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

2. 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

3. In the case of [Co(NH3)6]3+
4S2 3d7 Co [Ar]18
4S0 3d6 Co [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

4. Thus the geometrical shape is outer orbital octahedral structure
In the case of [CoF6]3-
4S2 3d7 Co [Ar]18
4S0 3d6 Co [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

5. 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] Mn [Ar]18
4S0 3d Mn [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

6. [Ni(CN)4]2- In the case of
4S2 3d8 Ni [Ar]18
4S0 3d8 Ni [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

7. 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 Fe [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-

8. 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.