Which of the following compounds is octahedra | vedclass

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(C) To determine the geometry,we calculate the hybridization of the central atom:$1$. For $SF_6$,the central atom is $S$ (Sulfur). It has $6$ valence

Vedclass · Accepted Answer

$SF_6$

Vedclass · Answer

(C) To determine the geometry,we calculate the hybridization of the central atom:$1$. For $SF_6$,the central atom is $S$ (Sulfur). It has $6$ valence electrons and forms $6$ bonds with $F$ atoms. The steric number is $6 + 0 = 6$,which corresponds to $sp^3d^2$ hybridization and an octahedral geometry.$2$. $TeF_4$ has a steric number of $5$ ($4$ bond pairs + $1$ lone pair),resulting in a see-saw shape.$3$. $SeCl_2$ has a steric number of $4$ ($2$ bond pairs + $2$ lone pairs),resulting in a bent shape.$4$. $SeF_4$ has a steric number of $5$ ($4$ bond pairs + $1$ lone pair),resulting in a see-saw shape.Therefore,$SF_6$ is the only octahedral molecule among the given options.

List-$I$	List-$II$
$a$. See-Saw Shape	$i$. $XeF_4$
$b$. Square Pyramidal	$ii$. $ClF_3$
$c$. $T$-Shape	$iii$. $PbCl_2$
$d$. Bent Shape	$iv$. $SF_4$
	$v$. $BrF_5$

Which of the following compounds is octahedral?

Similar Questions

Which of the following molecules has trigonal planar geometry?

The $\angle FAsF$ bond angle in the $AsF_3Cl_2$ molecule is:

Match the following:
List-$I$ List-$II$
$a$. See-Saw Shape $i$. $XeF_4$
$b$. Square Pyramidal $ii$. $ClF_3$
$c$. $T$-Shape $iii$. $PbCl_2$
$d$. Bent Shape $iv$. $SF_4$
$v$. $BrF_5$

The correct answer is

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

The geometry of $SF_6$ molecule is

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