The pair of xenon compounds,which have the sa | vedclass

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

Question

(A) To determine the number of lone pairs on the central $Xe$ atom,we use the formula: $	ext{Lone pairs} = \frac{1}{2} (V - M)$,where $V$ is the numb

Vedclass · Accepted Answer

$XeO_3, XeF_6$

Vedclass · Answer

(A) To determine the number of lone pairs on the central $Xe$ atom,we use the formula: $	ext{Lone pairs} = \frac{1}{2} (V - M)$,where $V$ is the number of valence electrons of the central atom ($8$ for $Xe$) and $M$ is the number of monovalent atoms bonded to it (Oxygen is divalent,so it does not contribute to $M$).$(1) XeO_3$: $V=8$,$M=0$. $	ext{Lone pairs} = \frac{1}{2} (8 - 0) = 4$ electrons,which is $1$ lone pair.$(2) XeF_6$: $V=8$,$M=6$. $	ext{Lone pairs} = \frac{1}{2} (8 - 6) = 2$ electrons,which is $1$ lone pair.$(3) XeF_2$: $V=8$,$M=2$. $	ext{Lone pairs} = \frac{1}{2} (8 - 2) = 6$ electrons,which is $3$ lone pairs.$(4) XeF_4$: $V=8$,$M=4$. $	ext{Lone pairs} = \frac{1}{2} (8 - 4) = 4$ electrons,which is $2$ lone pairs.$(5) XeOF_4$: $V=8$,$M=4$ (Oxygen is divalent). $	ext{Lone pairs} = \frac{1}{2} (8 - 4) = 4$ electrons,which is $1$ lone pair.Comparing the results,$XeO_3$,$XeF_6$,and $XeOF_4$ all have $1$ lone pair. Among the given options,the pair $(XeO_3, XeF_6)$ matches.

The pair of xenon compounds,which have the same number of lone pairs of electrons on the central atom is

Similar Questions

Which one of the following species has a plane triangular shape?

The sum of the number of lone pairs of electrons on each central atom in the following species is $[TeBr_6]^{2-}, [BrF_2]^+, SNF_3$,and $[XeF_3]^-$ (Atomic numbers: $N=7, F=9, S=16, Br=35, Te=52, Xe=54$)

Which one of the following compounds has the smallest bond angle?

What is the shape of $ClO_3^-$?

Identify the shape of the $IF_5$ molecule from the following options.

Vedclass Test Series

Exam Paper Generator

Online Exam Module