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

Q: Why does Sulfur form a triple bond with Nitrogen in $SNF_3$?

Nitrogen is in Group 15 and needs 3 electrons to complete its octet. Since Sulfur is the central atom and has empty d-orbitals (allowing it to expand its octet), it shares 3 of its valence electrons to form a triple bond with Nitrogen, ensuring Nitrogen becomes stable.

Q: Can Xenon really form bonds? I thought noble gases were completely inert.

While generally unreactive, heavier noble gases like Xenon have lower ionization energies because their outer electrons are further from the nucleus. This allows them to form stable compounds when reacting with highly electronegative elements like Fluorine and Oxygen.

Q: How do I know which atom is the central atom in these complex ions?

Usually, the least electronegative atom is the central atom (with Hydrogen being an exception as it can only form one bond). In these species, Te, Br, S, and Xe are the least electronegative atoms compared to the highly electronegative F, Br, and N atoms surrounding them.

Q: What if the remaining electrons after subtracting bonding electrons is an odd number?

If you get an odd number of remaining electrons, the molecule is a radical, meaning it has an unpaired electron. However, in standard VSEPR problems like these, the electrons usually pair up perfectly into lone pairs.

Q: Do I need to draw the exact 3D geometry to count the lone pairs?

No, you don't need to know the 3D geometry beforehand to count lone pairs. The formula $LP = (V \pm C - B)/2$ works purely on mathematical electron counting. In fact, you determine the geometry *after* finding the number of bond pairs and lone pairs.

Question

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 number : N = 7, F = 9, S = 16, Br = 35, Te = 52, Xe = 54]

Sigma Percentile · Accepted Answer

### The Art of Electron Counting

In the fascinating world of Chemical Bonding, determining the shape and reactivity of a molecule often boils down to a simple game of accounting: counting the electrons. The Valence Shell Electron Pair Repulsion (VSEPR) theory relies heavily on knowing exactly how many lone pairs reside on the central atom.

To do this flawlessly, we use a master formula. The number of lone pairs ($LP$) is given by taking the valence electrons ($V$) of the central atom, adjusting for any ionic charge ($C$), subtracting the electrons shared in bonds ($B$), and dividing the remainder by two:

$$LP = \frac{V \pm C - B}{2}$$

Let's apply this powerful tool to the four exotic species given in the problem.

### Decoding the Hexabromotellurate Ion

Our first candidate is $[TeBr_6]^{2-}$. The central atom is Tellurium (Te), which sits comfortably in Group 16 of the periodic table, granting it $6$ valence electrons.

The molecule carries a $-2$ charge, which means it has gained $2$ extra electrons. This brings our total available electron pool to $8$. Tellurium forms $6$ single bonds with the surrounding Bromine atoms, utilizing $6$ electrons.

Subtracting the bonding electrons from the total gives us $8 - 6 = 2$ remaining electrons. Dividing by two, we find exactly **$1$ lone pair** on the Tellurium atom.

### The Difluorobromonium Cation

Next, we examine $[BrF_2]^+$. Here, Bromine (Br) acts as the central atom. As a Group 17 halogen, it starts with $7$ valence electrons.

However, the $+1$ charge indicates the loss of an electron, leaving us with $6$ available electrons. Bromine forms $2$ single bonds with Fluorine, consuming $2$ electrons.

We are left with $6 - 2 = 4$ electrons. Pairing them up, we get **$2$ lone pairs**.

### The Thiazyl Trifluoride Trap

Now we arrive at $SNF_3$, a classic trap designed to test your fundamental understanding of valency. Sulfur (S) is the central atom, belonging to Group 16, meaning it has $6$ valence electrons.

It forms $3$ single bonds with Fluorine, using $3$ electrons. But what about Nitrogen? Nitrogen requires $3$ electrons to complete its octet. Because Sulfur can expand its octet, it generously forms a **triple bond** with Nitrogen, utilizing $3$ more of its valence electrons.

The total bonding electrons used by Sulfur is $3 + 3 = 6$. Subtracting this from its initial $6$ valence electrons leaves $0$. Therefore, Sulfur has **$0$ lone pairs** in this molecule. Don't fall for the trap of assuming Nitrogen only forms a single bond!

### The Trifluoroxenate Anion

Finally, we look at $[XeF_3]^-$. Xenon (Xe) is a noble gas from Group 18, possessing a full octet of $8$ valence electrons.

The $-1$ charge adds an extra electron, bringing the total to $9$. Xenon forms $3$ single bonds with Fluorine, using $3$ electrons.

Subtracting the bonding electrons leaves $9 - 3 = 6$ electrons. Dividing by two, we discover that Xenon holds **$3$ lone pairs**.

### The Grand Finale

We have successfully decoded the lone pairs for all four species: $1$, $2$, $0$, and $3$. The question asks for the sum of these lone pairs.

$$	ext{Sum} = 1 + 2 + 0 + 3 = 6$$

The final answer is **$6$**. By systematically applying our electron-counting formula and staying vigilant against valency traps, even the most complex molecules become easy to decipher.

Enter Numerical Value:

Visualized Solution

\text{The Lone Pair Formula}

\text{Analyzing } [\text{TeBr}_6]^{2-}

\text{Analyzing } [\text{BrF}_2]^+

\text{Analyzing } \text{SNF}_3

\text{Analyzing } [\text{XeF}_3]^-

\text{Final Summation}

The Sigma Insight: Hybridisation and VSEPR Theory

The Art of Electron Counting

Decoding the Hexabromotellurate Ion

The Difluorobromonium Cation

The Thiazyl Trifluoride Trap

The Trifluoroxenate Anion

The Grand Finale

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