Inorganic Chemistry (2)

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Inorganic Chemistry (2) Prepared by Dr. Hoda El-Ghamry Lecturer of Inorganic Chemistry Faculty of Science-Chemistry Department Tanta University

Where n is the number of the unpaired electron in the Now how can we deduce the geometrical shape of the metal complexes from the values of magnetic moment. Example 1: If you know that the magnetic moment value of the compound [Co(NH3)6]3+ is zero (diamagnetic) and the magnetic moment of the compound [CoF6]3- is 5.1 B.M., Deduce the geometrical shape and the type of hyperidization of the two compounds. Tim Remember that s The theoretical magnetic moment value s is given by s=[n(n+2)]1/2 Where n is the number of the unpaired electron in the metal center

In the case of [Co(NH3)6]3+ 4S2 3d7 Co27 [Ar]18 4S0 3d6 Co3+ [Ar]18 3d 4s 4p Co27 3d 4s 4p Co3+ As the complex is diamagnetic, this means that there is no unpaired electrons the metal center so all unpaired electrons should be paired as follow 3d 4s 4p XX XX XX d2sp3 hyberidization Thus the geometrical shape is inner orbital octahedral structure

Thus the geometrical shape is outer orbital octahedral structure In the case of [CoF6]3- 4S2 3d7 Co27 [Ar]18 4S0 3d6 Co3+ [Ar]18 3d 4s 4p Co27 3d 4s 4p Co3+ As the magnetic moment of the complex is 5.1 B.M., this means that the central metal ion contains four unpaired electrons, thus the electronic distribution is as follow: 3d 4s 4p XX XX 4d XX hyberidization is sp3d2 Thus the geometrical shape is outer orbital octahedral structure

4S2 3d5 For the compound [MnCl4]2- Mn25 [Ar]18 4S0 3d5 Mn2+ [Ar]18 Example 2 If you know that compound [Ni(CN)4]2-is diamagnetic while the magnetic moment of the compound [MnCl4]2- is 5.95 B.M. deduce the geomterical shape and the type of hyberidization of the two compounds Solution 4S2 3d5 For the compound [MnCl4]2- Mn25 [Ar]18 4S0 3d5 Mn2+ [Ar]18 3d 4s 4p Mn25 3d 4s 4p Mn2+ As the results of the magnetic moment of this complex is 5.95 B.M., this is indication of the presence of five unpaired electrons in the metal center, so the hyberidization is as follow 3d 4s 4p XX XX sp3 hyberidization The geometrical structure is tetrahedral

[Ni(CN)4]2- In the case of 4S2 3d8 Ni28 [Ar]18 4S0 3d8 Ni2+ [Ar]18 3d 4s 4p Ni28 3d 4s 4p Ni2+ As the complex is diamagnetic, this means that all the electrons are paired so the electronic distribution and the hyberidization is as follow 3d 4s 4p XX XX XX dsp2 hyberidization The geometrical structure is square planar

So the hyberidization d2sp3 is and the formula is Example 3: Iron ion forms an inner orbital diamagnetic octahedral complex with cyanide ligand, derive the formula of the complex. 4S2 3d6 Fe26 [Ar]18 3d 4s 4p Fe26 As the complex is diamagnetic, this means that Fe should be divalent not trivalent 3d 4s 4p Fe2+ The complex is diamagnetic, this means that all the electrons are paired so the hyberidization is as follow 3d 4s 4p XX XX XX So the hyberidization d2sp3 is and the formula is [Fe(CN)6]4-

Exercises 1- Tris( ethylenediamine) iron(II) chloride is an outer complex a- Calculate the magnetic moment value of the complex (Fe= 26) b- Write the formula of the complex. 2- Cr(II) ion forms an aqua complex with μeff 4.9 B.M., what is the expected geometry of the complex.