Draw all the resonance structures for the following ionic comp...

Question

Draw all the resonance structures for the following ionic compound:RbIO₂

Accepted Answer

here, it says to draw all the residents structures for the following Ionic compound, So we have our B i 02 It is an ionic compound, so you must break it up into its two ions. RBS in group one A. So it's plus one and then we have the IOD ICT ion I owed. I iodine is in Group seven A. So it's seven Valence electrons, Oxygen's and groups six A. So that's times two and then we have a minus charge, which means an outside electron has been added. So that's gonna be seven plus 12 plus one, which gives me 20 total valence electrons. Iodine goes in the center. It's going to be single bonded to the oxygen's. Remember the surrounding oxygen's have tohave There are terrible followed. So we've used 16 of our electrons. We have four meaning right now we're putting them on the iodine. But here's the thing. Remember, resident structures exist when there is at least one pipe on pi Bon present. Right now there isn't any pi bon. So that must mean that double bond exists somewhere within this structure, so we just need to show it. So here I'm going to use one of these lone pairs to make a pie bond here. Now, when I do that, let's calculate the formal charges. So we're gonna say here iodine is in Group seven A. We see it making three bonds, and it has four electrons, not bonding. So that zero, let's say this is oxygen. A that's double bonded in oxygen. Be that single bonded. So here's in Group six A. We see it making to bonds, and it has four electrons that are not bonding oxygen B. We see it's in Group six A. We have one bond and six electrons, so this is minus one when we add up all the formal charges, that gives me back the charge off the ion, which is minus one. So this year is the correct interpretation of the iodide ion. Now, if we show it this way, other residents structure would be instead of the oxygen on the left being double bonded, let's make it the oxygen on the right. That's double bonnet, so here we still have our two lone pairs on the iodine, but now it's going to be the oxygen on the right. That's double body and because they have charges. We put them in brackets with the charge on the outside. So these would be the two acceptable resident structures for the iodide ion.

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