Factors Affecting the magnitude of ∆

There are several factors that affect the magnitude of splitting (0)of d-orbitals by the surrounding ligands.

1. Oxidation State of the Metal Cation:

The higher the oxidation state of the metal cation, the greater will be the magnitude of ∆. The higher the oxidation state of the metal causes the ligand to approach more closely to it and, therefore the ligand causes more splitting of metal d-orbitals.
For example               ∆0 for [Co(H2O)6]2+ = 9200 cm-1
and                     ∆0 for [Co(H2O)6]3+ = 20760 cm-1

2. Same Oxidation State of Metal Cation but the number of d-electrons is                   Different:

In general, for a given series of transition elements(say 3d-series), in complexes having the metal cation with the same oxidation state but the different number of electrons in the d-orbitals, the magnitude of  decreases with increase in the number of d-electrons. It is due to the fact that the higher number of d-electrons prevents the ligands to come closer to the metal cation.
For example        0 for [Co(H2O)6]2+ = 9200 cm-1(3d7)
and                ∆0 for [Ni(H2O)6]2+ = 8500 cm-1(3d8)

3. Principal Quantum Number(n) of the d-orbital of the Metal Cation:

In case of complexes having the metal cation with the same oxidation states and the same number of d-electrons, the magnitude of for analogous complexes within a given group increases about 30% to 50% from 3d to 4d and by about the same amount from 4d to 5d. It is because:
(1) On moving from 3d to 4d and 4d to 5d, the size of the d-orbital increases and electron density              decreases in them. Therefore the ligands can approach the metal cation with larger d-orbital                more closely.
(2) There is less steric hindrance around a larger metal cation.
For example      ∆0 for [Co(NH3)6]2+ = 2300 cm-1
∆0 for [Rh( NH3)6]2+= 34100 cm-1
∆0 for [Ir( NH3)6]2+= 41200 cm-1

4. Nature of Ligands:

The ligands are classified as a weak and strong ligand. The ligand which causes a small degree of splitting of d-orbitals are called weak ligands and the ligands which cause a large splitting are called strong ligands. The common ligands have been arranged in order of their increasing crystal field splitting power to cause splitting of d-orbitals from a study of their effects on the spectra of transition metal ions.
(weak end)O22−< I < Br < S2− < SCN (S–bonded) < Cl− < N3 < F< NCO < OH < C2O42− < H2O < NCS (N–bonded) < CH3CN < gly (glycine) < py (pyridine) < NH3 < en (ethylenediamine) < bipy (2,2′-bipyridine) < phen (1,10-phenanthroline) < NO2 < PPh3 < CN < CO < CH2(strong end)
This order s usually called as Spectrochemical series
The order of the field strength of common ligands is independent of the nature of the metal cation and the geometry of the complex.

5. Number of Ligands:

The magnitude of crystal field splitting(∆)increases with the increase of the number of ligands.For example  > ∆t
Though the number of ligands in a square planar complex is smaller than that of octahedral complexes, the magnitude of  ∆sp is greater than ∆0. It is because of the fact that square planar complexes are formed by much strong ligands with dmetal cation of 3d-series transition metal cation and 4d and 5d series d8metal cation with either weak or strong ligand. The very strong ligands and 4d or 5d series transition metal cations are responsible for higher crystal field splitting. Also in square planar complexes of dmetal cation, the dZ2 orbital with two electrons is stabilized and the vacant dX2-y2 orbital is destabilized.

Crystal field splitting in Octahedral complex:

•  In a free metal cation all the five d-orbitals are degenerate(i.e.these have the same energy.In octahedral complex say [ML6]n+the metal cation is placed at the center of the octahedron and the six ligands are at the six corners. These six corners are directed along the cartesian coordinates i.e. along the x, y, and z-axis. When all the ligands are at an infinite distance from the metal cation, the five d-orbital of the metal cation will not be affected by the ligand electrostatic field and will remain degenerate. When the ligands move towards the metal cation, there are two electrostatic forces i.e.

(1)The attraction between metal cation and ligand
(2)Electrostatic repulsion between d-electrons of the metal cation and lone pairs of ligands

• Greater the repulsion between metal cation and ligands, ligands will be more closer to the metal cation and hence more will be the repulsion between the metal d-electrons and the lone pair of electrons on the ligand. When the ligands are closer to the metal cation an electrostatic force of repulsion also exists among the ligands.These two repulsion cause to adopt the octahedral geometry that locates the ligand at the internuclear distance from the metal cation and as far apart from one another as possible.
• The force of repulsion between metal d-electron and the ligand electrons cause to increase in potential energy of metal d-electrons. Remember that greater the force of repulsion higher will be the potential energy. If all the six ligands approaching the metal cation surrounds it spherically symmetric i.e. all the six ligands are at equal distance from each of the d-orbitals.
• The energy of each d-orbital will raised by the same amount and all the five d-orbital will remain degenerate. This is a hypothetical situation and has the average energy of a set of d-orbitals.In an actual octahedral complex, a spherically symmetric field is never obtained. Therefore all the five d-orbitals are not affected by the same extent.
• Since the two d-orbitals( dx2-y2 and  dz2 ) points directly towards the ligands and the three d- orbitals(  dxy ,dyz and dzx ) point in between the path of the approaching ligand. Therefore the  dx2-y2 and  dz2 orbitals will be more strongly repelled than the  dxy ,dyz and dzx orbitals. Therefore the energy of the dx2-y2 and  dz2 orbitals will be raised and that of the other three orbitals which lie far away from the ligand will be decreased relative to the hypothetical energy state.

• The five d-orbital which were degenerate in a free metal cation is now split into two sets of d-orbitals of different energies, a higher energy level with two orbitals(dx2-y2 and  dz2)having the same energy and a lower level with three equal energy orbitals(dxy,dyz, and dzx). The set of   dx2-y2 and dz2orbitals are referred to as eset which is doubly degenerate and the set of dxy,dyz, and dzx is referred to as t2g set which is triply degenerate.
• Since the distance between metal cation and the ligands has remained the same, the net potential energy(or average energy) of the system must remain the same as that of the spherical field before splitting. This state of average energy is called the barycentre.
• The separation of five d-orbitals of metal cation into two sets of different energies is called crystal field splitting. The energy difference between two sets of orbitals which arise from an octahedral field is measured in terms of the parameter ∆0 or 10Dq where o in ∆0 stands for octahedral.
• Since the energy of barycentre remains constant, the total energy decrease of the t2g set must be equal to the total energy increase of the eg set. Therefore since there are two eg orbitals, they must increase by 0.6∆0 or 6Dq and the three t2g orbitals must decrease by 0.4∆0 or 4Dq per electron. The decreased energy t2g of orbitals stabilizes the complex by 0.4∆0 and the increase in energy of eorbitals destabilizes the complex by 0.6∆0.

Crystal Field Theory:

Valence bond theory is useful to visualize the bonding in complexes, but it fails to explain the nature of ligands, colour and electronic spectra, effect of temperature on magnetic moment and magnetic susceptibilities, why some complexes are high spin and others are low spin, stability of complexes.
To explain these properties Bethe and van Vleck proposed the crystal field theory.This theory was originally applied to ionic crystals and is therefore called crystal field theory.
This theory is based on the following assumptions:
• Ionic ligands such as Cl,OH,CN are regarded as negative point charges(or simply point charges) and the neutral ligands such as  H20,NH3,Py are regarded as dipole(or simply dipoles) because these ligands are dipolar.If the ligand is neutral molecule like the negative end of the dipole is directed towards the metal ion.
• Metal-ligand bond is not covalent i.e. there is no overlapping of orbitals.Instead of bonding in complexes is purely electrostatic in nature.In complexes two types of  electrostatic forces come into account,

(1)One is the attraction between metal cation and the negatively charged ligand or the                                negative end of the polar ligand(i.e. dipole)

(2)The second type of electrostatic interaction is the electrostatic repulsion between the                            lone pairs of electrons on the ligands and the electrons in the d-orbital of the metal cation                    and the repulsion between nuclei of metal cation and the ligands but to a small extent.
• Another repulsion also come into account that occurs among the ligands.
• the five d-orbital in a free metal ion are degenerate(i.e.have same energy).When a complex is formed, the electrostatic field of ligands destroy the degeneracy of these d-orbitals i.e. these orbitals now have same energies.
• The orbital lying in the direction of the lidands are raised in energy more than those lying away from the ligands because of the repulsion between the d-electrons and the ligands.
• In order to understand CFT, it is necessary to know the geometry and orientation of the five d-orbitals.

Hybridisation in outer orbital octahedral complex

Outer orbital octahedral complex:

Let’s take an example   [Fe(H2O)5(NO)]2+
• In this complex ion oxidation state of Fe is +1, because NO exists in +1 oxidation state in a complex of Fe and it is valence shell electronic configuration is 3d6, 4s1
• Magnetic moment measurement indicates that its experimental magnetic moment is 3.89 B.M. which corresponds to three unpaired electrons in the d-orbital of complexion.
• The single NOstrong ligand has little tendency to pair up only two unpaired electrons. Since  H2O is a weak ligand, therefore, it has no tendency to pair up electrons and none of the five 3d orbitals is vacant.
• Therefore the  4s, 4p and two of the five 4d orbitals (i.e.4dx2-y2  and  4dz)combine to give six sp3d2 hybrid orbitals.
• These hybrid orbital form bonds with six ligands by accepting six pairs of electrons, one pair from each of the six ligands.

• Here as outer d-orbital is involved in hybridization it gives outer orbital octahedral geometry.

Note:

• The octahedral complex of  d1, d2, and d3 metal cation are always inner orbital octahedral complexes whether the ligands are strong or weak.
• The octahedral complex of d8, d9, and d10 metal cation are always outer orbital complexes either the ligands are strong or weak.
• The complexes of d4, d5,d6, and d7 metal cation are outer orbital complex if the ligands are weak.

Inner orbital octahedral complex

Let’s discuss inner orbital complexing taking an example, [Co(CN)6]3-  ion

• In this complex oxidation state of cobalt is +3

Valence shell electronic configuration is

• Magnetic moment measurement indicates that  [Co(CN)6]3-C ion is diamagnetic(given). So that all the d-electrons arranged in such a way, that no unpaired electron is left in the d-orbital.
• In other words as CN  ion is a strong ligand, it causes the pairing of 3d-electrons.

•  Here two vacant 3d-orbital combine with the vacant 4s and 4p orbital to form six  d2sp3

hybrid orbital.

• Then six hybrid orbital overlap with six filled orbital of  CN ligand, one on each of the ligand and thus six coordinate covalent bonds are formed which gives d2sp3  hybridization.
• As inner d-orbitals are involved in hybridization, it gives inner orbital octahedral geometry.

Explain hybridisation in,
[Co(NO2)6]4-   (one unpaired electron)
[Mn(CN)6]3-    (two unpaired electron)
[Cr(CN)6]3-C   (paramagnetic corresponding to three unpaired electron)

How to determine hybridization in coordination complex

To understand hybridization  let’s take an example,  [Co(NH3)6]3+
Here it is clear that the coordination number of this complex is 6. So the complex must adopt octahedral geometry. The only thing we have to predict is whether it’s hybridization is  sp3d2 ord2sp3

Steps to predict hybridization:

1. In the first step, we have to calculate the oxidation state of the metal ion. So the oxidation state of cobalt is +3.
2. Then predict whether the ligand is strong or weak and then according to this arrange electrons in the d-orbital.
3. If the ligand is strong, then pairing occurs from the initial condition(low spin complex) and if the ligand is weak then first all the d-orbital is singly filled and then pairing occur(High spin complex)
4. In the given example  NH is a strong ligand so that it will form a low spin complex.

5. From the above picture, we can see that  6 vacant orbitals of metal ion combine with 6   NHligands to give d2sp3  hybridization.
6. As the d-orbital present in the inner side, it is an inner orbital octahedral complex.

Hybridization in coordination compound

Hybridization in coordination compound:

The main application of Valence Bond Theory is to predict hybridization in a coordination compound.
• Hybridization is the concept of mixing of atomic orbital into a new hybrid orbital.
• For example, if one s-orbital combine with one p-orbital it will form a new sp hybrid orbital.

Like that, sp2hybrid orbital = one s-orbital + two p-orbital

sp3 hybrid orbital = one s-orbital + three p-orbital
sp2d hybrid orbital = one s-orbital + two p-orbital + three p-orbital
sp3d2 hybrid orbital =  one s-orbital + three p-orbital + two d-orbital
• The coordination number of the metal atom or ion decide the hybridization and geometry of the complex.

For example, if the coordination number is 6, then 6 hybrid orbitals combine among themselves and give  sp3dhybridization.

• If the coordination number is 6 two possibilities is there i.e.

1. Inner orbital Octahedral complex ( d2sp3hybridization)

2. Outer orbital Octahedral complex ( sp3d2hybridization)
But, both the hybridization leads to Octahedral geometry.
• If the coordination number is 4 there is the possibilities of two types of geometry i.e.

1. Tetrahedral complex ( sphybridization)

2. Square planar complex( dsp2hybridization)

High spin and Low spin complex

High spin and low spin complex are two possible classifications of spin states that occur in a coordination compound.

• Before going to this topic we must have an idea about strong ligand and weak ligand. To know which ligand is strong and which ligand is weak, we must go through spectrochemical series i.e.
• The spectrochemical series : (weak end)O22−< I < Br < S2− < SCN (S–bonded) < Cl− < N3 < F< NCO < OH < C2O42− < H2O < NCS (N–bonded) < CH3CN < gly (glycine) < py (pyridine) < NH3 < en (ethylenediamine) < bipy (2,2′-bipyridine) < phen (1,10-phenanthroline) < NO2 < PPh3 < CN < CO < CH2(strong end)
• The ligands which are present on the left of the series are consider to be strong ligahds and those which are present on the right of the series are consider to be weak ligand

High spin complex:

• It is also called spin free complex.
• The complex having a maximum number of unpaired electrons are called high-spin or spin-free complex.
• In the high spin complex, first all the d-orbital are singly filled and then pairing occour .
• Strong ligand i.e. ligands which are on the left of the spectrochemical series are always form high spin or spin free complex.

Low spin complex:

• It is also called spin paired complex.
• The complex having a minimum number of unpaired electron i.e. more number of paired electrons are called low spin or spin paired complex.
• In a low spin octahedral complex pairing of d electrons take place from the initial condition.
• Weak ligand i.e. ligands which are present on the right of the spectrochemical series always form low spin or spin paired complex.

State Function:

• In thermodynamics, a state function, state quantity, or a function of the state, is a property of a system that depends only on the initial and final state the system, not on the way in which the system acquired that state.
• A state function describes the equilibrium state of a system and thus also describes the types of system.
• The cyclic integral involving a state function is always zero.
• All the thermodynamics property satisfy the requirements of state function.
• U = q + w         change in thermodynamic energy
S = qrev/T            entropy
H = U = PV         enthalpy
G = H – TS          Gibb’s free energy
A = U – TS          Helmholtz free energy
• Internal energy, enthalpy, entropy are the example of state functions.

Path Function:

• Path function depends on the path taken to reach that specific value, not on the initial and final state of that value.
• Path function needs multiple integral and limits of the integration in order to integrate.
• It is based on how the state of a system was established.
• Work, heat, arc length are the example of path function.

Thermodynamic process:

It is the path or operation by which a system changes from one state to another state.

1. Isothermal process:

The temperature of the system remains constant during each step.
T = constant => dT = 0

Details:

• For such change system should be contained in a perfectly conducting container.
• Perfect isothermal change is impossible but when a change is carried out very slowly approximate isothermal change occurs.
• It follows Boyle’s law.
• Work done in the isothermal process is graphically given by the area under the P-V curve.
• ∆H = nCp∆T  and  ∆H = nCv∆T
In isothermal process ∆T = 0
=> ∆H = 0 and ∆E = 0
• Specific change at constant T is infinitely great in the isothermal process.

There is no heat exchange between the system and surroundings.
q = constant
=> dq = 0
Perfectly adiabatic change is impossible but when a process is carried out very rapidly fairly

3. Isobaric process:

It is the system in which pressure of the system remains constant during each step.
P = constant , dp = 0

4. Polytropic process:

In this process heat capacity of the body remains constant.
Cp = constant  and  Cv = constant

=> dCp = 0  and  dC = 0

5. Quasistatic process:

The process in which the deviation from thermodynamic equilibrium is infinitesimal and all the states through which the system passes can be considered as equilibrium states.

6. Isochoric process :

When there is no change in the volume of the system during various operations the change is said to be isochoric.
V = constant  => dv = 0
For example, the combustion of a substance in a bomb calorimeter is an isochoric process.

7. Cyclic process:

The process which brings back a system to its original state after a series of changes is called acyclic process.
As the E and H depend only on their states, E and H are constant.
So, dE = 0 and  dH = 0