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Add lone pairs to these Lewis structures of interhalogen compounds. BrF3 and BrF5..

Add lone pairs to these Lewis structures of interhalogen compounds. BrF3 and BrF5.

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Concepts and reason

The electrons that do not participate in bonding are represented as lone pair electrons. The lone pair electrons are also called valance electrons present in the outermost shell of an atom. To find the non – bonded pair of electrons, the lone pair electrons are indicated in the structure.

Fundamentals

An example to indicate lone pair electrons:

In the molecule ammonia, nitrogen is bonded with three hydrogens. The electronic configuration of nitrogen is 1s22s22p3{\rm{1}}{{\rm{s}}^{\rm{2}}}{\rm{2}}{{\rm{s}}^{\rm{2}}}{\rm{2}}{{\rm{p}}^{\rm{3}}} . The outermost shell contains 5 electrons. Hydrogen-3 shares its electron with 2p electron of nitrogen. The remaining 2 electrons are situated as lone pair electrons on the nitrogen atom.

Br=[Ar]3d104s24p5Br=7\begin{array}{l}\\{\rm{Br}}\,\,{\rm{ = }}\,\,{\rm{[Ar]}}\,{\rm{3}}{{\rm{d}}^{{\rm{10}}}}{\rm{4}}{{\rm{s}}^{\rm{2}}}{\rm{4}}{{\rm{p}}^{\rm{5}}}\\\\{\rm{Br}}\,\,{\rm{ = }}\,\,7\\\end{array}

F=[He]2s22p53F=3×73F=21\begin{array}{l}\\{\rm{F}}\,\,{\rm{ = }}\,\,\left[ {{\rm{He}}} \right]{\rm{2}}{{\rm{s}}^{\rm{2}}}{\rm{2}}{{\rm{p}}^{\rm{5}}}\\\\{\rm{3F = }}\,\,{\rm{3}}\,{\rm{ \times }}\,{\rm{7}}\,\,\\\\{\rm{3F = }}\,{\rm{21}}\\\end{array}

BrF3=21+7BrF3=28\begin{array}{l}\\{\rm{Br}}{{\rm{F}}_{\rm{3}}}{\rm{ = }}\,\,{\rm{21 + 7}}\\\\{\rm{Br}}{{\rm{F}}_{\rm{3}}}{\rm{ = }}\,{\rm{28}}\\\end{array}

The structure of BrF3{\rm{Br}}{{\rm{F}}_{\rm{3}}} :

Br=[Ar]3d104s24p5Br=7\begin{array}{l}\\{\rm{Br}}\,\,{\rm{ = }}\,\,{\rm{[Ar]}}\,{\rm{3}}{{\rm{d}}^{{\rm{10}}}}{\rm{4}}{{\rm{s}}^{\rm{2}}}{\rm{4}}{{\rm{p}}^{\rm{5}}}\\\\{\rm{Br}}\,\,{\rm{ = }}\,\,7\\\end{array}

F=[He]2s22p55F=5×75F=35\begin{array}{l}\\{\rm{F}}\,\,{\rm{ = }}\,\,\left[ {{\rm{He}}} \right]{\rm{2}}{{\rm{s}}^{\rm{2}}}{\rm{2}}{{\rm{p}}^{\rm{5}}}\\\\{\rm{5F = }}\,\,5\,{\rm{ \times }}\,{\rm{7}}\,\,\\\\{\rm{5F = }}\,\,35\\\end{array}

BrF5=35+7BrF5=42\begin{array}{l}\\{\rm{Br}}{{\rm{F}}_5}{\rm{ = }}\,\,35{\rm{ + 7}}\\\\{\rm{Br}}{{\rm{F}}_5}{\rm{ = }}\,\,42\\\end{array}

The structure of BrF5{\rm{Br}}{{\rm{F}}_5} :

Ans:

The structure of BrF3{\rm{Br}}{{\rm{F}}_{\rm{3}}} :

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