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Is ICl4- a square planar and polar?

Updated: 8/11/2023
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12y ago

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Yes, it is square planar. The central iodine atom exceeds the octet rule by bonding with all four chlorine atoms and having two lone pairs. A central atom with six electron pairs (d2sp3 hybridization) and two lone electron pairs by definition is square planar (see VSEPR theory for more information). Because of its symmetrical geometry, it will have no dipole moment.

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Q: Is ICl4- a square planar and polar?
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What is the molecular geometry for ICL4 -?

square planar


What is the electron geometry of ICl4-1?

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Is KrF4 sp3d2 octahedral square planar and nonpolar?

A KrF4 sp3D2 octahedral square, which has octahedral electron domain geometry two nonbonded pair, AB4E2, is non-polar. It also is planar, as is any shape that has sides (planes).


Is ICl4 bent?

First off, ICl4 doesn't really exist. It should be ICl4- as the lone electron on Iodine (the central atom) will instantly pull another electron to itself to make a lone pair. Therefore iodine has two lone pairs and four elemental bonds (one to each chlorine). Since we have six total bonds (including the lone pairs) the geometrical shape will be octahedral. However, since, the lone pairs repel each other they will go to the top and bottom of the structure, making ICl4 square planar. So no, it is not bent per se, but it does deviate from the expected shape of ICl4 (3+) (which would be trigonal bipyramidal) because of the lone pairs.


What shape would you expect forXeF4?

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It is non polar. The central Xe atom has 4 Fluorine atoms and two electron pairs attached giving it AB4E2 molecular structure. This leads to a square planar molecular shape. F is more electronegative than Xe and is thus a polar covalent bond, but the 4 bonds in a square cancel out, and the electron pair on top and bottom of the "square plane" cancel each-other out as well, leaving you with a non polar molecule


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What is true about the ion ICl4-?

ICl4-'s electron domain geometry is octahedral.