The type of oxide formed by the element among Li, Na, Be, Mg, B and Al that has the least atomic radius is: (1) A₂O₃$A_{2}O_{3}$ (2) AO₂$AO_{2}$ (3) AO (4) A₂O$A_{2}O$

Related Formula

Standard downward group variations typically follow predictable monotonic pathways:

Core Logic

Let's review the explicit values and trends across F, Cl, Br, I$\mathrm{F, Cl, Br, I}$:

• Covalent radius: F < Cl < Br < I$\mathrm{F < Cl < Br < I}$ (Perfect monotonic increase).
• Ionic radius: F^- < Cl^- < Br^- < I^-$\mathrm{F^- < Cl^- < Br^- < I^-}$ (Perfect monotonic increase).
• First ionization energy: F > Cl > Br > I$\mathrm{F > Cl > Br > I}$ (Perfect monotonic decrease).
• Electron affinity: Cl > F > Br > I$\mathrm{Cl > F > Br > I}$ (Irregular trend! Fluorine has an anomalously lower electron affinity than Chlorine due to high inter-electronic repulsions inside its exceptionally compact 2p$2p$ valence subshell).

Step 1: Conclusion

Therefore, only electron affinity (B) demonstrates an irregular trend line among the first four halogens.

Pattern Recognition

Fluorine anomalous properties are a classic JEE question archetype. Whenever a question asks about irregularities in halogens, immediately check Electron Gain Enthalpy (Electron Affinity) and Bond Dissociation Enthalpy, where Fluorine routinely breaks the monotonic descending order.

Chapter Mix

Class 11 Chemistry: Classification of Elements and Periodicity in Properties Class 12 Chemistry: The p-Block Elements

Related Formula

Effective nuclear charge scaling expression layout:

Core Logic

Let's analyze both properties systematically:

• Statement I: The metallic radius of Gallium (Ga$\mathrm{Ga}$) is anomalously smaller than Aluminum (Al$\mathrm{Al}$) due to the poor shielding effect of the 10$10$ d-electrons inserted before it. This increases Zeff$Z_{\text{eff}}$, pulling the outer shell inward tightly. Thus, Radius of Al > Radius of Ga$\text{Radius of Al} > \text{Radius of Ga}$, making Statement I incorrect.
• Statement II: When looking at completely stripped ionic configurations (Al³⁺$\mathrm{Al^{3+}}$ and Ga³⁺$\mathrm{Ga^{3+}}$), the extra shell layer in Gallium resumes its dominant role, meaning standard periodic increase down a group holds true: Al³⁺ < Ga³⁺$\mathrm{Al^{3+} < Ga^{3+}}$. (Statement II is correct).

Step 1: Evaluation

Thus, Statement I is incorrect but Statement II is correct.

Pattern Recognition

This is a classic trap: d-block contraction reverses the atomic/metallic radius trend line between Al and Ga, but does NOT reverse the ionic radius sequence line where Ga3+ is larger than Al3+.

Chapter Mix

Class 11 Chemistry: Classification of Elements and Periodicity in Properties Class 11 Chemistry: The p-Block Elements

Related Formula

The groups of the modern periodic table correspond to standard IUPAC group families:

• Group 15 (Pnictogens): Nitrogen family
• Group 16 (Chalcogens): Oxygen family
• Group 17 (Halogens): Fluorine family
• Group 18 (Noble Gases): Helium family

Core Logic

Map the heavy transactinide elements (Period 7) to their respective periodic groups using their atomic numbers:

• Moscovium (Mc, Z=115$Z=115$): Group 15 (Pnicogen)
• Livermorium (Lv, Z=116$Z=116$): Group 16 (Chalcogen)
• Tennessine (Ts, Z=117$Z=117$): Group 17 (Halogen)
• Oganesson (Og, Z=118$Z=118$): Group 18 (Noble Gas)

Step 1: Match the symbols

• A (Pnicogen) arrow$\rightarrow$ IV (Mc)
• B (Chalcogen) arrow$\rightarrow$ III (Lv)
• C (Halogen) arrow$\rightarrow$ I (Ts)
• D (Noble Gas) arrow$\rightarrow$ II (Og)

This maps to A-IV, B-III, C-I, D-II, matching Option (2).

Pattern Recognition

Modern IUPAC nomenclature adds heavy synthetic elements to complete Period 7. They correspond to group properties: 115$115$ is below bismuth (Pnicogen), 116$116$ is below polonium (Chalcogen), 117$117$ is below astatine (Halogen), and 118$118$ is below radon (Noble Gas).

Chapter Mix

Class 11 Chemistry: Classification of Elements and Periodicity in Properties

Core Logic

Let's list the identity of the elements given by their atomic numbers (Z$Z$):

• Z = 19 Potassium (K)$Z = 19 \implies \text{Potassium (K)}$, an alkali metal in Period 4.
• Z = 32 Germanium (Ge)$Z = 32 \implies \text{Germanium (Ge)}$, a metalloid in Group 14, Period 4.
• Z = 35 Bromine (Br)$Z = 35 \implies \text{Bromine (Br)}$, a halogen in Group 17, Period 4.
• Z = 87 Francium (Fr)$Z = 87 \implies \text{Francium (Fr)}$, an alkali metal in Period 7.

Ionization Enthalpy Trends:

• Ionization enthalpy increases across a period from left to right due to an increase in effective nuclear charge.
• Ionization enthalpy decreases down a group due to increasing atomic radius and increasing screening effects.

Comparing alkali metals (K$K$ and Fr$Fr$), Francium (Fr, [Rn]7s¹$Fr, [Rn]7s^1$) is located far lower in Group 1, possessing an immense atomic size and the highest shielding. This makes its outermost valence electron exceptionally loose and effortless to remove.

Pattern Recognition

Alkali metals always define the absolute minimum first ionization energy in their respective horizontal rows. Among alkali metals, values monotonically decrease downwards, validating Francium (Z=87$Z=87$) immediately.

Chapter Mix

Class 11 Chemistry: Classification of Elements and Periodicity in Properties