Metals typically have low electronegativity, which makes them form ions easily and prefer making metallic bonds to covalent. However, some of them seem to disprove that. Take, for example, gold, lead and phosphorus (for comparison). Their electronegativity values are 2.54, 2.33 and 2.19 respectively. Phosphorus is a typical non-metal, while gold and lead should have higher electron affinity and be even stronger non-metals that phosphorus. Why isn't that true?

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    $\begingroup$ Chemistry is about emergent properties from simple components. While we have general trends, it is a mistake to assume that the trends always hold or that outliers are easy to understand. $\endgroup$ – Zhe Oct 14 '17 at 18:37
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    $\begingroup$ There are many "electronegativities" - they aren't properties like density but invented parameters which, as you should already see, are rather lame. $\endgroup$ – Mithoron Oct 14 '17 at 20:22
  • $\begingroup$ The Van Arkel–Ketelaar triangle is relevant to this question. Not only does the absolute electronegativity of an element influence metallicity, but also the difference in electronegativity between the elements in a material. See the axis labels on this diagram. That said, electronegativity is a concept which extends rather poorly to transition metals under most definitions. $\endgroup$ – Nicolau Saker Neto Oct 15 '17 at 1:07

Whether an element adopts a metallic or nonmetallic structure does not depend on electronegativity alone. Aside from electronegativity at least two factors promote formation of a nonmetallic structure:

1) If the atoms are close to filling their valence shells they are more likely to do so through covalent bonds. Iodine needs only one more valence electron per atom and can get it by forming covalently bound diatomic molecules; gold needs seven valence electron per atoms to complete the valence shell and goes the metallic bonding route instead.

2) elements in earlier periods tend to form stronger covalent bonds favoring the nonmetallic structure. Silicon form the familiar tetrahedral semiconducting structure but lead, below it with similar electronegativity and valence electron configuration, is metallic.

Other factors may enter, too, and Group 14 is an excellent place to look for subtle effects. Above I said that silicon has the tetrahedral structure we all know and love, but that is under ambient conditions. The tetrahedral structure of silicon is energetically more stable than the metallic one, but the latter has more entropy and thus silicon may be converted to a metal upon heating. In the case of silicon the temperature for this switchover is high enough so that the metal is molten, so we have simultaneous metallization with melting -- and, therefore, contraction with melting. Germanium does the same thing, whereas with lead the metallic structure is more stable all the time. Tin is a transitional case where it's a semiconductor when cold but switches to a metal below the melting point.

So there is indeed more to what makes a metallic element or a nonmetallic one than just electronegativity.

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    $\begingroup$ Thanks! BTW, is there a strict formula/algorithm which lets one determine the type of a chemical element or we need a table for that? $\endgroup$ – Василий Свинко Oct 14 '17 at 23:24
  • $\begingroup$ Probably a table is better. As you see we are dealing with a complex phenomenon to describe theoretically. $\endgroup$ – Oscar Lanzi Oct 14 '17 at 23:32
  • $\begingroup$ @ВасилийСвинко Elements has no fixed properties. Allotropes only , see eg diamond vs graphite vs carbon nanotubes. $\endgroup$ – Greg Oct 15 '17 at 4:23
  • $\begingroup$ I have a question related to the answer above. If gold needs seven valence electron per atoms to complete the valence shell, why is the electronegativity of gold high? If it is easier for gold to just lose one electron than to gain seven, isn't it right to have a lower electronegativity? $\endgroup$ – grace May 5 '18 at 10:25
  • $\begingroup$ Losing just one electron does not give a highly stable configuration. There really is much more to electronegativity than electron count. $\endgroup$ – Oscar Lanzi May 5 '18 at 23:50

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