i) In increasing order of solubility in water:
\(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2},\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{NH}, \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}  \)
\(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}>\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}>\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{NH}\)
ii) In increasing order of basic strength:
a. Aniline, p - toluidine and p - nitro aniline
p - toluidine > aniline >p - nitro aniline

 b.  \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}, \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NHCH}_{3}, \mathrm{p}-\mathrm{Cl}-\mathrm{C}_{6} \mathrm{H}_{4}-\mathrm{NH}_{2} \)
\(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NHCH}_{3}>\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}>\mathrm{p}-\mathrm{Cl}-\mathrm{C}_{6} \mathrm{H}_{4}-\mathrm{NH}_{2}\\ 2^{o} \text { amine } \quad \quad \quad \quad   e^{\ominus} \text { with drawing (group) }\)
(iii) In decreasing order of basic strength in gas phase:
\(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2},\left(\mathrm{C}_{2} \mathrm{H}_{5}\right) \mathrm{NH},\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{3} \mathrm{~N} \text { and } \mathrm{NH}_{3}  \)
\(\mathrm{NH}_{3}<\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}<\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{~N}<\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{3} \mathrm{NH}\\ \quad \quad \quad 1^{0} \text { amine } \quad 2^{0} \text { amine } \quad \quad \quad 3^{0} \text { amine }\)
(iv) In increasing order of boiling point:
\(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH},\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}, \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2} \)
\(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}>\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}>\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH} \\ 2^{0} \text { amine } \quad \quad1^{0} \text { amine }\)
Generally amines have lower boiling point than alcohol. Due to comparable molecular mass and weaker H-bonds in Amines.
(v) In decreasing order of the pK_b values:
\( \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}, \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NHCH}_{3},\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{NH} \text { and } \mathrm{CH}_{3} \mathrm{NH}_{2} \)
\(\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{NH}<\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}<\mathrm{CH}_{3} \mathrm{NH}_{2}<\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NHCH}_{3} \)
\(\quad \quad \quad 2^{0} \text { amine } \quad 1^{0} \text { amine } 1^{0} \text { amine } \quad 2^{0} \text { amine } \)
\(\mathrm{PK}_{b}: \quad 3.00\quad < \quad 3.29 \quad < \quad 3.38 \quad<\quad 9.30\)
pK_b ,Due to + 1 effect of C₂H₅ group. Higher the value of pK_b lower is the basicity
(vi) Increasing Order of basic strength:
\(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}, \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{~N}\left(\mathrm{CH}_{3}\right)_{2},\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{NH} \text { and } \mathrm{CH}_{3} \mathrm{NH}_{2} \)
\(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}>\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{~N}\left(\mathrm{CH}_{3}\right)_{2}>\mathrm{CH}_{3} \mathrm{NH}_{2}>\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{NH} \\ \left(\mathrm{pK}_{6}: 9.38 \quad > \quad \quad 8.92 \quad \quad > \quad 3.38 \quad > \quad \quad 3.00\right)\)
Due to + 1 effect of C₂H₅ group.
In decreasing order of basic strength:

(i) To understand how a soap works as a cleansing agent, let us consider sodium palmitate an example of a soap. The cleansing action of soap is directly related to the structure of carboxylate ions (palmitate ion) present in soap. The structure of palmitate exhibit dual polarity. The hydrocarbon portion is non polar and the carboxyl portion is polar.

(ii) The nonpolar portion is hydrophobic while the polar end is hydrophilic. The hydrophobic hydro carbon portion is soluble in oils and greases, but not in water. The hydrophilic carboxylate group is soluble in water.
(iii) The dirt in the cloth is due to the presence of dust particles intact or grease which stick. When the soap is added to an oily or greasy part of the cloth, the hydrocarbon part of the soap dissolve in the grease, leaving the negatively charged carboxylate end exposed on the grease surface.
(iv) At the same time the negatively charged carboxylate groups are strongly attracted by water, thus leading to the formation of small droplets called micelles and grease is floated away from the solid object. When the water is rinsed away, the grease goes with it. As a result, the cloth gets free from dirt and the droplets are washed away with water. The micelles do not combine into large drops because their surfaces are all negatively charged and repel each other. The cleansing ability of a soap depends upon its tendency to act as a emulsifying agent between water and water insoluble greases.

The cleansing action of detergents are similar to cleansing action of soap. Eg: the structure of a cationic detergents is:

(i) Collision theory is based on the kinetic theory of gases. According to this theory, a chemical reaction occurs as a result of collisions between the reacting molecules.
(ii) Let us understand this theory by considering the following reaction.
A₂(g) + B₂(g) ⟶ 2AB(g)
(iii) If we consider that, the reaction between A₂ and B₂ molecules proceeds through collisions between them, then the rate would be proportional to the number of collisions per second.
(iv) Rate ∝ number of molecules colliding per litre per second (collision rate).
(v) The number of collisions is directly proportional to the concentration of both A₂ and B₂.
Collison rate ∝ [A₂][B₂]
Collision rate = Z [A₂][B₂]
(vi) Where, Z is a constant
(vii) A fraction of effective collisions (f) is given by the following expression
\(f={ e }^{ \frac { { -E }_{ a } }{ RT } }\)
(xiv) This fraction of collisions is further reduced due to orientation factor i.e., even if the reactant collides with sufficient energy, they will not react unless the orientation of the reactant molecules is suitable for the formation of the transition state.
(viii) The diagram illustrates the importance of proper alignment of molecules which leads to reaction.

(ix) The fraction of effective collisions (f) having proper orientation is given by the steric factor p.
⇒ Rate = p x f x collision rate
\(\Rightarrow Rate=p\times { e }^{ \frac { -Ea }{ RT } }\times Z\left[ { A }_{ 2 } \right] \left[ { B }_{ 2 } \right] \quad ...(1)\)
As per the rate law,
Rate = \(k=\left[ { A }_{ 2 } \right] \left[ { B }_{ 2 } \right] \quad ...(2)\)
Where k is the rate constant
On comparing equation (1) and (2), the rate constant k is
\(k=pZ{ e }^{ \frac { -Ea }{ RT } }\)

Metal excess defect:
(i) It arises due to the presence of more number of metal ions as compared to anions.
(ii) Examples: NaCl, KCl
(iii) The electrical neutrality of the crystal can be maintained by the presence of anionic vacancies equal to the presence of extra cation.

(iii) For example, when NaCI crystals are heated in the presence of sodium vapour, Na⁺ ions are formed and are deposited on the surface of the crystal.
(iv) Chloride ions (Cl^-) diffuse to the surface from the lattice point and combines with Na⁺ ion.
(v) The electron lost by the sodium vapour diffuse into the vacancy created by the Cl^- ions.
(vi) Such anionic vacancies which are occupied by unpaired electrons are called F centers. Hence, the formula of NaCl can be written as Na_1+xCl.
Metal deficiency defect:

(i) Metal deficiency defect arises due to the presence of less number of cations than the anions. This defect is observed in a crystal in which, the cations have variable oxidation states.
(ii) For example, In FeO crystal, some of the Fe²⁺ ions are missing from the crystal lattice. To maintain the electrical neutrality, twice the number of other Fe²⁺ ions in the crystal is oxidized to Fe³⁺ ions. In such cases, overall number of Fe²⁺ and Fe³⁺ ions is less than the O^2- ions.

Zone refining :
1. Zone refining method is based on the principles of fractional crystallisation.
2. When an impure metal is melted and allowed to solidify, the impurities will prefer to be in the molten region. In this process the impure metal is taken in the form of a rod.
3. One end of the rod is heated using a mobile induction heater which results in melting of the metal on that portion of the rod.
4. When the heater is slowly moved to the other end the pure metal crystallises while the impurities will move on to the adjacent molten zone.
5. As the heater moves further away, the molten zone containing impurities also moves along with it.
6. The process is repeated several times by moving the heater in the same direction again and again to get pure metal.
7. This process is carried out in an inert gas atmosphere to prevent the oxidation of metals.
8. Elements such as germanium (Ge), silicon (Si) and galium (Ga) that are used as semiconductor are refined using this process.

(ii) benzoic acid

(iii) α - hydroxyphenylaceticacid.

pH=-log[H⁺]
For weak acids,
\(= \sqrt{k_a \times C}\)
=\(\sqrt{1.8\times10^{-5}\times0.1}\)
=\(1.34 \times10^{-3}\) M
\(pH=-\log(1.34\times10^{-3})\)
= 3-log1.34
= 3-0.1271
= 2.8729 \(\simeq\) 2.87

The oxidation potential of M₁ is more +ve than the oxidation potential of Fe which indicates that it will prevent iron from rusting.

Electronic configuration of Ti = 3d²4s²
Electronic configuration of Ti³⁺ =3d¹
Hence number of unpaired electron = 1
Spin only magnetic moment \((\mu)=\sqrt{\mathrm{n}(\mathrm{n}+2)}\)
= \(\sqrt{1(1+2)} \)
= \(\sqrt{3}\) 
=1.732 BM
Electronic configuration of \(\mathrm{Mn}=3 \mathrm{~d}^{5} 4 \mathrm{~s}^{2}\)
Electronic configuration of \(\mathrm{Mn}^{2+}=3 \mathrm{~d}^{5}\)
Hence number of unpaired electrons = 5
Spin only magnetic moment 
\((\mu) =\sqrt{5(5+2)}\)
= 5.92 BM

The energy of the two e_g orbitals will increase by  \(\frac{3}{5} \Delta_{O}\) and that of the three t_2g will decrease by (2/5) \(\Delta_{O}\)

Argon prevents the oxidation of hot filament and prolongs the life in filament bulbs.

Linear arrangement of spheres in one direction is repeated in two dimension (i.e.) more number of rows can be generated identical to the one dimensional arrangement such that all spheres of different rows align vertically as well as horizontal.
If we denote the first row as A type arrangement, then the above mentioned packing is called AAA type, because all rows are identical as the first one. In this arrangement each sphere is in contact with four of its neighbours.

Chlorine is prepared by the action of conc. sulphuric acid on chlorides in presence of manganese dioxide
4NaCl + MnO₂ + 4H₂SO₄ \(\longrightarrow \)Cl₂+ MnCl₂ +4NaHSO₄ + 2H₂O

(i) In diborane two BH₂ units are linked by two bridged hydrogens.
(ii) It has eight B-H bonds.
(iii) Diborane has only 12 valance electrons.
(iv) The four terminal B-H- bonds is "2c - 2e" bond (two centre - two electron bond.)
(v) Two three centred B - H - B bonds two electrons each. "(3c - 2e)"
(vi) In diborane, the boron is "sp³" hybridised
(vii) Three of the four "sp³" hydridised orbitals contains single electron and the fourth orbital is empty.

(i) E is monobasic acid (RCOOH) having molecular weight 60 and it is formed from D on oxidation. So E must be acetic acid and D must be acetaldehyde.
(ii) (B) on oxidation gives CH₃CHO. So (B) must be alcohol (CH₃CH₂OH).
(iii) Acetaldehyde (D) on treating with aqueous alkali (NaOH) gives aldol which on heating gives 2- butenal (E).

(iv) Compound E on catalytic hydrogenation gives butyl alcohol.

(v) Hence compound (A) must be an ester. Ester (A) on reduction with LiAIH₄ yields two alcohols (B) and (C). (A) is ethyl butyrate.

'A' can also be CH₃COOCH₂CH₂CH₂CH₃. This structure will be answering all the above reactions.

(i) In lyophillic colloids or sols definite attractive force or affinity exists between dispersion medium and dispersed phase. Examples: sols of protein and starch. They are more stable and will not get precipitated easily. 
(ii) In a lyophobic colloids, no attractive force exists between the dispersed phase and dispersion medium. They are less stable and precipitated readily, but cannot be produced again by just adding the dispersion medium. 
Examples: sols of gold, silver, platinum and copper.

\({ CH }_{ 3 }-{ CH }_{ 2 }-OH\overset { { H }_{ 2 }{ SO }_{ 4 } }{ \underset { 443k }{ \longrightarrow } } { CH }_{ 2 }={ CH }_{ 2 }+{ H }_{ 2 }O\)
Mechanism
Primary alcohols undergo dehydration by E₂ mechanism

(a) If we apply DC current through the conductivity cell, it will lead to the electrolysis of the solution taken in the cell.
(b) So, AC current is used for this measurement to prevent electrolysis.

i) Reducing sugars:
1. Sugars which reduce Tollen's reagent or Fehling's solution or Benedict's solution are called reducing sugars.
2. These contain either α - hydroxyl ketone or cyclical hemi acetal or hemi ketal or structures in equilibrium with open chain forms having a free- CHO or C=O group.
3. E.g. a) All monosaccharide's like D - glucose, D - fructose (aldoses and ketoses)
b) Sugars like Lactose and maltose except sucrose.
ii) Non - reducing sugars:
1. Sugars which do not reduce either Tollen's reagent, Fehling's solution or Benedict's solution are called non-reducing sugars.
2. They contain a stable acetal or ketal structures which cannot be opened into a free carbonyl group.
E.g. Sucrose, starch, cellulose, glycogen, dextrin etc.

Considering Ba(OH)₂ to be a strong base:
\(\text { Normality } =\text { Molarity } \times \text { Acidity }  \)
\(=1.5 \times 10^{-3} \times 2\)
\({\left[\mathrm{OH}^{-}\right] } =3 \times 10^{-3}  \)
\(\mathrm{pOH} =-\log _{10}[\mathrm{OH}^-]  \)
\(=-\log _{10}\left(3 \times 10^{-3}\right)  \)
\(=-\left[\log _{10} 3+3 \log 10\right]  \)
= 3-log 3 
= 3-0.4771 
= 2.5229 
\(\mathrm{pH} =14-\mathrm{pOH} \)
= 14-2.5229 
pH = 11.4771 = 11.48

Given t_1/2= 0.6392 \(\times\)10⁴ s
To solve: when t = 100 min,
\(\frac { [{ A }_{ 0 }]-[A] }{ [{ A }_{ 0 }] } \times 100=?\)
We know that
For a first order reaction, \({ t }_{ 1/2 }=\frac { 0.6932 }{ k } \)
\(k=\frac { 0.6932 }{ 6.932\times { 10 }^{ 4 } } \)
\(k={ 10 }^{ -5 }{ s }^{ -1 }\)
\(k=\left( \frac { 1 }{ t } \right) In\left( \frac { [{ A }_{ 0 }] }{ [{ A }] } \right) \)
\({ 10 }^{ -5 }{ s }^{ -1 }\times 100\times 60s=In\left( \frac { [{ A }_{ 0 }] }{ [{ A }] } \right) \)
\(0.06-In\ \left( \frac { [{ A }_{ 0 }] }{ [{ A }] } \right) \)
\(\frac { [{ A }_{ 0 }] }{ [{ A }] } ={ e }^{ 0.06 }\)   ?(given : e^0.06 = 1.06)
\(\frac { [{ A }_{ 0 }] }{ [{ A }] } =1.06\)
\(\therefore \frac { [{ A }_{ 0 }]-[{ A }] }{ [{ A }_{ 0 }] } \times 100\%\)
\(=\left( 1-\frac { [{ A }] }{ [{ A }_{ 0 }] } \right) \times 100\%\)
=\(\left( 1-\frac { 1 }{ 1.06 } \times 100\% \right) \)  
= 5.66%

Mn³⁺ has large and negative standard electrode potential E⁰ (-1.18 V) than that of Cr²⁺ which has only -0.91 V. If the standard electrode potential of a metal is large and negative, the metal is a powerful reducing agent because it loses electrons easily. Hence Mn³⁺ is strongly oxidizing while Cr²⁺ is strongly reducing.