Which ion below has the largest radius? a. Cl+ b. K+ c. Br+ d. F- e. Na+
The Correct Answer and Explanation is :
Among the ions listed—Cl⁺, K⁺, Br⁺, F⁻, and Na⁺—the fluoride ion (F⁻) has the largest radius.
Explanation:
Ionic radii are influenced by an ion’s charge and its position in the periodic table. Anions (negatively charged ions) are generally larger than their parent atoms because they gain electrons, increasing electron-electron repulsion and expanding the electron cloud. Conversely, cations (positively charged ions) are smaller than their parent atoms due to the loss of electrons, which reduces electron-electron repulsion and allows the remaining electrons to be drawn closer to the nucleus.
Comparing the Ions:
- F⁻ (Fluoride Ion): Fluorine gains one electron to form F⁻, resulting in an electron configuration of [He] 2s² 2p⁶. This configuration is isoelectronic with neon, but the increased electron-electron repulsion in the same shell causes the radius to be larger than that of the neutral fluorine atom. The ionic radius of F⁻ is approximately 133 pm. (Wikipedia)
- Cl⁺ (Chloride Ion): Chlorine loses one electron to form Cl⁺, resulting in an electron configuration of [Ne] 3s² 3p⁶. This configuration is isoelectronic with argon, but the increased nuclear charge pulls the electrons closer, resulting in a smaller radius compared to the neutral chlorine atom. The ionic radius of Cl⁺ is approximately 181 pm. (Wikipedia)
- K⁺ (Potassium Ion): Potassium loses one electron to form K⁺, resulting in an electron configuration of [Ar] 4s² 3d¹⁰ 4p⁶. This configuration is isoelectronic with argon, but the increased nuclear charge pulls the electrons closer, resulting in a smaller radius compared to the neutral potassium atom. The ionic radius of K⁺ is approximately 138 pm. (Wikipedia)
- Na⁺ (Sodium Ion): Sodium loses one electron to form Na⁺, resulting in an electron configuration of [Ne] 3s² 3p⁶. This configuration is isoelectronic with neon, but the increased nuclear charge pulls the electrons closer, resulting in a smaller radius compared to the neutral sodium atom. The ionic radius of Na⁺ is approximately 102 pm. (Wikipedia)
- Br⁺ (Bromide Ion): Bromine loses one electron to form Br⁺, resulting in an electron configuration of [Kr] 4s² 3d¹⁰ 4p⁶. This configuration is isoelectronic with krypton, but the increased nuclear charge pulls the electrons closer, resulting in a smaller radius compared to the neutral bromine atom. The ionic radius of Br⁺ is approximately 196 pm. (Wikipedia)
Conclusion:
Among the ions listed, F⁻ has the largest ionic radius, followed by Cl⁺, K⁺, Na⁺, and Br⁺. This trend is due to the combination of the ions’ charges and their positions in the periodic table, which influence the balance between nuclear charge and electron-electron repulsion.