Determine the electron geometry (eg)and molecular geometry (mg)of ICl₂⁻. A) eg = tetrahedral, mg = trigonal bipyramidal B) eg = tetrahedral, mg = trigonal pyramidal C) eg = trigonal bipyramidal, mg = linear D) eg = trigonal bipyramidal, mg = tetrahedral E) eg = octahedral, mg = bent

Question

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First, we need to draw the Lewis structure of ICl₂⁻ molecule. The central atom is Iodine (I) with one lone pair and two bonded atoms (Cl and S). The two terminal S atoms also have one lone pair each. This gives us a total of 5 electron pairs around the central Iodine atom. The electron geometry is determined by the arrangement of these 5 electron pairs around the central atom. The lone pair and four bonding pairs repel each other, resulting in a trigonal bipyramidal arrangement. Next, we need to determine the molecular geometry by considering only the positions of the atoms (i.e. ignoring the lone pairs). In this case, we have two bonded atoms (Cl and S) and a lone pair around the central Iodine atom. The bonded atoms are in the axial positions of the trigonal bipyramidal arrangement, while the lone pair is in one of the equatorial positions. Since there are only two bonded atoms and one lone pair, the molecular geometry will be determined by the positions of the bonded atoms. In this case, the bonded atoms are in a linear arrangement, resulting in a linear molecular geometry. Therefore, the correct answer is C) eg = trigonal bipyramidal, mg = linear.

Determine the Electron Geometry (Eg)and Molecular Geometry (Mg)of ICl2⁻

Determine the Electron Geometry (Eg)and Molecular Geometry (Mg)of ICl₂⁻