3
$\begingroup$

As I know, the melting points of Alkali metals decrease down to the group due to decreasing of the energy of the metallic bond.

enter image description here

But according to my understanding, after it has become to liquid the metallic bonds are not affecting furthermore because of the lattice has destroyed, so the boiling point affects only the atomic mass. Then according to my logic the boiling points should be increase down to the group. But the above graph shows the opposite of that.

What is the wrong in my logic. Please help me.

$\endgroup$
  • 2
    $\begingroup$ A molten metal is still a metal, with (more or less) the same metallic interaction between the constituent atoms.What gets destroyed upon melting is the crystalline lattice. $\endgroup$ – Karl Nov 24 '19 at 8:34
1
$\begingroup$

You can explain the trend in terms of increased charge density, and ensuing charge repulsion between nuclei and between electrons as you go down the group.

An alternative explanation is provided in a libretext, which draws information from a popular educational chemistry site:

When any of the Group 1 metals is melted, the metallic bond is weakened enough for the atoms to move more freely, and is broken completely when the boiling point is reached. The decrease in melting and boiling points reflects the decrease in the strength of each metallic bond.

The atoms in a metal are held together by the attraction of the nuclei to electrons which are delocalized over the whole metal mass. As the atoms increase in size, the distance between the nuclei and these delocalized electrons increases; therefore, attractions fall. The atoms are more easily pulled apart to form a liquid, and then a gas. As previously discussed, each atom exhibits a net pull from the nuclei of +1. The increased charge on the nucleus down the group is offset by additional levels of screening electrons. As before, the trend is determined by the distance between the nucleus and the bonding electrons.

Although the two perspectives (my opening explanation and that in the libretext) would seem to be equivalent, they aren't identical, since a description in terms of Coulomb's law is more fundamental and does not require invoking concepts such as polarizability or moments. I would argue that appealing to nuclear screening to explain bonding is not the best way to understand the trend.

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.