Question

$XeF_4$ reacts with $PF_5$ as follows :

$XeF_4 + PF_5 \longrightarrow [XeF_3]^+ [PF_6]^-$

If total numbers of $90^\circ$ angles in reactants = x and if total number of $90^\circ$ angles in products = y, considering all effects of VSEPR theory, then calculate the value of (y -x).

• A. 2
• B. 4
• C. 6
• D. 8

Answer 1

Answer: (B)

4

Answer 2

The problem asks us to calculate the value of $(y-x)$, where $x$ is the total number of $90^\circ$ angles in the reactants ($XeF_4$ and $PF_5$) and $y$ is the total number of $90^\circ$ angles in the products ($[XeF_3]^+$ and $[PF_6]^-$), considering all effects of VSEPR theory. We assume that "total number of $90^\circ$ angles" refers to angles between bonded atoms only, unless otherwise specified.

Step 1: Determine the structure and bond angles of the reactants.

1. $XeF_4$:

   • Central atom: Xe. Valence electrons = 8. Number of bonded atoms (F) = 4.
   • Number of lone pairs on Xe = $(8 - 4 \times 1)/2 = 2$.
   • Total electron domains = 4 bond pairs + 2 lone pairs = 6.
   • According to VSEPR theory, 6 electron domains are arranged octahedrally. To minimize repulsion, the two lone pairs occupy opposite positions.
   • The molecular geometry is square planar. The four F atoms are in a plane around the Xe atom, and the two lone pairs are above and below the plane.
   • The angles between adjacent Xe-F bonds in the square plane are $90^\circ$. There are 4 such angles.
   • The angles between opposite Xe-F bonds are $180^\circ$.
   • Number of $90^\circ$ angles between bonded atoms in $XeF_4 = 4$.

2. $PF_5$:

   • Central atom: P. Valence electrons = 5. Number of bonded atoms (F) = 5.
   • Number of lone pairs on P = $(5 - 5 \times 1)/2 = 0$.
   • Total electron domains = 5 bond pairs + 0 lone pairs = 5.
   • According to VSEPR theory, 5 electron domains are arranged in a trigonal bipyramidal geometry.
   • In trigonal bipyramidal geometry, there are two types of bonds: 3 equatorial and 2 axial.
   • The angles between equatorial P-F bonds are $120^\circ$. There are 3 such angles.
   • The angles between axial and equatorial P-F bonds are $90^\circ$. Each axial bond is $90^\circ$ to the three equatorial bonds. Since there are two axial bonds, there are $2 \times 3 = 6$ such angles.
   • The angle between the two axial P-F bonds is $180^\circ$.
   • Number of $90^\circ$ angles between bonded atoms in $PF_5 = 6$.

Total number of $90^\circ$ angles in reactants, $x = (\text{number of } 90^\circ \text{ angles in } XeF_4) + (\text{number of } 90^\circ \text{ angles in } PF_5) = 4 + 6 = 10$.

Step 2: Determine the structure and bond angles of the products.

1. $[XeF_3]^+$:

   • Central atom: Xe. Valence electrons in neutral Xe = 8. Charge = +1.
   • Number of valence electrons in $[XeF_3]^+ = 8 - 1 = 7$. Number of bonded atoms (F) = 3.
   • Number of lone pairs on Xe = $(7 - 3 \times 1)/2 = 2$.
   • Total electron domains = 3 bond pairs + 2 lone pairs = 5.
   • According to VSEPR theory, 5 electron domains are arranged in a trigonal bipyramidal geometry. Lone pairs prefer equatorial positions. So, the two lone pairs occupy two equatorial positions, and the three F atoms occupy one equatorial and two axial positions.
   • The molecular geometry is T-shaped. The three F atoms form a T shape with Xe at the intersection.
   • In the T-shaped structure, the angle between the two axial Xe-F bonds is approximately $180^\circ$. The angles between the axial Xe-F bonds and the equatorial Xe-F bond are ideally $90^\circ$. There are 2 such angles. Due to lone pair repulsion, these angles might be slightly less than $90^\circ$, but VSEPR theory predicts the ideal angle based on the domain arrangement. We count the angles that are ideally $90^\circ$ in the predicted geometry.
   • Number of $90^\circ$ angles between bonded atoms in $[XeF_3]^+ = 2$.

2. $[PF_6]^-$:

   • Central atom: P. Valence electrons in neutral P = 5. Charge = -1.
   • Number of valence electrons in $[PF_6]^- = 5 + 1 = 6$. Number of bonded atoms (F) = 6.
   • Number of lone pairs on P = $(6 - 6 \times 1)/2 = 0$.
   • Total electron domains = 6 bond pairs + 0 lone pairs = 6.
   • According to VSEPR theory, 6 electron domains are arranged in an octahedral geometry.
   • In an octahedral geometry, all adjacent bond angles are $90^\circ$, and opposite bond angles are $180^\circ$.
   • The number of adjacent bond pairs in an octahedron is 12. (Consider one vertex/bond; it is adjacent to 4 others. With 6 bonds, this gives $6 \times 4 = 24$ pairs, but each pair is counted twice, so $24/2 = 12$).
   • Number of $90^\circ$ angles between bonded atoms in $[PF_6]^- = 12$.

Total number of $90^\circ$ angles in products, $y = (\text{number of } 90^\circ \text{ angles in } [XeF_3]^+) + (\text{number of } 90^\circ \text{ angles in } [PF_6]^-) = 2 + 12 = 14$.

Step 3: Calculate $(y - x)$.

$y - x = 14 - 10 = 4$.

The final answer is $\boxed{4}$.