5

Draw the Lewis structure for each of the following compounds. The carbon atoms are central in each one.(a) $mathrm{C}_{2} mathrm{H}_{6}$(b) $mathrm{C}_{2} mathrm{H}...

Question

Draw the Lewis structure for each of the following compounds. The carbon atoms are central in each one.(a) $mathrm{C}_{2} mathrm{H}_{6}$(b) $mathrm{C}_{2} mathrm{H}_{4}$(c) $mathrm{C}_{2} mathrm{H}_{2}$(d) $mathrm{C}_{2} mathrm{Cl}_{2}$

Draw the Lewis structure for each of the following compounds. The carbon atoms are central in each one. (a) $mathrm{C}_{2} mathrm{H}_{6}$ (b) $mathrm{C}_{2} mathrm{H}_{4}$ (c) $mathrm{C}_{2} mathrm{H}_{2}$ (d) $mathrm{C}_{2} mathrm{Cl}_{2}$



Answers

Draw Lewis structures for each molecular formula.
a. C$_2$H$_4$Cl$_2$ (two isomers) b. C$_3$H$_8$O (three isomers) c. C$_3$H$_6$ (two isomers)

All right. In this question, we are drawing Lewis structures for two different molecules. First up, we have C H 20 and the first step for drawing a little structure is to calculate how many total valence electrons we have. So we have four for the carbon to one for each of the hydrogen. So too, for the hydrogen and then six for the oxygen. That means we have a total of 12 valence electrons were working with. We'll put carbon in the center because we know that Carbon likes to form four bonds and likes to form or bonds and the other two components. So we'll put carbon in the center, put hydrogen on each of the sides, and then we'll put we'll put oxygen up top. Um, and I just drew three bones here. So both subtract the electrons from these bonds from our total Valence electron count. So we have three bonds, two electrons per bond. Then it's we have six of tons leftover Teoh distribute among the structure, so hydrogen recall hydrogen Onley needs to valence electrons. It's already got two from the bond, So this these hydrogen is don't need additional electrons. We can put, we can try putting six on the oxygen that would satisfy the up tough of the oxygen. But then this carbon only has six militants electrons. So we could try moving one of these repairs and turning it into a bond and see if that helps. So we drop down like that. Um, this is our two for six electrons that we used up so that, you see, are the rest of our valence electrons. And we just have Teoh check to make sure our Adams follow the up ted rule. So this oxygen now has eight male until a chance because it has four in, um, free electrons, and then it has to bonds, and there's two electrons per bond. So that's a total of eight valence electrons in that oxygen. On this carbon has four buns, and there's two electrons in each boned. So that means eight total electrons bonds air one to that, this is a double bond which councils to bonds. So the last step is to just double check to make sure that we have the right number of total electrons. So we have four. Look, Trump's over here. We have four over here you have to over here and we have to over here. That's a total of 12 which is what we completed. We need it. So this is the low structure for C two c h. 20 Next up we have you have seen to CEO for Okay, let's start by, um, talk leading are avail. It's about trunks. Actually, I'm going to live this down a little bit, so we have more room so that that's complete. Have any meal? It's electrons we need, so we have to Carpets. There's four fields of transfer carbon, and we have four. Chlorine is and there's seven minutes electrons per glory. So we have eight plus 20. I can do that. That's 28. Yep, that equals 36 total valence electrons. And let's start by drawing our structure. How did I reach the structure on this skeleton structure? Um, any time you see carbon, carbon likes to form a lot of bonds, so carbon will typically go in the center and the Koreans will typically go on the edges of the reason why I knew that these carbons had two Korean teach is because if we did a situation in which the carbon was spotted 24 Koreans, for example. Um, there's nowhere else for the other carbon to go. Um And then the other thing is, you could try drawing something where you do three coins on one of the carbons and then one chlorine on the other carbon. But you'll find that this carbon will have a hard time getting its full eight electrons. So this is a trial on air process. But my recommendation for drug low structure is to put the least electoral negative atoms like carbon in the center and the most Electra negative things that chlorine florian and roaming on the sides. And, um, that will help you get the right skeleton structure more often on the first guess So, um, let's start by calculating how many being loose electrons we just grew or how many electrons could just drew. We have 12345 bonds, and each bond has two electrons. So that's 10 electrons from our total that we just used up 26 left to distribute. I'm gonna play. I'll start by, um, putting virulence of transit each of the Koreans. So Koreans, um, they like to have eight males electrons like most Adams, and we know that we gave them a single bond already. So they just need six or chunks inch. And if I give each of the six each of the chlorine six electrons of theirs four Koreans, that means I just used 24 elections. That means you have till it turns left over. Okay, so these chlorine zehr set. But what about these carpets? Each of these carbons only has to balance. I mean, at six Valence electrons, it's got they have three bonds each two electrons per pound. Um, so into these governments have six, but they need eight, and we only have two electrons to distribute. If we give them to one of the carbon, the other carbon will have will be short of electrons. So we can turn this into a double bond that use up our last two electrons. And, um, all of the atoms in the structure have satisfied the architectural. His carbon, for example, has four bonds total because the stubble bond councils to bones. So this carbon has eight. Yeah, Valence electron students access to them seem with that carbon. And these Clarins also have eight and just to make sure we have a total 36 Vance about chunks in the diagram. We have eight over here. It over here. This double bond councils for being over here. And we have eight over here. So eight times for us 32 plus four more is 36. So this is our Lewis structure for a C to seal for.

For our first structure, we have C two h four. We should know that carbon is much more Electra negative than hydrogen. So carbon we placed from the center and we'll have a double bond between our two carbon atoms and two hydrogen coming off of each carbon atom. So our Lewis structure will look like this for our second molecule. We have ch three and H three repaired me and H two. We'll have a CH three group here on the left where we'll have a carbon bound to three hydrogen. And because of the number of valence electrons that carbon has, it could only hold four bombs. So on the fourth bond here it will have a nitrogen that has a lone pair. And this nitrogen will also be bound to two hydrogen atoms. And this is what our structure will look like for this organic molecule. And now, for our third compound, we have H h, C, H o, and it will look like this. Carbon and oxygen are our most Electra negative Adams in this compound, so they will be place from the center. We'll have a carbon that is a double bond to oxygen, oxygen will have two lone pairs because of its number of ailments. Electrons. Then we'll have two hydrogen atoms down to our carbon here on the left. So this is what our Lewis structure will look like for this molecule. And now moving on to ch three ch 20 H. We'll have a CH three group that is bound to a CH two group, which is These two are attached by a single bond between the carbons and then off of our second carbon Adam. We'll have a e hydroxide group with an oxygen that has too long pairs and a hydrogen again. We should notice in the structure that carbon and oxygen are the most electro negative Adams. So they are placed in the center. For our next structure. We have CH three ch o, and our final structure will look like this will have a carbon in our center because it's the most Electra negative. It will have one hydrogen that's bound to it, and then also it will be double bound to an oxygen that has two lone pairs, and then again we will have a hydroxide group where the oxygen is bound to the hydrogen and the oxygen will have two lone pairs as well. Notice here. How This is just saying. Oh, h, we don't have a line in between the and the H. That is perfectly OK to do as well. It is assumed that these two are bound to one another, which they are.

So for part A for 02 we know that we have two times six sixes for oxygen that gives you 12 founds electrons to play with. So now we have an action bonded to an oxygen. Don't have a choice there. We know that Asher Jin likes to bonds, and two lone pairs doesn't care how it gets to two bonds. If it's a double bond or two single bonds. In this case, we have to make it a double bond and the movie of each one to lone pairs. The reason why Action likes this is because in terms of formal electrons, it put in six. Now, if we count Thea's, you have one from each bond. So 1234 56 So six months six is equal zero. It has no formal charge, which is ideal in these cases, so it's just a double bonded oxygen, with two lone pairs on each. For Part B. As in boy, we have C as to which is four from the carbon plus two times six from the sulfur, which gives you 16 feelings like drones to mess with. So from here, we know that sulfur is just like oxygen because it's directly under it on the periodic table. So we knew that sulfur likes to bonds in this case, a double bond with two lone pairs. Same reason for oxygen. You have 123456 Formal electrons six minus 60 No formal charge. Carbon here is nice and simple. It put in four. So has one two 34 four minus 40 has no formal charge. It's nice and happy for part C. You have end to O so here for end to O, you do need some formal charges now and two. It's kind of interesting because there's no net charge. See how there's nothing up here. That means that there's no total charge. So start with the structure. We have to times five from the nitrogen plus six from the ash in 16 electrons. Nature is the least Electra negative out of these atoms. So we'll put a nightshirt in the middle. No, the rules for formal electrons are that you for formal charges are that you want your negative charges on your most elector negative atoms, which in this case, would be oxygen. So the easiest way to get a negative formal charge and oxygen is to give it three lone pairs in a single bond. Because that gives it 1234567 formal electrons. So six minus seven is equal the negative one, which is your formal charge. So there's a native on this oxygen. Now for this nitrogen. We need a positive formal charge on one of these two. The easiest one to give it to is the central one. So we're going to give a triple bond here, which then fills up the octet of this nitrogen, but doesn't give it enough form. Electrons has the one from the bottom of the oxygen, too. 34 five minus four is positive one. So this notion is a positive charge. You know, if you do the math here, we have two electrons left over. So this last night train gets a loan pair, which gives it 12345 um, formal electrons. So has no charge. We have net zero formal charge here, which is what we need up here. So this is your lewis dot structure for n tuo to bid in

Starting off with Kant counting the number of Valence electron sulfur has six feelings electrons and then hydrogen has one beyond Selectron. And since there are two hydrogen atoms, we're gonna have a total balance. Electrons. We have a total Beyonce electrons. So drawing sulfur in the center with six valence electrons. We have 123456 Hydrogen sulfur has to open positions for bonding. Hydrogen has one available position for bonding, as only has one Beyonce electron and then each pair of electrons conform. Ako, Baylin, Bond. This is how our final structure for Salford die hydride will look or sulfuric acid. In this case, you have to bonds to the hydrogen and two appears for sulfur dioxide. There are sulfur and oxygen each have six valence electrons. And since there are three total adams six times three is 24. No, totally forgot. Six times three is 18 huh? So we're gonna have sulphur in the center with six valence electrons. 123456 And then oxygen was six. Felix electrons. 123456 And 123456 So each of the lone pairs on sulfur can be shared with e oxygen atoms to form a cool Vaillant bond. So our final Louis structure would be s 02 However, we have two youto have, ah, a formal or the least amount of formal charges right now. And at this point, we have a formal charge of negative one on the oxygen, and we only have 246 electrons around. So first, we haven't We have yet to satisfy the architect rule, um, for sulfur. So we're going have to move one of the lone pairs on the oxygen to form a double bomb with our sulfur to in order for sulfur to complete its octet and to reduce the A total amount of formal charge. So our final structure, we'll be s oh, with two lone pairs on the right oxygen alone pier on the sulfur and three lone pairs on the left oxygen. So it's double check. We have 18 electrons. We have 2468 10 12 14 16 18. Do you have a full amount of 18 life draws in each? Adam Octa is satisfied having a peered eight shared of electrons. Okay for B F for three. Let's count the number of Ayla's electrons. Boron is in group. I mean, that would check. Boron is in Group three, says three electrons. Flooring is in group seven and there are four adams of flooring. So seven times four is 21 plus three is 24. However, we do have a formal charge of negative one. Whenever we have a negative formal charge, that means we're gaining electrons. So we're gonna add plus one, So this will be a total of 25,000,000,000 electrons. So Bram being in the center with three If. 123 and then we have flooring Adams around with seven valence electrons. We 1234567 1234567 1234567 and then a final flooring. 234567 However we do have, we do have an extra negative charge. So that means we have to add an additional electron somewhere. So I'm going to go ahead and add it to our boron. Since they're, um, since there are four high of four flooring Adams, then we'll make sense that there will be an additional bonding between the fourth flooring Adams and that we have oh Pavillon bonds between young period electrons on the boron with our flooring for selenium Die chloride. Selenium is in Group six was chlorine being in Group seven and there's two Adams. So seven times two is 14 less. Six is going to be 20 electrons, selenium having six villains. Electrons. We have one too. 3456 And then we have to. Corinne Adams with seven Valence Electrons 1234567 1234567 So I'm going to just kind of correct what I drawn. I wanted us to put one of the lone pairs at the bottom. So now we have two available positions on the selenium and two available positions on the chlorine. So these we're going to form our bond. Arco Vaillant, boss between the shared electrons. And then it's double check it that we have our total number of elect Valence electrons in our structure and the each Adam has its architect rules satisfied. So we have 2468 10 12 14 16 18 20 Valence electrons. We have to bonds and tool in pairs around the selenium and then for flooring. Um, Boron has an expanded act or 2468 Boring is complete and sore. The flooring Adams. And then, uh whenever we do have a formal negative charge, we just close it off with brackets and to note the formal charge on the outside of the bracket.


Similar Solved Questions

5 answers
Q1 Petrochemical industries produce various of product such as resins, mubber, rergent and many more:(a) Identify one product produced in the petrochemical industries.mark)(b) Identify the structure and its name for one of the chemical compounds in the product(2-mark)Describe ALL possible IR Absorption bands for compound in ()marks)Discuss FIVE (5) industries which apply IR Spectroscopy and how they apply (10 marks)
Q1 Petrochemical industries produce various of product such as resins, mubber, rergent and many more: (a) Identify one product produced in the petrochemical industries. mark) (b) Identify the structure and its name for one of the chemical compounds in the product (2-mark) Describe ALL possible IR Ab...
5 answers
008 10.0 points The tape in videotape cassette has a total length 182 m and can play for 1.7 h. As the tape starts to play; the full reel has a outer radius of 49 mm and an inner radius of 13 mm At some point during the play; both reels will have the same angular speed: What is this common angular speed? Answer in units of rad/s
008 10.0 points The tape in videotape cassette has a total length 182 m and can play for 1.7 h. As the tape starts to play; the full reel has a outer radius of 49 mm and an inner radius of 13 mm At some point during the play; both reels will have the same angular speed: What is this common angular s...
5 answers
Determine f" (1) for f(x) = Vxz+ 2.f_(UJa U1 O
Determine f" (1) for f(x) = Vxz+ 2 .f_(U Ja U 1 O...
5 answers
Complete the table by identifying $u$ and $d u$ for the integral.$$intleft(5 x^{2}+1ight)^{2}(10 x) d x$$
Complete the table by identifying $u$ and $d u$ for the integral. $$ intleft(5 x^{2}+1 ight)^{2}(10 x) d x $$...
5 answers
5.2.11Coreider binomial probability distribution with =0.3 and n = 10 What Is Ihe probability of the (ollbwing? exactly Inree succ0s505 less than thraa successes giqhl Or mare [email protected] = 31=(Round I0 four decimal placos 08 needod )
5.2.11 Coreider binomial probability distribution with =0.3 and n = 10 What Is Ihe probability of the (ollbwing? exactly Inree succ0s505 less than thraa successes giqhl Or mare SuCc05506 @ipix = 31= (Round I0 four decimal placos 08 needod )...
5 answers
The problem [6) The hciol ia clcrs of Phocnicia (emalcs is aproximated by h = 05 + 1.5 log 54 whcrc is dc fentale' ag3 in Years und 40420. Estimete Lhe hcighl {t Ihc ncurest hundredrh 0t [neler) Dl an &ycar old Phacuicidn SeriA)1.0S mB) /.80 InC1146 mDi 1.28 mn
the problem [6) The hciol ia clcrs of Phocnicia (emalcs is aproximated by h = 05 + 1.5 log 54 whcrc is dc fentale' ag3 in Years und 40420. Estimete Lhe hcighl {t Ihc ncurest hundredrh 0t [neler) Dl an &ycar old Phacuicidn Seri A)1.0S m B) /.80 In C1146 m Di 1.28 mn...
5 answers
Question 12Let A € Mzxs (R) be such that rank(A) = 3, and let B € Msxs (R) be such that rank(B) = 5_Not yetThen rank(AB) (select the appropriate answer):answeredis possibly equal to 1,2,3,4, or 5, but not moreMarked out of1.00is possibly equal to 1,2, or 3, but not moreFlag questionMUST be exactly equal to 5MUST be exactly equal to 3
Question 12 Let A € Mzxs (R) be such that rank(A) = 3, and let B € Msxs (R) be such that rank(B) = 5_ Not yet Then rank(AB) (select the appropriate answer): answered is possibly equal to 1,2,3,4, or 5, but not more Marked out of 1.00 is possibly equal to 1,2, or 3, but not more Flag ques...
5 answers
Perform the appropriate operations to simplify 10r+164A10. Solve each equation or inequalitv 3 $
Perform the appropriate operations to simplify 10r+16 4A 10. Solve each equation or inequalitv 3 $...
1 answers
Cone and sphere The cone $z^{2}=x^{2}+y^{2},$ for $z \geq 0,$ cuts the sphere $x^{2}+y^{2}+z^{2}=16$ along a curve $C.$ a. Find the surface area of the sphere below $C,$ for $z \geq 0.$ b. Find the surface area of the sphere above $C.$ c. Find the surface area of the cone below $C,$ for $z \geq 0.$
Cone and sphere The cone $z^{2}=x^{2}+y^{2},$ for $z \geq 0,$ cuts the sphere $x^{2}+y^{2}+z^{2}=16$ along a curve $C.$ a. Find the surface area of the sphere below $C,$ for $z \geq 0.$ b. Find the surface area of the sphere above $C.$ c. Find the surface area of the cone below $C,$ for $z \geq 0.$...
5 answers
Write each expression with positive exponents only. Then simplify, if possible.$$ rac{a^{-4} b^{7}}{c^{-3}}$$
Write each expression with positive exponents only. Then simplify, if possible. $$ \frac{a^{-4} b^{7}}{c^{-3}} $$...
5 answers
At a hospital maternity ward, there are an average of 8 babiesborn per day. what is the probability that exactly 6 babies will beborn a day
at a hospital maternity ward, there are an average of 8 babies born per day. what is the probability that exactly 6 babies will be born a day...
5 answers
An actress has a probability of getting offered a job after atry-out of 0.1. She plans to keep trying out for new jobs until shegets offered. Assume outcomes of try-outs are independent.Find the probability she will need to attend more than 6try-outs.
An actress has a probability of getting offered a job after a try-out of 0.1. She plans to keep trying out for new jobs until she gets offered. Assume outcomes of try-outs are independent. Find the probability she will need to attend more than 6 try-outs....
5 answers
W 1 MI 4sue WimL 1
W 1 MI 4sue WimL 1...
5 answers
23 Tyl2 Does lim(z,y)- (0,0) exist? x2 y6Yes:No:
23 Tyl2 Does lim(z,y)- (0,0) exist? x2 y6 Yes: No:...

-- 0.023826--