Question

Which of the following noble gas compounds has a square planar geometry: \(XeF_2\), \(XeF_4\), or \(XeF_6\)?

• A. \(XeF_2\)
• B. \(XeF_4\)
• C. \(XeF_6\)
• D. None of these

Hint:

Calculate the lone pairs on Xe for each compound using (valence electrons of Xe minus bonding pairs) divided by 2, then add bonding pairs and lone pairs to get the total electron domains. A square planar shape needs exactly 4 bonding pairs and 2 lone pairs sitting opposite each other.

Answer 1

The Correct Option is B

Concept: The geometry of molecules can be predicted using VSEPR theory, which states that electron pairs around the central atom arrange themselves to minimize repulsion.

Step 1: Count the valence electrons of Xenon. Xenon has 8 valence electrons.

Step 2: Determine bonding and lone pairs in \(XeF_4\). In \(XeF_4\): \[ \text{Number of Xe-F bonds} = 4 \] After forming four bonds, two lone pairs remain on Xenon. Thus, \[ \text{Total electron pairs} = 6 \] Hybridization: \[ sp^3d^2 \]

Step 3: Determine the molecular geometry. The six electron pairs arrange octahedrally, and the two lone pairs occupy opposite positions to minimize repulsion. Therefore, the four fluorine atoms lie in one plane forming a: \[ \text{Square planar geometry} \] Hence, the compound with square planar geometry is: \[ \boxed{XeF_4} \]

Answer 2

Concept:

• The number of lone pairs on the central Xe atom can be found directly from the formula: Lone pairs = (valence electrons of Xe minus number of Xe-F bonds) divided by 2, since Xe has 8 valence electrons and each bond uses one electron from Xe.
• Once the total electron domains (bonding pairs plus lone pairs) are known, each combination has a fixed, known molecular shape that can be looked up directly, without working through hybridization.

Step 1: Find the lone pairs on Xe in each compound.
For $XeF_2$: bonds $= 2$, lone pairs $= \frac{8-2}{2} = 3$. Total electron domains $= 2 + 3 = 5$.
For $XeF_4$: bonds $= 4$, lone pairs $= \frac{8-4}{2} = 2$. Total electron domains $= 4 + 2 = 6$.
For $XeF_6$: bonds $= 6$, lone pairs $= \frac{8-6}{2} = 1$. Total electron domains $= 6 + 1 = 7$.

Step 2: Match each domain count with its known shape.
5 domains with 3 lone pairs gives a linear shape, so $XeF_2$ is linear, not square planar.
7 domains with 1 lone pair gives a distorted shape, so $XeF_6$ is also not square planar.
6 domains with 2 lone pairs gives an octahedral electron arrangement, and this case needs a closer look.

Step 3: Fix the positions of the 2 lone pairs in the 6-domain case.
Lone pairs repel more strongly than bonding pairs, so they move as far apart as possible, taking up positions directly opposite each other.
This leaves the 4 fluorine atoms occupying the remaining 4 positions, all in one flat plane around Xe.

Final Answer: $XeF_4$ has the square planar geometry.