Explain in terms of forces between structural units why (a) \(\mathrm{Br}_{2}\) has a lower melting point than \(\mathrm{NaBr}\). (b) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) has a higher boiling point than butane, \(\mathrm{C}_{4} \mathrm{H}_{10}\). (c) \(\mathrm{H}_{2} \mathrm{O}\) has a higher boiling point than \(\mathrm{H}_{2} \mathrm{Te}\). (d) Acetic acid \(\mathrm{CH}_{3}-\mathrm{C}-\mathrm{OH}\) has a lower boiling point

First, we need to identify the type of forces acting between the structural units in each compound. These forces can be ionic, covalent, or van der Waals, and they directly impact the melting/boiling points. (a) Br₂: Intermolecular forces: van der Waals forces NaBr: Intermolecular forces: ionic bonding (b) C₂H₅OH: Intermolecular forces: hydrogen bonding and van der Waals forces C₄H₁₀: Intermolecular forces: van der Waals forces (c) H₂O: Intermolecular forces: hydrogen bonding H₂Te: Intermolecular forces: van der Waals forces (d) Acetic acid (CH₃C(O)OH): Intermolecular forces: hydrogen bonding

Next, we need to compare the relative strengths of the intermolecular forces in each compound. The stronger the forces, the higher the melting/boiling points. Ionic forces > hydrogen bonding > van der Waals forces (a) Br₂ has weaker van der Waals forces compared to the ionic bonding in NaBr, which leads to its lower melting point. (b) C₂H₅OH has stronger hydrogen bonding compared to the van der Waals forces in C₄H₁₀, which leads to its higher boiling point. (c) H₂O has stronger hydrogen bonding compared to the van der Waals forces in H₂Te, which leads to its higher boiling point. (d) Although Acetic acid has hydrogen bonding, it has more molecular flexibility due to larger molecular structure than water, which leads to its lower boiling point.