Question
Explain the shape of $I^{-3}$ ion.
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Answer
The central l-atom has the outer shell electronic configuration in the ground state as $\text{5s}^2\ \text{5p}_\text{x}^2\ 5\text{p}^2_\text{y}\ \text{5p}^1_\text{z}\ 5\text{d}^0.$ It undergoes $s p^3$ d hybridisation. Out of the five $s p^3$ d hybrid orbitals, one is half-filled, one is empty and the remaining three are fully-filled. The half-filled orbital forms covalent bond with iodine atom.
The empty orbital accepts electron pair from $I^-$ ion to form a coordinate bond. The remaining three fully-filled orbitals occupy equatorial positions. Thus, the geometry of three lone pairs and two bond pairs is trigonal bipyramidal and the shape of $\text{I}^-_3$ is linear as shown in the figure.
The empty orbital accepts electron pair from $I^-$ ion to form a coordinate bond. The remaining three fully-filled orbitals occupy equatorial positions. Thus, the geometry of three lone pairs and two bond pairs is trigonal bipyramidal and the shape of $\text{I}^-_3$ is linear as shown in the figure.
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