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The increasing d-character in hybridisat...

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`

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The correct Answer is:
To determine the increasing d-character in the hybridization of xenon in the compounds XeF2, XeF4, and XeF6, we can follow these steps: ### Step 1: Analyze XeF2 - **Valence Electrons**: Xenon (Xe) has 8 valence electrons. In XeF2, it forms bonds with 2 fluorine atoms. - **Bonding and Lone Pairs**: Each fluorine atom uses 1 electron to form a bond, using 2 of Xe's 8 electrons. This leaves 6 electrons, which can be arranged as 3 lone pairs. - **Hybridization**: The hybridization is calculated as the sum of sigma bonds and lone pairs: - Sigma bonds = 2 (from 2 Xe-F bonds) - Lone pairs = 3 - Total = 2 + 3 = 5 - Thus, hybridization = sp³d (1 d-orbital involved). ### Step 2: Analyze XeF4 - **Valence Electrons**: In XeF4, xenon again has 8 valence electrons. It forms bonds with 4 fluorine atoms. - **Bonding and Lone Pairs**: 4 electrons are used to form bonds with fluorine, leaving 4 electrons. These can be arranged as 2 lone pairs. - **Hybridization**: - Sigma bonds = 4 (from 4 Xe-F bonds) - Lone pairs = 2 - Total = 4 + 2 = 6 - Thus, hybridization = sp³d² (2 d-orbitals involved). ### Step 3: Analyze XeF6 - **Valence Electrons**: In XeF6, xenon uses all 6 of its available electrons to bond with 6 fluorine atoms. - **Bonding and Lone Pairs**: 6 electrons are used for bonding, leaving 2 electrons, which can be arranged as 1 lone pair. - **Hybridization**: - Sigma bonds = 6 (from 6 Xe-F bonds) - Lone pairs = 1 - Total = 6 + 1 = 7 - Thus, hybridization = sp³d³ (3 d-orbitals involved). ### Step 4: Conclusion on Increasing d-Character - From the analysis: - XeF2: sp³d (1 d-orbital) - XeF4: sp³d² (2 d-orbitals) - XeF6: sp³d³ (3 d-orbitals) - Therefore, the increasing d-character in hybridization is: - XeF2 < XeF4 < XeF6 ### Final Answer The increasing d-character in hybridization of Xe in XeF2, XeF4, and XeF6 is: **sp³d < sp³d² < sp³d³**. ---

To determine the increasing d-character in the hybridization of xenon in the compounds XeF2, XeF4, and XeF6, we can follow these steps: ### Step 1: Analyze XeF2 - **Valence Electrons**: Xenon (Xe) has 8 valence electrons. In XeF2, it forms bonds with 2 fluorine atoms. - **Bonding and Lone Pairs**: Each fluorine atom uses 1 electron to form a bond, using 2 of Xe's 8 electrons. This leaves 6 electrons, which can be arranged as 3 lone pairs. - **Hybridization**: The hybridization is calculated as the sum of sigma bonds and lone pairs: - Sigma bonds = 2 (from 2 Xe-F bonds) - Lone pairs = 3 ...
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NCERT FINGERTIPS ENGLISH-CHEMICAL BONDING & MOLECULAR STRUCTURE-Assertion And Reason
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