a. Draw a Lewis structure for each of the following:
b. ...

Question

a. Draw a Lewis structure for each of the following:

b. Draw a structure for each of the species that shows approximate bond angles.

c. Which species have no dipole moment?

2. HNO₃

Accepted Answer

Welcome back everyone. For the species given below. Number one, draw an appropriate Lewis structure. Number two draw a structure that shows approximate bond angles. And number three state. If it has a net zero dial moment, we're given the structure of H C L 03 which is chloric acid. We're going to start with the first part where we want to calculate where we actually want to draw an appropriate Louis structure. We have three different atoms in the given structure. Hydrogen chlorine and oxygen which have 17 and six valence electrons respectively because those are the group a numbers that each atom belongs to. So the total number of electrons in the structure would be one electron from hydrogen plus seven, from chlorine plus six from each oxygen. So that would be three multiplied by six. And if we calculate the result, we end up with 26 valence electrons, our next step is to assign the central atom, we are choosing between chlorine and oxygen because they form more bonds than hydrogen. Hydrogen can only form one bond. Chlorine is less lene than oxygen. So it must be the terminal atom. I'm sorry, it must be the central atom, right, since central atoms are less electron negative than terminal ones. So if oxygen is more electron negative, it will be the terminal atom. So we're going to assign three chlorine oxygen bonds, we are always drawing single bonds to begin with. And the hydrogen atom must be bonded to one of the oxygen atoms. That's because if we have a structure of an acid, the hygiene atom must be bonded to oxygen so that the bond is more polarized. And that makes sense, right? Whenever we have an oxo acid hygiene is always bonded to oxygen. Our next step is to satisfy the octet rule or the terminal atoms. So that means we are adding three lone pairs for two of our oxygen atoms to have eight total, right, two from the bonds and then six electrons from lone pairs for the final oxygen atom, we only need to add two lone pairs because there are two bonds already which pro provide for electrons. Now, our next step is to calculate the total number of electrons as signs we have eight eight, then we have our chlorine oxygen bond. So that'll be 12345678. So eight more eight multiplied by three gives us 24. So we need two more electrons to have 26. And based on the rules, those two electrons must be placed onto our central atom, which is in this case, chlorine, well done. Our next step is to try to minimize the formal charges. But before we do so, we need to calculate them. So the formal charge of chlorine, according to the formula, we are using the number of valence electrons of chlorine. So chlorine has seven valence electrons, we're subtracting the number of bonds. So we have three bonds and the number of lone pair electrons. So that be two seven minus three minus two, that gives us a formal positive two charge on chlorine. For each oxygen, specifically, for the first two oxygen atoms, we will have six V electrons minus one bond minus six lone pair electrons. And that gives us a negative formal charge for oxygen, one of them, the other one. And of course, the third oxygen has two bonds and two lone pairs. So it has no formal charge and hygiene also has no formal charge because it simply has one bond in its normal state. If we want to minimize the formal charges, we want to get rid of one of the loan pairs from the first oxygen and introduce a double bond. So that so now oxygen will have two lone pairs and two bonds as a result, its new formal charge will be zero. So we're going to get rid of negative one and make it zero. And similarly, we want to minimize the formal charge of the other oxygen performing the same step, we will introduce a double bond out of its loan pair. And that's how we will end up with a formal charge of zero for that oxygen. If we recalculate the formal charge of chlorine, it is now zero as well because seven minus five bonds minus two lone pair electrons gives us a formal charge of zero. Well, then, so we have our Lewis structure. Let's go ahead and draw it once again. So we have chlorine with a long pair, two double bonded oxygen atoms. Each oxygen has two long pairs than one single bonded oxygen that has hygiene attached to it with two long pairs. And that'd be our answer for the first part of the problem, we got the les structure of chloric acid. Our next step is to predict the bond angles and to do that, we are going to look at the previously structure and understand how many election regions there are on the central atom. So basically, we have one chlorine oxygen double bond, that's one electron region, another one, that's two electron regions. Then we have a single chlorine oxygen bond, that's the third electron region. And we have a lone pair. So the number of electron regions is four, which corresponds to an S P three hybridized chlorine atom. And S B three electronically corresponds to 109.5 degrees. But because instead of having a substituent, we actually have a loan pair for each pair, we can estimate that we are subtracting 2. degrees to end up with an estimated value of 107 degrees, right, because we have trigonal peral geometry similarly to what we see in ammonia. So the geometry is trigonal for a middle. However, for the oxygen atom, because we have an additional bond angle, we also have 1234 electron regions, two bonds and two lone pairs. So that that would also correspond to an S P three hybridization. But because there are two lone pairs, if we want to predict the correct bond angle, we're going to take a 109.5 degrees. And we're going to subtract 2.5 degrees multiplied by two, which gives us five degrees, right? Because each lone pair squeezes a our bond angle by roughly 2.5 degrees. So we have to subtract 5.0 degrees and we end up with 104.5 degrees, which makes sense, right. The bond angle is very similar to the bond angle in the water, which is 104.5 degrees. The geometry is bent, right? Because even though we have an S P three hybridization, which electronically is tetrahedral, but due to the presence of two lone pairs, we have ben geometry for oxygen, meaning we can draw the three dimensional structure now. So we have our chloral group, actually, we have our central chlorine atom double bondage to t oxygen atoms. We want to make sure that we draw a trigonal parameter geometry. That means one of the substituent specifically our oxygen atom is pointing away from the viewer and that oxygen atom is bonded to hydrogen. Of course, we can add lo pairs and neatly state the bond angles starting with chlorine. We know that the geometry is trigonal per and we have O C L O bond angles. So those angles, oxygen chlorine oxygen bond angles will be 107 degrees. In addition to that, we have O C L O bond angle for our single bonded oxygen atoms. So the first one was between double bonded oxygen atoms and now we have single bonded oxygen atoms. So O C L 01 of the auction atoms is double bonded, that would also be 107 degrees because each of the angles is 107 degrees. Now for the oxygen atom which corresponds to the bend geometry, we said that the hydrogen oxygen chlorine bond or H O C L corresponds to 104.5 degrees. Well done. So we have predicted our bond angles and we have drawn the three dimensional representation of the structure. And finally, we are ready to answer the final question in this problem. What about the net diaper moment? We have to recall that the net dile moment corresponds to the vector sum of the individual dial moments. Each oxygen atom is more electron negative than chlorine. So to illustrate the concept, we are just going to redraw the structure once again because each oxygen atom is more electron negative, we will have one dipole moment from chlorine to oxygen, another one from chlorine to oxygen. The third one from chlorine to oxygen. And we also have a dial moment from hygiene to oxygen because oxygen is more electron negative. But there's also a long pair which creates a dial moment pointing upwards. The diaper moments do not cancel out due to multiple reasons. The lone pair is not an oxygen atom. So its dipole moment does not have the same magnitude as the dipole moment between chlorine and oxygen. So that's one of the reasons why the dipoles do not cancel out. And in addition to that, we have a comparison between single and double bonds. So that means for the third part, the net diaper moment is not equal to zero and those will be our final answers. Thank you for watching. And I hope to see you in the next video.

a. Draw a Lewis structure for each of the following:
b. Draw a structure for each of the species that shows approximate bond angles.
c. Which species have no dipole moment?
2. HNO₃

Key Concepts

Lewis Structures

Molecular Geometry and Bond Angles

Dipole Moment

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