SbF_5 reacts with XeF_4 and XeF_6 to form ion | vedclass

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(B) The number of orbitals involved in hybridization $(H)$ is calculated as $H = \frac{1}{2} [V + M - C + A]$,where $V$ is the number of valence elect

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$T$-shaped,Square pyramidal

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(B) The number of orbitals involved in hybridization $(H)$ is calculated as $H = \frac{1}{2} [V + M - C + A]$,where $V$ is the number of valence electrons of the central atom,$M$ is the number of monovalent atoms,$C$ is the charge of the cation,and $A$ is the charge of the anion.For $[XeF_3]^+$,$H = \frac{1}{2} [8 + 3 - 1] = 5$,which corresponds to $sp^3d$ hybridization with $2$ lone pairs. This results in a $T$-shaped geometry.For $[XeF_5]^+$,$H = \frac{1}{2} [8 + 5 - 1] = 6$,which corresponds to $sp^3d^2$ hybridization with $1$ lone pair. This results in a square pyramidal geometry.Therefore,the correct option is $B$.

List-$I$	List-$II$
$A. ICl$	$i. \text{Linear}$

$B. ICl_3$	$ii. \text{T-Shape}$
$C. ClF_5$	$iii. \text{Square pyramidal}$
$D. IF_7$	$iv. \text{Pentagonal bipyramidal}$

$SbF_5$ reacts with $XeF_4$ and $XeF_6$ to form ionic compounds $[XeF_3]^+ [SbF_6]^-$ and $[XeF_5]^+ [SbF_6]^-$ respectively. What are the molecular shapes of the $[XeF_3]^+$ and $[XeF_5]^+$ ions,respectively?

Similar Questions

The $H_3O^{+}$ ion has the following shape:

Match List-$I$ with List-$II$.
List-$I$ List-$II$
$A. ICl$ $i. \text{Linear}$
$B. ICl_3$ $ii. \text{T-Shape}$
$C. ClF_5$ $iii. \text{Square pyramidal}$
$D. IF_7$ $iv. \text{Pentagonal bipyramidal}$

Choose the correct answer from the options given below:

Which of the following species is not tetrahedral?

According to Lewis theory,the total number of bond-pairs and lone pairs of electrons around the central atom of $XeO_6^{4-}$ ion is . . . . . . .

Which of the following species are isostructural,planar,and non-polar?

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