DRAGO’S RULE:NO NEED OF HYBRIDIZATION:

Keeping surrounding atom same if electronegativity of central atom increases bond angle increases.

According to Drag’s rule six molecules (PH₃, AsH₃, SbH₃ H₂S, H₂Se and H₂Te) are no Hybridisation having on their central atom.

If central atom is in third row of or below in periodic table then lone pair will occupy in stereo-chemically inactive S-orbital and bonding will be through p-orbital and near 90⁰ angles. If electronegativity of surrounding atom is less than or equal < 2.5.

ORBITAL ANALYSIS: 

Hence (theta) is bond angle between equivalent orbital % S and P-character is given by S and P

Note:

1- It is applicable of them is % S + % P = 100%

2- All the orbital must not be involve

3- It is applicable when

ILLUSTRATIVE EXAMPLE: Phosphine (PH₃) :  

About 600 k j mole^-1 more energy is required so hybridized the central phosphorous (P) atom. due to this energy factor more stable arrangement is, when pure orbital’s are involved in bonding and lone pair present into pure s-orbital.

ILLUSTRATIVE QUESTIONS:

Q:1 Explain why NH₃ is more soluble in water than PH₃?

Q:2 The formation of PH₄⁺ is very difficult but the formation of NH₄⁺ is easier than PH₄⁺ ?

Q:3 NH₃ is stronger Lewis base  in comparison PH₃?

Q:4 The complex formation Tendency of NH₃ is wisher than PH₃.

SOLUTION: In NH₃ lane pair is present in one of the sp³ Hybridized orbital. While in PH₃ lone pair is present in pure S orbital and hence lane pair donation capacity of NH₃ is stronger than PH₃

Paragraph for question nos. 5 & 6:

Drago suggested an empirical rule which is compatible with the energetic of hybridisation. It states that if the central atom is in the third row or below in the periodic table, the lone pair will occupy a stereo chemically inactive s-orbital, and the bonding will be through p-orbitals and bond angles with be nearly 90º if the electronegativity of the surrounding atom is  2.5.

Question (5): In which of the following molecule central atom has higher % s-character in its bond pair-

(A) AsH₃                    (B) GeH₄                  (C) P₄                 (D) H₂Se

Question (6): Correct order of bond angle is-

(A) PH₄ ⁺ > OF₂ > SF₂> SbH₃ > H₂Te     (B) OF₂ > SF₂ > PH₄ > SbH₃ > H₂Te

(C) PH₄ ⁺ > SF₂ > OF₂ > SbH₃ > H₂Te    (D) SF₂ > OF₂ > PH₄ ⁺ > SbH₃ > H₂Te

Paragraph for question nos. 7 & 8:

The more ionic character (i.e. poorer covalency) in a bond (due to the more electronegative substituent to the central atom) leads to the utilization of hybrid orbitals containing more p-character (i.e. less s-character) of the central atom. In other words, the multiple bonds encourage the central atom to utilize its hybrid orbitals which contains more s-character.

The percentage of s- or p-character between two adjacent and equivalent hybrid orbitals can determine from the knowledge of the corresponding bond angle () as follows:

Question (7): If x₁, x₂ and x₃ are S – S bond lengths in S₂O₄^-2, S₂O₅^-2 and S₂O₆^-2 respectively, then correct order for S – S bond length is -

(A) x₃ > x₂ > x₁          (B) x₁ > x₂ > x₃          (C) x₃ > x₁ > x₂        (D) x₁ > x₃ > x₂

Question (8): The percentage of p-character (approx) in hybrid orbital having the lone pair at central atom in SF₄ molecule will be:

 (Given F –S- F equatorial bond angle is 102º and cos102º = – 0.21)

(A) 65.5                    (B) 34.4                     (C) 50.5                        (D) 82.6

  

Paragraph for question nos. 9 & 10:

In all expected compounds each case central atom only uses its s and p orbitals in hybridisation. The relation between bond angle theta and decimal fraction of s and p character present in the equivalent hybrid orbitals is given by:

Here theta is bond angle between equivalent hybrid orbital (S and P)

Question (9): The correct order of % P character in bond pairs of central atoms in the following compounds:

(A) P>T>S>Q>R      (B) S>R>T>P>Q   (C) P>Q>S>R>T   (D) P>Q>S>T>R

Question (10): If Value of n is 2 for compound T, then number of lone pair present at central atom of compound T will be:

(A) 0                             (B) 1                              (C) 2                           (D) 3

Question (11): The correct statement is :

(A) The ratio of % p character to % s character is less than four, for the bond pair of central atom of compound S  

 (B) Central atom uses three hybrid orbitals to form compound R

 (C) Central atom uses four hybrid orbitals to form compound S

 (D) There are three compounds present between point C to E. according to % s character  in bond pair of central atom.

Answers Key:

TITRATIONS OF DIPROTIC ACID WITH STRONG BASE:

DIPROTIC ACID  Vs NaOH:  [H₂CO₃ and Oxalic acid H₂A]

ILLUSTRATIVE EXAMPLE: Give the answers of following questions when 100 ml of Malonic acid is titrated with 0.10 M NaOH the (Given that Ka₁=1.5×10^-3 and Ka₂=2.0×10^-6 for Malonic acid HOOC-CH₂-COOH represented as H₂A ).

(A) Write out the reactions and equilibrium expression associated with Ka₁ and Ka₂.

(B) Calculate the PH when:

(1)100 ml of 0.10M H₂A + 0.0 ml of 0.10 ml NaOH 

(2) 100 ml of 0.10M H₂A+ 50 ml of 0.10 ml NaOH
(3) 100 ml of 0.10M H₂A+ 100 ml of 0.10 ml NaOH
(4) 100 ml of 0.10M H₂A+ 150 ml of 0.10 ml NaOH
(5) 100 ml of 0.10M H₂A + 200 ml of 0.10 ml NaOH

(C) Sketch the titration curve for this titration.

(A) Write out the reactions and equilibrium expression associated with Ka₁ and Ka₂
SOLUTION:

(B) Calculate the PH when:

SOLUTION:

 (1) At Point-(1):100 ml of 0.10 M H₂A + 0.0 ml of 0.10 ml NaOH 

Note: Ignore the amount of [H+] coming from 2^nd dissociation of acid 

(2) At Point-(2): 100 ml of 0.10 M H₂A + 50 ml of 0.10 ml NaOH 

At Point -2- is the half of the first equivalent point   where [H₂A=HA^-]   hence pH of solution is due to best buffer

(3) At Point-(3):100 ml of 0.10M H₂A + 100 ml of 0.10 ml NaOH]

At point-3- 1^st equivalent point occurs and here major species is HA^- which is act as both acid and base (Amphoteric species).

(4) At Point-(4): 100 ml of 0.10 M H₂A + 150 ml of 0.10 ml NaOH

At Point -4- is the half of the 2^nd equivalent point   where [HA^- = A^-2]   hence pH of solution is due to best buffer .

(5) At Point-(5):100 ml of 0.10 M H₂A + 200 ml of 0.10 ml NaOH

At point-5- It is 2^nd equivalent point there is only A^-2 present which act as weak base and undergo polyvalent anion  hydrolysis;

(C) Sketch the titration curve for the titration point of question (B) titration.

S.N.	Given condition	comments	PH
1	100 ml of 0.1M H2A + 0.0 ml of 0.1 ml NaOH	H2A Diprotic acid	1.94
2	100 ml of 0.1M H2A + 50 ml of 0.1 ml NaOH	Acidic Best buffer PH= Pka1 (1st half E-Point)	2.82
3	100 ml of 0.1M H2A + 100 ml of 0.1 ml NaOH	Amphoteric salt Hydrolysis PH=1/2(Pka1+Pka2)	4.26
4	100 ml of 0.1M H2A + 150 ml of 0.1 ml NaOH	Acidic Best buffer PH= Pka2 (2st half E-Point)	5.70
5	100 ml of 0.1M H2A + 200 ml of 0.1 ml NaOH	Anionic salt Hydrolysis	9.11

Graphical representation: 

TITRATION OF TRIPROTIC ACID WITH STRONG BASE:

ACID-BASE TITRATION:

THEORY OF INDICATOR'S:

It is method to determine concentration of any Solution with the help of a solution of known concentration.
EQUIVALENT POINT:
When the complete reaction occurs between the solutions, it is called equivalent point of titration.
END POINT: 
When sudden change in colour of solution occurs, it is called end point of titration.
INDICATORS:
(1) They are very weak organic acids or bases that show different colours at different pH value.
(2) They exist in ionized and unionized form which have different colours .
(3) The point at which Number of equivalent of acid and base become equal is known as equivalent point / Stoichiometric point / neutralization point.
(4) We assure that indicators do affect the pH of Solution.
(5) Around the equivalent point pH change the drastic change.
(6) For the successful titration end point should be as close as possible to the equivalent point.
TYPE OF INDICATOR'S:
(1) ACIDIC INDICATOR'S
(2) BASIC INDICATOR'S

(1) ACIDIC INDICATOR'S:

Phenophthalene (HPh): It is weak acid and can represented as [HPh] If it ionised gives [H⁺ ] and [Ph^- ] 

Addition of [H⁺]: In addition of an acid the ionization of HPh is practically negligible and as the equilibrium shift left hand side due to high concentration of [H⁺] ion thus solution would remain colourless.

Addition of [OH^- ]: In the presence of a base H⁺]^[ ions  are removed by [OH^-] ions in the in the form of water molecules and the above equilibrium shift to right hand side .Thus the concentration of Ph- ions increases in solution and they impart a pink colours to the solution.

INDICATOR THEORY: Let consider acidic indicators [HPh] Phenophthalene. An indicator has two colouring parts.

(1) Unionized part of indicators 

(2) Ionized part of indicators

The relative concentration of these species will depend upon PH of medium .

Take negative log both sides

CASE (1): It is observed that 

Then colour of indicator decided by concentration of [Ph^-)]

CASE (2): It is observed that 

Then colour of indicator decided by concentration of [HPh]

PH RANGE OF INDICATORS:

The given indicators work between a pH range i.e.

Then working of indicators is best;

ILLUSTRATIVE EXAMPLE (1):  For an acidic indicator, dissociation constant, K_in is 2×10^-6 Calculate pH range of indicator.

SOLUTION:  

ILLUSTRATIVE EXAMPLE (2): The pH range of a basic indicator is 4 to 6.5 Calculate the dissociation constant of indicator?.

SOLUTION:  pK_In must be midpoint of pH range for acidic indicators and pOH range for basic indicators

The pH range = 4 to 6.5 so pOH range is 10 to 7.5

Hence Pk_In =   (10+7.5) /2 = 8.75 

ILLUSTRATIVE EXAMPLE (3):  For an indicator pKa is 6 Calculate pH of Solution having this indicator such that 40% indicator molecules remain in ionised form.

SOLUTION:  We know that  

(2) BASIC INDICATOR'S

Methyl orange (MeOH): It is weak base and can represented an MeOH If it ionised gives [ Me⁺] and[ OH^- ]  and methyl orange is an intensely coloured indicator that is red below pH 3.1 and orange-yellow above pH 4.4

The red (acid) form has an [H+] attached to one of the N atoms and the yellow (basic) form has lost the [H+]

Addition of [H⁺ ] ^: In the presence of an acid, OH^- ions are removed in the form of water molecules and the above equilibrium shift to right hand side .This effect Me⁺ ions are produced which impart red coloure to the Solution.

Addition of [OH^- ]^: In addition of alkali the concentration of OH^- increases in the solution and equilibrium shift left hand side i.e. the ionization of MeOH is practically negligible .thus solution acquired the colour of unionised methyl orange molecules i.e. yellow

DEGREE OF DISSOCISTION OF INDICATOR'S (DOD): Consider a general indicator dissociation;

ILLUSTRATIVE EXAMPLE (4):  For acidic indicator, pH range is 3 to 4.6 calculate the ratio of[ In^- ] and H⁺ for the appearance of solution in a single colour.

SOLUTION:

Given pH range 3.0 to 4.6 so pK_In= (3.0+ 4.6)/2 =3.6

ILLUSTRATIVE EXAMPLE (5): An indicator with Ka = 10^-5 is solution with pH = 6 Calculate % of indicator in ionised form?

SOLUTION:

ILLUSTRATIVE EXAMPLE (6): The pH of at which an acid indicator with Ka is 10^-15 changes colour when indicator concentration 1×10^-5 M is? 

SOLUTION:

TYPE OF TITRATION:

(A) ACID-BASE TITRATION: 

(A-1): Strong acid Vs Strong base:

(A-2): Weak acid Vs Strong base:
(A-3): Weak base Vs Strong base:
(A-4): Weak acid Vs Weak base: weak acid and weak base titration can not be carried out because due to very low PH change , their is no suitable indicator for this titration.
(A-5); Salt of SB and WA Vs Strong acid:

(A-6); Double Indicators Titration:

(A-7): Back Titration:
(B):TITRATION OF POLYPROTIC ACIDS:
(1) Titration of diprotic acid with strong base:
(2) Titration of triprotic acid with strong base:

(C) REDOX TITRATION:

PERCENTAGE (%) AVAILABLE CHLORINE IN BLEACHING POWDER:

Bleaching powder is not a true compound but it is a mixed salt of calcium hypochlorite [Ca (OCl) ₂].3H₂O and basic calcium chloride [CaCl₂Ca (OH) ₂H₂O]. [Simply it is represented by CaOCl₂.
The bleaching action of [CaOCl₂] is due to liberation of oxygen with limited quantity of dilute acid.                                                                                                        

Whereas disinfectant action available of Cl₂ on reaction with excess of acid.

The Chlorine liberates is called available Chlorine, and calculation of available Chlorine is called IODOMETRY.

IODOMETRIC-CALCULATION OF AVAILABLE % CHLORINE: Take a W (gm) sample of bleaching powder; and  when bleaching powder treated with dilute acid or water then liberates chlorine gas.

The chlorine produced in above reaction is titrated with KI solution which produced (I₂) Iodine which is further completely titrated with hypo solution.                                          

If number of millimoles of hypo solution consume is M V then millimoles of %available chlorine is calculated as ;

                            Millimoles of Iodine is =1/2 Millimoles Hypo solution

                            Millimoles of Iodine is = Millimoles chlorine

Weight of available chlorine is = Number of moles x molecular Wt of Chlorine 

ILLUSTRATIVE EXAMPLE(1): If trace of chlorine are not remove from pulp used for paper manufacturing , then on the long standing it weaken the paper and makes it yellowish , which of following antichlor can be used to remove chlorine from pulp ?.
(A) Na₂S₂O₃        (B) conc. HCl            (C) NaHCO₃           (D) Both (A) and (B)

SOLUTION:

An antichlor  is a substance used to decompose residual hypochlorite or chlorine after use of chlorine based bleaching , in order to prevent ongoing reactions .example s of antichlor – Sodiumbisulphite (NaHSO₃), Pottassiumbisulphite (KHSO₃) ,  Sodiummetasulphite (Na₂S₂O₃), Sodiumthiosulphate  (Na₂S₂O₃) and hydrogen peroxide (H₂O₂).

ILLUSTRATIVE EXAMPLE (2): Available chlorine in sample of bleaching powder can be calculated as per the following reactions.

If 4 gm of bleaching powder dissolved to give 100 ml solution 25 ml of it react with excess of CH₃COOH and KI .The Iodine liberation required 10 ml of 0.125 N Hypo solutions. Calculate % of available chlorine in the sample.
(A)  21%                       (B) 35 %                     (C) 45 %             (D) 4.4%

SOLUTION:

ILLUSTRATIVE EXAMPLE (3): Calculate the % of available chlorine in the sample in a sample of 3.35 gm of bleaching powder which was dissolved to 100 ml water. 25 ml of this solution on treatment react with KI and dilute acid. Required  20 ml 0.125 N Hypo solution (Sodiumthiosulphate- Na₂S₂O₃). 

SOLUTION:

ILLUSTRATIVE EXAMPLE (4): 25 ml of household bleach solution was mixed with 30 ml of 0.50 M KI and 10 ml of 4.0N acetic acid. In this titration of the liberated iodine 48 ml of 0.25 N Na₂S₂O₃ was used to reach the end point. The Molarity of household bleach solution is?
SOLUTION: