Cation exchange resins have large organic molecule with SO₃H group which are insoluble in water. Ion exchange resin (RSO₃H) is changed to RNa on treatment with NaCl. The resin exchange Na⁺ ions with Ca²⁺ and Mg²⁺ ions present in hard water and make it soft.
\(2RNa(s)+{ M }^{ 2+ }(aq)\longrightarrow { R }_{ 2 }M(s)+2{ Na }^{ + }(aq)\)
where, M = Mg, Ca.
The resins can be regenerated by adding aqueous NaCl solution.

Due to hyperconjugation, alkyl groups act as electron donors when attached to a \(\pi\)-system as shown below:

Large quantities of phosphates and nitrates increase the growth of phytoplanktons. These phytoplankton use so much oxygen that it is not sufficiently available for other organism to use in respiration. Moreover, a large consume the oxygen dissolved in water, Microorganisms may become so abundant that they from a mat covering on the water surface and thereby preventing photosynthesis. In all these processes concentration of dissolved oxygen in water decreases. When the concentration of dissolved oxygen of water is below 6 ppm, the growth of fish gets inhibited and they cannot survive.

The general formula of alkanes is C_nH_2n+2=12xn+1x(2n+2)=72 or 12n+2n+2=72 or n=5
Thus, the molecular formula of the alkane is C₅H₁₂. For structural isomers and their IUPAC names.

Producer gas is a mixture of CO and N₂, therefore, the tetravalent element is carbon and its monoxide and dioxide are CO and CO₂ respectively.
\(2C(s)+\underbrace { O_{ 2 }(s)+4N_{ 2 }(g) }_{ Air } \overset { 1273k }{ \rightarrow } \underbrace { 2CO(g)+4N_{ 2 }(g) }_{ Producer \ gas } \)
The carbon monoxide is a strong reducing agent and reduces ferric oxide to iron.
\(Fe_{ 2 }O_{ 3 }(s)+3CO(g)\overset { \Delta }{ \rightarrow } 2Fe(s)+3CO_{ 2 }(g)\)

\( According \ to \ hendrson's \ equation\)
\(PH={ Pk }_{ a }=log\cfrac { [salt] }{ [acid] } \)
\( pH=4.91,{ Pk }_{ a }=4.76\)
\(4.91=4.76+log\cfrac { [salt] }{ [acid] }\)
\(log\cfrac { [NaAc] }{ AcH } =antilog\)
\(V=\cfrac { 0.2\times 50 }{ 0.1\times 1.41 } =70.92ml\)
\( Volume \ of \ 0.1 \ m \ Acid \ required=70.92\)

\( V_{\text {initial }}=5 \mathrm{~L}, T=27^{\circ} \mathrm{C}=27+273 \mathrm{~K}=300 \mathrm{~K} \)
\( \mathrm{~V}_{\text {final }}=\frac{\mathrm{nRT}}{\mathrm{P}}=\frac{10 \times 0.0821 \times 300}{1.0}=246.3 \mathrm{~L} \)
\( \Delta \mathrm{V}=\Delta_{\text {firinal }}-\mathrm{V}_{\text {instial }}=246.3-5=241.3 \mathrm{~L} \)
\( \mathrm{w}_{\text {exp }}=-\mathrm{P} \Delta \mathrm{V}=-1 \times 241.3 \mathrm{~L} \text { atm }=-241.3 \times 101.3 \mathrm{~J} \)
\(=-24443.7 \mathrm{~J} \text {. } \)

volume of 1.00 m NaOH = 400 mL and mass and of NaCl produced = 23.4 g

(a) Moles of iron  =  \( \frac{ mass \ of \ iron }{atomic mass}\)
= \(\frac{7.85}{55.8}\)
= 0.141 mol
(b) Moles of silicon  = \(\frac{ mass \ of \ silicon}{atomic mass}\)
= \(\frac{4.68 \times10^{-3}}{28.1}    \)     
= 1.67\( \times\) 10^-4 mol
(c) Moles of carbon  =  \(\frac{mass \ of \ carbon }{atomic mass}\) 
=  \(\frac{65.6\times10^{-6}}{12}\)
= 5.47 \(\times\) 10^-6 mol

(a) It may be noted that for oxidation reactions, i.e., Eq. (i), the sign of the llectrode potential as given in  Table B.l is reversed. To get the equation for the overall reaction, the number of electrons lost in Eq. (i) and gained in Eq. (ii) must be cancelled. To do so, Eq. (ii) is multiplied by 2 and added to Eq. (i). Further, it may be noted that whenever any half reaction equation is multiplied by any integer, its electrode potential is not multiplied by that integer. Thus
Overall reaction: 2Fe³⁺(aq) + 2I^-(aq) \(\rightarrow\)  2Fe²⁺(aq) + 1₂(s);E^O= + 0.23 V
Since the EMF for the above reaction is positive, therefore, the above reaction is feasible.
(b) The possible reaction between Ag⁺(aq) and Cu(s) is
Cu(s) + 2Ag⁺(aq) \(\rightarrow\) Cu²⁺(aq) + 2Ag(s)
The above redox reaction can be split into the following two half reactions. Writing electrode potential for each half reaction from Table , we have
Oxidation Cu(s) \(\rightarrow\) Cu²⁺(aq) + 2e^-; E^o= -0.34 V
Reduction: Ag⁺(aq) + e^- \(\rightarrow\) Ag(s)] x 2; E^o= + 0.80 V
(c) Overall reaction: Cu(s) + 2Ag⁺(aq) \(\rightarrow\) Cu²⁺(aq) + 2Ag(s); po = +0.46 V
Since the EMF of the above reaction comes out to be positive, therefore, the above reaction is feasible.
(c) Suppose the reaction between Fe³⁺(aq) and Cu(s) occurs according to the following equation.
Cu(s) + 2Fe³⁺(aq) \(\rightarrow\) 3Cu²⁺(aq) + 2Fe²⁺(aq)
The above reaction can be split into the following two half reactions. Writing electrode potential for each half reaction from Table , we have,
Oxidation: Cu(s) \(\rightarrow\) Cu²⁺(aq) + 2e-; E^o = -0.34 V
Reduction: Fe³⁺(aq) + e^- \(\rightarrow\) Fe²⁺(aq)] x ²; E^o= +0.77 V
(d) Overall reaction: Cu(s) + 2Fe³⁺(aq) \(\rightarrow\) Cu²⁺(aq) + 2Fe²⁺(aq); E^o = +0.43 V
Since the EMF of the reaction is positive, therefore, the above reaction is feasible. Alternatively, if the reaction between Fe³⁺(aq) and Cu(s) occurs according to the following equation.
3Cu(s) + 2Fe³⁺(aq) \(\rightarrow\) 3Cu²⁺(aq) + 2Fe(s)
The EMF of the reaction comes out to be - Ie, i.e., -0.376 V (-0.34 V - 0.036 V) and hence this reaction is not feasible.
(d) Suppose the reaction between Ag(s) and Fe3+ (aq) occurs according to the following equation:
Ag(s) + Fe³ + (aq) \(\rightarrow\) Ag + (aq) + Fe²⁺(aq)
The above reaction can be split into the following two half reactions. Writing electrode potential for each half reaction from Table , we have,
Oxidation: Ag(s) \(\rightarrow\) Ag⁺(aq) + e^-; E^o = -0.80 V
Reduction: Fe + (aq) + e^- \(\rightarrow\) Fe²⁺(aq); E^o = +0.77 V
(e) Overall reaction: Ag(s) + Fe³⁺(aq) \(\rightarrow\) Ag⁺(aq) + Fe2+(aq); E^o= -0.03 V
Since the EMF of the reaction is negative, therefore, the above reaction is not feasible.
Alternatively, the reaction between Ag(s) and Fe³⁺(aq) may occur according to the following equation
3Ag(s) + Fe³⁺(aq) \(\rightarrow\) 3Ag⁺(aq) + Fe(s) 
On similar lines, we can calculate the e.m.f. of this reaction comes to be even more negative, i.e., -0.836 V, and hence this redox reaction is also not feasible.
(e) Suppose the reaction between Br₂(aq) and Fe²⁺(aq) occurs according to the following equation:
Br₂(aq) + 2Fe²⁺(aq) \(\rightarrow\) 2Br^-(aq) + 2Fe³⁺(aq)
The above reaction can be split into the following two half reactions. Writing electrode potential for each half reaction from the Table 8.1, we have
Oxidation: Fe₂₊(aq) \(\rightarrow\) Fe3+(aq) + e^-] x 2; E^o = -0.77 V
Reduction: Br₂(aq) + 2e^- \(\rightarrow\) 2Br-(aq); E^o = +1.09 V

Yes, polluted water is the water whose quality has been degraded by the addition of substance such as chemical effluents, metal residues, sewage, oil, fertilisers, detergents, etc.
It can be controlled by the following methods.
(i) Industrial waste discharge from paper, fertilisers, should not be allowed to get mixed in water bodies such as river, lakes, etc.
(ii) Non-biodegradable detergents should be used for cleaning of clothes
(iii) The ph of water should be checked.
(iv) Excessive use of fertilisers should be prevented.
(v) oils spills should be avoided as much as possible
(vi) Domestic waste water should be properly discharged and treated.
(vii) Avoid the use of DDT, malathion at home.
(viii) Waste water should be treated in sewage treatment plant.

Chlorophyll presents in green leafy vegotables, contains magnisium

The efficiency of fuel cells is 70% whereas other conventional cells are only 40% efficient.

(a) In acidic medium
\({ 2Fe }^{ 2+ }(aq)+2{ H }^{ + }(aq)+{ H }_{ 2 }{ O }_{ 2 }(aq)\rightarrow 2{ Fe }^{ 3+ }(aq)+2{ H }_{ 2 }O(l)\)
(b) In basic medium
\({ 2Fe }^{ 2+ }(aq)+{ H }_{ 2 }{ O }_{ 2 }(aq)\rightarrow 2{ Fe }^{ 3+ }(aq)+2{ OH }^{ - }(aq)\)

(Multiply \({ S }_{ 2 }{ O }_{ 3 }^{ 2- }\) by 2 because there are 4 S-atoms in \({ S }_{ 4 }{ O }_{ 6 }^{ 2- }\) ion). Increase and decrease in oxidation number is already balanced. Charge and oxygen atoms are also balanced. This represents a balanced redox reaction.

(a) \(\because { pK }_{ a }=-log{ K }_{ a }\)
\( \because 4.74=-log{ K }_{ a }\)
\(or \ { K }_{ a }= \ antilog \ \bar { 5 } .26=1.82\times { 10 }^{ -5 }\)
\( From \ \alpha =\sqrt { \frac { { K }_{ a } }{ C } }\)
\( \alpha =\sqrt { \frac { 1.82\times { 10 }^{ -5 } }{ 0.05 } } =\sqrt { 0.364\times { 10 }^{ -3 } } =\sqrt { 3.64\times { 10 }^{ -4 } }\)
\( \alpha =1.908\times { 10 }^{ -2 }\)
In the presence of 0.01 M H⁺
\({ CH }_{ 3 }COOH\rightleftharpoons { CH }_{ 3 }CO{ O }^{ - }+{ H }^{ + }\)
\( Initiaconc.\ 0.05M\ \quad 0 \quad 0\)
\( Equili.conc.0.05 \ -C\alpha \quad C\alpha \quad (C\alpha +0.01)\approx 0.01\)
\( \left[ { CH }_{ 3 }COOH \ is \ a \ weak \ acid \ and \ HCl \ is \ a \ strong \ acid,so \ we \ can \ assume \ that \ (C\alpha +0.01)\approx 0.01 \right] \)
\(\because \ { K }_{ a }=\frac { \left[ { CH }_{ 3 }CO{ O }^{ - } \right] \left[ { H }^{ + } \right] }{ \left[ { CH }_{ 3 }COOH \right] } \)
\(\therefore 1.82\times { 10 }^{ -5 }=\frac { C\alpha \times 0.01 }{ 0.05 }\)
\(or \ C\alpha =\frac { 1.82\times { 10 }^{ -5 }\times 0.05 }{ 0.01 }\)
\(or \ 0.05\times \alpha =9.1\times { 10 }^{ -5 }\)
\(\therefore \alpha =\frac { 9.1\times { 10 }^{ -5 } }{ 0.05 } =1.82\times { 10 }^{ -3 }\)
(b) \(\because { pK }_{ a }=-log{ K }_{ a }\)
\( \because 4.74=-log{ K }_{ a }\)
\( or \ { K }_{ a }= \ antilog \ \bar { 5 } .26=1.82\times { 10 }^{ -5 }\)
\(From \ \alpha =\sqrt { \frac { { K }_{ a } }{ C } }\)
\( \alpha =\sqrt { \frac { 1.82\times { 10 }^{ -5 } }{ 0.05 } } =\sqrt { 0.364\times { 10 }^{ -3 } } =\sqrt { 3.64\times { 10 }^{ -4 } } \)
\( \alpha =1.908\times { 10 }^{ -2 }\)
In the presence of 0.1 M HCl
\(Similarly,\ { K }_{ a }=\frac { \left[ { CH }_{ 3 }CO{ O }^{ - } \right] \left[ { H }^{ + } \right] }{ \left[ { CH }_{ 3 }COOH \right] } \)
\( 1.82\times { 10 }^{ -5 }=\frac { C\alpha \times 0.1 }{ 0.05 } \)
\( \left[ { H }^{ + } \right] =(C\alpha +0.1M)\approx 0.1M\)
\((because \ 0.1M \ HCl \ =0.1 \ M \ { H }^{ + }ions)\)
\(C\alpha =\frac { 1.82\times { 10 }^{ -5 }\times 0.05 }{ 0.1 } =0.91\times { 10 }^{ -5 }\)
\( \alpha =\frac { 0.91\times { 10 }^{ -5 } }{ 0.05 } =1.82\times { 10 }^{ -4 }\)
In the presence of strond acid, dissociation of weak acid i.e. CH₃COOH decreases due to common ion effect.

One decides the spontaneity of a reaction by considering
\(\triangle { S }_{ total }(\triangle { S }_{ sys }+\triangle { S }_{ surr })\). For calculating ∆Ssurr, we have to consider the heat absorbed by the surroundings which is equal to – \({ \triangle }_{ r }{ H }^{ \circleddash }\). At temperature T, entropy change of the surroundings is
\(\triangle { S }_{ surr }=\frac { { \triangle }_{ r }{ H }^{ \circleddash } }{ T } \)  (At constant pressure)
\( =\frac { (-1648\times { 10 }^{ 3 }J{ mol }^{ -1 } }{ 298K } \)
= 5530.20 JK^-1mol^-1
Thus, total entropy change for this reaction
\(\triangle { S }_{ total }=5530 \ J{ K }^{ -1 }{ mol }^{ -1 }+(-549.4 \ J{ K }^{ -1 }{ mol }^{ -1 })\)
\(=4980.6\ J{ K }^{ -1 }{ mol }^{ -1 }\)
This shows that the above reaction is spontaneous.

Initial condition     Final condition
Let,  V₁= V              \({ V }_{ 2 }=\frac { V }{ 2 } \)
p₁= 500 - 24 = 476 mm og Hg p₂ =?
According to Boyle's law (temperature is constant)
p₁V₁ = p₂V₂
\({ p }_{ 2 }=\frac { { p }_{ 2 }{ V }_{ 1 } }{ { V }_{ 2 } }\)  = \(\frac { 476XV }{ V/2 } m\)
= 2 x 476 mm of Hg
= 952 mm of Hg

In Bcl₃ molecule, B-atom acquire sp² hybridisation.

One s and one p-orbitals undergo hybridisation to form three equivalent sp² hybrid orbitals.
The three sp² hybrid orbitals are used to from \(\sigma \) -bonds with three Cl-atoms to give trigonal planar structure.

Anhydrous AlCl₃ exists in the form of dimer Al₂Cl₆. In Al₂Cl₆ each A1 atom is surround by four bond pairs. As a result they assumes tetrahedral disposition.

The applications of dipole moment are
(a) The dipole moment helps to predict whether a molecule is polar or non-polar. As μ = q × d greater is the magnitude of dipole moment, higher will be the polarity of the bond. For non-polar molecules, the dipole moment is zero.
(b) The percentage of ionic character can be calculated as Percentage of ionic character \(=\frac { { \mu }_{ observed } }{ { \mu }_{ ionic } } \times 100\)
(c) Symmetrical molecules have zero dipole moment although they have two or more polar bonds.
(d) It helps to distinguish between cis and trans-isomers. Usually cis-isomer has higher dipole moment than trans-isomer.
(e) It helps to distinguish between ortho, meta and para-isomers. Dipole moment of para-isomer is zero. Dipole moment of ortho-isomer is greater than that of.

(i) Potassium, K (Z = 19)
(ii) Strontium, Sr (Z = 38)
(iii) Ruthenium, Ru (Z = 44)
(iv) Praseodymium, Pr (Z = 59 ) 
(v) Xenon, Xe(Z = 54 )

Radius of nth orbit H like species, \({ r }_{ n }=\frac { 52.9({ n }^{ 2 }) }{ Z } pm\)
\({ r }_{ 1 }=1.3225nm=1322.5pm=\frac { 52.9{ n }_{ 1 }^{ 2 } }{ Z } \)
\( { r }_{ 2 }=211.6pm=\frac { 52.9{ n }_{ 2 }^{ 2 } }{ Z } \)
\(\frac { { r }_{ 1 } }{ { r }_{ 2 } } =\frac { 1322.5 }{ 211.6 } =\frac { { n }_{ 1 }^{ 2 } }{ { n }_{ 1 }^{ 2 } } \Rightarrow \frac { { n }_{ 1 }^{ 2 } }{ { n }_{ 1 }^{ 2 } } =6.25 \ or \ \frac { { n }_{ 1 } }{ { n }_{ 2 } } =2.5\)
If \({ n }_{ 1 }{ =5,n }_{ 2 }=2,\) then, the transition is from 5th orbit to 2nd orbit and it belongs to Balmer series.
\(\overline { v } =\frac { 1 }{ \lambda } =1.09677\times 10^{ 7 }\left( \frac { 1 }{ { n }_{ 1 }^{ 2 } } -\frac { 1 }{ { n }_{ 1 }^{ 2 } } \right) m^{ -1 }\)
\( \frac { 1 }{ \lambda } =1.09677\times 10^{ 7 }\left( \frac { 1 }{ { 2 }^{ 2 } } -\frac { 1 }{ 5^{ 2 } } \right) m^{ -1 }\)
\(\frac { 1 }{ \lambda } =1.09677\times 10^{ 7 }\times \frac { 21 }{ 100 } =2.303\times 10^{ -9 }m=434nm\)
It belongs to visible region.

Volume of a drop of water = 0.05 mL
Mass of a drop of water  = volume \(\times\) density
= ( 0.05 mL ) \(\times\)(1.0g / mL)
= 0.05 g
Gram molecular mass of water  ( H₂O ) = 2\(\times\)1 + 16 = 18 g ;
18 g of water = 1 mol
\(\therefore\) 0.05 g of water = \(\frac{1 mol}{(18 g)}\)\(\times\) ( 0.05 g )
= 0.0028 mol
\(\because\) 1 mole of water conmtains molecules = 6.022 \(\times\)10²³
0.0028 mole of water  will contain molecules
= 6.022 \(\times\)10²³\(\times\)0.0028 = 1.68 \(\times\)10²¹ molecules

They form blue coloured solution. The solution is paramagnetic in nature.

Reducing.

Monohydrate (Na₂CO₃·H₂O) is formed as a result of efflorescence.

The type of flow in which there is regular gradation of velocity in passing from one layer to the next is called laminar flow.

(i) The combining atomic orbitals should have comparable energies.
For example, Is orbital of one atom can combine with Is atomic orbital of another atom, 2s can combine with 2s.
(ii) The combining atomic orbitals must have proper orientations. So that they are able to overlap to a considerable extent.
(iii) The extent of overlapping should be large.

Molecular formula = (Empirical formula)_n
where n is positive integer.

For exhibiting cis-trans (or geometrical isomerism, a molecule must fulfil the following condition.
The groups attached to each double bonded carbon atom must be different.

Since, compound a of boron reacts with NMe₃ to form an adduct B. Thus, compound A is a Lewis acid. Since adduct B on hydrolysis gives an acid C and hydrogen gas, therefore, A is B₂H₆ and C is boric acid.
\(B_{ 2 }{ H }_{ 6 }+2NMe_{ 3 }\rightarrow 2BH_{ 3 }.NMe_{ 3 }\)
Diborate(A)           Adduct(B)
\(BH3.NMe_{ 3 }+3H_{ 2 }O\rightarrow H_{ 3 }BO_{ 3 }+NMe_{ 3 }+6{ H }_{ 2 }\)
                          Boric acid(C)

E^o_{AI³⁺ / AI} = -.166V, E^o_{Cu²⁺ / Cu} =+ 0.34 V
E_{Fe²⁺ / Fe} = - 0.44 V, E_{Mg²⁺ / Mg} = - 2.36 V
and  E^o_{Zn²⁺ / Zn} = - 0.76 V
A metal with more negative value of E^o_red is a stronger reducing agent than those which have less negative or positive value of E^o_red .Therefore, Mg can displace all the given metals from their aqueous salt solutions.AI can displace all metals from their aqueous salt solutions.Zinc can displace Fe and Cu from their aqueous salt solutions and Fe can displace only Cu from its aqueous salt solution. Hence, the orderin which they can display each other from the solution of their salts is as follows.
 Mg, AI, Zn, Fe, Cu

\(w=-2.303nRT \ log\frac { { V }_{ 2 } }{ { V }_{ 1 } } -4754\)
\(=-2.303\times n\times 8.314\times 300\times log\frac { 20 }{ 10 }\)
\(On \ solving, \ n=2.75mol\)

Given, wavelength, \(\lambda =242nm=242\times 10^{ -9 }m\)
Energy,
\(E=hv=\frac { hc }{ \lambda } =\frac { 6.626\times 10^{ -34 }Js\times 3.0\times { 10 }^{ 8 }ms^{ -1 } }{ 242\times { 10 }^{ -9 }m }\)
\(E=0.0821\times { 10 }^{ -17 }J/atom\)
This energy is sufficient for ionisation of one Na atom, so it is the ionisation energy of Na.
\(E=6.023\times { 10 }^{ 23 }\times 0.0821\times { 10 }^{ -17 }J/mol\)
\(\\ E=4.945\times { 10 }^{ 5 }J/mol\)
\(=4.945\times { 10 }^{ 2 }kJ/mol\)