Hybridization

Let us analyze each set: - I. SiH₄: - Hybridization: sp³ - Geometry: Tetrahedral - Number of lone pairs on central atom: 0 This is correct because silicon (Si) in SiH₄ has no lone pairs and forms a tetrahedral shape with sp³ hybridization. - II. BeCl₂: - Hybridization: sp² - Geometry: Linear - Number of lone pairs on central atom: 1 This is incorrect because beryllium (Be) in BeCl₂ forms two bonds and has no lone pairs. Hence, its geometry is linear with sp hybridization, not sp². - III. SF₄: - Hybridization: dsp³ - Geometry: Square planar - Number of lone pairs on central atom: 1 This is incorrect because SF₄ has a seesaw geometry with 1 lone pair, not square planar. Square planar geometry occurs with dsp² hybridization. - IV. SnCl₂: - Hybridization: sp - Geometry: Linear - Number of lone pairs on central atom: 0 This is incorrect because SnCl₂ has an sp³ hybridization and two lone pairs on the central tin atom, leading to a bent structure. Thus, the correct answer is option (1), which is I.

- When diborane burns in air, products are and , not and .
- Boron in orthoboric acid is trigonal planar with hybridization.

Therefore, Statement-I is incorrect, Statement-II is correct.