5 MARK ANSWERS 9994342531 1 .ATOMIC STRUCTURE · PDF file · 2017-10-109994342531 1...

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5 MARK ANSWERS S.JOSEPH SURESH,MSc.,BEd.

9994342531

1 .ATOMIC STRUCTURE –II (Q.NO:52) 1.Derive de-Broglie’s equation. Plank’s quantum theory E=hϒ ϒ= frequency of the wave, h = Plank’s constant Einstein equation E = mc2 m = mass of photon, c = velocity of light hϒ = mc2 h.c/λ = mc2; λ=h/mc Significance: 1. Construction of electron microscope. λ=h/mv (or) 2. Study of surface structure of solids. λ=h/p ( mv = p) 2. Explain Davisson and Germer’s experiments. a heated tungsten filament is accelerated by using a high positive potential. fall on a large nickel crystal , and scattered in different directions. Diffraction pattern similar to diffraction of X-rays. X-rays have wave character, therefore, electrons must have wave character 3. POSTULATES OF MO THEORY 1. Electrons are present in new orbital called molecular orbital 2. Combining atomic orbital = Formed molecular orbital 3. Shape of molecular orbital = Shape of atomic orbital 4. Bonding molecular orbitals are σ , π , δ , anti bonding MO are σ∗, π∗, δ*. 5. Electrons are filled with increasing order of energy level. 4. Explain the formation of Nitrogen molecule by using MO theory. Atom : Nitrogen Atomic number : 7 Electronic configuration : 1s22s22p3

Number of electrons in N2 : 14 Electronic configuration of : KK= σ 1s2 ,σ 1s2

Bond order = ½ (Nb – Na) = ½ (8-2) =6/2 = 3 Bond nature = Triple bond (N = N) Magnetic nature = Diamagnetic

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5. Explain the formation of Oxygen molecule by using MO theory. S.JOSEPH SURESH,MSc.,BEd Atom : Oxygen Atomic number : 8 Electronic configuration : 1s22s22p4

Number of electrons in O2 : 16 Electronic configuration of : KK= σ 1s2 ,σ 1s2

Bond order = ½ (Nb – Na) = ½ (8 - 4) =4/2 = 2 Bond nature = Double bond (O=O) Magnetic nature = Paramagnetic.

2. PERIODIC CLASSIFICATIONS-II { Q.No: 64 (a) } 1. Explain the factors affecting electron affinity 1. Atomic size:

Electron affinity 1/ Size of atom Smaller the size of an atom, greater is its electron affinity. As the size of atom increases, the effective nuclear charge decreases.

Electron affinity Effective nuclear charge. 2. Shielding or Screening Effect:

Electron affinity 1/ Shielding effect Electronic energy state, lying between nucleus and outermost state hinder the nuclear attraction for incoming electron 3. Electronic Configuration: Electron affinities of inert gases are zero. This is because their atoms have stable ns2np6

configuration Electron affinity of beryllium, magnesium and calcium is practically zero. Because they have fully filled or half filled orbitals, its electron affinity will be low. 2. Explain the calculation atomic radii from the covalent bond length. a. Homo nuclear diatomic molecules Molecules of A2 type (e.g. F2, Cl2, Br2, I2... etc.) d(A – A) = r(A) + r(A) d(A – A) = 2 × r(A) r(A) = d(A–A) / 2 b. Heteronuclear diatomic molecule Molecule of AB type, d(A – B) = r(A) + r(B) r(A) and r(B) are the covalent radii of A and B atoms.
































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3. Explain the Pauling’s Method to find the radii of an ion. Pauling has found four crystals namely NaF, KCl, RbBr and CsI. In each ionic crystal the cation and anions are isoelectronic with inert gas configuration. NaF crystal : Na+

: 2, 8 F-

:2, 8 Ne type configuration KCl crystal : K+

: 2, 8, 8 Cl - : 2, 8, 8 Ar type configuration i) The sum of the radii will be equal to the inter nuclear distance between them. r(C+) + r(A-) = d (C+–A–) (1) where r(C+) = radius of cation, r(A-) : radius of anion,

d(C+–A–) = internuclear distance between C+ and A– ions in C+A– ionic crystal ii) For a given noble gas configuration, the radius of an ion is inversely proportional to its effective nuclear charge. i.e. 4. Explain Pauling scale for finding electro negativity. *This scale is based on the energy of a bond

*Consider a bond A-B between two dissimilar atoms A and B of a molecule AB. *Let the bond energies of A-A, B-B and A-B are EA-A, EB-B and EA-B respectively.

Here, XA and XB are the electro negativities of A and B respectively. The electro negativity of hydrogen to be 2.1 5. Explain how the electro negativity value helps to find the nature of bond between atoms?
































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3. p-Block Elements { Q.No: 64 (b)} 1. Explain the uses of silicones. 1) Silicones act as excellent insulators for electric motors. 2) Straight chain polymers of 20 to 500 units are used in waterproofing textiles, as lubricants and as polish. 3) Silicone rubber are mixed with paints to make them damp-resistant. 4) Silicone resins, a cross-linked polymer used as non-stick coating for pans and are used in paints and varnish. 5) Silicone oils are used for high temperature oil bath, high vacuum pump etc. 2. Explain the extraction of lead . 1. Ore: Galena: PbS 2. Concentration: Froth floatation process. 3. Smelting in a Reverberatory furnace: Ore is heated under a controlled supply of air in moderate temperature

2PbS + 3O2 2 PbO + 2SO2

PbS + 2O2 PbSO4

4. Air supply is reduced

PbS + 2PbO 3Pb + SO2

PbS + PbSO4 2Pb + 2SO2

About 90% of lead is obtained as metal, the rest passes into slag. 5. Purification: a. Liquation b. Desilverisation by Pattinson’s process or Park’s process. c. Electrolytic refining

Anode – Impure lead Cathode – Very pure lead Electrolyte – Lead fluosilicate + Hydrofluosilicic Acid

(PbSiF6) (H2SiF6) 3. Explain Anamalous Nature of Fluorine: 1. Fluorine is the most reactive element among halogen.

2. Hydrofluoric acid is a weak acid whereas the other hydrohalic acids are strong acids.

...... H– F...... H– F..... H–F.

3. AgF is soluble in water while the other AgX are insoluble.

4. Being strongly electronegative it can have only a negative oxidation state.

5. HF attacks glass while others do not.

6. It does not form any polyhalides.

4.Exlplain Denni’s method for the isolation of Fluorine. Fluorine is prepared by the electrolysis of fused sodium or potassium hydrogen fluoride

KHF2 KF + HF

HF H+ + F¯

2H+ + 2e– H 2 (At cathode)

2F – – 2e– F2 (At anode)

NaF +HF NaHF2

5. Explain the uses of Fluorine. 1. Fluorine is as freons used as refrigerants in refrigerators, deep freezers and air conditioners.

2. CaF2 is used as flux in metallurgy.

3. NaF is used for preventing dental cavities.
































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4. SF6 is used as an insulating material in high voltage equipment.

5. Teflon is used as container to store hydrofluoric acid.

6. UF6 is used in the separation of U235 from U238.

6. Explain Ramsay - Raleigh’s method: *A mixture of air and oxygen is passed into a glass globe of about

50 litres capacity. *Two platinum electrodes about 6000 - 8000 volts

N2 + O2 → 2 NO

2 NO + O2 → 2NO2

2NO2 + 2NaOH → NaNO3 + NaNO2 + H2O

Oxygen if any is removed by introducing alkaline pyrogallol in the globe. 7. Explain the separation of noble gases by Dewar’s process. Principle: coconut charcoal which adsorbs different gases at different temperatures.

Diagram: Temperature Gas that adsorbed Gas get free 173 K Ar, Kr, Xe He,Ne 93K Ne He Liquid air Ar Kr,Xe Temperature 183K Xe Kr 8. Explain the structure of Interhalogen compounds.

Type Example Hybridisation Shape Diagram AX ClF sp3 Linear AX3 ClF3 sp3d Bipyramidal (without lone pair it is T-shaped)

AX5 IF5 sp3d2 Octahedral (without lone pair it is square pyramidal )

AX7 IF7 sp3d3 Pentagonalbipyramidal
































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4. d- Block elements ( Q.No: 53)

1. Explain the extraction of Zinc

1.Ore: Zinc blende, ZnS

2.Concentration : Froth-floatation process.

3.Roasting:

2 ZnS + 3O2 1200 K 2ZnO + 2SO2

4. Reduction

ZnO + C 1673K Zn + CO

5. Purification

Anode - Impure Zn

Cathode - Pure Zn

Electrolyte - ZnSO4 + dil -H2SO4

2. Explain the extraction of Silver

1. Ores: Argentite or silver glance, Ag2S

2. Concentration: Froth-floatation process.

3. Treatment of the ore with NaCN:

The concentrated ore is treated with 0.4-0.6% solution of NaCN for several hours.

Ag2S + 4NaCN 2Na [Ag(CN)2] + Na2S

Sodium argento cyanide (soluble)

4. Precipitation of silver:

Filtrate is treated with zinc dust, silver gets precipitated.

2Na [Ag(CN)2] + Zn → Na2[Zn(CN)4] + 2Ag↓

5. Electrolytic refining:

Anode - Impure Ag

Cathode - Pure Ag

Electrolyte - AgNO3 + dil – HNO3

3. Explain the extraction of Gold.

1. Ores: Alluvial sand or Gravel sand

2. Concentration: Froth-floatation process.

3. Roasting:

ore is roasted in excess of air impurities (S,As,Te) are oxidised and escape as their volatile oxides

4.Treatment with KCN

Ore is treated with a dilute (0.5%) solution of KCN to form a soluble complex.

4Au + 8KCN + 2H2O + O2 → 4K[Au(CN)2] + 4KOH

Pot. aurocyanide

5. Precipitation of gold

Filtrate is treated with zinc dust, Gold gets precipitated.

2K[Au(CN)2] + Zn →K2[Zn(CN)4] + 2 Au↓

6. Electrolytic refining:

Anode - Impure Au

Cathode - Pure Au

Electrolyte - AuCl3 + dil – HCl
































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4. Explain the extraction of Potassium dichromate from its ore. 1. Ore: Chromite - FeO.Cr2O3

2. Conversion of chrome iron ore to sodium chromate

4FeO.Cr2O3 + 8Na2CO3 + 7O2 → 8Na2CrO4 + 2Fe2O3 + 8CO2↑

3. Conversion of Na2CrO4 to Na2Cr2O7

2Na2CrO4 + H2SO4 → Na2Cr2O7 + Na2SO4 + H2O

4.Conversion of sodium dichromate into potassium dichromate

Na2Cr2O7 + 2KCl → K2Cr2O7 + 2NaCl

5. Explain the Aluminothermic process.

*Chromic oxide is mixed with powdered Aluminium

*A mixture of barium peroxide and Mg powder is placed over this.

* The crucible is surrounded by sand which prevents loss of heat

by radiation.

* It is ignited by a piece of Mg ribbon

Cr2O3 + 2Al →2Cr + Al2O3 + 468.6 kJ

6.Explain the extraction of Silver from Silver coin. Silver coins ( Ag-Cu ) Con- HNO3 AgNO3 + Cu(NO3)2. (The solution is boiled to expel excess of nitric acid) con.HCl AgCl AgCl is separated and converted to silver by fusing with excess Na2CO3. 4AgCl + 2Na2CO3 → 4Ag↓ + 4NaCl + 2CO2 + O2

Obtained silver is purified by fusion with borax and followed by electrolytic purification.

5. f-Block Elements ( Q.No: 54) 1. Explain Lanthanide contraction , causes and consequences. Lanthanide contraction: Regular decrease in the size of M3+ ion. Causes : 1. Due to poor shielding of 4f electron 2. increasing of nuclear charge by 1 unit
































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Consequences of lanthanide contraction i) Basicity of ions According to Fajan’s rule, decrease in size of ions increase the covalent character and decreases

the basic character

La3+ > Ce3+............... >Lu3+

ii) There is regular decrease in their ionic radii.

iii) Regular decrease in their tendency to act as reducing agent.

iv) 2nd and 3rd rows of d-block transition elements are quite close in properties.

v) These elements occur together in natural minerals and are difficult to separate.

2. Differentiate Lanthanides and Actinides Lanthanides Actinides i) Binding energies of 4f electrons are higher. i) Binding energies of 5f electrons are lower. ii) Maximum oxidation state exhibited by ii) They show higher oxidation states such as +4, lanthanides is +4 e.g. Ce4+ +5 and +6. Uranium exhibits +6 iii) 4f electrons have greater shielding effect. iii) 5f electrons have poor shielding effect. iv) Most of their ions are colourless. iv) Most of their ions are coloured U3+ (red) v) They do not form complexes easily. v) They have much greater tendency to form complexes. vi) Except promethium, they are vi) All of them are radioactive. non-radioactive. 3. Explain the extraction of Lanthanides from Monazite sand. Monazite sand Heat with 2100c Con-H2SO4

Gray mud Cold water Unreacted monazite Filtrate contain TiO2 ,SiO2,ZrSiO4 Ln3+,Th3+,H3O

+, SO42-,HSO4-

Added HF Precipitate of Th3(PO4)4 Filtrate containing lanthanide And phosphate ion NaOH oxalic Acid Lanthanide hydroxide/oxalate
































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4. Explain the uses of lanthanides and actinides: Lanthanides: 1. Pyrophoric alloys : Cigarette lighters , toys , tracer bullets, flame throwing tank.

2. Ceria ,Thoria : Gas lamp materials

3. Cerium salts : Catalyst, batteries, bleaching of cotton

4. Lanthinidothermic : Extraction of pure metals like Nb, Zr, Fe, Co, Ni, Mn, Y, W, U, B and Si.

5. Mish-metals : Production of brands of steel like heat resistant, stainless, and instrumental

steels.

Actinides: 1. U-235 is used as fuel in nuclear power plants and in nuclear weapons.

2. Plutonium - 238 is used as a power source in long mission space probes.

6. Co-ordination compounds { Q.No: 55 & 65 (a) }

1. Explain co-ordination and ionisation isomerism with example

Coordination isomerism In a bimetallic complex, the distribution of ligands between the two coordination spheres can vary, giving rise to isomers called the coordination isomers. 1. [CoIII(NH3)6] [Cr(CN)6] and [CrIII(NH3)6] [CoIII(CN)6] Hexammine hexacyano Hexamine hexacyano

cobalt(III) chromate(III) chromium (III) cobaltate (III)

Ionisation isomerism Coordination compounds having the same molecular formula but forming different ions in solution are called ionisation isomers. [Co(NH3)5Br]SO4 [Co(NH3)5SO4]Br pentaamminebromocobalt(III) sulphate pentaamminesulphatocobalt (III) bromide 2. Explain Hydrate (solvent) and Linkage isomerism with example Hydrate (solvent) isomerism: Isomerism occurs for chromium chloride “CrCl3.6H2O” which may contain 4, 5, (or) 6 coordinated water molecules. 1. [Cr(H2O)4Cl2]Cl.2H2O - Bright green Tetraaquadichlorochromium(III) chloride dihydrate 2. [Cr(H2O)5Cl]Cl2.H2O - grey-green Pentaaquachlorochromium(III) chloride monohydrate 3. [Cr(H2O)6]Cl3 - Violet Hexaaquachromium(III) chloride

Linkage isomerism: Linkage isomerism occurs with ambidentate ligands. These ligands are capable of coordinating in more than one way. (monodentate ligands SCN-/NCS-

and NO2-/ONO-

For example [Co(NH3)5ONO]Cl2 the nitrito isomer - red colour pentaamminenitritocobalt(III) chloride - O attached [Co(NH3)5 NO2]Cl2 the nitro isomer - yellow colour pentaamminenitrocobalt(III) chloride - N attached
































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3. Explain the postulates of Werner’s theory 1. Every central metal atom has two valency i) primary ii) secondary

2. Primary valency = oxidation state ; satisfied by negative ion .

3. Secondary valency = Co-ordination number satisfied by negative ion or neutral molecule.

4. The molecules or ion that satisfy secondary valencies are called ligands.

5. Secondary valencies are directional in nature

Primary valencies are non-directional in nature.

6. Ligand which donate the unshared pair of electron to central metal ion .

4. Explain the postulates of Valance Bond (VB) theory

1. The central metal atom/ion makes vacant orbitals = coordination number.

2. These vacant orbitals form covalent bonds with the ligand orbitals.

3. A covalent bond is formed by the overlap of a vacant metal orbital and filled ligand orbitals.

This complete overlap leads to the formation of a metal ligand,σ (sigma) bond.

4. A hybridised orbital has a better directional than an unhybridised one.

5. Coordination number Types of hybridisation Geometry

2 sp linear

4. Explain the function of Haemoglobin

* Haemoglobin in the red blood cells carries oxygen from the lungs to thetissues. *It delivers the oxygen molecule to myoglobin in the tissues. *When the oxygen has been released it loses its bright red colour and becomes purple. *It then combines with the waste CO2 and deposits in the lungs so that can be exhaled. Nature of haemoglobin and myoglobin *Myoglobin is a monomer and haemoglobin is a tetramer. *Both are iron-porphyrin complex. It contains (Fe2+). *Each haemoglobin molecule consists of four subunits, each unit is being a folded chain. *The working part of haemoglobin is a hemegroup. *Fe2+coordinated to four nitrogen atoms of porphyrin group and one nitrogen atom of histidine group. *The sixth octahedral site is available to bind oxygen molecule. 5. How chlorophyll is important in environmental chemistry? Mention its function. *Chlorophyll is a magnesium - porphyrin complex. *The magnesium is at the centre of the modified porphyrin ring septeon (corrin). *The oxidation state of magnesium is +2 (Mg2+). *Chlorophyll ‘a’ is the pigment directly responsible for the transformation of light energy to chemical energy. * Chlorophyll acts as a light sensitizers in this important process. *Photosynthesis requires four other metal complexes, a manganese complexes, two iron complexes and a copper complex.

6. Application of VB theory in Fe complex and Ni complex study in text book

( page number: 128 &129)
































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7. Explain the following for the complexs

Content K4[Fe(CN)6] [Cu(NH3)4]SO4 [Co(en)3]Cl3 [Cr(CN)6]3-

IUPAC name

Central metal

ion

Co3+

Ligand CN

Primary

valency

2

3

Co-ordination

number

4

Geomentry Octahedral

Charge of the

complex

2+

7. NUCLEAR CHEMISTRY [Q.No.65 (b) ]

1. Distinguish between Chemical and Nuclear reaction. Chemical Reaction Nuclear Reaction 1. loss, gain or overlap of outer orbital Emission of α, β and γ from the nucleus electrons of the reactant atoms.

2. Balanced by mass Balanced by mass and energy

3. Less energy changes More energy changes

4. Energy in KJ/mole Energy in MeV

5. No new element is formed New element is formed

2. Write a note on Radio carbon dating. * 7N

14 + 0n1 (from cosmic ray) → 6C

14 + 1H1

* 6C14 → 7N14 + -1e

0

*The C14 atoms thus produced are rapidly oxidized to 14CO2 which in turn is incorporated in

plants as result of photosynthesis.

* Half life period of fossil is 5700 years.

* Amount of C14 or the number of β-particles emitted per minute per gram of carbon at the initial

and final stages, the Age of carbon material is

2.303 x t1/2 Amount of C14 in fresh wood

t = log

0.693 Amount of C14 in dead wood

* Great tool for correlating facts of historical importance.
































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3. Give the uses of radioactive isotope in medicine.

S.No Isotope Uses

1. Tritium (1H3) Measure water content of the body

2. Carbon – 11 Brain scan

3. Carbon – 14 Radio immunology

4. Iron – 59 Diagnosis of anemia

5. Cobalt – 60 Treatment of cancer

4. Explain nuclear fusion reactions in star & sun. Sun is giving out energy equally in all possible directions at the rate of 3.7 × 1033 ergs/sec 5. Explain the uses of radioactive elements in the study of reaction mechanism i) Photosynthesis : The molecular oxygen comes from water not from CO2

` ii) Hydrolysis of Ester: Isotopic O18 Present in acid not in alcohol. 8.SOLID STATE-II { Q.No.66 (a) } 1. Mention the properties of Ionic crystals. 1. Melting and boiling points are very high. 2. Hard and brittle. 3. Insulators in the solid state. 4. Soluble in water & polar solvents. 5. Good conductors when dissolved in water. 6. Heat of vapourisation is high 7. Vapour pressure is low. 2. Explain the Bragg’s spectrometer method.

Studying crystals using X rays.

X-rays is allowed to fall on the crystal mounted on a rotating table with scale and vernier,

From which the angle of incidence, θ can be measured.

Crystal table carries an ionisation chamber. X ray ionize the gas present inside.

direct measure of intensity of reflected beam

from the crystal.

current is measured from the electrometer.

values are plotted in the form of graph

For NaCl, the maximum reflection for 100 plane Order

of reflection
































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Angle of reflection (θ) Sin values Ratio 5.9° 0.103 1:2:3 11.85° 0.205 18.15° 0.312 The ratio confirms the correctness of Bragg’s equation. 3. Explain Schottky&Frenkel defects. Schottky defects: Frenkel defects: Example:(NaCl) Ionic crystal Example:(AgBr) Ionic crystal 1. Lattice points are unoccupied. It is called An ion occupies an interstitial lattice vacancies. position between the lattice points. 2. Crystal remains neutral Crystal remains neutral. ( missing no.ofcation = missingno.of anion). (no. of cation = no.of anion ) 3. Size of anion = cation Size of anion > cation 4. Explain the nature of Glass. Chief characteristics Hardness, rigidity and ability to withstand shearing stresses.

Optically isotropic

high temperature glasses undergo phase transition.

form super cooled liquid or Amorphous solids.

Vitreous state - Substance lying between the solid and liquid states.

5. What is super conductivity and give its applications. Super conductors: Certain ultra-cold substances to conduct electricity without resistance is called super conductivity. Substance having this property are called super conductors Application of superconductors:

- It consumes low energy and save more energy. s - High efficiency ore separating machines. – Used in NMRI (whole body) scan equipment.
































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9.THERMODYNAMICS-II{Q.NO.56}

1. Give the various statements of II-law of thermodynamics. i) Kelvin- Planck statement “It is impossible to construct an engine which operated in a complete cycle will absorb heat from a single body and convert it completely to work without leaving some changes in the working system”. ii) Clausius statement: “It is impossible to transfer heat from a cold body to a hot body by a machine without doing some work”. iii) Entropy statement: ‘A process accompanied by increase in entropy tends to be spontaneous”. iv) “Efficiency of a machine can never be cent percent”. v) (T1 – T2 ) % Efficiency = x 100, T1 By II law, T2<T1 % efficiency less than 100. 2. Give any five characteristics of Entropy.

1) Entropy S = q /T

2) Entropy increase in all spontaneous process.

3) Energy of the universe remains constant , the entropy of the universe tends to be maximum.

4) ΔS > 0 = Spontaneous; ΔS = 0 - equilibrium; ΔS < 0 – Non-spontaneous.

5) Units of entropy: cgs units : cal.K-1 and SI unit : JK-1

3. Give the characteristics of Gibb’s free energy.

1) Free energy is defined as G = (H-TS). ‘G’ is a state function.

2) G-Extensive property. ΔG- become intensive property , when the system is closed.

3) G has a single value for the thermodynamic state of the system.

4) ΔG<0 - spontaneous , ΔG = 0 – equilibrium, ΔG > 0 - non- spontaneous

5) ΔG = ΔH – TΔS. ΔH = ΔE + PΔV and ΔE = q – w. But TΔS = q

ΔG = q - w + PΔV – q . ΔG = - w + PΔV = network.

4. What are spontaneous reactions? What are the essential conditions for spontaneity. Spontaneous reactions:

* All the natural processes are spontaneous.

* It occur its own and does not need to be induced.

* A spontaneous process is accompanied by increase in the Entropy.

* For a spontaneous process, the enthalpy change at constant pressure will be negative.

* Combining negative ΔH and positive ΔS, ΔG get negative sign.

Essential conditions : ΔG < 0 , ΔH < 0, ΔS > 0
































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5. State Trouton’s rule. What are the substances deviate from this rule?

The heat of vaporisation (ΔHvap) in calories per mole divided by the boiling point of the liquid in

Kelvin is a constant equal to 21 cal deg-1 mole-1 and is known as the entropy of vapourisaiton.

ΔSVap = ΔHvap / Tb =21 cal deg-1 mole-1

Substances deviate from this rule:

Low boiling liquids (H and He ) which boil only a little above 0K.

water, alcohol which form hydrogen bonded liquids .

Acetic acid - partially associated in the vapour phase and possess very low entropy vaporization.

10. Chemical Equilibrium-II ( Q.No: 57)

1. Explain the Haber’s process. N2 (g) + 3H2 (g) ⇌2NH3 (g) ; ΔHf = –22.0 kcal/mole

Pressure(atm) Temperature Catalyst Ratio Reaction type Yield

300 – 500 5000C-5500C Iron (Fe) 1:3 (N2: H2) Exothermic 37%

Steam is passed to remove away the ammonia.

High pressure and low temperature – Favour for forward reaction & more yields.

2. Explain the Contact process.

2SO2 (g) + O2 (g) ⇌2SO3 (g) ; ΔHf = –47.0 kcal/mole Pressure(atm) Temperature Catalyst Ratio Reaction type Yield

700 – 1200 4000C–4500 C V2O5 (porous) 2:1 (SO2 : O2) Exothermic 97%

It is used in the manufacture of H2SO4 & oleum.

High pressure and low temperature –Favour for forward reaction & more yield.
































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3. Explain the relation between Kp and Kc aA + bB + cC + ........ ⇋ lL + mM + nN + .....

P L

l PMm P N

n ..... [L] l [M]m [N]n .... K p = , Kc =

PAa P B

b P C c [A] a [B] b[C] c...... Ci = Pi / RT

` ∴ Kc = Kp 1/(RT)Δng or ∴ Kp = Kc(RT)Δng

Δng = total number of moles of gaseous products – total number of moles of gaseous reactants. 4. Derive Kp and Kc value for the dissociation of PCl5.

PCl5(g) ⇋ PCl3(g)+ Cl2(g)

5. Derive Kp and Kc value for the Formation of HI.

H2(g) + I2(g) ⇋ 2HI(g)
































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12. Surface Chemistry { Q.No: 66 (b) } 1. Distinguish between physical adsorption and chemical adsorption.

Physical adsorption Chemical adsorption

1. It is due to intermolecular 1. It is due to chemical bond formation.

Vander waal’s force.

2. Depends on the nature of gas. 2. More specific than the physical adsorption.

Easily liquefiable gases are adsorbed readily.

3. Heat of adsorption is small. 3. Heat of adsorption is large.

4. Reversible. 4. Irreversible.

5. Forms multimolecular layers 5. Forms unimolecular layer

2.Explain the General characteristics of catalytic reactions 1. The catalyst remains unchanged in mass and in chemical composition at the end of the reaction. 2. Only a small quantity of catalyst is generally needed. 3. A catalyst cannot initiate a reaction. 4. A catalyst does not alter the position of equilibrium. 5. The catalyst is generally specific in its action.

3. Explain the types of catalytic reactions with example.

Homogeneous catalysis: Reactant and catalyst remains in same phase, SO2 (g) + O2 (g)

NO (g) 2SO3 (g)

Heterogeneous catalyst: Reactant and catalyst are in different phase, SO2 (g) + O2 (g)

pt (s) 2SO3 (g)

4. Explain the following: (i) positive (ii) Negative (iii) Auto catalyst with example (i) positive catalyst: A substace which enhance ( increase) the speed of reaction is called positive catalyst.

Ex: Decomposition of H2O2 in presence of colloidal platinum Pt

2H2O2 2H2O + O2

(ii) Negative catalyst: A substace which retards ( decrease) the speed of reaction is called negative catalyst.

2H2O2 Glycerine

2H2O + O2

(iii) Auto catalyst: One of the product act as a catalyst for the same reaction is called auto catalyst.
































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5. Explain (i) Promoters , (ii) Catalytic poison.

(i) Promoters:

A substance which, though itself not a catalyst, promotes the activity of a catalyst is called a promoter. Fe N2+ 3H2 2NH3 +Mo (ii) Catalytic poison:

A substance which destroys the activity of the catalyst is called as catalytic poison Fe N2+ 3H2 2NH3 H2S (H2S act as a catalytic poison) 6. Explain Intermediate Compound Formation Theory * The catalyst first forms an intermediate compound with one of the reactants. *The compound is formed with less energy consumption than needed for the actual reaction. *The intermediate compound is unstable combines with other reactant to form product. C

A + B AB A + C AC Catalyst Intermediate compound

AC + B AB + C Product Catalyst

Ex: The catalytic oxidation of SO2 to SO3 in the lead chamber process probably takes place as; 2 NO + O2 2NO2

Catalyst Intermediate

NO2 + SO2 SO3 + NO Product Catalyst

7. Explain Adsorption Theory. This theory is for heterogeneous catalysis. Catalyst

A(g) + B(g) C(g) + D(g)

1 2 3 4

Step - 1. Adsorption of reactant molecules:

The reactant molecules A and B strike the surface by weak vanderwaal’s forces of the catalyst.

Step - 2. Formation of Activated complex:

The reactants join to form an intermediate complex (A-B).

Step - 3. Decomposition of Activated complex:

The activated complex breaks to form the products C and D.

Step - 4. Desorption of Products:The particles of the products are desorbed from the surface.
































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8. Explain Dispersion Methods for the preparation of colloids 1. Mechanical dispersion using colloidal mill The colloidal mill consists of two steel plates nearly touching each other and rotating in opposite directions with high speed. The solid particles are ground down to colloidal size and then dispersed in the liquid. Colloidal graphite and printing inks. 2. Electro-dispersion method: (Bredig’s Arc Method) *This method is suitable for the preparation of gold, silver, platinum colloidal solution . * An arc is struck between the metal electrodes under the surface of water containing some stabilising agent such as trace of alkali. *The water is cooled by immersing the container in a cold bath that prevents the vapourises of metal 3. Ultra-sonic dispersion : *The sound waves of high frequency are usually called ultra-sonic waves. *Ultrasonic waves are passed through the solution containing larger particles. *They break down to form colloidal solution. 4. Peptisation : The dispersion of a precipitated material into colloidal solution by the action of an electrolyte in solution is termed as peptisation. Ex: Ferric hydroxide yields a sol by adding ferric chloride 9. Explain chemical methods to prepare colloids. (i) Double decomposition: As2O3 + 3 H2S As2S3 + 3 H2O arsenious oxide Yellow Arsenic sulphide

(ii) Oxidation : 2H2S + SO2 2H2O + 3 S Sulphur sol (iii) Reduction: AgNO3 + tannic acid Ag Sol AuCl3 + tannic acid Au Sol (iv) Hydrolysis: FeCl3 + 3H2O Fe(OH)3 (red sol)+ 3HCl
































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10. Explain Dialysis. Animal membranes (bladder) or those made of parchment have very fine pores. These pores permit ions (or small molecules) to pass through but not the large colloidal particles. When a sol containing dissolved ions (electrolyte) or molecules is placed in a bag of semipermeable membrane dipping in pure water, the ions diffuse through the membrane. By using a continuous flow of fresh water, the concentration of the electrolyte outside the membrane tends to be zero. Thus diffusion of the ions into pure water remains brisk all the time. In thisway, practically all the electrolyte present in the sol can be removed easily. The process of removing ions (or molecules) from a sol by diffusion through a permeable membrane is called Dialysis. The apparatus used for dialysis is called a Dialyser. 11. Explain Electro Osmosis. Electro-osmosis: The movement of the dispersion medium under the influence of applied potential..

Experiment: The phenomenon of electroosmosis can be demonstrated by using a U-tube in which a plug of wet clay (a negative colloid) is fixed. The two limbs of the tube are filled with water to the same level. The platinum electrodes are immersed in water and potential applied across them. It will be observed that water level rises on the cathode side and falls on anode side. This movement of the medium towards the negative electrode, shows that the charge on the medium is positive. Similarly, for a positively charged colloid electro-osmosis will take place in the reverse direction. 12. Explain i)Induced Catalyst ii) Auto catalyst i)Induced Catalyst: When one reactant influences the rate of other reaction. Sodium arsenite + air No oxdation sodium arsenite + sodium sulphite + air Oxidation product of both ( induced catalyst) ii) Auto catalyst : (refer Qn.No:4)
































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17. ETHERS { Q.No: 60}

1. Write a note on the type of isomerism exhibited by ethers: i) Functional Isomerism

Ethers are functional isomers of alcohols as both have the same general formula CnH2n+2O.

Molecular formula

Ethers

Alcohols

C4H10O CH3–CH2–O–CH2–CH3

Diethyl ether

CH3– O–CH2–CH2–CH3

Methyl n-propyl ether

CH3 O CH CH3

CH3

Methyl isopropyl ether

CH3–CH2–CH2–CH2–OH n-butyl alcohol

CH3 – CH – CH2 –OH

CH3

Isobutyl alcohol

OH

CH3–CH2–CH–CH3

sec.butyl alcohol

CH3

CH3 – C–OH

CH3

tert.butyl alcohol

ii) Metamerism: Molecules with same formula, same functional group, differing only in the nature of the alkyl group attached to oxygen.

CH3 O CH2 CH2

CH3 CH3 CH2 O CH

2 CH3 CH3 O CH CH3

CH3

methyl n-propylether diethylether methyl isopropylether

2. Give the methods of preparing diethyl ether:

1. Intermolecular dehydration of alcohol C2H5 – O – H + H – O – C2H5 C2H5 – O – C2H5 + H2O

Ethyl alcohol 2. Williamson’s synthesis

C2H5Br + NaOC2H5 C2H5 – O – C2H5 + NaBr Ethyl bromide sodium ethoxide

3. From alkyl halides

C2H5 – I + Ag2O + I – C2H5 C2H5 – O – C2H5 + 2AgI
































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3. How does ether react with the following? a) Air or O2 / long contact b) dil.H2SO4 c) PCl5

a) Air or O2 / long contact

C2H5–O–C2H5 (C2H5)2O2

Diethylperoxide

b) dil.H2SO4

C2H5–O–C2H5 + H–O–H 2 C2H5OH Ethyl alcohol

c) PCl5

C2H5–O–C2H5 + PCl5 2C2H5 Cl + POCl3

Ethyl chloride

4. How does diethyl ether react with PCl5, one equivalent of HI and excess of HI?

a) PCl5

C2H5–O–C2H5 + PCl5 2C2H5 Cl + POCl3

Ethyl chloride

b) With one equivalent of HI

C2H5 – O – C2H5 + HI C2H5OH + C2H5I

Ethyl alcohol Ethyl iodide c) With excess of HI

C2H5– O – C2H5 +2HI 2C2H5I + H2O

Ethyl iodide

5. How do ethers react with HI? Give the significance of the reaction C2H5 – O – CH3 + HI C2H5OH + CH3I

With excess hot concentrated hydroidic acid

CH3 – O – C2H5 + 2HI CH3I + C2H5I + H2O

This reaction is used in the Zeisel’s method of detection and estimation of alkoxy

(especially methoxy) group in natural products like alkaloids.
































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6. Give the differences between aromatic ether and aliphatic ether

No Aromatic ether (Anisole)

Aliphatic ether (Diethyl ether)

1 Comparatively high boiling liquid Volatile liquid

2 Used in perfumery Used as anaesthetic

3 Not used as solvent Used as a solvent

4 Can not be used as a substitute for

petrol

Mixed with alcohol, used as a

substitute for petrol

5 On heating with HI forms phenol and

CH3I only

It forms C2H5OH, and C2H5I

6 With nitrating mixture forms nitro

anisoles

Nitration does not take place

7 Does not form peroxide easily Forms peroxide in air.

20. Organic Nitrogen compounds { Q.NO: 69 (a) }

1. Explain the preparation of Nitro benzene from benzene

Preparation: NO2

+ Conc.H2SO4 + HNO3 330K

+ H2O

Benzene Nitrobenzene Mechanism: Step 1: Formation of Nitronium ion H

H2SO4 + HONO2 H – O – NO2 + HSO4-

+

H | +

H – O – NO2 H2O + NO2+ (Nitronium ion)

Step2: Electrophilic attack on benzene

H

+ NO2+ + NO2

H Step3 : Deprotanation H NO2 NO2

+
































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2. Explain the reduction of Nitrobenzene in different medium.

(a) Strongly acidic medium Sn/con.HCl

C6H5NO2 + 6 [H] C6H5NH2 + 2H2O Aniline (b) Neutral medium Zn/NH4Cl

C6H5NO2 + 4 [H] C6H5NHOH + H2O Phenyl hydroxylamine

(c) Alkaline medium Nitro benzene on reduction forms the intermediate products nitrosobenzene (C6H5NO) and phenyl hydroxylamine (C6H5NHOH Na3AsO3/NaOH

C6H5 – N = N– C6H5

or glucose + NaOH

O SnCl2 + NaOH azoxy benzene

C6H5NO2 C6H5– N = N – C6H5

azo benzene Zn/NaOH

C6H5– NH – NH – C6H5

hydrazo benzene or (N, N' - diphenyl hydrazine)

(d) Catalytic reduction C6H5NO2 LiAlH4 C6H5NH2 [Aniline] 3. Explain Hoffman’s hypobromite reaction or Hoffman’s bromamide reaction Reaction : Br2/KOH

CH3CONH2 CH3NH2 + CO2

Mechanism: (i) CH3CONH2 +Br2 CH3CONHBr.

(ii) CH3CONHBr CH3CO – N – Br- + K

+

.. (iii) CH3CO – N – Br CH3– CO –N + Br–

.. Nitrene - an electron deficient specie. (iv) Migration of CH3- to electron deficient nitrogen forms methyl isocyanate. O

|| .. CH3 – C – N O = C = N – CH3

.. (v) CH3– N = C = O CH3NH2 + CO2 methyl isocyanate H2

– O methylamine
































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4. Explain how primary, secondary and tertiary amines reacts with nitrous acid? (a) Primary amines react with nitrous acid to form alcohols

CH3NH2 + O = N – OH [CH3–N=N – OH] CH3OH + N2

(b) Secondary amines react with nitrous acid to form N-nitroso amines

(CH3)2 N H + HO – N = O (CH3)2 N – N = O

Secondary amine N-nitroso dimethyl amine - yellow oil

(insoluble in water)

(c) Tertiary amines react with nitrous acid to form trialkyl ammonium nitrite

(CH3)3 N + HONO (CH3)3 NH+ NO2–

(salt soluble in water)

5.Differentiate primary, secondary and tertiary amines.

Reaction Primary Secondary Tertiary

1. With HNO2 Alcohol n-Nitroso amine Salt

2. With CHCl3/KOH Carbyl amine No reaction No reaction

3. With acetyl chloride N-alkyl acetamide N,N-dialkyl acetamide. No reaction

4. With CS2 and HgCl2 alkyl isothiocyanate No reaction No reaction

5. With Diethyl oxalate dialkyl oxamide N,N-dialkyl oxamic

ester

No reaction

6. How Benzene Diazonium chloride reacts with the following: i) H2O ii) CH3OH iii) KI (i) C6H5 N2 Cl C6H5OH + N2 + HCl H O H ( phenol) (ii) C6H5 N2 Cl C6H5OCH3 + N2 + HCl CH3O H (anisole) (iii) C6H5 N2 Cl C6H5 I + N2 + KCl I K ( phenyl iodide)
































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21. Bio-Molecules { Q.NO: 69 (b)} 1. Explain the classification of Carbohydrate ( Crystalline , sweet taste) ( C6H10O5)n ( Starch , Cellulose ,Inulin ) (Glucose) (Fructose) ( Sucrose) ( Raffinose) 2. Explain the structure of Glucose

1. From the Elemental analysis and molecular weight determination Glucose = C6H12O6

2. Glucose + HI /P n- Hexane ; Indicates the presence of 6 carbon in unbranched chain.

3. Glucose + water Dissolves easily ; Indicates the absence of –COOH group ( carboxyl)

4. Glucose +Hydroxyl amine Monoxime

+ cyanohydrin

HCN

Indicates the presence of either aldehyde or keto group ( -CHO Or –C =O )

5. Glucose + Bromine water Gluconic acid. Conforms the presence of aldehyde group.

6. Gluconic acid + Con – HNO3 Saccharic acid. Indicates the presence of a 10 alcoholgroup.

7. Glucose reduces Tollen’s reagent and Fehling’s solution conforms the presence of aldehyde

group.

8. Glucose + acetic anhydride pyridine Penta acetate . This indicates the presence of5 –OH

group.
































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3. Explain the structure of Fructose

1. From the Elemental analysis and molecular weight determination Fructose = C6H12O6

2. Fructose + HI /P n- Hexane ; Indicates the presence of 6 carbon in unbranched chain.

3. Fructose + water Dissolves easily ; Indicates the absence of –COOH group ( carboxyl)

4. Fructose +Hydroxyl amine Monoxime

+ cyanohydrin

HCN

Indicates the presence of either aldehyde or keto group ( -CHO Or –C =O )

5. Fructose + Bromine water Does not oxidised. Indicates the absence of aldehyde group.

6. Fructose + Con – HNO3 glycollic acid and tartaric acid. Indicates the presence of a

carbonyl group as a Keto group.

7. Fructose + Na –Hg + water Epimer ( sorbitol + Mannitol) ; keto group at C2.

8. Fructose + acetic anhydride pyridine Penta acetate . This indicates the presence of

5 –OH group.

(3)

4. Explain the formation of peptide bond with example.

The bond formed between two amino acids by the elimination of a water molecule is called a peptide linkage or bond. – C– OH + H –N – – C – N – + H2O || | || | O H O H linkages, – CO – NH –, is called a peptide. Peptides are further designated as di, tri, tetra or penta peptides accordingly as they contain two, three, four or five amino acid molecules, same or different, joined together. Example: CH3 CH3

| | (i) NH2 – CH2 – COOH + H2N – CH – COOH H2N–CH2–CO–NH–CH–COOH Glycine Alanine Glycylalanine (a dipeptide) CH3 CH3 | | (ii) H2N – CH – COOH + H2N–CH2–COOH H2N–CH–CO–NH–CH2–COOH

Alanine Glycine Alanyl glycine
































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5. Explain the functions of LIPIDS.

* Fats and oils act as storage of energy in plants and animals. In the animal Fat give about 2¼ times as much energy as carbohydrates or proteins. Fat is a poor conductor of heat. Fat layer under skin serves to prevent losses of heat from the body. * Wax acts as a protective agent on the surfaces of animals and plants. protects from excessive loss of moisture and becoming infected with fungi and bacteria. *Phospholipids like lecithins, and cephalins play a greater role in biosystem. The lecithins are required for normal transport and utilisation of other lipids, lecithins in the liver .And organisation of the cell structure. Cephalins are found in the brain. Cephalins have been implicated in blood coagulation. *Galactolipids occur in considerable amount in the white matter of the brain and of all nervous tissue. The presence of galactose in the glycolipids suggests the importance of milk sugar in the diet of infants and children during the development of the brain and nervous system.

22. Chemistry in action { Q.No : 62 }

1. Explain about Anaesthetics The drugs which produce loss of sensation are called anaesthetics. They (i) General anaesthetics are the agent, which bring loss of all modalities of sensation, particularly pain along with ‘reversible’ loss of conciousness. (ii) Local anaesthetics prevent the pain sensation in localised areas without affecting the degree of conciousness. (i) Nitrous oxide N2O : It is the safest of the anaesthetic agents. This is used after mixing general anaesthetics ii) Chloroform CHCl3 : With oxygen it forms a toxic carbonyl chloride. Hence it is not used now. (iii) Ether : This is mixed with stabilizer 0.002% propyl halide. After absorption by tissues it attacks the central nervous system and makes unconcious. 2. Explain chromophore auxochrome theory (or) Structure of Dye i) Chromophore : An organic compound appears coloured due to the presence of certain

unsaturated groups (= or triple ) called chromophore.

Ex: N=N ( azo) ,-( -CH=CH-)n polyene

ii) Chromogen : Compound that contain chromophore are as called chromogen.

Ex: CH3 – (CH = CH)6 – CH3 is yellow in colour.

iii) Auxochrome: A group which is not a chromophore but deepan the colour of chromogen is

called auxochrome.

Example:
































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3. Explain Explain briefly on characteristics of rocket propellants 1. Rocket motors are used both in space vehicles and in offensive weapons such as missiles.

2. Rocket engines are powered by chemicals called rocket propellants.

3. Propellants are combustible compounds which on ignition undergo rapid combustion to release

large quantities of hot gases.

4. Propellants = oxidiser and a fuel.

5. Rocket to move forward according to the Newton’s Third law of Motion

6. Example: Hydrazine, Liquid hydrogen.

4. Explain the preparation of Buna –S , Buna –N and Nylon 66.

Buna-S:

Butadiene + Styrene Na

Buna-s

n CH2=CH-CH=CH2 + n CH=CH2 -(-CH2-CH=CH-CH2 – CH-CH2 -)n-

C6H5 C6H5

Properties and uses:

*Extremely resistance to wear and tear.

*Used in the manufacture of tyre.

Buna-N:

Butadiene + Acrylonitrile Na

Buna-N

2 CH2=CH- CH= CH2 + CH= CH2 Na -(-

CH2-CH-CH2- CH- CH- CH2-)n-

CH CH CN

CH2 CH2


*Extremely resistance to Swelling action of oils.

* Used for the manufacture of storage tanks for solvents.

Nylon – 66:

Hexamethylene diamine + Adipic acid Nylon- 66

n H2N-(CH2)6-NH2 + HOOC-(CH2)4-COOH H-(HN-(CH2)6-NH-OC-(CH2)4-CO-)n-OH


*Very high tensile strength.

*Used for making elastic hosiery.

5. Write a note about (i) Antipasmodics and (ii) Antibiotics.

(i) Antipasmodics:

There are group of medicines that include natural bellodana alkaloids

( atropine , bellodona ) Ex: Anisotropine , Atropine.
































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(ii) Antibiotics.

Micro organisms produce certain chemicals which inhibit the growth or metabolism of some

other micro organism are called antibiotics.

Ex: penicillins.

HARD WORK NEVER FAILS

BY

S.JOSEPH SURESH,MSc.,BEd.,

9994342531

P.G.ASST in Chemistry

Akshaya Academy Mat.Hr.Sec.School,

Oddanchatram, Dindigul.
































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