Problem 2 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

The correct order of the ionic radii of \(\mathrm{O}^{2-}, \mathrm{N}^{3-}, \mathrm{F}^{-}, \mathrm{Mg}^{2+}, \mathrm{Na}^{+}\)and \(\mathrm{Al}^{3+}\) is : (a) \(\mathrm{N}^{3-}<\mathrm{O}^{2-}<\mathrm{F}^{-}<\mathrm{Na}^{+}<\mathrm{Mg}^{2+}<\mathrm{Al}^{3+}\) (b) \(\mathrm{Al}^{3+}<\mathrm{Na}^{+}<\mathrm{Mg}^{2+}<\mathrm{O}^{2-}<\mathrm{F}^{-}<\mathrm{N}^{3-}\) (c) \(\mathrm{Al}^{3+}<\mathrm{Mg}^{2+}<\mathrm{Na}^{+}<\mathrm{F}^{-}<\mathrm{O}^{2-}<\mathrm{N}^{3-}\) (d) \(\mathrm{N}^{3-}<\mathrm{F}^{-}<\mathrm{O}^{2-}<\mathrm{Mg}^{2+}<\mathrm{Na}^{+}<\mathrm{Al}^{3+}\)

Step-by-Step Solution

Answer

The correct order is (c): \(\mathrm{Al}^{3+}<\mathrm{Mg}^{2+}<\mathrm{Na}^{+}<\mathrm{F}^{-}<\mathrm{O}^{2-}<\mathrm{N}^{3-}\).

Step 1: Understand the Concept of Ionic Radius

The ionic radius is the measure of an atom's ion in a crystal lattice. It is the distance from the nucleus to the outermost electrons of the ion. Anions (negatively charged ions) are generally larger than their parent atoms, while cations (positively charged ions) are smaller.

Step 2: Examine the Ions Provided

Consider the ions given: \(\mathrm{O}^{2-}, \mathrm{N}^{3-}, \mathrm{F}^{-}, \mathrm{Mg}^{2+}, \mathrm{Na}^{+},\) and \(\mathrm{Al}^{3+}\). These ions have different charges and belong to different groups in the periodic table.

Step 3: Determine the Effect of Charge on Ionic Size

More negative charge usually implies larger size due to increased electron-electron repulsion in the outer shell. More positive charge usually implies smaller size due to the removal of electron(s) and greater effective nuclear charge on the remaining electrons.

Step 4: Compare Anions and Cations

Among \(\mathrm{O}^{2-}, \mathrm{N}^{3-}, \mathrm{F}^{-}\), \(\mathrm{N}^{3-}\) has the largest ionic radius due to the highest negative charge. Among cations, \(\mathrm{Al}^{3+}\) will have the smallest radius due to the highest positive charge.

Step 5: Construct Order Based on Ion Charges

Organize the ions from smallest to largest based on their charges. Since \(\mathrm{Al}^{3+}\) has the most positive charge, it will be smallest, followed by \(\mathrm{Mg}^{2+}\) and \(\mathrm{Na}^{+}\). The sequence will continue with the anions \(\mathrm{F}^{-}\), \(\mathrm{O}^{2-}\), and \(\mathrm{N}^{3-}\), where \(\mathrm{N}^{3-}\) is largest.

Step 6: Verify Against Provided Options

The correct order is \(\mathrm{Al}^{3+}<\mathrm{Mg}^{2+}<\mathrm{Na}^{+}<\mathrm{F}^{-}<\mathrm{O}^{2-}<\mathrm{N}^{3-}\), which matches option (c).

Key Concepts

Cation and Anion Sizes

When atoms gain or lose electrons to form ions, their sizes also change significantly. Anions, which are negatively charged because they gain electrons, tend to be larger than their parent atoms. This enlargement is due to increased electron-electron repulsion within the atom's outer shell. As electrons are added, these repulsions cause the electron cloud to expand.
On the other hand, cations are positively charged ions formed by losing electrons. The removal of electrons reduces the electron-electron repulsion. Furthermore, with fewer electrons, the effective nuclear charge (the net positive charge experienced by the electron cloud due to the protons in the nucleus) becomes stronger. This increase in effective nuclear charge pulls the remaining electrons closer to the nucleus, resulting in a smaller atomic radius.

• Anions are larger due to increased electron repulsion in the expanded outer shell.
• Cations are smaller because of reduced electron repulsion and increased effective nuclear charge.

Ionic Charge Effects

The magnitude of an ion's charge directly affects its size. For example, consider the ions with different charges as given in the problem:

• A higher positive charge, as in cations like \(\mathrm{Al}^{3+}\\), means more electrons are removed, causing a significant decrease in size. More positive charge results in a stronger pull from the nucleus, leading to compaction of the remaining electron cloud.
• More negative charge, as found in anions like \(\mathrm{N}^{3-}\\), results in a larger size. Extra electrons cause increased repulsion in the outer shell, making the ion more spread out.
• Differences in charges among ions merely from an electron difference (such as \(\mathrm{Na}^{+}\\) versus \(\mathrm{Mg}^{2+}\\)) can still lead to noticeable differences in ionic radius.

Thus, as a rule of thumb, greater positive charge (stronger nuclear pull) shrinks the ion, while greater negative charge (more repulsive forces) enlarges it.

Periodic Trends

Periodic trends provide an invaluable framework to understand the variations in atomic and ionic sizes across the periodic table. As you move across a period from left to right (for example moving from \(\mathrm{Na}^{+}\\) to \(\mathrm{Mg}^{2+}\\)), the ions typically become smaller for cations. This occurs because additional protons are added across the period, increasing the effective nuclear charge without a significant increase in electron shielding. The result is a tighter pull on the electron cloud.
For anions, such as moving from \(\mathrm{O}^{2-}\\) to \(\mathrm{N}^{3-}\\), the increased negative charge across a period can lead to larger sizes even though the nuclear pull is stronger. This is due to increasing numbers of electrons with relatively small increases in nuclear charge leading to increased repulsion. Thus,

• Within a particular group, as you go down the group, both cations and anions see increased radii due to added electron shells.
• Across a period, cations become smaller with increasing positive charge, whereas similarly charged anions become larger.

With these trends, we begin to identify patterns, allowing predictions of ionic sizes based on position within the periodic table and ionic charges.