# Role of φ ( ionic Potential )

1. Prediction of degree of covalency in an ionic compound.

2. Tendency of a cation to form complexes can be estimated.

3. Tendency of cations towards solvation

E.g.: [Li(H2O)6]+ , [Na(H2O)4]+ , [Cs(H2O)]+

Nature of oxides: Emperically for an oxide of the type

< 2.2 basic

∎ √φ lying between 2.2 – 3.2 amphoteric

∎ √φ > 3.2 acidic

i.e. higher is the value of φ greater is acidic nature of oxide.

Illustration : The melting point of KCl is higher than that of AgCl though the crystal radii of Ag+ and K+ions are almost same.

Solution: Now whenever any comparison is asked about the melting point of the compounds which are fully ionic from the electron transfer concept it means that the compound having lower melting point has got lesser amount of ionic character than the other one.
To analyse such a question first find out the difference between the 2 given compounds. Here in both the compounds the anion is the same. So the deciding factor would be the cation.

Now if the cation is different, then the answer should be from the variation of the cation.

Now in the above example, the difference of the cation is their electronic configuration. K+ = [Ar]; Ag+ = [Kr] 4d10.

This is now a comparison between a noble gas core and pseudo noble gas core, the analysis of which we have already done.

Exercise :
Na2CO3 does not decompose on heating whereas CaCO3 decomposes, why?

### Solubility of the hydroxides

Illustration : The solubility of the hydroxides of the alkaline earth metals increases i.e. Ba(OH)2 has got a higher solubility in water compared to Mg(OH)2 .

Solution: Here both the cations Ba2+ and Mg2+ have the same charge, but as the radius of Ba2+ is more therefore φ of Ba2+ is less which implies that Ba(OH)2 having higher degree of ionic character is more soluble in polar solvents like water.

But now if I ask to predict the solubility of MgSO4 & BaSO4. The answer seems to be quite similar to the earlier one and BaSO4 turns out to be the one having higher solubility.

But contrary to our expectation the trend is reversed here. BaSO4 is sparingly soluble in water.The question comes why ?

In case of hydroxide it is something. In case of sulfate it’s the other way around. Is there any way by which we can a predict the solubility trend ? The answer is yes.

When a lattice is dissolved in water , the ions are solvated and the solvated ions are more stable than a free ion and due to this stability energy is released.
This energy released is called solvation energy and if this overcomes the lattice energy then it is soluble.

The lattice energy of NaCl is 778 kJ mol-1 and the heats of hydration of Na+ and Cl is -787 kJ mol-1. As it is more than the lattice energy of NaCl therefore it is soluble.

Now we should focus our attention to the solubility trend in a given series. For a comparison of the solubility both the lattice energy and hydration energy factors have to be taken into account.

If in a series the decrease of lattice energy is more compared to the decrease in hydration energy then the substance becomes soluble.

Now the hydration enthalpy of a salt is given by

$\large \Delta H = \frac{k_1}{r_{+}} + \frac{k_2}{r_{-}}$ ;where k1 and k2 = constant

Lattice energy

$\large U = \frac{k_3}{r_{+} + r_{-}}$ ; k3 = constant

Case I:
When r+ << r
the contribution of the anion to the hydration enthalpy is small so the total ΔHhydration would be dominated by the cation alone.

In a series of salts of a large anion, the hydrational enthalpy will decrease in magnitude with increasing cation size.

Now how does the lattice energy respond to this changing cation radius ? The lattice energy is inversely proportional to (r+ + r ).

Since r >> r+, the sum will not change significantly as r+ increases. Consequently the lattice energy will not decrease as fast as the hydration energy with increasing cationic size.

The more quickly diminishing hydration energy results in a decrease in solubility.

E.g. Solubility of      LiI > NaI > KI

MgSO4 > CaSO4 > SrSO4 > BaSO4

Case II:
r+ ≈ r
Here the lattice energy decreases with increasing cationic size more rapidly than the hydration energy which therefore results in an enhanced solubility in a series.

E.g. Solubility of      LiF < NaF < KF

Mg(OH)2 < Ca(OH)2 < Ba(OH)2

### Deviation from Covalent to Ionic Character

In the previous section we discussed about those compounds which deviate from fully ionic to some degree of covalency. A similar trend can also be observed with pure covalent molecules which can change to a partially ionic bond.

This happens when the electronegativities of the two atoms which form the covalent bond are not the same.
The atom having higher electronegativity will draw the bonded electron pair more towards itself resulting in a partial charge separation. The distribution of the electron cloud in the bond does not remain uniform and shifts towards the more eletronegative one. Such bonds are called polar covalent bonds.

For example the bond formed between hydrogen and chlorine or between hydrogen and oxygen in water is of this type.

Molecules like HCl , H2O , NH3 i.e. molecules of the type H – X having two polar ends (positive and negative) are known as polar molecules.

The extent of polar character or the degree of polarity in a compound is given by it’s dipole moment which is defined as the product of the net positive or negative charge and the distance of separation of the charges i.e. the bond length. The symbol of dipole moment is μ

μ = electronic charge (e) × distance

The unit of dipole moment is Debye (D)

1D = 3.33 × 10-30 Cm = 10-18 esu cm

Dipole moment is indicated by an arrow having a symbol ( ) pointing towards the negative end. Dipole moment has both magnitude and direction and therefore it is a vector quantity.

To calculate the dipole moment of a molecule we should calculate the net dipole due to all bonds and for lone pair if any. Diatomic molecules like HCl , HF have the dipole moment of the bond (called bond dipole) equal to the molecular dipole as the structure has one bond only.

But for poly atomic molecules the net dipole is the resultant of the individual bond dipoles. A compound having a zero dipole moment indicates that the compound is a symmetrical one.

Illustration : CO2 has got dipole moment of zero why ?

Solutions: The structure of CO2 is .
This is a highly symmetrical structure with a plane of symmetry passing through the carbon. The bond dipole of C-O is directed towards oxygen as it is the negative end.

Here two equal dipoles acting in opposite direction cancel each other and therefore the dipole moment is zero

Illustration : Dipole moment of CCl4 is zero while that of CHCl3 is non zero.

Solution: Both CCl4 & CHCl3 have tetrahedral structure but CCl4 is symmetrical while CHCl3 is non-symmetrical.

Due to the symmetrical structure of CCl4 the resultant of bond dipoles comes out to be zero. But in case of CHCl3 it is not possible as the presence of hydrogen introduces some dissymmetry.

### Dipole Moment in Aromatic Ring System

The dipole moments of the aromatic compounds present a very good illustration of dipole moment. We all know when a substituted benzene is treated with a reagent different products namely ortho, meta and para products are formed.
The dipole moments of these products are different since the orientation of the groups is different at ortho, meta and para position. Let us take an example which will make it easily digestive for you.

Suppose we have three isomers of o-nitrophenol, m-nitrophenol and p-nitrophenol. We have also the e.g. o-aminophenol, m-aminophenol and p-aminophenol.

In the case X = Y, the para isomer becomes symmetrical and have zero dipole.
Now the obvious question that is peeping through your mind is that which isomer in which case has got higher dipole moment. The answer lies in the nature of the groups linked to the benzene ring.

In nitrophenol groups one group is electron pushing and the other is electron withdrawing while in the second case both the groups attached are electron pushing. So depending on the nature of the groups attached one of the isomer, o, m or p has the largest dipole moment

Case I: Now when X & Y are both electron pushing or electron withdrawing.

Suppose,
bond dipole of C – X = μ1

And that of C – Y = μ2

Here we have assumed a sign of + when groups are electron pushing and when groups are electron withdrawing. The net dipole is the resultant of two bond dipoles at different orientations. When both X & Y are electron pushing or electron withdrawing

$\large \mu_{ortho} = \sqrt{\mu_1^2 + \mu_2^2 + 2 \mu_1 \mu_2 cos60^o}$

$\large \mu_{ortho} = \sqrt{\mu_1^2 + \mu_2^2 + \mu_1 \mu_2 }$

$\large \mu_{meta} = \sqrt{\mu_1^2 + \mu_2^2 + 2 \mu_1 \mu_2 cos120^o}$

$\large \mu_{meta} = \sqrt{\mu_1^2 + \mu_2^2 – \mu_1 \mu_2 }$

$\large \mu_{para} = \sqrt{\mu_1^2 + \mu_2^2 + 2 \mu_1 \mu_2 cos180^o}$

$\large \mu_{para} = \sqrt{\mu_1^2 + \mu_2^2 – 2\mu_1 \mu_2 }$

From the above expression of μ0 , μm & μp it is clear that when both X & Y are of the same nature i.e. both are electron withdrawing or both are electron pushing the para product has the least dipole moment and ortho product has the highest.

Now when X = Y,

μp = μ1 – μ2

μp = μ1 – μ1 = 0

Case II: When X is electron pushing and Y is electron withdrawing or vice versa.

Let C – X dipole = μ1

& C – Y dipole = μ2

$\large \mu_{ortho} = \sqrt{\mu_1^2 + (-\mu_2)^2 + 2 \mu_1 (-\mu_2 )cos60^o}$

$\large \mu_{ortho} = \sqrt{\mu_1^2 + \mu_2^2 – \mu_1 \mu_2 }$

$\large \mu_{meta} = \sqrt{\mu_1^2 + (-\mu_2)^2 + 2 \mu_1 (-\mu_2) cos120^o}$

$\large \mu_{meta} = \sqrt{\mu_1^2 + \mu_2^2 + \mu_1 \mu_2 }$

$\large \mu_{para} = \sqrt{\mu_1^2 + (-\mu_2)^2 + 2 \mu_1 (-\mu_2) cos180^o}$

$\large \mu_{para} = \sqrt{\mu_1^2 + \mu_2^2 + 2\mu_1 \mu_2 }$

Now if you see the expressions, it is very clear that the para isomer has the highest dipole moment and ortho is the least.

So to calculate the dipole moments of disubstituted benzene one should consider about the nature of the groups linked and then only one can predict the dipole moment of the molecule.