The I_3^- ion possesses: | vedclass

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Question

(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.

List-$I$ (Molecule)	List-$II$ (Shape)
$A$. $NH_3$	$I$. Square pyramid
$B$. $BrF_5$	$II$. Tetrahedral
$C$. $PCl_5$	$III$. Trigonal pyramidal
$D$. $CH_4$	$IV$. Trigonal bipyramidal

The $I_3^-$ ion possesses:

Similar Questions

The number of species having a non-pyramidal shape among the following is:
$A$. $SO_{3}$
$B$. $NO_{3}^{-}$
$C$. $PCl_{3}$
$D$. $CO_{3}^{2-}$

Which of the following pairs is isostructural?

Which of the following pairs of species is not isostructural?

Match List-$I$ with List-$II$:

List-$I$ (Molecule) List-$II$ (Shape)

$A$. $NH_3$ $I$. Square pyramid

$B$. $BrF_5$ $II$. Tetrahedral

$C$. $PCl_5$ $III$. Trigonal pyramidal

$D$. $CH_4$ $IV$. Trigonal bipyramidal

Choose the correct answer from the options below:

In molecules of the type $AX_2L_n$ (where $L$ represents lone pairs and $n$ is its number),there exists a bond between element $A$ and $X$. The $\angle XAX$ bond angle:

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