How do you explain metallic bonding in terms of the sparsely populated outermost orbitals of metal atoms?

Question

Jeremiah Adams · Accepted Answer

By #"electronic delocalization"# and consequent #"metallic bonding"#...........

A metal consists of a close-packed array of metal atoms that EACH contribute ONE or TWO or THREE (or more) electrons to the lattice. The result? Metallic bonding, which is aptly described as #"positive ions in a sea of electrons"#. Now the positively charged nuclear cores of the metal atoms can move with respect to each other without disrupting the electrostatic bond with the electronic cloud. And this gives rise to (i) #"malleability"#, the ability to be beaten out into a sheet; (ii) #"ductility"#, the ability to be drawn out into a wire; and (iii) generally good #"thermal"# and #"electrical conductivities"#, which are common characteristics of metals. Alkali, and alkali earth metals with only ONE or TWO electrons to contribute the metallic lattice, tend to be soft, and of low melting point. Metals with many valence electrons to contribute tend to be very high melting, and possess good electrical and thermal conductivity.

Cooper Anderson · Answer

undefined Electrons from the sparsely populated outermost orbitals of metal atoms delocalize during metallic bonding, forming a "sea of electrons" that are free to move within the metallic lattice and forming a bond.