POLARITY IN BONDS AND DIPOLE MOMENT:

The covalent bond form between two or more having different electronegativity 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. and molecules of the type H – X having two polar ends (positive and negative) are known as polar molecules. For example the bond formed between hydrogen and chlorine or between hydrogen and oxygen in water.

                              

The extent of polar character or the degree of polarity in a compound is determined by the term dipole moment. 

IONIC NATURE IN COVALENT BOND:

(1) Polarity of any polar covalent bond or molecule is measured in terms of dipole moment(μ).

(2) For measurement of extent of polarity, Pauling introduced the concept of dipole moment .

"The product of positive or negative charge (q) and the distance (d) between two poles is called  dipole moment". Here

                            [Dipole moment= magnitude of charge x distance]

                                                  [μ = q x d] 

                                   Where μ= dipole moment

                                              q= Charge in esu unit 

                                              d= distance between two charges

Dipole moment is usually expressed in Debye (D)

CALCULATION OF DIPOLE MOMENT: 

IN C.G.S UNIT:

Let charge on one electron in esu unit = 4.8×10^-10 esu {(e=1.6×10^-19 × 3.0×10^-9) (1C=3.0×10^-9)}

Consider a standard dipole of length (d) 1A° or 10^-8 cm and charge (q) 4.8×10^-10 esu.

                   μ= 4.8×10-10×10^-8 cm

                  μ= 4.8×10^-18 esu cm

                  μ=4.8 D (1×10^-18 esu cm = 1debye or 1D)

IN S.I. UNIT:

Let charge on one electron in S.I. unit =1.6×10^-19 coulomb's or C 

Consider a standard dipole of length (d) 1A° or 10^-10 m and charge (q) 1.6×10^-19

                   μ= 1.6×10^-19 ×10^-10 Cm

                   μ= 1.6×10^-29 C m

On comparing value of dipole moment in both units  

             4.8 D=1.6×10^-29 Cm

                 1D= 3.3×10^-30 Cm

         
(3) Dipole moment is a vector quantity i.e. it has both magnitude as well as direction.

(5) In the diatomic molecule dipole moment (μ) depends upon difference of electronegativity i.e. dipole moment (μ) electronegativity order of dipole moment (μ)  H–F > H–Cl > H–Br > H–I. and dipole moment (μ) = 0 for  H–H, F–F, Cl–Cl, Br–Br, O=O

(6) For poly atomic molecule dipole moment (μ) depends on the vector sum of dipole moments of all the covalent bonds.

(7) For PCl₅ and SF₆, etc. dipole moment (μ) = 0 due to their regular geometry.

(8) Benzene, naphthalene, biphenyl have dipole moment (μ) = 0 due to planar structure.

(9) If the vector sum is zero, than compound is non-polar compound or symmetrical compound (and it is not essential that individual dipole moments (μ) of every bond should be zero).

 ILLUSTRATIVE EXAMPLE(A): BX₃, CCl₄, SiCl₄, CH₄, CO₂, CS₂, PCl₅, SiH₄ etc.  In these examples the bond B–F, C–Cl, C–H, C–O, P–Cl etc. are polar even though compounds are non-polar.

                                                         

ILLUSTRATIVE EXAMPLE(B): 

(10) Dipole moment of H₂O is 1.85 D which is resultant dipole moment (μ) of two O–H bonds. Dipole moment (μ) of H₂O is more than dipole moment (μ) of H₂S because electronegativity oxygen is higher than sulphur.

 

(11) Angular structure of molecule have greater dipole moment.

ILLUSTRATIVE EXAMPLE (1): CO₂ has got dipole moment of zero why?

SOLUTION: 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.

 

ILLUSTRATIVE EXAMPLE (2): 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.

ILLUSTRATIVE EXAMPLE (3): Compare the dipole moment of H₂O and F₂O.

SOLUTION: Let’s draw the structure of both two compounds and then analyses it.

                                

In both H₂O and F₂O the structure is quite the same. In H₂O as O is more electronegative than hydrogen so the resultant bond dipole is towards O, which means both the lone pair and bond pair dipole are acting in the same direction and dipole moment of H₂O is high. In case of F₂O the bond dipole is acting towards fluorine, so in F₂O the lone pair and bond pair dipole are acting in opposition resulting in a low dipole.

ILLUSTRATIVE EXAMPLE (4): Compare the dipole moment of NH₃ and NF_3.?

SOLUTION:

                              

Hence μ NH3 is greater than μ NF

ILLUSTRATIVE EXAMPLE (5): but -2- ene. It exists in two forms Cis and Trans

                      

SOLUTION: The Trans isomer is symmetrical with the 2-methyl groups in anti position. So the bond dipoles the two Me– C bonds acting in opposition cancel each other result in a zero dipole. Whereas in cis isomer the dipoles do not cancel each other resulting in a net dipole.

ILLUSTRATIVE EXAMPLE (6): Compare the dipole moment of Cis 1,2 trichloroethylene and trans 1,2 trichloroethylene. 

                                             

SOLUTION: In the Trans compound the C-Cl bond dipoles are equal and at the same time acting in opposition cancel each other while C is compound the dipoles do not cancel each other resulting in a higher value. Generally all Trans compounds have a lower dipole moment corresponding to C is isomer, when both the substituent’s attached to carbon atom 
are either electron releasing or electron withdrawing.

DIPOLE MOMENT IN AROMATIC RING SYSTEM:

To locate position of substituents in aromatic compounds

                         

( A)  If same substituents are present in the symmetrical position dipole moment (u) of benzene ring compounds will be zero.

     

(B) As angle between substituent’s decrease, value of dipole moment (u )increas.

CALCULATION AS BELOW:

                        

                  

ILLUSTRATIVE EXAMPLE (7):Out of Meta and ortho isomers 4-methylnitrobenzene which is having greater dipole moment?

SOLUTION: Para isomer is having highest dipole moment since two groups attached to benzene ring have dipole moment directed in the same direction thereby they reinforce one another in this case.

ILLUSTRATIVE EXAMPLE (8): The resultant dipole moment of water is 1.85 D ignoring the effects of lone pair. Calculate, the bond moment of each O-H bond (given that bond angle in H₂O = 104°, cos 104° = –0.25)

SOLUTION:

                                              R₂ = P₂ + Q₂ + 2PQ cos(theta)

SOME IMPORTANT ORDERS OF DIPOLE MOMENTS (u)

TO CALCULATE % IONIC CHARACTER: 

ILLUSTRATIVE EXAMPLE (9): Calculate the % of ionic character of a bond having length = 0.92 Å and 1.91 D as its observed dipole moment ?.

SOLUTION:   To calculate μ considering 100% ionic bond

                      = 4.8 × 10^–10 × 0.92 × 10^–8esu cm

                      = 4.8 × 0.83 × 10^–18 esu cm = 4.416 D

                         ∴ % ionic character = 1.91/4.416× 100 = 43.25                     

ILLUSTRATIVE EXAMPLE (10): 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

MULTIPLE CHOICE QUESTIONS(MCQ)

Q 1: Which of the following has been arranged in order of decreasing dipole moment?

Ans-(A)

Q 2: Which has maximum dipole moment?

Ans (A) Due to the symmetrical structure dipole moment of (C) & (D) are zero & (A) having maximum dipole moment.

EXERCISE:

Q. (1) the dipole moment of HBr is 7.95 Debye and the inter molecular separation is 1.94 x10-10 m Find the % ionic character in HBr molecule.

Q. (2) HBr has dipole moment 2.6x10-30 C-m. If the ionic character of the bond is 11.5%, calculate the inter atomic spacing.

Q.(3) Dipole moment of LiF was experimentally determined and was found to be 6.32 D. Calculate  percentage ionic character in LiF molecule Li--F bond length is 156 pm.

Q. (4) A diatomic molecule has a dipole moment of 1.2 D. If bond length is 1.0 Å, what percentage of an electronic charge exists on each atom.

 Ans:  (1).    85 %, (2)   1.4 Å, (3).     84.5% ,  (4).  25%   

                                                    

[5] POLYVALENT ION HYDROLYSIS:

(1) POLY ANION HYDROLYSIS:(Poly acidic base):

For example PO₄^-3 (n=3) , C₂O₄^-2 (n=2), S^-2 (n=2) , N^-3 (n=3) etc.

Na2S, NH3PO4, Na3PO4   (Hydrolysis is study just like hydrolysis of polybasic acids / polyacidic base) we know that dissociation constant of phosphoric acid is given as:

DISSOCIATION CONSTANT OF PHOSPHORIC ACID:

Step-1

Step-2

Step-3

HYDROLYSIS OF ANION (PO₄^-3):

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

Step-1

Experimentally we know that   Ka₁>>Ka₂>>Ka_3, hence x >>y>>Z   so y and z can be neglected with respect to x 

Special Case (1):  

Special Case (2)

          then x can be calculate by quadratic equation. 

Step-2

Experimentally we know that   Ka₁>>Ka₂>>Ka₃   hence   x>>y>>Z    So y and z can be neglected with respect to x and the x present in denominator and numerator both  are cancelled .

Step-3

Experimentally we know that   Ka₁>>Ka₂>>Ka₃   hence   x>> y>>Z  So y and z present in numerator can be neglected with respect to x and the z present in denominator is also neglected with respect to y.

Finally -  

And concentration of different species  

 (2) POLY CATION HYDROLYSIS: (Poly basic acid):
For example Ca⁺² , Fe⁺³ , Al⁺³ , Mn⁺² , Cd⁺²  , Zn⁺³  etc
ILLUSTRATIVE EXAMPLE:
Calculate pH and concentration of all species in 0.1 M solution of FeCl₃ given Fe(OH)₃ have  Kb₁=10^-3 , Kb₂=10^-7 and Kb₃=10^-12.

SOLUTION:

  Fe⁺³ àundergo hydrolysis while, Cl^-  à do not under goes hydrolysis.

   We know that 

Experimentally know that Kb₁>>Kb₂>>Kb₃      so x>>Y and Y>>Z

Hence y and z neglected with respect to x

HYDROLYSIS OF FeCl₃:

Experimentally we know that   Ka₁>>Ka₂>>Ka₃ ,   hence  x>>y>>Z    So  y and  z can be neglected with respect to x

TITRATION BASED POLYVALENT ION HYDROLYSIS:

(1) TITRATION OF Na₂CO₃ Vs HCl:

(2) TITRATION OF NA₃PO₄ Vs HCl:

(1) TITRATION OF Na₂CO₃ Vs HCl:

ILLUSTRATIVE EXAMPLE:
The following volume of 0.1M HCl Solution is added to the in20 ml 0.1M Na₂CO₃ Solution the pH of resulting solution in each case.
(H₂CO₃, Ka₁= 4×10^-6, Ka₂= 5×10^-11)
(1) 0.0 ml (No HCl added)
(2) 10 ml HCl is added
(3) 20 ml HCl is added
(4) 30 ml HCl is added
(5) 40 ml HCl is added 

                                                            

FACTER'S AFFECTING ON SOLUBILITY

TOPIC COVER:
(1) Effect of temperature on Solubility
(2) Effect of common ions on Solubility
(3) Effect of simultaneous Solubility
(4) Effect of solvent on Solubility
(5) Effect of pH on Solubility
(i) Effect of pH on Solubility of metal hydroxide
(ii) Effect of pH on Solubility of salt of weak acid
(iii) Effect of pH on Solubility of salt of strong acid
(6) Effect of buffer solution on Solubility
(7) Effect of complex  formation  on Solubility

(1) Effect of temperature on Solubility:
In general most cases Solubility increases on temperature . however we must follow two cases .
(1) In endothermic reaction Solubility Increases on increasing temperature.
(2) In exothermic reaction Solubility decrease on increasing temperature .

(2) Effect of common ions on Solubility:
(1) Solubility of a salt decrease in the presence of common ion.
(2) Higher the concentration of common ion smaller Solubility.
ILLUSTRATIVE EXAMPLE (1): Calculate the Solubility of AgCl in .. ( Kal AgCl is 10-10)
(1) In pure water
(2) In 0.1M NaCl
(3) In 0.01 M NaCl
(4) In 10-5 M NaCl
(5) In 0.05 M HgCl2
(6) In 0.01 M AgCl
(7) In 0.1M KNO3
SOLUTION :

ILLUSTRATIVE EXAMPLE (2): Arrange the following in increasing order of their Solubility.
(a) AgCl           Ksp = 2×10-10
(b) BaSO4       Ksp = 4×10-8
(C) CaF2.         Ksp = 1.08×10-10
(D) Hg2I2.       Ksp = 9×10-17
SOLUTION:

(3) Effect of simultaneous Solubility:
ILLUSTRATIVE EXAMPLE (1):
Calculate simultaneous Solubility of AgSCN and AgBr in water . ( Ksp AgSCN=10-12 Ksp AgBr =5×10-13)

ILLUSTRATIVE EXAMPLE (2):
CaCO3 and BaSO4 have Solubility products value 1.0×10-8 and 5.0× 10-9 respectively, if water is shaken up with both solids till equilibrium is reached calculate the concentration of CO3-2 ion is ?

ILLUSTRATIVE EXAMPLE (3):
The Ksp value of CaCO3 and CaC2O4 in water are 4.7 × 10-9 and 1.3 ×10-9, respectively , at  25°c .If mixture of two is washed with water , what is Ca2+ ion concentration in water ?
(Ans- 7.707×10-5 )
ILLUSTRATIVE EXAMPLE (4):

(4) Effect of solvent on Solubility:
Solubility of solute in solvent purely depends on nature of both solute and solvent , a polar solute dissolved in polar solvent and non polar solute dissolved in non polar solvent . A polar solute has low solubility or insoluble in a non polar solvent . For this reason if you want to decrease the Solubility of an inorganic salt (polar salt ) in water you mixed the water with an organic solvent (non polar )

PRIDICTING OF PRECIPITATION:
We know that:
IP= Ionic product  and Ksp= Solubility product
CASE(1): If IP < Ksp  then Solution is unsaturated
CASE(1):If IP > Ksp  then Solution is Oversaturated or ppt formation occurs.
CASE(1): If IP =Ksp  then Solution is saturated is No more solute dissolved.

ILLUSTRATIVE EXAMPLE (1):
A 200 ml  of 1.3×10-3 M AgNO3 is mixed with 100 ml of 4.5×10-5 M Na2S Solution will precipitatation occurs ?
(Ksp = 1.6×10-19) .

ILLUSTRATIVE EXAMPLE (2):
50ml of 6.9×10-3M CaCl2 mixed with 30 ml of  0.04 M NaF2. Will precipitatation of CaF2 occurs ?
( Ksp for CaF2= 4.0×10-11)
ILLUSTRATIVE EXAMPLE (3):
How much solid Pb(NO3)2 must be added to 1.0 L of 0.0010 M NaSO4 Solution for precipitatation of PbSO4 (Ksp=1.6×10-8) to form.
(assume no change in volume when the solid is added).

ILLUSTRATIVE EXAMPLE (4):
PbCl2 has Ksp=1.6×10-5 , If equal volume of 0.030 M PB(NO3)3 and 0.030M KCl are mixed , will precipitatation occurs ?
ILLUSTRATIVE EXAMPLE (5):

ILLUSTRATIVE EXAMPLE (6):

(5) Effect of pH on Solubility:
Many weak soluble ionic compound have Solubility which depends upon  the pH of the Solution for example metal hydroxide and  salt of weak acids.

(i) Effect of pH on Solubility of metal hydroxide
ILLUSTRATIVE EXAMPLE (1): Zince hydroxide ( Zn(OH)2 ) has Ksp 4.5×10-17 in pure water calculate it's molar Solubility and pH of resulting solution .
(Ans- S=2.2×10-6 M and pH = 8.6434 )
ILLUSTRATIVE EXAMPLE (2): At what pH the  Zince hydroxide will start precipitate (pHs) and at what pH precipitation is completed (pHc) from the Solution containing 0.1M Zn+2 ? ( Given Ksp of Zn(OH)2 is 4.5×10-17)
(Ans- pHs = 6.33 and pHc = 8.33 )

ILLUSTRATIVE EXAMPLE (3):  A solution containing 0.1 M Ca+2 and 0.02 M Mg+2 , is it possible to seperate one of these ions by precipitatating it as  hydroxide while keeping the other in Solution ?.
(Given Ksp of Ca (OH)2 is 5.5×10-6 and Ksp of Mg(OH)2 is 5.0×10-12)

(ii) Effect of pH on Solubility of salt of weak acid:
For salt of weak acids (eg Sulphides , Carbonate , Oxalates , and Phosphates ) the smaller the value of Ksp the lower the pH at which the salt precipitate (pH ---Ksp) exactly same as metal hydroxide .
It is noted that the salt having smaller Ksp , precipitate in more acidic medium and  other hand  the salts having Higher Ksp , precipitate in less acidic medium.
For example the precipitatation  of Ca +2 at low pH , CO3-2 will be turned to HCO3-1 Or  may be to H2CO3 , below at pH=< 8( see the Diagram)  while at pH >=13  all the carbonic acid species are present as CO3-2 , therefore CaCO3 will precipitate in basic medium and will dissolve in acidic medium .

ILLUSTRATIVE EXAMPLE (1):
Which one will precipitate in more acidic medium CaCO3 (Ksp=4.8×10-9) or MgSO4 (Ksp=1.0×10-5) ?

ILLUSTRATIVE EXAMPLE (2):
A solution containing 0.1 M Ti+ and 0.05 M Cd+2 . Is it possible to separate these two ions by precipitatating one of as Sulphides ?
( Ksp(CdS)=2.0×10-28 , Ksp(TiS)=2.0×10-22)

(iii) Effect of pH on Solubility of salt of strong acid :
Note that the pH has no effect on Solubility of  strong acids salts  eg Cl- Br- and SO4-2 etc because the concentration of of these conjugated base in the same either in acidic or basic medium . However metal ions can be separated by these anions  according to their Ksp value as the hydroxide or salts of weak acids.

ILLUSTRATIVE EXAMPLE (1):
Ca(OH)2 has Ksp = 7.9 ×10-6 , what is the pH of Solution made by equilibrating solid Ca(OH)2 with water ?
ILLUSTRATIVE EXAMPLE (2):
Cu(OH)2 has Ksp =1.6×10-19 calculate the
(1) What is the pH of a saturated solution of Cu(OH)2 ?
(2) What  is the maximum Cu+2 concentration possible in a  Neutral Solution ? (pH=7)
(3) What is maximum pH of Solution in which concentration of  Cu+2 is 0.50 M.

(6) Effect of buffer solution on Solubility:
Whenever a salt dissolved in a buffer solution it's pH remain same .
ILLUSTRATIVE EXAMPLE (1):
The Solubility of Pb(OH)2 in water =6.70×10-6. calculate it's Solubility in a buffer solution of pH=8 .   [ IIT-1999]
ILLUSTRATIVE EXAMPLE (2):
Calculate Solubility of AgCN in a buffer of pH=3 . ( Given Ksp=1.2×10-15 and Ka HCN is 4.8×10-10 ).
ILLUSTRATIVE EXAMPLE (3):
Calculate Solubility of AgCN in a buffer of pH=3 . ( Given Ksp=1.2×10-16 and Ka HCN is 4.8×10-10 ).
ILLUSTRATIVE EXAMPLE (4):
Calculate the molarity Solubility of Cu(OH)2 [Ksp=2.2×10-20] in
(1) Distilled water
(2) pH =13.0 NaOH (aq)
(3) pH= 4.0 buffer
ILLUSTRATIVE EXAMPLE (5):
ILLUSTRATIVE EXAMPLE (4):
Calculate Solubility of AgCN in a buffer of pH=3 . ( Given Ksp=8×10-10 and Ka HCN is 9×10-10 ).

(7) Effect of complex  formation  on Solubility:
The Solubility of many salt can be increased by addition of a species that can form complex ion with one of the ions (usually the cation ) formed when poorly soluble salt dissolved.
ILLUSTRATIVE EXAMPLE (1):
Find Solubility (s) of AgCl in 'C'M NH3 (aq) (given , Ksp of AgCl and Kf [Ag(NH3)2]+.
SOLUTION:
[Kf =complex formation constant=K solubility constant=1/K(insolubility)=1/K(dissociation)]

ILLUSTRATIVE EXAMPLE (2):
Calculate Solubility of  AgCl(s) in 2.7 M NH3(aq) solution . ( Ksp AgCl=10-10 and Kf[Ag(NH3)2]=1.6×10+7).
SOLUTION:
ILLUSTRATIVE EXAMPLE (3):
Calculate the Solubility of AgCN in 0.01M KCN Solution assuming
(1) No Complex formation
(2) Complex formation
(KspAgCN=8×10-8 and Kf[Ag(CAN)2]=4×10+7)

ILLUSTRATIVE EXAMPLE (4):
Find the Solubility of MX(s) ( Ksp=1.6×10-10) in 0.05M NaCN(aq).
(1/K dissociation [M(CN2)]=2.5×10-8 )

ILLUSTRATIVE EXAMPLE (5):
What is Molar solubility of AgBr in 0.1M NaS2O3 ? ( Given Ksp AgBr = 5×10-13 and Kf [Ag(S2O3)2]-3 =5×10+13)
ILLUSTRATIVE EXAMPLE (7):
Calculate the Molar Solubility of AgBr in 1.0 M NH3 at 25° ( Ksp AgBr =5×10-13 and Kf [Ag(NH3)2]+ =1.7×10+7)
ILLUSTRATIVE EXAMPLE (6):

ILLUSTRATIVE EXAMPLE (6):

Amphoteric salts hydrolysis:

Example of amphoteric salts  NaHS, NaHCO₃, Na₂HPO₄, NaH₂PO₄

(A) HCO3- act as conjugate acid as well as conjugate base:

Both reaction will support each other extent of hydrolysis and extent of dissociation is same.

(B) Here H₂PO₄^- and HPO₄^-2 are amphoteric anions. The pH of amphoteric salts anions is independent of concentration of salts.

Here HPO₄^-2 is conjugate base of H₂PO₄^- and H₃PO₄ is conjugate acid of H₂PO₄^-Similarly PO₄^-3 is conjugate base of HPO₄^-2 and H₂PO₄^-1 is conjugate  acid of HPO₄^-2

When these salts are dissolved in water [H₃O+] concentration can be determined as;

ILLUSTRATIVE EXAMPLE: Calculate pH of solution of

(1) 100 ml 0.1M H₃PO₄ + 100 ml 0.1M NaOH.

(2) 100 ml 0.1M H₃PO₄ + 200 ml 0.1M NaOH.

(3) 100 ml 0.1M H₃PO₄ + 300 ml 0.1M NaOH.

(4) 100 ml 0.1M H₃PO₄ + 400 ml 0.1M NaOH.