Question

1. (q-2,6b) etin c₂h₂ molekulasidagi uglerod atomlari orbitallari qanday gibridlanish holatida boladi? a) sp² b)sp c) sp² va sp³ d) sp³

Explanation:

To determine the hybridization of carbon atoms in \( \text{C}_2\text{H}_2 \) (ethyne, also known as acetylene), we analyze the bonding. In \( \text{C}_2\text{H}_2 \), each carbon atom forms a triple bond with the other carbon and a single bond with a hydrogen atom. The triple bond consists of one \( \sigma \) bond and two \( \pi \) bonds. For a carbon atom with a triple bond, the hybridization is \( \text{sp} \) because \( \text{sp} \) hybridization involves the mixing of one \( s \) orbital and one \( p \) orbital, resulting in two hybrid orbitals (used for \( \sigma \) bonds) and two unhybridized \( p \) orbitals (used for \( \pi \) bonds). This matches the bonding in \( \text{C}_2\text{H}_2 \), where each carbon has two \( \sigma \) bonds (one to the other carbon, one to hydrogen) and two \( \pi \) bonds (from the triple bond).

Answer:

B) \( \text{sp} \)