Nickel(II) complexes with the formula NiX 2 L 2 , where X is Cl - or N-bonded NCS - and L is the monodentate triphenylphosphine ligand P(C 6 H 5 ) 3 , can be square planar or tetrahedral. (b) If NiCl 2 L 2 is paramagnetic and Ni(NCS) 2 L 2 is diamagnetic, which of the two complexes is tetrahedral and which is square planar?

Hi, everybody. Welcome back. Here's our next question. Consider the following nickel two complexes and we have nickel then cl two and then two of the ligand triethyl phosphine, which is abbreviated capital P, capital E lowercase T subscript three. So nickel with two chloride and two triethyl phosphine ligands and nickel with two nitrate ligands and two of those pe T tri of phosphate ligands, the ligand triethyl phosphine pe T three or P. And then in parentheses, ch two chch three subscript three. So a phosphate with three ethyl groups is a mono dentate ligand. It was found that our first complex with the chloride ligands is diamagnetic. While our second complex with the nitrate ligands is paramagnetic. Determine which of the two complexes is tetrahedral and which is square planar. As our answer choices, we have a tetrahedral being the complex with the chloride ions. Square planar being the complex with the nitrate ions. Choice B tetrahedral is the complex with the nitrate ions. While square planar is the complex of the chloride ions c both complexes are tetrahedral or d both complexes are square planar. Well, before we can do anything, we have to think about our number of D electrons. However, we've been told that both of our complexes are nickel too. Therefore, we know that both complexes have the N two plus version. So they both must have the same number of D electrons since they have nickel and the same ox with the same oxidation number. So we can immediately rule out choice c both complexes are tetrahedral or both complexes are square planar C and D because we know that one is diamagnetic and the other is paramagnetic. So one has all its electrons paired and one has some unpaired electrons since they have the same number of D electrons, since they have nickel in the same oxidation state, they must be different. So let's figure out which is which, well, let's figure out how many electrons we have. So nickel just elemental is closest noble gasses, argon, argon in brackets four s two, 3d 8. So nickel, two plus remove two electrons and we have argon and brackets and 3d 8. So we have eight electrons, let's think about our energy crystal field splitting of the energy levels and how it looks different for tetrahedron square planar complexes to determine which is which, which leaves us with all paired electrons, which leaves us with unpaired electrons. So we know that our tetrahedral crystal field splitting is the opposite of octahedral. So we have the three higher energy levels and two lower energy level orbitals. And that the tetrahedral complexes always have a small delta since none of the orbitals point directly at any of our ligands, there's just a small difference in the energy level here. Meanwhile, our square planner complexes do have some energy levels that point directly or an energy level, excuse me, an orbital that points directly at our ligands. But then as compared to octahedral, you don't have any ligands on the Z axis. So those move down in energy. So our square planar splitting has our one very high energy level and then four lower energy levels, two by themselves. And then the lowest two at the same. And that the gap between our highest and then our four lower energy levels is a large delta. So when we fill in our eight electrons in a tetrahedral geometry, we put one electron in each of the five D orbitals and then start pairing with our next three. So the tetrahedral arrangement gives us two unpaired electrons. Now, if I'm on a test and time is of the essence, I've already got the paramagnetic arrangement. So two impaired electrons will make me paramagnetic. And I've been told that my complex with the nitrate ions, I'll just put no three in parentheses is paramagnetic. So I can go right to my answer choices and say that my complex of the nitrate ions must be tetrahedral. So choice B is going to be my correct answer and choice A will be ruled out since that has the complex of the nitrate ions being square planar. So that's if I'm needing to be speedy here, we'll be a little more thorough. So go ahead and look at our square planar arrangement where because we have four lower energy levels and a large gap we're going to fill in four electrons. And then rather than jump up high, since we have that large gap pair them up. So our square planar arrangements gives us zero unpaired electrons. And that is the complex with the chloride ions. Now, a note here is that you usually see uh configurations we have eight D electrons being a square planar geometry. And that's because you can see this is a pretty stable arrangement with a full set of lower energy levels and no electrons in the higher energy level. So it's unusual to have a tetrahedral complex with ad electrons but obviously not impossible because that's what we have here. So this confirms that indeed choice B is our correct answer. Our tetrahedral complex, which is paramagnetic since it has unpaired electrons is our complex with the nitrate ligands and the square planar complex which is diamagnetic since all electrons are paired is our complex with the chloride ions. So choice B is our correct answer here. See you in the next video.

Ch.21 - Transition Elements and Coordination Chemistry

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McMurry 8th Edition

Ch.21 - Transition Elements and Coordination Chemistry

Problem 21.131b

Chapter 21, Problem 21.131b

Nickel(II) complexes with the formula NiX₂L₂, where X is Cl^- or N-bonded NCS^- and L is the monodentate triphenylphosphine ligand P(C₆H₅)₃, can be square planar or tetrahedral.
(b) If NiCl₂L₂ is paramagnetic and Ni(NCS)₂L₂ is diamagnetic, which of the two complexes is tetrahedral and which is square planar?

Verified step by step guidance

Understand the terms: Paramagnetic substances have unpaired electrons, while diamagnetic substances have all electrons paired.

Recall that square planar complexes often result in low-spin configurations, leading to paired electrons, while tetrahedral complexes are usually high-spin, resulting in unpaired electrons.

Consider the electronic configuration of Nickel(II), which is [Ar] 3d^8.

In a square planar geometry, the d-orbitals split such that the electrons pair up, leading to a diamagnetic complex.

In a tetrahedral geometry, the d-orbitals split differently, often resulting in unpaired electrons, leading to a paramagnetic complex.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Coordination Geometry

Coordination geometry refers to the spatial arrangement of ligands around a central metal atom in a complex. Nickel(II) complexes can adopt different geometries, such as square planar or tetrahedral, depending on the number and type of ligands coordinated to the metal. The geometry influences the electronic properties and reactivity of the complex.

Magnetism in Coordination Complexes

The magnetic properties of coordination complexes are determined by the presence of unpaired electrons. A paramagnetic complex has unpaired electrons, resulting in a net magnetic moment, while a diamagnetic complex has all electrons paired, leading to no net magnetic moment. Understanding these properties helps in predicting the geometry of the complexes based on their magnetic behavior.

Ligand Field Theory

Ligand field theory explains how the arrangement of ligands around a metal ion affects its electronic structure and energy levels. In square planar complexes, the d-orbitals split in a way that can stabilize unpaired electrons, leading to paramagnetism, while tetrahedral complexes have a different splitting pattern that can lead to diamagnetism. This theory is crucial for understanding the electronic configurations of metal complexes.

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Guided course

02:40

Strong-Field Ligands result in a large Δ and Weak-Field Ligands result in a small Δ.

Related Practice

Textbook Question

There are two possible [M(OH)4]- complexes of first-series transition metals that have three unpaired electrons.

(a) What are the oxidation state and the identity of M in these complexes?

(b) Using orbital diagrams, give a valence bond description of the bonding in each complex.

(c) Based on common oxidation states of first-series transition metals (Figure 21.6), which [M(OH)4]- complex is more likely to exist?

<QUESTION REFERENCES FIGURE 21.6>-

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Textbook Question

Two first-series transition metals have three unpaired electrons in complex ions of the type [MCl4]2-.

(a) What are the oxidation state and the identity of M in these complexes?

(b) Draw valence bond orbital diagrams for the two possible ions.

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Textbook Question

Nickel(II) complexes with the formula NiX₂L₂, where X⁻ is Cl⁻ or N-bonded NCS⁻ and L is the monodentate triphenylphosphine ligand P(C₆H₅)₃, can be square planar or tetrahedral.

(a) Draw crystal field energy-level diagrams for a square planar and a tetrahedral nickel(II) complex, and show the population of the orbitals.

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Textbook Question

Nickel(II) complexes with the formula NiX₂L₂, where X is Cl^- or N-bonded NCS^- and L is the monodentate triphenylphosphine ligand P(C₆H₅)₃, can be square planar or tetrahedral.

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Textbook Question

The amount of paramagnetism for a first-series transition metal complex is related approximately to its spin-only magnetic moment. The spin-only value of the magnetic moment in units of Bohr magnetons (BM) is given by sqrt(n(n + 2)), where n is the number of unpaired electrons. Calculate the spin-only value of the magnetic moment for the 2+ ions of the first-series transition metals (except Sc) in octahedral complexes with (a) weak-field ligands and (b) strong-field ligands. For which electron configurations can the magnetic moment distinguish between high-spin and low-spin electron configurations?

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Textbook Question

Spinach contains a lot of iron but is not a good source of dietary iron because nearly all the iron is tied up in the oxalate complex [Fe(C₂O₄)₃]^3-.

(c) Draw a crystal field energy-level diagram for [Fe(C₂O₄)₃]^3-, and predict the number of unpaired electrons. (C₂O₄^2- is a weak-field bidentate ligand.)

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