Tetrahedral complex ions

C18-0122. Cadmium forms a tetrahedral complex ion with cyanide ions j2+... 

It dissolves in acetonitrile, CH3CN forming tetrahedral complex ion [Cu(CH3CN)4]+ which can be precipitated with large anions such as CIO4 or PFe". 

Give the crystal field diagram for the tetrahedral complex ion CoCl42-. Solution... 

Tetrahedral complexes of Co2+ are quite common. Use a d-orbital splitting diagram to rationalize the stability of Co2+ tetrahedral complex ions. 

Salts of the three metals form ammines, principally tetrahedral complex ions of the form [M(NH3)4]2+, but in addition mercury salts form linear, diammine complexes, and zinc and cadmium octahedral hexa-ammines, [M(NH3)g]2+. Ethylenediamine produces 6-co-ordinate complexes, [M(en)3]2+, with all three cations. 

Beryllium hydroxide is amphoteric and this sets it apart from the hydroxides of the other group 2 metals which are basic. In the presence of excess [OH] , Be(OH)2 behaves as a Lewis acid (equation 11.24), forming the tetrahedral complex ion 11.7, but Be(OH)2 also reacts with acids, e.g. reaction 11.25. 

Why do tetrahedral complex ions have a different crystal field diagram than octahedral complex ions What is the tetrahedral crystal field diagram Why are virtually all tetrahedral complex ions "high spin" ... 

In a solution of the complex HglJ", the Hg—I bond length is 0.28 nm, which, in agreement with the X-ray data on the various solid iodomercurates [Ha 55, Fe 66], corresponds to a regular tetrahedral complex ion. 

CoCLi forms a tetrahedral complex ion and Co(CN)6 forms an octahedral complex ion. What is wrong about the following statements concerning each complex ion and the d orbital splitting diagrams ... 

Figure 13.43 Effect of tetrahedral complex ion formation on the energies of the 3d electrons in the... 

One important aspect of the electronic structure of tetrahedral complex ions which was not discussed in detail in Chapter 7—although it was mentioned in Chapter 6 and is contained in Table 6.6—is that, unlike the octahedral case, in a tetrahedral complex the metal p orbitals have the same symmetry 12) as do the metal d, dy and d orbitals. Because they are of the same symmetry, the set of d orbitals will be mixed by the crystal field with the metal p orbitals. This in turn means that an e transition contains some d p component and this component is an allowed transition. Evidently, the intensity of the e t2 transition is related to the extent of d-p mixing so that this may be worked backwards and the extent of mixing assessed, at least qualitatively, from the intensities of d-d bands in the spectra. Notice both the similarity and the difference between the intensitygenerating mechanisms for d-d transitions in octahedral and tetrahedral complexes. Both depend on d-p mixing but only for the tetrahedral case does this mixing occur for the non-distorted molecule. 

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