The pair of species having identical shapes for molecules of both species, is
(a) CF4, SF4
(b) XeF2,

Q: Why do the lone pairs in $\text{XeF}_2$ go to the equatorial positions instead of axial?

In a trigonal bipyramidal geometry ($sp^3d$), the equatorial positions are $120^\circ$ apart, while axial positions are $90^\circ$ from the equatorial ones. Placing bulky lone pairs in the equatorial plane minimizes the severe $90^\circ$ repulsions, making the molecule more stable.

Q: Why is $\text{CO}_2$ considered $sp$ hybridized when it has double bonds?

In VSEPR theory and hybridization, we only count electron domains (sigma bonds and lone pairs). A double bond consists of one sigma and one pi bond, but it acts as a single region of electron density pushing away from others. Thus, 2 double bonds = 2 domains = $sp$ hybridization.

Q: Does the shape of a molecule include its lone pairs?

No! The *electron geometry* includes lone pairs, but the *molecular shape* (or molecular geometry) only describes the positions of the atoms. Think of lone pairs as invisible forces shaping the visible atoms.

Q: Why isn't $\text{SF}_4$ tetrahedral like $\text{CF}_4$?

Carbon in $\text{CF}_4$ has 4 valence electrons and forms 4 bonds, leaving 0 lone pairs (Tetrahedral). Sulfur in $\text{SF}_4$ has 6 valence electrons, forms 4 bonds, and has 1 lone pair left over. This lone pair pushes the bonds, creating a 'See-saw' shape.

Q: How can I quickly calculate the number of lone pairs on a central atom?

Use the formula: $\text{Lone Pairs} = \frac{\text{Valence Electrons} - \text{Electrons shared in bonds}}{2}$. For $\text{XeF}_2$: $\frac{8 - 2}{2} = 3$ lone pairs.

Question

The pair of species having identical shapes for molecules of both species, is

Options:

(A) $	ext{CF}_4, 	ext{SF}_4$

(B) $	ext{XeF}_2, 	ext{CO}_2$

(C) $	ext{BF}_3, 	ext{PCl}_3$

(D) $	ext{PF}_5, 	ext{IF}_5$

Sigma Percentile · Accepted Answer

### The Quest for Identical Shapes

Imagine you are a molecular architect, tasked with finding two buildings that look exactly the same from the outside, even if their internal support structures are completely different. This is exactly what we are doing when we look for molecules with identical shapes!

The shape of a molecule is dictated by the **Valence Shell Electron Pair Repulsion (VSEPR) theory**. This theory states that electron pairs around a central atom will arrange themselves as far apart as possible to minimize repulsion. However, there is a catch: while both bond pairs and lone pairs determine the overall *electron geometry*, only the positions of the atoms (the bond pairs) determine the final *molecular shape*.

### Analyzing Xenon Difluoride ($	ext{XeF}_2$)

Let's dive into our first candidate, $	ext{XeF}_2$. Xenon is a noble gas, which means it comes with a full octet of $8$ valence electrons. When it forms two single bonds with fluorine atoms, it uses up $2$ of those electrons.

This leaves us with $6$ non-bonding electrons, which pair up to form $3$ lone pairs.

To find the hybridization, we calculate the **Steric Number**:
$$ 	ext{Steric Number} = 	ext{Bond Pairs} + 	ext{Lone Pairs} = 2 + 3 = 5 $$

A steric number of $5$ corresponds to $sp^3d$ hybridization, which has a base electron geometry of a **trigonal bipyramid**.

Now, where do we put those $3$ bulky lone pairs? According to VSEPR theory, lone pairs demand more space and will occupy the **equatorial positions** (where they are $120^\circ$ apart) to minimize $90^\circ$ repulsions with other electron pairs. This forces the two fluorine atoms into the **axial positions** (top and bottom).

When we look at the molecule, we ignore the invisible lone pairs. What remains is a straight line connecting F-Xe-F. Therefore, the molecular shape of $	ext{XeF}_2$ is **Linear**.

### Analyzing Carbon Dioxide ($	ext{CO}_2$)

Now, let's look at $	ext{CO}_2$. Carbon has $4$ valence electrons. It forms two double bonds with two oxygen atoms. In VSEPR theory, a double bond counts as a single "super pair" or electron domain.

Since all $4$ of carbon's valence electrons are involved in bonding, there are $0$ lone pairs left on the central atom.

Let's calculate the Steric Number:
$$ 	ext{Steric Number} = 	ext{Bond Domains} + 	ext{Lone Pairs} = 2 + 0 = 2 $$

A steric number of $2$ corresponds to $sp$ hybridization. With only two electron domains pushing away from each other, they will settle at a perfect $180^\circ$ angle.

Since there are no lone pairs to hide, the electron geometry and the molecular shape are identical. The shape of $	ext{CO}_2$ is also **Linear**.

### The Final Verdict

Both $	ext{XeF}_2$ and $	ext{CO}_2$ result in a perfectly **Linear** shape, despite having completely different hybridizations ($sp^3d$ vs $sp$) and different numbers of lone pairs ($3$ vs $0$). This beautiful geometric coincidence makes option (b) the correct answer!

Always remember: don't just count the atoms. The invisible lone pairs are the secret puppet masters shaping the molecules in our universe.

Animated Solution for Chemistry - Chemical Bonding and Molecular Structure: The pair of species having identical shapes for molecules of both species, is

Select Answer:

Visualized Solution

Problem Analysis

Steric Number of $XeF_{2}$

Shape of $XeF_{2}$

Steric Number of $CO_{2}$

Shape of $CO_{2}$

Conclusion

Checking Other Options

The Sigma Insight: Hybridisation and VSEPR Theory

The Quest for Identical Shapes

Analyzing Xenon Difluoride ( $XeF_{2}$ )

Analyzing Carbon Dioxide ( $CO_{2}$ )

The Final Verdict

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