The correct statement about ICl_5 and ICl_4^- | vedclass

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Question

(B) For $ICl_5$: The central iodine atom has $7$ valence electrons. It forms $5$ bonds with $Cl$ atoms and has $1$ lone pair. The steric number is $5

Vedclass · Accepted Answer

$ICl_5$ is square pyramidal and $ICl_4^-$ is square planar

Vedclass · Answer

(B) For $ICl_5$: The central iodine atom has $7$ valence electrons. It forms $5$ bonds with $Cl$ atoms and has $1$ lone pair. The steric number is $5 + 1 = 6$,which corresponds to $sp^3d^2$ hybridization. Due to $5$ bond pairs and $1$ lone pair,the geometry is octahedral,but the shape is square pyramidal.For $ICl_4^-$: The central iodine atom has $7$ valence electrons,plus $1$ from the negative charge,totaling $8$ electrons. It forms $4$ bonds with $Cl$ atoms and has $2$ lone pairs. The steric number is $4 + 2 = 6$,which corresponds to $sp^3d^2$ hybridization. Due to $4$ bond pairs and $2$ lone pairs,the geometry is octahedral,but the shape is square planar.

The correct statement about $ICl_5$ and $ICl_4^-$ is

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