Question

Without doing a calculation, arrange the group in order of decreasing standard molar entropy (s o): PF₂Cl3(g), PF5(g), PF₃(g)?

1) PF5(g), PF₃(g), PF₂Cl3(g)
2) PF₃(g), PF5(g), PF₂Cl3(g)
3) PF₂Cl3(g), PF₃(g), PF5(g)
4) PF₂Cl3(g), PF5(g), PF₃(g)

Answer 1

To arrange the group in order of decreasing standard molar entropy, consider the molecular complexity of each molecule. Based on this, the correct arrangement is 3) PF₂Cl₃(g), PF₃(g), PF₅(g).

Step-by-step explanation:

To arrange the group in order of decreasing standard molar entropy, we need to consider the molecular complexity and the number of atoms in each molecule.

Generally, as the number of atoms and molecular complexity increase, the entropy also increases. So, we can determine the order of decreasing standard molar entropy based on the molecular complexity of each molecule.

1. PF₃(g) - Phosphorus trifluoride has one phosphorus atom and three fluorine atoms, making it the simplest molecule in the group. Therefore, it has the lowest entropy.
2. PF₂Cl₃(g) - Phosphorus dichloride trifluoride has one phosphorus atom, two fluorine atoms, and three chlorine atoms. It is more complex than PF₃(g) but less complex than PF₅(g).
3. PF₅(g) - Phosphorus pentafluoride has one phosphorus atom and five fluorine atoms, making it the most complex molecule in the group. Therefore, it has the highest entropy.

Based on this information, the correct arrangement of the group in order of decreasing standard molar entropy is

3) PF₂Cl₃(g), PF₃(g), PF₅(g).