Valence bond theory (VBT)

Report
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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
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1-Type of hybrid orbital's (sp3):
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:
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Therefore we expect that when that when hydrogen
reacts with the molecule will give a (CH2), and that
corner will be (90 º) is that the molecule (CH2) is
stable and that the simplest compound of carbon and
hydrogen is methane (CH4). 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 :
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This means that hybridization occurs
between the orbit (s) and three orbits of (p)
consists of four identical orbits, which tend
to the corners of the tetrahedral pyramid
and intermediate between capacity and
energy orbit (s) and orbit (p).
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Figure shows the process of hybridization
over (s) and orbits (p) to give a series of
four orbits of the hybrid type (sp3) an
atom of carbon.
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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 (sp3)
and the formation of links in methane.
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B) Interdependence in the ammonia and water :
Hybridization on the oxygen atom and
nitrogen in the water and ammonia is also of
the type (sp3) 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
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We can explain the interdependence of
the orbits planned ammonia by nitrogen
level equivalence follows
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2-Type of hybrid orbital's (sp2):
Interdependence in Boron
Can be represented by the type of hybridization
(sp2) for the boron atom as follows:
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Gives the type of hybridization (sp2) Plane triangle
structure so that the angles between the links
(120 º) as in (BF3).
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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:
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Hybridization of the type leads to a linear
structure in the sense that the angle of the
Association equal to (180 º) as in (BeCl2).
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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 (PCl5), 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 (sp3d).
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Hybridization of the type (sp3d) has a structure
of the Trigonal bipyramidal as in the picture .
4- Which element can expand its valence shell to ac
more than eight electrons?
A) O B) C C) P D) He
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To explain the correlation in (SF6) 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 (sp3d2)
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The six hybrid orbital's of the type (sp3d2)
structure has Octahedral, as in the form :
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In (CH4), for example the carbon atom
composition of four orbits of the hybrid
type (sp3) as follows:
Electron configuration of carbon:
Excited atom
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This theory is based mainly on the
hybridization of atomic orbital's of the atom
or ion Central.
To apply this theory to the transition
elements Complexes Pauling suppose the
following hypotheses :
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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%).
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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.
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4- The number of these hybrid orbital's is
the equal to the number of coordinate of
central atom in complexes .
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[Be(H2O)4]2+
Hybridization : (sp3)
Geometry : Tetrahydral
Diamagnetic (absence of single electrons) .
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Coordination
Hybridization
number
sp
Geometry
2
Linear
sp2
3
Plane triangle
sp3
4
Tetrahydral
dsp2
4
Square planar
dsp3
5
Trigonal
bipyramidal
sp3d
5
Squar pyramidal
d2sp3 (sp3d2)
6
Octahedral
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Example
HgCl2
BCl3
]NiCl4[2]Ni(CN)4[2[Fe(CO)5]
[V(acac)5]3[Fe(CN)6]324
[Cr(CO)6]
Hybridization : (d2sp3)
Geometry : Octahedral
Diamagnetic
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[Fe(CO)5]
Hybridization : dsp3
Geometry : Trigonal bipyramidal
Diamagnetic
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[Ni(CN)4]2-
Hybridization : dsp2
Geometry : Square planar
Diamagnetic
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[NiCl4]2-
Hybridization : (sp3)
Geometry : Tetrahydral
Paramagnetic
Contains two electrons alone (µ ≈ 2.8 β.M)
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[Ti(H2O)6]3+
Hybridization : (d2sp3)
Geometry : Octahedral
Paramagnetic
Contains a single electron (µ ≈ 1.7 β.M)
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[Fe(CN)6]4-
Hybridization : (d2sp3)
Geometry : Octahedral
Diamagnetic
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[Fe(H2O)6]2+
Hybridization : (sp3d2)
Geometry : Octahedral
Paramagnetic
Contains four electrons alone (µ ≈ 4.8 β.M)
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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 ..
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*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.
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The theory considers which atomic orbitals
on the metal are used for bonding.
From this the shape and stability of the
complex are predicted
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
*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.
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*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.
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* Hybrid orbitals do not actually exist. 
* Hybridization is a mathematical 
manipulation of the wave equations for
the atomic orbitals involved.
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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 3d7 4S2. Thus a Co+3
ion will have the structure 3d6, and the
electrons will be arranged as follows:
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
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
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 Co3+ can be assigned to the 3d
orbitals, following Hund's rule. Co3+ 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.
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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 Co3+ ion.
 The electrons from the fluoro ligands
simply occupy the upper vacant orbitals
on Co. With hybridization, this will be a
set of sp3d2 hybrid orbitals ( bond
formed) outer orbital complex.
 The d orbitals used are 4dx2-y2, 4dz2

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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 d6 arrangement,
and this is called a high–spin or a spin-free
complex.
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The same type of hybridization can be 
imagined in the hexaammine complex.
*In this case, however, the electrons 
originally with Co3+ 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 d2sp3.
* Low energy inner d orbitals are used to
form an inner orbital complex.
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
* 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
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


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.
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** The metal ion could also form fourcoorinate complexes.

Two different arrangements are possible: 
*sp3 hybridization – tetrahedral shape. 
*dsp2 hybridization- square planar shape 
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* Linear complexes:
[Cu(CN)2]Cu=Ar 4s1 3d10
,
Cu+1 = Ar 4s0 3d10
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[Fe(CN)6]3- Fe = Kr 4s2 3d6
Fe+3 = Kr 4s0 3d5
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When very strong ligands are present, fourcoordinated d8 metal ions usually form square
planar complexes (rather than tetrahedral).
[Ni(CN)4]2- Ni+2 = kr 4s0 3d8
Ni+2 = kr 4s0 3d8
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[Cd(NH3)4]2+
Cd+2 = kr 5s0 4d10
Cd° = kr 5s2 4d10
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Try to apply the theory to complexes following :
[Cu(Cl)4]2- ،
[Pt(Cl)4]2[Ni(Co)4]
،
[Co(NH3)6]3+
[Co(en)2Cl]+ ،
[Fe(CN)6]3[Ni(Cl)4]- ،
[Fe(H2O)6]3-
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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).
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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 :
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Copper ion (II) Is superimposed with ammonia has
a coordination number four On this basis, we can
apply the theory as follows:
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X
Predicted theoretical form spatial pyramid of four aspects
(sp3) due to the fullness of the cover (d) with elictrons
(d9), but proved to X-ray, this overlay is a four-level which
makes it imperative to have a hybridization of the type
(dsp2) 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.
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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)
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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.
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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.
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