bonding in electron deficient molecules

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    4. Bonding in electron deficient molecules

    4.1 Diborane

    Diborane, B2H6, is just one of a large series of boranes of the general formula BxHy. Many borane anions ofthe form BnHn

    2 also exist. There is an extensive chemistry of boranes and their anions, which includessubstitution of boron atoms by carbon and metals. These molecules are notable in that there are insufficientelectron pairs for every connection to be a conventional 2 electron bond. To understand and predict thestructures and chemical properties of these electron deficient molecules, it is necessary to investigate theseemingly unique nature of the bonding.

    The molecule possesses 12 valence electrons (3 from each B and 1 from each H) or 6 pairs. There are 8 B-Hconnections in the picture above so that there appears to be insufficient electrons to hold the moleculetogether.

    In the treatment below, it will be assumed that the B-H t bonds are 'normal' 2 electron bonds. The geometryaround B is approximately tetrahedral so that it will also be assumed that it is sp 3 hybridized. These

    approximations greatly simplify the MO treatment as only the bonding in the bridging region needs to beconsidered. Any overlap between the bridging and terminal fragments is neglected.

    (a) Symmetry of fragment orbitals

    The axes and valence orbitals to be considered in diborane are:

    The molecule has D2h symmetry. The representations generated by the four boron sp3 hybrids and by the

    two hydrogen 1s orbitals must be obtained:

    D2h E C2(z) C2(y) C2(z) i (xy) (xy) (yz)B sp3} 4 0 0 0 0 0 4 0{H 1s} 2 0 0 2 0 2 2 0

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    Both can be reduced quickly "by eye". Note that both have 0 for the i operation so that they must consist of

    an equal number of g and u irreps. For {B sp3}, the +4 for the (xz) operation means that the only

    reduction is,

    {B sp3} = ag + b2g + b1u + b3uFor {H 1s}, the behaviour under the C2 axes can only be produced by,

    {B sp3} = ag + b3u

    (The reduction formula also gives these results, of course, as may be readily checked).

    (b) Construction of SALCs

    The projection operator technique will be used to generate the combinations of the B hybrids. The effect ofthe symmetry operations on one of the orbitals is first determined:

    D2h E C2(z) C2(y) C2(z) i (xy) (xy) (yz)

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    B1 B1 B3 B4 B2 B4 B2 B1 B3

    The ag, b2g, b1u and b3u orbitals can now be generated by multiplying the characters for each by the effect ofeach operation in the table.

    (ag) = 1(B1) + 1(B3) + 1(B4) + 1(B2) + 1(B4) + 1(B2) +1(B1) + 1(B3)

    = 2B1 + 2B2 + 2B3 + 2B4

    (ag) (B1 + B2 + B3 + B4)(b2g) = 1(B1) -1(B3) + 1(B4) -1(B2) + 1(B4) -1(B2) +1(B1) -

    1(B3)= 2B1 - 2B2 - 2B3 + 2B4

    (b2g) (B1 - B2 - B3 + B4)(b1u) = 1(B1) +1(B3) -1(B4) -1(B2) -1(B4) -1(B2) +1(B1)

    +1(B3)= 2B1 - 2B2 + 2B3 - 2B4

    (b1u) (B1 - B2 + B3 - B4)(b3u) = 1(B1) -1(B3) -1(B4) +1(B2) -1(B4) +1(B2) +1(B1) -

    1(B3)

    = 2B1 + 2B2 - 2B3 - 2B4

    (b3u) (B1 + B2 - B3 - B4)

    These SALCs are drawn below.

    To the right of each is drawn an orbital of the appropriate symmetry on an imaginarycentral atom (s for ag,dxz for b2g, pz for b1u and px for b3u). The symmetries match. This illustrates that matching with orbitals on

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    the central atom can be used even when there is not a real atom at the centre.

    For the H combinations, this technique will be used in preference to the projection operator:

    (c) Construction of MO diagram

    An MO diagram for the bridging region can now be constructed:

    The ag and b3u SALCs of the two fragments overlap to form bonding and antibonding MOs. There are noorbitals in the {H} fragment to match the {B} b2g and b1u SALCs which thus remain unchanged and are non-

    bonding.

    Of the 12 valence electrons in B2H6, 8 are used to form the B-Ht bonds leaving 4 electrons for the bridging

    region. These occupy the 1ag and 1b3u bonding orbitals drawn above. Note that the molecule has just theright number of electrons to maximize the bonding for the shape it adopts. If the 'electron deficient'

    molecule possessed any more electrons, these would occupy the non-bonding orbitals and there would be noextra bonding. Each B achieves an octet through making 2 B-Ht bonds and the two bridging bonds. Theshape of diborane allows B to make maximum use of its four valence orbitals.

    The total bond order for the bridging region is 2. This may be divided up into one for each B-H-B bridge. Thebond is then described as a 3-centre, 2-electron (3c2e) bond:

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    where the usual symbol of a line has been used for the 2c2e B-H bonds and a 'Y' has been used for the 3c2eB-H-B bonds.

    4.2 B6H62-

    This is an example of a borane anion. It is octahedral in shape, with one (terminal) hydrogen attached toeach boron.

    sp-hybrid orbitals on each boron are used; one is directed at the terminal H to form a 2c2e B-Ht bond. Theother hybrid is directed towards the centre of the octahedron. The B also has two p-orbitals. The inwarddirected sp-hybrid and two p-orbitals on each B are used to bond the cage together:

    (For convenience, the p-orbitals are directed along the edges of the octahedron - although any twoorthogonal directions perpendicular to the sp hybrid could be used). These fall into two sets: {B sp}consisting of 6 hybrids and {B p} consisting of 12 orbitals (two on each of the B atoms). Therepresentations generated by these two sets must now be obtained. (Note that the {B sp} set behavesidentically to the {F sp3} set in SF6).

    Oh E8C33C2'6C43C2''i6S46S6h v

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    {Bsp}

    6 0 0 2 2 0 0 0 4 2 a1g + eg + t1u

    {B p} 12 0 0 0 -4 0 0 0 0 0 t1g + t2g + t1u + t2u

    The SALCs for each set are now obtained using the methods described above. (Note that those found for the{F sp3} hybrids in SF6 can be used for the {B sp}:

    One combination from each of the t representations is shown. The others may be found by rotating the oneshown by 90 in two directions. A qualitative MO diagram can then be constructed by noting that only thea1g, t2g and t1ucombinations are bonding. There are 14 electrons to be placed in these orbitals (3 per B, 1 perH and a 2 charge gives a total of 26 electrons and 12 are used for the 6 B-H tbonds). The 7 bondingmolecular orbitals are exactly filled.

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    4.3 Other BnHn2- anions

    Borane anions of this form are known for n = 6 12, and in each case, the anion has the structure of a n-vertex polyhedron. These complete, closed polyhedra are known as closo-boranes. An MO analysis showsthat are (n + 1) bonding type molecular orbitals for n atoms arranged in an n-vertex polyhedron. A B nHn

    2-

    anion has,

    3n electrons from the Bn atoms + 1n electrons from the Hn atoms + 2 electrons

    from the charge

    = 4n +2 electrons or 2n+1 electron pairs.

    Of these, n pairs are required for the B-Ht terminal bonds. This leaves n+1 electron pairs for bonding thecage. The shape adopted leads to the cage bonding molecular orbitals being exactly filled. This is what isbehind Wade's rule. This rule states that

    a borane with (n + 1) electron pairs for boron cage bonding will have a structurebased on an n-vertex polyhedron.

    Wades rule is equally applicable to other boron hydride species.

    4.4 Wades rule in action

    B5H9 has the structure,

    As can be seen, the shape adopted is an octahedron with one vertex missing. Clusters based on a n-vertexpolyhedron with one vertex missing are known as nido-boranes. The missing vertex has been replaced by4 bridging H atoms. The B5H9 cage has

    53 electrons from the B atoms + 91 electrons from the H atoms= 24 electrons or 12 pairs

    Each of the 5 B-Ht bonds requires one pair leaving 7 pairs to bond the cage (B-B and B-H-B).

    Wades rule predicts that the structure is based on a (7 1) = 6 vertex polyhedron. It is indeed based on anoctahedron.

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    B5H9 has C4v symmetry. The four bridging H atoms lie around the base of the pyramid (the "basal plane").The SALCs for the four H 1s orbitals can be generated in the normal way:

    The basis orbitals of the "missing" B-Ht unit are shown on the right-hand side. a1 and e SALCs of the "extra"

    hydrogen 1s orbitals in B5H9 are equivalent (in symmetry) to the basis provided by one of the B-Ht units inB6H6

    2- The b2 SALC of basal H 1s orbitals is unable to overlap with the orbitals of the rest of the cage and isessentially nonbonding. B5H9 and B6H6

    2- have equivalent basis orbitals and electron pairs for cage bonding.

    Wade's rule can be used to predicted borane shapes. The B4H10 cage has,

    43 electrons from the B atoms + 101 electrons from the H atoms= 22 electrons or 11 pairs

    Each of the 4 B-Ht bonds requires one pair leaving 7 pairs to bond the cage. If (n+1) =7 then the structure

    is based on an n=6 vertex poyhedron - the octahedron. There are only 4 B atoms so 2 vertices have toremoved:

    Note that the second BH unit removed is next to (rather than opposite) the first. The extra H atoms pointtowards the 2 missing vertices replacing the B orbitals:

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    Clusters based on an n-vertex polyhedron with two vertices removed are known as arachno-boranes.

    4.5 Other clusters

    These ideas can be extended to other clusters. The most important heteroboranes arethe carboranes which contain a number of C-H groups in place of B-H. For example, reaction of B5H9 withacetylene at 500 c gives C2B3H5. This molecule has,

    24 electrons from C + 33 electrons from B + 51 electrons from H= 22 electrons or 11 pairs

    Each of the two C-Ht and three B-Ht bonds requires a pair of electrons leaving 6 pairs for cage bonding.

    If (n+1) = 6 then the structure is based on a 5-vertex polyhedron. As the molecule has 5 B and C atoms,the predicted shape is a closo-trigonal bipyramid (A).

    In main group metal clusters such as P4 and Sn52-, the same method is used and it is assumed that each

    metal has a lone pair in place of a B-Ht bond:

    P4 has 45 electrons = 20 electrons or 10 pairs. If each P has a lone pair then this leaves 6 pairs for cluster

    bonding: n+1 = 6 implies a structure based on the n = 5 polyhedron. As there are only 4P atoms, thestructure is a trigonal bipyramid with one vertex removed (B).

    Sn52- has 54 electrons from Sn + 2 from the charge = 22 electrons or 11 pairs. If each Sn has a lone pair

    then there are 6 pairs for cluster bonding: n+1 = 6 again. As there are 5 Sn atoms, the structure is a closotrigonal bipyramid (C)