Dr-Najlaa alradadi
1

There are theories as to why the links in the complexes and transition elements, including :
Valence Bond Theory (V B T)
We will study the hybridization
Dr-Najlaa alradadi 2

1-Type of hybrid orbital's (sp 3 ):
A) Interdependence in methane :
If we tried to apply the valence bond method prior to the molecules containing many atoms get a big disappointment. In most cases, the bond angles will be derived from the valence bond method for different angles of observation. For example, carbon has a configuration of E (= latency idle state) is:
Dr-Najlaa alradadi 3

Therefore we expect that when that when hydrogen reacts with the molecule will give a (CH
2
), and that corner will be (90 º) is that the molecule (CH
2
) is stable and that the simplest compound of carbon and hydrogen is methane (CH
4
). To get the molecular formula and the way the valence bond, we need to plan orbital of carbon has four electrons Single even result in overlapping orbits to the formation of four links (C - H) to such a scheme to imagine that one of electrons from the orbit of (2s) in the carbon absorbed amount of and rose to power over the (2p) empty.
Distribution of the resulting electronic excited state is :
4 Dr-Najlaa alradadi

5 Dr-Najlaa alradadi

Figure shows the process of hybridization over (s) and orbits (p) to give a series of four orbits of the hybrid type (sp 3 ) an atom of carbon.
Dr-Najlaa alradadi 6

In the process of hybridization number of hybrid orbital's is equal to the total count of atomic orbital's United.
Figure shows the type of hybrid orbital's (sp 3 ) and the formation of links in methane.
Dr-Najlaa alradadi 7

B) Interdependence in the ammonia and water :
Hybridization on the oxygen atom and nitrogen in the water and ammonia is also of the type (sp 3 ) in the sense that we expect that the value of the angle Association (HOH) and the angle in the (HNH) are (109.5 º) and this value does not differ much from the values ​​measured in practice are (104.5 º) of water and (107.0 º) of ammonia
8 Dr-Najlaa alradadi

We can explain the interdependence of the orbits planned ammonia by nitrogen level equivalence follows
Dr-Najlaa alradadi 9

2-Type of hybrid orbital's (sp 2 ):
Interdependence in Boron
Can be represented by the type of hybridization
(sp 2 ) for the boron atom as follows:
Dr-Najlaa alradadi 10

Gives the type of hybridization (sp 2 ) Plane triangle structure so that the angles between the links
(120 º) as in (BF
3
).
Dr-Najlaa alradadi 11

3-Type of hybridization (sp):
Beryllium has four orbits in the level of parity, while only two electrons have
Therefore, the hybridization of beryllium compounds include round (s) and over (p) to give the kind of hybrid orbits (sp) and remain two of the orbits (p) without hybridization and can be represented as in the form:
12 Dr-Najlaa alradadi

Hybridization of the type leads to a linear structure in the sense that the angle of the
Association equal to (180 º) as in (BeCl
2
).
Dr-Najlaa alradadi 13

4-Orbits (d) hybrid:
Hybridization schemes are used which include orbits (d) to explain the thread that includes a number of electrons more than (8).
In (PCl
5
), there are five links and this means that there are five orbits of half-packed on the central atom of phosphorus. Hybridization is of course (s) and three orbits of (p) and one of the orbits (d) to give the five orbits of the hybrid type (sp 3 d).
14 Dr-Najlaa alradadi

Hybridization of the type (sp 3 d) has a structure of the Trigonal bipyramidal as in the picture .
4- Which element can expand its valence shell to accommodate more than eight electrons?
15

To explain the correlation in (SF
6
) you need to half-packed six orbits on the sulfur atom where the hybridization between the orbit (s) and three orbits (p) over the two (d) gives the six orbits of the hybrid type (sp 3 d 2 )
Dr-Najlaa alradadi 16

The six hybrid orbital's of the type (sp 3 d 2 ) structure has Octahedral, as in the form :
Dr-Najlaa alradadi 17

In (CH
4
), for example the carbon atom composition of four orbits of the hybrid type (sp
3
) as follows:
Electron configuration of carbon:
Excited atom
Dr-Najlaa alradadi 18

19 Dr-Najlaa alradadi

1 - atom central element (Element transition or Lewis acid) can configure orbital's hybrid empty of electrons receive a pair electrons of the group-giving in other words we can say that it will consist of bonds
(σ) overlap between orbital's empty hybrid of the atom central and between the orbital's of group-giving and containing a pair of electrons thus formed coordination bonds .
and that the bond between the central atom and ligand is covalent bond (100%).
20 Dr-Najlaa alradadi

2 -
The donor atom or the group-giving or
Lewis base must contain an atom where there is at least a pair of electrons.
3 In addition to the formation of bonds (σ) there is a possibility to create bonds (π) that are available atomic orbital's contain electrons in central atom overlap with the empty orbital's of the donor atom.
21 Dr-Najlaa alradadi

4-
The number of these hybrid orbital's is the equal to the number of coordinate of central atom in complexes .
Dr-Najlaa alradadi 22

[Be(H
2
O)
4
] 2+
Hybridization : (sp 3 )
Geometry : Tetrahydral
Diamagnetic ( absence of single electrons ) .
Dr-Najlaa alradadi 23

Hybridization
Coordination number sp 2 d 2 sp 3 sp sp dsp dsp sp 3
2
3 d
2
3
(sp 3 d 2 )
Geometry
Linear
3
Plane triangle
4
4
Tetrahydral
Square planar
5
5
Trigonal bipyramidal
Squar pyramidal
6
Dr-Najlaa alradadi
Octahedral
Example
HgCl
2
BCl
3
] NiCl
4
[ 2-
] Ni(CN)
4
[ 2-
[Fe(CO)
5
]
[V(acac)
5
] 3-
[Fe(CN)
6
] 3-
24

[Cr(CO)
6
]
Hybridization : (d 2 sp 3 )
Geometry : Octahedral
Diamagnetic
Dr-Najlaa alradadi] 25

[Fe(CO)
5
]
Hybridization : dsp 3
Geometry : Trigonal bipyramidal
Diamagnetic
Dr-Najlaa alradadi 26

[Ni (CN)
4
] 2-
Hybridization : dsp 2
Geometry : Square planar
Diamagnetic
Dr-Najlaa alradadi 27

[Ni Cl
4
] 2-
Hybridization : (sp 3 )
Geometry : Tetrahydral
Paramagnetic
Contains two electrons alone ( µ ≈ 2.8 β .
M )
Dr-Najlaa alradadi 28

[Ti(H
2
O)
6
] 3+
Hybridization : (d 2 sp 3 )
Geometry : Octahedral
Paramagnetic
Contains a single electron
( µ ≈
1.7 β
.
M
)
Dr-Najlaa alradadi 29

[Fe(CN)
6
] 4-
Hybridization : (d 2 sp 3 )
Geometry : Octahedral
Diamagnetic
Dr-Najlaa alradadi 30

[Fe(H
2
O)
6
] 2+
Hybridization : (sp 3 d 2 )
Geometry : Octahedral
Paramagnetic
Contains four electrons alone
( µ ≈ 4.8 β .
M )
Dr-Najlaa alradadi 31

Valence bond theory _ :
* Several theories currently are used to interpret bonding in coordination compounds.
* In the valence bond (VB) theory, proposed in large part by the American scientists Linus Pauling and John C. Slater, bonding is accounted for in terms of hybridized orbital's of the metal ion, which is assumed to possess a particular number of vacant orbitals available for coordinate bonding that equals its coordination number.
* Each ligand donates an electron pair to form a coordinate-covalent bond , which is formed by the overlap ..
Dr-Najlaa alradadi 32

* VB theory treats metal to ligand
(donor group) bonds as coordinate covalent bonds, formed when a filled orbital of a donor atom overlaps with an empty hybrid orbital on the central metal atom.
33 Dr-Najlaa alradadi

The theory considers which atomic orbitals
on the metal are used for bonding.

From this the shape and stability of the complex are predicted

Dr-Najlaa alradadi 34

* The molecular geometry is predicted using
VSEPR.
* The theory proposes that the number of metal-ion hybrid orbital's occupied by donor atom lone pairs determine the geometry of the complexes .
* Lone pairs of electrons are ignored . Since lone pairs are in the inner shell, they are believed to have little effect on molecular geometry.
Dr-Najlaa alradadi 35

* In the formation of covalent bonds, electron orbital's overlap in order to form "molecular" orbital's, that is, those that contain the shared electrons that make up a covalent bond .
Although
* the idea of orbital overlap allows us to understand the formation of covalent bonds, it is not always simple to apply this idea to polyatomic molecules .
Dr-Najlaa alradadi 36

Dr-Najlaa alradadi 37

Dr-Najlaa alradadi 38

* Hybrid orbitals do not actually exist.

* Hybridization is a mathematical
 manipulation of the wave equations for the atomic orbitals involved.
Dr-Najlaa alradadi 39

The formation of a complex may be
 considered as a series of hypothetical steps.
* First the appropriate metal ion is taken, e.g. Co +3 , a Co atom has the outer electronic structure 3d 7 4S 2 . Thus a Co +3 ion will have the structure 3d 6 , and the electrons will be arranged as follows:

Dr-Najlaa alradadi 40

Dr-Najlaa alradadi 41


First, with transition metals, as the ion is formed, the valence s and p orbitals ( 4s and 4p , in the case of a first row transition metal such as Co ) become higher in energy compared to the valence d orbital
( 3d in the case of Co). The six valence electrons of Co 3+ can be assigned to the 3d orbitals, following Hund's rule.
Co 3+ can coordinate to six ligands (each ligand donating a pair of electrons ) which leads to an octahedral complex.
Two examples are shown above, the fluoro complex is an example of a complex anion while the ammonia (ammine) complex is considered a complex cation.
Dr-Najlaa alradadi 42





The above diagram also shows two possibilities for the electronic configuration.
The fluoro complex keeps the Co electrons in the same configuration as the isolated Co 3+ ion.
The electrons from the fluoro ligands simply occupy the upper vacant orbitals on Co . With hybridization, this will be a set of sp 3 d 2 hybrid orbitals ( bond formed) outer orbital complex.
The d orbitals used are 4dx 2 -y 2 , 4dz 2
Dr-Najlaa alradadi 43

Note:

* The energy of these orbitals is quite high so the complex will be reactive or labile.

* The magnetic moment depends on the number of unpaired electrons .

*The 3d level contains the maximum
 number of electrons for a d 6 arrangement, and this is called a high–spin or a spin-free complex.
Dr-Najlaa alradadi 44

The same type of hybridization can be imagined in the hexaammine complex.

*In this case, however, the electrons
 originally with Co 3+ are forced to pair first and the electrons from the ligands will also occupy the orbitals vacated due to the pairing of the Co electrons to form d 2 sp 3 .
* Low energy inner d orbitals are used to form an inner orbital complex.

Dr-Najlaa alradadi 45

* Inner complex is more stable than outer complex.

* The unpaired electrons in the metal ion have been forced to pair up – to form low spin complex.

* In this case for Co +3 all electrons are paired and the complex will be diamagnetic

Dr-Najlaa alradadi 46


This happens because ammonia is known to be a "strong" ligand.
From the above example, a drawback of the valence bond theory becomes evident.

One needs to know first if the ligand is
 strong enough to cause electrons on the metal to pair first.
Dr-Najlaa alradadi 47

** The metal ion could also form fourcoorinate complexes.

Two different arrangements are possible:

* sp 3 hybridization – tetrahedral shape.

* dsp 2 hybridization- square planar shape

Dr-Najlaa alradadi 48

* Linear complexes:
[Cu(CN)
2
] -
Cu=Ar 4s 1 3d 10 , Cu +1 = Ar 4s 0 3d 10
Dr-Najlaa alradadi 49

[Fe(CN)
6
] 3Fe = Kr 4s 2 3d 6
Fe +3 = Kr 4s 0 3d 5
Dr-Najlaa alradadi 50

When very strong ligands are present, fourcoordinated d 8 metal ions usually form square planar complexes (rather than tetrahedral).
[Ni(CN)
4
] 2Ni +2 = kr 4s 0 3d 8
Ni +2 = kr 4s 0 3d 8
51 Dr-Najlaa alradadi

[Cd(NH
3
)
4
] 2+
Cd +2 = kr 5s 0 4d 10 Cd° = kr 5s 2 4d 10
Dr-Najlaa alradadi 52

Try to apply the theory to complexes following
:
[Cu(Cl)
4
] 2-
[Ni(Co)
[Co(en)
،
4
]
،
2
Cl] + ،
[Ni(Cl)
4
]
،
[Pt(Cl)
4
] 2-
[Co(NH3)
6
] 3+
[Fe(CN)
6
] 3-
[Fe(H2O)
6
] 3-
53 Dr-Najlaa alradadi

1 \ theory assumed that the orbits under the cover
(3d) all have the same energy in the complexes and this is not true.
2 \ used the theory of orbits (3d) , (4d) in the formation of links in spite of the large difference in capacity.
3 \ did not give an explanation of the spectrum of electronic vehicles (spectral properties) .
Dr-Najlaa alradadi 54

4 \ did not give a clear explanation of the magnetic measurements.
5 \ theory lacked a means to express the phenomenon and to predict a four-layered consistency
(Is it a
Tetrahydral pyramidal or
Square planar
)
"Why is no duplication or duplication occurs in the orbit (d) " Take the following example :
Dr-Najlaa alradadi 55

Copper ion ( II ) Is superimposed with ammonia has a coordination number four On this basis, we can apply the theory as follows:
Dr-Najlaa alradadi 56

X
Predicted theoretical form spatial pyramid of four aspects
(sp
3
) due to the fullness of the cover (d) with elictrons
(d
9
), but proved to X-ray , this overlay is a four-level which makes it imperative to have a hybridization of the type
(dsp
2
) and a solution to this problem suggested Pauling order e Next, where electron moving from the ninth orbit
(3d) to the orbit (4p), which makes it easy to remove this electron due to the presence in an orbit with a high energy.
Dr-Najlaa alradadi 57

We expect that it is any oxidation of copper (II)
To copper (III) - As mentioned in the case of cobalt, but Cu complexes (II) Stable and consistent as that complexes copper (III) Are strong oxidizing agents
6 \ theory failed to explain why not configured to form a regular octahedral in the case of Cu complexes (II)
Dr-Najlaa alradadi 58

7 \ did not address the theoretical excited state of the compounds, one of the most phenomena that are worth processing elements transitional compounds in all of this made it necessary to search for other theories can explain it.
Dr-Najlaa alradadi 59

8/ The theory provides no explanation for electronic spectra of coloured complexes.

9/ The theory does not explain why the
 magnetic properties vary with temperature.
Dr-Najlaa alradadi 60