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Classification of Elements and Periodicity in Properties - Ionic Radii

Grade 11CBSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Ionic radius is the effective distance from the nucleus of an ion up to which it exerts its influence on its electron cloud.

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A cation is always smaller than its parent atom (e.g., Na+<NaNa^+ < Na). This is because the loss of electrons reduces inter-electronic repulsion while the nuclear charge remains constant, increasing the effective nuclear charge (ZeffZ_{eff}).

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An anion is always larger than its parent atom (e.g., Cl−>ClCl^- > Cl). The addition of electrons increases inter-electronic repulsion, causing the electron cloud to expand and decreasing the ZeffZ_{eff} per electron.

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Isoelectronic species are atoms/ions having the same number of electrons (e.g., O2−,F−,Ne,Na+,Mg2+O^{2-}, F^-, Ne, Na^+, Mg^{2+} all have 1010 electrons).

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For isoelectronic species, the ionic radius decreases with an increase in atomic number (ZZ) because the increasing nuclear charge pulls the same number of electrons closer to the nucleus.

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Across a period, ionic radii follow the same trend as atomic radii (generally decreasing), though there is a jump when transitioning from cations to anions. Down a group, ionic radii increase due to the addition of new principal energy shells (nn).

📐Formulae

Zeff=Z−σZ_{eff} = Z - \sigma

Ionic Radius∝1Z (for isoelectronic species)Ionic\ Radius \propto \frac{1}{Z} \text{ (for isoelectronic species)}

Radiuscation<Radiusatom<RadiusanionRadius_{cation} < Radius_{atom} < Radius_{anion}

💡Examples

Problem 1:

Arrange the following isoelectronic species in increasing order of their ionic radii: N3−,Mg2+,Na+,O2−,F−,Al3+N^{3-}, Mg^{2+}, Na^+, O^{2-}, F^-, Al^{3+}.

Solution:

Al3+<Mg2+<Na+<F−<O2−<N3−Al^{3+} < Mg^{2+} < Na^+ < F^- < O^{2-} < N^{3-}

Explanation:

All these species have 1010 electrons. The atomic numbers (ZZ) are Al=13,Mg=12,Na=11,F=9,O=8,N=7Al=13, Mg=12, Na=11, F=9, O=8, N=7. As ZZ increases, the nuclear pull on the 1010 electrons increases, causing the radius to decrease. Thus, the ion with the highest positive charge (Al3+Al^{3+}) is the smallest, and the ion with the highest negative charge (N3−N^{3-}) is the largest.

Problem 2:

Explain why the radius of Fe2+Fe^{2+} is larger than that of Fe3+Fe^{3+}.

Solution:

Fe2+>Fe3+Fe^{2+} > Fe^{3+}

Explanation:

Both ions have the same nuclear charge (Z=26Z=26). However, Fe3+Fe^{3+} has lost one more electron than Fe2+Fe^{2+}. In Fe3+Fe^{3+}, the ratio of nuclear charge to the number of electrons is higher (26/2326/23 vs 26/2426/24), leading to a greater inward pull on the remaining electrons, making the Fe3+Fe^{3+} ion smaller.