1. We need the number of lone pairs on the central atom in each species:

   \\ SNF_3, \\ [XeF_3]^-$$

2. We use the steric number / valence electron counting method.

1) $[TeBr_6]^{2-}$

• Central atom: $Te$
• Valence electrons on $Te = 6$
• Because charge is $2-$, total electron count around central atom effectively increases by $2$.

So electrons available on central atom side: $6+2=8$

• There are $6$ $Te-Br$ bonds.
• In VSEPR counting, each bond uses one electron from the central atom.

Electrons left on $Te$: $8-6=2$

Thus lone pairs on $Te$: $\frac{2}{2}=1$

So, in $[TeBr_6]^{2-}$, lone pairs on central atom $=1$.

2) $[BrF_2]^+$

• Central atom: $Br$

• Valence electrons on $Br = 7$

• Charge is $+$, so subtract one electron: $7-1=6$

• Number of $Br-F$ bonds $=2$

Electrons left on $Br$: $6-2=4$

Lone pairs on $Br$: $\frac{4}{2}=2$

So, in $[BrF_2]^+$, lone pairs on central atom $=2$.

3) $SNF_3$

This is sulfur nitride trifluoride with structure: $F_3N\rightarrow S$

More standard bonding description: central atom is $N$, bonded to three $F$ atoms and to $S$.

• Central atom: $N$
• Valence electrons on $N = 5$
• Neutral molecule, so no charge correction.
• Number of bonds around $N = 4$ (three $N-F$ and one $N-S$)

Electrons left on $N$: $5-4=1$

This seems odd, since lone pairs must be integer pairs. So let us write the proper Lewis structure.

In $SNF_3$, the accepted Lewis structure is: $F_3N=S$

Thus nitrogen forms:

• three single bonds with $F$
• one double bond with $S$

Total bonding electrons contributed by $N$ side in formal charge counting = $5$ valence electrons, with no lone pair left.

Let us verify via formal charges:

• $N$ with 5 bonds has formal charge $0$ if it has no lone pair.
• $S$ with one double bond and two lone pairs is also reasonable.

Hence lone pairs on central atom $N = 0$.

So, in $SNF_3$, lone pairs on central atom $=0$.

4) $[XeF_3]^-$

• Central atom: $Xe$

• Valence electrons on $Xe = 8$

• Charge is $-$, so add one electron: $8+1=9$

• Number of $Xe-F$ bonds $=3$

Electrons left on $Xe$: $9-3=6$

Lone pairs on $Xe$: $\frac{6}{2}=3$

So, in $[XeF_3]^-$, lone pairs on central atom $=3$.

Total lone pairs

Adding all central-atom lone pairs: $1+2+0+3=6$

Therefore, the required sum is: $\boxed{6}$

Comparison with stored correct answer

Stored correct answer = $6$

Our derived answer = $6$

Hence, the answer agrees with the stored correct answer.