Based on VSEPR theory, the number of $$F-Br-F$$ angles which has close to $$90$$ $$0$$ possible in a molecule of $$BrF5$$ will be $$_________$$

Covalent Bonding

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Question

Based on VSEPR theory, the number of $F - Br - F$ angles which has close to 90⁰ possible in a molecule of BrF₅ will be _________

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Solution

Hybridisation of Br in : ${BrF}_{5} = \frac{1}{2} (7 + 5) = 6 = {sp}^{3} d^{2}$
Geometry = Octahedral
Due to one lp of e^- 's, it has square pyramidal in shape.
It has eight (F-Br-F) bonds which are at 90°. There are
$1. F^{1} - Br - F^{2} 2. F^{2} - Br - F^{3} 3. F^{3} - Br - F^{4}$
$4. F^{4} - Br - F^{1} 5. F^{1} - Br - F^{5} 6. F^{5} - Br - F^{4}$
$7. F^{5} - Br - F^{3} 8. F^{5} - Br - F^{2}$

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Covalent Bonding

Unlock the full solution & master the concept.

Based on VSEPR theory, the number of F-Br-F angles which has close to 90 0 possible in a molecule of BrF5 will be _________

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Based on VSEPR theory, the number of $F - Br - F$ angles which has close to 90⁰ possible in a molecule of BrF₅ will be _________