To solve this problem, we need to determine the hybridization of the carbon atoms in each compound and compare it with the hybridization of $$CO$_2. \\$First, let's find the hybridization of $$CO$_2: \\$• $$CO$_2$ has a linear structure with the central carbon atom forming two double bonds with oxygen atoms.$\\$• The hybridization of carbon in $$CO$_2$ is $sp. \\$Now, let's analyze each option:$\\$Option 1: Ethane (C$_2$H$_6$)$\\$• In ethane, each carbon atom forms four single bonds (three with hydrogen and one with another carbon).$\\$• The hybridization of carbon in ethane is $sp^3. \\$Option 2: Ethyne (C$_2$H$_2$)$\\$• In ethyne, each carbon atom forms a triple bond with another carbon and a single bond with hydrogen.$\\$• The hybridization of carbon in ethyne is $sp. \\$Option 3: Ethene (C$_2$H$_4$)$\\$• In ethene, each carbon atom forms a double bond with another carbon and two single bonds with hydrogen.$\\$• The hybridization of carbon in ethene is $sp^2. \\$Option 4: Ethanol (C$_2$H$_5$OH)$\\$• In ethanol, each carbon atom forms single bonds (with hydrogen, carbon, and oxygen).$\\$• The hybridization of carbon in ethanol is $sp^3. \\$Conclusion:$\\$• The hybridization of carbon in $$CO$_2$ is $sp. \\$• The only compound with the same $sp$ hybridization is ethyne.$\\$Therefore, the correct option is Option 2: Ethyne.