Question

One reason suggested for the instability of long chains of silicon atoms is that the decomposition involves the transition state shown below:

The activation energy for such a process is $\text{210kJ/mol}$, which is less than either the $\text{Si - Si}$or thebond energy. Why would a similar mechanism not be expected to be very important in the decomposition of long carbon chains?

Short answer

Due to lack of d orbitals carbon atom cannot form fifth bond so the transition state in the mechanism cannot be expected to be very important in the decomposition of long carbon chains.

Step-by-step solution

Concept Introduction

A chemical reaction's transition state is a specific arrangement along the reaction coordinate. It's the state that corresponds to the reaction coordinate with the maximum potential energy.

Decomposition of long carbon chains

Both carbon and silicon belong to the Group$\text{5A}$ .

They have same number of valence shell configuration ${\text{(ns}}^{\text{2}}{\text{np}}^{\text{2}}\text{)}$.

Carbon is the first element of the group; therefore, the second shell is the valence shell for it, and it can form only 4 covalent bonds, not 5 due to lack of d orbitals.

Silicon has the third shell, and therefore, it has d orbitals and can form a given transition state in the mechanism. The same mechanism is not possible for the decomposition of a long carbon chain because carbon atom cannot form fifth bond and does not have lower energy d orbitals to expand its octet.

Therefore, Carbon atom cannot form fifth bond.