If 1 mole of gaseous carbon atoms are converted into diamonds, calculate enthalpy change of process given that bond enthalpy of $\ce{C-C}$ bond is $\pu{400 kJ mol-1}.$

According to me, the answer should be that since C(diamond) has 4 bonds and C(gas) doesn't have any bonds, 4 bonds are being formed so enthalpy change should be

$$ΔH = 4\cdot (\pu{-400 kJ mol-1})$$

But in the correct solution it says each bond is counted twice, so answer will be

$$ΔH = \frac{4\cdot (\pu{-400 kJ mol-1})}{2}$$

I don't get why are we doing this. In other questions we don't consider the double counting of bonds. For example, another question I did had 1 mole phosphorus gas converting into $\ce{P4}$ molecule, and even though $\ce{P4}$ has 6 bonds, we did not consider what we did with diamond, so what is the reason for that?

  • $\begingroup$ Consider the simpler question: If 1 mole of hydrogen atoms (H) form dihydrogen (H-H), how many bonds are formed? Using your logic (each hydrogen makes one bond), there should be one mole bonds. But only 1/2 mole of dihydrogen was made. That would be 1 mole of single bonds in 1/2 mole of dihydrogen, not possible. $\endgroup$
    – Karsten
    Jul 22, 2019 at 13:36

1 Answer 1


$\ce{P4}$ forms a molecular solid, whereas diamond is a covalent-network solid (see eg definitions here).

In a molecular solid, molecules containing a discrete number of covalently bonded atoms are arranged in a lattice stabilized by weak interactions such as van der Waals forces or hydrogen bonds between the molecules. In a covalent-network solid, the "molecules" are extended with no finite stoichiometry definining a molecular size.

When estimating the enthalpy change, you consider only covalent bonds, not weak intermolecular forces. Therefore, for $\ce{P4}$ you consider only the 6 P-P bonds, whereas in diamond you consider covalent bonds to nearest neighbours, and correct for the fact that when you rupture a bond to a neighbour, that neighbour also loses a bond, therefore the division by 2.

  • $\begingroup$ Thanks ! , but if in future questions , I am unable to identify is the given substance if a covalent network solid or not , can we say that any molecule of the form Xy (like O2) where y is any number will not be a covalent-network solid ? $\endgroup$
    – bscripts
    Jul 22, 2019 at 14:59
  • 1
    $\begingroup$ @bscripts If you can envision the molecule existing on its own in solution or in the gas phase, then it is likely to be a molecular solid. I can't think of exceptions of the bat, but there might be some, particularly if you change pressure or temperature to the right values. $\endgroup$
    – Buck Thorn
    Jul 22, 2019 at 16:37
  • $\begingroup$ I saw a question where nitrogen atoms where present on all the corners of a truncated octahedron and the energy required to break the octahedron was asked , and something similar was done to what was done with diamond above , would you call nitrogen a covalent network solid ? $\endgroup$
    – bscripts
    Jul 22, 2019 at 17:07
  • $\begingroup$ @bscripts Apparently there are exotic phases of nitrogen, see en.wikipedia.org/wiki/Solid_nitrogen. I would not call solid phases of molecular nitrogen $\ce{N2}$ covalent network solids. $\endgroup$
    – Buck Thorn
    Jul 22, 2019 at 18:56

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