I was looking again at the periodic trends like Ionization Energy, Electron Affinity, Radius and so on. I understood all the general trends and the exeptions for these ones, but I can't fully understand the trends for Atomization Enthalpy/Energy.

Quoting my book ( Inorganic Chemistry - Atkins ) : "[...] enthalpies of atomization first increase and then decrease across Periods 2 and 3, reaching a maximum at C in Period 2 and Si in Period 3. The values decrease between C and N, and Si and P: even though N and P each have five valence electrons, two of these electrons form a lone pair and only three are involved in bonding. A similar effect is seen between N and O, where O has six valence electrons of which four form lone pairs and only two are involved in bonding."

Now, I understand why the enthalphy decreases from C ( a covalent solid in its standard state ) to N ( a biatomic gas in its standard state ), because of course it's easier to "atomize" the gaseous molecule. But I don't understand why this happens from Si ( covalent solid ) to P ( molecular solid, P4, in its standard state ): I'd think that for "atomizing" P4 we would need much more energy because firstly we need to break the intermolecular forces and then the covalent bonds of the tetratomic molecules (P4). Even if I considered P4 as a single molecular species, not a solid, I would expect that ( as the book implies ) P4 would be harder to "atomize" than Si, because every P has 3 unpaired electrons and it forms 3 bonds with other P atoms, while Si has only 2 unpaired electrons.

The trends in the d-block are clear to me. But this one bothers me.

Please excuse my english and thank you in advance.

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    $\begingroup$ Consider that the silicon crystal structure has 4 bonds per silicon atom. $\endgroup$ Commented May 21, 2022 at 19:40
  • $\begingroup$ @GeoffHutchison yeah you're right. How stupid I was for not considering that. But isn't the bond energy of P-P in P4 bigger than the one of Si-Si in silicon? I'd say so, because of the greater Zeff, in fact Si-Si distance is around 2,5A° and P-P distance is around 1.99A° so I would expect a bigger Atomization Enthalpy for P(s). Do I need to look deeper at the solid structure? $\endgroup$ Commented May 21, 2022 at 21:27

1 Answer 1


I don't think I can give a canonical answer, but here are some quick thoughts:

  1. Breaking intermolecular interactions almost always requires significantly less energy than breaking covalent or ionic bonds. Consider that a typical hydrogen bond is ~20 kJ/mol while the O-H covalent bond dissociation is ~400 kJ/mol.

  2. The silicon crystal structure is a network solid with four Si-Si bonds per atom.

  3. The $\ce{P4}$ tetrahedra as you note only have three (strained) P-P bonds per atom.


I previously mentioned the BDE for P-P and Si-Si, but the standard CRC tables mention offhand that the values are for diatomic molecules, even though they use the P-P single bond notation.

Using the canonical Comprehensive Handbook of Chemical Bond Energies:

  • $\ce{H3Si-SiH3}$ has a BDE of 321±4 kJ/mol
  • $\ce{H2P-PH2}$ has a BDE of 256.1 kJ/mol

The P-P dissociation energy is for an average P-P bond, not counting the high ring strain inherent in the $\ce{P4}$ tetrahedra, in which the P-P-P bond angle is 60°, not ~100° as you might expect in some sort of hypothetical P-P-P chain compound.

P4 tetrahedron

Consequently, it's this angle strain that makes $\ce{P4}$ fairly reactive.

So you have more bonds for silicon and they're a bit stronger, and the $\ce{P4}$ tetrahedra are strained.

On the other hand, to quote "The Thermodynamic Properties of Phosphorus and Solid Binary Phosphides." by Schlesinger:

Phosphorus is available in a bewildering variety of physical states, and the thermodynamic stabilities of most are understood poorly at best.

Most importantly:

Because of the difficulty in obtaining and performing experimental work with red and black phosphorus, the most reliable thermodynamic data has historically been obtained working with white phosphorus and the metastable liquid. That in turn has led to the choice of R-white/liquid phosphorus as the reference state for the element, despite experimental evidence showing that red (I), red (IV), red (V), and both black forms are all more thermodynamically stable.

In short, if you're confused by phosphorus enthalpies, you're not alone. It's been tricky to measure and understand as well.

When I perform a search for P-P bond dissociation energies, I find a huge range of values.

  • $\begingroup$ The 1., 2. and 3. points that you mentioned are all clear to me. The fact that the P-P bond dissociation energy is higher than that of a Si-Si bond makes me more confuse, but as you suggest I should consider the fact that the P-P bond in the P4 tetrahedra is surely a weaker bond than other types of P-P bonds. That might be the reason of the Atomization Enthalpy's values. Thanks for sharing your reasoning and spending your time for a stranger. Anyways, I was entitled to be confused if "the thermodynamic stabilities of most are understood poorly at best." ahahah $\endgroup$ Commented May 22, 2022 at 8:25
  • $\begingroup$ P-P bond has only 183 kJ/mol I think you perhaps wrote the value for triple bond? $\endgroup$
    – Mithoron
    Commented May 23, 2022 at 0:06
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    $\begingroup$ I've seen a wide range of P-P single bond dissociation energies. I've now revised, based on a reliable BDE for $\ce{H2P-PH2}$ which seems comparable. $\endgroup$ Commented May 23, 2022 at 0:49
  • $\begingroup$ @GeoffHutchison and what about the lower atomization enthalpy for Sulfur ( standard state S8 (s) ). In the cyclo-octasulfur every S atom forms 2 bonds, with a lower bond distance ( S-S: 2.05 A°, P-P: 2.21 A°) but with a lower BDE ( S-S: 226 kJ/mol, P-P: 256 kj/mol). I'd say that the reason is that every S atom forms less bonds with a lower BDE. I considered Intermolecular forces as negligible. $\endgroup$ Commented May 23, 2022 at 9:35
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    $\begingroup$ Boiling and melting temperatures don't involve breaking bonds, only intermolecular interactions. Consider ice, for example. Trends in atomization energy, however, do. $\endgroup$ Commented May 23, 2022 at 11:11

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