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AtldentAnatonwy & PhyscocumentUnnileddocumcrgooglecom/document/d/TeMDo XZ DnVelWOcAKyglAigkgkMdtALY1935KueditocumentFollowinglew Insert Formal Tools Add-ons Hel...

Question

AtldentAnatonwy & PhyscocumentUnnileddocumcrgooglecom/document/d/TeMDo XZ DnVelWOcAKyglAigkgkMdtALY1935KueditocumentFollowinglew Insert Formal Tools Add-ons Help LastedilWas 12mi:1003Normal textHamg Nem4 4Directions: For cach ofthe Six tests. record observations in the cycloherane layer fromn the vidzo a; follows:[areaction OcCUD:Record the initial and final color: Then Wzite the net ionic equation for tle reaction:[Eno Ieaction oCcl3:Record "No change color; Retained (Fill in initial c

atldent Anatonwy & Physc ocument Unnileddocumcr googlecom/document/d/TeMDo XZ DnVelWOcAKyglAigkgkMdtALY1935 Kuedit ocument Following lew Insert Formal Tools Add-ons Help LastedilWas 12mi: 1003 Normal text Hamg Nem 4 4 Directions: For cach ofthe Six tests. record observations in the cycloherane layer fromn the vidzo a; follows: [areaction OcCUD: Record the initial and final color: Then Wzite the net ionic equation for tle reaction: [Eno Ieaction oCcl3: Record "No change color; Retained (Fill in initial color) ' and "No eaction Note: Be careful t0 use the terms halogen and halide properly when recording your results and explanatious: 1) CL NaBr NBr + CL 2NACI+Brz Soditm bromide reacts with chlorine to produce endinati chloride aud bromiue: Sodium bromide heated. Nal C12-Nal-NaCH+2 Chloriue react with sodiun iodide produce sodium chloride and fodine. Chlorine- diluted Counou Sodittu iodiue -cold solutiou Observcd in the cyclohexaue Iayer Net Ionic Equation(s) No reaction Cotuncnt onthc_ability of chlotitc toclisplace brominc iLiodine from thcic lalide Jalte; Chlorine has tlie ability to take clectrons from both bramide iOns and iodide ions. Chlorine las the #bility tnke Icthon from both bromide iOni and iodide iOns: Chlorine Ja4 powcrful axidising #Rcnt tlan citlict broinine or iodline



Answers

Consider the following organic reaction, in which one halogen replaces another in an alkyl halide:
$$
\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Br}+\mathrm{KI} \longrightarrow \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{I}+\mathrm{KBr}
$$
In acetone, this particular reaction goes to completion because
KI is soluble in acetone but $\mathrm{KBr}$ is not. In the mechanism, I- approaches the carbon opposite to the Br (see Figure 16.20 , p. $704,$ with $\mathrm{I}^{-}$ instead of $\mathrm{OH}^{-} ) .$ After $\mathrm{Br}^{-}$ has been replaced by $\mathrm{I}^{-}$ and precipitates as $\mathrm{KBr}$ , other $\mathrm{I}^{-}$ ions react with the ethyl iodide by the same mechanism.
(a) If we designate the carbon bonded to the halogen as $\mathrm{C}-1$ what is the shape around $\mathrm{C}-1$ and the hybridization of $\mathrm{C}-1$ in ethyl iodide?
(b) In the transition state, one of the two lobes of the unhybridized 2$p$ orbital of $\mathrm{C}-1$ overlaps a $p$ orbital of I, while the other lobe overlaps a $p$ orbital of Br. What is the shape around $\mathrm{C}-1$ and the hybridization of $\mathrm{C}-1$ in the transition state?
(c) The deuterated reactant, $\mathrm{CH}_{3} \mathrm{CHDBr}$ (where $\mathrm{D}$ is deuterium, 'H), has two optical isomers because $\mathrm{C}-1$ is chiral. If the reaction is run with one of the isomers, the ethyl iodide is not optically active. Explain.

Here's a coordination compound, sir. Cuban on the Cushing is right there for black Signed present in it and order the counter islands. Then electronics, contribution of the transition metals and geology of the complex saying in debt Garnishee compact. So when we're when we're doing this. So in case of a a Cornish component which is given these Coburn tablets combined mid water Monday, June that is six working monkeys and this is the coordination component with the clothing outside duplicate. Now here we have to decide what is the complex I in present so complex. And here is this part which is inside the blanket. This is the complex. Same He should mean blessed Second important thing is over the counter tried contra trying is the iron which is present outside the bracket and it is unnoticeable. So it should be grain minus curry Ease electrical conflagration of the transition. Actually, now what is here edge The world is atomic number 27 and he urges at plus two exhibition state select turning contribution off this place to world is immunised noble Guess conflagration And after that region rate, we should be full s you three d seven So this is that really attorney contribution off by darkness. And after that, at the end, ever the complex me up to date. What is just jumping? See, when six is the Kardashian number that Mr Geology is uptight. Hedren. Mr Jumped. Yeah. So this is about the economy. Lex now becomes big. Jobless is and a T Then a gene then turns and fruit a stupid T choice and insect. This is the big complex, sir Coordination component Skillman and ever. This we have delayed first, is what is the complex I'm presenting. It can rate the thing which is inside direct it so as to be to, uh this wrong My ministry. Mr Corbyn saying Now, what is the underlying? That is energy, whether it is outside the bracket. So jin range, it is sodium the same and it is a little over three and third a thing. The verdict is Forrester Elektronik contribution off dissented, matter election So here it is edgy Plus center matter London And you're saying Johnny contribution should be here is the nearest well This confirmation after that can write. It should be Friday s CEO for Dejan So this should be the electronic contribution and forth question ever complexes. What is your junkie? So when coalition number is to that means ladies Virginia Joe, you should So this show this is dancer for the B. Now come to see So for does she accomplice? Beavan is Kardashian compound Human is go bird which is inside the wreckage. Well, blood than manage to four. And this whole thing is inside the blockage. And now said the activities seal too. No one weatherman, all the cushions first for shame. What is the complex? I am here soon. Should me The thing just inside the wreckage. Energy for range Eating the bracket and the judgment close to now Come to second point. They just What is the culture? I am so here The counter drying ins upset The flippant finishing one This is two C minus. No going to turn one. It is good Is that you're turning contribution? Oh, the sentry battler Doom open is edge Oh, I don't know s contribution So they're silly of me A And after that region raid, you treat me for as she Now, when this treating name and all the electrons are feeling the Dior and last one is only one in John that is unfairly drawn. Soon the joint should be, as with the hibernation and get your head in jumps. So it is, uh, given in the question also that as young, this copper complex science have tried a junkie, and accordingly, we can rate the drunk on tradition and job Children.

This is a great problem because this allows us to become more familiar with terms like coordination, compound, complex ion and counter ions. And more specifically, we have to determine all of these compounds from a reaction. And that is what this whole problem is going to be centered around and from there. Once we know our compounds, we confined the electron configuration of our metal. So for part A, our coordination compound is going to be cobalt that's bound to six waters. This will be our cat ion, and this will be countered by to an ion chlorine. Our reaction will look like this. We'll start with cobalt that is bound to six waters and our counter ion will be die atomic, chlorine. And then when we separate this into our various compounds and ions, we'll have cobalt with six waters. It has a positive to charge plus two chlorine, an ions. So from here we can determine our complex. Ion is our cobalt complex, and our counter ion would be our two chlorine. And now from here we have to determine the oxidation state off cobalt in our complex Ion and we know from this positive to charge at cobalt must be positive, too, because water is neutral so we can find the electron configuration of cobalt two plus which is are gone three d seven. And now we have to determine the geometry around our complex ion lift. You notice we have six waters. And if you know if we have six Liggins, that means that we have an octahedron geometry and now on to part B will do the exact same process. But now it's just different compounds. Essentially, we start with the coordination compound where we have the counter ion of sodium and we have three of these sodium atoms and this will be countering silver. It's bound to s 203 and we have two of those groups so we can write this reaction. We'll start with our coordination compound. We can split this up and associate it into three positively charged sodium plus our silver bound as +203 This would have a negative three charge notice here. Now that are conflicts. Ion is our an ion and our counter ion is our cat ion are complex. Ion will be our silver complex and our counter ion would be our three sodium atoms. And then what we have to do is find the oxidation state of silver which we would do the exact same ways we did before. And we find that silver is positively charged. So the electron configuration will be crypt on for D 10 and now we have to determine its molecular geometry. And this is a little bit confusing because based on our formula, we only have two Liggins. But what we need to know is as 203 is by identity. So we'll have toe Add those two Liggins from our by Dent eight ligand, Which means that our geometry for this complex ion would be Tetra hydro. Now moving on Thio sie Pardon me, that should be a c Not to be, um but we start with our coordination compound of platinum bound to amine groups and a die atomic chlorine. And what we need to notice here is that our coordination compound is completely neutral and because it's neutral, we don't have any counter ion. We don't have any complex ion. Our coordination compound is the Onley formula we're dealing with. So from here we can find the pardon me, the oxidation state of platinum and how we do that is we know that a mean groups are neutral and our chlorine zehr negatively charged. So that means we're dealing with platinum two. Plus, we could find the electron configuration off Zen in five d eight. Then we have to determine our geometry notice. Here we have four Liggins. So that means we're dealing with, obviously something that is tetra huge or square cleaner because we have four regions of electron density Orleans In this case, now the question becomes, is a Tetra Hegel or two square player. What we need to know is we need to look at the Siris of Ligon strength and we see that we have a mean here which we have to compare on our strength of our Liggins and because of the strength of this mean group that we see, we should know that our square pardon me are geometry would be squared plainer. I hope that this problem helped you understand a little bit more about essentially comparing coordination compounds, complex ions and counter ions. And hopefully you now understand how we can figure out the oxidation states of the medals and figure out the geometry of each of our coordination compounds

From the list of chemical reactions, let's identify the precipitation reaction. So, a precipitation reaction where we take an Aquarius species, together with in Aquarius species to produce a precipitate, which would be a solid and another a creates species here. So this would be our precipitate, which is our solid. So out of the list of reactions, only one is a precipitation reaction, which is H. Which would be a g. L. 03 and HcL. Yeah, to produce a precipitate of gcl in solid state And a channel three. Uh huh Aquarius. So our precipitation reaction is H. Out of the given list.


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