Match List-I with List-II.
List-I (Species)
A. XeF2
B. XeO2F2
C. XeO3F2
D. XeF4
List-II (Number of l

Q: Why does Oxygen form double bonds with Xenon here?

Oxygen needs 2 electrons to complete its octet, so it typically forms double bonds with the central atom in these oxofluorides, utilizing 2 of Xenon's valence electrons.

Q: Can Xenon expand its octet? I thought noble gases don't react.

Yes! Xenon is in the 5th period and has empty d-orbitals available, allowing it to accommodate more than 8 electrons in its valence shell and form these hypervalent compounds when reacted with highly electronegative elements like F and O.

Q: Why are the lone pairs in $XeF_2$ in the equatorial positions?

According to VSEPR theory, lone pairs occupy positions that minimize repulsion. In a trigonal bipyramidal geometry (5 electron domains), the equatorial positions offer 120-degree angles, which minimizes lone pair-lone pair repulsions compared to the 90-degree axial positions.

Q: Is $XeO_3F_2$ a stable molecule?

While it can be synthesized, it is highly reactive. The fact that all 8 valence electrons are involved in bonding makes it a fascinating example of Xenon reaching its maximum oxidation state (+8).

Q: How do I quickly determine the shape from the lone pairs?

First, find the steric number (Bond Pairs + Lone Pairs). This gives the base geometry (e.g., 5 = trigonal bipyramidal, 6 = octahedral). Then, remove the positions occupied by lone pairs to see the actual molecular shape.

Question

Match List-I with List-II.

\begin{array}{ll}
	ext{List-I (Species)} & 	ext{List-II (Number of lone pair of electrons on the central atom)} \
	ext{A. } XeF_2 & 1.\ 0 \
	ext{B. } XeO_2F_2 & 2.\ 1 \
	ext{C. } XeO_3F_2 & 3.\ 2 \
	ext{D. } XeF_4 & 4.\ 3
\end{array}

Choose the most appropriate answer from the options given below :

Options:

(A) A-4, B-1, C-2, D-3

(B) A-3, B-4, C-2, D-1

(C) A-3, B-2, C-4, D-1

(D) A-4, B-2, C-1, D-3

Sigma Percentile · Accepted Answer

### Unveiling the Lone Pairs of Xenon Compounds

Xenon, a noble gas, was long thought to be completely unreactive. However, under the right conditions, it forms a fascinating array of compounds, particularly with highly electronegative elements like fluorine and oxygen. In this problem, we are tasked with finding the number of lone pairs on the central Xenon atom for four different molecules.

To solve this, we need to rely on a fundamental concept: **valence electrons**. Xenon, being in Group 18, has a full octet, meaning it has $8$ valence electrons in its outermost shell.

### The Master Formula

Finding the number of lone pairs is straightforward if you keep track of how many electrons are used in bonding. The formula is simple:

$$ 	ext{Lone Pairs} = \frac{	ext{Valence } e^- - 	ext{Bonding } e^-}{2} $$

Remember the bonding rules for the surrounding atoms:
- **Fluorine (F)** forms single bonds, utilizing $1$ of Xenon's electrons per bond.
- **Oxygen (O)** forms double bonds, utilizing $2$ of Xenon's electrons per bond.

Let's break down each molecule systematically.

### Molecule by Molecule Breakdown

**1. Xenon Difluoride ($XeF_2$)**
Here, Xenon is bonded to two fluorine atoms. Since each fluorine forms a single bond, $2$ valence electrons are used up.

Subtracting this from the initial $8$ leaves us with $6$ non-bonding electrons. Dividing by $2$, we get exactly **$3$ lone pairs**. According to VSEPR theory, these three lone pairs occupy the equatorial positions of a trigonal bipyramidal geometry, giving the molecule a linear shape.

**2. Xenon Dioxide Difluoride ($XeO_2F_2$)**
This molecule is a bit more complex. It has two oxygen atoms and two fluorine atoms. The two oxygens will form double bonds, using up $4$ electrons. The two fluorines will use another $2$ electrons for their single bonds.

That's a total of $6$ electrons used in bonding. $8 - 6$ leaves us with $2$ electrons, which makes exactly **$1$ lone pair**. This lone pair sits in the equatorial plane, resulting in a "see-saw" molecular geometry.

**3. Xenon Trioxide Difluoride ($XeO_3F_2$)**
In this highly oxidized compound, we have three oxygen atoms forming double bonds, consuming $6$ electrons. The two fluorine atoms use the remaining $2$ electrons for their single bonds.

All $8$ valence electrons are now involved in bonding! This means there are zero non-bonding electrons left. Hence, there are **$0$ lone pairs**. The molecule adopts a perfect trigonal bipyramidal geometry.

**4. Xenon Tetrafluoride ($XeF_4$)**
Finally, let's look at Xenon tetrafluoride. Xenon is bonded to four fluorine atoms, using up $4$ electrons for four single bonds.

Subtracting $4$ from $8$ leaves us with $4$ non-bonding electrons. Dividing by $2$ gives us **$2$ lone pairs**. These lone pairs occupy the axial positions of an octahedral electron geometry, pushing the fluorine atoms into a square planar shape.

### Final Calculation

Matching our findings with the given lists:
- **A ($XeF_2$)** $ightarrow$ $3$ lone pairs (Matches with 4)
- **B ($XeO_2F_2$)** $ightarrow$ $1$ lone pair (Matches with 2)
- **C ($XeO_3F_2$)** $ightarrow$ $0$ lone pairs (Matches with 1)
- **D ($XeF_4$)** $ightarrow$ $2$ lone pairs (Matches with 3)

This exact sequence (A-4, B-2, C-1, D-3) corresponds perfectly to option **(d)**. By simply keeping an accounting of the valence electrons, even the most intimidating noble gas compounds become easy to decipher!

Select Answer:

Visualized Solution

$L P = \frac{Valence e ^{-} - Bonding e ^{-}}{2}$

$X e F_{2}$

$X e O_{2} F_{2}$

$X e O_{3} F_{2}$

$X e F_{4}$

\text{Final Matching}

The Sigma Insight: Group 18 Elements

Unveiling the Lone Pairs of Xenon Compounds

The Master Formula

Molecule by Molecule Breakdown

Final Calculation

Similar Questions

The type of hybridisation and number of lone pair(s) of electrons of Xe in $XeOF_{4}$ , respectively, are

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Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } LP=2Valence e−−Bonding e−​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } XeF2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } XeO2​F2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } XeO3​F2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } XeF4​