Question

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 from the options given below:

• A. Both Statement I and Statement II are false
• B. Statement I is true but Statement II is false
• C. Both Statement I and Statement II are true
• D. Statement I is false but Statement II is true

Hint:

For p-block elements, remember the trend in the nature of oxides. For non-metals (like C, Si), oxides are acidic. For metalloids (like Ge), they are amphoteric/acidic. For metals (like Sn, Pb), they are amphoteric. Also, the properties of BF₃ as a classic Lewis acid are a fundamental concept in chemical bonding.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:

We are given two statements. Statement I deals with the acidic or amphoteric nature of oxides of Group 14 elements. Statement II describes the chemical properties and structure of boron trifluoride (BF₃).

Step 2: Detailed Explanation:

Analysis of Statement I:

The statement claims that there are exactly two pairs of Group 14 oxides in which both oxides are amphoteric.

General Trend in Group 14 Oxides:
As we move down Group 14 (C → Si → Ge → Sn → Pb), the acidic character of oxides decreases and amphoteric character increases.

CO₂ – Acidic
SiO₂ – Acidic
GeO₂ – Amphoteric (mainly acidic)
SnO, SnO₂ – Amphoteric
PbO, PbO₂ – Amphoteric

Evaluation of Given Pairs:

Pair 1: SiO₂ and CO₂
Both are acidic oxides. Hence, this pair does not qualify.

Pair 2: SnO and SnO₂
Both oxides are amphoteric and react with both acids and bases. This pair qualifies.

Pair 3: PbO and PbO₂
Both oxides are amphoteric in nature. This pair also qualifies.

Pair 4: GeO and GeO₂
GeO₂ is amphoteric but GeO is not commonly classified as amphoteric. Hence, this pair does not qualify.

Thus, there are exactly two qualifying pairs: (SnO, SnO₂) and (PbO, PbO₂).

Conclusion for Statement I:
Statement I is true.

Analysis of Statement II:

Electron-deficient molecule:
In BF₃, boron has only six electrons in its valence shell, making it electron deficient. This is true.

Lewis acid behavior:
Because of its electron deficiency, BF₃ can accept an electron pair. Hence, it acts as a Lewis acid. This is true.

Formation of adduct with NH₃:
Ammonia donates its lone pair to BF₃, forming the adduct \[ F_3B \leftarrow NH_3 \] This is true.

Geometry of BF₃:
BF₃ has three bond pairs and no lone pairs on boron. According to VSEPR theory, its geometry is trigonal planar with bond angles of 120°. This is true.

Conclusion for Statement II:
Statement II is true.

Step 3: Final Answer:

Both Statement I and Statement II are true. Correct option: (C)