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Molecular bromine on reaction with three moles of molecular fluorine produces an inter halogen compound. The total number of lone pairs at the central halogen atom is
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Molecular bromine $\left(\mathrm{Br}_2\right)$ on reaction with 3 moles of molecular fluorine $\left(\mathrm{F}_2\right)$, it will produced 2 moles of $\mathrm{BrF}_3$.
$\mathrm{BrF}_3$ is interhalogen compound i.e. trifluorobromide.
$$
\mathrm{Br}_2(g)+3 \mathrm{~F}_2(g) \longrightarrow \underset{\text { Trifluoro-bromide }}{2 \mathrm{BrF}_3(g)}
$$
$\mathrm{BrF}_3=\frac{7 e^{-}+3 e^{-}}{2}=\frac{10}{2} e^{-}=5 e^{-}$i.e. $s p^3 d$ hybridisation.
$\mathrm{BrF}_3$ contain $3 \mathrm{Br}-\mathrm{F}^{-}$covalent bonds and 2 lone pairs.
$\mathrm{BrF}_3$ is $s p^3 d$ hybridised molecule having T-shape.
Hence, total number of lone pairs at central halogen atom is 2.
$\mathrm{BrF}_3$ is interhalogen compound i.e. trifluorobromide.
$$
\mathrm{Br}_2(g)+3 \mathrm{~F}_2(g) \longrightarrow \underset{\text { Trifluoro-bromide }}{2 \mathrm{BrF}_3(g)}
$$
$\mathrm{BrF}_3=\frac{7 e^{-}+3 e^{-}}{2}=\frac{10}{2} e^{-}=5 e^{-}$i.e. $s p^3 d$ hybridisation.
$\mathrm{BrF}_3$ contain $3 \mathrm{Br}-\mathrm{F}^{-}$covalent bonds and 2 lone pairs.
$\mathrm{BrF}_3$ is $s p^3 d$ hybridised molecule having T-shape.
Hence, total number of lone pairs at central halogen atom is 2.
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