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Shape of the compound depend on type and...

Shape of the compound depend on type and number of electron pair around central atom . These electron pair repel each other and stay as far as possible . The repulsion sequence is as
`L.P - L.P. gt B.P. - L.P. gt B.P - B.P.`
Which of the given compound is planar ?

A

`XeF_(5)^(-)`

B

`XeF_(4)`

C

`I Cl_(4)^(-)`

D

All of these

Text Solution

AI Generated Solution

The correct Answer is:
To determine which of the given compounds is planar, we need to analyze the molecular geometry of each compound based on the number of bond pairs (B.P.) and lone pairs (L.P.) around the central atom. The repulsion between these electron pairs influences the shape of the molecule. ### Step-by-Step Solution: 1. **Understanding Electron Pair Geometry**: - The shape of a compound is determined by the arrangement of electron pairs (both bond pairs and lone pairs) around the central atom. - The repulsion sequence is: L.P - L.P > B.P - L.P > B.P - B.P. - This means lone pairs repel each other more than bond pairs do, and bond pairs repel each other the least. 2. **Analyzing Option 1: XeF5⁻**: - Central atom: Xenon (Xe) - Bond pairs: 5 (from 5 F atoms) - Lone pairs: 2 - Steric number = Bond pairs + Lone pairs = 5 + 2 = 7. - Hybridization: sp³d³. - Geometry: Pentagonal bipyramidal. - Shape: The lone pairs are positioned 180° apart, and the fluorine atoms form a pentagonal plane. Thus, XeF5⁻ has a planar structure. 3. **Analyzing Option 2: XeF4**: - Central atom: Xenon (Xe) - Bond pairs: 4 (from 4 F atoms) - Lone pairs: 2 - Steric number = 4 + 2 = 6. - Hybridization: sp³d². - Geometry: Octahedral. - Shape: The two lone pairs are 180° apart, and the four fluorine atoms form a square plane. Thus, XeF4 is also planar. 4. **Analyzing Option 3: ICl4⁻**: - Central atom: Iodine (I) - Bond pairs: 4 (from 4 Cl atoms) - Lone pairs: 2 - Steric number = 4 + 2 = 6. - Hybridization: sp³d². - Geometry: Octahedral. - Shape: The two lone pairs are 180° apart, and the four chlorine atoms form a square plane. Thus, ICl4⁻ is also planar. 5. **Conclusion**: - All three compounds (XeF5⁻, XeF4, and ICl4⁻) exhibit planar structures due to their respective arrangements of bond pairs and lone pairs. ### Final Answer: All of the given compounds (XeF5⁻, XeF4, and ICl4⁻) are planar.
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