Wurtz synthesis: Higher alkanes are prepared by heating an alkyl halide (RX) with sodium metal in dry ether solution.
R - X + 2Na + XR \(\overset { ether }{ \longrightarrow } \) R - R + 2NaX
CH₃Br + 2Na + BrCH_{3\(\longrightarrow \)} CH₃-CH₃ + 2NaBr
Limitations: Use of two different alkyl halides in Wurtz reaction always leads to a mixture of alkanes. The separation of these alkanes is difficult because there is only a little difference in their boiling points. Thus only symmetrical alkanes can be prepared by this method.

(a) When organic compound is fused with sodium metal and then extracted by water, it is called Lassaigne's extract. Yes.
(b) Blood red colour.
(c) Lassaigne's extract is prepared in distilled water since tap water contains Cl^- ions. No, oxygen cannot be detected by Lassaigne's test.

(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

(a) Conc. H₂SO₄ cannot be used because it acts as an oxidizing agent also and gets reduced to SO_2.
Zn + 2H₂SO₄ (Conc.) \(\rightarrow\) znSO₄ + 2H₂O + SO₂
Pure Zn is not used because it is non-porous and reaction will be slow. The impurities in Zn help in constitute of electrochemical couple and speed up reaction.
(b) water is amphoteric in nature and it behaves both as an acid as well as base. With acids stronger than itself (e.g., H₂S) it behaves as a base and with bases stronger than itself (e.g., NH₃) it acts as an acid.
(i) As a base: H₂O(1) + H₂S(aq) \(\rightarrow\) H₃O(aq) + HS^-(aq)
(ii) As an acid: H₂O(1) + NH₃(aq) \(\rightarrow\) OH^-{aq) + \(NH^+_4\)(aq)
(c) The decomposition of H₂O₂ occurs readily in the presence of rough surface (acting as catalyst). It is also decomposed by exposure of light. Therefore, wax-lined smooth surface and coloured bottles retard the decomposition of H₂O₂.

CaCO₃ (s)\(\overset { heat }{ \longrightarrow } \)Cao + CO₂
NH₃ + H₂O \(\longrightarrow \)NH₄⁺ + OH^-
NaCl + NH₄OH + CO₂\(\longrightarrow \)NaHCO₃ + NH₄Cl
2NaHCO₃ \(\overset { heat }{ \longrightarrow } \)Na₂CO₃ + CO₂ + H₂O
Na₂CO₃ + 10H₂O\(\longrightarrow \)Na₂CO₃ + 10H₂O

(a) Being acidic B₂O_3,Sio₂ and CO₂ react with alkalis to form salts
\(B_{ 2 }O_{ 3 }+2NaOH\rightarrow 2NaBO_{ 2 }+H_{ 2 }O\\ Boric\quad \quad \quad \quad \quad \quad sodium\quad metaborate\\ anhydride\)
\(SiO_{ 2 }+2NaOH\underrightarrow { D } Na_{ 2 }CO_{ 3 }+H_{ 2 }O\)
\(\\ Silica\quad \quad \quad \quad \quad \quad \quad \quad \quad Sodium\quad carbonate\\ CO_{ 2 }+2NaOH\rightarrow Na_{ 2 }CO_{ 3 }+H_{ 2 }O\\ Carbon\quad dioxide\quad \quad \quad sodium\quad carbonate\quad \quad \quad \quad \)
(b) Being amphotric, Al₂O₃ and PbO₂ react with both acids and bases.
\(Al_{ 2 }O_{ 3 }+3H_{ 2 }SO_{ 4 }\rightarrow Al_{ 2 }(SO)_{ 4 }+3H_{ 2 }O\)
\( Al_{ 2 }O_{ 3 }+2NaOH\overset { fuse }{ \rightarrow } \ 2NaAlO_{ 2 }+H_{ 2 }O\)
\(\\ Alumina\quad \quad \quad \quad \quad \quad \quad sodium\quad metaaluminate\\ 2PbO_{ 2 }+2H_{ 2 }SO_{ 4 }\rightarrow 2PbSO_{ 4 }+2H_{ 2 }O+O_{ 2 }\)
\(\\ PbO_{ 2 }+2NaOH\rightarrow Na_{ 2 }PbO_{ 3 }+H_{ 2 }O\\ Lead \ dioxide\quad \quad Sodium \ plumbate\)
(c) Being basic,Tl₂O₃ reacts with acid
\(Tl_{ 2 }O_{ 3 }+6HCl\rightarrow 2TICl_{ 3 }+3H_{ 2 }O\)

Diamond has a three-dimensional network with strong C---C bonds, which are very difficult to break and thus, diamond has high melting point.

It is unsuitable for laundry as it does not form enough lather with soap
It is harmful to boilers due to deposits of salts in the form of scale which reduces the efficiency of the boilers.

\({ C }_{ 6 }{ H }_{ 5 }{ NH }_{ 2 }+{ H }_{ 2 }O\rightleftharpoons { C }_{ 6 }{ H }_{ 5 }{ NH }_{ 3 }^{ + }+{ OH }^{ - }\)
\({ K }_{ b }=\frac { \left[ { C }_{ 6 }{ H }_{ 5 }{ NH }_{ 3 }^{ + } \right] \left[ { OH }^{ - } \right] }{ \left[ { C }_{ 6 }{ H }_{ 5 }{ NH }_{ 2 } \right] } =\frac { \left[ { OH }^{ - } \right] ^{ 2 } }{ \left[ { C }_{ 6 }{ H }_{ 5 }{ NH }_{ 2 } \right] } \)
\(\left[ { OH }^{ - } \right] =\sqrt { { K }_{ a }.C } =\sqrt { 4.27\times { 10 }^{ -10 }\times 0.001 }\)
\( \left[ { OH }^{ - } \right] =6.534\times { 10 }^{ -7 }\)
\(pOH=-log\left[ { 0H }^{ - } \right] =-log\left[ 6.534\times { 10 }^{ -7 } \right]\)
\( pOH=-0.8152+7=6.18\)
\(From,\ pH+pOH=14\)
\( pH=14-6.18=7.82\)
\({ C }_{ 6 }{ H }_{ 5 }{ NH }_{ 2 }+{ H }_{ 2 }O\rightleftharpoons { C }_{ 6 }{ H }_{ 5 }{ NH }_{ 3 }^{ + }+{ OH }^{ - }\)
\(Initial \ conc.\ C \ \ \quad 0\quad 0\)
\( Equili \ conc.C-C\alpha \quad C\alpha \quad C\alpha\)
\( { K }_{ b }=\frac { C\alpha .C\alpha }{ C(1+\alpha ) } \ [(1-\alpha )\approx 1 \ for \ weak \ base]\)
\( { K }_{ b }=C{ \alpha }^{ 2 }\)
\( or \ \alpha =\sqrt { \frac { { K }_{ b } }{ C } } \)
Degree of ionisation,
\(\alpha =\sqrt { \frac { 4.27\times { 10 }^{ -10 } }{ 0.001 } } =6.53\times { 10 }^{ -4 }\)
\( { K }_{ a } \ of \ conjugate \ acid \ of \ aniline,\)
\( { K }_{ a }=\frac { { K }_{ w } }{ { K }_{ b } } =\frac { { 10 }^{ -14 } }{ 4.27\times { 10 }^{ -10 } } =2.34\times { 10 }^{ -5 }\)

\({ N }_{ 2 }\left( g \right) +3{ H }_{ 2 }\left( g \right) \rightleftharpoons 2{ NH }_{ 3 }\left( g \right) \)
\({ Q }_{ c }=\frac { { { \left[ { NH }_{ 3 } \right] }^{ 2 } } }{ { \left[ { N }_{ 2 } \right] \left[ { H }_{ 2 } \right] }^{ 3 } } \)
Given, \(\left[ { NH }_{ 3 } \right] =\frac { 8.13 }{ 20 } M=0.4065M\)
\(\left[ { N }_{ 2 } \right] =\frac { 1.57 }{ 20 } M=0.0785M\)
\(\left[ { H }_{ 2 } \right] =\frac { 1.92 }{ 20 } M=0.096M\)
\({ Q }_{ c }=\frac { { { \left[ 0.4065M \right] }^{ 2 } } }{ { \left[ 0.0785M \right] \left[ 0.096M \right] }^{ 3 } } =2.379\times { 10 }^{ 3 }{ M }^{ -2 }\)
\({ Q }_{ c }\neq { K }_{ c }\) so the reaction mixture is not in equilibrium.
\({ Q }_{ c }>{ K }_{ c }\) it indicates that the reaction will proceed in the direction of reactants i.e. the reverse direction.

(i) The second law of thermodynamics states that complete conversion of energy of one kind into another is not possible as some energy is always lost in the form of some other energy.
(ii) Because some energy may be used up in overcoming friction of wheels.
(iii) In refrigerator, heat flows from higher temperature to lower temperature in a compressor with mechanical work which is done by the compressor on the system.
The system consists of four components: compressor, condenser, expansion, value and evaporator. the following cycle is repeated over and again.
Refrigerant \(\overset { compressed }{ \underset { in \ compressor }{ \longrightarrow } } \) Gas form \(\overset { condensed }{ \underset { in \ condenser }{ \longrightarrow } } \)
Liquid form \(\overset { Allowed }{ \underset { to \ throttle \ through \ expansion \ valve }{ \longrightarrow } } \)
Gas form \(\overset { Allowed }{ \underset { to \ evaporate \ in \ evaporator }{ \longrightarrow } } \)Refrigerant. 
(iv) Mr.sharma is scientific and practical

100g of glucose is equivalent to 1560 kJ of energy 
Energy utilised in the event = \(\frac { 1560\times 50 }{ 100 } \) = 780
Energy left unutilised = 1560 - 780 = 780 kJ
Enthalpy of vaporisation of water = 44kJ/mol = \(\frac { 44 }{ 18 } \)kJ/g
Water needed to perspire =\(\frac { 44 }{ 18 } \) x 780 = 1906.66g

To calculate the empty space, let us first find the total volume of 1 mole\((6.02\times { 10 }^{ 23 })\)  molecules) of N₂
Volume \(6.02\times { 10 }^{ 23 }\) of molecules 
\({ N }_{ 2 }=3.35\times { 10 }^{ -23 }\times 6.02\times { 10 }^{ 23 }=20.17{ cm }^{ 3 }\)
Now, volume occupied by 1 mole of gas at STP
= 22.4 litre = 22400 cm³
Empty volume = Total volume of gas-volume occupied by molecules
\(=(22400-20.17){ cm }^{ 3 }=22379.83{ cm }^{ 3 }\)
Percentage empty space
=\(\frac { empty \ space }{ total \ volume } \times 100\)
\(=\frac { 22379.83 }{ 22400 } \times 100=99.9%\)
Thus, 99.9% of space of 1 mole N₂ at STP is empty.

From ideal gas equation we have,
\(pV=nRT \ or \ p=\frac { m }{ v } \times \frac { RT }{ M } =\frac { dRT }{ M } \)
\( d \ =\frac { pM }{ RT } \)
\( or \ d\alpha \frac { p }{ T } \ [R \ and \ M \ constant \ for \ a \ given \ gas]\)
\( \frac { { d }_{ 1 } }{ { d }_{ 2 } } =\frac { { p }_{ 1 } }{ { p }_{ 2 } } \times \frac { { T }_{ 2 } }{ { T }_{ 1 } } or \ { d }_{ 2 }=\frac { { d }_{ 1 }\times { p }_{ 2 }\times { T }_{ 1 } }{ { p }_{ 1 }\times { T }_{ 2 } }\)
\( { d }_{ 1 }=6.46 \ g \ { dm }^{ -3 },\ { p }_{ 1 }=5bar,\)
\({ T }_{ 1 }=273+25=298k\)
\( { d }_{ 2 }=?,\ { p }_{ 2 }=1 \ bar,{ T }_{ 2 }=27k\)
\( { d }_{ 2 }=\frac { 6.46g{ dm }^{ -3 }\times 1 \ bar\times 298k }{ 5 \ bar\times 273k }\)
\( =1.41 \ g{ dm }^{ -3 }\)

The amount of energy released for the conversion of 1million, Le.1 x 10⁶ atoms of iodine is 4.9 x 10^-13 J According to the reaction,
\(\mathrm{I}(\mathrm{g})+e^{-} \longrightarrow \mathrm{I}^{-}(\mathrm{g})\)
The amount of energy released for the conversion of one mole (6.02 x 10²³) of atoms of iodine into 1 ions can be calculated. This corresponds to electron gain enthalpy. Thus, amount of energy released for1 x10⁶ atoms of iodine
\(=4.9 \times 10^{-13} \mathrm{~J}\)
Amount of energy released for 6.02 x 10²³ atoms of iodine
\(=\frac{4.9 \times 10^{-13}}{1 \times 10^{6}} \times 6.02 \times 10^{23}=29.5 \times 10^{4} \mathrm{~J}\)
= 295kJ/mol
Now, leV/atom = 96.3k] mol^-1
∴ EIectron gain enthaIpy = \(-295 \mathrm{~kJ} \mathrm{~mol}^{-1}=\frac{295}{96.3}=-3.06\) eV/atom

Given \(\Delta x\) = 0.0002 nm = 2.00 x 10^-12 m
From Heisenberg;s uncertainty principle,
\(\Delta xX\Delta p\ge \frac { h }{ 4\pi } \)
\( \Delta p=\frac { h }{ 4\pi \Delta x } =\frac { 6.6\times10^{ -34 } \ Kg \ m^{ 2 }s^{ -1 } }{ 4\times3.14\times0.2.00\times10^{ -12 }m } \)
\( \Delta p=2.638\times10^{ -23 } \ kg \ ms^{ -1 }\)
Actual momentum = \(\frac { h }{ 4\pi \times0.05 \ nm } \)
=\(\frac { 6.6\times10^{ -34 } \ Kg \ m^{ 2 }s^{ -1 } }{ 4\times3.14\times0.2.00\times10^{ -19 }m } \)
= 1.055 x 10^-24 kg ms^-1
It cannot be defined as the actual value of momentum is smaller than uncertainty.

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

A cation is highly polarising if its charge/size ratio is very high. Li⁺ion has the highest polarising power.

\(Na_{ 2 }B_{ 4 }{ O }_{ 7 }+2HCl+5{ H }_{ 2 }O \ \rightarrow \ 2NaCl+4H_{ 3 }BO_{ 3 }\)
Borax(A)                                               Orthoboric acid(X)
\(H_{ 3 }BO_{ 3 } \ \underrightarrow { \Delta ,370K } \ HBO_{ 2 }+{ H }_{ 2 }O\)
(X)                         Metaboric acid
\(4 \ HBO_{ 2 }\xrightarrow [ \Delta ,>370K ]{ -H_{ 2 }O } [{ H }_{ 2 }B_{ 4 }{ O }_{ 7 }]\underrightarrow { Red \ heat } \ 2{ B }_{ 2 }{ O }_{ 3 }+H_{ 2 }O\)
                             Tetraboric acid            Boron trioxide

In order to solve such problems, following steps are involved.
Step I: Set up a balance sheet using symbols for the expected equilibrium amount of reactants and products.
\(2NO+Br_{ 2 } \longrightarrow 2NOBr\) 
Initial moles : 0.087   0.0437   0 (as no reaction occur)
At equilibrium moles : (0.087-2x) (0.0437-x)  2x
[Remember! multiply the value of x with the stiochiome3tric coefficient of the molecules].
Step II: Compare with the given information
Given equilibrium moles of NOBr = 0.0518
\(\therefore \ 2x=0.0518 \ and \ x=\frac { 0.0518 }{ 2 } =0.0259\) 
Step III: Put the value of x to obtained equilibrium concentration of other species.
Moles of NO at equilibrium = (0.087 - 2x)
= 0.087 - 0.0518 = 0.0352 mol  
Moles of Br₂ at equilibrium = (0.0437 - x)
= 0.0437 - 0.0259 = 0.0178 mol

(i) We know that \(w=-2.303nRTlog\frac { { V }_{ 2 } }{ { V }_{ 1 } } \)
\(10\times 10^{ 3 }J=2.303\times 1\times 8.314\times T\times log\frac { 10{ V }_{ 1 } }{ { V }_{ 1 } } \)
\(or \ T=522.3K\)
\(For \ initial \ conditions,\ { p }_{ 1 }{ V }_{ 1 }={ n }_{ 1 }RT,\)
\( i.e \ (10^{ 7 }Pa){ V }_{ 1 }=1\times 8.314\times 522.3\)
\(or \ { V }_{ 1 }=4.342\times 10^{ -4 }m^{ 3 }=4.342\times 10^{ 2 }cm^{ 3 }\)
\( =434.2cm^{ 3 }\)
We cannot apply the formula \(-w=p\triangle V\) because expansion is not against constant pressure.)
(ii) If there were 2 moles of the gas, applying \({ p }_{ 1 }{ V }_{ 1 }={ n }_{ 1 }RT,\) we get \(({ 10 }^{ 7 }Pa)(4.342\times 10^{ -4 }m^{ 3 })=2\times 8.314\times T \ or \ T=261.1K,\) i.e. half of the first value.

Number of H₂ molecules = 4.879 x 10¹⁷

When n = 5, l = 0, 1, 2, 3. The order in which the energy of the available orbitals 4d, 5s and 5p increases is 5s < 4d < 5p. The total number of orbitals available are 9. The maximum number of electrons that can be accommodated is 18; and therefore 18 elements are there in the 5^th period.

Isoelectronic species have the same number of electrons but different atomic numbers. Number of positive charge shows the number of electrons lost and number of negative charges shows the number of electrons gained by an atom. Calculation of number of electrons have been shown below.
\(_{ 11 }Na^{ + }=11-1=10{ e }^{ - }, \ _{ 19 }K^{ + }=19-1=18{ e }^{ - },\)
\(_{ 12 }Na^{ 2+ }=12-2=10{ e }^{ - }, \ _{ 20 }Na^{ 2+ }=20-2=18{ e }^{ - },\)
\(_{ 16 }S^{ 2- }=16+2=18{ e }^{ - }, \ _{ 18 }Ar=18{ e }^{ - }\)
Hence, isoelectronic species are
\(Na^{ + }and \ Mg^{ 2+ } \ { k }^{ + },{ Ca }^{ 2+ },{ s }^{ 2- } \ and \ Ar\)

CH₃CH₂C(CH₃) = CHCH₃

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.

On adding dilute H₂ SO₄ for testing halogens in an organic compound with AgNO₃ white precipate of Ag₂SO₄ Mmay be mistaken for white precipate of chlorine as AgCI.Hence dilute HNO₃ is used instead of dilute H₂SO₄.

Since pH of solution A = 6.Hence, [H⁺] = 10^-6 mol L^-1
pH of solution B = 4.Hence ,[H⁺] = 10^-4 mol L^-1
On mixing 1L of each solution, molar concentration of total H⁺ is halved.
Total, [H⁺] = \(\frac { { 10 }^{ -6 }{ +10 }^{ -4 } }{ 2 } \)  mol L^-1
[H⁺] = 5.05 x 10^-5 mol L^-1
pH = log[H⁺]
pH = -log(5.0 x 10^-5) =  4.2967 \(\approx \)  4.3

-75 KJ

For ideal gas,compressibility factor, Z = 1