$XeF_4$ is square planar whereas $CCl_4$ is tetrahedral because
- Ain $XeF_4$,'$Xe$' is $sp^2$ hybridised and in $CCl_4$ '$C$' is $sp^3$ hybridised
- Bin both $XeF_4$ and $CCl_4$ the central atom is $sp^3$ hybridised
- Cin $XeF_4$,'$Xe$' is $sp^3 d^2$ hybridised but due to the presence of $2$ lone pairs of electrons shape is square planar whereas in $CCl_4$ '$C$' is $sp^3$ hybridised
- D$Xe$ is noble gas,whereas $C$ is a non-metal
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Fill in the blanks:
$(i)$ Based on the type of overlapping,covalent bonds are classified into ........ and .......... .
$(ii)$ The bond formation involving hybrid orbitals is called ............. .
$(iii)$ The type of hybridization observed in the $C_2H_6$ molecule is ........... .
$(iv)$ The type of hybridization observed in the $BrF_5$ complex is ............... .
$(i)$ Based on the type of overlapping,covalent bonds are classified into ........ and .......... .
$(ii)$ The bond formation involving hybrid orbitals is called ............. .
$(iii)$ The type of hybridization observed in the $C_2H_6$ molecule is ........... .
$(iv)$ The type of hybridization observed in the $BrF_5$ complex is ............... .
Medium
View SolutionWhich one of the following has the maximum number of hybrid orbitals?
Considering the state of hybridization of carbon atoms,find out the molecule among the following which is linear?
MediumAIPMT 2011
View SolutionIn which one of the following pairs do the central atoms exhibit $sp^2$ hybridization?
The order of electronegativity of carbon in $sp$,$sp^{2}$ and $sp^{3}$ hybridised states follows:
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