Question:
The molecular shapes of SF4,CF4 and XeF4 are
The molecular shapes of SF4,CF4 and XeF4 are
Updated On: Apr 11, 2025
- the same with 2,0 and 1 lone pairs of electrons on the central atoms, respectively.
- the same with 1,1 and 1 lone pairs of electrons on the central atoms, respectively.
- the same with 0,1 and 2 lone pairs of electrons on the central atoms, respectively.
- the same with 1,0 and 2 lone pairs of electrons on the central atoms, respectively.
Hide Solution
Verified By Collegedunia
The Correct Option is D
Solution and Explanation
To determine the number of lone pairs and molecular shapes for SF4, CF4, and XeF4, let's examine each molecule:
1) SF4:
- Central atom = Sulfur (S), atomic number = 16
- Valence electrons = 6
- SF4 has 4 bonding pairs (from 4 F atoms) and 1 lone pair
- Electron pair geometry = Trigonal bipyramidal
- Molecular geometry = See-saw
2) CF4:
- Central atom = Carbon (C), atomic number = 6
- Valence electrons = 4
- CF4 has 4 bonding pairs and 0 lone pairs
- Electron pair geometry = Tetrahedral
- Molecular geometry = Tetrahedral
3) XeF4:
- Central atom = Xenon (Xe), atomic number = 54
- Valence electrons = 8
- XeF4 has 4 bonding pairs and 2 lone pairs
- Electron pair geometry = Octahedral
- Molecular geometry = Square planar
Conclusion:
- SF4 → 1 lone pair
- CF4 → 0 lone pairs
- XeF4 → 2 lone pairs
Correct answer: (D) the same with 1, 0 and 2 lone pairs of electrons on the central atoms, respectively.
Was this answer helpful?
0
0
Top Questions on Chemical bonding and molecular structure
- Show the formation of magnesium chloride by electron transfer. Write the name of the cation and anion present in the compound formed. (Atomic Number of Mg = 12, Cl = 17)
- CBSE Class X - 2025
- Science
- Chemical bonding and molecular structure
- A group 15 element forms \( d\pi - d\pi \) bond with transition metals. It also forms a hydride, which is the strongest base among the hydrides of other group members that form \( d\pi - d\pi \) bonds. The atomic number of the element is ________________________.
- JEE Main - 2025
- Chemistry
- Chemical bonding and molecular structure
Identify the correct orders against the property mentioned:
A. H$_2$O $>$ NH$_3$ $>$ CHCl$_3$ - dipole moment
B. XeF$_4$ $>$ XeO$_3$ $>$ XeF$_2$ - number of lone pairs on central atom
C. O–H $>$ C–H $>$ N–O - bond length
D. N$_2$>O$_2$>H$_2$ - bond enthalpy
Choose the correct answer from the options given below:- NEET (UG) - 2025
- Chemistry
- Chemical bonding and molecular structure
- Identify the geometry and hybridization of the central atom in SF₄.
- BITSAT - 2025
- Chemistry
- Chemical bonding and molecular structure
- What is the bond order for diatomic carbon according to molecular orbital theory?
- AIIMS Paramedical - 2025
- Chemistry
- Chemical bonding and molecular structure
View More Questions
Questions Asked in WBJEE exam
- The minimum force required to start pushing a body up a rough (having coefficient of friction \( \mu \)) inclined plane is \( F_1 \), while the minimum force needed to prevent it from sliding is \( F_2 \). If the inclined plane makes an angle \( \theta \) with the horizontal such that \( \tan\theta = 2\mu \), then the ratio \( \frac{F_1}{F_2} \) is:
- WBJEE - 2025
- Friction
- The variation of displacement with time of a simple harmonic motion (SHM) for a particle of mass \( m \) is represented by: \[ y = 2 \sin \left( \frac{\pi}{2} + \phi \right) \, \text{cm} \] The maximum acceleration of the particle is:
- WBJEE - 2025
- simple harmonic motion
- Consider a particle of mass 1 gm and charge 1.0 Coulomb at rest. Now, the particle is subjected to an electric field \( E(t) = E_0 \sin(\omega t) \) in the x-direction, where \( E_0 = 2 \, \text{N/C} \) and \( \omega = 1000 \, \text{rad/sec} \). The maximum speed attained by the particle is:
- WBJEE - 2025
- Electric Field
- The equation of a stationary wave along a stretched string is given by \[ y = 5 \sin \left( \frac{\pi x}{3} \right) \cos (40 \pi t) \] Additional Information Here, \(x\) and \(y\) are in cm and \(t\) in seconds. The separation between two adjacent nodes is:
- Ruma reached the metro station and found that the escalator was not working. She walked up the stationary escalator with velocity \( v_1 \) in time \( t_1 \). On another day, if she remains stationary on the escalator moving with velocity \( v_2 \), the escalator takes her up in time \( t_2 \). The time taken by her to walk up with velocity \( v_1 \) on the moving escalator will be:
- WBJEE - 2025
- Relative Motion
View More Questions
Concepts Used:
Chemical Bonding and Molecular Structure
Such a group of atoms is called a molecule. Obviously, there must be some force that holds these constituent atoms together in the molecules. The attractive force which holds various constituents (atoms, ions, etc.) together in different chemical species is called a chemical bond.
Types of Chemical Bonds:
There are 4 types of chemical bonds which are formed by atoms or molecules to yield compounds.
- Ionic Bonds - Ionic bonding is a type of chemical bonding which involves a transfer of electrons from one atom or molecule to another.
- Covalent Bonds - Compounds that contain carbon commonly exhibit this type of chemical bonding.
- Hydrogen Bonds - It is a type of polar covalent bonding between oxygen and hydrogen wherein the hydrogen develops a partial positive charge
- Polar Bonds - In Polar Covalent chemical bonding, electrons are shared unequally since the more electronegative atom pulls the electron pair closer to itself and away from the less electronegative atom.
Factors Affecting Bond Enthalpy in Chemical Bonding:
- Size of the Atom
- Multiplicity of Bonds
- Number of Lone Pair of Electrons Present
- Bond Angle