The compound that is treated with NaNH2 to give a sodium salt is option A, C2H2.

Explanation:
When a compound is treated with NaNH2, it undergoes a process called deprotonation, where a proton (H+) is removed from the compound and replaced by a sodium ion (Na+). This results in the formation of a sodium salt.

Let's analyze each option to determine which compound is likely to undergo deprotonation with NaNH2:

a) C2H2 (Acetylene): Acetylene is an alkyne compound with the molecular formula C2H2. It contains two carbon atoms and two hydrogen atoms. Alkynes are known to undergo deprotonation with NaNH2 to form sodium alkynides. Therefore, option A, C2H2, is a valid choice.

b) C6H6 (Benzene): Benzene is an aromatic compound with the molecular formula C6H6. It contains six carbon atoms and six hydrogen atoms. Aromatic compounds, like benzene, do not undergo deprotonation with NaNH2. Therefore, option B, C6H6, is not a valid choice.

c) C2H6 (Ethane): Ethane is an alkane compound with the molecular formula C2H6. It contains two carbon atoms and six hydrogen atoms. Alkanes do not undergo deprotonation with NaNH2. Therefore, option C, C2H6, is not a valid choice.

d) C2H4 (Ethylene): Ethylene is an alkene compound with the molecular formula C2H4. It contains two carbon atoms and four hydrogen atoms. Alkenes do not undergo deprotonation with NaNH2. Therefore, option D, C2H4, is not a valid choice.

In conclusion, the compound that is treated with NaNH2 to give a sodium salt is option A, C2H2.