Iodine Lewis structure

Using Lewis structures and VSEPR, give the VSEPR formula for each of the following species and predict its shape (a) sulfur tetrachloride (b) iodine trichloride (c) 1F4 ... 

C09-0084. Iodine forms three compounds with chlorine ICl, ICI3, and ICI5. Determine the Lewis structure, describe the shape, and draw a ball-and-stick model of each compound. 

B The Lewis structure for NI3 is similar to that of NH3. The central nitrogen atom is attached to each iodine atom by a single covalent bond. All of the atoms in this structure get a closed valence shell. 

The Lewis structure reveals a VSEPR number of 5 for the central iodine atom, two bonded neighbors and three unshared pairs. To determine which corners of the trigonal bipyramid are occupied by the terminal iodine atoms, find the arrangement which maximizes the angles between the unshared pairs. The preferred arrangement, Fig. 9-45(a), must be the one in which the unshared pairs are all at 120° because any other alternative [Fig. 9-45(b) and (c)] would have two sets of pairs at 90°. Therefore, the two terminal iodine atoms must occupy the axial positions (180° to each other), making the molecule linear. 

Use the periodic table to draw Lewis structures for the following elements barium (Ba), gallium (Ga), tin (Sn), bismuth (Bi), iodine (I), cesium (Cs), krypton (Kr), xenon (Xe). 

There are also examples of molecules whose existence is beyond question, but for which no satisfactory Lewis structures can be written. Two examples are the triiodide ion I3 , and the bifluoride ion HF2 . The triiodide ion is a well known species found in aqueous solutions containing iodine and iodide ions ... 

Two chlorine atoms and one iodine atom in total have 21 electrons. However, we have a cationic species at hand, so we have to remove one electron and start with a total of 20 valence electrons for ICI2. That gives a Lewis structure in which iodine is a central atom (being the more electropositive of the two) and is bonded to two chlorines with a single bond to each. All atoms have a precise octet. Looking at iodine, there are two bonding... 

E2.7 The Lewis structure of ICV and its geometry based on VSEPR theory is shown below. The central iodine atom is sunounded by six bonding and one lone pair. However, this lone pair in the case of ICU" is a slereochemically inert lone electron pair. This is because of the size of iodine atom—large size of this atom spread around all seven electron pairs resulting in a minimum repulsion between electron pairs. As a consequence all bond angles are expected to be equal to 90° with overall octahedral geometry. 

In order to obtain octets of electrons, these atoms tend to form compounds in which they have one bond and three lone pairs. Note how the Lewis structures of hydrogen fluoride, HF (used in the refining of uranium), hydrogen bromide, FiBr (a pharmaceutical intermediate), and hydrogen iodide, Fil (used to make iodine salts) resemble the structure of hydrogen chloride. 

Both aluminum and iodine form chlorides, ALClg and LClg, with bridging Cl atoms. The Lewis structures are... 

Practice Problem A Draw the Lewis structure of the iodine tetrachloride ion (ICI4). Practice Problem B Draw the Lewis structure of krypton difluoride (KrF2). 

It must form three covalent bonds to do so. Iodine, a group 7A element, has seven valence electrons and needs one more electron to achieve a complete octet. It must form only one bond to do so. The Lewis structure for NI3 is shown below. 

The interhalogen compounds obey the expectations based on the VSEPR theory, and typical structures are giver in Chapter 6. Ore compound not included there is the dimeric iodine trichloride, in which the iodine atom of the monomeric species appears to act as a Lewis acid and accept an additional pair of electrons from a chlorine atom (Fig. 17.5). 

Coal and many coal-derived liquids contain polycyclic aromatic structures, whose molecular equivalents form radical cations at anodes and radical anions at cathodes. ESR-electrolysis experiments support this (14). Chemically, radical cations form by action of H2SO4 (15,19), acidic media containing oxidizing agents (15,20,21,22), Lewis acid media (18,23-35) halogens (36), iodine and AgC104 (37,38), and metal salts (39,40). They also form by photoionization (41,42,43) and on such solid catalytic surfaces as gamma-alumina (44), silica-alumina (45), and zeolites (46). Radical anions form in the presence of active metals (76). 

Simple cyclooctatetraenediyls [M(C8H8)] (M = Eu, Yb) can be made by reaction of cyclooctatetraenewith solutions ofthe metal in liquid ammonia. Another route to [M(CsH8)] (Sm, Yb) is the reaction of the metals with cyclooctatetraene using iodine as catalyst. These compounds are not monomers, but the Lewis base adduct [M(C8H8)(py)3] has a piano-stool structure. 

The number of species isoeiectronic with XeF, is quite limited. The anions SbBr, TeCl , and TeBr are octahedral. Both IF and XeF are nonoctahedral." Iodine heptafluoride and rhenium heptafluoride may be considered isoeiectronic with these species if they are all considered to have 14 valence shell electrons of approximately equal steric requirements. Both have a pentagonal bipyramidal structure (see Fig. 6.12). Most interestingly, the XeF anion, formed by the Lewis acid XeF ... 

The fact that cinnamyl methyl ether underwent cyclopropanation under similar conditions to afford the corresponding cyclopropane as an almost racemic mixture indicates that the free hydroxy group is necessary for producing an effective chiral environment. This can presumably take place through complexation as a zinc alkoxide. The authors propose that a trinuclear zinc complex 89, in which both the oxygen atom of a zinc alkoxide and iodine atom of iodo-methylzinc contribute to the Lewis character, must be close to the transition structure of the reaction. This might account also for the distinct enantioselectivity between the allylic alcohol and its methyl ether. 

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