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Draw a crystal field energy-level diagram, and predict the number of unpaired electrons for the following complexes:(a) $left[mathrm{NiCl}_{4}ight]^{2-}$ (tetrahedr...

Question

Draw a crystal field energy-level diagram, and predict the number of unpaired electrons for the following complexes:(a) $left[mathrm{NiCl}_{4}ight]^{2-}$ (tetrahedral)(b) $left[mathrm{Ni}(mathrm{CN})_{4}ight]^{2-}$ (square planar)

Draw a crystal field energy-level diagram, and predict the number of unpaired electrons for the following complexes: (a) $left[mathrm{NiCl}_{4} ight]^{2-}$ (tetrahedral) (b) $left[mathrm{Ni}(mathrm{CN})_{4} ight]^{2-}$ (square planar)



Answers

For each of the following metals, write the electronic configuration of the atom and its $2+$ ion: (a) $\mathrm{Mn},(\mathbf{b}) \mathrm{Ru},(\mathbf{c}) \mathrm{Rh}$. Draw the crystal-field energy-level diagram for the $d$ orbitals of an octahedral complex, and show the placement of the $d$ electrons for each $2+$ ion, assuming a strong-field complex. How many unpaired electrons are there in each case?

So here we have a question. 23.6 psi, too. For each of the following medals, right, Elektronik configuration at three. Plus ayan. And then we need to draw the crystal field energy level diagram for the D orbital's oven. Octo. He drew complex and also show the placement of the deep elections for each three. Plus I it right. So question a we have, aren't you, Kathy? So the electric configuration for F E is Oregon 3 86 for us to. So if we are looking for every three plus, then we will have electric configuration three d five. All right, so now we draw the energy level for this will be they are two higher key orbital and three lo ver or there t two. Or and there are five electrons. How feeling this 53 the organelles and the impaired intellectuals or the placement of electoral st D orbital is five. So that is the party part B. We have m o. And for oh metal. The electron configuration is or d 55 us one. So if we're looking at m 03 plus we live three valence electrons, they will have k R for d three. So the first the field energy level diagram it will look like E. This is 12 and three t two orbital's and we have three. Elektra. Lt's filling lower energy level. So there are three I'm paired electrons feeling the Dior bottle. That is the part B and part C. We have Coble. Electric configuration for Coble is our three. He's seven for us too. So if we're looking at Kobel three plus, the electron configuration will be three d six. So now we draw the field Energy level diaper lower. Issa T to hires e So we have six elections in total and we're going to start drawing the elections from the lower energy level. Juan two, three, four, five, six. So there are four on paired electrons.

This question gives us six different complexes and asks us to draw the crystal field energy level diagrams and to show the placement of the D electrons. So this first complex I this two plus, we're all complex shows us that this CR has is a two plus because this cold water molecules, uh, don't have any charge, and it tells us that they're four unpaid electrons. So we know that this system, it's gonna be a d four. And the only way that we can get four unpaid electrons in this energy level diagram is if we have one, 23 four. And so we know that this complex is high spin because there's the's Warren pair of electrons. Um, And if it were low spin, we would have put this this upper level energy electrons down here, which would only give us two unpaid electrons. Part B gives this, uh, also to plus complex with mn is our mettle. Uh, and it tells us that this is a high spin complex. And so because this water does not, uh, give any charge the molecule, we know that the MN is going to be a tube course. Ah, where that electron configuration for the two plus, it's gonna be a r and then three d five. So they're five electrons in this d orbital. It told us that this was a high spin complex. And so we know that all five of these electrons are gonna be uncared. That's in one. You three for five curtsy has this much longer complex Ah, which tells us that it's a low spin complex in the question where this are you has a two plus. So our electron configuration here is going to be They are for D six. This is low spin. So we know we're gonna entirely fill this tea to orbital before we feel the higher Angelo orbital. So we have 123 45 offer parts he threw up all over the second page. Well, we have this, uh, two minus complex with an I r. Uh, which tells us it's the question. It's a low spin complex. And so we have to determine the number of electrons in the D orbital for I R. Our electron configuration starts with 60 and then we have the huh? Were F 14 and then we have five t guys we know that because the ah, this I r is going to be a plus four complex. The Corinne Zara minus one. There, six of them. The total charge for the Affleck, the ion or the complex is minus two. So are my our needs to be a plus four. So we're gonna have five electrons in the D orbital, and we've been told that it is a most been complex. And so we're going to entirely fill this tea to orbital with our five electrons. So we're gonna have one, 23 four Part E. We have, ah, this three plus complex that we are told. Ah, we're just given. Given the name of the complex, we know that the the CR is going to be a three plus and so that electron configuration is going to be a r three d three. Ah, and this is why they didn't give us any information about whether it's high spinner low spin, because no matter no matter which one it is, we're just gonna fill the three orbital's in this teach you for part f We part just given the This is complex where we know that the nickel is gonna have to be a plus two because F is minus one There, six of them for a minus six charge. The total charges minus four. So the nickel needs to be a plus two. This is going to be all right. A r for electron configuration three d eight. And again, they didn't indicate whether it's high spirit, loose pin. And that doesn't matter in this case, because no matter how we feel it, 123 45678 And if we feel that the other way, it would have been 123456 feet. So no matter how you feel these electrons Hi spinner Lisbon, you'll get the same, uh, same configuration.

So here we have the problem. 23.64. We're going to draw the crystal field energy level diagram and also show the placement of electrons for the following complexes. Some part A, we have palladium C 06 three minus. So first we want to find the oxidation number of the metal palladium in this case. So we assume it's X. And we know that the oxidation number for chloride is negative. One that is equal to the total oxidation state, which is negative three. And with off for the X equal to positive three. So the oxidation number for palladium is positive. Three. So we'll be three. Plus, we'll have an electron configuration as they are. Three D, too. So now we draw the lecture. Now we draw the energy level. Diagram one two, this is the three d orbital. All right. Next one is part B. We have of E uh six three. Minus this case, we assume the iron has oxidation. Stayed up x, and we know the floor. Reid has oxidation number off. Negative one that is equal to negative. Three. So itself for axe, that is a quantum positive three. So the station. Number of iron in the complex is three plus. So for iron three plus, the electron configuration would be a R three D five. So now we draw the energy level diagram. For this we have. This is the D five orbital. There are five electrons, three for in five. All right, so that is party part. See, we have the complex rubidium I period ing three three plus. And this is a low Spain complex. All right, so now a first look for the Arc station number on the rubidium, so we know that the by puritan has a zero oxidation state, so the total oxidation stayed up. The complex is three plus, so we know the rubidium has a three plus oxidation state. So for a medium three plus, the electron configuration would be 45. So there are five elections in the 40 orbital. So if we draw the four d orbital right here, this isn't the five since Is a Lowe's being complex? It will be preferably Perry at the lower energy level. So two pairs in one, um, parent electrons. But that's for the part C party. We have legal C 04 two minus, and this is a tetra. He draw. All right, So assuming Niko has an observations, they'd of X close four times. Chloride is negative. One that is equal to negative too. So we saw further Number X is equal to positive too. So the oxidation instead of Nico in the complex is positive too. So nickel to plus will have electric configuration miss a r three t eight. All right, since is a tetra hydro and the configuration will look a little different from before and the energies reverse. So we have two orbital's at lower state and will be start feeling with eight Elektra lt's one, two, three for five, 67 eight. So there are three pairs electrons into our impaired Elektra lt's that is the party. So this is the key. Ain't Tantra hinder orbital? All right, now look at Part E. We have a complex PT br six two minus. So assuming the platinum has no oxidation number off X plus six time bromine is negative for equal to number two. So solving for the number ofthe X, we get a number of positive for cool. So the plot No, in this is complex, will have a charge of positive for so PT four plus we'll have electron configuration us Taxi four of 14 5 36 So if we drop the electron so now if we draw the energy level diagram, we will have two Orbital's on top three right here. So we have five electric one 23 for 56 So this will be the E six High's being orbital. Looks like her next one is question. We have to Tanya B. N three two plus. So we know the oxidation number off the leak and e and is zero. So that Italian we'll have in oxidation. Number of positive, too. So for detainee in two plus, the electric configuration is Oregon three D too, and we will draw the energy level. Was too electrons on the lower energy level? That is too deed too. So that is, They answered part off.

In this question, we give the villains bone description off the bowling off the following complexes and then find out the number off on paired electrons and the hybridized orbital that are involved. So first complexes, F E cyanide 63 Negative. So first off or we write that electronique configuration off every tree plus sign which come out Toby Argon and 35 Because here, atomic number of ironies 26 we remove three electoral. So we get it in place. Five at 23. Now this complex is lost in complex and we know that cyanide is a Ah, cyanide is which type of flick and cyanide is a strong feeling and so Sinai's it in strong for illegal. So there is a painting can take place in this complex So we're right, So we grow there bonding So, like, this is the treaty orbited 12345 So this is true the and this is forests and this is for people but four b So this is four case off every three. Plus I'm now do go the way we see that cyanide is the strong field, so pairing can take place. So we fill these five electrons. 1234 in the pair form. Now see, the six electrons are like feeling this way. 12345 and six. Now when we see that So this is the I realized orbital that are involved. That means how many orbital in world did do that went toe s Petri. So this is the hour answer for the party did us Petri Because we feel these are six llegan cyanide legal So like this week with that? Now we see that unburned electron So see, there is only look at this So there is only one unpaid electoral so we can't right There is only one un paired electrons 100. So in this complex, there is only when unpaid Electron no, for part B off the ocean is part B that is given ISS you at school six No policy. So this complex is the hi It's been complex and we know that water is a big free legal. So there is a no parent can take us so far. Stoffel were right. The electronic configuration off mobile do place. So this come out Toby Argon and three d seven. That means getting plus seven. So that women 25. So this is the because while these atomic number 27 So we removed. So now, brother. So this is the seals sea or hopeless. So this is we can write. 12345 So this is the 30 then he had four s and this is the 41 toe. 3 40 Now see, we see if you see that water. So see what? Arista Wakefield. So there is no Perry can take place in the water. So we feel data seven electoral like 12345 67 So in this way we feel the seven elected Now what we have to do is we now offer the league and so water I s o c. That What? There is six so six water. We have to feel it. But we have here only four. So we again this 12345 That meant for the orbital are involved in order to feel now, fill that So one, 234 five and six So deadman thes our total are involved in this case So we can write. This is like how many so s B three and deep. So this is over. I produced orbiting that are in the world. And when we see that the unpretty left So see there is one entry so there is three and pretty little So we can write three un bird electron in this complex for part c off the ocean. So see, party off the Chinese So that is given as v c l for minus. So this is in the tetrahedron. So this is that federal that is given Tried. Really? Okay now, first of all right, they like Tony configuration off vanadium triplets. So this come out Toby Argon and Tito Credito. So here argon is 18 plus 2 20 we know that vanadium atomic number is 23. So remote three letter which come out we readyto Argh! Until now this thesis chloride I'm so we know that chloride and is a weak field And it in Provo high spin and there is no betting can take place in this case, so we can write as a V three plus iron. This is 12345 And this is for us. And this is for the be so this gun, right? Does it really orders? And this is 40. So no building. So we that we can write like this. So there is no better. Now we have two friendlies for are vital for Allied in these orbiting. So this can be free less like this one. So 34 Because it under what? I go ahead and we see that So these are involved so that in oneness Petri So it andro sp tree And when we see that on pretty late point so see that this is the unpredictable. So how many and pretty left wing in this case? So there is too unfair electoral so two on pretty like terms in this complex now but body off the question is, that is a party electing them c l four tu minus so that it's scare planner. So we write Other scare planet that is given in the ocean is now we write as a Laconi configuration for stuff all platinum plus sign this Come out, Toby over argon and this is equal Do that is five p it not we write Desert Beauty two plus like this. So 12345 So this is five d and this is our six s and this is our six Speak like this. Now we see that see seek glory in the platinum BSO so it undergoes care planner complex So we see that Do do that Diffuse five day orbital So there is a due to defuse five The orbital there is a pairing can take place where week religion. So here, bearing Baird The elections of 12 3456 and seven So this is a painting can take place Now we have to fill these glorying chloride I'm so like in this way So one do three and four So see, this is the like this So that means deep as speed. So the DSP took that men it will undergo Scare Planner complex And we see that here all our weird electoral So we can write zero so we can write zero un bird electron in this complex Zero on pretty locked down in this complex. So this is a one answer soc! There we have to You have to know that dough diffused five the or biter they paired up through big free ligand So that's why we pay our in this case


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