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

Step 1: Understand the Concept of Isoelectronic Species

Isoelectronic species are atoms, ions, or molecules that have the same number of electrons, meaning they have identical electron configurations. In this exercise, we are comparing the isoelectronic species \[ \mathrm{Ca}^{2+}, \mathrm{K}^{+}, \mathrm{Cl}^{-}, \mathrm{S}^{2-} \]which all have 18 electrons, same as the electronic configuration of Argon ([Ar]).

Step 2: Determine the Parameter Affecting Ionic Radius

For isoelectronic species, the size of the ion primarily depends on the effective nuclear charge, which is the net positive charge experienced by an electron in a multi-electron atom/ion. The greater the positive charge of the nucleus (higher atomic number), the smaller the ionic radius, because more protons increase the attraction on the electrons, pulling them closer.

Step 3: Arrange in Order of Decreasing Effective Nuclear Charge

First, write the nuclear charge (atomic number) of each ion:- \(\mathrm{Ca}^{2+} (Z=20)\)- \(\mathrm{K}^{+} (Z=19)\)- \(\mathrm{Cl}^{-} (Z=17)\)- \(\mathrm{S}^{2-} (Z=16)\)An ion with a higher atomic number has a greater effective nuclear charge, thereby having a smaller ionic radius.

Step 4: Arrange the Ions in Increasing Order of Ionic Radii

Since ions with a higher atomic number have a greater effective nuclear charge and smaller size, the increasing order of ionic radii will be:1. \(\mathrm{Ca}^{2+}\)2. \(\mathrm{K}^{+}\)3. \(\mathrm{Cl}^{-}\)4. \(\mathrm{S}^{2-}\)

Step 5: Match the Order with Given Options

Refer back to the given options:- (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) \(\mathrm{K}^{+}, \mathrm{S}^{2-}, \mathrm{Ca}^{2+}, \mathrm{Cl}^{-}\)The answer that matches our calculated order is **(a)**.

Key Concepts

Isoelectronic Species

Isoelectronic species refer to different atoms, ions, or molecules that possess the same number of electrons, leading them to share identical electron configurations. In this particular exercise, you'll see that

• \( \mathrm{Ca}^{2+} \)
• \( \mathrm{K}^{+} \)
• \( \mathrm{Cl}^{-} \)
• \( \mathrm{S}^{2-} \)

All have the same number of electrons, precisely 18, which is similar to Argon's electron configuration \([Ar]\).
Understanding the concept of isoelectronic species is the first step in comparing their properties like ionic radii. When different elements have the same electron count, it is fascinating to see how their chemical behavior changes due to differing nuclear charges.

Effective Nuclear Charge

The effective nuclear charge is a critical factor that influences the size of ions, especially for isoelectronic species. It represents the net positive charge an electron experiences in a multi-electron atom or ion.

For isoelectronic species, the concept can be explained as follows:

• The more protons present in the nucleus (higher atomic number), the greater the attractive force exerted on the electrons.
• This increased positive pull draws electrons closer to the nucleus, thus reducing the ionic radius.

To put this in the context of the exercise, consider the species

• \( \mathrm{Ca}^{2+} \) with \(Z=20\)
• \( \mathrm{K}^{+} \) with \(Z=19\)
• \( \mathrm{Cl}^{-} \) with \(Z=17\)
• \( \mathrm{S}^{2-} \) with \(Z=16\)

Here, \( \mathrm{Ca}^{2+} \) will exert the maximum nuclear charge on its electrons, and it will therefore have the smallest radius due to the strong attractive forces pulling the electrons close.
This concept helps clarify why ions with a greater nuclear charge often end up being smaller in size, despite sharing the same number of electrons.

Radius-order Calculation

When arranging isoelectronic species by ionic radii, understanding the effective nuclear charge helps significantly. Higher nuclear charge typically means smaller ionic size. Let's look at how to order these ions based on their radii.
1. Begin by listing the ions' atomic numbers:

• \( \mathrm{Ca}^{2+} \) has \( Z = 20 \)
• \( \mathrm{K}^{+} \) has \( Z = 19 \)
• \( \mathrm{Cl}^{-} \) has \( Z = 17 \)
• \( \mathrm{S}^{2-} \) has \( Z = 16 \)

2. Next, arrange them in order of increasing nuclear charge, which translates to decreasing ionic radii:

• \( \mathrm{S}^{2-} \)
• \( \mathrm{Cl}^{-} \)
• \( \mathrm{K}^{+} \)
• \( \mathrm{Ca}^{2+} \)

This means \( \mathrm{S}^{2-} \) has the largest ionic radius, while \( \mathrm{Ca}^{2+} \) has the smallest. Understanding this calculation ensures that you can match species based on size, linking it back to how nuclear charge governs ionic dimensions.