Problem 28 - Chemistry: The Central Science

Question

Explain the following variations in atomic or ionic radii: (a) \(\mathrm{I}^{-}>\mathrm{I}>\mathrm{I}^{+}\) (b) \(\mathrm{Ca}^{2+}>\mathrm{Mg}^{2+}>\mathrm{Be}^{2+}\) (c) \(\mathrm{Fe}>\mathrm{Fe}^{2+}>\mathrm{Fe}^{3+}\)

Step-by-Step Solution

Answer

The atomic radius of the given species varies due to differences in electron-electron repulsion. In case (a), the larger radius of \(\mathrm{I}^-\) is due to more electrons causing greater repulsion, while \(\mathrm{I}^+\) has fewer electrons, lesser repulsion, resulting in a smaller radius. In case (b), the atomic radius decreases with an increasing number of protons in the nucleus since the effective nuclear charge increases, pulling electrons closer. For case (c), the \(Fe\) atom has the largest atomic radius due to the highest number of electrons, and the positive ions have smaller radii due to fewer electrons and reduced repulsion.

Step 1: Case (a): \(\mathrm{I}^{-}>\mathrm{I}>\mathrm{I}^{+}\)

In case of iodine (I), we have to compare the radii of its neutral atom (I), negative ion (I⁻), and positive ion (I⁺). 1. Number of protons: In all three species, the number of protons in the nucleus is the same (53 protons). 2. Number of electrons: - Iodine atom (I) has 53 electrons. - Iodine negative ion (I⁻) gains one electron and has 54 electrons. - Iodine positive ion (I⁺) loses one electron and has 52 electrons. 3. Effective nuclear charge: As the number of protons is the same in all cases, we have to focus on the effect of the number of electrons to explain the variation in atomic radii. With more electrons, the I⁻ ion experiences more electron-electron repulsion, causing its atomic radius to be larger than that of the neutral I atom. In the case of I⁺, it has fewer electrons, so less repulsion occurs, and its atomic radius is smaller than that of the neutral I atom.

Step 2: Case (b): \(\mathrm{Ca}^{2+}>\mathrm{Mg}^{2+}>\mathrm{Be}^{2+}\)

In this case, we compare the radii of Ca²⁺, Mg²⁺, and Be²⁺ ions. 1. Number of protons: - Calcium ion (Ca²⁺) has 20 protons. - Magnesium ion (Mg²⁺) has 12 protons. - Beryllium ion (Be²⁺) has 4 protons. 2. Number of electrons: All three ions have the same number of electrons (10 electrons). As the number of electrons is the same in all cases, we need to assess the effect of the increasing number of protons in the nucleus. With more protons, the effective nuclear charge increases, pulling the electrons closer to the nucleus, which results in a decrease in atomic radius. Therefore, among these ions, Ca²⁺ has the largest atomic radius and Be²⁺ the smallest.

Step 3: Case (c): \(\mathrm{Fe}>\mathrm{Fe}^{2+}>\mathrm{Fe}^{3+}\)

In this case, we are comparing the atomic radius of iron atom (Fe), its di-positive ion (Fe²⁺), and its tri-positive ion (Fe³⁺). 1. Number of protons: In all three species, the number of protons in the nucleus is the same (26 protons). 2. Number of electrons: - Iron atom (Fe) has 26 electrons. - Iron di-positive ion (Fe²⁺) has 24 electrons. - Iron tri-positive ion (Fe³⁺) has 23 electrons. Fe³⁺ has the least number of electrons and, therefore, experiences less electron-electron repulsion compared to Fe²⁺ and Fe. This results in a smaller atomic radius for Fe³⁺. Similarly, Fe²⁺ has fewer electrons than Fe, so it is smaller in size as well. As a result, Fe has the largest atomic radius, followed by Fe²⁺ and Fe³⁺.