Triels

Group 13 elements They all have ₙs² ₙp¹ configuration

• 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

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

• 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)

• 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

• 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)

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

• 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)

Central boron has a tetrahedral shape and is sp³ hybridised

• 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

• 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

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

• 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

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

• 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

• 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

• 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

• 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

• 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)

• 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

• 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

• 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

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