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Hc 5uI Uce the appropriale leti What the Identity cach of the three orhitals belaw! with the correct orientation indicator (* L-) for the orbital (e & ,' a...

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Hc 5uI Uce the appropriale leti What the Identity cach of the three orhitals belaw! with the correct orientation indicator (* L-) for the orbital (e & ,' alongWhat set of orbitals are represented below? Provide only the letter indicator (e & ,Write out the orbital box diagram and the condensed electron configuration tor Nitrogen:Write out the orbital box diagram andthe condenseo ectron configuration for Silicon:Write out the orbital box diagram and the conderised lectron configurati

Hc 5uI Uce the appropriale leti What the Identity cach of the three orhitals belaw! with the correct orientation indicator (* L-) for the orbital (e & ,' along What set of orbitals are represented below? Provide only the letter indicator (e & , Write out the orbital box diagram and the condensed electron configuration tor Nitrogen: Write out the orbital box diagram andthe condenseo ectron configuration for Silicon: Write out the orbital box diagram and the conderised lectron configuration for Iron: Which quantum number represents the energy level of particular lectron? What is the name of the rule that says no twO electrons inan atom can be represented witl the same set of quantum numbers? 12. What is the name of the rule that says every orbital in - subleve (for example, within the orbitals) is singly occupied before any orbital is doubly occupied, and moreover that all of the electrons in singly occupied orbitals have the same spin?



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The orbital occupancies for the d orbitals of several complex ions are diagrammed below.
(a) Which diagram corresponds to the orbital occupancy of the cobalt ion in $\left[\mathrm{Co}(\mathrm{CN})_{6}\right]^{3-2} ?$
(b) If diagram D depicts the orbital occupancy of the cobalt ion in $\left[\mathrm{CoF}_{6}\right]^{n}$ , what is the value of $n ?$
(c) $\left[\mathrm{NiCl}_{4}\right]^{2-}$ is paramagnetic and $\left[\mathrm{Ni (\mathrm{CN})_{4}\right]^{2-}$ is diamagnetic. Which diagrams correspond to the orbital occupancies of the nickel ions in these species?
(d) Diagram $\mathrm{C}$ shows the orbital occupancy of $\mathrm{V}^{2+}$ in the octahedral complex $\mathrm{VL}_{6}$ Can you determine whether L is a strong- or weak-field ligand? Explain.

Cycle cycle. Octa to train has eight carbon, an eight hydrogen and a negative to charge. There are four double bombs and hydrogen coming off each of the carbon. Each carbon is SP two hybridized and if we look at bonding, non bonding, an anti bonding, we can represent the electrons In this way. The P electrons are assigned to the orbital's properly and because they are all paired, this molecule is dia magnetic.

So in this question, were given little summer equation asked to solve all the parts. So for a we have the quantum number and describes the blank oven atomic orbital. Now we know that n is the principal quantum number and it tells us how far how high and energy and atomic orbital is. It also tells us this size of the atomic orbital were also given that l describes its blinked. Now we know l can range from zero to n minus one and actually describes the type of subsoil we have in our orbital. So this exceptional next we have when a equals three the positive values of L o R. So we just said that l ranges from zero to n minus one to the possible ranges. Possible values are 01 or two. What type of war bill corresponds to illegals three. Now, this is straight memorization. You should know by now that zero is an s orbital one is a P orbital to was a D orbital and three is an F orbital for a 40 orbital. The value of and his blank and a possible values of amsa bell is blank. Now the value of n is always the number proceeding the letter of the sub show. So in this case, the value of N is four possible value of M sub l I'm civil ranges from negative l to positive it. So with the D orbital l is equal to two, so am Cybele can be negative too, to positive to any number in between. So we're just gonna go with zero in this case. Next, we have the shapes of two orbital's and arrest to find the letter the L value and the number of notes. So for the letter for the 1st 1 we see a dumb bell shaped and by now you should know a dumbbell shapes. Orbital was a P orbital with a P orbital R l value is one and the nodal surfaces that we have in an orbital is equal to the value of l. So because of this, we have one nodal surface and I can actually outline it right here. Here is our note. Next, we have something that looks like to dumbbells put together. So this is a D orbital with an L value of two. So which means it has to notes the notes are right here and right here. An atomic or bill with three nuttall services is so we only have three nuttall services. So l has to be three. If we took a look above again, we can see that Zeros s one p two d and three corresponds to f next report. Age were asked which of these is not a valid set of quantum numbers. So let's go ahead and take a look. We know l has a range from zero two and minus one and amsa Belkin only range from negative l to positive up. So for the 1st 1 we have an eagles three and l equals two, which is correct so far, M sabbatical one, which is all correct. So this is fine. Next, we have an equals two and l equals one, which is correct. But the Isabel equals two. So if amsa bell ranges from negative lt Positivo, the only possible values for this are negative 10 or one. So this here is not a valid set, and this one here is next. We're has to determine the maximum number of orbital's in each of these quantum number designations. So we have an equals two and l equals one so we can figure out I'm so Bell is negative 10 or one so we can have three possible orbital's. Next we have an equals three and we can figure out the maximum number of orbital's by simply taking and squared. So you have nine orbital's and this one next we have an equals three in l equals three. Now, this should catch your eye because l ranges from zero to n minus one. So this is actually not a possible quantum number designation. So because of this, we have no orbital's. Lastly, we haven't equals two Illegals one and M. Sabella equals zero were given all three quantum number designations.

Because you're doing problem. One ten of chapter nine in chemistry Center science using the Valance of Atomic Orders of Ha Jin at Ha Jin and flooring. And following the model figure nine point four six How may molecular corporals would you expect from the H f molecule? So Heidrich Jin has one Valance orbital and Florine has four Valance Orbital's So Hodgins gonna have with one ass orbital and florins going tohave two p t. P x tupi. Why? To Pete Easy and to us. Therefore, you would have five or rules those and this could create five molecular organelles. Now for B, how many of the military rebels from party will be occupied by electrons? So Ha Jin is one of the islands, like trying for flooring has seven valence electrons. So there will be eight or electrons and any atomic order for H f. So eight electrons would be filled, filled in. That's it. Eight electrons would be filled in in four of the five like nobles because since two electrons can fil am election but orbital, then us then you're going to feel all for beer or broads before you hit that fifth orbital. Okay, Now see, it turns out different. And energy between the Vaillant atomic orbital's of agent death are significantly different that we could make, like the interaction of one ass of hydrogen and one two s of flooring. One is a wardrobe only mixed with one two p horrible of flooring draw a door pictures showing the different orientation patterns of the three p or those and on interacting interacting with a one. We have a one with a one s orbital on hydrogen. So this is going to be it's kind of difficult to draw on on a two. Yes, US honey us on a two on a two d absolute. There's going to be so there's gonna be a bit tricky. So So let's say this this No. And this like, yes, this is there's going to be our p What? This is going to be our I guess we got our S p y. There's going to be S p X and it's going to be S p Z. So our hydrogen is going to interact with this. It's gonna have this kind of overlap between the P orbital and the S orbital. Now which of now they won't know which of the two pupils will actually make a bond on with for one as Sobral assuming that the Adams lie on the Z axis. So let me draw like an X Y Z axis. This is our why this is R X and this is R Z. So it would look something like this. Thank me juice do Green Teo. So our Ha Jin is going to look like something like this. It's goingto act interest with the piece with our P Orgel on the Z axis. It's a deal. In most exempted pictures of a Jeff, all of the atomic corporals on flooring move over at the same energy into the molecular energy. So in most accepted in the most accepted picture of H f, all other atomic or present flooring would move over at the same energy onto the molecular energy or Bill diagram for H F. These air, called non binding moguls, sketch the energy a little diagram using for each of you this information and casually the bond order. So here's our age orbital. Here's our P. Here's our floor in orbital. So since she isn't so since flooring is ely at I'm using p orbital's. That means we could theoretically on ly form one pie bond. So now we're gonna put our electrons from are nonbinding overalls into our bonding or Brazil ripping our Sigmund rpai pons. And then those last two electrons were going to put our anti pylons. So H f has a single box like you literally can't form a double bond. You literally can't form a double bond with hydrogen. So those two So so the fact that we have two electrons and our end try a little of that cancels out the pie bond. So you so that bratty gates, any double bond performing. So that means we're going to have a bond order of one now looking at e look at the Louis structure for H f. What were the non binding electrons? So this is our Louis structure. So the non binding electrons are going to be on our are more letter negative, Adam, which is going to be a floor? These lone pairs? These are our non bonding electrons

Quantum numbers can tell us about both the energy and location of electrons in an atom. There are different quantum numbers describing different properties. When we look at the quantum number and this tells us the energy level, the quantum number l describes the shape of the probability of finding an electron or the shape of sub level, leading to the orbital when n equals three R l values range from zero to the number before three, so there are a total of three values, or 012 This is true for all energy levels. They have the same number of sub levels, or else starting at zero, she rising up. Each sub level corresponds to a different type of orbital. We see here that if l equals zero, it's an S. L equals one p two. It's a D. And if l equals three, it's an F orbital. Quantum numbers can be used to describe specific orbital's. So if we're describing a four d orbital, we know that the value of N is equal to the energy level, which is the number So Ennis four l corresponds to the letter D l equals two is D and there are several possible values for m subscript l the strangers from the negative l value to the positive l value. So any of those air possible, um, quantum numbers for the four d orbital? Yeah, when are given pictures of each of the orbital's three correspond capacity specific letters and held values. So, for example, when we see this diagram here, this is a P orbital P Orbital's have a quantum member of l equals one. This is a D orbital of l equals two, and we can talk about the planner nodes. A planner node is if you could draw planes straight through where would it not touch any electron density? So we see there's one plainer node for the P orbital for the D orbital. There are two. We could slice it this direction, or we could slice it this direction if we're to talk about something with three plainer nodes, though, even though we don't have a picture of it, the next option is our chat orbital. Because the L value is equal to the number of planner nodes. Looking at our quantum numbers, we can see that there some orbital designations that cannot exist according to modern quantum theory to s indicates the second energy level with an L equals zero. So that exists. Three p similarly hasn't n equals three hell equals one, which is a possibility. So that also exists if we consider to D This has an energy level of to in an l value for D of two. We see that that is not possible because my l values range from 0 to 1 for two, so this one does not exist. Similarly, if we consider three f and equals three l for the F values should equal three. We see that that's not a possibility, because the L ranges from 0 to 2. So this is not a possible orbital combination. Five p and equals five l equals one and the range of l's for the energy level 501 to three or four. So this is a possibility. And finally, if we consider six p, we can see that this is also a possibility because N equals six and l equals one on my possible l values range from 1 to 5. So this is also a possibility. We can also see that there are only certain quantum numbers that can happen. Forgiven energy levels if we have an equals three l equals two ml equals one. And there's 1/4 quantum number called M s, which is this spin is a negative 1/2 M s could have two values. It's either a spin up positive 1/2 or a negative 1/2 so we can see that this is a possibility and close three l equals two m l equals one and m s his negative 1/2. So that's a possible set of quantum members. If we consider n equals two l equals one m l equals to M s People's plus 1/2 anythings to l equals one. But my ml values can Onley range from your negative one to positive one. So this does not exist. Okay, and for and equals four l equals three m l equals zero m s h equals zero. We see that it's possible to have a four and a three and a zero, but we can't have an M s values zero. So that's not a possible combination of quantum numbers. And finally, we can predict how maney orbital's are present in each associated with each different quantum number if we know that n equals two and l equals one. That means we're talking about the two p and there are three possible orbital's or three possible smells. The Milken range from zero negative one to positive one. So there are three orbital's with those quantum numbers for n equals three tell can equal 01 to And for each of those quantum numbers we see that we can have zero for my m l a negative 10 or one for my l equals one and negative two, ranging two to for l equals two for a total of nine possible orbital's. If anything was three and l equals three. We see this is not a possibility, because l can Onley range to help to two. So there are no orbital's and finally, of n equals two l equals one an m l equals zero. We see that this is a specific orbital in the second energy level and there is only one for Biddle


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