Which of the following molecules has trigonal | vedclass

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(D) To determine the geometry,we look at the hybridization of the central atom:$1$. $CH_4$: The central carbon atom is $sp^3$ hybridized,resulting in

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$BF_3$

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(D) To determine the geometry,we look at the hybridization of the central atom:$1$. $CH_4$: The central carbon atom is $sp^3$ hybridized,resulting in a tetrahedral geometry.$2$. $C_2H_2$: The carbon atoms are $sp$ hybridized,resulting in a linear geometry.$3$. $NH_3$: The central nitrogen atom is $sp^3$ hybridized with one lone pair,resulting in a trigonal pyramidal geometry.$4$. $BF_3$: The central boron atom is $sp^2$ hybridized with no lone pairs,resulting in a trigonal planar geometry.Therefore,the correct molecule with trigonal planar geometry is $BF_3$.

List-$I$ (species)	List-$II$ ($O-N-O$ angle)
$A$. $NO_2^+$	$1$. $180^{\circ}$
$B$. $NO_2$	$2$. $132^{\circ}$
$C$. $NO_2^-$	$3$. $115^{\circ}$
$D$. $NO_3^-$	$4$. $120^{\circ}$

$List-I$ (Compound)	$List-II$ (Shape)
$A$. $BrF_5$	$I$. bent
$B$. $[CrF_6]^{3-}$	$II$. square pyramidal
$C$. $O_3$	$III$. trigonal bipyramidal
$D$. $PCl_5$	$IV$. octahedral

Which of the following molecules has trigonal planar geometry?

Similar Questions

Match List-$I$ with List-$II$ and select the correct answer:

List-$I$ (species) List-$II$ ($O-N-O$ angle)

$A$. $NO_2^+$ $1$. $180^{\circ}$

$B$. $NO_2$ $2$. $132^{\circ}$

$C$. $NO_2^-$ $3$. $115^{\circ}$

$D$. $NO_3^-$ $4$. $120^{\circ}$

In which of the following molecules,the maximum number of lone pairs is present on the central atom?

Match $List-I$ with $List-II$.
$List-I$ (Compound) $List-II$ (Shape)
$A$. $BrF_5$ $I$. bent
$B$. $[CrF_6]^{3-}$ $II$. square pyramidal
$C$. $O_3$ $III$. trigonal bipyramidal
$D$. $PCl_5$ $IV$. octahedral

Choose the correct answer from the options given below.

The bond angle $H-X-H$ is the greatest in the compound

$A$ molecule of the type $AX_5$ has square pyramidal geometry. Hence,the number of lone pairs on '$A$' is:

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