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304 Boron Hydrides

TABLE III Formulas and Nomenclature for
Boron Hydrides Described as Polyhedra with
Vertices Removed
Missing vertices Formula Designation
0 B x H 2− closo
x
1 B x H x+4 nido
2 B x H x+6 arachno
FIGURE 2 Removing a vertex from an octahedron produces a 3 B x H x+8 hypho
square pyramid.

in the structure and by each negative charge if the species is pentagon of this pentagonal pyramid, the formula would
2−
a negative ion. The number of electrons needed for boron be B 5 H 11 . All of these species, B 7 H ,B 6 H 10 , and B 5 H 11 ,
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hydride deltahedra and deltahedral fragments of formula exist. However, the hypho species, B 4 H 12 , has never been
B x H y is given in Table II. observed.
If no vertices are missing, there are no open edges, and
each boron atom bears only a single terminal hydrogen
atom. Thus there will be only x electrons from terminal III. BONDING THEORIES
hydrogen atoms for bonding the boron structure together
and a charge of −2 will be necessary to provide the nec- The bonding in the boron hydrides is most simply exem-
essary electrons. This result means that the chemical for- plified in B 2 H 6 . As mentioned above, there are 12 valence
mula of any species with a complete deltahedron of boron electrons, 3 from each boron atom and 1 from each hydro-
atoms will be B x H , and a neutral molecule containing gen, and this is insufficient to bond every pair of adjacent
2−
x
only boron and hydrogen atoms in a closed deltahedron atoms with ordinary single bonds. Each boron atom is
would violate Wade’s rules. connected to two terminal hydrogen atoms and these four
If one vertex is missing there will be sufficient elec- single bonds will each require two electrons leaving only
trons for the boron framework if the chemical formula four valence electrons to bond the two boron atoms to each
is B x H x+4 , if two vertices are missing the formula will other and to the two bridging hydrogen atoms. These four
be B x H x+6 , and if three are missing the formula will be remaining electrons participate in two boron-hydrogen-
B x H x+8 .ThisinformationissummarizedinTableIIIalong boron bridge bonds.
with the designation used to describe each of the four Consider one of the boron-hydrogen-boron bridge
situations. bonds. As shown in Fig. 4, the bond results from orbital
As an example, these rules can be applied to boron hy- overlap of the spherical orbital on the hydrogen atom and
drides derived from the pentagonal bipyramid (Fig. 3). a directed orbital from each of the two boron atoms. Since
With no boron atoms missing from the deltahedron there we are combining three atomic orbitals for the bridge, it
would be seven boron atoms and the formula would be is necessary that we form three molecular orbitals from
2−
B 7 H .Iftheboronatomdepictedatthebottomofthepen- them. The energy diagram for the atomic and molecular
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tagonal bipyramid is removed, the boron atom structure orbitals is given in Fig. 5, which shows a bonding molec-
is a pentagonal pyramid, and the chemical formula would ular orbital at an energy lower than the atomic orbitals,
be B 6 H 10 . If one boron atom is removed from the base and antibonding molecular orbital at a higher energy than
the atomic orbitals, and a nonbonding molecular orbital at
about the same energy as the atomic orbitals. A bonding
TABLE II Calculated Numbers of Electrons Needed to be situation occurs when two electrons occupy the lowest en-
Supplied by Hydrogen Atoms or Negative Charges for Dif- ergy orbital, and the drop in potential energy can result in a
ferent Geometries of the Boron Atom Structure Described
as Polyhedra with Vertices Removed stable molecule. Occupation of the nonbonding molecular
Number of missing vertices Electrons needed from H atom
in boron hydride B x H y and negative charges
0 x + 2
1 x + 4
2 x + 6
3 x + 8
FIGURE 3 Removing vertices from the pentagonal bipyramid.