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

Answer: (B)

A significant decrease from $O_2^–$ to $Al^{3+}$

Answer 2

To determine the trend in ionic radii, let's analyze the situation:
Oxygen ion ($O_2^–$) has 10 electrons, fluorine ion ($F^–$) has 10 electrons, sodium ion ($Na^+$) has 10 electrons, magnesium ion ($Mg^{2+}$) has 10 electrons, and aluminum ion ($Al^{3+}$) has 10 electrons.
All these ions have the same number of electrons, so the main factor affecting their ionic radii will be the effective nuclear charge. The effective nuclear charge increases as you move across a period (from left to right) in the periodic table.
As you move from $O_2^–$ to $Al^{3+}$, you are going across the periodic table from left to right. This means that the effective nuclear charge is increasing, which generally leads to a decrease in ionic radii.

So, the correct option is (B): A significant decrease from $O_2^–$ to $Al^{3+}$