In the silicon diamond structure the bond distance is about two times the bond distance of the carbon diamond structure. Is this caused only by the greater dimension of $\mathrm{Si}$ atoms with respect to $\mathrm{C}$ atoms? Or are there other reasons for a greater bond distance in the case of silicon?

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    $\begingroup$ How could you conclude that is not? It is tricky to isolate bond length from atom size. $\endgroup$
    – Alchimista
    Commented Jan 2, 2018 at 15:37

1 Answer 1


Yes, both structures belong to the same diamond cubic type, and the interatomic distances for diamond and silicon (you can use "Measurements" tool from the JSmol menu)

$$ \begin{align} d(\ce{C-C}) &= \pu{1.54 Å} \\ d(\ce{Si-Si}) &= \pu{2.35 Å} \\ \frac{d(\ce{Si-Si})}{d(\ce{C-C})} &\approx 1.53 \end{align} $$

are in a good agreement with the corresponding covalent radii

$$ \begin{align} r_\mathrm{cov}(\ce{C}) &= \pu{0.76 Å} \\ r_\mathrm{cov}(\ce{Si}) &= \pu{1.11 Å} \\ \frac{r_\mathrm{cov}(\ce{Si})}{r_\mathrm{cov}(\ce{C})} &\approx 1.46 \end{align} $$

which would set the silicon atoms further apart from each other.


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