Complex Compound (Valence Bond Theory).pptx

ppt

1. Octahedral Complexes ( or hybridization)   Octahedral complexes are those in which the C.N. of the central metal atom or ion is 6. These complexes are formed as a result of or hybridization of the central metal atom or ion. What type of hybridization (i.e., whether or will occur depends on the number of unpaired or paired electrons present in the complex ion. or hybridization is called octahedral hybridization .   Octahedral complexes in which the central metal atom is hybridized are called inner-orbital octahedral complexes   Octahedral complexes in which the central metal atom is hybridized are called outer -orbital octahedral complexes.  

2. hybridization in Inner Orbital Octahedral Complexes   This type of hybridization takes place in those octahedral complexes which contain strong ligands. On the basis of the orientation of the lobes of d-orbitals in space, these orbitals have been classified into two sets ( i ) or   (ii) or   or set consists of , , and orbitals while   or set consists of and orbitals   In the formation of six hybrid orbitals, two (n-1) d-orbitals of set [i.e., (n-1) and (n-1) orbitals], one ns and three np (n , n , n ) orbitals combine together and form six hybrid orbitals.   It is clearly seen that the two d-orbitals used in hybridization are from penultimate shell [i.e., (n-1) th shell] while s and three p-orbitals are from ultimate shell (i.e., nth shell)   This discussion shows that in case of octahedral complex ions of 3d transition series elements two d-orbitals used in hybridization are and 3 orbitals ( set of orbitals) while s- and p-orbitals are 4s and 4p orbitals. Thus hybridization taking place in such complexes can be represented as 3 4s. .4 .4 ( ).  

3. Since two d-orbitals used in hybridization belong to inner shell [i.e., (n-1) th shell], the octahedral complex compounds resulted from hybridization are called inner orbital octahedral complexes.   Since these complexes have comparatively lesser number of unpaired electrons than outer-orbital octahedral complexes, these complexes are also called low spin or spin paired octahedral complexes. It is due to the presence of strong ligands in inner-orbital octahedral complexes of 3d transition series that the electrons present in and ( set) are forced to occupy , , and orbitals ( set) and thus 3d orbitals of set become vacant and hence can be used in 3 4s. .4 .4 ( ) hybridization.   Now we will discuss the structure of some octahedral complex ions of transition series elements which result from hybridization  

4. 1. Ferricyanide Ion, [Also called Hexacyanoferrate (III) Ion]   In this ion the coordination number of Fe is six and hence the given complex ion is octahedral in shape. In this ion, Fe is present as ion whose valence configuration is or (Fe = , = = ) as shown below:   3d 4s 4p (a) Fe-atom ( or ) ↓↑ ↑ ↑ ↑ ↑ ↓↑ − − − (b) ion ( or ) ↑ ↑ ↑ ↑ ↑ − − − − (c) ion in ion ( ) ions are strong ligand) ↓↑ ↓↑ ↑ − − − − − − (d) ion involving hybridization ↓↑ ↓↑ ↑ ∗∗ ∗∗ ∗∗ ∗∗ ∗∗ ∗∗ 3d 4s 4p (a) ↓↑ ↑ ↑ ↑ ↑ ↓↑ − − − (b) ↑ ↑ ↑ ↑ ↑ − − − − (c) ↓↑ ↓↑ ↑ − − − − − − (d) ↓↑ ↓↑ ↑ ∗∗ ∗∗ ∗∗ ∗∗ ∗∗ ∗∗

5. In order to explain the formation of octahedral metal complexes two suggestions have been put forward: ( i ) Pauling’s Suggestion – Inner-Orbital Octahedral Metal Complexes: Pauling originally suggested that in the complex ion such as ion the metal-ligands are primarily ‘ionic’ and not covalent (as in the inner-orbital complexes) and the d-electrons in the complex ion are, therefore, not required to pair but are allowed to remain unpaired just as they are in the free ion. Thus he assumed that the metal ion d-orbitals are left free to accommodate only the metal ion electrons and the ligand electrons are contained in separated orbitals located primarily on the ligand. According to him when electronegative ligands are involved, the electrostatic bonding takes place which does not disturb the electrons of the d-orbitals. The use of the term “ionic” for such strongly paramagnetic complexes was abandoned, when it becomes known that many of such paramagnetic complexes such as Fe behave like typical covalent compounds.  

6. (ii) Huggin’s Suggestion – Outer-Orbital Octahedral Metal Complexes: To account for the paramagnetic nature of the complexes like Huggin assumed that the d-orbitals involved in the hybridization process to form six hybrid orbitals are not 3d-orbitals (as in case of inner-orbital octahedral complexes ) but are 4d-orbitals so that the four 3d-orbital electrons remain unpaired which explain the paramagnetic character of ion.   Thus ion can be represented as shown below: 3d 4s 4p 4d ion involving hybridization ↓↑ ↑ ↑ ↑ ↑ ∗∗ ∗∗ ∗∗ ∗∗ ∗∗ ∗∗ (n = 4) 3d 4s 4p 4d ↓↑ ↑ ↑ ↑ ↑ ∗∗ ∗∗ ∗∗ ∗∗ ∗∗ ∗∗ (n = 4) Six hybrids having 12 electrons donated by six ligands. 6 ; outer orbital complex  

7. Complexes such as are called “outer-orbital” complexes, since the d-orbitals involved in the octahedral hybridization scheme are from the same shell as the s- and p- orbitals as is evident from the fact that in the present case the principal quantum number, n for all the three orbitals (s-, p- and d-orbitals) being used in the formation of six hybrids in four. In general complexes which use one ns-, three np- and two nd -orbitals ( n orbitals) to form the six hybrids are called outer-orbital complexes. Thus in the outer orbital complexes of the cations derived from the elements of the firs tow of transition series elements the d-orbitals are 4d-orbitals while in the inner-orbital complexes of the same cations the d-orbitals are 3d orbitals . In both types of complex d-orbitals are or orbitals which are called set of orbitals.   At present it has become usual to call the “ionic complexes” as outer-orbital complexes and “the covalent” complexes as inner orbital complexes . These two cases of complexes are also called spin free and spin paired (by Nyholom ), high spin (abbreviated as HS) and low-spin (abbreviated as LS) (by Orgel) and hypo-ligand and hyper ligand (by Pauling ) complexes.

8. The formation of outer-orbital octahedral complexes (i.e., ionic complexes) of , , and ions of 3d-series elements is explained in the following table.  

9. Tetrahedral complexes are those in which the C.N. of the central metal atom or ion is 4. These complexes are formed as a result of hybridization of the central metal atom or ion. Here we shall discuss the structure of the following complex compounds which have tetrahedral geometry.   1. Molecule:   In this complex compound Ni is in zero oxidation state and has its valence-shell configuration as . This compound has tetrahedral geometry which arises due to hybridization of Ni atom.   The magnetic studies of have indicated that this molecule is diamagnetic (n=0), showing that the two 4s electrons are forced to pair up with 3d orbitals. This results in hybridization and molecule has tetrahedral structure. See in the following slide   Tetrahedral Complexes ( hybridization)  

10. 3d 4s 4p (a) Ni-atom ( ) ↓↑ ↑↓ ↑↓ ↑ ↑ − − − − (n=2) (n=2) (b) Ni- atom in ion ( ) ↓↑ ↓↑ ↓↑ ↓↑ ↓↑ − − − − (n=0) (n (c) Molecule involving hybridization ↓↑ ↓↑ ↓↑ ↓↑ ↓↑ ∗∗ ∗∗ ∗∗ ∗∗ CO CO CO CO 3d 4s 4p (a) ↓↑ ↑↓ ↑↓ ↑ ↑ − − − − (n=2) (n=2) (b) ↓↑ ↓↑ ↓↑ ↓↑ ↓↑ − − − − (n=0) (n (c) ↓↑ ↓↑ ↓↑ ↓↑ ↓↑ ∗∗ ∗∗ ∗∗ ∗∗ CO CO CO CO hybridization tetrahedral geometry of moleucle   Figure. hybridization of Ni-atom in molecule which has tetrahedral shape.  

11. 2. Ion:   In this complex compound Ni is in +2 oxidation state and has its valence-shell configuration as . Magnetic measurements reveal that the given ion is paramagnetic and has two unpaired electrons (n=2). This is possible only when this ion is formed by hybridization and has tetrahedral structure. See figure below   3d 4s 4p (a) Ni-atom ( ) ↓↑ ↑↓ ↑↓ ↑ ↑ − − − − (n=2) (n=2) (b) ion in ( ) ↓↑ ↓↑ ↓↑ ↑ ↑ − − − − (n=0) (d) ion involving hybridization ↓↑ ↓↑ ↓↑ ↓↑ ↓↑ ∗∗ ∗∗ ∗∗ ∗∗ 3d 4s 4p (a) ↓↑ ↑↓ ↑↓ ↑ ↑ − − − − (n=2) (n=2) (b) ↓↑ ↓↑ ↓↑ ↑ ↑ − − − − (n=0) (d) ↓↑ ↓↑ ↓↑ ↓↑ ↓↑ ∗∗ ∗∗ ∗∗ ∗∗ hybridization tetrahedral geometry of Ion   Figure. hybridization of Ni-atom in Ion which has tetrahedral shape.  

12. Square Planar Complexes ( hybridization)   Square Planar complexes are those in which the C.N. of the central metal atom or ion is 4. These complexes are formed as a result of d hybridization of the central metal atom or ion. In case of the square planar complexes of the metals of 1 st transition series d hybridization is , 4s hybridization. The formation of some square planar complexes has been explained below.   1. Ion:   In this complex compound Ni is in zero oxidation +2 state and has its valence-shell configuration as . (n=2). Since C.N. of is 4, ion can have tetrahedral or square planner geometry.   Tetrahedral Geometry: This geometry is due to hybridization of ion. This geometry gives n=2 and hence tetrahedral ion is paramagnetic.   Square Planar Geometry: This geometry arises from hybridization of ion. For making possible the two unpaired electrons present in 3d orbitals in the V.S.E.C. of ion get paired against Hund’s rule. The pairing of electrons takes place due to the fact that ligands are strong ligands and hence as they approach the ion, the unpaired electrons get paired, hybridization gives n=0 and hence square planar ion is diamagnetic.  

13. 3d 4s 4p (a) Ni-atom ( ) ↓↑ ↑↓ ↑↓ ↑ ↑ ↓↑ − − − (n=2) (n=2) (b) ion in ( ) ↓↑ ↓↑ ↓↑ ↑ ↑ − − − − (n=0) (c) ion involving hybridization (Formation of square planar geometry) ↓↑ ↓↑ ↓↑ ↓↑ ↓↑ ∗∗ ∗∗ ∗∗ ∗∗ (d) ion involving hybridization (Formation of square planar geometry) ↓↑ ↓↑ ↓↑ ↓↑ ∗∗ ∗∗ ∗∗ ∗∗ − 3d 4s 4p (a) ↓↑ ↑↓ ↑↓ ↑ ↑ ↓↑ − − − (n=2) (n=2) (b) ↓↑ ↓↑ ↓↑ ↑ ↑ − − − − (n=0) (c) ↓↑ ↓↑ ↓↑ ↓↑ ↓↑ ∗∗ ∗∗ ∗∗ ∗∗ (d) ↓↑ ↓↑ ↓↑ ↓↑ ∗∗ ∗∗ ∗∗ ∗∗ − Figure.   Experimentally it has been found that ion is diamagnetic. Thus this ion is formed hybridization and has square planar geometry.   hybridization tetrahedral geometry of Ion   hybridization tetrahedral geometry of Ion  

14. 2. Ions:   The tetrahalide complexes of Cu(II) have two different geometries. In ( , ion has square planar geometry with n=1 (paramagnetic) but in Cs and Cs where (X= Cl, Br) has a slightly squashed tetrahedral geometry. Square planar ions are yellow in color while tetrahedral ions are orange in color,   Let us explain the square planar geometry of ion present in . In this complex ion, Cu is present as ion whose V.S.E.C is as (n=1) . We have said above that ion has square planar geometry which results from hybridization of ion. For hybridization , one 3d orbital should be vacant. To get one 3d orbital vacant one 3d electron gets excited to 4p orbital. Since square planar geometry of ion has one unpaired electron, this ion is paramagnetic.  

15. 3d 4s 4p (a) Cu-atom ( ) ↓↑ ↑↓ ↑↓ ↑↓ ↑↓ ↑ − − − (n=0) (n=2) (b) ion in ( ) ↓↑ ↓↑ ↓↑ ↑↓ ↑ − − − − (n=1) (c) ion involving hybridization ↓↑ ↓↑ ↓↑ ↓↑ ∗∗ ∗∗ ∗∗ ∗∗ ↑ 3d 4s 4p (a) ↓↑ ↑↓ ↑↓ ↑↓ ↑↓ ↑ − − − (n=0) (n=2) (b) ↓↑ ↓↑ ↓↑ ↑↓ ↑ − − − − (n=1) (c) ↓↑ ↓↑ ↓↑ ↓↑ ∗∗ ∗∗ ∗∗ ∗∗ ↑ hybridization square planar geometry of Ion   Figure. Square planar geometry of ion in  

16. 3.   It’s a neutral molecule. In this molecule, Ni is present ion whose V.S.E.C is as (n=2) . Dmg ion is dimethylglyoximate ligand which is a bidentate ligand. Hence each of the two ions present in donated two electrons. ion is strong ligand. Magnetic measurements have shown that has no unpaired electrons (n=0) and has square planar geometry which arises because of hybridization.   3d 4s 4p (a) Ni-atom ( ) ↓↑ ↑↓ ↑↓ ↑ ↑ ↑↓ − − − (n=2) (n=2) (b) ion in ( ) ↓↑ ↓↑ ↓↑ ↑↓ − − − − − (n=1) (c) involving hybridization ↓↑ ↓↑ ↓↑ ↓↑ ∗∗ ∗∗ ∗∗ ∗∗ − 3d 4s 4p (a) ↓↑ ↑↓ ↑↓ ↑ ↑ ↑↓ − − − (n=2) (n=2) (b) ↓↑ ↓↑ ↓↑ ↑↓ − − − − − (n=1) (c) ↓↑ ↓↑ ↓↑ ↓↑ ∗∗ ∗∗ ∗∗ ∗∗ − hybridization square planar geometry of molecule   Figure. hybridization of in molecule which has square planar geometry. Note that ion is a bidentate ligand and hence donates two electron pair.  

17. In this ion, Cu is present ion whose V.S.E.C is as (n=1) .   2. Ions:   3d 4s 4p (a) Cu-atom ( ) ↓↑ ↑↓ ↑↓ ↑↓ ↑↓ ↑ − − − (n=1) (b) ion in ( ) ↓↑ ↓↑ ↓↑ ↑↓ ↑ − − − − ( n=1) (c) ion involving hybridization ↓↑ ↓↑ ↓↑ ↓↑ ↓↑ ∗∗ ∗∗ ∗∗ ∗∗ ↑ 3d 4s 4p (a) ↓↑ ↑↓ ↑↓ ↑↓ ↑↓ ↑ − − − (n=1) (b) ↓↑ ↓↑ ↓↑ ↑↓ ↑ − − − − ( n=1) (c) ↓↑ ↓↑ ↓↑ ↓↑ ↓↑ ∗∗ ∗∗ ∗∗ ∗∗ ↑ hybridization square planar geometry with n=1.   Figure. Square planar geometry of ion in   Now if the square planar geometry for ion is supposed to be correct the unpaired electron present in the higher energy 4p orbital should be expected to be easily lost, i.e., ion (Cu = +2) ion should be easily oxidized to ion (Cu = +3) by losing the unpaired electron residing in 4p ornital .   +   (− e)

18. However, experiments have shown that the oxidation of to as shown above does not occur. Then how to explain the geometry of ion.   Huggin’s suggestion: Huggin has suggested that ion has square planar geometry and ion is d hybridized as shown in figure below. The unpair electron resides in 3d orbital.   3d 4s 4p 4d (a) Cu-atom ( ) ↓↑ ↑↓ ↑↓ ↑↓ ↑↓ ↑ − − − (n=1) (b) ion in ( ) ↓↑ ↓↑ ↓↑ ↑↓ ↑ − − − − (n=1) (c) ion involving hybridization ↓↑ ↓↑ ↓↑ ↓↑ ↓↑ ↑ ∗∗ ∗∗ ∗∗ − ∗∗ 3d 4s 4p 4d (a) ↓↑ ↑↓ ↑↓ ↑↓ ↑↓ ↑ − − − (n=1) (b) ↓↑ ↓↑ ↓↑ ↑↓ ↑ − − − − (n=1) (c) ↓↑ ↓↑ ↓↑ ↓↑ ↓↑ ↑ ∗∗ ∗∗ ∗∗ − ∗∗ hybridization of ion: square planar geometry of ion with n=1.   Figure. hybridization of ion in square planar ion with n=1. (According to Huggin’s)  

19. 5. and Ions:   5d 6s 6p (a) Pt-atom ( ) ↓↑ ↑↓ ↑↓ ↑↓ ↑ ↑ − − − (n=1) (b) ion ( ) ↓↑ ↓↑ ↓↑ ↑ ↑ − − − − ( n=2) (c) ( ions are strong ligand) ↓↑ ↓↑ ↓↑ ↓↑ ↓↑ ∗∗ ∗∗ ∗∗ ∗∗ − n=0 (Diamagnetic) n=0 Diamagnetic 5d 6s 6p (a) ↓↑ ↑↓ ↑↓ ↑↓ ↑ ↑ − − − (n=1) (b) ↓↑ ↓↑ ↓↑ ↑ ↑ − − − − ( n=2) (c) ↓↑ ↓↑ ↓↑ ↓↑ ↓↑ ∗∗ ∗∗ ∗∗ ∗∗ − n=0 (Diamagnetic) n=0 Diamagnetic hybridization square planar geometry with n=0.   hybridization of ion in these complexes has been shown below in figure. All these ions are diamagnetic and have square planar geometry.   Figure. Square planar geometry of   ( i ) Ions:  

20. ( compound:   5d 6s 6p (a) Pt-atom ( ) ↓↑ ↑↓ ↑↓ ↑↓ ↑ ↑ − − − (n=1) (b) ion ( ) ↓↑ ↓↑ ↓↑ ↑ ↑ − − − − ( n=2) (c) ( ions are strong ligand) ↓↑ ↓↑ ↓↑ ↓↑ ↓↑ ∗∗ ∗∗ ∗∗ ∗∗ − n=0 (Diamagnetic) n=0 5d 6s 6p (a) ↓↑ ↑↓ ↑↓ ↑↓ ↑ ↑ − − − (n=1) (b) ↓↑ ↓↑ ↓↑ ↑ ↑ − − − − ( n=2) (c) ↓↑ ↓↑ ↓↑ ↓↑ ↓↑ ∗∗ ∗∗ ∗∗ ∗∗ − n=0 (Diamagnetic) n=0 hybridization square planar geometry with n=0   Figure. Square planar geometry of due to dsp 2 hybridization of Pt 2+ ion   hybridization of ion in these complexes has been shown below in figure. This ion is diamagnetic and has square planar geometry.  

21. ( ion :   5d 6s 6p (a) Pt-atom ( ) ↓↑ ↑↓ ↑↓ ↑↓ ↑ ↑ − − − (n=1) (b) ion ( ) ↓↑ ↓↑ ↓↑ ↑ ↑ − − − − ( n=2) (c) ( NH 3 are strong ligand) ↓↑ ↓↑ ↓↑ ↓↑ ↓↑ ∗∗ ∗∗ ∗∗ ∗∗ − n=0 (Diamagnetic) n=0 5d 6s 6p (a) ↓↑ ↑↓ ↑↓ ↑↓ ↑ ↑ − − − (n=1) (b) ↓↑ ↓↑ ↓↑ ↑ ↑ − − − − ( n=2) (c) ↓↑ ↓↑ ↓↑ ↓↑ ↓↑ ∗∗ ∗∗ ∗∗ ∗∗ − n=0 (Diamagnetic) n=0 hybridization square planar geometry with n=0   Figure. Square planar geometry of due to dsp 2 hybridization of Pd 2+ ion   hybridization of ion in this complex has been shown below in figure. This ion is diamagnetic and has square planar geometry.  

22. 5. Ions:   5d 6s 6p (a) Pd-atom ( ) ↓↑ ↑↓ ↑↓ ↑↓ ↑↓ − − − − (n=0) (b) ion ( ) ↓↑ ↓↑ ↓↑ ↑↓ − − − − − ( n=0) (c) ( ions are strong ligand) ↓↑ ↓↑ ↓↑ ↓↑ ↓↑ ∗∗ ∗∗ ∗∗ ∗∗ − n=0 (Diamagnetic ) n=0 Diamagnetic 5d 6s 6p (a) ↓↑ ↑↓ ↑↓ ↑↓ ↑↓ − − − − (n=0) (b) ↓↑ ↓↑ ↓↑ ↑↓ − − − − − ( n=0) (c) ↓↑ ↓↑ ↓↑ ↓↑ ↓↑ ∗∗ ∗∗ ∗∗ ∗∗ − n=0 (Diamagnetic ) n=0 Diamagnetic hybridization square planar geometry with n=0.   This ion has square planar geometry and is hybridization of ion in this complexes has been shown below in figure.   Figure. Square planar geometry of   ( i ) Ions:  

23. Examples of Square Planar Complexes (dsp 2 hybridization) Square Planar Complexes Central metal atom No. of unpaired electrons Fe(III) Phthalocyanide ) 2 Co(II) Phthalocyanide ) 1 [Ni(dmg) 2 ] ) [Ni( ) ) ) [( ) 1 ) 1 (in 4p) (sp 2 d) ) 1 (in 3d) Square Planar Complexes Central metal atom No. of unpaired electrons Fe(III) Phthalocyanide 2 Co(II) Phthalocyanide 1 [Ni(dmg) 2 ] 1 1 (in 4p) 1 (in 3d)

24. Trigonal Bipyramidal Complexes ( hybridization)   Iron pentacarbonyl, is an important example of complex compound which has trigonal bipyramidal geometry. In molecule, Fe is present as neutral atom whose V.S.E.C is as (n=4) . Since C.N. of Fe is 5, molecule has trigonal bipyramidal geometry and hence is formed by hybridization. This compound has been found to be diamagnetic (n=0)   Explanation. Since has been found to be diamagnetic (n=0), all the eight electrons present in the V.S. of Fe are paired up in 3d orbitals to give n=0. Thus the V.S.E.C. of Fe atom in is as shown in the figure in the next slide. The pairing of electrons makes on 3d orbital and one 4s orbital vacant. Three 4p orbitals are already vacant. These five vacant orbitals mix together and form five hybrid orbitals. Thus in the formation of , Fe atom undergoes hybridization and hence molecule has trigonal bipyramidal geometry.  

25. 3d 4s 4p (a) Fe-atom ( ) ↓↑ ↑ ↑ ↑ ↑ ↓↑ − − − (n=4) (b) Fe in ( ) (CO molecules are strong ligand) ↓↑ ↓↑ ↓↑ ↑↓ − − − − − ( n=0) (c) ↓↑ ↓↑ ↓↑ ↓↑ ↓↑ ∗∗ ∗∗ ∗∗ ∗∗ ∗∗ n=0 (Diamagnetic) n=0 Diamagnetic 3d 4s 4p (a) ↓↑ ↑ ↑ ↑ ↑ ↓↑ − − − (n=4) (b) ↓↑ ↓↑ ↓↑ ↑↓ − − − − − ( n=0) (c) ↓↑ ↓↑ ↓↑ ↓↑ ↓↑ ∗∗ ∗∗ ∗∗ ∗∗ ∗∗ n=0 (Diamagnetic) n=0 Diamagnetic hybridization trigonal bipyramidal geometry with n=0.  

26. Application of Valance Bond Theory (VBT) 1. To find out the number of unpaired electron in a given complex ion The number of unpaired electron (n ) present in a given complex ion depends on the geometry of the complex ion and the type of hybridization undergone by the central metal atom or ion. Ther e is a relation between the number of unpaired electrons and geometries of the complex ion 2. To Predict the type of hybridization in octahedral complexes. This can be done with the help of magnetic moment values ( 𝛍 ) of the complexes. 3. To Predict the geometry of a complex. This can also be done with the help of magnetic moment values ( 𝛍 ) of the complexes. Configuration of the central metal atom/ion Inner- orbital octahedral complex ion ( hybridization) Outer-orbital octahedral complex ion ( hybridization) Tetrahedral complex ion hybridization Square planar complex ion (d or ( hybridization) Configuration of the central metal atom Inner- orbital octahedral complex ion ( hybridization) Outer-orbital octahedral complex ion ( hybridization) Tetrahedral complex ion hybridization Square planar complex ion (d or ( hybridization) 1 1 1 1 1 5 5 3 2 2 2 2 4 4 2 3 3 3 3 1 3 3 1 2 4 4 4 − 2 2 − 1 1 1 Configuration of the central metal atom/ion Configuration of the central metal atom 1 1 1 1 1 5 5 3 2 2 2 2 4 4 2 3 3 3 3 1 3 3 1 2 4 4 4 − 2 2 − 1 1 1