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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?

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    $\begingroup$ 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 band gap, and hence the elements are metals. $\endgroup$
    – Ian Bush
    Apr 24 at 8:27
  • $\begingroup$ Possible duplicate of How does metallic bonding in alkaline earth metals work? $\endgroup$
    – andselisk
    Apr 25 at 14:33
  • $\begingroup$ @andselisk the question seems to look like a duplicate but it isn't. Because both the contents in the question and the answer are different. I was searching for the missing concepts in MOT that I hadn't learned. $\endgroup$
    – Infinite
    Apr 25 at 14:41
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    $\begingroup$ The fundamental problem is that solid state band structures are not molecular orbitals, period. The band structure of beryllium is described in journals.aps.org/pr/abstract/10.1103/PhysRev.133.A819 and looks nothing like what you think (particularly the 'cigar and cornet' Fermi surface). $\endgroup$
    – Jon Custer
    Apr 25 at 17:31

1 Answer 1

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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:

enter image description here

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:

enter image description here

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
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  • $\begingroup$ I didn't understand what these graphs are. But okay I got the main point i.e., p band overlaps with s band. So what actually happens after overlapping ? $\endgroup$
    – Infinite
    Apr 24 at 15:27
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    $\begingroup$ @Infinite I have a feeling VBT might be more accessible for you. TL:DR is because of s-p mixing, p orbitals are also used for bonding, while for He energy difference is too big for that. You could imagine Be wants to have octet - is this simplified enough, or too much? $\endgroup$
    – Mithoron
    Apr 24 at 17:39
  • $\begingroup$ Observation: even though the overlap at the Fermi level looks small, beryllium has a higher electrical conductivity than any other alkaline earth metal; actually the heavier congeners strontium and barium are the ones where conductivity drops significantly. $\endgroup$ Apr 25 at 12:42
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    $\begingroup$ @Infinite Thing is VBT is actually equivalent to MOT in practice - both are bases for computational chemistry. Now that I think of it, you should probably rather read about electronic band structure $\endgroup$
    – Mithoron
    Apr 25 at 15:04
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    $\begingroup$ @IanBush - the classic paper would be journals.aps.org/pr/abstract/10.1103/PhysRev.133.A819 $\endgroup$
    – Jon Custer
    Apr 25 at 17:31

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