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Y2 Inorganic Chemistry > Triels > Flashcards

Triels Flashcards

1
Q

The Triels are

A

Group 13 elements
They all have ₙs² ₙp¹ configuration

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2
Q

Why is the chemistry of triels dominated around their electron deficiency?

A
  • They all have 3 valence electrons
  • So these elements can only ever have a minimum valance of three (3 covalent bonds)
  • So can only have 6 electrons around them
  • Which is short of a full ocetet
  • Therefore they are electron pair acceptors or lewis acids
  • e.g. BF₃ gaining a LP from diethyl ether
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3
Q

What is the term for the following reaction?
3MX → 2M + MX₃

A
  • Disproportionation
  • two of the atoms are reduced (+1 to 0)
  • one atom is oxidised (+1 to 3+)
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4
Q

Why does the +1 oxidation state become more stable as the group descended

A
  • Inert pair effect
  • (the tendency of the electrons in the outermost atomic s orbital to remain unionised or unshared in compounds of the group 13-16 elements)
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5
Q

What happens to the size of the atom radius (pm) as you go down group 13?

A
  • The atomic radii increases going down group 13
  • This effect results from the fact outer electrons are being places in orbitals with increasing principial quantum numbers
  • Hence lie further away from the nucleus
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6
Q

What happens to the first ioniation energy (kJ/mol) as you go down group 13?

A
  • Aluminium is lower than boron due to the outer electrons being further away from the nucleus, hence experiencing less nuclear charge
  • However it stabilises from aluminum
  • Ga, In and Tl are proceeded by d-block and f-block electrons which do not provide effective shielding (d-block and f-block contraction)
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7
Q

What are the most common oxidation states for group 13 elements?

A
  • +3 is the most common oxidiation state
  • Apart from Thallium where it is +1 due to the inert pair effect
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8
Q

The complex shown below is borax
What is the hybridisation of the boron unit on the left and how does this enable extra stability of the complex?

A
  • the boron has a trigonal planar shape, hence is sp² hybridised
  • This means it has a empty p-orbital
  • This allows for oxygen to back donate some of its electron density (pπ-pπ interaction)
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9
Q

The complex shown below is borax
What is the hybridisation of the central boron unit

A

Central boron has a tetrahedral shape and is sp³ hybridised

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10
Q

How can BF₃ participate in π-bonding

A
  • The boron is sp² hybridised
  • uses all 3 valance electrons to form covalent bonds with F
  • The LP on fluorine can back donate to boron (pπ-pπ) giving some partial pi-bond character
  • This strengths the bond and locks boron into an sp³ planar geometry
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11
Q

The π-bonding in BF₃ is not strong (large electronegativity difference between B and F) it still has a major influence on the properties
How?

A
  • Fluorine is the most electronegative compared to other halogens
  • Hence should be the strongest lewis acid (electron reciever) based on electron pulling ability
  • However: BI₃>BBr₃>BCl₃>BF₃ (it is the weakest)
  • This is due to π-bonding, which other larger halides cannot do due to large diffuse orbitals = poor overlap
  • The π-bonding means the boron in BF₃ is less electron deficient
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12
Q

How do the non-fluoride halides for group 13 bond, if they cannot π-bond well?

A
  • The non-fluoride halides for the other elements dimeries, e.g.AlCl₃ (to Al₂Cl₆)
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13
Q

What is the product of the following reaction?

A
  • We’ll firstly form a lewis acid-based addut (now tetrahdral with 4 bond, however a valency of 3)
  • HX can then be eliminated (e.g. HF) and form a trigonal planar trivalent boron again
  • If there is an excess of alcohol, X can be substituted multiple times for the alcohol
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14
Q

What halide can this following reaction not occur for?

A

BF₃ resists
Due to the strong B-F bonds
There is no thermodynamic driving force for the second part of the reaction (breaking the B-F bond)
But will still act as a lewis acid like in the first step

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15
Q

Across Al, Ga, In, which has the greater stability of the acid-base (Ligand) complex for fluoride

A
  • MF₃L > MCl₃L > MBr₃L > MI₃L
  • (this is opposite if M was boron)
  • The reason for this is pπ-pπ back-donation can no longer occur as the valance orbitals are too far away (hence longer M-X bonds)
  • Therefore the trend is based soely on electronegativity
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16
Q

What is Borane?
And what is the bonding and structure like?

A
  • It is a boron hydride complex
  • B cannot π-bond with H (no lone pairs) - and the lack of electron density at boron because of this makes it a strong lewis acid
  • Hence forms a dimer
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17
Q

What is special about the planar B-H bonds in Borane

A
  • The B-H-B (banana) bonds have 3 centres with 2 electrons
  • (ususally there should be 2 centres with 2 electrons)
  • AND valance bond theory fails here. It cannot account for this bonding are there is not enough valance electrons
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18
Q

For each Boron atom in Borate, which orbitals do we have left to bond together the bridging B-H-B units?

A
  • Each boron atom is sp³ hybridised
  • For the dashed and wedged bonds 8 valence electrons are needed
  • Which 2x sp³ hybrids are used to form those bonds
  • Therefore remaining for the B-H-B bonds: 2x sp³ and 2x 1s orbitals from the bridging hydrogens
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18
Q

Which of the following orbitals from Boron and hydrogen can lead to a bonding interaction?

A
  • B₁ and S₁ are of the same symmetry (in-phase and out-of-phase additions lead to bonding and antibonding MOs)
  • B₃ and S₂ are of the same symmetry (in-phase and out-of-phase additions lead to bonding and antibonding MOs)
  • B₂ and B₄ do not have complimentary symmetry to either S₁ or S₂ group orbitals (leading to non-bonding MOs)
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19
Q

How many electrons are left to feed into the B-H-B planar (banana) bonds
(hint: 12 valance electrons in total (6 from B’s; 6 from H’s))

A
  • valance electrons used up in forming the four terminal B-H bonds
  • Therefore 4 electrons to feed into the B-H-B units
  • 4 B-H bonds - each has a bond order of 1/2 - 2 electrons delocalised across each B-H-B bond
  • 3-centre 2-electron bond
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20
Q

What happens when you add a BH₃ dimer to other BH₃ units?

A
  • Weak B-H-B bonds make the BH₃ dimer a good starting material for the preperation of BH clusters, or boranes
  • With the clusters having both regular and ‘banana’ B-H bonds
21
Q

There are extensive covalent cluster chemistry of the boron hydrides that is not paralleled with heavier group 13 elements
This is because hydrides become less stable as the group is descended
Why?

A
  • Larger valence orbitals = less effective sharing electron density between nuclei = weaker bond
  • e.g. gallene (GaH₃) decomposes at room temperature and InH₃ and TlH₃ are too unstable to exist unless coordinated
22
Q

The following compound shown below is more stable than Al₂H₆, why?

A
  • The Et groups are electron releasing (+I effect)
  • Which stabilises electron deficient Aluminium
23
Q

Balance the following equation

A
24
Q

Balance the following equation

A
25
Q

What are Wade’s Rules?

A

A system based on Molecular Orbital Theory for predicting the structure of a boron hydride BₙHₘ (or anions BₙHₘ⁻) by working out the number of pairs of electrons for cluster bonding
- known as polyhedral skeletal electron pairs (PSEPs)

26
Q

We use Wade’s rules to explain the structure of Boron hydrides:
1) How many electrons does each BH unit donate to cluster bonding
2) How many electrons does each bridging hydrogen atom (bridging) donate to cluster bonding

A

1) Each BH unit donates two e- (one pair) to cluster bonding
2) Each hydrogen atom (bridging) donates one e- to cluster bonding

27
Q

In Wades rules:
If we have n pairs of electrons the structure is based around…

A

…a polyhedron with n-1 vertices
(e.g. if we have 7 PSEPs then the structure is based around a 6 vertex polyhedron = octehedron)

28
Q

In Wade’s rules:
If the number of B-H units is equal to the number of vertices then….

A

Place on B-H unit at each vertex → the structure is described as CLOSO (closed)

29
Q

In Wade’s rules:
If the number of B-H units is one fewer than the number of vertices then…

A

Place B-H units at all but one of the vertices → the structure is described as NIDO (nest)

30
Q

In Wade’s rules: If the number of B-H units is two fewer than the number of verticies then…

A

Fill all but two of the vertices (the two vertices being adjacent) → the structure is described as ARACHNO (cobweb)

31
Q

For B₆H₆²⁻ it has 7 PSEPS
What is the shape of the polyhedron formed and the name?

A

7 PSEP = 7 -1 vertices = 6 vertices means cluster shape based on an octahedron (8 faces)

32
Q

For B₅H₉ it has 7 PSEPS
What is the shape of the polyhedron formed and the name?

A

7 PSEP = 7-1 vertices = 6 vertices means cluster shape based on an ocetahedon (8 faces)

33
Q

For B₄H₁₀ it has 7 PSEPS
What is the shape of the polyhedron formed and the name?

A

7 PSEP = 7-1 vertices = 6 vertices means cluster shape based on an ocetahedon (8 faces)
This complex is most electron deficient

34
Q

What is the shape of the polyhedra formed when n-1 = 4 vertices

A

Tetrahedron

35
Q

What is the shape of the polyhedra formed when n-1 = 5 vertices

A

Trigonal bipyramid

36
Q

What is the shape of the polyhedra formed when n-1 = 6 vertices

A

Ocetahedron

37
Q

What is the shape of the polyhedra formed when n-1 = 7 vertices

A

Pentagonal bipyramid (most common polyhedra for cluster chemistry)

38
Q

What is the shape of the polyhedra formed when n-1 = 8 vertices

A

Dodecahedron

39
Q

What is the shape of the polyhedra formed when n-1 = 9 vertices

A

Tricapped trigonal prism

40
Q

What is the shape of the polyhedra formed when n-1 = 10 vertices

A

Bicapped sqaure antiprism

41
Q

What is the shape of the polyhedra formed when n-1 = 11 vertices

A

Octadecahedron

42
Q

What is the shape of the polyhedra formed when n-1 = 12 vertices

A
43
Q

How do you work out the number of PSEPs for B₆H₆²⁻

A
  • Work out how many B-H units can be formed: in this case 6x B-H = (BH)₆²⁻
  • 2 electrons per B-H unit = 2x6 = 12
  • +2 from the 2- charge
  • 12+2 = 14 electrons = 7 pairs (7 PSEPs)
44
Q

How do you work out the number of PSEPS for B₄H₁₀

A
  • Work out how many B-H units can be formed: 4x B-H = (BH)₄(H)₆
  • 2 electrons per B-H unit = 2x4 = 8
  • 1 electron per H = 1x6 = 6
  • 8+6 = 14 electrons = 7 pairs (7 PSEPs)
45
Q

B₄H₁₀ has 7 PSEPs but only 4x B-H units
What is the shape of the polyhedron?

A

This is two fewer B-H units than the number of vertices so we place one BH fragment at each of four vertices and leave two vertices empty
The structure is described as ARACHNO
The two vertices chose to be vacent are adjacent (less strained)

46
Q

Wade’s rules can be extended to predict the structures of carboranes - molecules in which one or more of the B atoms has been replaced by carbon
Because of this B-H and ….

A

⁺C-H are interchangeable in clusters

47
Q

How do you work out the PSEPs for C₂B₃H₅

A
  • Work how many B-H and ⁺C-H units can be formed: 2x C-H and 3x BH = (CH⁺)₂(BH)₃²⁻
  • 2- charge as 2x (CH⁺) units
  • 2 electrons per (CH⁺) unit = 2x2 = 4
  • 2 electron per (BH) unit = 3x2 = 6
  • +2 from -2 charge
  • 4+6+2 = 12 = 6 pairs = (6 PSEPs)
48
Q

What is the shape of the polyhedron formed for C₂B₃H₅ if there is 6 PSEPS

A
  • Counting up the BH and CH units in C₂B₃H₅ we have 5
  • This is the same as the number of vertices (6-1) so we place one fragement at each vertex and the structure is CLOSO
  • But due to having different units there are 3 distinct ways to place the units
49
Q

Where is it most stable two place two carbon units in a carborane

A

For most carboranes, the most thermodynamically stable isomer is generally that with two carbons apart

Y2 Inorganic Chemistry (19 decks)

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