Question

Which of the following sets is correct? 

#	Molecule	Hybridization	Geometry	No. of lone pairs on central atom
I	\(\mathrm{SiH_4}\)	\(\mathrm{sp^3}\)	tetrahedral	0
II	\(\mathrm{BeCl_2}\)	\(\mathrm{sp^2}\)	linear	1
III	\(\mathrm{SF_4}\)	\(\mathrm{dsp^3}\)	seesaw	1
IV	\(\mathrm{SnCl_2}\)	\(\mathrm{sp^2}\)	bent	2
V	\(\mathrm{CH_4}\)	\(\mathrm{sp^3}\)	tetrahedral	0

• A. I
• B. II
• C. III
• D. IV

Hint:

Always check the number of lone pairs and the hybridization before deciding on the geometry of a molecule.

Answer 1

The Correct Option is A

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.

Answer 2

Step 1: Analyze Each Molecule's Hybridization and Geometry
- I. \(\mathrm{SiH_4}\): Silicon forms four single bonds with hydrogen, with no lone pairs. Hybridization is \(\mathrm{sp^3}\) and geometry is tetrahedral. This is correct.

- II. \(\mathrm{BeCl_2}\): Beryllium dichloride is linear with \(\mathrm{sp}\) hybridization and no lone pairs on Be. Given \(\mathrm{sp^2}\) and 1 lone pair is incorrect.

- III. \(\mathrm{SF_4}\): Sulfur tetrafluoride has \(\mathrm{dsp^3}\) hybridization, geometry is seesaw, and has 1 lone pair on sulfur. This is correct.
However, the question asks for the correct set, so partial incorrectness in other options matters.

- IV. \(\mathrm{SnCl_2}\): Tin dichloride is bent with \(\mathrm{sp^2}\) hybridization and 1 lone pair, not 2.

- V. \(\mathrm{CH_4}\): Methane is \(\mathrm{sp^3}\) hybridized, tetrahedral, with zero lone pairs, which is correct.

Step 2: Conclusion
Only set I correctly matches hybridization, geometry, and lone pairs. Other options have errors in hybridization or lone pairs.
Hence, the correct answer is I.