POLARITY AND FAJAN'S RULE:

IONIC CHARACTER IN COVALENT COMPOUNDS:

(1) Pauling used dipole moment for depicting percentage of polarity and ionic character of the bond. According to Pauling, a bond can never be 100% ionic.

ILLUSTRATIVE EXAMPLE (1): Calculate the % of ionic character of a bond having length = 0.83 Å and 1.82 D as its observed dipole moment?

SOLUTION: To calculate μ considering 100% ionic bond

                      = 4.8 × 10^–10 × 0.83 × 10^–8esu cm

                      = 4.8 × 0.83 × 10^–18 esu cm = 3.984 D

                         ∴ % ionic character = 1.82/3.984 × 100 = 45.68%

(2) When electronegativity difference between two atoms is 2.1, there is 50% ionic character in the bond.

(3) When electronegativity difference is zero (identical atoms), the bond will be 100% covalent.

HANNEY AND SMITH EQUATION:

According to Haney and Smith, the percentage of ionic character of a polar covalent bond can be calculated with help of the following expression.
%ionic character:

POULING EQUATION:

ILLUSTRATIVE EXAMPLE (2) : Calculate the & ionic character in H-X bond of

Halogen acids. Given that electronegativity of  H ,F, Cl, Br. and I  are 2.1, 4.0, 3.0, 2.8,

and 2.5 respectively.

FAJAN'S RULE:

Consider an ionic bond formed between two oppositely charged ions of unequal size of A^{+  and}

B^-. In this bond cation and anion remains bonded by electrostatic force of attraction by the valence electrons of larger anion (B^-) are attracted by small cation (A⁺) and so the shape of valence shell is deformed and electron cloud of valence shell of anion remains no longer symmetrical. This process is called Polarisation of anion by cation. The ability of cation to polarise anion is called polarising power and the tendency of anion to get polasied, is called polarisability.

Due to polarisation of anion, an ionic bond acquires some partial covalent character and

greater is the extent of polarisation, more is the covalent character incorporated in ionic bond.

The magnitude of polarization depends upon following factors :

(1)SIZE OF THE CATION: Polarization of the anion increases as the size of the cation decreases i.e., the electrovalent compounds having smaller cations show more of the covalent nature. For example, in the case of halides of alkaline earth metals, the covalent character decreases as we move down the group. Hence melting point increases in the order of

                  BeCl₂ < MgCl₂ < CaCl₂ < SrCl₂ < BaCl₂ 

  

(2)SIZE OF ANION: The larger the size of the anion, more easily it will be polarized by the cation i.e., as the size of the anion increases for a given cation, the covalent character increases. For example, in the case of halides of calcium, the covalent character increases from F^– anion to I^– anion i.e. increasing covalent character

                               CaF₂ < CaCl₂ <CaBr₂ < CaI₂

Similarly, in case of trihalides of aluminum, the covalent character increases with increase in size of halide anion i.e.

                                  AlF₃ <AlCl₃< AlBr₃ < AlI₃

Covalent character increases as the size of halide ion increases

(3) CHARGE ON EITHER OF THE IONS: As the charge on the cation increases, its tendency to polarize the anion increases. This brings more and more covalent nature in the electrovalent compound. Whereas with the increasing charge of anion, its ability to get polarized, by the cation, also increases. For example, in the case of NaCl, MgCl₂ and AlC_l3 the polarization increases, thereby covalent character becomes more and more as the charge on the cation increases. Similarly, lead forms two chlorides PbC_l2 and PbCl₄ having charges +2 and +4 respectively. PbCl₄ shows covalent nature. Similarly among NaCl, Na₂S, Na₃P, the charge of the anions are increasing, therefore the increasing order of

Covalent character.  NaCl < Na₂S < Na₃P

(4) NATURE OF THE CATION: Cations with 18 electrons (s² p⁶ d¹⁰) in outermost shell polarize an anion more strongly than cations of 8 electrons (s² p⁶) type. The d electrons of the 18 electron shell screen the nuclear charge of the cation less effectively than the s and p electrons of the 18- electron shell. Hence the 18-electron cations behave as if they had a greater charge. Copper (I) and Silver (I) halides are more covalent in nature compared with the corresponding sodium and potassium halides although charge on the ions is the same and the sizes of the corresponding ions are similar. This illustrates the effect of 18- electron configuration of Cu+ (3s² 3p⁶ 3d¹⁰) and Ag+ (4s2, p6, d¹⁰) ions.

ILLUSTRATIVE EXAMPLE:

The decomposition temperature of Li₂CO₃ is less than that of Na₂CO₃. Explain.

SOLUTION: As Li+ ion is smaller than Na+ ion, thus small cation (Li+) will favour more covalent character in Li₂CO₃ and hence it has lower decomposition temperature than that of Na₂CO₃.

                                                                  @@@

TITRATION OF TRIPROTIC ACID WITH STRONG BASE:

TRIPROTIC Vs NaOH: [H₃PO₄ Vs NaOH]
ILLUSTRATIVE EXAMPLE: Phosphoric acid (H₃PO₄), is a Triprotic acid with Ka₁= 10^-3 , Ka₂=10^-8 and Ka₃= 10^-12 .Consider the titration of 0.10M 100 ml H₃PO4 with 0.1M NaOH Solution and answers the following questions.

(A) Write out the reactions associated with Ka₁, Ka₂ and Ka₃.

(B) Calculate the pH after the following titration points:

(1) 100 ml of 0.1M H₃PO₄ + 0.0 ml of 0.1 ml NaOH
(2) 100 ml of 0.1M H₃PO₄ + 50 ml of 0.1 ml NaOH
(3) 100 ml of 0.1M H₃PO₄ + 100 ml of 0.1 ml NaOH
(4) 100 ml of 0.1M H₃PO₄ + 150 ml of 0.1 ml NaOH
(5) 100 ml of 0.1M H₃PO₄ + 200 ml of 0.1 ml NaOH
(6) 100 ml of 0.1M H₃PO₄ + 250 ml of 0.1 ml NaOH
(7) 100 ml of 0.1M H₃PO₄ + 300 ml of 0.1 ml NaOH 

(C) Sketch the titration curve for this titration.

(D) What weak acid and what conjugate base make the best phosphate Buffer at pH ~7.0?

(A) Write out the reactions associated with Ka₁, Ka₂ and Ka₃.

 SOLUTION:

Step-1    

Step-2  

 Step-3    

(B)  Calculate the pH after the following titration points;

(1) 100 ml of 0.1M H₃PO₄ + 0.0 ml of 0.1 ml NaOH 

SOLUTION:

By approximation we know that if

(2) 100 ml of 0.1M H₃PO₄ + 50 ml of 0.1 ml NaOH 

SOLUTION:

Given mill moles (M x V) of acid= 0.1x100= 10 and base 0.1x50 = 5.0

Here H₃PO₄/NaH₂PO₄ remain same in solution and which are act acidic buffer

(3) 100 ml of 0.1M H₃PO₄ + 100 ml of 0.1 ml NaOH 

SOLUTION:

Given mill moles (M x V) of acid= 0.1x100 = 10 and base 0.1x100 = 10

Here only NaH₂PO₄ remain same in solution and which are under go amphoteric hydrolysis. PH of amphoteric salts is independent of concentration of salt.

(4) 100 ml of 0.1M H₃PO₄ + 150 ml of 0.1 ml NaOH

SOLUTION:

Given millimoles (M x V) of acid= 0.1x100 = 10 and base 0.1x150 = 15

Here only NaOH (5 millimoles) and NaH₂PO₄ both are remain present in solution hence which are further go titration.

Here NaH₂PO₄ and Na₂HPO₄ remain same in solution and which are act acidic buffer

(5) 100 ml of 0.1M H₃PO₄ + 200 ml of 0.1 ml NaOH

SOLUTION:

Given mill moles (M x V) of acid= 0.1x100 = 10 and base 0.1x 200 = 20

Here only NaOH (10 millimoles) and NaH₂PO₄ (10 millimoles) both are remain present in solution hence which are further go titration.

Here only Na₂HPO₄ remain same in solution and which are under go amphoteric hydrolysis. pH of amphoteric salts is independent of concentration of salt.

(6) 100 ml of 0.1M H₃PO₄ + 250 ml of 0.1 ml NaOH 

SOLUTION:

Given mill moles (M x V) of acid= 0.1x100 = 10 and base 0.1x 250 = 25

Here NaOH (15 millimoles) and NaH₂PO₄ (10 millimoles) both are remain present in solution hence which are further go titration.

Here Na₂HPO₄ (10 millimoles) and (5.0millimoles) both remain same in solution and which are further go titration.

Here Na₂HPO₄ and Na₃PO₄ remain same in solution and which are act acidic buffer

(7) 100 ml of 0.1M H₃PO₄ + 300 ml of 0.1 ml NaOH 

SOLUTION:
Given millimoles (M x V) of acid= 0.1x100 = 10 and base 0.1x 300 = 30

Here NaOH (20 millimoles) and NaH₂PO₄ (10 millimoles) both are remain present in solution hence which are further go titration.

Here NaOH (10 millimoles) and Na₂HPO₄ (10 millimoles) both remain same in solution and which are further go titration.

Finally 10 millimoles (Molarity=10/400) N₃PO₄ is formed which undergo polyvalent salt hydrolysis 

HYDROLYSIS OF ANION (PO₄^-3):

Step wise illustration of hydrolysis of poly basic acids and polyacidic base is given as -

Na+ ion do not under goes hydrolysis while PO₄^-3 undergoes step wise hydrolysis

Experimentally we know that   Ka₁>>Ka₂>>Ka_3, hence x >>y>>Z   so y and z can be neglected with respect to x   it mean total OH^-  ions count from only first step OH- ions coming from 2^nd and 3^rd  hydrolysis is ignored 

Special Case (1):  

Special Case (2) 

This is the quadratic equation solve by following formulae

So approximation not valid hence follows 2^nd case

(C) Sketch the titration curve for this titration:

SOLUTION:    

S.N.	Given condition	comments	PH
1	100 ml of 0.1M H3PO4 + 0.0 ml of 0.1 ml NaOH	H3PO4 Polyprotic acid	02.02
2	100 ml of 0.1M H3PO4 + 50 ml of 0.1 ml NaOH	Acidic Best buffer PH= Pka1 (1st half E-Point)	03.00
3	100 ml of 0.1M H3PO4 + 100 ml of 0.1 ml NaOH	Amphoteric salt Hydrolysis PH=1/2(Pka1+Pka2)	05.50
4	100 ml of 0.1M H3PO4 + 150 ml of 0.1 ml NaOH	Acidic Best buffer PH= Pka2 (2st half E-Point)	08.00
5	100 ml of 0.1M H3PO4 + 200 ml of 0.1 ml NaOH	Amphoteric salt Hydrolysis PH=1/2(Pka2+Pka3)	10.00
6	100 ml of 0.1M H3PO4 + 250 ml of 0.1 ml NaOH	Acidic Best buffer PH= Pka3 (3st half E-Point)	12.00
7	100 ml of 0.1M H3PO4 + 300 ml of 0.1 ml NaOH	Poly anionic salt Hydrolysis 3rd Equivalent point	12.07

 Graphical representation of titration: