Question

The ratio of total number of lonepairs
in XeF2
and XeF4
are

Answer 1

Explanation:

9 : 14

As the question asked total pairs , count the lone pairs of fluorine also

XeF2

3+6

 XeF4

2+12

Answer 2

Answer:

The ratio of the total number of lone pairs in $XeF_{2}$ and $XeF_{4}$ is $3:14$

Explanation:

In the case of $XeF_{2}$ :

$Xe$ has $8$ number of valence electrons.

$F$ has $7$ number of valence electrons.

Now, the total number of valence electrons on $XeF_{2}$ ,

= (No. of Xe atoms $\times$ No. of valence electrons of Xe) + (No. of F atoms $\times$ No. of valence electrons on F)

= $(1 \times 8) + (2 \times 7)$

= $22$

There are $5$ electron pairs and $2$ number of $XeF_{2}$ bonds.

∴ The number of lone pairs on the molecule $XeF_{2}$ is 3.

In the case of $XeF_{4}$ :

$Xe$ has two lone pairs as it is the central atom.

$F$ has three lone pairs on each of its atoms.

∴ The number of lone pairs on the molecule $XeF_{4}$ is

= $2 + (3 \times 4)$

= $2 + 12$

= $14$

Hence, the ratio of the total number of lone pairs in $XeF_{2}$ and $XeF_{4}$ is $3:14$

#SPJ2