When iodine is dissolved in aqueous potassium | vedclass

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(A) When $KI$ and $I_2$ react,they form $KI_3$ as follows:$KI + I_2 \rightarrow KI_3$The species $KI_3$ dissociates into $K^+$ and $I_3^-$ ions.The ce

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linear

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(A) When $KI$ and $I_2$ react,they form $KI_3$ as follows:$KI + I_2 \rightarrow KI_3$The species $KI_3$ dissociates into $K^+$ and $I_3^-$ ions.The central iodine atom in the $I_3^-$ ion has $3$ lone pairs and $2$ bond pairs.According to $VSEPR$ theory,the hybridization is $sp^3d$,which corresponds to a trigonal bipyramidal electron geometry.Since there are $3$ lone pairs occupying the equatorial positions,the remaining $2$ bond pairs are at the axial positions,resulting in a linear shape.Therefore,the shape of the $I_3^-$ ion is linear.

List-$I$	List-$II$
$A. BrF_5$	$I. AB_4E$,see-saw
$B. SF_4$	$II. AB_4E_2$,square planar
$C. XeF_4$	$III. AB_5E$,square pyramidal
$D. ClF_3$	$IV. AB_3E_2$,$T$-shape

When iodine is dissolved in aqueous potassium iodide,the shape of the species formed is

Similar Questions

The incorrect geometry is represented by

Match the following:

List-$I$ List-$II$

$A. BrF_5$ $I. AB_4E$,see-saw

$B. SF_4$ $II. AB_4E_2$,square planar

$C. XeF_4$ $III. AB_5E$,square pyramidal

$D. ClF_3$ $IV. AB_3E_2$,$T$-shape

The correct answer is:

The correct increasing order for bond angles among $BF_3$,$PF_3$ and $ClF_3$ is:

Which of the following is not isostructural with $SiCl_4$?

State the shapes of molecules that contain only bonding electron pairs around the central atom.

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