The correct statement about $\mathrm{ICl}_{5}$ and $\mathrm{ICl}_{4}^{-}$is:

(B) $\mathrm{ICl}_{5}$ is square pyramidal and $\mathrm{ICl}_{4}^{-}$is square planar

(A) $\mathrm{ICl}_{5}$ is trigonal bipyramidal and $\mathrm{ICl}_{4}^{-}$is tetrahedral

(C) $\mathrm{ICl}_{5}$ is square pyramidal and $\mathrm{ICl}_{4}^{-}$is tetrahedral

JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Chemical Bonding and Molecular Structure

The correct statement about

ICl_{5}

and

ICl_{4}^{-}

is:

(A) $ICl_{5}$ is trigonal bipyramidal and $ICl_{4}^{-}$ is tetrahedral
(B) $ICl_{5}$ is square pyramidal and $ICl_{4}^{-}$ is square planar✓ Correct
(C) $ICl_{5}$ is square pyramidal and $ICl_{4}^{-}$ is tetrahedral
(D)Both are isostructural

Answer: (B)

Solution

Count the electron domains on iodine in each. In

ICl X_{5}

iodine has five bonding pairs and one lone pair, six domains in all, so the electrons arrange octahedrally and, with one position taken by the lone pair, the molecule is square pyramidal. In

ICl X_{4} X^{-}

iodine has four bonding pairs and two lone pairs, again six domains; the two lone pairs take opposite axial positions, leaving the four chlorines in a plane, so the ion is square planar. So

ICl X_{5}

is square pyramidal and

ICl X_{4} X^{-}

is square planar.

JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Single correct option

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JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Chemical Bonding and Molecular Structure

Solution

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