Among the following compounds,the one that is | vedclass

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(A) $1$. $BF_3$ has a trigonal planar geometry with $sp^2$ hybridization,but it is non-polar due to the cancellation of dipole moments.$2$. $SiF_4$ ha

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$H_2CO_3$

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(A) $1$. $BF_3$ has a trigonal planar geometry with $sp^2$ hybridization,but it is non-polar due to the cancellation of dipole moments.$2$. $SiF_4$ has a tetrahedral geometry with $sp^3$ hybridization and is non-polar.$3$. $HClO_2$ has a central $Cl$ atom with $sp^3$ hybridization.$4$. $H_2CO_3$ (carbonic acid) has a central carbon atom with $sp^2$ hybridization. Due to the asymmetric arrangement of atoms and the difference in electronegativity between $C$,$O$,and $H$,the molecule is polar.

List-$I$ (Hybridisation)	List-$II$ (Shape)
$A$. $dsp^2$	$I$. Square planar
$B$. $sp^3$	$II$. Tetrahedral
$C$. $d^2sp^3$	$III$. Octahedral
$D$. $sp^3d$	$IV$. Trigonal bipyramidal

Among the following compounds,the one that is polar and has a central atom with $sp^2$ hybridization is:

Similar Questions

Match the following List-$I$ (Hybridisation) with List-$II$ (Shape):
List-$I$ (Hybridisation) List-$II$ (Shape)
$A$. $dsp^2$ $I$. Square planar
$B$. $sp^3$ $II$. Tetrahedral
$C$. $d^2sp^3$ $III$. Octahedral
$D$. $sp^3d$ $IV$. Trigonal bipyramidal

If $p_x$ and $p_y$ orbitals participate in $sp^2$ hybridisation of the sulphur atom in $SO_3$,then the $d$ orbitals which are used in $p\pi-d\pi$ bonds are

As a result of $sp$ hybridization,we get

If pure unhybridised orbitals are used in forming $CH_4$ then which statement is incorrect :-

In which of the following hybridizations is the maximum bond angle of $90^\circ$ observed between bond pair-bond pair electrons?

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