Problem 28 - Chemistry: An Atoms-Focused Approach

Question

Which ion is larger: \(\mathrm{Cl}^{-}\) or \(\mathrm{I}^{-} ?\)

Step-by-Step Solution

Answer

Answer: The larger ion is I-.

Step 1: Recall the periodic table trend for atomic size

In the periodic table, atomic size generally increases from top to bottom within a group and decreases from left to right across a period. This trend occurs because as atomic number increases, the number of electrons and protons also increases. As a result, electrons are added to the outermost energy levels (shells), and the nucleus attracts these electrons more strongly due to the increase in protons, causing them to move closer to the nucleus. However, the shielding effect of inner shell electrons also increases, and this counteracts the increased attraction from the protons. The overall trend is that atomic size increases moving down a group because the shielding effect becomes more significant.

Step 2: Identify the group and period for each ion

Both \(\mathrm{Cl}^{-}\) and \(\mathrm{I}^{-}\) are anions belonging to Group 17, also known as the halogens, in the periodic table. The neutral Cl atom has an atomic number of 17, while the neutral I atom has an atomic number of 53. Therefore, Cl belongs to Period 3, and I belongs to Period 5.

Step 3: Compare the atomic size of the neutral atoms

Since Cl is located above I in the periodic table, Cl is smaller than I.

Step 4: Consider the effect of gaining an electron

Upon gaining an electron to form the anions \(\mathrm{Cl}^{-}\) and \(\mathrm{I}^{-}\), both anions now have an extra electron in their outermost energy levels (shells). This increases the electron-electron repulsion, leading to a larger size compared to their respective neutral atoms. As such, the trend in atomic size observed in the periodic table remains the same for these anions.

Step 5: Determine the larger ion

Since Cl is smaller than I and the trend in atomic size remains the same upon gaining an electron, \(\mathrm{Cl}^{-}\) is smaller than \(\mathrm{I}^{-}\). Therefore, the larger ion is \(\mathrm{I}^{-}\).