# Greater bond distance for silicon diamond structure than for carbon diamond structure?

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?

• How could you conclude that is not? It is tricky to isolate bond length from atom size. – Alchimista Jan 2 '18 at 15:37

## 1 Answer

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.