To Prepare a Pure Sample of the Complex Potassium Trioxalatoferrate (III), Kg[Fe(C₂O₄)₃l . 3H₂0 

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Theory
The complex potassium trioxalatoferrate(III) can be prepared by dissolving freshly prepared ferric hydroxide in a solution of potassium oxalate and oxalic acid.

Requirements
Three beakers (250 mL), china dish, funnel, funnel-stand, glass-rod, wash bottle, tripod stand and wire gauze. Ferric chloride, oxalic acid hydrated, potassium oxalate and potassium hydroxide.

Procedure

1. Dissolve 3.5 g of anhydrous ferric chloride 50 mL of distilled water in a 250 mL beaker.
2. In another beaker dissolve 4 g of potassium hydroxide in 50 mL of water.
3. Add KOH solution to FeCl₃ solution in small portions with constant stirring. Filter the precipitates of ferric hydroxide so formed through a buchner funnel. Wash the ppt. with distilled water.
4. In another beaker (250 mL) take 4 g of hydrated oxalic acid and 5.5 g of hydrated potassium oxalate. Add about 100 mL of water and stir thoroughly to get a clear solution.
5. Add the freshly prepared Fe(OH)₃  ppt. in small amounts to the above solution with constant stirring. The ppt. get dissolved. If ppt. does not dissolve then warm it and leave the contents for sometime.
6. Filter and transfer the filtrate to china dish and heat on a sand bath or wire-gauze to obtain crystallisation point.
7. Now place the china dish on a beaker full of cold water and keep it aside for crystallisation. China dish should be covered with a black paper as the complex is sensitive to light.
8. Decant off the mother liquor, wash the crystals with a small amount of ethyl alcohol and dry them between the folds of filter paper.
9. Find out the weight of the crystals.

Observations
Weight of the crystals obtained = ………..g
Colour of the crystals is………..

Precautions

1. Do not concentrate the solution too much.
2. Let the concentrated solution cool slowly and undisturbed to get large crystals.

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Preparation of Inorganic Compounds Viva Questions with Answers

Question.1. Define the term ‘crystallisation’.
Answer. The substances when present in well-defined geometrical shapes are called crystals. These are formed when a hot saturated solution of the salt is allowed to cool slowly and undisturbed. The process of obtaining crystals is termed as crystallisation.

Question.2. What is meant by equimolar proportions ?
Answer. Proportion of the substances in the ratio of their molecular masses, i.e. 1: 1 mole ratio.

Question.3. Why is the hot saturated solution not cooled suddenly ?
Answer. If the solution is cooled suddenly, crystals of smaller size are formed. By allowing saturated solution to cool slowly, crystals grow in size.

Question.4. What is the term ‘seeding” ?
Answer. Sometimes on cooling the saturated solution, crystallisation does not occur. A crystal of same substance is placed in the saturated solution which induces crystallisation. This process is known as seeding. It helps in quick separation of crystals from saturated solution.

Question.5. Does lithium sulphate combine with aluminium sulphate to form alum ?
Answer. No, lithium ion being very small in size does not form alums.

Question.6. What is green vitriol ?
Answer. It is hydrated ferrous sulphate (FeSO₄.7H₂0).

Question.7. What is mother liquor ?
Answer. The liquid left behind after the separation of crystals from a saturated solution is known as mother liquor. It contains soluble impurities.

Question.8. What are alums ?
Answer. Alums are double sulphates having general formula X₂SO₄.M₂(SO₄)₃.24H₂0, where X = monovalent cation such as Na⁺, K⁺ etc. and M = trivalent cation such as Al⁺³, Cr⁺³, etc.

Question.9. In the preparation of Mohr’s salt can concentrated H₂SO₄ be used in place of dilute H₂SO₄?
Answer. No, because it would oxidize ferrous ions to ferric ions.

Question.10. What is the action of heat on potash alum ?
Answer. It loses water of crystalization and becomes light and fluffy.

Question.11. Give the names of some alums where cations are other than Al³⁺.
Answer. Ferric alum, (NH₄)₂SO₄. Fe₂(SO₄)₃.24H₂0; Chrome alum, K₂So₄ . Cr₂(S0₄)₃.24H₂0.

Question.12. What are isomorphous substances ?
Answer. The substances having similar crystal structure are known as isomorphous.

Question.13. Why is dilute sulphuric acid added to the solution during the preparation of Mohr’s salt crystals ?
Answer. It prevents hydrolysis of ferrous sulphate.

Question.14. What are the uses of potash alum ?
Answer. It is used for purification of impure water. It is also used to stop bleeding from a wound and in dyeing industry.

Question.15. Why is water, used for the preparation of Mohr’s salt solution, boiled for 5 minutes ?
Answer. To expel dissolved oxygen from the water which otherwise will oxidize ferrous salt to ferric salt.

Question.16. What happens when potash alum is heated ?
Answer. It first melts and then swells up and is known as burnt alum. At this point it loses its water of crystallisation.

Question.17. Why dilute sulphuric acid is added during the preparation of aluminium sulphate solution ?
Answer. To prevent the hydrolysis of aluminium sulphate.

Question.18. How does potash alum help in purification of water ?
Answer. When potash alum is added to impure water, it causes the coagulation of colloidal impurities present in water. The precipitated impurities can be removed by filtration or decantation.

Question.19. How does potash alum help in stopping bleeding ?
Answer. Blood is a negatively charged sol, in the presence of potash alum it gets coagulated.

Question.20. Is aqueous solution of potash alum acidic or basic ?
Answer. It is acidic, it turns blue litmus paper red. The solution is acidic due to hydrolysis of the salt.

Question.21. What is the geometry of the complex K₃[Fe(C₂O₄)₃]. 3H₂0 ?
Answer. Octahedral.

Question.22. Why is the complex K₃[Fe(C₂0₄)₃]. 3H₂0 paramagnetic ?
Answer. It is paramagnetic due to the presence of five unpaired electrons in 3d-orbitals of Fe atom.

Question.23. What is the IUPAC name of the complex K₃[Fe(C₂O₄)₃]. 3H₂0 ?
Answer. Potassium trioxalatoferrate (III)-3-water.

Question.24. What is meant by high spin and low spin complexes ?
Answer. When the unpaired electrons pair up prior to complex formation, the complex is known as low spin complex and if prior to complex formation no electron pairing takes place, the complex is known as high spin complex.

Question.25. What is the difference between a complex and a double salt ?
Answer. In complex salt, the properties of all individual ions of the constituent salts may not be exhibited. Whereas in double salts properties of the ions of the constituent salts are exhibited in solution.

Question.26. What is the difference between iron compounds given below ?
K₄[Fe(CN)₆] and FeSo₄ .(NH₄)₂ So₄.6H₂0.
Answer. K₄[Fe(CN)₆] is a coordination complex whereas FeSo₄.(NH₄)₂ So₄.6H₂0 is a double salt.

Question.27. What is the coordination number of iron is potassium trioxalatoferrate (III) ?
Answer. Six

Question.28. List two examples of bidentate ligands other than oxalate ion.
Answer. (i) Ethylene diamine (en), H₂N-CH₂-CH₂-NH₂ (ii) Glycinate, H₂N-CH₂-COO^–.

Question.29. Why does the compound, potassium trioxaloferrate (III), not give tests for ferric ions ?
Answer. Because it contains Fe(III) as complex ion, [Fe(C₂0₄)₆]^3- and not as Fe³⁺ ions.

Question.30. What are chelates ?
Answer. Multidentate ligands are known as chelates.

Chemistry Lab ManualNCERT Solutions Class 12 Chemistry Sample Papers

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To Prepare a Sample of Acetanilide from Aniline

Chemistry Lab ManualNCERT Solutions Class 12 Chemistry Sample Papers

Theory
Acetanilide is prepared by acetylating aniline with acetic anhydride in the presence of glacial acetic acid. The chemical equation can be written as :

Apparatus
Round bottom flask (100 ml), water condenser, wire-gauze, tripod stand, burner, iron-stand, clamp, measuring cylinder, etc.

Chemicals Required
Aniline = 5 ml
Acetic anhydride = 5 ml
Glacial acetic acid = 5 ml.

Procedure

1. Take 5 ml of acetic anhydride in a clean dry 100 ml conical flask and add 5 ml of glacial acetic acid and shake the contents thoroughly.
2. To this mixture taken in the flask, add 5 ml of aniline and fit a water condenser.
3. Place the flask on a wire-gauze placed on a tripod stand as shown in Fig.
4. Boil the mixture for 10-15 minutes.
5. Allow the mixture to cool. Detach the condenser and pour the liquid into about 150 ml ice-cold water contained in a beaker. During addition, stir vigorously the contents of the beaker with the help of glass-rod.
6. Filter the white precipitates which separate out and wash with cold water.
   
7.  Recrystallise from hot water containing a few drops of ethyl alcohol. Weigh the crystals and record the yield.
8. Determine the melting point of the compound.

Result
Weight of acetanilide obtained =…………g
Melting point of acetanilide = ……..°C
Note: Acetanilide has white flaky crystals. Its melting point in 113°C.

Precautions

1. Freshly distilled aniline should be used in order to get good results or small amount of zinc can be added in the reaction mixture. Zinc reduces the coloured impurities in the aniline and also prevents its oxidation during the reaction.
2. Prolonged heating and use of excess of acetic anhydride should be avoided.
3. Reaction mixture should first be cooled and then poured in ice-cold water otherwise hydrolysis of acetanilide may take place.

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Semiconductor Diodes and Transistors

Electronic Devices
The devices whose action is based on the controlled flow of electrons through it is called an electronic devices. Electronic devices sire the building blocks of all the electronic circuits.
The electronic devices are two types :

1. Vacuum tubes: Vacuum diodes (cathode, anode) triode (three electrodes) and pentode (five electrodes) etc.
2. Solid state electronic devices: The semiconductors are the basic materials for designing the solid state electronic devices such as junction diode (2-electrodes), transistor (3-electrodes) and integrated circuits (IC).
   The solid state devices are better than vacuum tubes because vacuum tubes are bulky, consume high power and voltage (≈100 V), low life and low reliability while solid state devices are small in size, consume low power and low voltage, long life and high reliability.

Energy levels and energy bands
(a) Energy Levels in Atoms: In an atom electrons revolve around the nucleus in almost circular orbits of different radii representing shells and subshells. Energy of electrons in each subshell is definite. These definite energy values, are called energy levels of the atoms. Hence, each electrons will have a different energy levels.
They are represented by straight lines drawn parallel to each other at distances proportional to the energy they represent.
(b) Energy Bands in Solids (Crystals): In a crystal, atoms are packed very closely. A crystal of volume 1 c.c. has about 10²³ closely packed atoms. The energy levels (values) of electrons in outer shells are influenced in the presence of nuclei of other atoms. Due to interatomic interaction, the different energy levels with continuous energy variation form is called energy band.

Different types of energy bands

1. Conduction Band (C): The topmost partially filled band, is called Conduction band. It is so called because by accommodating more electrons in itself, the band provides mobility to electrons and helps in conduction.
2. Valence Band (V): The energy band which includes the energy levels of the valence electrons is called the valence band. It is lower most completely filled band.
3. Forbidden Band (F): The band separating a valence band and conduction band, is called Forbidden band. The difference in the extreme energy levels of the forbidden band, is called energy gap (E_g).

Conductors, insulators and semiconductors

1. Conductors: In case of conductors, the partially filled conduction band (C) and the valence band (V) overlap. Electrons of valence band move freely in partially filled conduction band.
   Free electrons are available for movement even when small electric field is applied. Electric current flows.
2. Insulators: In case of insulators, the empty conduction band (C) and the valence band (V) have an energy gap (E ) of about 6 eV  Due to large energy gap, no electrons jumped from valence band to empty conduction band. Valence band remains completely filled.
   
   No electron movement takes place even when large electric field is applied. No electric current flows.
   An important example of insulator is diamond with energy gap of about 5.4 eV.
3. Semiconductors: In case of semiconductors, the partially filled conduction band (C) and the valence band (V) have an energy gap (E_g) of about leV . Due to small energy gap some electrons may gain sufficient thermal energy at ordinary room temperatures and enter the empty conduction band to make it partially filled. Electrons reaching conduction band, leave electron vacancy (hole) in valence band.
   Due to vacancy in valence band, some electron movement may take place when a moderate electric field is applied. A weak electric current flows.
   Common examples of semiconductors are Silicon (14) and Germanium (32) with energy gaps of about 1.2 eV and 0.73 eV respectively.

Intrinsic and extrinsic semiconductors
(a) Intrinsic Semiconductor: A pure semiconductor material, is called intrinsic semiconductor. Crystalline form of Germanium (Ge) and Silicon (S_i) are examples of intrinsic semiconductors. Concentration of holes (n_h) = concentration of electrons (n_e) in pure semiconductor.
(b) Extrinsic Semiconductor: A semiconductor material made deliberately impure by adding suitable impurity atoms through doping, is called an extrinsic semiconductor.
It is of two types :

1. n-type semiconductor: The re-type semiconductor is obtained by adding a small quantity (one millionth part) of a pentavalent impurity like Phosphorus (15), Arsenic (33), Antimony (51) or Bismuth (83) to a pure semiconductor crystal.
   Generally Arsenic (As) is taken for this purpose. N_e >> n_h
2. p-type semiconductor: Ap-type semiconductor is obtained by adding a small quantity (one millionth part) of a trivalent impurity like Boron (5), Aluminium (13), Gallium (31), Indium (49) or Thalium (81) to a pure semiconductor crystal.
   Generally Indium (In) is taken for this purpose. N_h >> ne
   The process of adding impurity deliberately, is called doping.

Semiconductor diode (Junction Diode)
(a) Preparation: When a p-type semiconductor crystal is grown over an n-type semiconductor crystal (or vice-versa), the common surface of the two types, is called a junction. The compound crystal forms a semi-conductor device, called junction diode

(b) Description: Two important processes occur during the formation of a p-n junction. Diffusion and drift. Due to higher concentration of holes in p-section, holes diffuse from p-side
to re-side and leaves befind an ionised negative charge which is immobile. As the holes continue to diffuse, a layer of negative charge on the p-side of the junction is developed. Similarly, electrons diffuse due to its high concentration from re-side to p-side and leaves befind the positive charge which is immobile. As the electrons continue to diffuse, a layer a positive on re-side developed. This space charge region on either side of the junction together is known as depletion layer. Due to diffusion, an electric field set-up directed from positive charge (re-side) to negative charge (p-side). Due to this field, electrons drift from p-side to re-side and holes from re-side to p-side, making a drift current which is opposite to diffusion current. Initial, diffusion current is large and drift current is small, but when these current become equal, the p-n junction is formed. The loss of electrons from re-region and gain of electrons in p-region causes a potential difference across the junction of the two region. It is called potential barrier. This potential barrier tends to prevent further flow of electrons from re-region to p-region.

Biasing of junction
(a) Definition: Applying an external potential difference on the faces of a junction, is called biasing of the junction.
It is done by connecting the outer ends of the two sections of the junction diode to the positive and the negative terminals of a battery (source of potential difference).
(b) Type: It has two types :
(1) Forward biasing (2) Reverse biasing

1. Forward Biasing: When outer end of re-type section is connected to the negative terminal and that of p-type section is connected to the positive terminal, the biasing of junction is called forward biasing.
   The free majority charge carriers from each section are made to move forward towards the junction. IF forward bias potential is more than potential barrier, the charges from both sections cross the junction and a current flows through the junction and the circuit. It is called forward current. It is formed due crossing the junction by majority carriers. It is order of 10-3 A. The size of depletion layer decreases in forward biasing and hence resistance decreases.
2. Reverse Biasing: When outer end of re-type section is connected to the positive terminal and that of p-type section is connected to the negative terminal, the biasing of junction is called reverse biasing.
   The free majority charge carrier from each section are made to move in reverse direction, away from the junction. The majority carrier charges do not cross the junction and no current flows through the junction due to flow of majority carriers. The minority carrier charge are made to move toward the junction due to reverse biasing. The minority carrier charge cross the junction and there is very small current order to micron across the junction. The size of depletion layer increases in reverse biasing and hence resistance increases.

Characteristics of a junction diode
(a) Definition: Graphs drawn between bias voltage and current of a junction diode, are called characteristics of the diode. They reveal the character (way of behaviour) of the junction diode.
(b) Type: These are of two types :
(1) Forward Bias Characteristic: This is obtained by plotting a graph between forward bias voltage and forward current. Junction resistance for forward bias is about 10 ohm.
(2) Reverse Bias Characteristic: This is obtained by plotting a graph between reverse bias voltage and reverse circuit current. Junction resistance for reverse bias is about 10,000 ohm.

Zener diode
A specially designed junction diode with a heavily doping to operate in the reverse breakdown region is called a zener diode.
(a) Principle: It works on phenomenon of Zener breakdown at reverse voltage, for which large changes in diode current produce only a small change in diode voltage. This makes the zener diode to use as the voltage regulator.
(b) Zener Breakdown: When the junction diode is reverse biased, free charge carriers are attracted away, (in reverse direction) from junction. As the charges do not cross the junction, no current flows in external circuit.
As bias voltage is increased, covalent bonds between atoms break, setting more electrons and holes free in each section. The free holes in n-type section and free electrons in p-type section, are called minority carriers.
The reverse bias makes these minority carriers move towards junction and cross it. Thus a current flows through junction and in opposite direction. The current is very small.
At a certain reverse bias voltage, a breakdown takes place and the current rises suddenly. The breakdown is called Zener breakdown and the voltage is called Zener voltage (V_z). When the reverse bias voltage V = Vz, then the electric field strength is high enough to pull valence electrons from the host atoms on the p-side which are accelerated to n-side These electrons account fer high current observed at the breakdown. The emission of electrons from the host atoms due to high electric field is known as internal field emission or field ionisation. The situation is shown in Fig. 10.04. There is a large change in diode current for a small change in diode voltage.

Semiconductor triode (junction triode) or transistor
Introduction. When a thin layer of one type lightly doped semiconductor is grown between two comparatively broad sections of other type heavily doped semiconductor, the arrangement forms a semiconductor device called Junction Triode or Transistor.
There are two types of transistors

1. n-p-n transistor
2. p-n-p transistor.

n-p-n transistor. It is formed by growing a thin layer of p-type semiconductor over n-type semiconductor crystal. Then a thicker layer of n-type is grown over p-type thin layer.
The first thick layer section is called Emitter (E), the second thin layer is called Base (B), the third layer is called Collector (C)  Emitter, base and collector from the three important terminals, hence the name ‘triode’.
The level of doping is more in emitter than in collector. But collector section is larger than emitter section. The thin base layer is lightly doped.
The symbol of an n-p-n transistor is given in is the symbol of a p-n-p transistor. Arrow in emitter represents direction of flow of positive charge in the emitter.
Symbols

Characteristic of a transistor
(a) Definition: Graphs drawn between bias voltage and current in the circuit, are called characteristics of the transistor. They reveal the character (way of behaviour) of the transistor.
(b) Types: These are of two types :

1. Input Characteristics: In common base circuit, these are obtained by plotting graphs between emitter voltage (V_e) and emitter current (I_e) for different constant collector voltage (V_c).
   In common emitter circuit, these are obtained by plotting graphs between base voltage (V_b) and base current (I_b) for different constant collector voltage (V_c).
2. Output Characteristics: In common base circuit, these are obtained by plotting graphs between collector voltage (V_c ) and collector current (I_c) for different constant emitter current (I_e).
   In common emitter circuit, these are obtained by plotting graphs between collector voltage (V_c) and collector current (I_c) for different constant base current (I_b).

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To Prepare a Sample of Dibenzalacetone

Theory
The preparation of dibenzal acetone is an example of Claisen-Schmidt reaction. This reaction takes place between aromatic aldehydes and aliphatic ketones in presence of sodium hydroxide. Two moles of benzaldehyde condense with one mole of acetone to give dibenzalacetone. The chemical equation can be written as :

Apparatus
Conical flask (100 ml), beaker (250 ml), test-tube, funnel, filter-papers, etc.

Chemicals Required
Benzaldehyde = 2.5 ml
Acetone = 1.0 ml
10% NaOH solution = 5 ml
Rectified spirit = 25 ml

Procedure

1. Take a conical flask (100 ml) and add 2.5 ml benzaldehyde, 1.0 ml of acetone and 25 ml of methylated spirit. Cork the flask and shake to obtain a clear solution.
2. Take 5 ml of 10% NaOH solution in a test-tube and add this to conical flask drop by drop with shaking of the flask. Maintain the temperature of the reaction mixture between 20-25°C during addition of sodium hydroxide solution.
3. Cork the flask again and shake vigorously for about 10 minutes, releasing pressure from time to time.
4. Allow it to stand for about 20 minutes at room temperature and then cool in ice water for a few minutes.
5. Filter the yellow coloured solid and wash it with water to remove traces of alkali.
6. Recrystallization of dibenzalacetone.
   Dissolve the above yellow coloured crude solid in minimum amount of hot rectified spirit and then allow it to cool slowly. Pale yellow crystals of dibenzalacetone separate out. Filter the crystals and dry.
7. Weigh and record its yield and melting point.

Result
Weight of dibenzalacetone obtained =…………g
Melting point of dibenzalacetone is………°C
Note: (Approximate expected yield of dibenzalacetone is 1.5 g)
The melting point of dibenzalacetone is 112°C.

Precautions

1. Add NaOH dropwise to the reaction mixture with constant shaking and maintaining the temperature around 20°C.
2. Wash the ppt. with water to remove traces of sodium hydroxide sticking to them.
3. Use minimum amount of rectified spirit to dissolve crude sample for crystallisation.

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To Prepare a Sample of p-Nitro acetanilide from Acetanilide

Chemistry Lab ManualNCERT Solutions Class 12 Chemistry Sample Papers

Theory
The nitration of aniline is difficult to carry out with nitrating mixture (a mixture of cone. H₂SO₄ ,and cone. HN0₃ ) since —NH₂ group gets oxidised which is not required. So the amino group is first protected by acylation to form acetanilide which is then nitrated to give p-nitroacetanilide as a major product and o-nitroacetanilide as a minor product. Recrystallisation from ethanol readily removes the more soluble ortho-compound and the pure p-nitroacetanilide is obtained. The chemical equation can be written as :

Apparatus
Conical flask (100 ml), beaker (250 ml), measuring cylinder (100 ml), funnel, glass-rod, test-tube, filter-papers, etc.

Chemicals Required
Acetanilide = 5g
Glacial acetic acid = 5 ml
Cone. H2S04 =10 ml
Fuming HN03 = 2 ml
Methylated spirit = 20 ml.

Procedure

1. Take a 100 ml conical flask and add 5 g of powdered acetanilide in it. Add 5 ml of glacial acetic acid and stir the mixture by the use of glass-rod.
2. Place 2 ml of fuming nitric acid in a clean test-tiibe and cool it in a freezing mixture (ice + salt) taken in a beaker. Carefully add drop by drop 2 ml of cone, sulphuric acid with constant shaking and cooling.
3. Add the remaining 8 ml of cone. H2S04 drop by drop (with cooling under tap water) to the conical flask containing acetanilide and glacial acetic acid. Place the conical flask in a freezing mixture (Fig). Stir the contents and wait until the temperature becomes less than 5°C.
4. To the cooled contents in the flask add nitrating mixture prepared in step (2) drop by drop with constant stirring. During addition temperature of the mixture should not rise above 10°C. This operation should take about 15 minutes (Fig).
5. Remove the conical flask from the freezing mixture and allow it to stand for 30 minutes at room temperature.
6. Pour the contents of the flask on the crushed ice taken in a beaker. Stir it and filter the crude product. Wash thoroughly with cold water to remove acid.
   
7. Recrystallisation of p-nitroacetanilide. Dissolve the crude product obtained above in about 20 ml of methylated spirit. Warm to get a clear solution. Filter while hot and cool the filtrate in ice. o-Nitroacetanilide goes in the filtrate while p-nitroacetanilide is obtained as colourless crystals on the filter paper. Wash the solid on the filter paper with cold water. Dry the solid, weigh it and record its yield.

Result
Weight of p-nitroacetanilide is obtained =………g
Melting point of the compound is……….°C
Note: Approximate expected yield is 4 g.
The melting point of p-nitroacetanilide is 214°C.

Precautions

1. During addition of nitrating mixture, the temperature of the reaction mixture should not rise above 10°C.
2. Addition of fuming nitric acid should be done drop wise.
3. Do not inhale the vapours of nitric acid as they are very corrosive in nature. Addition of nitrating mixture may preferrably be done in a fume-cupboard.

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To Prepare 2-Naphthol Aniline or Phenyl-azo-β-Naphtholdye 

Theory
2-Naphthol aniline dye or Phenyl-azo-β-naphthol is an orange-red dye. It belongs to a large class of azo-compounds, all of which contain the characteristic grouping

Azo compounds are all coloured compounds. For the preparation of this dye, aniline is diazotised and then diazonium salt thus obtained is subjected to coupling reaction with 2-naphthol.

Apparatus
One 100 mL conical flask, one 100 mL beaker, one 250 mL beaker, ice-bath, glass-rod, buchner funnel, water pump.

Chemicals Required
Aniline = 4.5 ml
Sodium nitrite = 4g
2-Naphthol = 7 g
Cone, hydrochloric acid = 10 ml
Glacial acetic acid = 40 ml

Procedure

1. Take a 100 ml conical flask and add 4.5 ml of aniline, 10 ml of cone. HCl and 20 ml of water. Cool this solution to 5°C by placing the conical flask in a trough containing ice- cold water.
2. In a 100 ml beaker dissolve 4 g of sodium nitrite in 20 ml of water and cool this solution also to 5°C.
3.  Now slowly add sodium nitrite solution to the solution of aniline in cone. HCl.
4. Dissolve 7.0 g of 2-naphthol in 60 ml of 10% NaOH solution taken in a 250 ml beaker and cool this solution to 5°C by placing in an ice bath. Some crushed ice may be added directly to fecilitate cooling.
5. Now add the diazotised solution very slowly to the naphthol solution with constant stirring. The mixed solutions immediately develop a red colour and the phenyl-azo-β- naphthol rapidly separates as orange-red crystals.
6. When the addition of diazo solution is complete, allow the mixture to stand in ice-salt mixture for 30 minutes, with occasional stirring. Filter the solution through a buchner funnel under suction from the pump. Wash the phenyl-azo-β-naphthol with water and dry the crystals obtained by pressing between the folds of filter paper.
7. Recrystallise the product from glacial acetic acid. Filter the crystals obtained at the pump. Wash with a few ml of ethanol to remove acetic acid. Phenyl-azo-β-naphthol is obtained as orange-red crystals. Expected yield is 3 g and melting point is 133°C.

Result
Weight of phenyl-azo-β-naphthol obtained as orange-red crystals = …….g.
Melting point of phenyl-azo-β-naphthol is…….°C.

Precautions

1. The solution of the aniline hydrochloride should be cooled to 5°C, and this temperature should be maintained throughout the addition of the sodium nitrite solution.
2. Addition of sodium nitrite should be very slow because the reaction is exothermic and may cause the temperature to rise.
3. Always add diazonium chloride solution to β-naphthol solution for dye formation and not vice-versa.

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Preparation of Organic Compounds Viva Questions with Answers

Chemistry Lab ManualNCERT Solutions Class 12 Chemistry Sample Papers

Question.1. What is Claisen-Schmidt reaction ?
Answer. It is the condensation of aromatic aldehydes with aliphatic ketones (or aryl alkyl ketones) in the presence of dil. NaOH solution.

Question.2. What is IUPAC name of acetanilide ?
Answer. Phenylethanamide.

Question.3. What is the function of zinc dust in the preparation of acetanilide ?
Answer.

1. It reduces the coloured impurities present in aniline.
2. It prevents oxidation of aniline during the reaction.

Question.4. Give the formula of 2-naphthol aniline dye.
Answer.

Question.5. What is the colour of 2-naphthol aniline dye ?
Answer. Orange-red.

Question.6. What is diazotisation reaction ?
Answer. It is the reaction of primary aromatic amines with nitrous acid to form diazonium salt. The reaction is carried out at low temperature (below 5°C)

Question.7. What is coupling reaction ?
Answer. It is the reaction of diazonium salts with highly activated benzene rings such as phenolic ring. The reaction involves electrophilic substitution and the product obtained is generally a dye.

Question.8. Which of the following compounds on diazotisation followed by compling with β-naphthol will form an azo dye ?
(i) p-Toluidine (ii) Benzylamine (iii) N-Methylaniline
Answer. p-Toluidine, because it is a primary aromatic amine.

The post Preparation of Organic Compounds Viva Questions with Answers appeared first on Learn CBSE.

To draw the I-V characteristic curve of a p-n junction in forward bias and reverse bias

Aim
To draw the I-V characteristic curve of a p-n junction in forward bias and reverse
bias.
Apparatus
A p-n junction (semi-conductor) diode, a 3 volt battery, a 50 volt battery, a high resistance rheostat, one 0-3 volt voltmeter, one 0-50 volt voltmeter, one 0-100 mA ammeter, one 0-100 μA ammeter, one way key, connecting wires and pieces of sand paper.

Theory
Forward bias characteristics. When the p -section of the diode is connected to positive terminal of a battery and n-section is connected to negative terminal of the battery then junction is said to be forward biased. With increase in bias voltage, the forward current increases slowly in the beginning and then rapidly. At about 0.7 V for Si diode (0.2 V for Ge), the current increases suddenly. The value of forward bias voltage, at which the forward current increases rapidly, is called cut in voltage or threshold voltage.
Reverse bias characteristics. When the p -section of the diode is connected to negative terminal of high voltage battery and n-section of the diode is connected to positive terminal of the same battery, then junction is said to be reverse biased.
When reverse bias voltage increases, initially there is a very small reverse current flow, which remains almost constant with bias. But when reverse bias voltage increases to sufficiently high value, the reverse current suddenly increases to a large value. This voltage at which breakdown of junction diode occurs (suddenly large current flow) is called zener breakdown voltage or inverse voltage. The breakdown voltage may^tarts from one volt to several hundred volts, depending upon dopant density and the depletion layer.

Diagram

Procedure
For forward-bias

1. Make circuit diagram as shown in diagram.
2. Make all connections neat, clean and tight.
3. Note least count and zero error of voltmeter (V) and milli-ammeter (mA).
4. Bring moving contact of potential divider (rheostat) near negative end and insert the key K. Voltmeter V and milli-ammeter mA will give zero reading.
5. Move the contact a little towards positive end to apply a forward-bias voltage (V_F) of
   0. 1 V. Current remains zero.
6. Increase the forward-bias voltage upto 0.3 V for Ge diode. Current remains zero, (It is due to junction potential barrier of 0.3 V).
7. Increase V_F to 0.4 V. Milli-ammeter records a small current.
8. Increase V_F in steps of 0.2 V and note the corresponding current. Current increases first slowly and then rapidly, till VF becomes 0.7 V.
9. Make V_F = 0.72 V. The current increases suddenly. This represents “forward break-down” stage.
10. If the V_F increases beyond “forward breakdown” stage, the forward current does not change much. Now take out the key at once.
11. Record your observations as given ahead.
    For reverse-bias
12. Make circuit diagram as shown in diagram.
13. Make all connections neat, clean and tight.
14. Note least count and zero error of voltmeter (V) and micro-ammeter (μA).
15. Bring moving contact of potential divider (rheostat) near positive end and insert the key K Voltmeter V and micro-ammeter μA will give zero reading.
16. Move the contact towards negative end to apply a reverse-bias voltage (V_R) of 0.5 V, a feebly reverse current starts flowing.
17. Increase V_R in steps of 0.2 V. Current increases first slowly and then rapidly till V_R becomes 20 V. Note the current.
18. Make V_R = 25 V. The current increases suddenly. This represents “reverse break-down” stage. Note the current and take out the key at once.
19. Record your observations as given ahead.

Observations
For forward-bias
Range of voltmeter                        = …..V
Least count of voltmeter              = …..V
Zero error of voltmeter                = …..V
Range of milli-ammeter              = …..mA
Least count of milli-ammeter    = …..mA
Zero error of milli-ammeter      = …..mA

1.                         Table for forward-bias voltage and forward current

Note. The readings are as a sample.
For reverse-bias
Range of voltmeter                     = …..V
Least count of voltmeter           = …..V
Zero error of voltmeter              = …..V
Range of micro-ammeter          = …..μA
Least count of micro-ammeter = …..μA
Zero error of micro-ammeter    = …..
2.                        Table for reverse-bias voltage and reverse current

Note. The readings are given as a sample.

Calculations
For forward-bias
Plot a graph between forward-bias voltage V_F (column 2) and forward current I_F (column 3) taking V_F along X-axis and I_F along Y-axis.
This graph is called forward-bias characteristic curve a junction diode.


For reverse-bias
Plot a graph between reverse-bias voltage V_R (column 2) and reverse current I_R (column 3) taking V_R along X-axis and I_R along Y-axis.
This graph is called reverse-bias characteristic curve of a junction diode.

Result
Junction resistance for forward-bias = 40 ohms
Junction resistance for reverse-bias = 2 x 10⁶ ohms.

Precautions

1. All connections should be neat, clean and tight.
2. Key should be used in circuit and opened when the circuit is not being used.
3. Forward-bias voltage beyond breakdown should not be applied.
4. Reverse-bias voltage beyond breakdown should not be applied.

Sources of error
The junction diode supplied may be faulty.

Viva Voce

Question. 1. Define energy level in an atom.
Answer. Definite energy value of an electron in the subshell of the atom, is called energy level of the atom.

Question. 2. Define energy band in a crystal.
Answer. Broadened energy level (line) in a crystal, is called energy band of the crystal.

Question. 3. Name different types of energy bands.
Answer. Different energy bands are :
1. Conduction band (C), 2. Valence band (V), 3. Forbidden band (F).

Question. 4. Define different energy bands.
Answer. Read Art. 10.03 (1, 2, 3, 4).

Question.5. Name different types of substances.
Answer. Different types of substances are :
1. Conductors, 2. Insulators, 3. Semiconductors.

Question. 6. Distinguish between a conductor, an insulator and a semiconductor.
Answer. Read Art. 10.04 (1, 2, 3).

Question.7. How are electrical conductivity and resistivity related?
Answer. Electrical conductivity is reciprocal of resistivity

Question. 8. What is S.I. unit of conductance?
Answer. S.I. unit of conductance is siemen (S).

Question. 9. What is order of conductivity of conductors, semiconductors and insulators?
Answer.10²-10⁸, 10⁵-10^-6 and 10^-11-10^-19 S m^-1 respectively.

Question. 10. Define a hole.
Answer. A place vacated by an electron, is called a hole. It is associated with a positive charge.

Question.11. Define an intrinsic semiconductor.
Answer. A pure semiconductor material, is called an intrinsic semiconductor, it n_e = n_n

Question. 12. Which materials are commonly used as semiconductors?
Answer. Silicon and germanium are commonly used as semiconductors.

Question. 13. Which of the upper two materials has less energy gap?
Answer. Energy gap has value 0.72 eV for germanium and 1.12 eV for silicon.

Question.14. Define an extrinsic semiconductor.
Answer. A semiconductor material made deliberately impure, is called an extrinsic semiconductor.

Question.15. Describe an n-type semiconductor (Ge).
Answer. An n-type Ge is obtained by adding a small quantity (one millionth part) of a pentavalent impurity to its crystal.

Question.16. Name the pentavalent impurities which make Ge n-type.
Answer. The pentavalent impurities are :
1. Phosphorus (15), 2. Arsenic (33), 3. Antimony (51), 4. Bismuth (83).
Generally, Arsenic (As) is taken for this purpose.

Question.17. Describe a p-type semiconductor (Ge).
Answer. A p-type Ge’ is obtained by adding a small quantity (one millionth part) of a trivalent impurity to its crystal.

Question.18. Name the trivalent impurities which make Ge p-type.
Answer. The trivalent impurities are :
1. Boron (5), 2. Aluminium (13), 3. Gallium (31), 4. Indium (49), 5. Thalium (81).
Generally Indium (In) is taken for this purpose.

Question.19. What is doping?
Answer. The process of adding a suitable impurity to pure semiconductor, deliberately, is called
doping.

Question.20. What is order of doping in an extrinsic semiconductor?
Answer. It is one part in one million.

Question.21. Why is n-type semiconductor so called?
Answer. Because it contains free electrons with negative charge, as charge carriers.

Question.22. Why is p-type semiconductor so called?
Answer. Because it contains holes with positive charge, as charge carriers.

Question. 23. What is a junction?
Answer. It is a common surface of n-type and p-type semiconductor.

Question.24. What is a depletion layer?
Answer. It is a layer with junction in the middle, having no free charge carriers. The opposite j charge carriers have become neutralized. (It is shown shaded in Fig. 10.02).

Question.25. What is junction potential barrier?
Answer. The potential difference between junction ends of the two types of semiconductors, is
called junction potential barrier.

Question.26. Why is junction potential barrier so called?
Answer. Because it prevents free charge carriers from entering the depletion layer by themselves.

Question.27. What is biasing of a junction?
Answer. Applying an external potential difference more than potential barrier on the faces of the junction, is called biasing of the junction.

Question.28. Give names of the two types of the biasing.
Answer. The two types of biasing are :
(i) forward biasing (ii) reverse biasing.

Question.29. Why is forward bias so called?
Answer. Because it makes free charge carriers to move forward towards junction.

Question.30. Why is reverse bias so called?
Answer. Because it makes free charge carriers to move reverse away from junction.

Question.31. How does the bias effect the thickness of the depletion layer?
Answer. Forward bias decreases the thickness of depletion layer.
Reverse bias increases the thickness of depletion layer.

Question.32. How does the bias effect the junction resistance?
Answer. The forward bias makes junction resistance less.
The reverse bias makes junction resistance more.

Question.33. Define characteristic of a junction diode.
Answer. Graph drawn between bias voltage and circuit current of a junction diode, is called characteristic of the diode. It reveals the character (way of behaviour) of the junction diode.

Question.34. Describe different types of characteristics of a junction diode.
Answer.

1. Forward bias characteristic. It is obtained by plotting a graph between forward bias
   voltage and circuit current. Junction resistance comes to be about 10 ohm.
2. Reverse bias characteristic. It is obtained by plotting a graph between reverse bias voltage and circuit current. Junction resistance comes to be about 10,000 ohms.

The post To draw the I-V characteristic curve of a p-n junction in forward bias and reverse bias appeared first on Learn CBSE.

To draw the characteristic curve of a Zener diode and to determine its reverse breakdown voltage

Aim
To draw the characteristic curve of a Zener diode and to determine its reverse breakdown voltage.

Apparatus
A Zener diode (with small reverse breakdown voltage of about 6 volts), [i.e., V_z = 6 V], a ten volt battery, a high resistance rheostat, two 0-10 V voltmeter, one 0-100 mA ammeter, one 20 Ω. resistance, one way key, connecting wires.

Theory
Zener Diode. It is a semiconductor diode, in which the n-type and the p-type sections are heavily doped, i.e., they have more percentage of impurity atoms. This heavy doping results in a low value of reverse breakdown voltage (BV_R). This value can be controlled during manufacture.
The reverse breakdown voltage of a Zener diode, is called Zener voltage (V_z). The reverse current that results after the breakdown, is called Zener current (I_z).

At breakdown, increase of V_I increases I_I by large amount, so that V₀ = V_I– R_I I_I becomes constant. This constant value of V₀ which is the reverse breakdown voltage, is called Zener voltage.
Formula used

Diagram

Procedure

1. Arrange apparatus as shown in circuit diagram.
2. Make all connections neat, clean and tight.
3. Note least count and zero error of voltmeters and milli-ammeter. (micro-ammeter)
4. Bring moving contact of potential divider (rheostat) near negative end and insert the key K. Voltmeters and milli-ammeter will give zero reading.
5. Move the contact a little towards positive end to apply some reverse bias voltage (V_I). Milli-ammeter reading remains zero. Voltmeters give equal readings.
   [i.e.,V₀ =V_I ._.. I_I = 0 (eqn. 2)]
6. As V_I is further increased, I_I starts flowing. Then V₀ becomes less than V_INote the values of V_p I_I and V₀.
7. Go on increasing V_I in small steps of 0.5 V. Note corresponding values of I_I and V₀ which will be found to have increased.
8. As V_I is made more and more, I_I and V₀ are found to increase. Values are noted.
9. At one stage, as V_I is increased further, I_I increases by large amount and V₀ does not increase. This is reverse breakdown situation.
10. As V_I is increased further, only I_I is found to increase, V₀ becomes constant. Note values of V_I, I_I and V₀.
11. Increase V₁ to a value of 10 V, noting corresponding values.
12. Record your observations as given ahead.

Observations

Calculations
Plot a graph between input voltage V_I (column 2) and input current (column 3), taking V_I along X-axis and I_I along Y-axis.

Result
The reverse’breakdown voltage of given Zener diode is

Precautions

1. All connection should be neat, clean and tight.
2. Key should be used in circuit and opened when the circuit is not being used.

Viva Voce

Question. 1. What is reverse current?
Answer. The current due to reverse bias voltage, is called reverse current.

Question. 2. What constitutes the reverse current?
Answer. The reverse current is due to minority carriers.

Question.3. What are minority carriers?
Answer. Free electrons in p-type semiconductor and holes in re-type semiconductor, are called minority carriers. They are present due to breaking of covalent bonds.

Question.4. What is reverse breakdown?
Answer. At certain stage of increased reverse bias voltage, the reverse current increases suddenly. This situation is called reverse breakdown. This phenomenon is called Zener effect.

Question.5. What causes reverse breakdown?
Answer. The rupture of all covalent bonds causes reverse breakdown.

Question. 6. Is the reverse breakdown recoverable?
Answer. Yes. The decrease of reverse bias voltage restores the condition. The broken bonds are reassembled.

Question.7. What is reverse breakdown voltage?
Answer. The reverse bias voltage which causes breakdown, is called reverse breakdown voltage. It is represented by the symbol BV_R.

Question. 8. On which factor does the reverse breakdown voltage depend?
Answer. It depends upon the level of doping of re-type and p-type section of the diode.
General purpose diodes have each section lightly doped. They have high value of reverse breakdown voltage.
Zener diodes have each section heavily doped. They have low value of reverse break¬down voltage.

Question.9. What is Zener voltage?
Answer. The reverse breakdown voltage of Zener diodes, is called Zener voltage. It is represented by the symbol V_z.

Question. 10. How does Zener voltage differ for Germanium and Silicon?
Answer. For same order of doping, it is less for Germanium and more for silicon.

Question.11.What is Zener current?
Answer.The reverse current after breakdown, is called Zener current. It is represented by the symbol I_z.

Question.12.How does a Zener diode work as a voltage regulator?
Answer. At breakdown and after, output voltage (V₀) becomes constant at value of Zener voltage (V_z) even when input voltage (V_I) increases.
Thus, the Zener diode will give same output voltage for all input voltage of higher values. It becomes a voltage stabilizer for voltage equal in value of Zener voltage (V_z). The current drawn does not affect the zener voltage.

Question.13.What is the Knee voltage?
Answer.The forward voltage beyond which the current starts to increase rapidly with voltage is called the cut-in or Knee voltage of the diode.

Question.14.What happens to the potential barrier and depletion layer when a reverse bias is applied to a p-n junction diode?
Answer. Both increases.

Question.15.What is fermi level?
Answer.It is the highest energy level in the conduction band occupied by the electrons at the absolute zero of temperature.

Question.16.What is Zener breakdown?
Answer.Due to small junction width, the junction field is high. Due to this internal high field, there is large production of electron-hole pairs. The corresponding breakdown is called breakdown.

Question.17.What is Zener diode?
Answer. It is specially designed p-n diode whose both sides are heavely doped and work only in the reverse breakdown region.

Question.18.Give one application of a diode.
Answer.Rectifier.

Question.19.What is ideal diode?
Answer.It is a diode which offers zero resistance in forward biasing and infinite resistance in reverse biasing.

Question.20.How the energy gap changes with (i) doping (ii) temperature?
Answer. Decrease with increase in doping and temperature.

The post To draw the characteristic curve of a Zener diode and to determine its reverse breakdown voltage appeared first on Learn CBSE.

To study the characteristics of a common emitter npn (or pnp) transistor and to find out the values of current and voltage gains

Aim
To study the characteristics of a common emitter npn (or pnp) transistor and to find out the values of current and voltage gains.

Apparatus
An n-p-n transistor, a three volt battery, a 30 volt battery, two high resistance rheostats, one 0-3 volt voltmeter, one 0-30 volt voltmeter, one 0-50 μA micro-ammeter, one 0-50 mA milli-ammeter, two one way keys, connecting wires.

Theory
In common-emitter circuit of a transistor, emitter-base make input section and emitter- collector make output section. As usual, base junction (input junction) is forward biased and collector junction (output junction) is reverse biased. ,
Resistance offered by base junction, is called input resistance (R_I) It has a very small resistance due to forward biasing.
Resistance offered by collector junction, is called output resistance (R₀). It has a high value due to reverse biasing.
Due to high output resistance (resistance in output section), a high resistance can be used as load resistance (R_L). Generally R_L = R₀.

Also emitter current (I_e) divides itself into base current (I_b) and collector current (I_c). In n-p-n transistor, I_c is about 98% of I_e, base current I_b remains only 2% of I_e. A little change in I_b causes a large change (about 49 times) in I_c. The ratio of change in collector current to the corresponding change in base current, measures current gain in common emitter transistor. It is represented by symbol β.

For the example given above, P becomes 49.
The product of current gain and the resistance gain measures voltage gain of the common emitter transistor. It is about fifty times the resistance gain.

Diagram

Procedure

1. Make circuit diagram as shown in figure.
2. Make all connections neat, clean and tight.
3. Note least count and zero errors of voltmeters and ammeters.
4. Make voltmeter readings zero in V₁ and V₂ and insert the keys.
   For input characteristics
5. Apply forward bias voltage on base junction. Read base voltage ( V_b) from and base
   current (I_b) from μA.
6. Go on increasing V_b till I_b rises suddenly. Note corresponding values of I_b for each value of V_b.
7. Make collector voltage V_c = 10 V and repeat steps 5 and 6.
8. Repeat step 7 with V_c = 20 V and 30 V.
9. Make all readings zero.
   For output characteristics
10. Keep collector voltage (V_c) zero. Adjust base voltage Vb to make base current I_b = 10 μA. Though collector voltage V_c is zero ; but there is collector current I_c. Note it.
11. Make collector voltage 10 V, 20 V and 30 V and note the corresponding collector currents.
12. Repeat steps 10 and 11 with I_b = 20μA, 30 μA, and 40 μA.
13. Record your observations as given below :

Observations

Calculations

1. Calculation for input resistance (R_I )
   Plot a graph between base voltage V_b (column 2—table 1) and base current I_b (column 3a—table 1) for zero collector voltage V_c, taking Vb along X-axis and I_b along Y-axis. Plot graphs for different values of V_c. The graphs come as shown.
   These graphs are called ‘input characteristics’ of the transistor.
   
   
2. Calculation for output resistance (R₀)
   Plot a graph between collector voltage V_c (column 2—table 2) and collector current I_c (column 3a – table 2) for 10 μA base current I_b, taking V_c along X-axis and I_c along Y-axis. Plot graphs for different values of I_b.
   
   These graphs are called ‘output characteristics’ of the transistor.
   
3. Calculation for current gain (β)
   Plot a graph between base current I_b (columns 3a, 36, 3c and 3d—table 2) and corre-sponding collector current I_c (from same columns) for 30 volts collector voltage V_c, taking I_b along X-axis and I_c along Y-axis. The graph comes to be a straight line. The graph is called current gain characteristic of the common emitter transistor.
   
   
4. Calculation for voltage gain A_v
   

Result

Precautions
Same as given in Experiment 8.

Sources of error
Same as given in Experiment 8.

Viva Voce

Question. 1. What is a semiconductor junction triode or transistor?
Answer. It is a semiconductor device having three sections which are (i) emitter (E), (ii) base (B) and (iii) collector (C).

Question. 2. How are the three sections arranged in a transistor?
Answer. The base is a thin layer of one type extrinsic semiconductor between two other sections of second type extrinsic semiconductors on either side of it.

Question.3. Are the emitter and the collector sections (made of same type of extrinsic semi-conductor) alike?
Answer. No. Emitter section is heavily doped, narrow and longer. It is done to provide more free charge carriers.
Collector section is moderately doped, broad and shorter. It is done for providing easy passage to free charge carriers from emitter section.

Question.4. What is the order of thickness of base section and what is reason for it?
Answer. The base thickness is of the order of 10 micron (10^-5 m). The small thickness allows the recombination of only 2 to 5 per cent of charge carriers during their passage through it.

Question.5. In how many ways a transistor can be used?
Answer. A transistor can be used in three ways :

1. common base circuit;
2. common emitter circuit and
3. common collector circuit.

Question.6. How are the two junctions used?
Answer. One junction is used as input junction and the other as output junction.

Question.7.How are the two junction biased?
Answer. Generally, input junction is forward biased to offer less resistance and output junction is reverse biased to offer more resistance. Reverse is also possible.

Question.8.Why is the semiconductor junction triode called a transistor?
Answer. It can be used to transform a low resistance of a forward-biased junction into a high resistance of a reverse-biased junction or vice-versa. Thus, it works as a Transformer of resistor, which has been shortened to transistor.

Question.9.What is current gain of a common base transistor?
Answer. It is the ratio of change in collector current to the corresponding change in emitter current. It is represented by the symbol α.

Question.10.What is current gain of a common emitter transistor?
Answer. It is the ratio of change in collector current to the corresponding change in base current. It is represented by the symbol β.

Question.11.What makes ‘p’ so much more than ‘α’?
Answer.

Question.12.What is the order of magnitude of emitter current and base current?
Answer. Emitter current has magnitude upto 50 mA. Base current has magnitude upto 100 μA.

Question.13.Why a common emitter circuit is preferred over a common base circuit in amplifiers?
Answer. It is due to large current gain from common emitter circuit.

Question.14.What is resistance gain of a transistor?
Answer. The ratio of load resistance (R_L ) used in output circuit to the input resistance (R_I ) of input junction, is called resistance gain of a transistor.
Since load resistance used has value equal to output resistance (R_g ) of output junction, the resistance gain is measured as R_o /R_I

Question.15.What is input characteristic of a common emitter transistor?
Answer. A graph between base voltage (V_b ) and base current (I_b) for fixed value of collector voltage (V_c), is called the input characteristic of the common emitter transistor.

Question.16.What is the importance of the input characteristic? ,
Answer. It helps in calculating input resistance of the transistor.

It is equal to the reciprocal of the slope of the input characteristic and can be found.

Question.17.What is output characteristic of a common emitter transistor?
Answer. A graph between collector voltage (V_c) and collector current (I_c) for a fixed value of base current (I_b), is called the output characteristic of the common emitter transistor.

Question.18.What is the importance of the output characteristic?
Answer. It helps in calculating output resistance of the transistor.

It is equal to the reciprocal of the slope of the output characteristic and can be found.

Question.19.How to determine current gain of a common emitter transistor?
Answer. A graph can be plotted between base current (I_b) and collector current (I_c) taking I_b

Question.20.How do we calculate resistance gain?
Answer. We find input resistance Rb from input characteristics and output resistance R_o from output characteristics.

Question.21.How do we calculate voltage gain?
Answer. Voltage gain is calculated by multiplying current gain by resistance gain.

Question.22.Write the three application of a transistor.
Answer. (i) Amplifier (ii) Switch (iii) Oscillator.

Question.23.Define the transconductance of a transistor.
Answer. It is the ratio of change in collector current to the change in base-emitter voltage. It is denoted by g_m .

Question.24.How does the collector current changes in a junction transistor of the base region has (i) large size (ii) large doping.
Answer. The collector current shall decrease in both cases.

Question.25.Why a common emitter transistor amplifier is preferred over a common base transistor amplifier ?
Answer. Because the current gain in CE mode is much larger than CB mode.

Question.26.Can two p-n junction diodes placed back to back work as a p-n-p transistor?
Answer. No. In this device, the base region will be quite thick and highly doped. The base current will become equal to emitter due to neutralisation of majority carriers coming from emitter to collector.

Question.27.Why is the base region of a transistor made thin and lightly doped?
Answer. It is to reduce the base current and to increase both collector current and current gain.

Question.28.Why a transistor cannot be used as a rectifier?
Answer. To use a transistor as a rectifier, either its emitter-base portion or collector base portion has to be used. As the base is thin and lightly doped, then it will not work as a p-n junction.

Question.29.Which one of the transistors p-n-p and n-p-n is more useful and why?
Answer. n-p-n transistor is better than p-n-p transistor. It is because that electrons are majority carrier in n-p-n transistor while holes are majority carrier in p-n-p transistor. The mobility of electrons is much larger than that of holes due to small mass.

Question.30.How would you test in a simple way whether the transistor is spoiled or in working order?
Answer. The resistance of input section of a transistor is low as compared the output sector. In a spoiled transistor, the resistance is low for both sections.

The post To study the characteristics of a common emitter npn (or pnp) transistor and to find out the values of current and voltage gains appeared first on Learn CBSE.

To study effect of intensity of light (by varying distance of he source) on an LDR

Aim
To study effect of intensity of light (by varying distance of the source) on an LDR.

Apparatus
Light source, light dependent resistors (L.D.R.s) of different variety, a multimeter (or meter bridge), a source of intense light (a lamp bulb with battery eliminator) and a convex lens.
Light Dependent Resistor
The light dependent resistance are the devices for detecting and measuring electromagnetic waves (light etc.). Its working is based upon the principle of variation of the photoconductivity when radiation is incident upon it and absorbed by it.
A light dependent resistor is prepared from cadmium sulphide. Its resistance depends upon the intensity and duration of light incident on it.
A good quality LDR shows a resistance variation from 1 MΩ in complete darkness to about 10 Ω in full day light. The intensity of light decreases inversely with increase the square of distance.

Procedure

1. Turn the selector switch and set it on R for the measurement of resistance, in multimeter.
2. Plug the metallic ends of black probe in terminal marked common in multimeter and that of red in terminal marked as P (or +). Short the other metallic ends and adjust the ‘R adjusting’ to get full scale deflection reading at zero ohm in the meter.
3. Touch the metallic probes to the two metal ends of the L.D.R. [Fig. (a)] and read the value of resistance when (a) the source is kept at a distance of 2 cm, fixing the source of light in a stand and keeping the L.D.R. vertically below it.
   (i) Moving the source to 4 cm distance from the L.D.R. and
   (ii) Moving the source to 6,8 and 10 cm from L.D.R. and repeating observation three more times.

Observation record

Conclusion
When the distance between light source and L.D.R. increases the resistance of L.D.R. decreases.
Note: Same activity can be done by varying the exposure time in steps for same source of light, same LDR and for same distance.

Precautions

1. No stray light should fall on the L.D.R. It is better to work in a dark room.
2. Connect L.D.R. carefully to the voltage source.

Viva Voce

Question. 1. Mention names of three basic logic gates.
Answer. (i) OR gate, (ii) AND gate, (iii) NOT gate.

Question. 2. What do you mean by logic gate in electronics?
Answer. The circuit which is used to perform the switching action is called a logic gate.

Question.3. What are n-p-n and p-n-p transistors?
Answer. An npn transistor consists of a thin p-type layer sandwitched between two thick n-type layers. A p-n-p transistor consists of a thin n-type layer sandwitched between two thick n-type layers.

Question. 4. In a transistor, base is made thin and is doped very lightly. Why?
Answer. Base is made very thin and is doped lightly so that most of the carriers are attracted straight into the collector and very few combine in the base. Therefore, to reduce the I_b and to increase I_c .

Question. 5. When does a diode work as an open switch?
Answer. In reverse bias, a diode works as an open switch.

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Identify the Functional Group Present in the Given Organic Compound

Chemistry Lab ManualNCERT Solutions Class 12 Chemistry Sample Papers

Result

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To observe refraction and lateral deviation of a beam of light incident obliquely on a glass slab

Aim
To observe refraction and lateral deviation of a beam of light incident obliquely on a glass slab.

Apparatus
Glass slab, drawing board, white paper sheet, drawing pins, office pins, protractor.

Theory
When a ray of light (PQ) incident on the face AB of glass slab, then it bends towards the normal since refraction takes place from rarer to denser medium. The refracted ray (QR) travel along straight line and incident on face DC of slab and bends away from the normal since refraction takes place from denser to rarer medium. The ray (RS) out through face DC is called emergent ray.
From the following diagram

1. The incident ray is parallel to the emergent ray i.e. i = e.
2. The emergent ray is laterally deviated from its original path (incident ray) by a distance d = t sec r sin (i – r).

Diagram

Procedure
Fix a white paper sheet by drawing pins on a drawing board.
Take a glass slab and put it symmetrically in the middle of the paper and mark its boundary ABCD.
Draw a normal at point Q on face AB and draw a line PQ making an angle i with the normal. PQ will represent an incident ray.
Fix two pins at points 1 and 2 on the line PQ at distances 1 cm or more between themselves.
See images of these pins through face DC and fix two more pins at points 3 and 4 (1 cm or more apart) such that these two pins cover the images of first two pins, all being along a straight line.
Remove the glass slab. Draw straight line RS through points 3 and 4 to represent emergent ray. Join QR to represent refracted ray.
Draw normal at point R on face DC and measure angle e. It comes to be equal to angle i. Produce PQ forward to cut DC at T. Draw TU perpendicular to RS. TU measures lateral displacement d.
Now take another set for different angle of incident and measure the lateral displacement.

Conclusions

1. Angle of incidence (i) = Angle of emergence (e).
2. The lateral displacement increases with the increase in the thickness of the slab.
3. The lateral displacement increases with the angle of incidence (i).

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To observe polarisation of light using two polaroids

Aim
To observe polarisation of light using two polaroids.

Apparatus
Thin glass sheet, a source giving monochromatic light beam with parallel rays, a polaroid.

Theory
When an unpolarised light is made incident on the interface of two transparent media at polarising angle, the refracted and reflected rays depart from each other at an angle of 90°. The reflected ray is completely plane polarised. It can be tested by a polaroid.

Diagram

Procedure
Keep the than glass sheet in a horizontal plane surface with a hole under the sheet. Take a beam of monochromatic light having parallel rays and make it incident on the upper face of the glass sheet.
Adjust the angle of incidence to 57.5°.
Observe the reflected rays and the refracted rays. They must make an angle of 90° with each other.
Testing of Polarisation

1. Take a polaroid (P) and keep it in between incident light and your eyes. Rotate it about an axis along incident ray. No change of intensity of light will be detected. It is so because the incident light is unpolarised.
2. Take the second polaroid (A) and place it at a proper distance between polaroid (P) and eye and parallel to it. Light is visible through them.
3. Now rotate the polaroid (A) ranging from 0° to 360°. Keeping the polaroid (P) fix and note the intensity of transmitted light.
4. When polaroid (A) and polaroid (P) at 90° than transmitted light through polaroid (A) will be zero.

Result
When the two polaroids are parallel to each other light transmitted through it. But when they are perpendicular, there is no transmitted light. The light obtained through polaroid (P) is plane polarised. The light has transverse nature.

Precautions

1. Two polaroids and source of light should be in a straight line.
2. Rotate only second polarised from 0° to 360°. It is so, because the reflected light is completely plane polarised.
   Figure shows the arrangement for testing the polarisation of light by tourmaline crystals which act as polaroids.
   

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Tests for the Functional Groups Present in Organic Compounds Viva Questions with Answers

Question.1.What is a functional group?
Answer. The group of atoms that largely determines the properties of an organic compound is called functional group.

Question.2.Name any four functional groups.
Answer. Hydroxyl group —OH
Amino group —NH₂ Carboxyl group —COOH Aldehydic group —CHO.

Question.3.Name the functional groups present in alkenes and alkynes.
Answer. Alkenes are unsaturated hydrocarbons with C = C bond present in them. Alkynes are un-saturated hydrocarbons with C = C bond present in them.

Question.4.What is Baeyer’s test for unsaturation?
Answer. When Baeyer’s reagent (alkaline potassium permanganate) is added to unsaturated com-pound, its colour gets discharged indicating presence of C = C or C = C in the compound.

Question.5.Do alkynes turn blue litmus paper red?
Answer. No.

Question.6. Which is more acidic: an alcohol or a phenol?
Answer. A phenol.

Question.7. Why is alcohol dried before carrying out sodium metal test?
Answer. Because water also reacts with sodium and gives hydrogen gas with brisk effervescence.

Question.8. What is the use of Lucas reagent?
Answer. It is used to distinguish between primary, secondary and tertiary alcohols.

Question.9. Which of the two is more acidic: phenol or carboxylic acid?
Answer. Carboxylic acid.

Question.10. Name a test by which you can distinguish between hexylamine (C₆H₁₃NH₂) and aniline. (C₆H₅NH₂).
Answer. Dye test.

Question.11.Name two tests which distinguish aldehydes from ketones?
Answer. Tollen’s test and Fehling’s test.

Question.12. Name a reagent used to detect carbonyl group in a compound.
Answer. DNP (2, 4-dinitrophenylhydrazine).

Question.13, What is Tollen’s reagent?
Answer. It is ammonical silver nitrate solution.

Question.14. What is the use of Schiff’s reagent?
Answer. Schiff’s reagent is used to detect aldehyde group.

Question.15. Give one test to distinguish between an aldehyde and a ketone.
Answer. Tollen’s test can be used to distinguish between an aldehyde and a ketone.

Question.16. What is Rochelle’s salt?
Answer. Sodium potassium tartarate is called Rochelle’s salt.

Question.17. What is Fehling’s solution?
Answer. It is a solution obtained by mining equal volumes of copper sulphate solution (Fehling A) and a solution of sodium hydroxide containing sodium potassium tartarate (Fehling B).

Question.18. How is nitrous acid is prepared?
Answer. When sodium nitrite is reacted with dil. HCl at a temperature below 5°C, nitrous acid is produced.

Question.19. What is application of carbylamine reaction?
Answer. it is used to detect primary amine.

Question.20. How can phenol and aniline be distinguished chemically?
Answer. Phenol is soluble in aqueous NaOH solution whereas aniline is not.
Aniline is soluble in dilute HCl whereas phenol is not.

Question.21. In contrast to aromatic primary amines, aliphatic primary amines do not form stable diazonium salts. Why?
Answer. Because alkyl carbocation formed on decomposition of diazonium salt is more stable than phenyl carbocation.

Question.22. Why is aniline weaker base than ammonia?
Answer. Because lone pair of nitrogen in aniline is delocalized over benzene ring and is not fully available for sharing with acids.

Question.23. How can you distinguish between methanol and ethanol chemically?
Answer. Methanol and ethanol can be distinguished by iodoform test. Ethanol gives yellow ppt. of iodoform in this test whereas methanol does not give this test positive.

Chemistry Lab ManualNCERT Solutions Class 12 Chemistry Sample Papers

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To observe diffraction of light due to a thin slit

Aim
To observe diffraction of light due to a thin slit. .

Apparatus
Two razor blades, adhesive tapes, a screen a source of monochromatic light (laser pencil) black paper and a glass plate.

Theory
Diffraction is a phenomenon of bending of light around the comers or edges of a fine opening or aperture. Diffraction takes place when order of wavelength is comparable or small to the size of slit or aperture. The diffraction effect is more pronounced if the size of the aperture or the obstacle is of the order of wavelength of the waves. The diffraction pattern arises due to interference of light waves from different symmetrical point of the same wavefront. The diffraction pattern due to a single slit consists of a central bright band having alternate dark and weak bright bands of decreasing intensity on both sides.

Procedure

1. Fix the black paper on the glass plate by using adhesive.
2. Place two razor blades so that their sharp edges are parallel and extremely close to each other to form a narrow slit in between.
   
3. Cut the small slit in between the sharp edges of blades and place at a suitable distance from a wall or screen of a dark room.
4. Throw a beam of light on the slit by the laser pencil.
5. A diffraction pattern of alternate bright and dark bands is seen on the wall.
   

Conclusion
When light waves are incident on a slit or aperture then it bends away (spread) at the comers of slit showing the phenomena of diffraction of light.

Precaution

1. Air gaps should not be left between glass plates and black paper.
2. The razor blades should be placed extremely closed as possible.
3. Diffraction pattern should be seen on a wall of a dark room.
4. A point source of monochromatic light like laser torch should be used.

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(a) To study the nature and size of the image formed by a convex lens on a screen by using a candle and a screen (for different distances of the candle from the lens)

Aim
To study the nature and size of the image formed by a convex lens on a screen by using a candle and a screen (for different distances of the candle from the lens).

Apparatus
An optical bench with three uprights, a convex lens with holder, a burning candle, a card-board screen.

Theory


As the object (burning candle) is moved from infinity towards the convex lens, its image (position of screen) moves from lens focus towards infinity.
For candle distance less than focal length, image becomes virtual and does not come on screen.

Diagram
Similar to Ray diagram (Experiment 3 : Section B) having a burning candle in place of object needle and cardboard screen in place of image needle.

Procedure
Find rough focal length of the convex lens by usual method.
Mount the convex lens in holder in central upright and keep it in the middle of the optical bench.
Mount the card-board screen on another upright and keep it at distance equal to rough focal length of the lens, from the central upright.
Mount the burning candle in third upright and keep it on the other side of the central upright and near the end of the optical bench.
Adjust heights so that the inverted image of erect flame of burning candle is formed on screen. Move the screen to make the image sharp. The screen will be nearly at the focus of the convex lens.
The image will be real inverted and much more diminished.
As the burning candle is moved towards the lens on one side, the screen has to be moved away from the lens on other side, for getting sharp flame image. The inverted image size increases.
When the position of the candle is at distance 2f from the lens, the screen is also at same distance on the other side. The image size will be equal to the actual flame size.
Move the candle further nearer to the lens. The screen has to be moved away for getting an enlarged inverted real image on screen.
As the candle reaches the focus of the lens, the screen may not be able to get its image which will be at infinity i.e. beyond the ends of the optical bench.

Conclusion
This change in position, nature and size of the image is according to theoretical predictions.

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(b) To study the nature and size of the image formed by a concave mirror on a screen by using a candle and a screen (for different distances of the candle from the mirror)

Aim
To study the nature and size of the image formed by a concave mirror on a screen by using a candle and a screen (for different distances of the candle from the mirror).

Apparatus
An optical bench with three uprights, a concave mirror with holder, a burning candle, a card-board screen.

Theory

Hence as the object (burning candle) is moved from infinity towards the concave mirror, its image (position of screen) moves from mirror focus towards infinity. The two cross each other at distance 2f i.e., at the centre of curvature of the mirror.
For candle distance less than focal length, image becomes virtual and does not come on screen.

Diagram
Similar to ray diagram as shown in Experiment 1: Section B having a burning candle in place of object needle and card-board screen in place of image needle.

Procedure
Find rough focal length of the concave mirror by usual method.
Mount the concave mirror in holder in first upright and keep it near one end of the optical bench, keeping mirror face inward.
Mount the card-board screen on a second upright and keep it at distance equal to rough focal length of mirror, from first upright.
Mount the burning candle in third upright and keep it near other end of the optical
bench.
Adjust heights so that the inverted image of erect flame of burning candle is formed on screen. Move the screen to make the image sharp. The screen will be nearly at the focus of the concave mirror.
The image will be real, inverted and much more diminished.
As the burning candle is moved towards the mirror, the screen has to be moved away from it for getting a sharp flame image. The inverted image size increases.
When the position of the candle approaches centre of curvature of the mirror, the screen also approaches the same position. The image size will be equal to the actual flame size.
Now interchange the uprights. Bring candle upright nearer to mirror than the screen upright.
Move the candle further nearer. The screen has to be moved away for getting an enlarged inverted real image on screen.
As the candle reaches the focus of the mirror, the screen may not be able to get its image which will be formed at infinity i.e. beyond the length of the optical bench.

Conclusion
This change in position, nature and size of the image is according to theoretical predictions.

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To obtain a lens combination with the specified focal.length by using two lenses from the given set of lenses

Aim
To obtain a lens combination with the specified focal length by using two lenses from the given set of lenses.

Apparatus and material
Apparatus. No particular apparatus is needed.
Material. A set of thin convex lenses, one of these is of given focal length (say 15 cm), (we have to select a second lens such that the combination gives a single lens of focal length  f_c = 10 cm), lens holder with stand, a white painted vertical wooden board with broad stand, half metre scale.

Theory

1. The reciprocal of focal length in metre is called power of lens in dioptre (D).
   
2. With a convex lens, the real image of a distant object is formed at a distance equal to its focal length.
3. If  f₁ and  f₂ be the focal lengths of the two lenses and F be the focal length of the combination.
   

Diagram

Procedure

1. Keep the white painted vertical wooden board to serve as a screen.
2. The convex lens (known focal length f₁ = 15 cm), fixed into a holder stand is put on the left of the screen. There are sunlight illuminated green trees at large distance on the left of the lens.
3. The lens is moved towards and away from the screen till a sharp, inverted image of
   trees is formed on the screen.
4. Distance between central lines of the screen and holder stand is measured by a half
   metre scale.
5. The distance gives the focal length of the convex lens about 15 cm.
6. Replace first lens by second convex lens of required power and repeat the steps from 2 to 5. This gives the focal length of second convex lens.
7. Now bring both lenses in contact and repeat the steps from 2 to 5. This gives the
   combined focal length.
8. Determine the focal length with other given lens. Determine the focal length of about six of the convex lenses.

Calculations

Following combinations will be suitable.

Verification
The above combinations may be tried and result verified.

Precautions

1. Thin lenses should be taken.
2. Lenses should have same aperture.

Sources of error

1. Lenses may not be thin.
2. Lens apertures may not be same.

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