Which of the following molecules has trigonal | vedclass

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(D) To determine the geometry,we calculate the number of electron pairs around the central atom using the formula: $H = \frac{1}{2}(V + M - C + A)$,wh

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$BF_3$

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(D) To determine the geometry,we calculate the number of electron pairs around the central atom using the formula: $H = \frac{1}{2}(V + M - C + A)$,where $V$ is the number of valence electrons of the central atom,$M$ is the number of monovalent atoms,$C$ is the cationic charge,and $A$ is the anionic charge.For $BF_3$:Central atom $B$ has $3$ valence electrons.Number of monovalent atoms $(F)$ = $3$.$H = \frac{1}{2}(3 + 3) = 3$.Since there are $3$ bond pairs and $0$ lone pairs,the hybridization is $sp^2$ and the geometry is trigonal planar.For $IF_3$: $H = \frac{1}{2}(7 + 3) = 5$ ($sp^3d$,$T$-shaped).For $PCl_3$: $H = \frac{1}{2}(5 + 3) = 4$ ($sp^3$,trigonal pyramidal).For $NH_3$: $H = \frac{1}{2}(5 + 3) = 4$ ($sp^3$,trigonal pyramidal).Therefore,the correct option is $D$.

List-$I$	List-$II$
$I$. $XeF_4$	$A$. See-saw
$II$. $I_3^-$	$B$. Tetrahedral
$III$. $XeO_2F_2$	$C$. Bond angle $90^o$
$IV$. $SO_4^{2-}$	$D$. Linear

Which of the following molecules has trigonal planar geometry?

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In $BrF_3$ molecule,the lone pairs occupy equatorial positions to minimize

Match List-$I$ with List-$II$ and select the correct answer using the codes given below the lists.

List-$I$ List-$II$

$I$. $XeF_4$ $A$. See-saw

$II$. $I_3^-$ $B$. Tetrahedral

$III$. $XeO_2F_2$ $C$. Bond angle $90^o$

$IV$. $SO_4^{2-}$ $D$. Linear

Are all the five bonds in $PCl_{5}$ molecule equivalent? Justify your answer.

Which of the following is planar?

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