Question

The shapes of \(XeF_2\), \(XeF_4\), and \(XeO_3\) respectively are

• A. Linear, Tetrahedral, Pyramidal
• B. Angular, Square Planar, Pyramidal
• C. Linear, Tetrahedral, Planar
• D. Linear, Square planar, Pyramidal

Hint:

Using VSEPR theory: \[ AX_2E_3 \rightarrow \text{Linear} \] \[ AX_4E_2 \rightarrow \text{Square planar} \] \[ AX_3E \rightarrow \text{Pyramidal} \] where \(E\) represents lone pairs.

Answer 1

The Correct Option is D

Step 1: Determine the structure of XeF₂.
Xe has 8 valence electrons. Two bonds to F use 2 electrons; 3 lone pairs remain on Xe. Total electron pairs = 2 (bond) + 3 (lone) = 5. VSEPR: electron geometry = trigonal bipyramidal. Lone pairs occupy equatorial positions. With 2 bond pairs and 3 lone pairs, the molecular shape is LINEAR.
Step 2: Determine the structure of XeF₄.
Xe has 8 valence electrons. Four bonds to F use 4 electrons; 2 lone pairs remain on Xe. Total electron pairs = 4 (bond) + 2 (lone) = 6. VSEPR: electron geometry = octahedral. Lone pairs occupy opposite axial positions (trans). Molecular shape = SQUARE PLANAR.
Step 3: Determine the structure of XeO₃.
Xe has 8 valence electrons. Three double bonds to O. One lone pair on Xe. Total electron pairs around Xe = 3 (bond) + 1 (lone) = 4. VSEPR: electron geometry = tetrahedral. With 3 bonds and 1 lone pair, molecular shape = PYRAMIDAL (trigonal pyramidal).
Step 4: Summarize the three shapes.
XeF₂: Linear. XeF₄: Square planar. XeO₃: Pyramidal.
Step 5: Match with options.
Option 4 states: XeF₂ = Linear, XeF₄ = Square planar, XeO₃ = Pyramidal. This matches our analysis exactly.
Step 6: Confirm hybridization.
XeF₂: sp$^3$d (linear arrangement of bonds). XeF₄: sp$^3$d$^2$ (square planar). XeO₃: sp$^3$ (pyramidal, one lone pair). All consistent.
\[ \boxed{XeF_2: \text{Linear},\ XeF_4: \text{Square planar},\ XeO_3: \text{Pyramidal (option 4)}} \]