MCQEasyJEE 2026VSEPR Theory & Shapes of Molecules

JEE Chemistry 2026 Question with Solution

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 22.

Statement II: BF3BF_3 is an electron deficient molecule, can act as a Lewis acid, forms adduct with NH3NH_3 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

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: Two statements are to be checked. Statement I concerns the nature of Group 1414 oxides, and Statement II concerns the properties of BF3BF_3.

Find: Which option correctly describes the truth values of Statement I and Statement II.

For Statement I, examine each pair:

  • [SiO2SiO_2, CO2CO_2]: both are acidic oxides, so this pair does not qualify.
  • [SnOSnO, SnO2SnO_2]: both are amphoteric oxides, so this pair qualifies.
  • [PbOPbO, PbO2PbO_2]: both are amphoteric oxides, so this pair qualifies.
  • [GeOGeO, GeO2GeO_2]: GeO2GeO_2 is amphoteric, but GeOGeO is not typically classified as amphoteric, so this pair does not qualify.

Thus, the number of pairs in which both oxides are amphoteric is exactly 22. Therefore, Statement I is true.

For Statement II:

  • BF3BF_3 is electron deficient because boron has only six electrons around it.
  • Therefore, BF3BF_3 can act as a Lewis acid.
  • It forms an adduct with NH3NH_3 as F3BNH3F_3B \leftarrow NH_3.
  • With three bond pairs and no lone pair on boron, its geometry is trigonal planar.

Hence, Statement II is true.

Therefore, both Statement I and Statement II are true. The correct option is C.

Common mistakes

  • Treating GeOGeO and GeO2GeO_2 as both amphoteric is incorrect because only GeO2GeO_2 is commonly taken as amphoteric in this context. Check the standard trend of Group 1414 oxides instead of extending the pattern too loosely.

  • Assuming BF3BF_3 cannot be a Lewis acid because fluorine shows back-bonding is incorrect. Despite back-bonding, boron remains electron deficient and can accept a lone pair. Focus on the incomplete octet on boron.

  • Confusing molecular shape of BF3BF_3 with pyramidal geometry is incorrect. Boron in BF3BF_3 has three bond pairs and no lone pair, so VSEPR predicts trigonal planar geometry, not trigonal pyramidal.

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