Class 12 Study Material Chemistry SA-1

CHAPTER 1 SOLID STATE 1 mk ‘Q’ Q 1. ‘Crystalline Solids are anisotropic in nature’. What does this Statement mean? A 1. The statement means that some of the physical properties like electrical resistance or refractive index of Crystalline Solids show different values when measured along different directions in the same crystal. Q 2.

Why does the presence of excess of lithium make LiCl crystals pink?

A 2. The presence of excess of lithium makes LiCl crystals pink due to e -s trapped in anionic vacancies (F centers). These electrons absorb some energy of the white light giving pink colour to LiCl crystal.6 Q3. What is meant by anti-ferromagnetism? What type of substances exhibit anti ferromagnetism? A3. Substance like MnO, MnO2 in which magnetic domains are oppositely oriented and cancel out each other’s magnetic moment exhibit anti ferromagnetism. Magnetic. Alignment of magnetic moments in antiferromagnetic substance:OR Q3 What type of substance would make better magnets, ferromagnetic or ferromagnetic? A3. Ferromagnetic substance would make better magnets because when ferromagnetic substance is placed in magnetic field all the domains get oriented in the direction of the magnetic field and a strong magnetic effect is produced eg : Co, Ni Q4. In a compound nitrogen atoms (N) make ccp and metal atoms (M) occupy one third of the tetrahedral voids present. Determine the formula of the compound formed by M & N? A4.

Let the no of nitrogen atoms (N) be x No of tetrahedral voids = 2x No of metal atoms = 2/3 x.

Ratio of M:N = 2/3x : x Therefore the formula of the compound is M2 N3 Q 5. Classify each of the following as being either a p type or n type semi conductor? (i) (ii) A 5.

Ge doped with In. B doped with Si.

(i) P type semiconductor because when group 14 element is doped with group 13 element, an electron deficit hole is created. (ii) n type semiconductor because when group 13 element is doped wih group 14 element , free electrons will become available.

Q 6.

Write a distinguishing feature between a metallic solid and an ionic solid?

A 6. Ionic Solids

Metallic Solids

In solid state ionic solids are Metallic Solids are good electrical electrical insulator :- ions are not conductors in solid state because of the free to move. Eg NaCl, CuSO4 etc presence of free electrons , Eg :- Copper, Iron etc 2 mk Q 1. Analysis shows that nickel oxide has the formula Ni fractions of nickel exist as Ni2+ and Ni3+ ions? A1.

The formula of the oxide is Ni 0.98 0 1.00 Let the number of O2- ions be 100. Then number of nickel ions = 98 Let the number of Ni2+ be x Then number of Ni3+ = 98 - x Since total charge on cations = total charge on anions. X x (2) + (98-x) x (3) = 100 x 2 2x + 294 – 3x = 200 x = 94 % of Ni2+ =

94 98

x 100 = 96%

0.98

0

1.00

. What

% of Ni3+ = 100 -96 = 4% Q 2. A compound forms hcp structure. What is the total number of voids in 0.5 mol of it? How many of these are tetrahedral voids? A2.

No of atoms = 0.5 mol = 0.5 x 6.022 x 1023 = 3.011 x 1023 No of octahedral voids = no of atoms in hcp structure = 3.011 x 1023 No of tetrahedral voids = 2 x no of atoms in hcp structure = 2 x 3.011 x 1023 = 6.022 x 1023 Total no of voids

Q3.

= 6.022 x 10 23 + 3.011 x 10 23 = 9.033 x 10 23

Examine the given defective crystal:-

A+ BA+ BBA+ BA+ BA+ BAnswer the following questions

BA+

A+ BA+ B-

A+

(i)

What type of stoichiometric defect is shown by the crystal? (1/2)

(ii)

How is the density of crystal affected by this defect?

(iii)

What type of ionic substance show such defect?

(1/2) (1)

A 3. (i) Schotty defect is shown by the crystals, since equal number of cations and anions are missing from the crystal lattice. (ii)

Due to this defect, the density of the crystal decreases.

(iii)

This defect is shown by those ionic substance in which cations and anions are of almost similar size eg :- NaCl, KCl etc.

Q 4. If NaCl crystals are doped with 2 x 10 cation vacancies per mole?

-3

mol percent of SrCl2 , calculate the

A 4. Doping of NaCl with 2 x 10 -3 mol percent of SrCl 2 means 100 moles of NaCl is doped with 2 x 10 -3 mole of SrCl2 or 1 mole of NaCl is doped with 2 x 10-5 mole of SrCl2

Each Sr2+ will occupy the place of Na+ and displace one Na+ from crystal lattice to create cation vacancies. Cation vacancies = Number of Sr 2+ ion added. = 2 x 10 -5 mol = 2 x 10 -5 x 6.022 x 10 23. = 12.046 x 10 18 mol -1

Q 5. Calculate the packing efficiency of a metal crystal for a simple cubic lattice? A 5.

Packing efficiency in simple cubic lattice

rr r

a Volume of one atom x 100 Volume of cubic unit cell (a3) Since a= 2r for simple cubic Vol of one atom =4/3 π r3 =

4/3 π r3 x 100 (2 r) 3

=

4 x 3.14 x r3 x 100 3 x 8 x r3

=

52.36%

≈

52.4%

Q 6 (a) In reference to a crystal structure, explain the meaning of coordination number? (b) What is the number of atoms in a unit cell of (i)

Face centered cubic structure?

(ii)

Body centered cubic structure?

A6 (a) The number of nearest neighbours of any constituted particle in the crystal lattice is called its coordination number. (b)

Number of atoms in a unit cell of (i)

Face centered cubic cell structure is 4.

(ii)

Body centered cubic structure is 2.

3 mk Q1.

In terms of band theory. What is the difference? (a)

between a conductor and an insulator.

(b)

between a conductor and a semi conductor. Empty conduction band

Filled (valence band)

_____

Partially filled band

Over lappling band

Conductors

A1

(a) In conductors, energy band is partially filled or it overlaps with a higher energy unoccupied conduction band. Due to this electrons can flow easily under an applied electric field and show conductivity In insulators, the gap between filled valence band and the next higher unoccupied band is large, hence electrons cannot jump to it and substance has very small conductivity and behaves as an insulator. Empty band (conduction band)

Filled band (Valence band)

(b) In semiconductors the gap between valence band and conduction band is small. Therefore some electrons may jump from valence band to conduction band and show some conductivity eg ‘Si’ and ‘Ge’

Q2 )

A 2)

Aluminum crystallizes in a cubic close packed structure. Its metallic radius is 125 pm. (a)

What is the length of the side of the unit cell?

(b)

How many unit cells are there in 1 cm3 of aluminum?

(a)

In a cubic close packed structure :4r=

√ 2a

Given r = 125 x 10 -12 m a=

4r √2

or 2

√2 r

a = 125 x 10 -12 x 2 x 1.414 m = 354 pm.

(c)

a3 = (354)3 x (10-12) 3 = 44.21 x 10-30 m 3 No of unit cells in 1 cm3

= Total Volume = Vol of one unit cell

10-6 44.21 x 10-30

= 2.261x 10 22 unit cells. Q3

How will you distinguish between the following pairs of terms :(a)

Hexagonal close packing and Cubic close packing?

(b)

Crystal lattice and unit cell?

(c)

Tetrahedral void and Octahedral void?

Hexagonal close packing AB AB—type packing is called hexagonal close packing. i.e spheres of the third layer are exactly aligned with those of the first layer . eg : ‘Mg’ and ‘Zn’

Q4

Cubic close packing ABC ABC—type packing is called cubic close packing. i.e spheres of the third layer are not aligned with those of 1 st or 2nd. When 4th layer is placed its spheres are aligned with those of 1st layer. eg :- ‘Cu’ and ‘Ag’ Tetrahedral void Octahedral void A tetrahedral void lattice is surrounded An octahedral void surrounded by Crystal Unit is cell The three whichdimensional crystal by lattice by 4 spheres lie at the The six smallest spheres portion and of formed a arrangement of tetrahedron. constituent which whenof 2repeated different vertices of a regular combination triangular in voids of the st nd particles in the space which directions, generates the entire lattice. 1 and 2 layer There are 2 how tetrahedral voids per There is one octahedral void per atom represents the constituent atom in a crystal particles (atoms, ions or in a crystal molecules) are arranged in a Account crystalfor the following :(i)

Table salt, NaCl sometimes appears yellow in colour.

(ii)

FeO(s) is not formed in stoichiometric composition.

(iii)

Some of the very old glass objects appear slightly milky instead of being transparent.

A4.

(i) Yellow colour in NaCl is due to metal excess defect due to which unpaired electrons occupy anionic vacancies. These sites are called F centers. These electrons absorb energy from the visible region and transmits yellow colour.

(ii) In the crystal of FeO, some of the Fe2+ cations are replaced by Fe3+ ions. Three Fe2+ ions are replaced by two Fe 3+ ions to make up for the loss of positive charge. Thus there would be less amount of metal as compared to stoichiometric properties. (iii) Very old glass objects become slightly milky, because of heating during the day & cooling at nights i.e annealing. Over a number of years, glass acquires some crystalline character. Q5.

The density of copper is 8.95 gcm -3. It has a face centered cubic structure. What is the radius of copper atom? (Atomic mass Cu = 63.5 g mol -1 . NA = 6.022 x 1023 mol -1) ?

A5.

Mass per unit cell

=

Atomic mass of Cu x 4 NA

=

______63.5__x 4______ 6.022 x 1023

=

4.22 x 10 -22 g

Volume of unit cell =

__Mass__ Density

=

4.22 x 10 -22 8.95

=

4.7 x 10-23 cm3

=

(Volume )1/3

=

(4.7 x10 -23)1/3

=

3.61 x 10-8 cm or 361pm

For fcc, r

=

___a___ 2 √2

r

=

r

=

Edge

361 2 x 1.41 128 pm

CHAPTER 2 SOLUTIONS 1 mk Q 1.

Define an ideal solution and write one of its characteristics?

A1

The solutions which obey Raoult’s law over the entire range of concentration are known as ideal solutions. i.e P = PoA x and P = PoB x A

For ideal solutions

A

H mix = 0 and

B

B.

V mix = 0.

eg ;- Solution of n – hexane and n- heptane Q 2.

What is meant by reverse osmosis ?

A2 The process in which the solvent flows from the solution into the pure solvent through the semi permeable membrane when a pressure higher than the osmotic pressure is applied on the solution is called reverse osmosis. Application :- The technique is used in the desalination of sea water. Q3.

Define mole fraction.

A3 The mole fraction of a component is the ratio of the number of moles of that component to the total number of moles of all the components present in the solution. For a binary solution consisting of 2 components A and B if n A is the number of moles of A and nB is the number of moles of B then. x

A

= __nA___ nA + nB

Q 4.

x

B

= _ nB___

nA + nB

Define the term azeotrope?

A4 The constant boiling mixture which distill out unchanged in their composition are called azeotropes. Eg :- A mixture of ethanol and water containing 95.4% of ethanol forms an azeotrope with boiling point 351.15 K.

Q5. Explain boiling point elevation constant for a solvent / Define ebullioscopic constant? A5

Since

Tb = Kb.m where m is molality

When m = 1 Kb =

Tb

Therefore Ebullioscopic constant is defined as the elevation in boiling point of a solution when 1 mole of a solute is dissolved in 1 Kg of solvent . Q6

What are isotonic solutions ?

A6 The solutions of equimolar concentrations having same osmotic pressure at given temperature are called isotonic solutions. eg :- A 0.9% solution of pure NaCl is isotonic with human red blood cells.

Q7 Explain why aquatic species are more comfortable in cold water rather than in warm water? A7 Aquatic species need dissolved oxygen for breathing. As solubility of gases decrease with increase of temperature, less oxygen is available in summers in lakes. Hence they feel more comfortable in winter when the solubility is higher. OR Q7 Why do gases nearly always tend to be less soluble in liquid as the temperature is raised? A7 Dissolution of gas in liquid is an exothermic process. Gas + Solvent Solution + heat. Therefore as per Lechatlier’s principle if the temperature is increased, equilibrium shifts backward i.e the solubility decreases. 2 mk Q1 How is the vapour pressure of a solvent affected when a non volatile solute is dissolved in it?

A1 When a non volatile solute is added to a solvent, its vapour pressure decreases because some of the surface sites are occupied by solute molecules. Thus less space is available for the solvent molecules to vaporize. Q2. The depression in freezing point of water observed for the same molar concentrations of acetic acid, trichloro acetic acid and trifluoroacetic acid increases in the order as stated above. Explain? A2 As depression in f.pt ( Tf) is dependent on degree of dissociation (α) and fluorine exerts the highest – I effect. So trifluoro acetic acid is the strongest acid and ionizes to a greater extent while acetic acid ionises to the minimum extent. Thus greater the number of ions produced, greater is the depression in freezing point.

Q3. Differentiate between molarity and molality for a solution. How does a change in temperature influence their values? Or Q3. State the main advantage of molality over molarity as the unit of concentration?

Molarity It is defined as the number of moles of solute dissolved in one litre of the solution Mathematically Molarity (M) =No of moles of solute x 1000 Vol of Solution (ml)

Molality It is defined as the number of moles of solute dissolved in 1 Kg of the solvent Mathematically Molality (M) =No of moles of solute x 1000 Mass of Solvent (g)

It decreases with increase in temperature It does not change with change in (as V α T) temperature Since molality does not change with a change in temperature therefore it is a better method to express the concentration of a solution. Q4. What is meant by colligative property. List any four factors on which colligative properties of a solution depend?

A4. The properties of solutions which depend upon the number of solute particles and not upon the nature of the solute are known as colligative properties eg :- Osmotic pressure. Factors :(i) (ii) (iii) (iv)

Number of particles of solute. Concentration of solution. Temperature. Association or dissociation of solute.

An aqueous solution of sodium chloride freezes below 273K. Explain the lowering in freezing point of water with the help of a suitable diagram? Q5.

A5. Freezing point of a substance is the temperature at which solid and liquid phases of a substance coexist i.e they have the same vapour pressure. As the vapour pressure of the solution is less than that of pure solvent For solution its vapour pressure will become equal to Vapour that of a solid solvent only at a lower temperature.

liquid solvent

………………… solution pressuresolid solvent

Tf

Tfo

Temperature

3 mk Q1. A sample of drinking water was found to be severely contaminated with chloroform CHCl3 , supposed to be carcinogen. The level of contamination was 15ppm (by mass)?

A1.

(i)

Express this in percent by mass.

(ii)

Determine the molality of CHCl3 in water sample.

15 ppm means 15 parts in 106 parts by mass in the solution. i.e

106 parts by mass of solution = 15 parts of solute. or 100 parts (Mass %) =

Mass of solvent

15 x 100 106

=

15 x 10 -4

=

(106 -15) g

≈ Therefore Molality

=

106 g nB wA

where nB = no of moles of solute wA = Mass of the solvent

Molality

=

15/119.5 X 1000 106

= 1.25 x 10-4m Q2 State Raoult’s law for a solution containing volatile components. How does Raoult’s law becomes as special case of Henry’s law? A2 For a solution of volatile liquids, the partial vapour pressure of each component in the solution is directly proportional to its mole fraction? i.e

PA α

x

and

PB α

x

A

or

PA = P0A

x

B

or

PB

P0B

x

=

A B

Where PA and PB are the partial pressure of A and B, P 0A , P0B are the vapour pressure of pure component and x and x are their mole fractions. A

B

If gas is the solute and liquid is the solvent then according to Henry’s law PA = K H x

A

Thus Raoult’s laws and Henry’s law become identical except that their proportionality constants are different. Q3 Some ethylene glycol is added to your car’s cooling system along with 5kg of water. If the freezing point of water-glycol solution is -15 0c, what is the boiling point of the solution? Kb = 0.52 K Kgmol-1 and Kf = 1.86 K Kg mol-1 A3

Tf = 15oc Kf = 1.86 k/m Molality =

Tf Kf

=

15/1.86 = 8.06 m

Tb

= = =

Kb x m 0.52 x 8.06 4.190c

Boiling point of pure water = 100oc Therefore Tb = Tb + Tbo Tb = 100 + 4.19 = 104.19oC Q4 Assuming complete dissociation. Calculate the expected freezing point of a solution prepared by dissolving 6 g of Glauber’s salt, Na 2SO4.10 H2O in 0.1 Kg mol-1 of water. Kf of water = 1.86 K Kg mol -1. A4

Kf WB MB

= = =

1.86 K Kgmol-1 6g WA = 0.1 Kg mol-1 322 g mol-1

Since there is a complete dissociation Therefore i = 3 Tf =

i K f WB MB x W A

=

3 x 1.86 x 6 322 x 0.1

=

1.04o

Q5. Calculate the mass of a nonvolatile solute (molar mass 40g mol -1) which should be dissolved in 114 g octane to reduce its vapour its pressure to 80%? A5.

Ps= 80% of Po = 0.80 Po No of moles of solute

=

W 40

No of moles of solvent (octane)

=

114 114

Po – Ps Po

=

XB

Po – 0.8 Po Po

=

W /40 W /40 + 1

=

W 40

therefore

or

0.2

W +1 40

0.8 W 40 or

=

=

1

0.2

W = 10g

Q6 19.5g of CH2FCOOH is dissolved in 500g of water. The depression in freezing point observed is 1oc. Calculate the Van’t Hoff factor and dissociation constant of fluoroacetic acid .Kf for water is1.86 K Kg mol-1. A6

Kf= 1.86K Kgmol-1

WB = 19.5g WA = 500g (ATf) obs = 1o. MB (obs) =

1000 Kf WB W ∆ T A

=

f

= 72.54 g mol-1

1000 X1.86 X 19.5 500 X 1

MB (cal) = 14 + 19 + 45 = 78 g mol-1 L = M(cal) MB (Obs)

= _78__ 72.54

=

1.0753

Calculation of dissociation const (K) Ka = _Cα2 1-α CH2FCOO -+ H+

CH2 FCOOH Initial C MOL L-1 C(1-α) i = C (1+α) C

O Cα = =

O Cα

1 + α or α = i - 1 1.0753 – 1 = 0.0753

Taking volume of solution as 500 ml C =19.5 x _1_ x 1000 78 500 Ka = Cα2 1-α

= 0.5 M

= 0.5 x (0.0753)2 1-0.0753

Ka = 3.07 x 10-3 Q7. Non ideal solutions exhibit either positive or negative deviations from Raoult’s law. What are these deviations and why are they caused? Explain with one example for each type? A7 For non ideal solutions, vapour pressure is either higher or lower than that predicted by Raoult’s law. If it is higher the solution exhibits positive deviation and if it is lower it exhibits negative deviation from Raoult’s law. Positive Deviation When solute – solvent interactions are weaker than solute – solute or solvent solvent interactions, vapour pressure increases which result in positive deviation. Eg :- Ethanol +Acetone. for a solution showing +ve deviation PA > PoA XA and PB > PoB XB H mix = +ve Vmix = +ve

Negative Deviation When solute solvent interactions are stronger than solute – solute or solvent – solvent interactions, vapour pressure decreases which result in negative deviation. eg :- Chloroform + Acetone. for a solution showing –ve deviation PA < PoA XA PB < P0B XB H mix = - ve Vmix = - ve

For plots of non ideal solution showing +ve and –ve deviation refer NCERT pg 56

5 mk Q1. (i) What is Van’t Hoff factor? What types of values can it have if in forming the solution the solute molecules undergo

(ii)

(a)

Dissociation?

(b)

Association?

How many ml of 0.1M HCl solution are required to react completely with 1g of a mixture of Na 2CO3 containing equimolar amounts of both ? (Molar Mass Na2CO3 = 106g and NaHC03 = 84g)

A1. (i) Van’t Hoff factor (i) is defined as the ratio of the experimental value of colligative property to the calculated value of colligative property . i Also, i =

=

Observed Colligative property Calculated Colligative property

Total number of moles of particles after dissociation/ association Number of moles of particles before association/ dissociation

Therefore , For (a) dissociation i > 1 And (b) Association i < 1 (ii)

Let x g of Na2CO3 be present in the mixture So amt of Na HCO3 = 1 –x Moles of Na2CO3 in x g =

x_ 106

Moles of NaHCO3 in (1-x) = _1-x_ 84 As mixture contains equimolar amounts of the two. x_ = 1-x_ 106 84 106 – 106 x = 84x X = _106__ = 0.558g 190 Moles of Na2 CO3 = __0.558__

= 0.00526

106 Moles of NaHCO3 = 1- 0.558 = 0.442 84 84

= 0.00526

To calculate the moles of HCl required. Na2CO3 + 2 HCl NaHCO3 + HCl

2 NaCl + H2O + CO2 NaCl + H2O + CO2

1 mole of Na2CO3 requires 2 moles of HCl Therefore 0.00526 mole of 0.0152 moles of HCl.

Na 2 CO3

requires 2 x 0.00526 moles or

1 mole of NaHCO3 requires 1 mole of HCl Therefore 0.00526 mole of NaHCO3 requires 0.00526 moles of HCl. Total moles of HCl required = 0.01052 + 0.00526 = 0.01578 moles To calculate volume of 0.1M HCl 1.1 mole of 0.1M HCl are present m 1000ml HCl Therefore 0.01578 mole of 0.1M HCl will be present in 1000 x 0.01578 0.1 = 157.8 ml Q2. (i) The molecular masses of polymers are determined by osmotic pressure method and not by measuring other colligative properties. Give two reasons? (ii) At 300k, 36g of glucose C6H12O6 present per litre in its solution has a pressure pf 4.98 bar. If the osmotic pressure of another glucose solution is 1.52 bar at the same temperature, calculate the concentration of the other solution? A2. (i) because :-

The osmotic pressure method has the advantage over other methods (a) It uses molarities instead of molalities and it can be measured at room temperature. (b) Its magnitude is large as compared to other colligative properties.

(ii)

π1 = C1 RT, π2 = C2 RT Therefore _π1 = C1 π2 C 2

Q3

(i)

or

4.98 1.52

= 36/180 C2

or

C2

= 0.061 moll-1 = 0.061 x 180 gl-1 = 10.98 gl-1

Define the terms osmosis and osmotic pressure?

(ii) An aqueous solution containing 12.48g of barium chloride in 1.0 kg of water boils at 373.0832 k. Calculate the degree of dissociation of barium chloride? (Given Kb for H2O = 0.52 Km-1 molar mass of Ba Cl2 = 208.34 gmol-1) A3. (i) The net spontaneous flow of the solvent molecules from the solvent to the solution or from a less concentrated solution to a more concentrated solution through a semipermeable membrane is called osmosis. Osmotic pressure :- The minimum excess pressure that has to be applied on the solution to prevent the entry of the solvent into the solution through the semipermeable membrane is called osmotic pressure. (iii) (iv)

Given W2 = 12.48g , W1 = 1 Kg = 1000g Tb (solution) = 373.0832 K Kb for H2O = 0.52 Km-1 and M2 = (Ba Cl2) = 208.34 Tb = Tb - Tbo = 373.0832 – 373 = 0.0832 K M2 (Observed) = Kb x W2 x 1000 Tb x W1 = 0.52 x 12.48 x 1000 0.0832 x 1000 = 78 g mol-1 i

= M2 (calculated) =

208.34 = 2.67

M2 (observed) For BaCl2

m

78

= 3 as it gives 3 ions on dissociation

α

=

i-1 m-1

= 2.67 -1 = 1.67 = 0.835 3-1 2

= 83.5% Q4. (a) What type of deviation is shown by a mixture of ethanol and acetone? Give reason. (b)

What do you expect to happen when RBC’s are placed in (i)

1% NaCl solution (ii) 0.5 % NaCl solution

(c) Calculate the molarity of 68% (w/w) solution of nitric acid , if the density of the solution is 1.504 g ml-1

A4. (a) Ethanol and acetone shows positive deviation because on mixing the two the forces of attraction decreases and the vapour pressure increases. (b)

(c)

Since RBC’s are isotonic with 0.9% NaCl solution therefore (i)

In 1% solution of NaCl they will shrink due to plasmolysis

(ii)

In 0.5% solution of NaCl they will swell or may even burst.

Molarity

=

% x density x 10 Molar mass

Mass % = 68 , d = 1.504 Molar mass of HNO3 = 63 gmol-1 Therefore Molarity = 68 x 1.504 x 10 63 = 16.23 M Q5.

(a)

State Henry’s law and mention two of its important applications

(b) The partial pressure of ethane over a saturated solution containing 6.56 x 10-2g of ethane is 1bar. If the solution were to contain 5 x 10 -2 g of ethane, then what will be the partial pressure of the gas.

A5.

(a)

Henry’s law states that at a constant temperature , the solubility

x of a

gas in a liquid is directly proportional to the pressure of the gas P = KH

x

Applications :-

To increase thje solubility of CO 2 in soda water, the bottle is sealed under high pressure

-

There is a low concentration of oxygen in the blood and tissues of the people living at high altitudes due to which they feel weak and are unable to think clearly (anoxia).

-

(b) M = KH x P For the 1st case 6.56 x 10-2 = KH x 1 bar or KH = 6.56 x 10-2 g bar -1 In the 2nd case 5 x 10 -2 = (6.56 x 10-2) x P or P = 5 x 10 -2 6.56 x 10 -2 P = 0.762 bar

CHAPTER 5

SURFACE CHEMISTRY 01 MARK Q1).

What is meant by shape selective catalysis?

A1) The catalytic reaction that depends upon the pore structure of the catalyst and the size of the reactant and product molecules is called shape selective catalysis. eg:-

zeolites (eg ZSM 5) converts alcohol directly into gasoline.

Q2)

What are the dispersed phase and dispersion medium in milk?

A2)

Dispersed phase : Fat Dispersion medium : water

Q3)

What is Collodion?

A3)

Collodion is 4% solution of nitro cellulose in a mixture of alcohol and ether.

Q4)

Define Kraft temperature?

A4) The formation of micelles from the ionic surfactant can take place only above a certain temperature which is called Kraft temperature. Q5)

Why is adsorption always exothermic ? Or What is the sign of

A5)

H&

S when a gas is adsorbed by adsorbent ?

When a gas is adsorbed in the surface of solid, its entropy decreased i.e. S =-ve. From Gibbs Helmholtz equation : G = H – T S for the process to be spontaneous , G must be negative, which is possible only when H = -ve . Hence adsorption is always exothermic.

Q6) Why is it essential to wash the precipitate with water before estimating it quantitatively? A6) Some amount of the electrolytes mixed to form the precipitate remains adsorbed on the surface of the particles of the precipitate. Hence it is essential to wash it with water before it quantitatively. Q7)

What happens when persistent dialysis of colloidal solution is carried out?

A7) The stability of a colloidal sol is due to the presence of a small amount of the electrolyte. On persistent dialysis, the electrolyte is completely removed, so the colloidal sol becomes unstable and gets coagulated.

Q8)

What causes Brownian movement in a colloidal solution?

A8) Brownian movement ie zigzag movement of the colloidal particles is due to hitting of these particles by the molecules of the dispersion medium with different forces from different directions. Q 9)

Why it is important to have clean surface in surface studies?

A 9)

It facilitates the adsorption of species on the adsorbent.

Q 10) What happens when gelatin is mixed with gold sol? A 10) Gold sol is a lyophobic sol. On addition of gelatin, the sol is stabilized.. Q 11) Gelatin which is a peptide is added in ice – creams. What can be it s role? A 11) Ice –creams are emulsion which get stabilized by emulsifying agents like gelatin. Q 12) What is the role of activated charcoal in gas mask used in coal mine? A 12) Activated charcoal adsorbs poisonous gases present in coal mine. Q 13) In what way is a sol different from a gel? A 13) Sols are colloidal solutions of solid dispersed in liquid while gels are colloidal solutions of liquid dispersed in solid. Q 14) Of NH3 and N2, which gas will be adsorbed more readily on the surface of charcoal and why? A 14) NH3 is adsorbed more readily as it is more easily liquefiable compared to N2, Moreover, NH3 molecule has greater molecular size. Q 15) How does it become possible to cause artificial rain by spraying silver iodide on the clouds? A 15) Clouds are colloidal in nature and carry a charge. On spraying silver iodide which is an electrolyte, the charge on the colloidal particles is neutralized. Clouds coagulate to form rain. Q 16) What is the role of diffusion in heterogeneous catalysis? A 16) The gas molecules diffuse onto the surface of the catalyst and get adsorbed. After the chemical change, the products formed diffuse away from the surface of the catalyst setting the surface free for other reactant molecules to adsorb on the surface and give the product.

02 MARKS Q1) Write the differences between physisorption and chemisorption with respect to the following :(i) Specificity

(ii)

Temperature dependence

(iii) Reversibility

(iv)

Enthalpy change

Criteria

Chemisorption

Specificity

It is not specific in nature

It is highly specific in nature

Temperature dependence

It decreases with increase in temperature. Thus, low temperature is favorable for physisorption Reversible in nature

It increases with increase in temperature. Thus, highly temperature is favorable for chemisorption Irreversible in nature

Low enthalpy of adsorption

High enthalpy of adsorption

Reversibility Enthalpy change

Q2)

Physisorption

Distinguish between homogeneous and heterogeneous catalysis.

A2) When reactants and the catalysts are in the same phase (i.e. liquid or gas) the catalysis is known as homogeneous catalysis. 3 O2 ( g )  NO   ( g )  O3 ( g ) 2

When reactants and the catalysts are in the different phase (i.e. liquid or gas) the catalysis is known as heterogeneous catalysis. 4NH3(g) + 5O2(g)

 Pt   ( s )  4 NO( g )  6 H 2 O ( g )

Q3) What is meant by coagulation of a colloidal solution? Describe briefly any three methods by which coagulation of lyophobic sols can be carried out? A3) The process of aggregating together the colloidal particles is called coagulation of the sol. It is also known as precipitation. Following are the three methods by which coagulation of lyophobic sols can be carried out.

Q4)

(i)

Electrophoresis. In this process, the colloidal particles move towards oppositely charged electrodes and get discharged resulting in coagulation.

(ii)

Mixing of two oppositely charges sols. When equal proportions of positively charges sols are mixed, they neutralize each other resulting in coagulation.

(iii)

Dialysis. By this method, electrolytes present in sol are removed completely and colloid becomes unstable resulting in coagulation colloid is a heterogeneous system, e.g gold sol, sulphur sol, soap, etc.

What are emulsions ? Discuss the role of an emulsifier in forming emulsion.

A4) Emulsions are one of the types of colloidal system, in which both the dispersed phase and dispersion medium are liquids, e.g milk.

Role of emulsifier Emulsifying agents are added to emulsions to stabilize them. The emulsifying agent forms an interfacial film between suspended particles and the medium. For oil in water emulsions, the principal emulsifying agents are gums, proteins, natural and synthetic soaps. Q 5)

A5)

How are the following colloidal solutions prepared? (i)

Sulphur in water

(ii)

Gold in water

(i)

Sulphur sol is prepared by the oxidation of H 2S with SO2. SO2 + 2H2S

(ii)

3S + 2H2O (Sol) Gold sol is prepared by Bredig’s arc process or by the reduction of AuCl 3 with HCHO. oxidation

2AuCl3 + 3HCHO + 3H20

Reduction

2Au + 3HCOOH + 6HCI (Sol)

Q6)

What are enzymes? Write in brief the mechanism of enzyme catalysis.

A6)

Enzymes are biochemical catalysts whih are globular proteins and form macromolecular colloidal solution in water.

The mechanism of enzyme catalysis may be explained on the basis of lock and key theory. Acc to the theory , Enzymes are highly specific due to the presence of active sites on their surface. The shape of active site of any given enzyme is such that only a specific substrate can fit in to it just as one key can fit into a particular lock enzyme catalyzed reactions takes place in two steps as follows :Step I :E

Formation of Enzyme – substrate complex +

Enzyme Step II :-

S

ES

(fast and reversible)

Substrate

Enzyme substrate complex

Dissociation of enzyme substrate complex to form the products ES

[EP]

E

+

P (slow and rate

determining) Enzyme Substrate Complex Q7)

Enzyme Product association

Enzyme (Regenerated)

Product

What do you mean by activity and selectivity of catalysts?

A7) Activity : Ability of a catalyst to accelerate chemical reactions is known as its activity . For example, Pt catalyses the combination of H 2 and O2 to form water. It has been found that for hydrogenation reactions, the catalytic activity increases from Group 5 to Group 11 metals with maximum activity being shown by Group 7-9 elements of the periodic table. Selectivity : The ability of a catalyst to direct a reaction to yield a particular product is called its selectivity. Combination of CO and H 2 yields different products with different catalysts as given below :CO (g) + 3H2 (g)

Ni

CH4 (g) + H2O (g)

CO (g) + H2 (g)

Cu

CO (g) + H2 (g)

Cu I ZnO – Cr O 2 3

HCHO (g) CH3OH (g).

Q8)

Describe some features of catalysis by zeolites?

A8)

(i) Zeolites are hydrated alumino-silicates which have a three- dimensional network structure containing water molecules in their pores.

(ii) On heating, water of hydration present in the pores is lost and the pores become vacant to carry out catalysis. (iii) The size of the pores varies from 260 to 740 pm. Thus, only those molecules can be adsorbed in these pores and catalyzed whose size fits these pores. Hence, they act as molecular sieves or shape selective catalysts. An important catalyst used in petroleum industry is ZSM -5 (Zeolite sieve of molecular porosity – 5). It converts alcohols into petrol by dehydrating them to form a mixture of hydrocarbons. ZSM -5 Alcohols Hydrocarbons Dehydration

03 MARKS Q1)

Account for the following :(a) Medicines are more effective in the colloidal form/ colloidal gold is used for intramuscular injection. (b)

Sky appears blue in colour.

(c)

Alum is added to purify water. OR

Bleeding stops on rubbing moist alum on the cut surface. A1)

(a) Medicines are more effective in the colloidal form because they have a large surface area and are easily assimilated in this form. (b) There are dust particles in the atmosphere. These dust particles are of colloidal size and scatter the light. Blue light coming from the sun is scattered and the sky appears blue. (c)

Alum coagulates colloidal impurities present in water. or Alum brings about and coagulation of blood and stops further bleeding.

Q2)

Explain clearly how the phenomenon of adsorption finds application in (i)

Production of vacuum in a vessel

(ii)

Heterogeneous catalysis

(iii)

Froth floatation process in metallurgy.

A 2)

(i) Production of high vacuum. The traces of air can be adsorbed by charcoal from a vessel evacuated by a vacuum pump to give a very high vacuum. (ii) Heterogeneous catalysis. Adsorption of reactants on the solid surface of the catalyst increases the rate of reaction. (iii) Froth floatation process. A low grade sulphine ore is concentrated by this method using pine oil and frothing agent. The mineral particles become wet by oils while the gangue particles by water.

Q3)

What is an adsorption isotherm? Describe Freundlich adsorption isotherm.

A 3) Adsorption isotherm. It is the variation in the amount of gas adsorbed by the adsorbent with pressure at constant temperature. Freundlich’s adsorption isotherm. It is an empirical relationship between the quantity of gas adsorbed by unit mass of solid adsorbent and pressure at a particular temperature. 1 x  kp n (n  1) m

n  1,

When,

x  kp m

or

…. (i) x  m

p

Where x is the mass of gas adsorbed on mass m of the adsorbent at pressure p.k and n are constants which depend on the nature of the adsorbent and the gas at a particular temperature. Taking log in Eq. (i), gives log

x 1  log k  log p m n

log

x m

The validity of Freundlich isotherm can be verified by plotting on Y-axis and log p on X-axis. If, it comes to be a straight line, the Freundlich isotherm is valid.

Q4)

Explain the terms with suitable examples :(i)

A4)

Alcosol

(ii)

Aerosol

(iii)

Hydrosol

(i) Alcosol :It is a colloidal dispersion having alcohol as the dispersion medium. For eg :- Collodion, which is a colloidal sol, of cellulose nitrate in ethyl alcohol. (ii)

Aerosol :-

It is a colloidal dispersion of a liquid in a gas eg:- fog.

(iii) Hydrosol :- IT is a colloidal dispersion of a solid in water as the dispersion medium.eg :- starch sol. Q5)

Explain the following observations :-

(a) Cotrell’s smoke precipitator is fitted at the mouth of the Chimneys use is factories. (b)

Physical adsorption is multilayered, while chemisorption is monolayered.

(c) A white precipitate of silver halide becomes coloured in presence of the dye eosin. A5)

(a) The charged collodial particles of carbon after coming in contact with oppositely charges electrode in Cottrell precipitator lose their charge and settle down at the bottom. (b) In physical adsorption, there are Weak Vander Waals forces. Therefore it forms multilayers. In Chemisorption, adsorbate is attached by chemical bond. There is a strong force of attraction. Therefore, only one layer is obtained. (c) Eosin is adsorbed on the surface of silver halide precipitate making it look coloured.

CHAPTER – 6 GENERAL PRINCIPLES OF EXTRACTION 01 MARK Q1.

Name the chief ores of aluminium and zinc.

A1.

Chief ores of – Aluminium – Bauxite. A

l0 x

(OH) 3- 2x [Where O

NH 3

>

O

Therefore splitting produced bythem in d-orbitals will also be in the same oreder (or O NO2 NH 3 H2 Will be maximum in , followed by and O). Hence , energy 1 absorbed will be in the same order. Since E α λ opposite order. i.e NO2 NH 3 H2 [Ni( )6]4- < [Ni( )6]2+< [Ni( O)6]2+.

, wavelength absorbed will be in the

CHAPTER – 10 HALOALKANES AND HALOARENES 1 Mk Questions: 1)

Write the IUPAC name of the following compound : CH3 H3C – C – CH2Cl CH3

Ans

1 – Chloro – 2, 2 – dimethylpropane

2) Ans

Draw the structure of 2-bromopentane H3C – CH2 – CH2 – CH – CH3 Br

3) Write a chemical reaction in which iodide ion replaces the diazonium group in a diazonium salt. warm + Ans N2 Cl + KI(aq) > I + N2 + KCl . 4) Out of CH3 CH Cl CH2 CH3 and CH3 CH2 CH2 CH2 Cl which one is hydrolysed more easily by aq. KOH? Ans CH3 CH – CH2 CH3 , as it’s a secondary halide | Cl 5) Ans

How can you convert methylbromide to methylisocyanide in a single step? Ag CN CH3Br alc . > CH3NC + Ag Br

2 Mks Questions 1) Ans

What are ambident nucleophiles? Explain with an example. Ambident nucleophiles have two nuclophilic sites through which they can attack.



Eg. [ΘC N ↔: C = NΘ] linking through C results in alkyl cyanides and through N results in isocyanides.

2)

Write chemical equations when i) ethylchloride is treated with aq.KOH ii) chlorobenzene is treated with CH3COCl in presence of anhyd.AlCl3.

Ans

i) C2H5Cl

aq . KOH > C2H5OH + KCl ❑ COCH3

ii) H3COC

CH 3 COCl Anhyd AlCl 3 >

Cl

Cl +

Cl + HCl 2-chloroacetophenone

4-chloroacetophenone

3)

Write the mechanism of the following reaction: E+ OH , H 2 O N – BuBr + KCN > n – BuCN ❑

Ans

CN- is an ambident nucleophile  it can attack through C and N

C-C bond is stronger than C-N bond, cyanide is formed. 4)

Ans

5) Ans

Give reasons: a) The order of reactivity of haloalkanes is RI>RBr>RCl b) Neopentyl chloride, (CH3)3C – CH3 Cl doesn’t follow SN2 mechanism. a) Larger the size of halogen atom, weaker the C-X bond, R group remaining same i.e; bond strength follows the order R-I CH3CH2OH .

b)

c)

5)

What are enantiomers? Identify the asymmetric carbons in the following molecule:a b c d HOOC CH (OH) – CH (OH) COOH

Ans

Enantiomers are non-superimposable mirror images of an optically active compound. b and c are asymmetric carbon atoms as they are bonded to 4 different groups.

6)

Write short notes on : a) Gatterman reaction b) Wurtz reaction c) Peroxide effect or Kharasch effect

Ans a) Aromatic 1o amines produce benzene diazonium salts with HNO2 (produced in situ) at 273 K. Chloro arenes and bromoarenes can be prepared using Cu/HCl or Cu/HBr from these diazonium salts.

b) Alkyl halides, when treated with sodium in presence of dry ether, produce an alkane with even number of C atoms. dry 2 R-X + Na ether > R _ R + 2NaX E.g. 2 CH3 Br + Na

dry . ether > CH3 – CH3 + 2NaBr

If a mix of two different alkyl halides is taken, then a mixture of alkanes is obtained which is difficult to separate. Na E.g. CH3 Br + C2H5Br dry ether > CH3-CH3 + C2H5 + CH3-C2H5 c) The addition of Hydrogen bromide to unsymmetrical alkenes in presence of a peroxide takes place in such a way that H goes to that C which has lesser H atoms and Br goes to the C with greater number of H atoms. ( C6 H 5 COO ) 2 CH3 – CH = CH2 + H Br > CH3 – CH2 – CH2 Br ❑ This rule doesn’t apply to addition of HCl or HI 7)

Explain the following with an example each a) Swarts reaction b) Finkelstein reaction c) Hunsdiecker reaction Ans a) Flourination of hydrocarbons directly with F 2 occurs explosively due to the large amount of energy released. Hence, they can be conveniently prepared indirectly by halogen exchange with chloro and bromoalkanes. CH3Br + AgF → CH3F + AgBr 2 C2H5Cl + .Hg2F2 → 2C2H5F + Hg2Cl2 b) Iodoalkanes can be easily prepared from chloroalkanes or bromoalkanes by heating with NaI in acetone. acetone R-Cl + NaI > R – I + NaCl Δ

R –Br + NaI

acetone > R – I + NaBr Δ

c) Bromoalkanes can be prepared by refluxing the silver salt of a carboxylic acid with Bromine in CCl4.

CH3CH2COOAg + Br2

CCl 4 Reflux > CH3CH2Br + CO2 + AgBr

The yield of alkyl bromide is 1o>2o>3o

8)

Explain why. a) Vinyl chloride is unreactive in nuceophilic substation reaction b) 3-Bromocyclohexene is more reactive than 4-Bromocyclohexene in hydrolysis with aq.NaOH c) tert-butyl chloride reacts with aq.NaOH by SN 1 mechanism, while n-butyl chloride reacts with aq.NaOH by SN2 mechanism. Ans a) Vinyl chloride is unreactive in nucleophikic substitution reactions due to reasanance which results in a partial double bond character of C-Cl bond which is difficult to break.

b) 3-bromocyclohexene forms allyl carbocation which is more stable than carbocation formed by 4-Bromocycohexene c) Tertiary carbocation is stable, so tert-butyl chloride follows SN 1 mechanism. Nbutyl chloride would form 1o carbocation which isn’t that stable. Hence it undergoes SN2 mechanism through formation of transition state. 9)

Identify the products A and B formed in the following reaction: CH3CH2CH = CH –CH3 + HCl → A + B Ans Since both the doubly-bonded C atoms have same number of H atoms, Markonikov’s rule becomes irrelevant. So, products will be → CH3 – CH2 – CH – CH2 – CH3 CH3CH2CH2CH CH3 | (A) and | (B) Cl Cl 3-chloropentane 2-choropentane 10)

What happens when: a) n-butyl chloride is treated with alc.KOH

Ans

b) c) a) a)

Bromobenzene is treated with Mg in presence of dry ether. Chlorobenze is subjected to hydrolysis. CH3CH2CH2CH2Cl - alc.KOH → CH3-CH2-CH=CH2+KCl + H2O But – 1 – ene dry ether Br + Mg > Mg Br ❑

a)

Cl2 + aq.NaOH → no reaction.

CHAPTER – 11 ALCOHOLS, PHENOLS AND ETHERS 1 Mark Questions 1)

A)

Write the IUPAC name of CH3 CH= CH – CH- CH2 – CH3 | OH Hex – 4 –en – 3 –ol

2) A)

Why is the boiling point of C2 H5 OH higher than that of CH3 OH ? Due to more number of C atoms, van der waal’s forces increase.

3) A)

Name the reagent used in bromination of phenol to 2,4, 6 – tribromophenol. Aqueous solution of bromine (Bomine water)

4) A)

Name the alcohol used to prepare the ester CH3 – C OOCH(CH3)CH3 Propan-2-ol CH3-CH(OH)-CH3

5) A)

Which is more volatile – o-nitrophenol or p-nitrophenol? o-nitrophenol,, as it has intramolecular H-bonding, whereas p-nitrophenol has intermolecular H-bonding.

O-Nitrophenol with intra molecular hydrogen bonding

02 MARKS 1)

A)

How will you convert: a) Propene to propan – 2 – ol b) Ethyl chloride to ethanol a) CH3-CH = CH2 H2O/H+ CH3 – CH – CH3  | OH Propan – 2 – ol b)

2)

A)

CH3 -CH2- Cl aq – NaOH CH3 CH2 OH [o] CH3 - CHO ----------- ------- Ethanal CrO3/PCC Write the mechanism for the following reaction: HBr → CH3CH2OH CH3CH2Br + H2O 1o alc undergoes the reaction by SN2 mechanism

Protonated Alcohol 3)

A)

How would you obtain i) Picric acid from phenol? ii) 2- Methylpropan-2-ol from 2- Methylpropene? i)

ii)

4)

A)

5) A)

Give reasons : a) Boiling points of alcohols decrease with increase in branching of the alkyl chain. b) Phenol does not give protonation reaction readily a) Because increase in branching of the alkyl chain reduces surface area, so intermolecular forces of attraction decrease. b) Because the lone pair on oxygen is delocalized over the benzene ring due to resonance, hence not available for protonation easily. Explain why cleavage of phenyl alkyl ether with HBr always gives phenol and alkyl bromide. Due to resonance, the O-C bond in phenyl alkyl ether has a partial double bond character, hence it is difficult to break. Also, phenoxide ion is stabilized by resonance, hence we get phenol and alkyl bromide.

03 MARKS 1) A)

Write a short note on Williamson’s Synthesis. An alkl halide, on treating with a suitable sodium alkoxide gives an ether RONa + RX - R– O – R + NaX Symmetric and unsymmetric ethers can be prepared by this method Aryl alkyl ethers can be prepared as shown

But there is no reaction if arylhalide and sodium alkoxide are taken Best yields of unsymmetrical ethers are obtained when alkyl halides are primary with 2 o and 3o alkyl halide, dehydrohalogenation occurs to give alkene.

2)

Convert: a) Methyl magnesium chloride --2-methyl propan-2-ol b) Benzyl chloride --- Benzyl alcohol c) Phenol -- benzoquinone

A) a)

propano ne

b)

c)

3)

A)

Give reasons : a) The boiling point of ethanol is higher than methoxymethane. b) Phenol is more acidic than ethanol c) O- and p-nitrophhenols are more acidic than phenol. a) Ethanol has intermolecular H-bonding while methoxymethane only has dipoledipole forces.

b) by effect of c)

4) A)

In phenols, O atom acquires a partial positive charge due to resonance, this weakens O-H bond, release of H+ is easy. Also, phenoxide ion is stabilised resonance. On the other hand, alkoxide ion is destabilized due to the +I alkyl group. -NO2 group, being electron-withdrawing, stabilizes the o- and p- nitrophenoxide ions.

Describe the mechanism of hydration of ethene to yield ethanol.

Protonation of Alkene

Nucleophilic attack by water on the carbocation

Deprotonation 5) Give equations for Reimer Tiemann reaction and Kolbe’s reaction. A) Reimer Tiemann reaction – Phenol reacts with chloroform in the presence of alkali to give o- and p-hydroxy aldehyde.

Kolbe’s reaction – Phenoxide ion generated by treating phenol with NaOH undergoes electrophilic substitution with CO2

6)

A)    

7)

A)

Arrange the following compounds in increasing order of their acid strength. Propan-1-ol, 2, 4, 6-Trinitrophenol, 3 – Nitrophenol, 3, 5- Dinitrophenol, phenol, 4-Methylphenol, 4-Methoxyphenol Propan-1-ol