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Q3.13:
Explain why cation are smaller and anions larger in radii than their parent atoms?

Solution :

Step 1: Analyzing Cationic Radii (Formation of Positively Charged Ions)

A cation is formed when a neutral atom loses one or more electrons. This process leads to two primary structural changes within the atomic configuration:

  • Reduced Electron-Electron Repulsion: As electrons are removed, the mutual repulsion between the remaining electrons in the valence shell decreases, allowing the electron cloud to contract [Per Coulombic interaction principles].
  • Increased Effective Nuclear Charge ($Z_{eff}$): Because the number of protons in the nucleus remains constant while the number of electrons decreases, the remaining electrons experience a stronger net pull from the positively charged nucleus.

Consequently, the valence shell is drawn significantly closer to the nucleus. Mathematical Representation: If $r$ is the radius, then $r_{cation} < r_{neutral\_atom}$.

Step 2: Analyzing Anionic Radii (Formation of Negatively Charged Ions)

An anion is formed when a neutral atom gains one or more electrons. This leads to the following structural modifications:

  • Increased Electron-Electron Repulsion: The addition of an electron to the valence shell increases the total inter-electronic repulsion. To maintain stability, the electrons redistribute themselves further apart to minimize potential energy.
  • Decreased Effective Nuclear Charge ($Z_{eff}$): The positive charge of the nucleus is now shared among a larger number of electrons. The nuclear pull per electron decreases, weakening the binding force on the outermost shell.

Consequently, the valence shell expands, resulting in a larger atomic radius. Mathematical Representation: $r_{anion} > r_{neutral\_atom}$.

Step 3: Visual Representation of Ionic Expansion and Contraction

Atom Cation Anion

Step 4: Comparative Summary Table

Parameter Cation ($M \rightarrow M^+ + e^-$) Anion ($X + e^- \rightarrow X^-$)
Electron-Electron Repulsion Decreases Increases
Effective Nuclear Charge ($Z_{eff}$) Increases Decreases
Ionic Radius Smaller than parent Larger than parent

Final Solution: Cations are smaller than their parent atoms because the loss of electrons reduces electron-electron repulsion and increases the net effective nuclear charge per electron, drawing the shell inward. Conversely, anions are larger because the addition of electrons increases inter-electronic repulsion and decreases the effective nuclear charge, causing the valence shell to expand outward.


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