Home
Class 12
CHEMISTRY
The correct geometry and hybridisation f...

The correct geometry and hybridisation for `XeF_(4)` are

A

octahedral, `sp^(3)d^(2)`

B

trigonal bipyramidal, `sp^(3)d`

C

planar triangle, `sp^(3)d^(3)`

D

square planar, `sp^(3)d^(2)`

Text Solution

AI Generated Solution

The correct Answer is:
To determine the correct geometry and hybridization for \( \text{XeF}_4 \), we can follow these steps: ### Step 1: Determine the Valence Electrons Xenon (Xe) is a noble gas and has 8 valence electrons. Each fluorine (F) atom has 7 valence electrons, and since there are 4 fluorine atoms, they contribute a total of \( 4 \times 7 = 28 \) electrons. Therefore, the total number of valence electrons in \( \text{XeF}_4 \) is: \[ 8 + 28 = 36 \text{ electrons} \] ### Step 2: Draw the Lewis Structure In the Lewis structure of \( \text{XeF}_4 \): - Xenon forms 4 single bonds with 4 fluorine atoms, using 4 of its 8 valence electrons. - This leaves 4 electrons remaining, which will be placed as 2 lone pairs on the xenon atom. The Lewis structure can be represented as: ``` F | F - Xe - F | F ``` With two lone pairs on the xenon atom. ### Step 3: Determine the Number of Electron Pairs Now we count the number of bonding pairs and lone pairs: - **Bonding pairs**: 4 (from the 4 Xe-F bonds) - **Lone pairs**: 2 (on the xenon atom) ### Step 4: Calculate the Steric Number The steric number is the sum of the number of bonding pairs and lone pairs: \[ \text{Steric Number} = \text{Number of Bonding Pairs} + \text{Number of Lone Pairs} = 4 + 2 = 6 \] ### Step 5: Determine Hybridization The hybridization can be determined from the steric number: - A steric number of 6 corresponds to \( \text{sp}^3\text{d}^2 \) hybridization. ### Step 6: Determine the Geometry With a steric number of 6 and 2 lone pairs, the molecular geometry is derived from the arrangement of the bonding pairs: - The electron pair geometry is octahedral. - The molecular geometry, considering the positions of the lone pairs, is square planar. ### Conclusion Thus, the correct geometry and hybridization for \( \text{XeF}_4 \) are: - **Geometry**: Square planar - **Hybridization**: \( \text{sp}^3\text{d}^2 \) ---

To determine the correct geometry and hybridization for \( \text{XeF}_4 \), we can follow these steps: ### Step 1: Determine the Valence Electrons Xenon (Xe) is a noble gas and has 8 valence electrons. Each fluorine (F) atom has 7 valence electrons, and since there are 4 fluorine atoms, they contribute a total of \( 4 \times 7 = 28 \) electrons. Therefore, the total number of valence electrons in \( \text{XeF}_4 \) is: \[ 8 + 28 = 36 \text{ electrons} \] ...
Doubtnut Promotions Banner Mobile Dark
|

Similar Questions

Explore conceptually related problems

Geometry and hybridisation of Xe in XeOF_(4) molecule is

Specify the coordination geometry around and the hybridisation of N and B atoms in 1 : 1 complex of BF_(3) and NH_(3) .

Knowledge Check

  • The increasing d-character in hybridisation of Xe in XeF_(2), XeF_(4), XeF_(6) is

    A
    `sp^(2) , sp^(3) d, sp^(3) d^(2)`
    B
    `sp^(3) d, sp^(3) d^(2), sp^(3) d^(3)`
    C
    `sp^(3) d^(2), sp^(3) d, sp^(3) d^(3)`
    D
    `sp^(2), sp^(3), sp^(3) d`
  • The geometry of a complex species can be understood from the knowledge of type of hybridisation of orbitals of central atom. The hybridisation of orbitals of central atom in [B(OH_(4))]^(-) and the geometry of the complex are respectively.

    A
    `sp^(3)`, tetrahedral
    B
    `sp^(3)`, square planar
    C
    `sp^(3), d^(2)` octahedral
    D
    `dsp^(2)` square plannar.
  • The geometry of XeF_6 molecule and the hybridization of Xe atom in the molecule is :

    A
    Distorted octahedral and `sp^3 d^3`
    B
    Square planar and `sp^3d^2`
    C
    Pyramidal and `sp^3`
    D
    Octahedral and `sp^3 d^3`
  • Similar Questions

    Explore conceptually related problems

    Specify the coordination geometry around and the hybridisation of N and B atoms in 1 : 1 complex of BF_(3) and NH_(3) .

    Which reactions involves a change in the electron - pair geometry ( i.e., hybridisation ) for the under lined geometry ?

    The geometry of a complex species can be understood from the knowledge of type of hybridisation of orbitals of central atom. The hybridisation of orbitals of central atom in [B(OH_(4))]^(-) and the geometry of the complex are respectively.

    In which of the following pairs, both the species have the same hybridisation ? (I) SF_(4),XeF_(4) " " (II) I_(3)^(-),XeF_(2) " " (III) ICI_(4)^(-),SiCl_(4) " " (IV) ClO_(3)^(-),PO_(4)^(3-)

    Write the geometry of XeF_(4) and SOF_(4) and clearly indicate the position of lone pair of electrons and hybridization of the central atom.