Trends In Periodic Table

The correct answer is (B) : Both A and R are correct but R is NOT the correct explanation of A
N-atom has a half-filled p-sub shell which provides it extra stability, O-atom on the other hand has four electrons in the p-subshell , causing greater electron-electron repulsion and lesser stability.
Due to this it’s 1^st ionization energy is more than oxygen.

Statement I is incorrect because the decrease in ionization enthalpy from B to Al is influenced by the addition of a new electron shell, which increases shielding, causing a significant drop in ionization energy. However, from Al to Ga, the decrease is smaller due to the poor shielding effect of -electrons, leading to only a slight decrease in ionization enthalpy.
Statement II is correct because in Ga, the orbitals (3d) are completely filled, contributing to the slight decrease in ionization energy.

Step 1: Analyze the Atomic Numbers and Their Corresponding Blocks
The blocks of the periodic table are determined by the type of orbital being filled:
- 37 (K): Potassium is an alkali metal, and its valence electron is in the s-orbital. Hence, it belongs to the s-block.
- 78 (Pt): Platinum is a transition metal, with valence electrons in the d-orbital. Hence, it belongs to the d-block.
- 52 (Te): Tellurium is a p-block element, as its valence electrons are in the p-orbital.
- 65 (Tb): Terbium is a lanthanide, with electrons filling the f-orbital. Hence, it belongs to the f-block.
Step 2: Match List – I with List – II
The correct matches are:
- A – IV (37, s-block)
- B – II (78, d-block)
- C – I (52, p-block)
- D – III (65, f-block)
Conclusion: The correct answer is (4) A – IV, B – II, C – I, D – III.

The given question revolves around understanding the existence of different forms of sulfur and oxygen, and the nature of bonding in these elements. Let's analyze the Assertion (A) and Reason (R) provided:

1. Assertion (A): Both rhombic and monoclinic sulfur exist as S₈ while oxygen exists as O₂.
   • Sulfur can form long chains and rings due to its ability to form single bonds with itself. The most common and stable allotrope of sulfur is the S₈ ring, which is true for both rhombic and monoclinic forms of sulfur.
   • Oxygen, on the other hand, typically forms a double bond with itself, resulting in the diatomic O₂ molecule.
   • Thus, Assertion (A) is correct.
2. Reason (R): Oxygen forms - multiple bonds with itself and other elements having small size and high electronegativity like C, N, which is not possible for sulfur.
   • Oxygen's ability to form strong - bonds is due to its small size and high electronegativity, allowing for efficient side-to-side overlap of p orbitals, which is not possible for sulfur due to its larger size and lower electronegativity.
   • While the reasoning regarding oxygen's bonding is correct, it does not directly explain why sulfur exists as S₈. The ability of sulfur to form S₈ rings is more a factor of its valence electron configuration allowing for single bonds rather than an inability to form - bonds.
   • Thus, Reason (R) is not the correct explanation of why sulfur exists as S₈.
3. Conclusion: Given that Assertion (A) is correct but Reason (R) is not the correct explanation, the most appropriate answer is:

(A) is correct but (R) is not correct.

The problem presents two statements comparing the metallic (atomic) and ionic radii of Aluminum (Al) and Gallium (Ga). We need to evaluate the correctness of each statement to determine the most appropriate answer.

Concept Used:

The solution requires understanding periodic trends in atomic and ionic radii, along with a key exception known as the d-block contraction (or transition metal contraction).

1. Atomic/Metallic Radius Trend: Generally, the atomic radius increases as we move down a group in the periodic table. This is because each subsequent element has an additional principal energy level (electron shell), which increases the distance of the valence electrons from the nucleus.
2. d-Block Contraction: This is an exception to the general trend. For elements that come immediately after the first d-block series (like Gallium), the 10 electrons in the intervening d-orbitals (e.g., the 3d orbitals for Ga) are poor at shielding the outer valence electrons from the nuclear charge. The increase in nuclear charge is not effectively screened, leading to a higher effective nuclear charge ( ) than expected. This pulls the valence electrons closer to the nucleus, causing the atom to be smaller than predicted by the general trend.
3. Ionic Radius Trend: The trend for ionic radii of ions with the same charge generally follows the atomic radius trend (increases down a group). This is because the ions of elements further down the group still have more electron shells.

Step-by-Step Solution:

Step 1: Evaluation of Statement I: "The metallic radius of Al is less than that of Ga."

Aluminum (Al) is in Period 3 and Gallium (Ga) is in Period 4, both belonging to Group 13. Based on the general periodic trend, one would expect the radius of Ga to be larger than that of Al because Ga has an additional electron shell (n=4 vs n=3).

However, we must consider the d-block contraction. Gallium (atomic number 31) has the electron configuration [Ar] 3d 4s 4p . The 10 electrons in the 3d orbital are filled just before the 4p orbital. These 3d electrons provide very poor shielding for the outer 4s and 4p electrons.

Due to this poor shielding, the effective nuclear charge experienced by Gallium's valence electrons is unusually high, causing a contraction in its size. As a result, the metallic radius of Gallium is actually slightly smaller than that of Aluminum.

The experimentally determined metallic radii are approximately:

• Metallic radius of Al 143 pm
• Metallic radius of Ga 135 pm

Thus, the radius of Al is greater than the radius of Ga (Radius(Al) > Radius(Ga)). The statement claims that the radius of Al is less than that of Ga. Therefore, Statement I is incorrect.

Step 2: Evaluation of Statement II: "The ionic radius of Al is less than that of Ga ."

Let's consider the formation of the ions Al and Ga by removing their three valence electrons.

• Al atom: [Ne] 3s 3p Al ion: [Ne]
• Ga atom: [Ar] 3d 4s 4p Ga ion: [Ar] 3d

After ionization, the outermost electrons in the Al ion are in the n=2 principal energy level. For the Ga ion, the outermost electrons are in the n=3 principal energy level (the 3d shell). Since Ga has an additional electron shell compared to Al , its ionic radius is expected to be larger. The d-block contraction effect is less pronounced in the ions as the valence shells have been removed.

The experimentally determined ionic radii (for coordination number 6) are approximately:

• Ionic radius of Al 53.5 pm
• Ionic radius of Ga 62.0 pm

The radius of Al is indeed less than the radius of Ga . Therefore, Statement II is correct.

Final Conclusion:

Based on the analysis, Statement I is incorrect and Statement II is correct. The correct option is the one that reflects this finding.