The I_3^- ion possesses: | vedclass

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(A) The central iodine atom in $I_3^-$ has $7$ valence electrons. It forms one bond with each of the other two iodine atoms,using $2$ electrons. This

Vedclass · Accepted Answer

Three equatorial lone pairs and two axial bond pairs on the central $I$ atom in a trigonal bipyramidal geometry.

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(A) The central iodine atom in $I_3^-$ has $7$ valence electrons. It forms one bond with each of the other two iodine atoms,using $2$ electrons. This leaves $5$ electrons,plus the $1$ negative charge,totaling $6$ electrons,which form $3$ lone pairs. Thus,the steric number is $2$ (bond pairs) $+ 3$ (lone pairs) $= 5$. $A$ steric number of $5$ corresponds to trigonal bipyramidal geometry. According to $VSEPR$ theory,to minimize repulsion,the $3$ lone pairs occupy the equatorial positions,and the $2$ bonding iodine atoms occupy the axial positions,resulting in a linear molecular shape.

The $I_3^-$ ion possesses:

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