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I really don't understand why metals form metallic bonds. I mean, it makes no sense. It would make much more sense for them to form covalent bonds with themselves and have a 'pseudo-full' outer shell. How does freeing off electrons make them any more stable?

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This is due to the low ionization energies of the metals. It's easier for them to release few electrons from the outer shell to obtain a noble gas configuration rather than consuming several ones. However, the difference between an ionic and a polar covalent bond is always fringe.

Again, it's not true that metals don't form covalent bonds at all. I guess you never heard of quadruple bond or δ-bond. There are several examples such as $\ce{K2[Re2Cl8]·2H2O}$ and Chromium(II) acetate hydrate.

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    $\begingroup$ >This is due to the low ionization energies of the metals || This is BS. silicon: ionisation energy 8 eV (nonmetallic bond); mercury: ionisation energy 10 eV(metallic bond). $\endgroup$ – permeakra Dec 1 '15 at 8:17
  • $\begingroup$ Silicon is not entirely nonmetallic, it becomes a metal when melted. The point is, what makes things metallic is not as simple and brute force as ionization energy. $\endgroup$ – Oscar Lanzi Feb 24 '18 at 14:02
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  1. Bulk d-metals and especially intermetallic compounds often do have significant covalent bonding.
  2. Strictly speaking, metal bonding is a kind of covalent bonding in a sense. It is a common knowledge, that 3- or more atoms may be bound by one pair of electrons, like $\ce{H3+}$ ion. In bulk metals, similar bonding bond is working for valence electron.
  3. Even more strictly speaking, to fully understand the matter you have to consider the matter from position of theory of molecular orbitals.
    Let's assume we have a nanocrystal of 100 3d row element atoms. From 400 orbitals of said atoms 400 molecular orbitals are formed, with strict distribution of their energies depending on shape and size of the crystal and crystal cell. Then, the only difference between metals and non-metals is that metals have semi-full group of orbitals of same energy (so called conductivity zone, while non-metals have only full and vacant orbitals.
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  • $\begingroup$ 3 is rather misleading, and in this form does not explain why metals would have such orbitals or what is special about them $\endgroup$ – Greg Nov 12 '16 at 9:08
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Certain metals under certain conditions form highly covalently bonded structures.

Certain allotropes (phases) of metals have no metal qualities such as alpha-tin. Formed at about 13.2 degrees celcius with pure tin- it has no metallic qualities and pure covalent bonds.

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The difference between covalent and metallic bonding looks much clearer when written on paper than it does in the real world.

If you take the elements from group 15 from phosphorus downwards and analyse them, you have ‘true’ covalent bonding for one allotrope of phosphorus (the white one, $\ce{P4}$) and ‘true’ metallic bonding for bismuth. The entire transition downwards from phosphorus — technically already including phosphorus’ red and black allotropes — is a transition from covalent to metallic bonding.

If you look at an orbital picture of metallic bonding, you see the clear similarity to covalent bonding, except that you aren’t dealing with discreet molecules but a huge network.

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First of all! Metal do form covalent bond. It is very common in transition metal like platinum, palladium.

However, it is not the way you are talking about. Typically, when pure metal atoms bond together, they prefer metallic bond.

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