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Topic: Why P4 but not P8?  (Read 5047 times)

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Offline shelanachium

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Why P4 but not P8?
« on: April 15, 2008, 10:23:10 AM »
In most P(III) compounds like PH3 the bond angles are close to 90 degrees. Why then is P4 with 60-degree angles known but not cubic P8 with 90-degree angles?

Offline sjb

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Re: Why P4 but not P8?
« Reply #1 on: April 15, 2008, 11:22:28 AM »
I don't know for absolutely sure, so this is perhaps a bit hand-wavy.

In both P4 and P8 (assuming you mean as a cube, rather than the ring you observe in S8) all the bonds are P-P. It's true that the bond angle and strain argument may favour P8 (angle of 90°, rather than 60°), but does the fact that you'd get 2 P4 molecules rather than 1 P8 suggest more an entropy driven breakdown of the solid state ?

S

Offline shelanachium

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Re: Why P4 but not P8?
« Reply #2 on: April 16, 2008, 09:10:15 AM »
I think you have a point there. At the very high temperatures used for making P4, most of the phosphorus probably first appears as P2. This can easily dimerise to P4 in a 2-body reaction; whilst the 3- and 4-body reactions needed to make trigonal-prismatic P6 or cubic P8 are strongly entropy-disfavoured.

C60 forms in C vapour, but no smaller molecular poly-C molecule remotely approaches it in thermodynamic stability; whilst P2 is not THAT much less stable than P4.

S2 in hot sulfur vapour presumably first generates S4 as P2 does P4, but S4 (either a simple ring or diradical) easily dimerises to cyclic S8. There is presumably no facile route from tetrahedral P4 to cubic P8.

Maybe P8 is in fact very stable, but difficult to produce because of no facile route from smaller molecules. Situation similar to dodecahedrane C20H20 - very stable but no easy route to it; necessary intermediates too unstable or too entropically disfavoured.

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