SbF_5 reacts with XeF_4 and XeF_6 to form ion | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$T$-shaped,Square pyramidal

Vedclass · Answer

(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 $B$.

$SbF_5$ 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

Which molecule $AX_3$,$AX_4$,$AX_5$,$AX_6$ is most likely to have a trigonal bipyramidal structure if $A$ has no lone pair?

Consider the following reactions:
$2 Cu_2S + 3 O_2 \rightarrow 2 Cu_2O + 2 SO_2$
$2 Cu_2O + Cu_2S \rightarrow 6 Cu + SO_2 \uparrow$
Here,$X$ is $Cu_2S$ and $Y$ is $SO_2$. The shape of the molecule $Y$ is:

The shapes of nitrite $(NO_2^-)$ and nitrile $(NO_2^-)$ respectively are

The geometry of $ClO_3^-$ according to valence shell electron pair repulsion $(VSEPR)$ theory will be

Which of the following molecules has no lone pair of electrons on the central atom?

Vedclass Test Series

Exam Paper Generator

Online Exam Module