Amphoteric Oxides of Group 14 and Lewis Acid Character of Boron Halides | JEE NEET Chemistry
Given below are two statements:

Statement I: The number of pairs among [SiO₂, CO₂], [SnO, SnO₂], [PbO, PbO₂] and [GeO, GeO₂], which contain oxides that are both amphoteric is 2.

Statement II: BF₃ is an electron deficient molecule, can act as a Lewis acid, forms adduct with NH₃ and has a trigonal planar geometry.

In the light of the above statements, choose the correct answer:
Correct Answer: C
Both Statement I and Statement II are true

Step-by-Step Logic & Calculation

1. Evaluating Statement I (Oxide Nature):
We need to examine the pairs of Group 14 oxides:
  • [CO₂, SiO₂]: Carbon dioxide is a molecular acidic oxide. Silica (SiO₂) is a covalent network acidic oxide. (Not the pair)
  • [GeO, GeO₂]: GeO is amphoteric, but GeO₂ is primarily acidic. (Not the pair)
  • [SnO, SnO₂]: Both Tin(II) and Tin(IV) oxides are classic Amphoteric oxides. (Pair #1)
  • [PbO, PbO₂]: Both Lead(II) and Lead(IV) oxides are Amphoteric. (Pair #2)
Since there are 2 such pairs, Statement I is True.
2. Evaluating Statement II (BF₃ Characteristics):
  • Electron Deficient: Boron (Z=5) has 3 valence electrons. In BF₃, it forms 3 bonds, totaling 6 electrons. It lacks an octet. (True)
  • Lewis Acid: Because it is electron-deficient, it seeks an electron pair. (True)
  • Adduct with NH₃: \(BF_3\) reacts with \(NH_3\) (Lewis base) to form a coordinate bond. (True)
  • Geometry: \(sp^2\) hybridization leads to a trigonal planar structure with 120° angles. (True)
Statement II is True.

Exhaustive Chemistry Theory & Concepts

Section 1: The Group 14 Oxide Landscape

Group 14 elements (Carbon, Silicon, Germanium, Tin, and Lead) show a remarkable transition in the chemical nature of their oxides as we move from top to bottom. This transition is governed by the increase in atomic size and decrease in electronegativity, which shifts the bonding character from purely covalent to increasingly ionic.

A. Acidic Oxides (CO₂ and SiO₂):
Carbon dioxide exists as discrete molecules with \(p\pi-p\pi\) multiple bonding. It dissolves in water to form carbonic acid (\(H_2CO_3\)). Silicon dioxide, however, cannot form stable \(p\pi-p\pi\) bonds due to the larger size of Silicon's 3p orbitals. Instead, it forms a massive three-dimensional covalent network where each Si is tetrahedrally bonded to four Oxygen atoms. Despite the structural difference, both are strictly acidic.

B. The Amphoteric Shift (Sn and Pb):
As we reach Tin (Sn) and Lead (Pb), the oxides become amphoteric. An amphoteric substance is one that can react as both an acid and a base.
Example for SnO₂:
1. As a Base: \(SnO_2 + 4HCl \rightarrow SnCl_4 + 2H_2O\)
2. As an Acid: \(SnO_2 + 2NaOH + H_2O \rightarrow Na_2[Sn(OH)_6]\) (Sodium stannate)
Lead oxides (PbO and PbO₂) show similar behavior. This duality is a hallmark of elements situated near the "metalloid line" in the periodic table.

Key Insight: GeO₂ is often debated in textbooks. While it shows very faint amphoteric tendencies in concentrated reagents, for the purpose of JEE/NEET, it is categorized as acidic, whereas GeO is considered amphoteric.

Section 2: Boron Trifluoride and the Lewis Concept

Boron is the first element of Group 13. Its small size and 3 valence electrons lead to unique bonding scenarios, most notably electron-deficient trivalent compounds like BF₃.

A. Hybridization and Geometry:
In BF₃, the Boron atom is in an \(sp^2\) hybridized state. One 's' and two 'p' orbitals mix to form three hybrid orbitals directed towards the corners of an equilateral triangle. The remaining 2p orbital on Boron remains empty and perpendicular to the molecular plane. This results in a Trigonal Planar geometry.

B. The Lewis Acid-Base Interaction:
A Lewis acid is defined as an electron-pair acceptor. Because Boron in BF₃ has only 6 electrons in its valence shell (an "incomplete octet"), it is highly electrophilic. When it encounters Ammonia (NH₃), which has a lone pair on the Nitrogen atom, a coordinate covalent bond is formed.
Reaction: \(F_3B + :NH_3 \rightarrow F_3B \leftarrow NH_3\)
In this adduct, the hybridization of Boron changes from \(sp^2\) (trigonal planar) to \(sp^3\) (tetrahedral). This is a critical point for advanced level questions.

C. The Role of Back-Bonding:
Interestingly, the Lewis acidity of Boron halides follows the order: \(BF_3 < BCl_3 < BBr_3 < BI_3\). This seems counterintuitive because Fluorine is the most electronegative. However, in BF₃, the empty 2p orbital of Boron overlaps with the filled 2p orbitals of Fluorine (p\(\pi\)-p\(\pi\) back-bonding). This partial double bond character stabilizes the Boron atom and reduces its tendency to accept external electron pairs.

Section 3: Periodic Trends and Electronegativity

The concepts in this question are tied together by Electronegativity and Oxidation States.

  • As electronegativity of the central atom decreases (down the group), the oxide shifts from acidic to basic. Amphoteric is the midpoint.
  • Higher oxidation states (+4) are generally more acidic than lower oxidation states (+2) for the same element due to increased charge density.
  • The "Inert Pair Effect" explains why Pb prefers the +2 state and Sn is stable in both +2 and +4, directly affecting their oxide chemistry.

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Syllabus Topics Covered

IIT JEE Inorganic Chemistry p-block Elements (Group 13 & 14) Amphoteric vs Acidic Oxides Lewis Acid-Base Theory Chemical Bonding and Molecular Structure Hybridization (sp2 vs sp3) Back-Bonding (pπ-pπ) Octet Rule Exceptions Coordinate Covalent Bonds Trends in Periodic Table JEE Advanced Problem Solving NEET Chemistry Concepts
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