Problem 159 - A Complete Resource Book in Chemistry for JEE Main

Question

The increasing order of the ionic radii of the given isoelectronic species is: (a) \(\mathrm{Ca}^{2+}, \mathrm{K}^{+}, \mathrm{Cl}^{-}, \mathrm{S}^{2-}\) (b) \(\mathrm{S}^{2}, \mathrm{Cl}^{-}, \mathrm{Ca}^{2+}, \mathrm{K}^{+}\) (c) \(\mathrm{Cl}^{-}, \mathrm{Ca}^{2+}, \mathrm{K}^{+}, \mathrm{S}^{2-}\) (d) \(\mathrm{K}^{+}, \mathrm{S}^{2}, \mathrm{Ca}^{2+}, \mathrm{Cl}^{-}\)

Step-by-Step Solution

Answer

The correct answer is (a) \(\mathrm{Ca}^{2+}, \mathrm{K}^{+}, \mathrm{Cl}^{-}, \mathrm{S}^{2-}\).

Step 1: Understand Isoelectronic Species

Isoelectronic species are atoms, ions, or molecules that have the same number of electrons. In this problem, the species \(\mathrm{Ca}^{2+}, \mathrm{K}^{+}, \mathrm{Cl}^{-}, \mathrm{S}^{2-}\)all have 18 electrons. Understanding that they are isoelectronic allows us to compare ionic radii based on their nuclear charge.

Step 2: Analyze the Nuclear Charge

The more protons (nuclear charge) an ion has, the smaller its radius due to the increased attraction pulling electrons closer to the nucleus. Let's list the nuclear charges of each species: - \(\mathrm{Ca}^{2+}: Z = 20\)- \(\mathrm{K}^{+}: Z = 19\)- \(\mathrm{Cl}^{-}: Z = 17\)- \(\mathrm{S}^{2-}: Z = 16\)

Step 3: Arrange by Increasing Nuclear Charge

Since more nuclear charge means a smaller ionic radius, we arrange the ions by increasing nuclear charge to determine the order of increasing ionic radii:1. \(\mathrm{S}^{2-}\) (smallest nuclear charge, largest radius)2. \(\mathrm{Cl}^{-} \)3. \(\mathrm{K}^{+} \)4. \(\mathrm{Ca}^{2+} \) (largest nuclear charge, smallest radius)

Step 4: Match with Given Choices

Now we match the order we derived with the choices provided. Our order is \(\mathrm{Ca}^{2+}, \mathrm{K}^{+}, \mathrm{Cl}^{-}, \mathrm{S}^{2-}\), which corresponds to choice (a).

Key Concepts

Isoelectronic Species

In chemistry, isoelectronic species are those atoms and ions that have the same number of electrons. This shared electron count leads to some interesting properties and behaviors. For example,

• In this particular problem, these species are \( \mathrm{Ca}^{2+}, \mathrm{K}^{+}, \mathrm{Cl}^{-}, \mathrm{S}^{2-} \). Though they each have different charges, all possess 18 electrons.
• They are isoelectronic not only in count but also in their electron configuration.

Understanding that these species are isoelectronic allows one to focus not on how many electrons they each have but how these electrons are affected by other characteristics, like the nuclear charge of each ion.

Nuclear Charge

Nuclear charge is determined by the number of protons in an atom's nucleus. It directly influences the size of an ion because electrons are attracted to the nucleus.

• In isoelectronic species, a higher nuclear charge means more protons, which translates to a greater pull on the electron cloud.
• This results in a smaller ionic radius.

For instance:

• \( \mathrm{Ca}^{2+} \), with 20 protons (nuclear charge \( Z = 20 \)), exerts a strongest pull, resulting in the smallest radius.
• Similarly, \( \mathrm{S}^{2-} \), having the lowest nuclear charge \( Z = 16 \), has the largest radius due to the weakest pull.

Electron Configuration

The electron configuration of an atom or ion refers to the arrangement of electrons. For isoelectronic species, this remains constant, but let's delve deeper into its importance:

• Knowing the electron configuration helps us understand the structural similarities and differences between ions.
• For the discussed species, with 18 electrons, the electron configuration is \([ \text{Ar} ] = 1s^2 2s^2 2p^6 3s^2 3p^6 \). This configuration is shared by all because they are isoelectronic.

Maintaining consistent configurations aids in focusing on other factors that differentiate these species, such as their nuclear charge, which directly affects their ionic sizes. Therefore, while their electron configurations don’t change, their physical sizes do.