Final answer:
Methanol (CH₃OH) and Ethanol (C₂H₅OH) can exhibit hydrogen bonds because both have hydrogens bonded to highly electronegative oxygen atoms. Carbon tetrachloride (CCl₄) and Dichloromethane (CH₂Cl₂) cannot form hydrogen bonds as they lack the necessary -OH group.
Step-by-step explanation:
The student has asked which pure liquids can exhibit hydrogen bonds among CH₃OH, CCl₄, C₂H₅OH, and CH₂Cl₂. Hydrogen bonding occurs when there is a hydrogen atom bonded to a highly electronegative atom such as oxygen, fluorine, or nitrogen. These bonds result in the existence of a slight positive charge on hydrogen which can attract the lone pair electrons of electronegative atoms on other molecules, forming a hydrogen bond.
a) CH₃OH (Methanol) and c) C₂H₅OH (Ethanol) both have -OH groups and thus have hydrogens bonded to oxygen, making them capable of forming hydrogen bonds. On the other hand, b) CCl₄ (Carbon tetrachloride) and d) CH₂Cl₂ (Dichloromethane) do not contain hydrogen atoms bonded to highly electronegative atoms, and therefore cannot exhibit hydrogen bonding. Thus, the correct answers are a) CH₃OH and c) C₂H₅OH for exhibiting hydrogen bonds.