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) \(\mathrm{Fe}>\mathrm{Fe}^{2+}>\mathrm{Fe}^{3+}\)

Short answer

The variations in atomic or ionic radii can be explained as follows: (a) I⁻ has one extra electron compared to I and experiences more repulsion, making it larger than I. I⁺ has one less electron and experiences a higher effective nuclear charge, making it smaller than I. Therefore, \(\mathrm{I}^{-}>\mathrm{I}>\mathrm{I}^{+}\). (b) As we move down a group in the periodic table, atomic size increases due to more electron shells. Since Ca is in the fourth period, Mg is in the third period, and Be is in the second period, their size order is \(\mathrm{Ca}^{2+}>\mathrm{Mg}^{2+}>\mathrm{Be}^{2+}\). (c) The size difference between Fe, Fe²⁺, and Fe³⁺ is due to electron shielding and effective nuclear charge. In Fe²⁺, there are fewer electrons and in Fe³⁺, even lesser electrons, hence smaller ion sizes. Thus, \(\mathrm{Fe}>\mathrm{Fe}^{2+}>\mathrm{Fe}^{3+}\).

Step-by-step solution

Understanding Electron Configuration

First, we should look into the electron configuration of Iodine (I). Iodine has an atomic number of 53, and its electron configuration is \([Kr] 4d^{10} 5s^2 5p^5\). When iodine gains one electron, it forms I⁻ with a configuration of \([Kr] 4d^{10} 5s^2 5p^6\), while when it loses an electron, it forms I⁺ with a configuration of \([Kr] 4d^{10} 5s^2 5p^4\).

Explaining the Size Difference

The size of the iodine atom or ions is determined by the balance between the shielding effect of inner electron shells and the effective nuclear charge experienced by the outermost electrons. In I, there are 5 electron shells with a +53 charge in the nucleus. In I⁻, an extra electron is added, and the same 5 electron shells shield the positive charge, while in I⁺, one electron has been removed, and the same 5 electron shells shield the positive charge. Since the negatively charged electrons repel each other, I⁻, with the extra electron, will be larger in size than the neutral I. Meanwhile, I⁺ will be smaller than the neutral I because it has one less electron repelling others and because the remaining outermost electrons experience a higher effective nuclear charge, drawing them closer to the nucleus. So, the order of size is: \(\mathrm{I}^{-}>\mathrm{I}>\mathrm{I}^{+}\) #Case (b) Ca²⁺ > Mg²⁺ > Be²⁺#

Understanding Periodic Trends

For this comparison, all three ions have a +2 charge. However, they are from different periods of the periodic table. Calcium has an atomic number of 20, Magnesium has an atomic number of 12, and Beryllium has an atomic number of 4.

Explaining the Size Difference

In this case, the explanation for the size difference mainly comes down to periodic trends. As we move down a group in the periodic table, the atomic size generally increases due to the increase in the number of electron shells. More electron shells mean more shielding and repulsion between electrons, leading to increased atomic size. Since Calcium is in the fourth period, Magnesium is in the third period, and Beryllium is in the second period, we expect the size order to be: \(\mathrm{Ca}^{2+}>\mathrm{Mg}^{2+}>\mathrm{Be}^{2+}\) #Case (c) Fe > Fe²⁺ > Fe³⁺#

Understanding Electron Configuration

In this case, we need to look at the electron configuration of Iron (Fe). Iron has an atomic number of 26, and its electron configuration is \([Ar] 3d^6 4s^2\). When iron loses two electrons, it forms Fe²⁺ with a configuration of \([Ar] 3d^6\), while when it loses three electrons, it forms Fe³⁺ with a configuration of \([Ar] 3d^5\).

Explaining the Size Difference

The size difference between the iron atom and its ions can again be attributed to electron shielding and effective nuclear charge. In neutral Fe, there are 26 protons in the nucleus and four complete electron shells. In Fe²⁺, there are only 24 negatively charged electrons, which cause less repulsion between electrons but still face the same nuclear charge, resulting in a smaller ion. In Fe³⁺, the ion has only 23 negatively charged electrons, and the remaining outer electrons are experiencing an even stronger effective nuclear charge, causing the ion to be even smaller than Fe²⁺. So, the order of size is: \(\mathrm{Fe}>\mathrm{Fe}^{2+}>\mathrm{Fe}^{3+}\)