In the molecule I_2Cl_6,choose the CORRECT st | vedclass

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(A) The structure of $I_2Cl_6$ is a dimer of $ICl_3$,which exists as a planar molecule.Each $I$ atom is $sp^3d$ hybridized with two lone pairs,making

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The molecule is planar.

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(A) The structure of $I_2Cl_6$ is a dimer of $ICl_3$,which exists as a planar molecule.Each $I$ atom is $sp^3d$ hybridized with two lone pairs,making the geometry around each $I$ atom square planar (if we consider the bridging atoms).There are $2$ bridging $Cl$ atoms and $4$ terminal $Cl$ atoms.Regarding the angles: $	heta_1$ is the $Cl-I-Cl$ angle involving bridging atoms,$	heta_2$ is the $Cl-I-Cl$ angle for terminal atoms,and $	heta_3$ is the $Cl-I-Cl$ angle between a terminal and a bridging atom.The correct order of angles is $	heta_3 > 	heta_2 > 	heta_1$.There are $20$ lone pairs in total ($3$ on each terminal $Cl$,$2$ on each bridging $Cl$,and $2$ on each $I$ atom: $4 	imes 3 + 2 	imes 2 + 2 	imes 2 = 12 + 4 + 4 = 20$).Thus,statement $(A)$ is the only correct statement among the options provided.

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

In the molecule $I_2Cl_6$,choose the $CORRECT$ statement$(s)$.

Similar Questions

The shape of the $XeF_4$ molecule is:

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:

Which among the following represents the correct order of variation of bond angles in the given molecules?

Which of the following pairs of compounds is isoelectronic and isostructural?

Which of the following has the largest bond angle?

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