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
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 |
Show Hint
Always check the number of lone pairs and the hybridization before deciding on the geometry of a molecule.
Updated On: Mar 18, 2026
- I
- II
- III
- IV
Show Solution
Verified By Collegedunia
The Correct Option is A
Approach Solution - 1
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.
Was this answer helpful?
0
0
Show Solution
Verified By Collegedunia
Approach Solution -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.
- 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.
Was this answer helpful?
0
0
Top AP EAPCET Chemistry Questions
- The four quantum numbers of the valence electron of potassium are :
- The volume of $0.02\, M$ acidified permanganate solution required for complete reaction of $60\, mL$ of $0.01\, MI^{-}$ ion solution to form $I_2$ in $mL$ is
- In which of the following the election gain enthalpy of elements is correctly arranged?
- The radii of $ \text{F,}{{\text{F}}^{-}}\text{O} $ and $ {{\text{O}}^{2-}} $ are in the order
- Which of the following has the highest first ionization potential?
View More Questions
Top AP EAPCET Hybridization Questions
Top AP EAPCET Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- The four quantum numbers of the valence electron of potassium are :
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The volume of $0.02\, M$ acidified permanganate solution required for complete reaction of $60\, mL$ of $0.01\, MI^{-}$ ion solution to form $I_2$ in $mL$ is
View More Questions