Which of the following ions has the greater charge density? (a) \(\mathrm{Na}^{+} ;\) (b) \(\mathrm{F}^{-} ;\) (c) \(\mathrm{K}^{+} ;\) (d) \(\mathrm{Cl}^{-}\).

Short Answer

Expert verified
The ion with the greatest charge density is \(\mathrm{F}^{-}\) (Fluorine ion).

Step by step solution

01

Understanding the Concept and Given Ions

Charge density is the measure of charge per unit volume. For ions with the same charge, those with smaller atomic radii will have a greater charge density because the same amount of charge is spread over a smaller volume. The given ions are \(\mathrm{Na}^{+}\), \(\mathrm{F}^{-}\), \(\mathrm{K}^{+}\), and \(\mathrm{Cl}^{-}\). \(\mathrm{Na}^{+}\) and \(\mathrm{K}^{+}\) are cations (positive ions) and \(\mathrm{F}^{-}\) and \(\mathrm{Cl}^{-}\) are anions (negative ions). All these ions carry a single charge.
02

Reference to the Periodic Table for Atomic Radii

From the periodic table, we know that the atomic radii increase as we move down a group and decrease as we move from left to right across a period. Given that Fluorine (F) and chlorine (Cl) are from the same group (Group 17 or VIIA), Fluorine being above Chlorine, has a smaller atomic radius. Similarly, Sodium (Na) and potassium (K) are in the same group (Group 1 or IA) with Sodium being above Potassium, thus, Sodium has a smaller atomic radius. Between Fluorine and Sodium, as we move from Group 1 to Group 17, the atomic radius decreases, so Fluorine's radius is the smallest.
03

Determining the Ion with the Greatest Charge Density

After comparing these atomic radii, we can conclude that Fluorine, with the smallest atomic radius, has the greatest charge density among the given ions.

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