Final answer:
The compounds are arranged in order of decreasing pKa by considering the electron-withdrawing effects. Compound b. ClCH₂CH₂SH with a chloro substituent is stronger acid than c. CH₃CH₂SH. Therefore, ClCH₂CH₂SH has a lower pKa and comes first followed by CH₃CH₂SH.
Step-by-step explanation:
To arrange the compounds in order of decreasing pKa values, we need to consider the electron-withdrawing or electron-donating effects of the substituents attached to the sulfur atom in the thiol groups. The more electron-withdrawing groups are attached, the stronger the acid will be (lower pKa), as they stabilize the negative charge on the sulfur in the conjugate base.
b. ClCH₂CH₂SH has a chloro (Cl) substituent, which is an electron-withdrawing group through inductive effect. This makes the thiol group attached to it a stronger acid compared to a thiol group without such a substituent. On the other hand, c. CH₃CH₂SH lacks such an electron-withdrawing substituent and is thus a weaker acid. So the compound b. ClCH₂CH₂SH has a lower pKa and is stronger acid than c. CH₃CH₂SH.