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5

This doesn't appear to be a rare case and there are plenty $\ce{Cr^2+}$ square planar complexes. I found 15 square planar complexes with well-determined crystal structures $(R_\mathrm{int} < 10\%)$ by applying the following geometrical restrains: Chromium must have C.N. exactly 4 (T4). Chromium is allowed to be coordinated by virtually any element of ...

4

An article by Bradley et al. [1] documents $\ce{[Cr(N(SiMe3)2)2(THF)2]}$ as a square planar complex. However, the $\ce{[Cr(N(SiMe3)2)3NO]}$ complex is pseudotetrahedral $(C_\mathrm{3v}).$ Figures below are provided by andselisk, the are from the reference above. References Bradley, D. C.; Hursthouse, M. B.; Newing, C. W.; Welch, A. J. Square Planar and ...

1

A structure of a coordination complex with properly assign wedged bonds for depicting stereochemical features should be be either fully enclosed in square brackets or denoted with an L-shaped brace on the top right corner with charge being superscripted. Since here both chloride anions are not ligands and don't belong to coordination sphere, they should be ...

2

Your formula, $\ce{Rb[Pt(NH3)Br2Cl]}$ is correct for the named complex, rubidium amminedibromochloroplatinate(II) (For clear understanding of naming, you may read Naming Coordination Compounds-1 and Naming Coordination Compounds-2). Many coordination compounds have distinct geometric structures. Two common forms are the square planar, in which four ...

3

I think it should be - tetraamminechloridoiodoplatinum(IV) chloride iodide. The rules of the naming of counter ions are pretty much the same as in general Inorganic Chemistry. I do know of a compound AgClI - named as Silver chloride iodide. So, I think the counter ions here - chloride and iodide should be named as by the nomenclature rules of anions i.e ...

2

While the vast majority of the compounds and complexes of the heavier transition metals are low spin (where that concept has any meaning) $\ce{PdF2}$ is one exception, which is unusual not only this way but also having $\ce{Pd^2+}$ in an octahedral environment (see e.g. Wikipedia). Thus your source is not quite correct. As for the reason why 2nd and 3rd row ...

-1

Due to improper shielding by 4d and 5d elements ligand electron get strongly attracted by the nucleus and hence the pairing energy become lower and the complex form low spin complex irrespective of strength of ligand

5

There is a key difference between the $\mathrm d^8$ state and the $\mathrm d^9$ state which lies in exactly that extra electron. The result of the extra electron is that instead of nickel(II)’s $\mathrm{(t_{2g})^6\ (e_g^*)^2}$ state copper(II) has a $\mathrm{(t_{2g})^6\ (e_g^*)^3}$ state. There are two $\mathrm{e_g^*}$ orbitals, thus occupying each with a ...

4

You are right. I took this two pictures from Wikipedia to compare EDTA (a tetradentate) to dmg (a bidentate). For EDTA you got this: For dmg, on the other hand, you got this structure: You can notice that for both of the cases, the metal complex is connected to the ligands through five-membered rings. This observation is general in chemistry: to have ...

4

It is not always as easy to see from a standard octahedral depiction. Optical activity in octahedral complexes can be understood better, if you draw the octahedron as looking onto a triangular face rather than along an edge; this results in a hexagonal arrangement of ligands around the central atom with ligands alternately pointing into and out of the paper ...

1

(I wish I could find pictures of the UV/Vis spectra to support this answer but alas …) In the very narrow context of your question that only considers $\mathrm{d{\rightarrow}d}$ transitions, I would like to refer entirely to Oscar Lanzi’s answer. However, there is an important aspect to the colours of both ions that $\mathrm{d{\rightarrow}d}$ transitions ...

4

The chance of a transition being allowed or 'forbidden', and hence its intensity, depends on both the symmetry of the states involved and the spin change that occurs, if any. The first step, however, is to find out how the orbitals $e_g,\; t_{2g}$ are occupied with the electrons. The Fe(II) is $d^6$ and has a 'strong' ligand in CN$^-$ compared to H$_2$...

5

You see that both cases are Laporte-forbidden, now how about spin-forbidden? Assuming the complex vibrates so it's no longer exactly octahedral, can you make a spin-allowed transition for $\ce{[Fe(CN)_6]^{4-}}$? How about $\ce{[Fe(H_2O)_6]^{3+}}$? When you answer this you will answer your question.

-5

It is seen that Au(III) and Au(I) complexes are more stable than Au(II).I have an explanation but that is my observation- Let's consider their electronic configuration. Au → 5d10 6s1 Au+ → 5d10 Au2+ → 5d9 Au3+ → 5d8 So, if we split d-orbital into t2g and eg, we get For Au → For Au+ → For Au2+ → For Au3+ → So, As you can see, Au (I) - ...

2

The hexaaquacobalt(II) complex $\ce{[Co(H2O)6]^2+}$ is octahedral. The tetrachloridocobaltate(II) complex $\ce{[CoCl4]^2-}$ is tetrahedral. knowing the oxidation state does not allow prediction of the coordination number or vice-versa.

-2

There is no relation between the oxidation state of the central atom and its coordination number. The complex compound you just mentioned is a good example. It was [Co(NH3)6]Cl3 and it contains Cobalt at the oxidation state +3. In water the corresponding ion will be formed : [Co(NH3)6]^3+. But a similar ion may be formed with Cobalt(II) and the same number ...

-3

I don't think it makes sense to compare Gold with Silver and Copper. In the 6th line of the periodic table, relativistic effects become important. And one of the relativistic effect is a contraction of the lengths. This effect is proportional to the sum of the two first quantum numbers. Gold outer orbitals are 4f 5d and 6s. Its configuration is [Xe] 4f14 ...

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