster is electron precise.

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10
Q

What is the TVEC, SEC and SEP of an electron deficient cluster? (Page 8)

A

TVEC < 5n, SEC < 3n, SEP < 3n/2, there are too few electrons for 2 centre 2 electron bonds. In general electron deficient clusters have more closed structures.

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11
Q

What is the TVEC, SEC and SEP of an electron rich cluster? (Page 8)

A

TVEC > 5n, SEC > 3n, SEP > 3n/2, have more electrons than needed for 2 centre 2 electron bonds. In general electron rich clusters have open structures

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12
Q

What is Wade’s rule?
(Page 9)

A

For an n vertex cage with n + 1 SEP then a closo structure (such as octahedral geometry) will be adopted

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13
Q

What is the ratio of sigma orbitals to pi orbitals of the bonding orbitals in [B6H6]2- ?
(Page 10)

A

1 strongly bonding symmetric combination of sigma orbitals and 6 (= n) less strongly bonding combinations of the pi orbitals. In general for an n-vertex, closo polyhedron there are n + 1 bonding orbitals, with the + 1 originating from sigma orbitals

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14
Q

What structures would you expect for clusters with n + 1, n + 2, n + 3 and n + 4 SEP
(Page 11)

A

n + 1 = closo
n + 2 = nido
n + 3 = arachno
n + 4 = hypno

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15
Q

How many vertices does a nido cluster have? (Page 12)

A

SEP - 1

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16
Q

To obtain the structure of nido, arachno and hypno clusters, how many vertices do you remove from the parent cluster? (Page 12)

A

One vertex is removed if it is a nido cluster, 2 if it’s an arachno and 3 if it’s a hypno cluster

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17
Q

If vertices are different how do you choose which vertices are removed when creating the new cluster from the parent closo cluster? (Page 12)

A

The most connected vertex is removed

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18
Q

Where are hydrogen atoms placed in the new cluster deriving from the parent closo cluster? (Page 12)

A

In bridging sites along B-B edges and in terminal sites, eg. BH –> BH2 if available

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19
Q

How is cluster geometry determined? (Page 14)

A
  • Determine vertices (n), TVEC and SEP
  • Assess if closo, nido, arachno or hypno
  • Determine parent closo structure
  • Remove vertices as appropriate
  • Add hydrogen atoms
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20
Q

Why is the isoarachno structure not favoured even though it derives from removal of the most connected vertices? (Page 14)

A

Because the isoarachno structure that is formed prefers a planar structure as it’s aromatic, however it can’t achieve a planar structure so it is unfavourable to form the isoarachno structure

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21
Q

Why does BH- act the same as CH in a cluster? (Page 15)

A

Because they have the same number of electrons lying in bonding orbitals:

C gives 4 e-, H gives one = 5, - 2 for C-H bond
B gives 3 e-, H gives 1, -1 charge gives 1 = 5, -2 for B-H bond

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22
Q

What is an isolobal relationship? (Page 18)

A

Two fragments are isolobal if they share the same symmetry, approx. same energy and the same number of electrons in their frontier orbitals

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23
Q

If you remove CO+ from a complex, what must be done to negate the negative charge? (Page 19)

A

The metal must be replaced with the one space along in the period. This metal has one more proton so the negative charge is negated. Also it doesn’t need the negative charge to obtain 17VE: Cr(CO)5- = 6 + 10 + 1 = 17, Mn(CO)5 = 7 + 10 , it can achieve 17VE without the negative charge because it has an extra d electron

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24
Q

Why is BH isolobal with Os(CO)3? (Page 20)

A

CH is isolobal with Co(CO)3, hence CH+ is isolobal with Co(CO)3+, we know that BH- behaves identically to CH hence BH can be replaced with CH+, Co(CO)3+ behaves the same as Fe(CO)3 as Fe has one less electron than Co, so losing an electron from Co gives either Co+ or Fe. It also has one less proton so the positive charge is negated.

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25
Q

What is the best way to calculate TVEC on complexes that have singular electrons shown in orbitals on metal? (Page 20)

A

Ignore the electrons in the orbitals and calculate the TVEC as normal. The electrons in the orbitals tell you how many electrons there are in the frontier orbitals.

26
Q

What is the rule for determining SEC for a transition metal cluster? (Page 22)

A

TVEC - 12n

27
Q

What structures would you expect from clusters with n, n - 1, n - 2 and n - 3 SEP? (Page 23)

A

n = monocapped (based on closo structure with n - 1 vertices)
n -1 = bicapped (based on closo structure with n - 2 vertices)
n - 2 = tricapped (based on closo structure with n - 3 vertices)
n - 3 = tetracapped

28
Q

What happens to the bond strength of transition metal-ligand bonds as you go down the group? (Page 26)

A

They generally get stronger as you go down the group, 3d orbitals are quite core like hence limited overlap with ligand orbitals so M-L bonds are quite weak

4d/5d orbitals have greater radial distributions and hence a better overlap with ligands = stronger bonds

29
Q

Where in the periodic table would you generally find elements involved in M-M bonding? (Page 26)

A

Towards the bottom

30
Q

Do M-M bonds affect oxidation state? (Page 26)

A

No, electrons are evenly shared

31
Q

How do you calculate bond order? (Page 27)

A

(Number of electrons in bonding orbital - number of electrons in antibonding orbitals) / 2

32
Q

Why is the orientation of ML4 units in [Re2Cl8]2- eclipsed? (Page 28)

A

Because one of the bonds between the elements is a delta bond which requires the ML4 units to be in line for the overlap to occur hence the eclipsed conformation is essential

33
Q

Why is the bond length of [Os2Cl8]2- shorter than that of [Re2Cl8]2- despite the only difference in their electron configurations being two additional electrons in a δ* orbital in the Os complex? (Page 28)

A

The electrons in the δ* orbital cancel out any δ character of the metal-metal bond. You would expect this to weaken (lengthen) the Os-Os bond. This isn’t the case though, δ orbitals have a very small contribution to bonding and so the loss of δ character would only weaken the bond very slightly.

You have to take into account the decreased atomic radius of Os, this outweighs the effect of decreased bond order and gives rise to a shorter stronger bond.

34
Q

Why is the staggered conformation favoured for most M2LX6 compounds? (Page 29)

A

Steric reasons, the eclipsed conformation actually has better orbital overlap.

35
Q

Why do orbitals appear to look like dx^2 - y^2 despite being derived from dxz orbitals? (Page 29)

A

They are hybrid orbitals, they have some dx^2 - y^2 mixed in.

36
Q

Why is [Cr2Cl9]3- paramagnetic when it is predicted to be diamagnetic from the MO diagram? (Page 30)

A

Orbital overlap is small in Cr complex as d orbitals increase in size down the group and Cr is at the top. This leads to longer (weaker) M-M bond and hence unpaired electrons.

37
Q

What type of bonds are the bridging CO bonds in Fe2(CO)9? (Page 31)

A

3c 2e bonds, there are not enough electrons for the bridging carbonyls to be 2 centre 2 electron bonds.

38
Q

Why are pi and delta orbitals not degenerate in the MO diagram for RCr-CrR? (Page 32)

A

Due to the position of the R group

39
Q

How do you determine the principal quantum number of an element? (Page 33)

A

The number of shells in the atom = n (principal quantum number)

40
Q

How can sterically bulky groups be used to stabilise double bonds? (Page 33)

A

If R isn’t a sterically bulky group, the pi bond will break down when R2P=PR2 comes into contact with another R2P=PR2 molecule to form a sigma bond with that molecule as this is more thermodynamically favourable.

If R is sterically bulky it inhibits two R2P=PR2 molecules from coming into contact with each other and hence the double bond is not broken meaning that it is stabilised by the bulky groups.

41
Q

Can double bonds be formed between different group 14 elements to carbon? (Page 34)

A

Yes, here it is shown that double bonds can be formed between two Sn atoms or two Si atoms

42
Q

Are double bonds between group 14 elements planar? (Page 35)

A

C=C bonds are but all other group 14 double bonds show a so-called trans-bent structure

43
Q

Can triple bonds be formed between group 14 elements other than carbon? (Page 36)

A

Yes, if the substituents are extremely bulky (to stabilise the molecule)
A Pb≡Pb bond can be formed when it is encased within organic groups, this means it can’t come into contact with other groups to break down

44
Q

How is an Si≡Si bond prepared? (Page 37)

A

It can’t be prepared in the same way as other group 14 elements so has to be prepared from a pre-existing Si-Si framework

45
Q

What is the odd thing about a Pb≡Pb bond? (Page 37)

A

It is longer than a Pb=Pb bond

46
Q

Why are the structures of compounds with double bonds between group 14 elements distorted (other than carbon)? (Page 38)

A

Mixing of π and σ* orbitals that doesn’t occur in carbon R2E=ER2 molecules.

It doesn’t occur when E = C because the energy gap between these orbitals is too large, as you go down the group it decreases and hence mixing occurs.

The mixing distorts the structure from the planar one observed when E = C hence we see non-planar R2E=ER2 molecules for all group 14 elements bar carbon.

47
Q

What is the reason for the distortion of planar structure of RE≡ER for elements other than carbon? (Page 39)

A

Identical reasoning to that for R2E=ER2

48
Q

What is a semi-conductor? (Page 40)

A

When band gap is large, we have an insulator (no electrons can be promoted) when it is small we have a conductor (electrons can move between the valence and conduction band).

Semi-conductor have a band gap in between small and large

49
Q

What is an extrinsic semi-conductor? (Page 40)

A

A semi conductor that has been doped with a donor band near to the conduction band or an acceptor band near to the valence band. This enables electrons to make the jump as energy gap is small enough

50
Q

What is the light emitted from a semi-conductor equivalent in energy to? (Page 41)

A

The band gap

51
Q

What are some properties/applications of semi-conductors? (Page 41)

A

Solar cells, blue LEDs etc. (table on this page showing extensive list of applications)

52
Q

How is pure silicon (electronic-grade) formed? (Page 42)

A

SiHCl3 is formed (process shown on this page) and this decomposes on hitting the hot stage, gaseous HCl is eliminated and deposits pure Si

53
Q

What are the requirements for the precursors of metal-organic chemical vapour deposition (MOCVD)? (Page 43)

A

Must be volatile (need to be in gas phase in CVD chamber)
Must be pure and decompose to give materials with no contamination
Stable enough to handle and prepare but decompose at correct temp in CVD chamber

54
Q

What is the general process for the formation of III/V materials? (Page 43)

A

MMe3 + EH3 → ME + 3CH4

55
Q

What is an example of a good substitute for AsH3? (Page 44)

A

AsH2tBu, bulky tBu group prevents unwanted reactions (makes it more stable)

When the molecule breaks down into 2 radicals the tBu group is one of the radicals and the AsH2 section is another one, the carbon based group being lost early on helps prevent carbon contamination in the product.

56
Q

Out of group 13 (III) alkyl compounds (MMe3) and group 15 (V) compounds (ER3) which is a Lewis acid and which is a Lewis base? (Page 44)

A

Group 13 compounds are Lewis acids and group 15 are Lewis bases.

57
Q

What are single source precursors? (Page 45)

A

Precursors containing both elements and several element-metal bonds, this increases stability and makes these precursors safe to handle.

58
Q

Is a Si-O (seen in polysiloxanes) bond strong? (Page 46)

A

Yes, it is very strong, stronger than a C-C bond

59
Q

How are polysilanes formed? (Page 47)

A

Wurtz coupling of dichlorosilanes

60
Q

Are polyethene and polysilane conductors or insulators? (Page 48)

A

Polyethene is an insulator with a large band gap. Polysilane has a much smaller band gap and can act as a semi-conductor

61
Q

Why is the band gap of polysilane much smaller than polyethene? (Page 48)

A

Sigma conjugation, orbitals are larger in Si than C and hence there is more overlap between orbitals and we begin to get bands

Year 3 Chem (5 decks)

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