Why does free silicon react with water?

Free silicon is not found in nature (in water, it might be present in the form $\ce{Si(OH)_{4}}$. Some compounds of silicon, such as silane ($\ce{SiH_{4}}$) will react spontaneously in an aqueous environment, while the analogous methane won't.

Is there an explanation for the tendency of silicon compounds to react with water?

Is there an explanation for the tendency of silicon compounds to react with water?

Reaction with water, oxygen, etc., the answer is the same and has to do with the strength of the $\ce{Si-O}$ bond. Let's compare the following two reactions:

$$\ce{CH4 + \frac{1}{2}O2 -> CH3OH}$$ $$\ce{SiH4 + \frac{1}{2}O2 -> SiH3OH}$$

In both reactions we break an $\ce{X-H}$ bond and an $\ce{O-O}$ bond, while we make an $\ce{X-O}$ bond and an $\ce{O-H}$ bond. The $\ce{O-O}$ bond breaking and the $\ce{O-H}$ bond making are common to both reactions and therefore cancel out. Let's compare the only differences, the $\ce{X-H}$ bond breaking and the $\ce{X-O}$ bond making.

The following Table contains the bond energies for these bonds.

\begin{array}{|c|c|c|c|} \hline \ce{X} & \text{X-H Bond Strength} & \text{X-O Bond Strength} & \text{Overall Change} \\ & \text{(kcal/mol)} & \text{(kcal/mol)} & \text{(kcal/mol)}\\ \hline \ce{C} & 99 & 86 & -13\\ \hline \ce{Si} & 75 & 110 & 35\\ \hline \end{array}

Looking at the numbers it becomes clear that the $\ce{Si-H}$ bond is a lot weaker than the $\ce{C-H}$ bond, and the $\ce{Si-O}$ bond is a lot stronger than the $\ce{C-O}$ bond. Consequently, the reaction involving silicon is more exothermic than the analogous reaction involving carbon. This greater stability in the $\ce{Si-O}$ product is reflected in the transition state leading to it. It lowers the energy of the transition state and consequently the reaction involving silicon is much more facile than the same reaction involving carbon.

• This is interesting. As carbon and silicon are in the same group (14), do you know why the Si-O bond is stronger? I would expect it to be weaker since Si has a greater radius (and less superposition). May 3 '15 at 21:36
• Most textbooks invoke $\ce{d \pi - p\pi}$ bonding to explain the strength of the $\ce{Si-O}$ bond. I'm not a fan of involving d-orbitals in non-transition metal bonding, but there must be some type of back bonding (electron transfer from the electronegative oxygen back to the electropositive silicon) that produces the high bond strength.
– ron
May 3 '15 at 22:15