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# When iodine is dissolved in aqueous potassium iodide, the shape of the species formed is:A) linearB) angularC) triangularD) see-saw

Last updated date: 19th Jun 2024
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Hint: We know that, for determining the shape and structure of any compound we have to know about the number of electrons which are bonded and participate. The lone pair count is also needed for the exact hybridization of the given compound.

Step by step answer: As we all know, when iodine is dissolved in aqueous potassium iodide they generally form a compound that is potassium triiodide.
${{\rm{I}}_{\rm{2}}} + {\rm{KI}} \to {\rm{K}}{{\rm{I}}_3}$

These molecules exist in ionic form, one is cationic and the other is anionic. The ${{\rm{K}}^{\rm{ + }}}$ ion is obtained as cation and ${{\rm{I}}_{\rm{3}}}^ -$ ion is obtained as anion.
The total valence electron of potassium triiodide is $3 \times 7 + 1 = 22$.
The bond pair present in potassium triiodide is $2$.

The electrons which participate to complete the octet is $2 \times 8 = 16$.
So, the lone pair of potassium triode is calculated from valence electrons and electrons which are used in octet.

${\rm{Lone}}\;{\rm{pair}} = \dfrac{{{\rm{Valence}}\;{\rm{electrons}} - {\rm{Electrons}}\;{\rm{used}}\;{\rm{in}}\;{\rm{octet}}}}{2}$
Substitute all the values in the above equation. We get,
$\begin{array}{c} {\rm{Lone}}\;{\rm{pair}} = \dfrac{{22 - 16}}{2}\\ = 3 \end{array}$

The number of lone pairs in potassium iodide is $3$.

So, the hybridization of potassium triiodide is ${\rm{s}}{{\rm{p}}^{\rm{3}}}{\rm{d}}$ and structure is trigonal bipyramidal. Thus, it is linear in shape because it has two bond pairs only.
Hence, the correct option for this question is A that is linear.

Note: The shape of the species can generally be affected by the change in bond pair of the respective molecule. The hybridization is necessary for identifying the other characters of any compound