What theory accurately explains metallic bonding in beryllium?

Question

Since beryllium is an alkaline earth metal, the bonds between beryllium atoms could be considered metallic and we can use molecular orbital theory (MOT) to explain metallic bonds in metals.

Consider metallic bonding in lithium metal. Say, $n$ lithium atoms combine to form $\ce{Li_n}$. Here $n$ atomic orbitals (AOs) combine to from $n$ molecular orbitals (MOs). Since there is only one valence electron per atom in lithium, and a MO can hold two electrons, only half the MOs are filled. As the result, all bonding molecular orbitals (BMOs) are occupied, and antibonding molecular orbitals (ABMOs) left unfilled.

I tried to explain metallic bonding in beryllium metal in a similar way. But the results are weird. This is because all the $n$ MOs (formed after combining $n$ beryllium atoms) get completely filled with electrons. As there are equal number of electrons in BMOs and ABMOs, this type of bonding is not possible in beryllium metal. A similar explanation is used to describe the instability of the $\ce{He2}$ molecule.

According to MOT, beryllium metal $\ce{Be_n}$ shouldn't exist. But we know that beryllium is metal exists with strong metallic bonds. Isn't a contradiction to MOT?

How are beryllium atoms in a metal are linked? Is there any new theory to explain this or is there any concept I missed in MOT?

Answer 1

The usual MOT explanation for why the group IIA elements are metals is that the band caused by the overlap of the np orbitals is sufficiently wide to overlap that due to the ns orbitals (the one you describe above). As the bands overlap there is zero gap, and hence the elements are metals.

To test this proposal I have done a quick calculation for Be with CRYSTAL, an ab initio code that uses a local basis set (i.e. something like atomic orbitals) for calculations on periodic systems such as crystals. It's not the most wonderful calculation ever, it's only LDA for instance, but it should be enough to show what is going on.

Here is the Density of States in the region of the Fermi Energy (the energy scale is shifted so the Fermi Energy is at zero), along with the projected density of states onto the s type functions in the calculation, and a similar projection onto the p type functions:

You can clearly see that the lower band is predominantly of s type character, and the upper is mostly p, with a little mixing. There is also clearly a dip at the Fermi energy.

Zooming into the region of the Fermi energy shows us the below:

It is clear that the Total DoS is not zero at the Fermi energy because the "p band" just overlaps with the "s band". Thus there are unoccupied states at and infinitesimally higher energy than occupied ones. Thus Be is a metal because the unoccupied p band is wide enough to overlap with the fully occupied s band.

The input file for the SCF calculation that I used is below. The basis set is Be_pob_TZVP_rev2

BERYLLIUM 
CRYSTAL
0 0 1
194
2.29  3.59
1
4   0.333333333333 0.666666666667 0.25
END
4 5
0 0 6 2.0 1.0
 4700.2365626        0.00023584389316
 704.82845622        0.00182437910190
 160.43110478        0.00939661482240
 45.425347336        0.03690892415900
 14.798334125        0.10897561281000
 5.3512452537        0.21694284551000
0 0 2 2.0 1.0
 2.1542044819        0.44695408857000
 0.9336374440        0.20866985771000
0 0 1 0.0 1.0
 0.3017450800        1.00000000000000
0 0 1 0.0 1.0
 0.1411145200        1.00000000000000
0 2 1 0.0 1.0
 0.6526215700        1.00000000000000
 99 0
END
DFT
XLGRID
EXCHANGE
BECKE
CORRELAT
PZ
END
SHRINK
12 24
TOLINTEG
12 12 12 12 16
MAXCYCLE
150
NODIIS
FMIXING
50
TOLDEE
8
END

And for the properties calculation

BASISSET
0
NEWK
12 24
1 0
DOSS
2 200 -1 -1 1 12 0
-0.5 0.75
8 1 2 3 4 8 9 10 11
6 5 6 7 12 13 14 
END