Question

Arrange the following in order of decreasing boiling point. (a) \(\mathrm{I}_{2}\) (b) \(\mathrm{F}_{2}\) (c) \(\mathrm{Cl}_{2}\) (d) \(\mathrm{Br}_{2}\)

Short answer

Question: Arrange the following molecules in the order of decreasing boiling points: \(\mathrm{I}_{2}\), \(\mathrm{F}_{2}\), \(\mathrm{Cl}_{2}\), and \(\mathrm{Br}_{2}\). Answer: The order of decreasing boiling points is \(\mathrm{I}_{2}\) > \(\mathrm{Br}_{2}\) > \(\mathrm{Cl}_{2}\) > \(\mathrm{F}_{2}\).

Step-by-step solution

Identify the atomic number of each element

We can find the atomic number (which represents the number of electrons in a neutral atom) for each element in the periodic table: Iodine (I): Atomic number = 53 Fluorine (F): Atomic number = 9 Chlorine (Cl): Atomic number = 17 Bromine (Br): Atomic number = 35

Calculate the total number of electrons in each molecule

Each molecule consists of two atoms, so we need to multiply the atomic number of each element by 2 to get the total number of electrons in each molecule: \(\mathrm{I}_{2}\): 53 x 2 = 106 electrons \(\mathrm{F}_{2}\): 9 x 2 = 18 electrons \(\mathrm{Cl}_{2}\): 17 x 2 = 34 electrons \(\mathrm{Br}_{2}\): 35 x 2 = 70 electrons

Arrange the molecules in order of decreasing boiling points

Recall that molecules with more electrons have stronger London dispersion forces which lead to a higher boiling point. Thus, we can arrange the molecules in decreasing order based on the number of electrons calculated in step 2: 1. \(\mathrm{I}_{2}\) (106 electrons) 2. \(\mathrm{Br}_{2}\) (70 electrons) 3. \(\mathrm{Cl}_{2}\) (34 electrons) 4. \(\mathrm{F}_{2}\) (18 electrons) Therefore, the order of decreasing boiling points is: \(\mathrm{I}_{2}\) > \(\mathrm{Br}_{2}\) > \(\mathrm{Cl}_{2}\) > \(\mathrm{F}_{2}\).