Praganya Prakashan

Bihar Board Class-12th

Chemistry Model Paper 1

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Section A

Objective Type Questions

In the following Question Nos. 1 to 35 there is only one correct answer against each question. For each question, mark the correct option on the answer sheet. 3 35 1 5# =

MCQ 1.1

Faraday’s law of electrolysis is related to(a) Atomic number of cation(b) Speed of cation(c) Speed of anion(d) Equivalent weight of element

Ans (d) Equivalent weight of element

EXPLANATION :

According to Faraday’s law,Mass of any element, w , \ Charge Q .i.e., for 1 mol of charge, if given to any element, 1 equivalent weight of that substance is obtained. Hence Faraday’s law of electrolysis is related with the equivalent weight of electrolytes.

MCQ 1.2

The shape of XeF4 is(a) tetrahedral (b) square planar(c) pyramidal (d) linear

Ans (b) square planar

EXPLANATION :

In XeF4, Xe has eight electrons in its valency-shell; out of which four electrons will form four single σ ^ h bonds with four fluorine-atoms, while remaining four electrons exist as two lone pair of electrons over xenon atom.Thus structure of XeF4 is:

To minimize the repulsion, the structure of XeF4 is square-planar.

MCQ 1.3

The number of P O P− − bond in cyclic metaphosphoric acid is(a) two (b) zero(c) three (d) four

Ans (c) three

EXPLANATION :

Cyliemetaphophsic acid has the formula and structure:

Thus, cyclic metaphosphoric acid has three ( )3 P O P− − − bonds in its structure.

MCQ 1.4

H PO3 3 is a(a) monobasic acid (b) dibasic acid(c) tribasic acid (d) none of these

Ans (b) dibasic acid

EXPLANATION :

H PO3 3 is a dibasic acid because H PO3 3 has three hydrogen atoms only two of these three hydrogen atoms are joined through the oxygen atoms and are ionisable. The third H-atom is directly linked to P-atom and is not ionisable In oxy-acid of P, the H atom joined to P atom confers reducing character and is not ionisable as a proton.

MCQ 1.5

Brown ring test is used to detect(a) Iodine (b) Nitrate(c) Iron (d) Bromide

Ans (b) Nitrate

EXPLANATION :

Brown ring test is done for confirmation of NO3− ions.

( ) ( ) ( ) ( )NaNO H SO NaHSO HNOaq aq aq aq3 2 4 4 3$+ +

( )FeSO HNO H SO Fe SO NO H O6 2 3 3 2 44 3 2 4 2 4 3 2$+ + + +

[ ( )]FeSO NO Fe NO SO

( )Ferrous nitroso sulphate

Brown ring

4 4$+−

MCQ 1.6

Williamson synthesis is used to prepare(a) Alcohols (b) Amines(c) Aldehydes (d) Ethers

Ans (d) Ethers

EXPLANATION :

The reaction of alkyl halides with sodium alkoxide or sodium phenoxide to form ethers is called Williamson synthesis. For example.

R X R ONa.Alkyl halide Sod alkoxide

− + −l R O R NaXEther

$ − − +l

MCQ 1.7

Which of the following oxides will be the least acidic?(a) As O4 6 (b) As O14 0

(c) P O4 10 (d) P O4 6

Ans (a) As O4 6

EXPLANATION :

As the Oxidation Number of the central atom of the compounds increases acidic atrength of that compound also increases and on moving from top to bottom in groups acidic strength of oxides also decrease due to decreasing electronegativity in groups.

P O P O As O As O> > >4

5

10 4

3

6 4

5

10 4

3

6

+ + + +

MCQ 1.8

Which of the following concentration unit is independent of temperature?(a) Normality (b) Molarity(c) Formality (d) Molality

Ans (d) Molality

EXPLANATION :

Volume is temperature dependent, hence expression involving volume term (normality, molarity and formality) varies with temperature.

Normality ..

vol of solution in Lof soluteEq=

Molarity .vol of solution in Lmoles of solute=

Formality .vol of solution in Lformula mass=

Molality mass of solvent in Kgmoles of solute=

Since, molality does not include the volume term, it is independent of temperature.

MCQ 1.9

Which of the following is the most basic oxide?(a) Sb O2 3 (b) Bi O2 3

(c) SeO2 (d) Al O2 3

Ans (b) Bi O2 3

EXPLANATION :

More the oxidation state (O.S.) of the central atom (metal) more is its acidity. Hence, ( . . )SeO O S of Se 42 =+ is acidic. Further for a given O.S., the basic character of the oxides increases with the increasing size of the central atom. Thus, Al O2 3 and Sb O2 3 are amphoteric and Bi O2 3 is basic.

MCQ 1.10

The IUPAC name of K PtCl2 66 @ is(a) hexachloroplatinate potassium(b) potassium hexachloroplatinate (IV)(c) potassium hexachloroplatinate(d) potassium hexachloroplatinum (IV)

Ans (b) potassium hexachloroplatinate (IV)

EXPLANATION :

K PtCl2 66 @ Potassium hexachloroplatinate (IV)Oxidation state of Pt is 4+ in the complex and anion is present in form of complex.

MCQ 1.11

Acetaldehyde reacts with(a) Electrophiles only(b) Nucleophiles only(c) Free radicals only(d) Both electrophiles and nucleophiles

Ans (b) Nucleophiles only

EXPLANATION :

Acetaldehyde reacts only with nucleophiles. Since, the mobile p electrons of carbon-oxygen double bond are strongly pulled towards oxygen, carbonyl carbon is electron-deficient and carbonyl oxygen is electron-rich. The electron deficient (acidic) carbonyl carbon is most susceptible to attack by electron rich nucleophilic reagents, that is, by base. Hence, the typical reaction of aldehydes and ketones is nucleophilic addition.

MCQ 1.12

If the rate of the reaction is equal to the rate constant, the order of the reaction is(a) 3 (b) 0(c) 1 (d) 2

Ans (b) 0

EXPLANATION :

Since, r [ ]k A n=

If, n 0=

r [ ]k A 0=

or r k=Thus, for zero order reactions rate is equal to the rate constant.

MCQ 1.13

Which of the following forms a colloidal solution in water?(a) NaCl (b) Glucose(c) Starch (d) Barium nitrate

Ans (c) Starch

EXPLANATION :

Starch molecules have colloidal dimensions whereas NaCl, glucose and ( )Ba NO3 2 are crystalloids and soluble in water.

MCQ 1.14

The unit of rate constant for a zero order reaction is(a) mol L s1 1− − (b) L mol s1 1− −

(c) L mol s2 2 1− − (d) s 1−

Ans (a) mol L s1 1− −

EXPLANATION :

Rate ( )K A 0=

Unit of K mol L s1 1= − −

MCQ 1.15

Aluminium displaces hydrogen from acids but copper does not. A galvanic cell prepared by combining /Cu Cu2+ and /Al Al3+ has an e.m.f of 2.0 V at °K298 . If the potential of copper electrode is +0.34 V, that of aluminium is (a) +1.66 (b) -1.66(c) +2.34 (d) -2.3 V

Ans (b) -1.66

EXPLANATION :

The given statement explains that Al is placed above Cu in E.C.S. hence it can reduce Cu from its salt solution (reduction). Thus Al acts as a reducing agent now,Electrode potential of copper,

( )E°cell . V0 34=+

e.m.f of cell V2=We know that,emf of a galvanic cell ( )E°cell

= Electrode potential of Cu (cathode)

−electrode potential of Al (anode)

.2 0 . E0 34 °Al= −

E°Al . .0 34 2 00= −

. V1 66=−

MCQ 1.16

Dry distillation of calcium formate gives(a) HCHO(b) HCOOH(c) CH COOH3

(d) CH CHO3

Ans (a) HCHO

EXPLANATION :

Calcium formate COOH Ca2^ h8 B gives form aldehyde on dry

distillation.

MCQ 1.17

Correct name of K Fe CN4 6^ h6 @ is(a) Potassium ferricyanide(b) Potassium ferrocyanide(c) Potassium hexacyanoferrate (II)(d) Potassium hexacyanoferrate (III)

Ans (c) Potassium hexacyanoferrate (II)

EXPLANATION :

Oxidation state of iron is 2+ in K Fe CN4 6^ h6 @

So, its correct name is potassium hexacyanoferrate (II).

MCQ 1.18

On mixing concentration NH OH4 to Cu2+ salt, the following blue complex is formed

(a) Cu NH4 42+

^ h6 @ (b) Cu NH3 22+

^ h6 @

(c) Cu NH3 42+

^ h6 @ (d) Cu NH4 22+

^ h6 @

Ans (c) Cu NH3 42+

^ h6 @

EXPLANATION :

On mixing concentration NH OH4 to a Cu2+ salt, theNH OH Cu Cu OH

.blue ppt4

22$+ +

^ h Cu NHNH OH

dark blue solution3 4

24 +^ h6 @

MCQ 1.19

Which alkyl halide follows only SN2 hydrolysis mechanism?(Chlorobenzene give DDT when it reacts with(a) charcoal (b) chloral(c) naphthalene (d) benzenoid

Ans (b) chloral

EXPLANATION :

DDT is synthesised by heating a mixture. of chloral (1 mol) with chlorobenzene (2 mol) in the presence of concentrated H SO2 4.

MCQ 1.20

The reaction A B$ follows first order kinetics. The time taken for 0.8 mole of A to produce 0.6 mole of B is 1 hour. What is the time taken for conversion of 0.9 mole of A to produce 0.675 mole of B?(a) 2 hours (b) 1 hour(c) 0.5 hour (d) 0.25 hour

Ans (b) 1 hour

EXPLANATION :

A B$ for a first order reaction.Given,

a .0 8= mol, ( ) . . .a x 0 8 0 6 0 2− = − =

k ...log1

2 3030 20 8=

or k . log2 303 4=

Again, a .0 9=

( )a x− . . .0 9 0 675 0 225= − =

k ...logt

2 3030 2250 9=

. log2 303 4 . logt2 303 4=

MCQ 1.21

One desires to prepare a positively charged sol of silver iodide. This can be achieved by(a) adding small amount of AgNO3 solution to KI solution in

slight excess(b) adding small amount of KI solution to AgNO3 solution in

slight excess(c) mixing equal volumes of equimolar solutions of AgNO3 and

KI(d) none of these

Ans (b) adding small amount of KI solution to AgNO3 solution in slight excess

EXPLANATION :

KI AgNO3+ (slight excess) AgI KNO3$ +

AgNO3 Ag NO3$ ++ −

( )AgI Ags + + [ ]AgI Ag$ +

MCQ 1.22

CuSO4 when reacts with KCN forms CuCN, which is insoluble in water. It is soluble in excess of KCN due to formation of the following complex(a) K Cu CN2 4^ h6 @

(b) K Cu CN3 4^ h6 @

(c) CuCN2

(d) Cu K Cu CN 4^ h6 @

Ans (b) K Cu CN3 4^ h6 @

EXPLANATION :

Copper sulphate react with KCN to give white ppt of Cu CN 2^ h and cyanogen gas. The insoluble copper cynaide dissolve in excess of KCN and give soluble potassium cuprocyanide.

CuSo KCN24 + K SO Cu CN2 4 2$ + ^ h

Cu CN2 2^ h CuCN CN CN2insoluble cyanogen

$ + −

CuCN KCN3+ K Cu CNSoluble

3 4$ ^ h6 @

MCQ 1.23

Boron shows diagonal relation with(a) Al (b) C(c) Si (d) Sn

Ans (c) Si

EXPLANATION :

Elements of second-period (mainly first three) show a diagonal-relation with the members of third-period elements in many chemical and physical properties.Silicon (Si) is placed at diagonal-position third period w.r.t. The boron in the second-period. Hence, shows diagonal relation with (Si).

MCQ 1.24

A solution contains non-volatile solute of molecular mass M2. Which of the following can be used to calculate the molecular mass of solute in terms of osmotic pressure?

(a) M m VRT22

π = a k (b) M Vm RT

22

π = a k

(c) M Vm RT2

2 π = a k (d) M Vm

RT22 π = a k

Ans (b) M Vm RT

22

π = a kEXPLANATION :

Vπ nRT= Mm RT=

or M2 Vm RT2

π = a k

MCQ 1.25

If 96500 coulomb of electricity is passed through CuSO4 solution, it will liberate(a) 63.5 g Cu(b) 31.76 g Cu(c) 96500 g Cu(d) 100 g Cu

Ans (b) 31.76 g Cu

EXPLANATION :

Molar mass of copper .63 5=As Copper (Cu) contain ( )2+ charge in ,CuSO4 it requires

( )F C2 2 96500#= charge to give one mole i.e. 63.5 g of copper. Thus, on giving 96500 C of electricity, we get . / . g63 5 2 31 76= of copper.Hence, (b) is the correct option.

MCQ 1.26

the X is

(a) LiAlH3

(b) /Sn HCl(c) /Na S NH S2 4 2^ h

(d) all of the above

Ans (b) /Sn HCl

EXPLANATION :

Sn and HCl is reducing reagent which is used to convert nitroalkanes and nitroarenes to the corresponding primary amines.

MCQ 1.27

Reaction CH CONH NaOBr3 2 gives:

(a) CH Br3

(b) CH4

(c) CH COBr3

(d) CH NH3 2

Ans (d) CH NH3 2

EXPLANATION :

This is Hofmann bromaide reaction in which 1o amine is obtained having one C atom less than the starting amide.

MCQ 1.28

Iodoform test is not given by(a) 2-Pentanone (b) Ethanol(c) Ethanal (d) 3-Pentanone

Ans (d) 3-Pentanone

EXPLANATION :

Iodoform test is exhibited by ethyl alcohol acetaldehyde, acetone methyl ketones and those alchols which possess ( )CH CH OH3

-group . As 3-pentanone does not contain CH CO3 -group as therefore it does not give iodoform test.

MCQ 1.29

The unit cell with the structure given below represents is ........... crystal system.

(a) cubic(b) orthorhombic(c) tetragonal(d) trigonal

Ans (b) orthorhombic

EXPLANATION :

Here, a b C! ! , 90cα β γ= = = . It belongs to orthorhombic system.

MCQ 1.30

A binary solid ( )A B+ − has a rock salt structure. If the edge length is 400 pm and radius of cation is 75 pm, the radius of anion is−(a) 100 pm (b) 125 pm

(c) 250 pm (d) 325 pm

Ans (b) 125 pm

EXPLANATION :

For NaCl type structure, distance between A+ and B−

21#= edge length

21 400#= pm200=

Radius of cation pm75=

So, radius of anion 200 75= − pm125=

MCQ 1.31

In aldehydes and ketones, carbon of the carbonyl group is(a) sp3-hybridized(b) sp2-hybridized(c) sp-hybridized(d) unhybridized

Ans (b) sp2-hybridized

EXPLANATION :

Aldehydes and ketones have C=O in which C is sp2 hybridised.

MCQ 1.32

Which of the following statement is not true about Mohr’s Salt?(a) It de-colourises KMnO4

(b) It is a primary standard(c) It is a double salt(d) Oxidation state of iron is 3+ in it.

Ans (d) Oxidation state of iron is 3+ in it

EXPLANATION :

Mohr’s salt is ( ) .FeSO NH SO H O64 4 2 4 2 The oxidation state of iron is +2 in it.

1. MnO Fe H MnO Fe H O5 8 5 4Coloured Colourless

4

72

4

73

2$+ + + +−+

+ + −+

+

2. It is a primary standard. Since, it is available in pure state & can be stored without decomposition.

3. It is a double salt since in solution it shows properties of Fe2+, SO4

2−, .NH4+

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MCQ 1.33

Which one of the following is least basic?(a) NCl3 (b) NBr3

(c) NI3 (d) NF3

Ans (d) NF3

EXPLANATION :

The basic nature of any compound depends on the ability to donate to lone pair of electrons. Among the given options, Fluorine atom in NF3 is directly bonded with Nitrogen atom.

As fluorine is the most electronegative element and pull the lone pair of electrons towards itself, thus are less available for donation.Hence, NF3 is least basic among the given options.

MCQ 1.34

Difference between S and S2− as S2− has(a) larger radii and large size(b) smaller radii and large size(c) larger radii and small size(d) smaller radii and small size

Ans (a) larger radii and large size

EXPLANATION :

The size of anion is always larger than corresponding neutral atom. So, S2− has larger radii and larger size than S.

MCQ 1.35

When electrons are trapped into the crystal in anion vacancy, the defect is known as−(a) Schottky defect (b) Frenkel defect(c) Stoichiometric defect (d) F -centres

Ans (d) F -centres

EXPLANATION :

When electrons are trapped in anion vacancies, these are called F -centres.

Section B (Non-objective Type)

In Section-B, there are 15 short answer type questions (each carrying 2 marks) out of which any 10 questions are to be answered. A part from this there are 3 long answer type questions (each carrying 5 marks). Each question has an alternative option.

QUE 1.1

What is salt bridge? Give its functions.

Ans :

A salt bridge is a U - shaped tube containing concentrated solution of an inert electrolyte like KCl, KNO3, K SO2 4, etc.(An inert electrolyte is one whose ions do not take part in the redox reaction and also do not react with electrolyte used.)Functions1. It prevents mixing of two electrolytes.2. It completes the electrical circuit.3. It maintains the electrical neutrality of the solutions in

both half cells.

QUE 1.2

Explain, why the valency of inert gases is zero.

Ans :

Valency of inert gases is zero due to their fully-filled outer most (valency shell) configuration As they have eight electrons (except helium), in their valency shell, they have no tendency to give or accept any electron. Thus they are inert and do not participate in chemical reactions. The electronic configuration of inert gas elements are-1. Helium (He), ( )Z s2 1 2"=2. Neon (Ne), ( ) [ ], .HeZ s p10 2 22 6"=3. Argon (Ar), ( ) [ ], ,NeZ s p18 3 32 6"=4. Krypton Kr, ( ) [ ] ,ArZ s p36 4 42 6"=5. Xenon (Xe), ( ) [ ] ,KrZ s p54 5 52 6"= .6. Radon (Rn), ( ) [ ] , , ,XZ f d s p86 4 5 6 6e

14 10 2 6"=

QUE 1.3

Explain the Henry’s law about dissolution of a gas in a liquid.

Ans :

Henry’s law states that, the partial pressure of the gas in vapour phase ( )p is directly Proportional to the mole fraction of the gas χ ^ h in the solution.

p χ KH $=

Here, KH = Henry’s law constant.Different gases have different KH values at the same temperature.

QUE 1.4

Why does O3 act as a powerful oxidising agent?

Ans :

On heating, ozone readily decompose to give dioxygen and nascent oxygen.

O3 O OHeat2 + (nascent oxygen)

Since, nascent oxygen is very reactive, therefore O3 act as a powerful oxidising agent.

QUE 1.5

1. How will you distinguish between isopropyl alcohol and ethyl alcohol.

2. How will you distinguish between isopropyl alcohol and t-butyl alcohol.

Ans :

1. Distinguish between Isopropyl Alcohol and Ethyl Alcohol :(a) Lucas Test : Isopropyl alcohol ( )2c produces turbidity

after five minutes ethyl alcohol ( )1c do not produce turbidity at room temperature.

(b) Victor Meyer’s Test : Blood red colour indicates ethyl alcohol ( )1c . Blue colouration indicates Isopropyl alcohol ( )2c .

2. Distinguish between Isopropyl Alcohol and t-butyl Alcohol (a) Lucas Test : Isopropyl alcohol ( )2c produces turbidity

after five minutes t-butyl alcohol ( )3c produces turbidity immediately.

(b) Victor Meyer’s Test : Blue colouration indicates isopropyl alcohol ( )2c . Colourless solution indicates t-butyl alcohol ( )3c .

QUE 1.6

Write chemical reaction to obtain the following:1. Methane to chloroform2. Chloroform to ethyne.

Ans :

1. Methane to Chloroform : CH CH Cl CH Cl

( )

/ /

methane HCl

Cl h

HCl

Cl h4 3 2 2

2 2ν ν

− − CHCl/

( )HCl

Cl h

chloroform3

2 ν

−

2. Chloroform to Ethyne : CHCl Ag H C C H Ag Cl2 6 6

( )chloroform Ethyne3 $ /+ − − +

QUE 1.7

C H A B,FeCl

Cl

AlCl

CH Cl6 6

3

2

3

3

T

−6 6@ @

Write the name and structure of A6 @ and B6 @.

Ans :

QUE 1.8

Write difference between EMF and Potential difference.

Ans :

E.M.F. Potential Difference

1. It is the potential difference between the two electrodes when no current is flowing through the circuit i.e., in the open circuit.

It is the difference between electrode potentials of two electrodes when current is flowing through the circuit.

2. It is the maximum voltage obtainable from the hell.

It is always less than the emf of the hell.

3. The work calculated from emf is the maximum work obtainable from a cell.

The work calculated from potential difference is less than the maximum work obtainable from a cell.

4. it is responsible for the flow of steady current through the circuit.

It is not responsible for the flow of steady current through the circuit.

QUE 1.9

The rate constant for a first order reaction is s60 1− . How much time will it take to reduce the initial concentration of the reactant to its /1 16th value?

Ans :

k s60 1= − , t .[ ]

[ ]logk A

A2 303161

0

0=

t . log602 303 16= . .60

2 303 1 2042#=

. s4 62 10 2#= −

QUE 1.10

Explain the Rate law or Rate Equation.

Ans :

For a general reaction : aA bB Products$+According to law of Mass Action,

Rate [ ] [ ]A Ba bα

Rate [ ] [ ]K A Ba b=Suppose, experimentally the rate of the reaction is found to depend upon x concentration terms of A and y concentration terms of B . Then,

Rate [ ] [ ]A Bx yα

or Rate [ ] [ ]K A Bx y=Where [ ]A and [ ]B are the molar concentrations of A and B respectively and K is rate constant. The above expression is called Rate law.The representation or reaction in terms of concentration of the reactants is known as rate law. It is also called as rate equation or rate expression.

QUE 1.11

Derive equilibrium constant from Nernst equation

Ans :

Consider the following cell reaction :

aA bB+ cC dD+Nernst equation for this cell reaction at 298 K is,

Ecell .

[ ] [ ][ ] [ ]

logFRTE n A B

C D2 303cello

a b

c d

= −

. logE n Q0 0591cello

C= −

where, QC is concentration quotient.

At equilibrium, Ecell 0=

and QC KC=

or Ecellc . ( )log at Kn K0 0591 298C=

QUE 1.12

HF is a weaker acid than HI. Explain.

Ans :

HF is the weakest acid because of its strong H F− bond. Fluorine being small in size overlaps better with s1 orbital of hydrogen leading to a strong bond. Hence, cannot give proton easily. But in case of HI, size of I is large. This result in poor overlap of it’s orbital with s1 orbital of hydrogen making H I− bond weaker, so, it losses proton easily and becomes strong acid.

QUE 1.13

Give the reason for the bleaching action of Cl2.

Ans :

In presence of moisture or in aqueous solution, Cl2 liberates nascent oxygen.

Cl H O2 2+ HCl O2Nascent Oxygen

$ + 6 @

This nascent oxygen brings about the oxidation of coloured substances present in vegetable and organic matter to colourless substances:Coloured substances O "+ Colourless substancesThus, the bleaching action of Cl2 is due to Oxidation.

QUE 1.14

Why noble gases are mostly inert? Give three reasons?

Ans :

Noble gases (group-18 elements) are mostly inert, it is due to the following facts.1. The atoms of noble gases have stable closed shell electronic

configuration ns np2 6^ h.2. Exceptionally high ionisation energies.3. Positive electron gain enthalpies i.e. energy is needed when

an atom of a noble gas takes up an electron.

QUE 1.15

How will you convert Aniline into Benzoic acid?

Ans :

Conversion of Aniline to Benzoic Acid

Long Answer Type Questions

Answer all the three questions :

QUE 1.16

How does Arrhenius equation explain the effect of temperature on the rate constant.

Ans :

Quantitatively the effect of temperature on the rate of a reaction and hence on the rate constant K was proposed by Arrhenius (1889). The equation, called Arrhenius equation.

K Ae /Ea RT= − ...(1i)Where A is constant known as frequency fact (It gives the frequency of binary collisions of the reacting molecules per second per litre), Ea is the energy of activation, R is gas constant and R is the absolute temperature. The two quantities A and Ea are collectively called Arrhenius parameters.Taking logarithm equation (1),

ln K = ln A + ln e /E RTa−

= ln lnA RTE ea−

= ln A RTEa− [Hence, ln e 1= ] ...(2)

If K1 and K2 are the values of rate constant at temperature T1 and T2 then,

ln K1 = ln A RTEa

1− ...(3)

ln K2 = ln A RTEa

2− ...(4)

Subtracting equation (4) from equation (3), we get

ln K2 −ln K1 RTE

RTEa a

2 1=− − −b l

ln K2 −ln K1 RTE

RTEa a

1 2= −

ln KK

1

2 RE

T T1 1a

1 2= −: D

Converting to common logarithm,

log KK

1

2 . RE

T TT T

3 303a

1 2

2 1= −: D

OR QUE

For a first order reaction, calculate the ratio between the time taken to complete three-fourth of the reaction and the time taken to complete half of the reaction.

Ans :

For first order reaction,

t .[ ][ ]

logk AA2 303 0=

For 3/4 of a reaction to take place,

t t /3 4=

[ ]A [ ] [ ]A A43

0 0= −

[ ]A41

0=

Thus, t /3 4 .

/ [ ][ ]

logk AA2 303

1 4 0

0=

. logk2 303 4= ...(1)

Now, for half of a reaction to take place,

t t /1 2= ;

[ ]A [ ] [ ]A A21

0 0= −

[ ]A21

0=

Thus, t /1 2 .

[ ] /[ ]

logk AA2 303

20

0=

. logk2 303 2= ...(2)

Dividing equation (1) by equation (2),

tt

/

/

1 2

3 4 loglog

24= .

.0 30100 6020= 2=

Thus, the time required for th43

^ h of the reaction to occur is twotimes that required for half of the reaction.

QUE 1.17

Write the test by which following are distinguished:1. Ethanol and Acetaldehyde2. Phenol and Carboxylic acid3. Aldehyde and Ketone4. Formic acid and Acetic acid5. Primary, secondary and tertiary alcohols

Ans :

1. Ethanol fruity smell of ester, on adding ethanoic acid. While acetaldehyle does not give any smell.

CH CH OH CH COOH CH COOCH CH( )

[ ]

H SO

EsterFruity smell

3 2 3 3 2 32 4+

2. Phenols and carboxylic acids can be easily distinguished by the sodium bicarbonate test. Sodium bicarbonate is a weak base that readily reacts with carboxylic acids but not with phenols (Carboxylic acids are stronger acids than phenols).

3. Aldehyde and ketone can be distinguished by tollen’s test, fehling’s test and Iodoform test.

Tollen’s Test : aldehyde can be reduces Tollen’s reagent. But, ketone does not reduce Tollen’s reagent.

CH CH CHO Ag NH OH2 3'Propanal Tollen s reagent

3 2 3 2+ ++ −^ h6 @

CH CH COO Ag NH H O4 2Pranoate ion silver mirror3 2 3 2$ .+ + +−

Fehling’s Test : aldehyde reduces Fehling’s solution to a red-brown precipitate of Cu O2 , but a ketone does not.

CH CH CHO Cu OH2 5Propanal

3 22+ ++ −

CH CH COO Cu O H O3Pranoate ion brown ppt3 2 2 2$ .+ +−

4. Both, formic acid and acetic acid are carboxylic acids. Formic acid, because of its structure, also acts as an aldehyde, and reduces Tollen’s reagent to silver. Thus, formic acid gives silver mirror test. Acetic acid does not give this test.

HCOOH Ag NH OH2'formic acid Tollen s reagent

3 2+ ^ h6 @

Ag H O CO NH2 2 4silver mirror

2 2 3$ + + +

5. Primary Alcohol : It is an alcohol which has hydroxyl group connected to a primary carbon atom. Or a molecule containing of a CH OH2− − group.

Secondary Alcohol : it is an alcohol which has hydroxyl group connected to a secondary carbon atom or a molecule containing a group.

Tertiary Alcohol : It is an alcohol which has hydroxyl group connected to a tertiary carbon atom or a molecule containing a group e.g.

These can be distinguished by ‘Lucas-test’.(a) 3c alcohol produce turbidity immidiathy with Lucas-

reagent ZnCl HCl2 +^ h.(b) 2c alcohol take about 5 min. to produce turbidity with

Lucas-reagent.(c) 1c alcohol does not produce turbidity with Lucas-reagent

at room-temperature.

OR QUE

Outline the principles of refining of metals by the following methods:1. Zone refining2. Electrolytic refining3. Distillation

Ans :

1. Zone Refining : This method is based on the principle that the impurities are more soluble in the melt than in the solid state of the metal. A circular mobile heater is fixed

at one end of the rod of the impure metal. The molten zone moves along with the heater which is moved forward. As the heater moves forward, the pure metal crystallises out of melt and the impurities pass on into the adjacent molten zone. The process is repeated several times and the heater is moved in the same direction at one end, impurities get concentrated. This method is very useful for producing semiconductor based other metals of very high purity like germanium, silicon, boron etc.

2. Electrolytic Refining : In this method, the impure metals in made to act as anode. A strip of the same metal in pure form is used as cathode. They are put in a suitable electrolyte both containing soluble slat of the same metal. The move basic metal remains in the solution and the less basic ores go to the anode mud. The reactions are:

Anode : M M nen$ ++ −

Cathode : M nen ++ − M$ Copper and zinc are refined by this method.3. Distillation : In this method, the impure metal is evaporated

to obtain the pure metal as a distillate. This method is very useful for low boiling metals like zinc and mercury.

QUE 1.18

Account for the following:1. NH3 has higher boiling point than PH3.2. H PO3 3 is a diprotic acid.3. Of noble gases only xenon forms real chemical compounds.4. Write the structural formula of XeOF4.

Ans :

Account for the following:1. NH3 has higher boiling point than PH3 : The highly electronegative element like nitrogen when

bounded with the hydrogen, it will form intermolecular hydrogen bonds with other NH3-molecules due to development of polarity.

Thus the attractive forces between the NH3-molecules becomes very strong, as a result the boiling point of NH3 is high.

On the other hand phosphorus in PH3 is not able to form hydrogen bonds between its molecules, also due to small polarity, these are only held together by a weak vander wall’s force. Hence has low boiling point.

2. H PO3 3 is a diprotic acid : Number of protons furnished by any OXO-acid, depends

on the number of OH− groups directly bonded with the central atom.

Since, Structure of H PO3 3 is,

It has two OH− groups, directly bonded with the phosphorus atom.

Thus, H PO3 3 will give two protons (H+) and therefore is a deprotic acid.

3. Of noble gases only xenon forms real chemical compounds

Ionisation potential of Xe is than comparable that of oxygen (O2

+). Since oxygen can form many stable compounds. Mr. Bartlett suggested that Xe can also form stable compounds. Later on may compounds of Xe are obtained, e.g. ,XeF2 XeF4, XeF6, XeO3, XeOF4 etc.

Another reason for the formation of stable compounds by Xe ions is the ability to use its d-orbitals of suitable energy.

4. Structural formula of XeOF4 is as follows:

XeOF4 Show square pyramidal shape in which Xe is in sp d3 2 hybridization, with one lone pair of electrons. Oxidation state of Xe in XeOF4 is (+)6.

OR QUE

State and explain Kohlrausch law. Also give applications of Kohlrausch law.

Ans :

Kohlrausch Law of Independent Migration of IonsLimiting molar conductivity of an electrolyte is the sum of the individual contributions of the anion and cation of the electrolyte. If an electrolyte on dissociation gives v+ cations and v− anions then its limiting molar conductivity is given by :

om/ v v_ _

o oλ λ= ++ +

Here, oλ + = Limiting molar conductivities of cations

oλ − = Limiting molar conductivities of anionsFor example : If Na

oλ + and Cloλ − are limiting molar conductivities

of the sodium and chloride ions respectively, then the limiting conductivity for sodium chloride is given by the equation :

( )m NaClo/ Na

oClo/ /= ++ −

Application of Kohlrausch Law1. Calculation of Limiting Molar Conductivity of Weak

Electrolytes : It is not possible to find mo/ for a weak

electrolyte like CH COOH3 by extrapolation, we can find it as following using Kohlrausch’s law.

( )m CH COOHo

3/ ( ) ( )CH COO Hmo

mo

3λ λ= +− +

This equation can be arrived at by knowing the molar conductivities at infinite dilution for the strong electrolytes KCl, CH COOK3 and HCl

( )m KClo/ K

oCloλ λ= ++

−

( )m CH COOHo

3/ CH COO ko

3cλ λ= +−+, ( )m HCl

oΛ H Clo oλ λ= ++ −

Hence, we have

CH COO Ho o

3λ λ+− + CH COO H Clo o o o

3λ λ λ λ= + + +κ − + + −^ ^h h

Clo oλ λ− +κ + −^ h

i.e., ( )m CH COOHo

3Λ ( ) ( ) ( )m CH COOK m HCl m KClo o o

3Λ Λ Λ= + −2. Calculation of the Degree of Dissociation : Degree of dissociation,

( )α momc

ΛΛ=

Here, mcΛ = Molar conductivity of a solution

at any concentration c

moΛ = Limiting molar conductivity

3. Calculation of Dissociation Constant of a Weak Electrolyte Dissociation constant,

( )KC c1

2

αα= −

Here, c = Concentration

α = Degree of dissociation

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