Based upon VSEPR theory,match the shape (geom | vedclass

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(B) To determine the shape of the molecules,we use the $VSEPR$ theory:$1$. $ClF_3$: The central atom $Cl$ has $7$ valence electrons. It forms $3$ bond

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$A-III, B-IV, C-I, D-II$

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(B) To determine the shape of the molecules,we use the $VSEPR$ theory:$1$. $ClF_3$: The central atom $Cl$ has $7$ valence electrons. It forms $3$ bonds with $F$ atoms and has $2$ lone pairs. The steric number is $3 + 2 = 5$ ($sp^3d$ hybridization). Due to $2$ lone pairs in equatorial positions,the shape is $T$-shaped $(A-III)$.$2$. $BF_3$: The central atom $B$ has $3$ valence electrons. It forms $3$ bonds with $F$ atoms and has $0$ lone pairs. The steric number is $3$ ($sp^2$ hybridization). The shape is trigonal planar $(B-IV)$.$3$. $XeF_4$: The central atom $Xe$ has $8$ valence electrons. It forms $4$ bonds with $F$ atoms and has $2$ lone pairs. The steric number is $4 + 2 = 6$ ($sp^3d^2$ hybridization). Due to $2$ lone pairs in axial positions,the shape is square planar $(C-I)$.$4$. $SF_4$: The central atom $S$ has $6$ valence electrons. It forms $4$ bonds with $F$ atoms and has $1$ lone pair. The steric number is $4 + 1 = 5$ ($sp^3d$ hybridization). Due to $1$ lone pair in an equatorial position,the shape is see-saw $(D-II)$.Thus,the correct matching is $A-III, B-IV, C-I, D-II$.

List-$I$ (Shape)	List-$II$ (Molecules)
$A$. $T$-shaped	$I$. $XeF_4$
$B$. Trigonal planar	$II$. $SF_4$
$C$. Square planar	$III$. $ClF_3$
$D$. See-saw	$IV$. $BF_3$

Similar Questions

State the shapes of molecules that contain only bonding electron pairs around the central atom.

The increasing order of the number of lone pairs of electrons on the central atom of the following molecules is:
$I) \ ClF_3$
$II) \ XeF_2$
$III) \ SF_4$
$IV) \ SiH_4$

The shape of $SO_4^{2-}$ ion is

What is the shape of $XeF_4$?

Which of the following shapes is not possible for any possible value of $n$ in the $XeF_n$ molecule?

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