Question

The sum of the number of lone pairs of electrons on each central atom in the following species is:

${\left[Te{Br}_6\right]}^{2-}$, ${\left[Br{F}_2\right]}^{+}$, $SN{F}_3$, and ${\left[Xe{F}_3\right]}^{-}$.

[Atomic numbers: $N=7$, $F=9$, $S=16$, $Br=35$, $Te=52$, $Xe=54$]

Answer 1

The sum of the number of lone pairs of electrons on each central atom in the given species is $1+2+0+3=6$.
Te atom in ${\left[Te{Br}_6\right]}^{2-}$ has 1 lone pair of electrons.

Br atom in ${\left[Br{F}_2\right]}^{+}$ has 2 lone pairs of electrons.

S atom in $SN{F}_3$ has 0 lone pair of electrons.

Xe atom in ${\left[Xe{F}_3\right]}^{-}$ has 3 lone pairs of electrons.