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6.) FeCls'- is known to be high spin complex: Determine the following: The crystal field energy-leve) diagram Label all orbitals and fill with the appropriate ...

Question

6.) FeCls'- is known to be high spin complex: Determine the following: The crystal field energy-leve) diagram Label all orbitals and fill with the appropriate number of electrons_ Is the complex paramagnetic Or diamagnetic?? Circle one Calculate the crystal field stabilization energy (CSFE) in terms of "Dq" (where 1ODq is the energy by which the d-orbitals are split) and "P" (the pairing energy). Circle answer

6.) FeCls'- is known to be high spin complex: Determine the following: The crystal field energy-leve) diagram Label all orbitals and fill with the appropriate number of electrons_ Is the complex paramagnetic Or diamagnetic?? Circle one Calculate the crystal field stabilization energy (CSFE) in terms of "Dq" (where 1ODq is the energy by which the d-orbitals are split) and "P" (the pairing energy). Circle answer



Answers

The following are low-spin complexes. Use the ligand field model to find the electron configuration of the central metal ion in each ion. Determine which are diamagnetic. Give the number of unpaired electrons for the paramagnetic complexes. (a) $\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{4-}$ $(c)\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}$ (b) $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{3}$ $(d)\left[\operatorname{Cr}(e n)_{3}\right] S O_{4}$

In the given complex. The central Artem is a transition metal iron which is a dodge with six nitrite irons. The oxidation state off Every is assumed to be x by the oxidation state off night. Right is negative solving for the oxidation state the oxidation number off if he turns out to be plus two. The electronic configuration off Effie is three d six for us too. In plus two oxidation state, it will lose Tau Elektron from its four s shell. The outermost configuration is represented as followed Looking, looking at the innermost orbital Onda splitting as it is shown over here, it can be observed that the complex has zero on beard electrons. Hence it is di magnetic compound and also has a low spin complex. In second complex, the central atom is again transition. Adam attached with six fluoride ions having oxidation state off negative one bye oxidation stayed off Iron is X solving the oxidation state the oxidation number off a free turns out to be blustery. We know every consist off 26 at 26 electrons. The electronic configuration is three d six for us too. In plus three oxidation state, it will lose total of three electrons. So the Newark city, the new electronic configuration, will be 354 sdo We can see the electronic electron pair donated by the light gone toe Goto the outermost orbital as Shona. What? This diagram. It can be observed that the complex has 500 electrons. It'll it will have ah, high magnetic induction, that is it is para magnetic compound along with high spin complex.

Okay in complex iron, which is ball titanium 93 5 H 202 positive. We have NH three and H 20. Which are neutral. Likens hence routine, Ium will have a charge of to positive. The electronic configuration of routine Ium itself is one is to to s to to be six, three years two, three P six three deep. Then for us to four p 6, 47 five s. Well, what the electronic configuration of routine iem to positive. This one will be your routine. And routine ium to positive will be Yeah, almost same to positive means that Putin iem to positive have to less electron as compared to its original configuration. So they lost subtracting two electrons. It will be 46 and five S 0, which we can cancel. Yeah, So we have 46. In routine ium in low spin complex the pairing of the electrons require less energy as compared to distribution so routinely, um can be represented us, this is the lower orbital, fully filled and no electrons in the upper orbital which is E g, the D O G E G. This routine iem to positive have no um paired electrons

This question gives us a complex and tells us that the complex has five UNP aired electrons. You minus uh then asks us to sketch the energy level diagram for the D Orbital's and indicate the place the placement of the electrons for the ion. Ah, it also asks us to determine if the eye on his high spin or Lisbon. Now it's given some some context clues for how we're supposed to figure this out. So the complex itself has six Legans indicating that it is an octave federal complex. And this is the energy level diagram for a knock cathedral complex. Uh, it also tells us that it has five unpaid electrons. Now, if we were to draw out five on Kurt Electrons, we put one in each of these. So with that here, 12345 Now, in order to have these five impaired electrons, you need for this molecule to be high spin. If it was a low spin you would have, it's just quickly this again, 19 3 42 If it was a low spin, you would have 123 for five electrons, and that does not leave us with five unpaid electrons on, and therefore the molecule cannot be a little spin. Um, because of these 500 electrons, you're gonna have three in this tea to space and two in this higher energy level space.

This problem focuses on the idea of the crystal field splitting diagram where we use our knowledge of the electron configuration of metal ions to draw how their D orbital's would fill. And from there using these diagrams, we can understand the magnetism of these metal ions. The first ion were given is magnesium. Pardon me, I'm agonies to plus and what do you look on? The periodic table We find them agonies to plus has an electron configuration of three D five. So when we draw our crystal fields putting diagram remember that this axis is energy are lower energy orbital's T to G and are higher energy orbital's e g. And remember, we're also told that all of these ions are low spin. So this is what our T tuju partner or t to G Orbital would look like once we have five electrons in it. And when you what you can see is we have one unpaid electron here. That means that I'm agonies to plus is para magnetic now on to part B. Now that you've seen this process a little bit, this should hopefully start making sense now for part B were given chromium three plus, we find its electron configuration. It has a three D six electron configuration. So we start by filling our TTG orbital will have three pairs of paired electrons. So that means that chromium three plus is dia magnetic. And now, for part C, we're given iron three plus iron three plus has a three d five electron configuration. So it's gonna going to look very similar to what we saw before. We're going to fill our TTG orbital and we'll have one leftover electron that's not paired. So we'll have one unpaid electron, which means iron three plus is para magnetic and finally aren't depart d were given chromium two plus chromium two plus has a three D four electron configuration. So again we'd fill our ttg orbital and we'd have to unpaid electrons here, which means that chromium two plus is para magnetic. I hope that this problem helped you understand how weakened your aw, the crystal field splitting diagrams Ah, various metal ions. And using that knowledge, we can determine its magnetism of either para magnetic or die a magnetic


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