In groups 1, 2, 13 and 14, the melting and boiling point decreases down the group with a few exceptions. In group 15 the melting/boiling point increases up to Arsenic and then started decreasing. In groups 16, 17 and 18 it increases down the group, but with few exceptions.

Groups 1 and 2 are held by metallic bonding and the strength of metallic bonding decreases down the group, so this explains the first two trends. Carbon doesn't actually melt or boil, it sublimes but the point still stands that it's sublimation point is much higher than Silicon's boiling point.

Oxygen exists as $\ce{O2}$ and sulphur as $\ce{S8}$, so this explains the large difference in boiling points, but atomic orbitals become more diffuse down the group, so the bond strength and thus, the melting points should decrease but that doesn't happen.

Almost all other periodic trends like ionisation enthalpy or density are fairly consistent across the groups. Why is this different? I couldn't find more information anywhere. What's happening here?

  • $\begingroup$ Describe in details how you "couldn't find more information anywhere.", as it is often just a figure of speech for "I have not thought nor searched thoroughly enough, but saying that would look bad." // Who says transition of declining slopes to inclining ones cannot go through a peak shape? Who says chemistry of elements cannot affect it? $\endgroup$
    – Poutnik
    Aug 28 at 11:51
  • $\begingroup$ @Poutnik, I did search for previous questions on this site and I got vague answers like van Der Waals forces, nothing related to thisunfortunately $\endgroup$
    – Sunaina
    Aug 28 at 13:03
  • $\begingroup$ Why just this site? You know, effort not shown can be considered as effort not done. You did not think/search deeply, if chemical bonds and element chemical character did not come to your mind. For group 15, it comes from triple bond of N2 through covalent single bonds of P, As to metallic bonds of Sb, Bi. Why should one even consider monotonous trends ? $\endgroup$
    – Poutnik
    Aug 28 at 13:10
  • $\begingroup$ @Poutnik, edited to add more information like you had suggested. $\endgroup$
    – Sunaina
    Aug 28 at 13:29
  • 1

The problem is, we want to see a trend in everything. There are various factors that govern melting point and boiling point which is the reason of perceived anomalies. "The equation which has more number of variables, is harder to solve."

For melting point, few of them are: Crystal system, size of atom, atom-atom distance, distance between two layers of atoms, metallic character, metallic bond strength, inert pair effect, relativistic effect, etc.

Note that, down a group, metal-metal bond strength decreases due to increase in size but metallic character increases because now, $\ce{e-}$ abstraction is easy from that metal.

Group 1 and 2: All have metal-metal bond but the strength of bond decreases. Thus, m.p. decreases.

Group 13: Except $\ce{B}$ all are metals. The m.p. decreases from $\ce{B}$ to $\ce{Al}$ then decreases for $\ce{Ga}$ and then increases till $\ce{Tl}$.

The low m.p. of $\ce{Ga}$ is due to existance of $\ce{Ga2}$ dimers in the orthorhombic crystal. The higher m.p. of $\ce{In}$ and $\ce{Tl}$ may be due to body centered tetragonal and hexagonal close packing structures, respectively.

This explains the boiling point trend in group 13.

Group 14: Except $\ce{C}$, $\ce{Si}$ and $\ce{Ge}$, all ($\ce{Sn}$ and $\ce{Pb}$) are metal. The m.p. decreases till $\ce{Sn}$ and then increases for $\ce{Pb}$.

$\ce{Ge}$ has face centered diamond cubic lattice, $\beta~\ce{Sn}$ has body centered tetragonal lattice and $\ce{Pb}$ has face centered cubic lattice. The higher m.p. of $\ce{Pb}$ than that of $\ce{Sn}$ may be due to this factor and a lower van der Waals radius.

In group 15 the melting/boiling point increases up to Arsenic and then started decreasing.

Group 15: The m.p. increases upto $\ce{Sb}$ then $\ce{Bi}$ has a low melting point. $\ce{As}$ sublimes at $\pu{887 K}$ and $\pu{1 atm}$.

$\ce{As}$, $\ce{Sb}$ and $\ce{Bi}$ all have layered structure crystalising in rhombohedral lattice. The atomic radius of $\ce{Sb}$ and $\ce{Bi}$ are very similar. The low m.p. may be due to higher atom-atom and layer-layer distance than those in $\ce{Sb}$ as explained here.

Group 16: The m.p. increases upto $\ce{Te}$ then $\ce{Po}$ has a low melting point.

$\ce{S}$ is orthorhombic, grey $\ce{Se}$ is hexagonal, $\ce{Te}$ is rhombohedral and $\alpha~\ce{Po}$ is simple cubic.

Another reason for low m.p. and b.p. of $\ce{Po}$ is explained here.

Group 17 and 18: None of them are metals. The m.p. and b.p. increases uniformly with the increase of van der Waals forces.

I know that this is not a complete answer but you can do your own findings based on this.


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