Question:
Bromine trifluoride autoionizes to form BrF$_2^+$ and BrF$_4^-$. The shapes of the cation and anion are respectively ________, and ________
Bromine trifluoride autoionizes to form BrF$_2^+$ and BrF$_4^-$. The shapes of the cation and anion are respectively ________, and ________
Updated On: Apr 12, 2026
- bent, square planar
- linear, square planar
- bent, see-saw
- linear, tetrahedral
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept:
Shape is determined by VSEPR theory. We calculate the steric number ($SN = \text{Bond Pairs} + \text{Lone Pairs}$).
Step 2: Detailed Explanation:
1. $BrF_2^+$: Central atom $Br$ has 7 valence $e^-$. After losing 1 $e^-$ for the charge, it has 6. It forms 2 bonds with F. Remaining $e^- = 6 - 2 = 4$ (2 lone pairs). $SN = 2 \text{ BP} + 2 \text{ LP} = 4$. Geometry: Tetrahedral. Shape: Bent/V-shaped. 2. $BrF_4^-$: Central atom $Br$ has 7 valence $e^-$. After gaining 1 $e^-$, it has 8. It forms 4 bonds with F. Remaining $e^- = 8 - 4 = 4$ (2 lone pairs). $SN = 4 \text{ BP} + 2 \text{ LP} = 6$. Geometry: Octahedral. Shape: Square Planar (lone pairs at axial positions).
Step 3: Final Answer:
The cation is bent and the anion is square planar.
Shape is determined by VSEPR theory. We calculate the steric number ($SN = \text{Bond Pairs} + \text{Lone Pairs}$).
Step 2: Detailed Explanation:
1. $BrF_2^+$: Central atom $Br$ has 7 valence $e^-$. After losing 1 $e^-$ for the charge, it has 6. It forms 2 bonds with F. Remaining $e^- = 6 - 2 = 4$ (2 lone pairs). $SN = 2 \text{ BP} + 2 \text{ LP} = 4$. Geometry: Tetrahedral. Shape: Bent/V-shaped. 2. $BrF_4^-$: Central atom $Br$ has 7 valence $e^-$. After gaining 1 $e^-$, it has 8. It forms 4 bonds with F. Remaining $e^- = 8 - 4 = 4$ (2 lone pairs). $SN = 4 \text{ BP} + 2 \text{ LP} = 6$. Geometry: Octahedral. Shape: Square Planar (lone pairs at axial positions).
Step 3: Final Answer:
The cation is bent and the anion is square planar.
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