Question

The ions \(O_2^–\), \(F^–\),  \(Na^+\), \(Mg^{2+}\) and \(Al^{3+}\) are isoelectronic. Their ionic radii show:

• A. A significant increase from \(O_2^–\) to \(Al^{3+}\)
• B. A significant decrease from \(O_2^–\) to \(Al^{3+}\)
• C. An increase from \(O_2^–\) to \(F^–\) and then decrease from \(Na^+\) to \(Al^{3+}\)
• D. An decrease from \(O_2^–\) to \(F^–\) and then increase from \(Na^+\) to \(Al^{3+}\)

Answer 1

The Correct Option is B

The ions \(O_2^–\), \(F^–\), \(Na^+\), \(Mg^{2+}\), and \(Al^{3+}\) are all isoelectronic, meaning they have the same number of electrons. They each have 10 electrons because the electronic configuration of each is that of neon, \([\text{Ne}]\). However, they have different nuclear charges (number of protons), which affects their ionic sizes.  

The ionic radius is influenced by the effective nuclear charge. The more positive the charge on the ion, the smaller its radius because the electrons are held more tightly by the increased nuclear charge. Conversely, a more negative charge means fewer protons for the same number of electrons, resulting in a larger ionic radius due to less effective nuclear attraction.

Let's look at each ion:

• \(O_2^–\): Has the least nuclear charge relative to the number of electrons, so it has the largest ionic radius.
• \(F^–\): One less negative charge compared to oxygen, with a slightly smaller radius.
• \(Na^+\): Neutral sodium loses one electron, increasing the effective nuclear charge per electron, thus reducing the ionic radius from the corresponding anions.
• \(Mg^{2+}\): With an even higher positive charge, this ion pulls electrons closer, further decreasing the ionic radius compared to the ions before it.
• \(Al^{3+}\): Has the highest nuclear charge among these ions, resulting in the smallest ionic radius.

Therefore, as we move from \(O_2^–\) to \(Al^{3+}\), the effective nuclear charge increases leading to a significant decrease in ionic radius. The trend is of decreasing ionic size: \(O_2^– > F^– > Na^+ > Mg^{2+} > Al^{3+}\).

Hence, the correct answer is:

A significant decrease from \(O_2^–\) to \(Al^{3+}\).