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Which structure for XeO3 and XeF4 are co...

Which structure for `XeO_3` and `XeF_4` are consistent with the VSEPR model

A

`XeO_(3)`, trigonal pyramidal , `XeF_(4)`, square planar

B

`XeO_(3)`, trigonal planar, `XeF_(4)` square planar

C

`XeO_(3)` trigonal pyramidal , `XeF_(4)` tetrahedral

D

`XeO_(3)`, trigonal planar , `XeF_(4)` tetrahedral

Text Solution

AI Generated Solution

The correct Answer is:
To determine the structures of \( \text{XeO}_3 \) and \( \text{XeF}_4 \) that are consistent with the VSEPR (Valence Shell Electron Pair Repulsion) model, we will follow these steps: ### Step 1: Determine the Valence Electrons - **For \( \text{XeO}_3 \)**: Xenon (Xe) has 8 valence electrons, and each oxygen (O) has 6 valence electrons. Thus, for \( \text{XeO}_3 \): \[ \text{Total valence electrons} = 8 + (3 \times 6) = 26 \text{ electrons} \] - **For \( \text{XeF}_4 \)**: Each fluorine (F) has 7 valence electrons. Thus, for \( \text{XeF}_4 \): \[ \text{Total valence electrons} = 8 + (4 \times 7) = 36 \text{ electrons} \] ### Step 2: Draw the Lewis Structures - **For \( \text{XeO}_3 \)**: - Place xenon in the center and bond it to three oxygen atoms. - Distribute the remaining electrons to satisfy the octet rule for oxygen. - The structure will show that xenon has one lone pair and three bonding pairs, leading to a trigonal pyramidal shape. - **For \( \text{XeF}_4 \)**: - Place xenon in the center and bond it to four fluorine atoms. - Distribute the remaining electrons. Xenon will have two lone pairs in addition to the four bonding pairs, resulting in a square planar shape. ### Step 3: Determine Hybridization and Molecular Geometry - **For \( \text{XeO}_3 \)**: - The presence of three bonding pairs and one lone pair leads to sp³ hybridization. - The molecular geometry is trigonal pyramidal. - **For \( \text{XeF}_4 \)**: - The presence of four bonding pairs and two lone pairs leads to sp³d² hybridization. - The molecular geometry is square planar. ### Conclusion - The structure of \( \text{XeO}_3 \) is trigonal pyramidal, and the structure of \( \text{XeF}_4 \) is square planar, both of which are consistent with the VSEPR model. ---

To determine the structures of \( \text{XeO}_3 \) and \( \text{XeF}_4 \) that are consistent with the VSEPR (Valence Shell Electron Pair Repulsion) model, we will follow these steps: ### Step 1: Determine the Valence Electrons - **For \( \text{XeO}_3 \)**: Xenon (Xe) has 8 valence electrons, and each oxygen (O) has 6 valence electrons. Thus, for \( \text{XeO}_3 \): \[ \text{Total valence electrons} = 8 + (3 \times 6) = 26 \text{ electrons} \] ...
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