The branch of physical chemistry which study about the relation between electricity and chemical process involved is called electrochemistry.

Electrolytes

Electrolytes are the aqueous solution of chemical substance like acid, base and salt which conduct electricity in aqueous medium.

For examples:- Aqueous solution of H₂SO₄, NaOH, NaCl, etc.

Electrolytes are ionized into charged particles (ie. Cation and anion) when electricity is passed through them.

Depending upon the strength of electrolytes, they can be classified into two types.

1. Strong electrolytes
2. Weak electrolytes

Strong electrolytes:

The electrolytes which are completely ionized in aqueous medium are called strong electrolytes. It can conduct electricity easily.

Examples- Solution of H₂SO₄, HNO₃, HCl, NaOH, NaCl,etc.

Weak electrolytes:

The electrolytes which are partially ionized in aqueous medium are called weak electrolytes. It conducts electricity partially.

Examples- Solution of CH₃COOH, NH₄OH, H₂CO₃, etc.

Non – Electrolytes:

Non – electrolytes are the aqueous solution of chemical substance which do not conduct electricity in aqueous medium.

Examples – Solution of glucose, sugar, urea etc.

Electrolysis

The process of chemical decomposition of an electrolyte in solution by using electric current is called electrolysis.

This process is carried out in a vessel called electrolytic cell or voltameter. The two metallic rods are connected to two terminals of battery in electrolytic solution with the help of electric wire. These metallic rods are called electrodes. The electrode connected to positive terminal of battery is called anode and the electrode connected to negative terminal of battery is called cathode. In cathode reduction takes place whereas in anode oxidation takes place.

If the electrolyte is NaCl solution, then

At cathode : Reduction

At anode : Oxidation

Applications of Electrolysis

• Electrolysis can be used to extract the pure metals in electroplating, electro refining, etc.
• It can be used to manufacture oxygen and hydrogen gas from water.

Faraday’s laws of electrolysis

Faraday’s first law of electrolysis : 

This law states that, “The mass of the substance deposited (discharged) at the electrode during electrolysis is directly proportional to the quantity of electricity passed through the solution.”

Faraday’s Second law of electrolysis

This law states that “the mass(weight) of different substances deposited or liberated at electrode by the same amount of electricity is directly proportional to the equivalent weight of substances”.

It is found that when 1.008 gram of H₂ is evolved from acidified water, then the masses of copper and silver deposited are 31.75 gram and 108 gram respectively, which are the equivalent weight of copper and silver respectively. Hence, this verifies Faraday’s second law of electrolysis.

Numerical Problems

1. How many Faraday of electricity are there in 6.0 X 10¹² electrons?

2. How many coulombs of electricity is required for the following changes?

a. 1 mole of Al³⁺ to Al

b. One mole of H₂O to O₂

c. One mole of MnO₄^– to Mn²⁺

d. 1 mole of FeO to Fe₂O₃

e. 1 mole of Cr₂O₇^{– –} to Cr³⁺

Ans. (d) 96500 (e) 5.79X10⁵

3. Find the charge in coulomb in 1 g ion of N^3-.

A gram-ion means the number of ions that are present in 1 mole of the substance. NN has only 1 gram-ion since it forms only one ion i.e., N³⁻ at a time.
Therefore, 1g-ion of N³⁻ means 1 mole of N³⁻ .

There are 3 electrons in this ion, hence, 3 moles of electrons are present in 1g-ion or 1 mole of N³⁻.
Charge present on 1 mole of electrons = 96500 C
Charge present on 3 moles of electrons = 96500×3 C
Therefore, charge in 1g ion of N^3- is 2.895×10⁵C

4. What current strength in amperes will be required to liberate 10g iodine from potassium iodide solution in one hour? (at.wt. of iodine = 127).

5. A current of 2.5 ampere passes through the solution of zinc sulphate for 30 minutes and deposits 1.52 g of zinc metal at cathode. What is the equivalent weight of zinc metal?

6) 0.2964 g of copper was deposited by the passage of a current of 0.5 amperes for 30 minutes through a solution of CuSO₄. Calculate the atomic weight of copper.

7) How long a current of 3 amperes has to be passed through a solution of AgNO3 to coat a metal surface of 80cm2 with 0.005mm thick layer.(Density of silver = 10.5g/cc and at.wt. of Ag = 107.7)

8) A metallic spoon is coated with silver by passing a current of 5 amperes through AgNO₃ solution for 5 hours. What is the thickness of silver plating if the area of spoon is 12 cm². [Given, density of silver is 10.5 g/cc, atomic weight of Ag is 107.92]

9) Silver is deposited on a metallic vessel of surface area 800 cm² by passing a current 0.2 amperes for 3 hours. Calculate the thickness of silver deposited. Given the density of silver as 10.47g/cc. [atomic mass of Ag = 107.92]

{Ans. 2.88 x 10^-4 cm}

10) Iron slab having dimension 50cm x 30cm x 20cm was electroplated all around the surface with a uniform thickness of copper by passing 5 ampere of current for 2 hours in CuSO₄ solution. Find the thickness of Cu deposited. Given the specific gravity of Cu = 8.9 and its atomic mass = 63.5

[Hint: Here, total surface area = 2[50×30+50×20+30×20]

Specific gravity = density]

11) Find out the number of coulombs needed to plate a rectangular area of 25cm x 4cm with thickness of 10^-2 cm using CuSO₄ as electrolyte. Density of Cu is 8.94 g/ml.

12) How many molecules of chlorine should be deposited from molten sodium chloride in one minute by a current of 300 milli amperes ?

13. If 5A current is supplied through zinc sulphate solution till 12 minutes deposits 1.2g of zinc, find the current efficiency.

14) A solution of CuSO₄ is electrolyzed by passing a current of 360mA for 35.3 minutes. Calculate the mass of metal deposited if current efficiency was 60% (at. Mass of Cu=63.5).

15) If the cost of electricity to deposit 5 g Mg from its salts solution is Rs. 25, find the cost of the electricity to deposit equal quantity of Zn from its salt solution.

16) Calculate the volume of Cl₂ at NTP produced during electrolysis of MgCl₂ which produces 6.50g Mg. (At.wt. of Mg=24.3)

17) An electric current is passed through three cell in series containing solution of copper sulphate, silver nitrate and potassium iodide respectively. What weight of silver and iodine will be liberated when 1.25g of copper is being deposited.

18) An electric current is passed between platinum electrode through a solution of silver nitrate, copper sulphate and dilute sulphuric acid solution in series. If one gram of hydrogen is evolved by this current in the third cell, calculate (i) the weight of silver and copper deposited in the first and second cell (ii) the volume of hydrogen measured at STP which is liberated from the third cell. [ eq.wt. of H = 1, Cu = 31.5 and Ag = 108]

19) Equal amount of current was passed through an aqueous solution of trivalent metallic salt and dil. H₂SO₄. The volume of H₂ liberated was 96.5ml at 27⁰C and 765mmHg and weight of metal deposited was 0.74gm. Calculate the atomic weight of metal.

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Redox reaction can be balanced by either of the following two methods:

• Oxidation number method
• Ion-electron method

Oxidation number method to balance redox reaction

The oxidation number of an oxidizing agent decreases and that of the reducing agent increases in a redox process. This change in oxidation number should be balanced to balance a redox reaction by the oxidation number method.

Balancing of redox reaction by oxidation number method involves the following steps:

Step I : Identifying the oxidizing and reducing agents.

Oxidizing and reducing agents are identified by observing the change in oxidation number of main elements present in oxidizing and reducing agents on reactant and product sides. Eg.

Here, Zn is oxidized and HNO₃ is reduced to Zn(NO₃)₂ and N₂O respectively.

So, Zn is reducing agent and HNO₃ is oxidizing agent.

Step II : Calculating the change in oxidation number.

The total change in oxidation number in oxidation and also in reduction is calculated after equalizing the number of atoms undergoing O.N. change on both sides. Eg.

Oxidation: total change in O.N. = 2-0 = 2

Reduction: Total change in O.N. = 1-5 = 4×2 =8

Step III : Balancing of oxidizing and reducing agents.

Number of atoms undergoing oxidation number change are balanced by criss-cross multiplication with change in oxidation number on reactant as well as product side. Eg.

Step IV : Balancing of other elements.

All the other atoms or radicals are balanced by hit and trial method.

• Balancing the equation with respect to all other atoms except hydrogen and oxygen.

• Finally, balancing hydrogen and oxygen.

It is balanced chemical equation.

• Note:

• For reactions taking place in acidic solutions, add H⁺ ions to the side deficient in hydrogen atoms.

• For reactions taking place in basic solutions, add H₂O molecules to the side deficient in hydrogen atoms and simultaneously add equal number to OH^– ions on the other side of the equation.

More examples:

Example – 1

Step I : Identifying the oxidizing and reducing agents.

Step II : Calculating the change in oxidation number.

Step III : Balancing of oxidizing and reducing agents.

Step IV : Balancing of other elements.

• Balancing the equation with respect to all other atoms except hydrogen and oxygen.

• Finally, balancing hydrogen and oxygen.

It is balanced chemical equation.

Example – 2

Step I : Identifying the oxidizing and reducing agents.

Step II : Calculating the change in oxidation number.

Step III : Balancing of oxidizing and reducing agents.

Step IV : Balancing of other elements.

It is balanced chemical equation.

Example-3 :

Step I : Identifying the oxidizing and reducing agents.

Here, Cl₂ undergoes self oxidation and reduction (i.e. disproportionation reaction). So, Cl₂ is written twice to calculate the change in oxidation number.

Step II : Calculating the change in oxidation number.

Step III : Balancing of oxidizing and reducing agents.

Step IV : Balancing of other elements.

It is balanced chemical equation.

Example-4 :

Step I : Identifying the oxidizing and reducing agents.

Step II : Calculating the change in oxidation number.

Here, more than two elements have changed in the oxidation number.

Change in O.N. of As = +2 X 2 = +4

Change in O.N. of S = +2 X 3 = +6

Change in O.N. of N = – 3

Thus, total increase in O.N. = 4+6=10

Total decrease in O.N. = 3

Step III : Balancing of oxidizing and reducing agents.

Step IV : Balancing of other elements.

It is balanced chemical equation.

Ion-Electron method [ Half reaction method] to balance reaction

This method is based on the fact that number of electrons lost during oxidation half reaction of redox reaction is equal to number of electrons gained during reduction half reaction.

Following steps are involved in balancing redox rection by ion-electron method:

• Write the given equation and identify the oxidized and reduced species by analyzing change in oxidation number.
• Write the oxidation and reduction half reactions (ionic if required)
• Balance two half reactions separately.

1. Balance main element (i.e. atom whose O.N. is changed) by hit and trial.
2. Balance oxygen by adding H₂O.
3. Balance hydrogen by adding H⁺.
4. Balance charge by adding electrons.
5. In basic medium, OH^– ions are added on either side to cancel out H⁺ ions.

• Add two half reactions by making electrons gained equal to electrons lost in two half reactions.
• Finally, convert the balanced ionic equation to balanced molecular equation (if required).

Some Examples:

Example-1 :

Counting of oxidation number:

Identification and balancing of two half reactions:

Oxidation half reaction:

Balancing charge by adding electrons:

This is balanced oxidation half reaction.

Reduction half reaction:

Balancing main element by hit and trial:

Balancing oxygen by adding H₂O:

Balancing hydrogen by adding H⁺:

Balancing charge by adding electrons:

This is balanced reduction half reaction.

Adding two half reactions:

Balanced two half reactions are added after multiplying oxidation half by 4 to cancel electrons.

This is balanced ionic equation. To convert it into molecular form, 8 NO₃^– ions are added on both side of equation.

This is balanced molecular equation.

Example-2:

Counting of oxidation number:

Identification and balancing of two half reactions:

Oxidation half reaction:

Balancing charge by adding electrons:

This is balanced oxidation half reaction.

Reduction half reaction:

Balancing main element by hit and trial:

Balancing oxygen by adding H₂O:

Balancing hydrogen by adding H⁺:

Balancing charge by adding electrons:

This is balanced reduction half reaction.

Adding two half reactions:

Balanced two half reactions are added after multiplying oxidation half by 6 to cancel electrons.

This is balanced ionic equation. To convert it into molecular form, suitable number of K⁺ and SO₄^2- have to be added on both side of equation.

This is balanced molecular equation.

Example -3:

Counting of oxidation number:

Identification and balancing of two half reactions:

Oxidation half reaction:

Balancing main element by hit and trial:

Balancing charge by adding electrons:

This is balanced oxidation half reaction.

Reduction half reaction:

Balancing oxygen by adding H₂O:

Balancing hydrogen by adding H⁺:

Balancing charge by adding electrons:

This is balanced reduction half reaction.

Adding two half reactions:

This is balanced equation.

Example-4: [ Basic medium ]

Counting of oxidation number: Here, chlorine is oxidized as well as reduced.

Identification and balancing of two half reactions:

Oxidation half reaction:

Balancing oxygen by adding H₂O:

Balancing hydrogen by adding H⁺.

The reaction is carried out in basic medium, so, all H⁺ ions must be neutralized by adding same number of OH^– ions on both sides.

Balancing charge by adding electrons:

This is balanced oxidation half reaction.

Reduction half reaction:

Balancing oxygen by adding H₂O:

Balancing hydrogen by adding H⁺:

The reaction is carried out in basic medium, so, all H⁺ ions must be neutralized by adding same number of OH^– ions on both sides.

Balancing charge by adding electrons:

This is balanced reduction half reaction.

Adding two half reactions:

This is balanced ionic equation. In order to convert it into molecular form, 8K⁺ ions are added on both sides of equation.

This is balanced molecular equation.

Example-5:

Counting of oxidation number: Here, chlorine is oxidized as well as reduced.

Identification and balancing of two half reactions:

Oxidation half reaction:

Balancing main element by hit and trial:

Balancing oxygen by adding H₂O:

Balancing hydrogen by adding H⁺:

The reaction is carried out in basic medium, so, all H⁺ ions must be neutralized by adding same number of OH^– ions on both sides.

Balancing charge by adding electrons:

This is balanced oxidation half reaction.

Reduction half reaction:

Balancing main element by hit and trial:

Balancing charge by adding electrons:

This is balanced reduction half reaction.

Adding two half reactions:

This is balanced ionic equation. In order to convert it into molecular form, 6 Na⁺ ions are added on both sides of equation.

This is balanced molecular equation.

Self Practice 

Balance the following equations by the oxidation number method or ion-electron method.

References

• https://chemicalnote.com/oxidation-and-reduction-oxidants-and-reductants-and-redox-reaction/
• Atkins, Peter, Paula, de Julio, Atkin’s Physical Chemistry, Seventh Edition, Oxford University Press, (Printed in India, 2002).
• Gurtu, J.N., Snehi, H., Advanced Physical Chemistry, Seventh Edition, Pragati Prakashan India, 2000.
• Sthapit, M.K., Pradhananga, R.R., Foundations of Chemistry, Vol 1 and 2, Fourth edition, Taleju Prakashan, 2005.

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Here is new grid and new model question paper of NEB class 11 Chemistry. Questions will be asked in class 11 final exam according to this grid from 2077. A set of model question paper of class 11 chemistry given by Curriculum Development Center (CDC) is attached here with detailed explanation of each question in video. Not only the solution of this model question, you will get the more ideas to solve other similar questions too. See the questions and watch the video to develop your chemistry knowledge.

Watch the video to get detailed explanation of each question [Detailed explanation]

Click here to watch the video…

Thank you….

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Alkanes (saturated hydrocarbons)

Preparation of alkanes

1. From haloalkanes

a. By Wurtz reaction: When an alkyl halide (haloalkane) is heated with sodium metal in presence of dry ether, an alkane containing double number of carbon atoms than in haloalkane is formed. This reaction is called Wurtz reaction.

Q) Identify A.

b. By reduction: Alkyl halides when reduced with Zn/HCl , H₂/Ni, etc. give alkanes. Eg.

2. By catalytic hydrogenation of alkenes and alkynes:

Hydrogenation of unsaturated hydrocarbons (i.e. alkenes and alkynes) in presence of nickel or platinum as catalyst results in the formation of alkanes. Eg.

3. By decarboxylation of (salt of) carboxylic acid:

When sodium salt of carboxylic acid is heated with soda-lime (NaOH+CaO), a molecule of carbondioxide is eliminated from the molecule to give alkane. This reaction is called decarboxylation reaction. Eg.

Chemical properties of alkanes

1. Halogenation: This reaction involves the substitution of hydrogen atoms of alkanes by halogen atoms. For example, chlorine reacts with methane in presence of sunlight or heat to form four different halogen derivatives.

2. Nitration: Alkane reacts with nitric acid at high temperature to form nitroalkane. Eg.

3. Sulphonation : When alkane is heated with fuming sulphuric acid, alkane sulphonic acid is formed. Eg.

4. Oxidation: If burnt in air (Oxygen), alkanes are completely oxidized to carbon dioxide and water with large amount of heat.

When burnt in insufficient supply of oxygen, alkane forms carbon monoxide and carbon (carbon black). Eg.

Lower alkanes when heated with limited supply of air at 350-500⁰C form aldehydes. Eg.

Alkenes

Preparation of alkenes

1. By dehydrohalogenation of alkyl halide ( elimination reaction):

When alkyl halide is heated with alcoholic solution of sodium or potassium hydroxide, hydrogen and halogen atom is eliminated from adjacent carbon atoms to give alkene. Eg.

If there is possibility of the formation of two alkenes, major product is formed according to Saytzeff’s rule. This rule states that when there is a chance of formation of more than one alkene, then the more substituted alkene is formed as major product.

Eg. In the dehydrohalogenation of 2-bromobutane, but-2-ene is the major product over but-1-ene.

2. By dehydration of alcohols:

Removal of water molecule from a molecule is called dehydration. Alcohol undergoes dehydration to form alkene when it is heated with dehydrating agent like sulphuric acid(H₂SO₄), phosphoric acid(H₃PO₄), alumina(Al₂O₃) etc. eg.

If there is possibility of the formation of two alkenes, major product is formed according to Saytzeff’s rule. eg.

3. By controlled hydrogenation of alkynes:

When alkyne is treated with hydrogen in presence of catalyst Pd on BaSO₄ poisoned by quinoline, alkene is formed. Eg.

Note: Lindlar’s catalyst = Pd+BaSO₄+quinoline

Chemical properties of alkenes

1. Addition of hydrogen (Catalytic hydrogenation):

When alkenes are heated with hydrogen gas in presence of metal catalyst like Ni, Pt or Pd, alkanes are formed. This reaction is called catalytic hydrogenation.

2. Addition of halogens:

Halogens react with alkene in presence of inert solvent like carbon tetrachloride to give dihaloalkane.

Eg. ethene reacts with Br₂ in presence of CCl₄ to give 1,2-dibromoethane. In this reaction reddish brown colour of bromine is discharged. Hence this is a test reaction of ethene (alkene).

3. Addition of hydrogen halides ( halogen acids)(HCl, HBr, HI):

Alkene reacts with halogen acids to give alkyl halide (haloalkane). Eg.

When alkene is unsymmetrical then the addition takes place according to Markovnikov’s rule.

Markovnikov’s rule:

This rule states that when an unsymmetrical reagent is added to an unsymmetrical alkene, the negative part of the reagent goes to that double bonded carbon which has lesser number of hydrogen atoms.

For example: The addition of HBr to propene gives 2- bromopropane instead of 1- bromopropane.

Other example:

Peroxide effect or anti-Markovnikov’s rule:

When HBr is added to an unsymmetrical alkene in presence of organic peroxide, bromine goes to the double bonded carbon atom having more number of hydrogen. This phenomenon of anti- Markovnikov’s addition of HBr caused by the presence of peroxide is known as peroxide effect or anti- Markovnikov’s rule.

4. Addition of water [Catalytic hydration]:

Alkenes react with water in presence of dilute mineral acid as catalyst to form alcohol. Eg.

5. Addition of sulphuric acid:

Alkenes react with concentrated sulphuric acid to give alkyl hydrogen sulphate. Eg.

6. Ozonolysis:

Alkene reacts with ozone to give ozonide. On warming ozonide with Zn in water, it breaks down to give two molecules of carbonyl compounds (aldehyde or ketone). This process of formation of ozonide and it’s decomposition to give carbonyl compounds is called ozonolysis.

7. Polymerization:

The process of making polymers from monomers is known as polymerization. Smaller molecules undergoing polymerization are called monomers. The polymers are high molecular weight large molecules made by the polymerization of monomers.

Ethene polymerizes to form polyethene.

Alkynes

Preparation of alkynes

1. By direct combination of elements (i.e. carbon and hydrogen):

Ethyne (acetylene) gas is formed when an electric spark is struck between two carbon electrodes in an atmosphere of hydrogen gas.

2. By dehydrohalogenation of vicinal dihalides:

When vicinal dihalides are treated with alcoholic KOH, alkynes are formed by dehydrohalogenation. Eg.

Note: Vicinal dihalide– Compounds that contain two hydrogen atoms on adjacent carbon atoms.

3. By heating trihalogen derivatives with silver powder:

Trihaloalkanes like chloroform and iodoform when heated with silver powder form alkynes. Eg.

Chemical properties of alkynes

1. Addition of hydrogen ( Reduction):

When alkyne is heated with hydrogen in presence of Ni, Pt or Pd catalyst, alkane is formed. Eg.

However, alkyne reacts with hydrogen in presence of palladium catalyst deposited over barium sulphate poisoned by sulphur to give alkene. Eg.

2. Addition of halogen acids(HX):

Alkynes react with two molecules of halogen acids according to Markovnikov’s rule to give dihaloalkane. Eg.

3. Addition of water : Catalytic hydration:

Alkynes react with water in presence of mercuric sulphate and sulphuric acid to give vinyl alcohol which rearranges to give aldehyde or ketone.

For example, ethyne gives ethanal (i.e. aldehyde).

Propyne gives propanone (i.e. ketone).

4. Reaction with bromine solution:

Alkynes react with bromine in water or CCl₄ to give tetrabromo compound. Here, red colour of bromine is discharged. This is test reaction of alkyne (unsaturated compound).

5. Polymerization reaction:

When alkynes are passed through a red hot iron or copper tube, they polymerize to form aromatic compounds.

Eg. Three molecules of ethyne (acetylene) polymerize to give benzene.

6. Formation of acetylides (Acidic nature of acetylene):

Acetylene is acidic in nature because it releases H⁺ easily.

a. Action with sodium metal:

Acetylene reacts with molten Na metal to form sodium acetylide.

b. Action with ammonical silver nitrate solution (Tollen’s reagent):

Acetylene reacts with ammonical silver nitrate solution ( i.e. Tollen’s reagent) to give silver acetylide which have white ppt.

c. Action with ammonical cuprous chloride solution:

Acetylene reacts with ammonical cuprous chloride solution to form copper acetylide which have red ppt.

Test of unsaturation ( i.e Test of alkenes and alkynes)

1. Bromine decolorization test:

Red colour of bromine is discharged when Br₂ solution in water or carbon tetrachloride is added to unsaturated compounds (alkenes or alkynes). Therefore, this reaction is used to detect the presence of multiple bond in a molecule. Eg.

2. Baeyer’s test ( Oxidation with alkaline solution of KMnO₄):

Alkaline solution of potassium permanganate is known as Baeyer’s reagent.

Alkene reacts with Baeyer’s reagent to form glycol, where pink colour of the potassium permanganate is discharged. Therefore this reaction is used as test reaction of alkenes.

Similarly, alkynes also discharge the pink colour of Baeyer’s reagent.

* Baeyer’s reagent oxidizes ethyne to oxalic acid and the pink colour of KMnO₄ is discharged.

* Other alkynes react with Baeyer’s reagent to give two molecules of carboxylic acids. Eg.

Q) Give a suitable test to distinguish following pairs.

a. ethyne and ethane

b. ethene and ethyne.

Kolbe’s electrolysis method for the preparation of Alkanes, alkenes and Alkynes

Alkanes, alkenes and alkynes are prepared by electrolysis of salt of monocarboxylic acid, dicarboxylic acid and unsaturated dicarboxylic acid respectively.

1. Preparation of alkanes:

An alkane is obtained by the electrolysis of sodium or potassium salt of a carboxylic acid in aqueous solution. Eg.

Ethane is produced at anode during the electrolysis of an aqueous solution of sodium or potassium acetate as follows:

2. Preparation of alkenes:

An alkene is obtained by the electrolysis of sodium or potassium salt of a dicarboxylic acid in aqueous solution. Eg.

Ethene is produced at anode during the electrolysis of an aqueous solution of sodium or potassium succinate as follows:

3. Preparation of alkynes:

An alkyne is obtained by the electrolysis of sodium or potassium salt of an unsaturated dicarboxylic acid in aqueous solution. Eg.

Ethyne is produced at anode during the electrolysis of an aqueous solution of sodium or potassium maleate (i.e. salt of maleic acid) as follows:

References

• Finar, I. L., Organic Chemistry, Vol. I and Vol. II, Prentice Hall, London, 1995.
• Ghosh, S.K., Advanced General Organic Chemistry, Second Edition, New Central Book Agency Pvt. Ltd., Kolkatta, 2007.
• Sharma, M.L., Chaudary, P.N., A Text Book of B.Sc. Chemistry, Second edition (Volume one and two), Ekta Books, Kathmandu, 2004.
• Sthapit, M.K., Pradhananga, R.R., Foundations of Chemistry, Vol 1 and 2, Fourth edition, Taleju Prakashan, 2005.
• https://en.wikipedia.org/wiki/Acetylene
• https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/Chapter_01%3A_Structure_and_Bonding/1.9%3A_Ethane%2C_Ethylene%2C_and_Acetylene
• https://www.britannica.com/science/hydrocarbon/Nomenclature-of-alkenes-and-alkynes

The post Hydrocarbons : Alkanes, Alkenes and Alkynes – Preparation and Properties appeared first on Online Chemistry notes.

The two or more forms of same element having similar chemical properties but different physical properties are called allotropes and the phenomenon is called allotropy.

Sulphur exists on following allotropic forms:

Uses of sulphur

• It is used for the manufacture of sulphuric acid, sulphur dioxide and other sulphur compounds.
• It is used in medicine to manufacture sulpha drugs and ointments for skin disease.
• It is used for the vulcanization of rubber.

Note: Vulcanization is a process of heating raw rubber with sulphur or its compounds to destroy its sticky nature and to increase the durability of rubber products.

• It is used for the manufacture of black gun powder which is a mixture of carbon (charcoal), sulphur and nitre (KNO₃).
• It is used in the manufacture of fungicides and insecticides.

Compounds of sulphur

Hydrogen sulphide (H₂S)

Working principle of Kipp’s apparatus : Intermittent supply of H₂S gas by Kipp’s apparatus

Intermittent = Stopping and starting at irregular intervals

H₂S gas is frequently required for salt analysis in the laboratory. The special apparatus which is used for the preparation of H₂S is called Kipp’s apparatus.

Kipp’s apparatus consist of three bulbs A, B and C. The bulbs are interconnected with each other. The upper bulb C contains a long stem which reaches upto bulb A. When dil. H₂SO₄ is poured from bulb C, liquid level rises in bulb A and reaches upto bulb B just to cover iron sulphide. Then the reaction between iron sulphide and H₂SO₄ takes place to form H₂S gas. As H₂S gas is formed, the pressure inside bulb B increases and forces the dil. H₂SO₄ up into the upper bulb C. Then the contact between FeS and dil. H₂SO₄ breaks and the formation of gas is stopped (ceased). When the tap is opened H₂S comes out and the pressure decreases. As a result liquid level again rises at bulb B and H₂S gas start to evolve again. In this way H₂S gas is prepared intermittently by Kipp’s apparatus.

Phsical Properties of hydrogen sulphide

1. It is colourless gas with rotten egg smell.

2. It is soluble in water.

3. It is poisonous gas.

{In its low concentration it causes nausea, dizziness and headache. At high concentration it is fatal (i.e. deadly poisonous). Conc. of H₂S in air 1 part in 1000 parts of air may be fatal.}

4. It is heavier than air{with vapour density 17}.

Chemical Properties of hydrogen sulphide

1. Acidic character : It acts as a weak diprotic(dibasic) acid. It ionizes in two steps to give two protons.

H₂S reacts with base to form two series of salts. Eg.

H₂S reacts with basic oxides to give salt. eg.

2. H₂S as reducing agent :

Sulphur in H₂S bears -2 oxidation state which in minimum oxidation state of sulphur. In chemical reaction, H₂S tends to get oxidized. So H₂S acts as reducing agent.

* It reduces halogens into halogen acids.

* It reduces sulphur dioxide to sulphur.

* It reduces ferric salt to ferrous salt.

* It reduces conc. HNO₃ to NO₂

* It reduces conc. H₂SO₄ to SO₂.

* It reduces acidified potassium permanganate solution and pink colour of KMnO₄ is discharged.

* It reduces acidified K₂Cr₂O₇ solution and orange color of K₂Cr₂O₇ changes to green.

3. H₂S as analytical reagent in salt analysis ( Precipitation of metal sulphide) :

H₂S gas is used as analytical agent to detect group II basic radicals in acidic medium i.e. HCl and group IIIB basic radicals in in alkaline medium i.e. NH₄Cl and NH₄OH.

Group II basic radicals :

Group II basic radicals consists of Cu ⁺⁺ , Hg ⁺⁺, Pb⁺⁺, Cd⁺⁺, Sn⁺⁺, As⁺⁺⁺, Sb⁺⁺⁺, Bi⁺⁺⁺. In qualitative salt analysis this group II radicals are precipitated in the form of their sulphide in acidic medium. By noting the colour of ppt., we can identify the metal cations.

Group IIIB basic radicals:

Group IIIB basic radicals consist of Zn⁺⁺, Co⁺⁺, Ni⁺⁺, Mn⁺⁺. These are precipitated by passing H₂S in alkaline medium ( i.e. NH₄Cl+ NH₄OH).

Structure of H₂S:

Uses of H₂S:

• It is used as analytical reagent in laboratory.
• It is used as reducing agent.
• It is used to prepare metallic sulphide, which can be used as pigment.

Sulphur dioxide (SO₂)

Chemical Properties of sulphur dioxide:

1. Weak acidic nature: It dissolves in water to give sulphurous acid, a weak diprotic (dibasic) acid.

Sulphurous acid ionizes in two steps:

It produces two series of salts if reacted with base. Eg.

2. SO₂ as a reducing agent: Aqueous sulphurdioxide shows reducing character and SO₂ itself gets oxidized to H₂SO₄.

*  It reduces halogens (Cl₂, Br₂, I₂) in aq. Solution to respective halides. Eg.

* It reduces acidified potassium permanganate solution and pink colour of KMnO₄ is discharged.

* It reduces acidified K₂Cr₂O₇ solution and orange color of K₂Cr₂O₇ changes to light green.

* It reduces ferric salt to ferrous salt and colour changes from yellow to light green.

* It reduces potassium iodate (KIO₃) solution to iodine.

3. SO₂ as an oxidizing agent:

SO₂ oxidizes powerful reducing agents like H₂S, HI, Mg, Fe, etc. and itself get reduced to sulphur or sulphide.

4. SO₂ as bleaching agent:

Formation of colourless product from coloured substance is called bleaching. Chemical substances such as SO₂, H₂O₂, CaOCl₂, Cl₂, etc. show bleaching action.

SO₂ can act as a bleaching agent in presence of moisture. It can bleach coloured wool, silk, flower, hair, etc. The bleaching action of sulphur dioxide is due to the formation of nascent hydrogen in presence of moisture, which reduces colouring substance to colourless reduced product.

In some cases, bleaching action of SO₂ is due to formation of colourless addition product.

→ The bleaching action of SO₂ is temporary. The bleached colourless compound will regain its original colour slowly on standing in air due to oxidation by air.

Comparison of bleaching action of SO₂ and Cl₂:

Test for SO₂ :

Potassium dichromate paper test : When SO₂ comes in contact with acidified potassium dichromate paper, the orange coloured paper turns to green due to formation of chromium sulphate.

Uses of SO₂:

• It is used as bleaching agent.
• It is used in the manufacture of sulphuric acid.
• It can be used as disinfectant, fungicide and food preservative.

Sulphuric acid (H₂SO₄)

Sulphuric acid is widely used chemical in industries. It is used in the manufacture of fertilizers, drugs, dyes, polymers, etc. thousands of tons of sulphuric acid is manufactured every year worldwide. So it is called king of chemicals.

Manufacture of sulphuric acid by contact process:

• In early days, sulphuric acid used to be manufactured by lead chamber process.

• Contact process is modern method. The acid obtained is pure ( free from impurities) and is quite concentrated (96-99%). The contact process, it’s name is mainly from the fact that the conversion of sulphur dioxide into sulphur trioxide is carried out in contact with surface of catalyst.

Principle :

1. Production of sulphur dioxide: Sulphur dioxide gas can be prepared either by burning sulphur or roasting of iron pyrites.

2. Catalytic oxidation of sulphur dioxide: Sulphur dioxide is oxidized to sulphur trioxide in presence of catalyst vanadium pentoxide at about 450⁰C temperature and 2 atm pressure.

3. Conversion of SO₃ into H₂SO₄: sulphur trioxide obtained is absorbed in 98% H₂SO₄ to produce oleum or fuming sulphuric acid. The oleum is diluted with calculated amount of water to get desired concentration of H₂SO₄.

Conditions for optimum yield of H₂SO₄:

Formation of SO₂ to SO₃ is one of the must important steps in the manufacture of sulphuric acid. The production of H₂SO₄ entirely depends on the amount of SO₃ formed. As reaction is reversible, exothermic and proceeds with decrease in volume, Le- Chatelier’s principle can be applied for the maximum yield of sulphur trioxide.

• Low temperature : This reaction is exothermic . So, low temperature is required for maximum yield. But too low temperature is too slow to attain equilibrium. So an optimum temperature of about 450⁰C is supplied.
• High concentration of reactants : High concentration of SO₂ and O₂ is used for more production of SO₃ .
• High pressure : High pressure favours the reaction because the product formed has less volume than reactant . But the acid resistant tower which are able to withstand high pressure are difficult to build. Hence, an optimum pressure of about 2 atm is applied.
• Use of catalyst: Rate of reaction is increased by the use of positive catalyst. So, vanadium pentoxide is used as catalyst for higher yield of H₂SO₄.

Details of the plant or process:

1. Sulphur or pyrite burner: SO₂ gas is obtained by burning sulphur or iron pyrite with air in sulphur or pyrite burner.

2. Purification unit: The impure SO2 gas obtained is purified by the purification unit.

a. Electrical dust precipitator: Dust particles present in sulphur dioxide gas is precipitated in electrical dust precipitator by the influence of high potential difference applied between the metallic conductors fitted in the chamber.

b. Steam chamber: The lighter dust particles are settled down by using steam in steam chamber.

c. Cooler: The gases coming out from the steam chamber are passed through cooler. The gases get cooled down to about 100⁰C.

d. Washing tower(scrubber): The cooled SO₂ gas is then passed into a tower called scrubber which is packed with quartz and water is sprayed from the top. The water soluble impurities are washed away.

e. Drying tower: The moisture present in sulphur dioxide is absorbed by the spray of conc.H₂SO₄ in drying tower.

f. Arsenic purifier: Arsenous oxide ( As₂O₃) present in gas is absorbed by ferric hydroxide( because it may causes catalytic poisoning).

3. Testing box: The purity of sulphur dioxide is checked in testing box by passing through a darkened box.

4. Preheater: The purified mixture of SO₂ and O₂ is heated upto 450⁰C.

5. Contact chamber: The SO₂ gas is catalytically oxidized to sulphur trioxide (SO₃) in the presence of V₂O₅ as catalyst at 450⁰C temperature and 2 atm pressure.

6. Absorption tower: This tower is packed with quartz (or acid proof stone) in which 98% H₂SO₄ is sprayed from the top of the tower. Concentrated H₂SO₄ absorbs sulphur trioxide to form oleum (or pyrosulphuric acid). Oleum is then treated with calculated amount of water to form sulphuric acid of desired concentration.

Physical Properties of H₂SO₄:

i. It is colourless, hygroscopic, syrupy liquid.

ii. It is highly soluble in water (due to formation of intermolecular hydrogen bond). It dissolves in water with the liberation of large quantity of heat(19Kcal/mol) which may causes the explosion or spurt the acid out of the container.

H₂SO₄ + nH₂O H₂SO₄.nH₂O + Heat

Therefore, sulphuric acid is diluted by adding the acid slowly to water with constant stirring and not by adding water to the acid.

iii. It’s melting point is 10⁰C and boiling point is 338⁰C.

iv. Pure H₂SO₄ is covalent compound and bad conductor of electricity. However aqueous H₂SO₄ conducts electricity.

v. Litmus test : Sulphuric acid is covalent compound. In pure state it does not change the colour of the blue litmus paper. However, aqueous acid gets ionized and produces hydrogen(or hydronium) ion into solution. Due to which it can change the colour of blue litmus paper to red.

Chemical properties of H₂SO₄:

1. Decomposition: On heating H₂SO₄ decomposes into SO₂, H₂O and O₂.

2. Acidic nature: H₂SO₄ is a strong dibasic (diprotic) acid and ionizes in two steps.

It gives two series of salts like bisulphate and sulphate when reacted with base.Eg.

3. H₂SO₄ as an oxidizing agent:

i. Action with metals: More electropositive metals ( i.e. metals lying above hydrogen in electrochemical series) like Zn, Fe, Mg, Al, etc. react with dil H₂SO₄ to produce hydrogen gas. Here H₂SO₄ reduces to H₂ and metals oxidize to metal sulphates.

Conc. H₂SO₄ oxidizes Zn, Cu, Ag, Hg, etc. to respective metal sulphates and H₂SO₄ gets reduced to SO₂.

ii. Action with non metals:

* Carbon is oxidized to carbondioxide.

* Sulphur is oxidized to sulphurdioxide.

* Phosphorus is oxidized to phosphoric acid.

iii. Action with some other reducing agents:

* H₂SO₄ oxidizes H₂S to S and itself reduces to SO₂.

* HBr and HI are oxidized to Br₂ and I₂ respectively.

{Note : HCl can not be oxidized to Cl₂ because of strong bond.}

4. Sulphuric acid as dehydrating agent:

H₂SO₄ is a good dehydrating agent. It absorbs water molecules from sugar, cellulose, copper sulphate crystals, etc.

* With sugar (charring action):

When conc. H₂SO₄ is treated with sugar, wood, paper, etc. it absorbs water and a black mass of carbon is formed. This process is called charring.

* With copper sulphate crystals:

It removes water of crystallization from hydrated salts.

* With oxalic acid crystals and formic acid:

Lewis structure of H₂SO₄

Uses of sulphuric acid:

• It is used for the manufacture of fertilizers.
• It is used in manufacture of HCl, HNO₃, H₃PO₄, etc.
• It is used as an oxidizing agent.
• It is used as a drying and dehydrating agent.

Test of H₂SO₄:

Sulphuric acid gives H+ ions and SO₄^{– –} ions in aqueous solution. The presence of H⁺ ion can be detected by litmus paper. The presence of SO₄^{– –} can be detected by BaCl₂ solution which gives white precipitate.

Sodium thiosulphate

Molecular formula: Na₂S₂O₃. 5H₂O

It is commonly called hypo.

Uses of sodium thiosulphate:

• It is used as a fixer in photography. { i.e. for fixing silver bromide in photographic plate}
• It is used in (iodometric) titration for the estimation of iodine.
• It is used as an antichlor agent to remove excess of chlorine from bleached textiles/fibers.

References

• Lee, j. D., Concise Inorganic Chemistry, Fifth Edition, Joh, Wiley and Sons, Inc., 2007.
• Cotton, F. A., and Wilkinson, G., Advanced Inorganic Chemistry, Fifth edition, John Wily and Sons, Singapore, 1995.
• Day, C.M., Selbin, J., Theoritical inorganic Chemistry, second edition, Affiliated East-West Press Pvt. Ltd., New Delhi, 2002.
• Mitra, L.A. , A Text Book of Inorganic Chemistry, Ghos and Company, 61^st edition, 1996.
• https://pubchem.ncbi.nlm.nih.gov/compound/Sulfuric-acid
• https://www.sciencedirect.com/topics/earth-and-planetary-sciences/sulphuric-acid
• https://www.britannica.com/technology/contact-process
• https://en.wikipedia.org/wiki/Kipp%27s_apparatus
• https://www2.humboldt.edu/scimus/HSC.54-70/Descriptions/Kipp%27sGasApp.htm
• https://www.environment.gov.au/protection/publications/factsheet-sulfur-dioxide-so2
• https://en.wikipedia.org/wiki/Sulfur_dioxide

The post Sulphur and it’s compounds – Basic note appeared first on Online Chemistry notes.

New bonds are formed and old bonds are broken down during chemical reactions which causes the change in valence electrons of atoms or radicals. The chemical reactions in which the electronic states of the atoms or radicals change in reactants and products are termed as oxidation and reduction reactions.

Oxidation and reduction reactions are very common in our daily lives, eg. digestion of food, burning of fuel, rusting of iron articles, extraction of metals, electrolysis, electroplating, etc.

Green plants prepare carbohydrates by photosynthesis which is also oxidation and reduction. Here, reduction of carbon dioxide into glucose and oxidation of water into oxygen takes place.

These reactions can be defined with classical and electronic concepts.

Classical concept of Oxidation and Reduction

Oxidation :

Oxidation is a chemical process which involves addition of oxygen ( or any other electronegative species) or removal of hydrogen ( or other electropositive species). Examples :

Here, C is oxidized to CO₂.

Conversion of C into CO₂ is oxidation.

Here, H₂S is oxidized to S.

Conversion of H₂S into S is oxidation.

Reduction :

Reduction is a chemical process which involves the addition of hydrogen (or any other electropositive species) or removal of oxygen ( or any electronegative species). Examples:

Here, ZnO is reduced to Zn.

Conversion of ZnO into Zn is reduction.

Here, Cl₂ is reduced to HCl.

Conversion of Cl₂ into HCl is reduction.

Modern or Electronic concept of Oxidation and Reduction

Oxidation :

Oxidation is a process which involves loss of electrons. Examples:

Here, Na is oxidized to Na⁺ by losing one electron.

Here, Cl^– is oxidized to Cl by losing one electron.

Reduction :

Reduction is a process which involves gain of electrons. Examples:

Here, Cl is reduced to Cl^– by gaining one electron.

Here, Fe⁺⁺⁺ is reduced to Fe⁺⁺ by gaining one electron.

Oxidation = increase in positive charge or decrease in negative charge.

Reduction = increase in negative charge or decrease in positive charge.

Oxidation and Reduction occurs simultaneously

Oxidation involves loss of electron and reduction involves gain of electron. A species gains electron only when the electron is lost by other species and a species loses electron only when electron is gained by some other species. There is no oxidation without reduction and there is no reduction without oxidation. This means oxidation and reduction takes place simultaneously. For example:

This equation is written in ionic form as follows:

The oxidation half reaction is:

The reduction half reaction is:

The reaction involving both reduction and oxidation process is called redox reaction.

Oxidation number or oxidation state

Oxidation number (state) is the charge which an atom of the element has in its ion or appears to have when present in the combined state with other atoms.

Rules for determining oxidation number

• Oxidation number of elements in free or uncombined state is always zero. For example- oxidation number of H₂, O₂, Mg, Al, etc. is zero.

• Algebraic sum of oxidation numbers of all elements in a chargeless molecule is equal to zero. Eg. In NaCl , O.N. of Na + O.N. of Cl = zero

• Algebraic sum of oxidation numbers of all elements in an ion is equal to the charge on the ion. Eg. In OH^– , O.N. of Oxygen + O.N. of H = -1

• Oxidation number of radicals is equal to the charge on them. Eg. oxidation number of Fe²⁺ = +2 , NH₄⁺ = +1 , OH^– = -1 , NO₃^– = -1 , CN^– = -1, SO₄^2- = -2 , etc.
• Oxidation number of alkali metal (Li, Na, K, Rb, Cs, Fr) is +1 in all compounds.
• Oxidation number of alkaline earth metal ( Be, Mg, Ca, Sr, Ba and Ra ) is +2 in all compounds.
• Oxidation number of metals in alloy is zero. Eg. O.N. of Na in sodium amalgam (Na.Hg) is zero.
• Oxidation number of hydrogen in compound is +1, except in metallic hydrides. In metallic hydrides, O.N. of hydrogen is -1 (eg. in NaH, CaH₂, etc. O.N. of H is -1)
• The oxidation number of halogens is always -1 in metal halides like MgBr₂, KF, AlCl₃, etc.
• Oxidation number of oxygen is -2 in compounds except in peroxides, suboxides and OF₂.
• In peroxides like H₂O₂, Na₂O₂, etc. the O.N. of oxygen is -1.

• In superoxides like KO₂, NaO₂, RbO₂, SnO₂, etc. the O.N. of oxygen is -1/2.

• In OF₂, oxidation number of oxygen is +2.

Q) Calculate the oxidation number (oxidation state) of N in HNO₃.

Here, oxidation number of H = +1

Oxidation number of O = -2

Suppose oxidation number of N = x

So, +1 + x + (3 x -2) = 0

x = +6 – 1 = +5

Therefore, oxidation number of N is +5.

Q) Calculate the oxidation number of P in PO₄^{– – –}

x + (4 x -2) = -3

Thus, x = -3+8 = +5

Q) Calculate the oxidation state of Fe in Fe₂(SO₄)₃.

Here, O.N. of SO₄ = -2

Suppose O.N. of Fe = x

Then, 2 x + (3 x -2) =0

Therefore, x = +3

Q) Calculate the oxidation number of underlined element.

(1) KMnO₄  (2) NH₄⁺ (3) Cr₂O₇^{– –}

(4) K₂Cr₂O₇  (5) HNO₃

(6) H₃PO₄ (7) H₂SO₄ (8) C₆H₁₂O₆

(9) Cu(OH)₂  (10) K₄[Fe(CN)₆]

(11) NH₄NO₃

Answer :

(1) +7  (2) -3  (3) +6  (4) +6

(5) +5  (6) +5  (7) +6  (8) 0

(9) +2  (10) +2  (11) -3

Note : NH₄NO₃ = NH₄⁺  NO₃^–

Therefore, x + (4 x +1) = +1

   or, x = -3.

Q) Calculate the oxidation number of Ag in [Ag(NH₃)₂] Cl.

NH₃ is a molecule. So, oxidation number of NH₃ = 0

Let, oxidation number of Ag = x

x + (0x2) + (-1) = 0

x + 0 – 1 = 0

Therefore , x = +1

Differences between oxidation number and valency

Oxidation and reduction in terms of oxidation number

• Oxidation is a chemical process which involves increase in oxidation number.
• Reduction is a chemical process which involves decrease in oxidation number. Example :

Oxidizing agents and Reducing agents

Oxidizing agents or oxidants :

An oxidizing agent or oxidant is a substance which oxidizes the other substance and itself gets reduced. Eg.

Here, FeCl₃ is oxidizing agent because it has oxidized SnCl₂ into SnCl₄ and itself gets reduced to FeCl₂.

Reducing agents or reductants :

A reducing agent or reductant is a substance which reduces the other substance and itself gets oxidized. Eg.

Here, SnCl₂ is reducing agent because it has reduced FeCl₃ into FeCl₂ and itself gets oxidized to SnCl₄.

Note : Oxidizing agent is a species whose oxidation number decreases and Reducing agent is a species whose oxidation number increases.

Minimum and maximum oxidation number

Minimum possible oxidation number of an element is the number of electrons required to complete octet whereas the maximum oxidation number is the number of valence electrons. For example: in case of nitrogen,

Oxidizing agent or Reducing agent or both

• A species having an element in maximum oxidation state can act as oxidizing agent only.
• A species having an element in minimum oxidation state can act as reducing agent only.
• A species having an element neither in maximum nor in minimum oxidation state can act as oxidizing as well as reducing agent. Examples :

• A good reducing agent have a good capacity of losing electrons.
• A good oxidizing agent have a good capacity of gaining electrons.

Redox reaction

• A chemical reaction in which oxidation and reduction process take place simultaneously is called a redox reaction. Redox reaction involves both loss and gain of electrons.
• Redox reactions are the chemical reactions in which the reactants undergo a change in their oxidation states. Example :

In this reaction, C is oxidized to CO and ZnO is reduced to Zn. So this reaction is a redox reaction.

Role of redox reaction in daily life :

Redox reactions occur in each step of our daily lives. Majority of natural phenomena involve redox reactions. Eg. digestion of food; combustion of fuels like wood, gas, petrol, diesel, etc.; photosynthesis in green plants where CO₂ is reduced into carbohydrates and H₂O is oxidized into oxygen; extraction of metals where metal oxides are reduced into metals; generation of energy from different electrochemical cells; electrolysis; etc.

Disproportionation reaction

A chemical reaction in which a same species is oxidized as well as reduced is called disproportionation. Example:

When chlorine gas is passed through water, Cl₂ undergoes disproportionation reaction.

Objective Questions and Answers

1. Which of the following can act as both oxidizing as well as reducing agent ?

a. NH₃       b. HNO₃

c. HNO₂    d. NaNO₃

2. Oxidation state of oxygen in ozone is

a. +3   b. +1

c. -2    d. 0

3. Which of the following is the most powerful oxidizing agent ?

a. F₂     b. Cl₂

c. Br₂   d. I₂

4. Hydrogen sulphide is

a. Oxidizing agent only

b. Reducing agent only

c. Both

d. None

5. Oxidation number of manganese in potassium permanganate is

a. +7   b. -7

c. +5   d. +3

6. Oxidation number of phosphorus in H₃PO₄ is

a. +7   b. -7

c. +5   d. 0

7. In which of the following reactions, SO₂ acts as oxidizing agent ?

a. With H₂S

b. With acidified KMnO₄

c. With acidified K₂Cr₂O₇

d. With acidified C₂H₅OH

8. In the reaction, P₄ + 3KOH + 3H₂O → 3KH₂PO₂ + PH₃. Phosphorus is

a. Reduced

b. Oxidized

c. Oxidized as well as reduced

d. Neither oxidized nor reduced

9. When K₂Cr₂O₇ is converted to K₂Cr₂O₄, the change in oxidation number of Cr is

a. 0   b. 4

c. 6   d. 3

10. SO₂ acts as

a. Oxidizing agent only

b. Reducing agent only

c. Both oxidizing and reducing agent

d. Neither oxidizing nor reducing agent

11. Oxygen has an oxidation state of +2 in

a. H₂O   b. H₂O₂

c. SO₂    d . OF₂

12. The strongest oxidizing agent of the following is

a. O₃    b. O₂

c. Cl₂   d. Br₂

13. The strongest reducing agent is

a. HNO₂    b. H₂S

c. H₂SO₄   d. SnCl₂

14. Oxidation number of Fe in K₃[Fe(CN)₆] is

a. +1   b. +2

c. +3   d. +4

15. The highest oxidation state of Mn is shown by

a. K₂MnO₄   b. KMnO₄

c. MnO₂   d. Mn₂O₃

16. Oxidation state of oxygen atom in potassium superoxide is

a. 0    b. -1

c. -2   d. -1/2

17. Better reducing agents have a good capacity of :

a. Losing electrons

b. Gaining electrons

c. Being reduced

d. All of the above

18. Oxidation number of Zn in Cu-Zn is :

a. +2   b. -2

c. 0   d. +4

19. Oxidation number of C in HCN is :

a. +2   b. -2

c. 0     d. +4

20. Oxidation number of Na in sodium amalgam is

a. +5   b. +1

c. -1   d. 0

Answers :

1-c   2-d   3-d  4-b

5-a  6-c  7-a  8-c

9-a  10-c 11-d 12-a

13-b  14-c  15-b  16-d

17-A  18-C  19-A  20-d

References

• Sthapit, M.K., Pradhananga, R.R., Foundations of Chemistry, Vol 1 and 2, Fourth edition, Taleju Prakashan, 2005.
• Gurtu, J.N., Snehi, H., Advanced Physical Chemistry, Seventh Edition, Pragati Prakashan India, 2000.
• Madan, R.L., Tuli, G.D., Physical Chemistry, S. Chand and company, New Delhi, 2012.
• Negi, A.S., Anand, S.C., A Text Book of Physical Chemistry, Seventh Edition, New Age International Pvt. Ltd. Publishers, 1999.
• http://www.chemistry-assignment.com/oxidation-and-reduction-in-terms-of-oxidation-number/
• https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Electrochemistry/Redox_Chemistry/Oxidation-Reduction_Reactions
• https://www.khanacademy.org/science/ap-chemistry/chemical-reactions-ap/types-of-chemical-reactions-ap/a/oxidation-number

The post Oxidation and Reduction , Oxidants and Reductants and Redox reaction appeared first on Online Chemistry notes.

Purpose of preparing sodium extract

Organic compounds are covalent compounds. Common laboratory reagents are ionic compounds and handled as aqueous solution. Hence, organic compounds do not interact (react) with common laboratory reagents easily. For example,

When fused with highly reactive sodium metal, at high temperature, covalently bonded atoms of the organic compounds are converted into ionic compounds and can be detected by simple chemical tests using common laboratory reagents.

On fusion following ionic compounds are formed:

If nitrogen is present in organic compounds,

If sulphur is present in organic compounds,

If nitrogen and sulphur both are present together in organic compounds,

If halogens are present in organic compounds,

Preparation of sodium extract

Small piece of dry sodium metal is taken in a fusion tube (ignition tube) and sodium metal is melted and cooled. To this cooled Na- metal, a pinch of organic sample is added then fusion tube is further heated up to red hot. This red hot tube is plunged into distilled water in a porcelain basin. The content is grinded, boiled and filtered to obtain clear filtrate which is known as Na- extract or Lassaigne’s extract.

Q) Sodium extract is alkaline in nature, why?

Sodium metal used during preparation of sodium extract isn’t totally utilized (consumed) by foreign elements and remaining sodium metal combines with water forming sodium hydroxide. Sodium metal combines with water forming sodium hydroxide. Therefore, sodium extract is alkaline in nature.

Q) Organic compounds are combustible, why?

Organic compounds contain covalent bond which are difficult to break at ordinary temperature. They need huge amount of energy to break and get combusted in presence of oxygen. But once the combustion starts, it produces such a huge amount of energy that the combustion becomes spontaneous i.e. the energy needed is continuously provided by the energy produced. So, organic compounds are highly combustible.

Detection of Nitrogen (N)

If organic sample contain nitrogen, sodium extract has following composition :

If this sodium extract is not alkaline, make it alkaline by adding dil.NaOH then 1 or 2 drop of freshly prepared FeSO₄ solution is added. The content is boiled, cooled and acidified with dil.HCl . Finally 1, 2 drops of ferric chloride (FeCl₃) is added. Appearance of prussian blue or green colour indicates the presence of nitrogen in organic compound.

Note : The role of HCl is to dissolve dirty green ppt. of ferrous hydroxide which otherwise masks the colour.

Detection of sulphur

If organic sample contains sulphur, sodium extract has following composition :

Two types of tests can be done for detection of sulphur.

• Lead acetate test :

Sodium extract is acidified with acetic acid solution and 1,2 drops of lead acetate solution is added to little sodium extract . Formation of black ppt. indicates the presence of sulphur in the organic compound.

• Sodium nitroprusside test :

Few drops of sodium nitroprusside solution is added to little sodium extract. Appearance of violet colour indicates the presence of sulphur in the organic compound.

Detection of both nitrogen and sulphur

If both nitrogen and sulphur are present in organic sample, sodium extract has following composition:

To this sodium extract, ferric chloride (FeCl₃) solution is added, appearance of blood red colour indicates the presence of both N and S.

Detection of halogen 

If halogens are present in organic sample, sodium extract has following composition :

A little sodium extract is taken in a test tube, few drops of nitric acid is added, boiled, cooled and finally few drops of silver nitrate solution is added.

1. If chlorine is present in the organic compound, white ppt. is obtained which is soluble in ammonia solution (i.e. NH₄OH) and reappears on adding HNO₃.

2. If bromine is present in organic compound, pale yellow ppt. is obtained which is partially soluble in NH₄OH.

3. If iodine is present in the organic compound, dark yellow ppt. is obtained which is insoluble in NH₄OH.

Q) Why Lassaigne’s extract is boiled with concentrated nitric acid while testing for halogen ?

Sometimes the organic compounds may contain N and S along with halogens.

If nitrogen is present in the organic compound, NaCN is formed in the sodium extract, which also gives white ppt. with silver nitrate solution.

If sulphur is present in the organic compound, Na₂S is formed in the sodium extract which gives black ppt. with silver nitrate solution which masks the other colour.

Hence, before carrying out the test for halogens, the sodium extract is boiled with conc. HNO₃ in order to remove cyanide(CN^–) and sulphide (S ^{– –} ) as their volatile hydrogen acids, otherwise they would produce their own precipitate with AgNO₃ and interfere with the result.

Objective Questions

1. In sodium extract solution, nitrogen of an organic compound is converted into :

a. Sodalime    c. Sodium nitrate

b. Sodamide   d. Sodium cyanide

2. During the detection of organic compound sodium extract is prepared to

a. Dissolve it in water

b. Convert it into ionic form

c. Make it more reactive

d. Make the reaction slow

3. The Prussian blue colour is formed in Lassaigne’s test if the nitrogen is present in an organic compound, the prussian blue colour is due to the formation of :

a. Na₄[Fe(CN)₆]    c. Fe(OH)₃

b. Fe₄[Fe(CN)₆]₃    d. Fe₄[Fe(CN)₆]

4. Lassaigne’s test is not applied for the detection of :

a. Sulphur    c. Nitrogen

b. Halogens   d. Phosphorus

5. In Lassaigne’s test, if both N and S are present, blood red colour is seen due to formation of :

a. Ferric thiocyanate    c. Ferric sulphide

b. Ferric cyanide     d. Ferric ferrocyanate

6. Sodium extract is heated with conc. HNO₃ before testing for halogens because :

a. Silver halides are insoluble in water.

b. Ag₂S is soluble in HNO₃

c. AgCN is soluble in HNO₃

d. Na₂S and NaCN are decomposed by conc. HNO₃.

7. In Lassaigne’s test, if sulphur is present in organic sample then acidified sodium extract forms ____ ppt. with lead acetate solution.

a. Prussian blue    c. Black

b. Blood red      d. Violet

Answer :

1 – d,  2 – b,  3 – b,  4 – d

5 – a (also called- Ferric sulphocyanide)

6 – d, 7 – c

References

• Ghosh, S.K., Advanced General Organic Chemistry, Second Edition, New Central Book Agency Pvt. Ltd., Kolkatta, 2007.
• Morrison, R.T. , Boyd, R.N., Organic Chemistry, Sixth edition, Prentice-Hall of India Pvt. Ltd., 2008.
• March, j., Advanced Organic Chemistry, Fourth edition, Wiley Eastern Ltd. India, 2005.
• Sthapit, M.K., Pradhananga, R.R., Foundations of Chemistry, Vol 1 and 2, Fourth edition, Taleju Prakashan, 2005.
• https://www.pinterest.com/pin/834291899699006381/
• https://www.coursehero.com/file/29198433/Detection-of-foreign-elements-Copydoc/

The post Lassaigne’s test to detect foreign elements present in organic compounds appeared first on Online Chemistry notes.

Organic compounds are defined as the hydrocarbons (compounds containing carbon and hydrogen) and their derivatives in which covalently bonded carbon is an essential constituent.

Organic compounds are classified as :

1. Open chain organic compounds :

Organic compounds in which the terminal C-atoms are not joined together are called open chain compounds. Eg.

The open chain organic compounds can be further classified as,

Alkanes : Alkanes are the saturated hydrocarbons with general formula C_nH_2n+2. They contain only carbon-carbon and carbon-hydrogen single bonds in their molecules. For example:

• Alkanes are also called paraffins (because they have a little affinity towards a general reagent. In other words, alkanes are less reactive substances. They undergo reactions under drastic conditions)

Note : Alkanes are called saturated because all the possible sites (i.e. 4) are bonded with other atoms.

But in alkenes and alkynes there is a possibility of addition of atoms or groups so they are called unsaturated.

Alkenes : Alkenes are unsaturated hydrocarbons with general formula C_nH_2n. They contain at least one carbon to carbon double bond in their molecules. For example:

• Alkenes are also called olefins ( i.e. oil forming because they form oily liquids on reaction with chlorine gas.

Alkynes : Alkynes are unsaturated hydrocarbons with general formula C_nH_2n-2. They contain at least one carbon to carbon triple bond in their molecules. For example:

2. Closed chain (cyclic) organic compounds :

Organic compounds in which the terminal carbons are joined together to form a cyclic structure are called closed chain or cyclic organic compounds. For exampne:

Cyclic organic compounds are further classified as – homocyclic and heterocyclic organic compounds.

Homocyclic compounds : Cyclic organic compounds in which the ring forming atoms are only carbon are called homocyclic or more specifically carbocyclic compounds.

Homocyclic compounds can be further classified as – Alicyclic and Aromatic compounds.

Alicyclic compounds :

Closed chain organic compounds whose properties are similar to open chain aliphatic compounds are called alicyclic compounds. For example:

Aromatic compounds :

Benzene and those cyclic compounds that chemically behave as benzene are called aromatic compounds. For example:

Aromatic compounds obey Huckel’s rule.

Note : Huckel’s rule :

Huckel’s rule states that a cyclic and planar molecule is aromatic if it contains 4n+2 delocalized π electrons, where n = 0,1,2,3,4,etc.

Examples : Benzene

Benzene is cyclic and planar and has cyclic overlap of p-orbitals. There are 3 double bonds i.e. 6 delocalized π – electrons, which is consistant with Huckel’s rule.

i.e. 4n+2 = 6

4n= 4

n = 1(which is an integer)

Therefore, benzene is an aromatic compound. It will show aromaticity.

Heterocyclic compounds :

Cyclic organic compounds in which at least one heteroatom (i.e. atom other than carbon eg. N, O or S ) is present as one of the ring forming atoms are called heterocyclic compounds. Examples :

Formula of organic compounds

• Molecular formula :

It represents actual number of atoms of all the elements present in one molecule of the compound. For example:

methane = CH₄

ethane = C₂H₆

ethene = C₂H₄

benzene = C₆H₆ , etc.

• Empirical formula :

It represents simple whole number ratio of atoms of all the elements in one molecule of the compound. For example:

ethane = CH₃

ethene = CH₂

benzene = CH , etc.

• Electron-dot formula : In this formula valence electrons are represented by dots placed around the chemical symbol. It is also called Lewis formula. For example:

• Structural formula : It indicates how the atoms are bonded in a molecule of the compound. For example:

• Contracted or condensed formula :

It is the structural formula in contracted form to save space and time.

• Bond – line formula :

In this type of formula, carbon and hydrogen atoms are not shown and only hetero atoms are shown. The point of intersection represents carbon along with required number of hydrogen to satisfy the valency of carbon. For example:

• Spatial formula :

This formula represents the three dimensional shape or arrangement of atoms in the molecule. For example :

Functional group

An atom or group of atoms in a molecule which largely determines the chemical properties of the organic compounds is known as functional group. All the compounds having same functional group show similar properties and constitute a class or a family.

For example: organic compounds having – OH as functional group constitutes a class of compounds called alcohol.

Some other examples of functional group are :

Homologous series

The series of organic compounds having same general formula and similar chemical properties but different physical properties in which one member differs from other member by single – CH₂ unit is known as homologous series. For example:

Each member of homologous series is called homologue and phenomenon of making homologous series is called homology.

Characteristics of homologous series :

• All the members of homologous series have same functional group.
• All the members of homologpous series have same chemical properties.
• All the members of homologous series can be prepared by a common method of preparation. Eg.

• All the members of homologous series can be represented by same general formula. Eg.

C_nH_2n+2 = Alkane

C_nH_2n = Alkene

C_nH_2n-2 = Alkyne

C_nH_2n+1OH = Alcohol, etc.

• Their molecular masses increases gradually hence their physical properties (eg. melting point and boiling point) changes gradually.
• Each member differs from the adjacent member by methylene (-CH₂-) unit.

Q) Write down the 1^st, II^nd, III^rd and IV^th homologue of aldehyde homologous series.

I^st homologue = HCHO

II^nd homologue = CH₃CHO

III^rd homologue = CH₃CH₂CHO

IV^th homologue = CH₃CH₂CH₂CHO

See the IUPAC Nomenclature of Organic compounds ……

Also see the Isomerism …

Objective questions

1. Cyclic organic compounds possessing the properties of aliphatic compounds are called ___ compounds.

a. Aromatic        c. Carbocyclic

b. Homocyclic    d. Alicyclic

2. The formula which represents the simple whole number ratio of different atoms present in one molecule of a compound is known as :

a. Molecular formula   c. Condensed formula

b. Empirical formula    d. Electron dot formula

3. An organic compound has empirical formula CH₂O and molecular weight 90. It’s molecular formula will be :

a. C₆H₁₂O₆    c. C₂H₄O₂

b. C₃H₆O₃      d. C₃H₉O₆

4. General formula of an alkene is :

a. C_nH_2n       c. C_nH_2n-2

b. C_nH_2n+2    d. C_nH_2n-1

5. A hydrocarbon is found to contain 81.80% carbon and 18.20% hydrogen. It’s empirical formula will be :

a. C₄H₈    c. C₃H₈

b. C₂H₆    d. C₃H₆

6. If two compounds have same empirical formula but different molecular formula, they must have :

a. Different percentage composition

b. Different molecular weight

c. Same viscosity

d. Same vapour density

7. Which of the following is an aromatic compound :

a. Benzene hexachloride   c. Cyclobutane

b. Cyclohexane     d. Toluene

8. Which of the following is a heterocyclic aromatic compound :

a. Nephthalene     c. Furan

b. Benzene hexachloride   d. Toluene

9. Which of the following belongs to a homologous series ?

a. Methanol, Ethanol, ethanoic acid

b. Propane, Propene, propyne

c. Butane, 2-methylpropane, 2-methylbutane

d. Chloroathane, 2-chloropropane, 1-chlorobutane

10. Which of the following is not true about homologous series ?

a. Adjacent members of group differ by one –CH2 group.

b. Adjacent members of group differs by a mass of 14 amu.

c. Members of a group have same chemical and physical properties.

d. Members of a group can be prepared by same general methods.

a. Isomers    c. allotropes

b. Homologues   d. None

12. Alkanes are also called :

a. Paraffins    c. Acetylene

b. Olefins     d. Both ‘a’ and ‘b’

Answer :

1 – d   2 – b   3 – b

4 – a   5 – c   6 – b

7 – d   8 – c   9 – d

10 – c   11 – b   12 – a

References

• Sthapit, M.K., Pradhananga, R.R., Foundations of Chemistry, Vol 1 and 2, Fourth edition, Taleju Prakashan, 2005.
• Bahl, B.S., A., Advanced Organic Chemistry, S. Chand and company Ltd, New Delhi, 1992.
• Finar, I. L., Organic Chemistry, Vol. I and Vol. II, Prentice Hall, London, 1995.
• Ghosh, S.K., Advanced General Organic Chemistry, Second Edition, New Central Book Agency Pvt. Ltd., Kolkatta, 2007.
• https://socratic.org/organic-chemistry-1/ways-to-draw-and-represent-molecules/condensed-structure
• https://www.britannica.com/science/homologous-series

The post Organic compounds Classification, Functional group and Homologous series appeared first on Online Chemistry notes.

Ozone is a tri-atomic molecule of oxygen having molecular formula O₃ and molecular weight 48.

There is a layer of ozone in the atmosphere from 12 to 24 km from the surface of the earth. Ozone is formed naturally in the stratosphere by the action of ultraviolet radiation from the sun on oxygen.

Ozone layer absorbs ultraviolet radiations from the sun and protects the living beings on the earth from the harmful effect of ultraviolet radiation.

Preparation of ozone from oxygen

Ozone is prepared by passing silent electric discharge through pure and dry oxygen in an apparatus called ozonizer. The produced gas is called ozonized oxygen that contains 10-15% ozone.

Structure of ozone

Experimental evidences (like electron diffraction, microwave studies) indicate that ozone molecule is angular with bond angle 116⁰4’’. The bond lengths are equal (1.279 Å ). This indicates ozone molecule is resonance hybrid of structure (I) and (II).

Ozone layer depletion (ozone hole)

Ozone hole means a severe depletion of ozone in the region of ozone layer through which UV rays from the sun can penetrate to the earth causing devastating effects.

Causes of ozone layer depletion :

The main causes responsible for ozone layer depletion are as follows:

Chlorofluorocarbons :

The main cause of ozone layer depletion is manmade chlorofluorocarbons (CFCs). CFCs also known as Freons are compounds containing chlorine, fluorine and carbons. They are extensively used by human beings as coolants in refrigerator, air conditioner etc.

The common CFC’s are CFCl₃ (Freon 11), CCl₂F₂ (Freon 12), CHClF₂ (Freon 22) etc.

Because of high stability, freons remain in the environment for a long time and reaches to the stratosphere. In stratosphere, CFCs absorbs UV light and decomposes to give chlorine atom which destroys the ozone molecules by chain reaction as follows:

It is estimated that one molecule of CFC can destroy 100,000 molecules of ozone. Therefore, the use of CFC’S is banned nowadays.

Unregulated rocket launches and nuclear tests :

Researchers say that the unregulated launching of rockets and nuclear tests result in much more depletion of ozone layer than the CFCs do. If not controlled, this might result in a huge loss of the ozone layer.

Nitrogenous Compounds :

The nitrogenous compounds such as NO₂, NO, N₂O are highly responsible for the depletion of the ozone layer.

Cleaning products and fire extinguishers :

Cleaning products and fire extinguishers contain or produce ozone depleting substances like halons (eg. Bromotrifluoromethane – CBrF₃), carbon tetrachloride, hydrofluorocarbons (organic compounds containing hydrogen and fluorine eg. CHF₃), etc.

Natural Causes :

The ozone layer has been found to be depleted by certain natural processes such as stratospheric winds. But it does not cause more than 1-2% of the ozone layer depletion.

The volcanic eruptions are also responsible for the depletion of the ozone layer.

Effects of ozone layer depletion:

• Direct exposure to ultraviolet radiations causes health problems such as skin and eye cancer, weekend immune system, etc. in human and other animals.
• Strong ultraviolet rays may lead to minimal growth, flowering and photosynthesis in plants.
• Synthetic polymers, naturally occurring biopolymers, as well as some other materials of commercial interest are adversely affected by UV radiation.

Control measures of ozone layer depletion:

Some points that would help to prevent ozone layer depletion are as follows:

• Avoid the use of dangerous gases like CFCs (chlorofluorocarbons), halogenated hydrocarbons, nitrogen oxides, etc.
• Minimize the use of vehicles – The vehicles emit a large amount of green house gases that lead to the ozone layer depletion. Therefore, the use of vehicles should be minimized as much as possible. Promote the use of electric vehicles and bicycles should be promoted.
• Do not use cleaning products containing halogens (eg.CHF₃). We can replace these dangerous substances with safe compounds like bicarbonates.
• Maintain air conditioners, as their malfunctions cause chlorofluorocarbons to escape into the atmosphere.
• Rocket lunching and nuclear tests should be regulated .

Test of ozone

These two reactions can help to identify ozone:

Action with mercury (Mercury tailing) :

When mercury comes in contact with ozone, ozone oxidizes mercury to mercurous oxide. The meniscus of mercury is lowered and it leaves a tail when allowed to fall through an inclined surface. This property of mercury is called tailing of mercury.

Action with KI and starch :

When ozone is passed into the solution of KI and starch, ozone oxidizes KI to iodine and iodine gives blue colour with starch.

These reactions can be used for the test of ozone.

Uses of ozone

• It is used as germicide and disinfectant for the sterilization of water due to its oxidizing nature.

• It is used to prepare organic compounds like aldehydes and ketones.
• It is used for bleaching delicate articles like silk, starch, wax, etc.

• It is used for air purification at the crowded places like cinema halls and tunnel railways.

Objective Questions

1. O₂ and O₃ are:

a. Isotopes    c. Allotropes

b. Isobars     d. Isomers

2. Ozone is tested by:

a. Ag     c. Hg

b. Au     d. Zn

3. Ozone is in high concentration in :

a. Troposphere     c. Mesosphere

b. Stratosphere     d. Thermosphere

4. Tailing of mercury is caused by :

a. Cl₂      c. O₂

b. H₂      d. O₃

5. One molecule of chlorofluorocarbons can destroy upto ____ molecules of ozone.

a. 100     c. 10000

b. 1000   d. 100000

6. Montreal protocol is related to the:

a. Global warming

b. Ozone layer depletion

c. Sustainable development

d. Food security

7. Which of the following is known as freon?

a. CCl₂F₂     c. CF₄

b. CHCl₃     d. CHF₂

Answers :

1.c   2. c   3. b    4. d

5. d    6. b    7. a

References

• Agrawal, S. K., Lal, K., Advanced Inorganic Chemistry, Fifth Revised Edition, Pragati Prakashan, Meerut, 2001.
• Cotton, F. A., and Wilkinson, G., Advanced Inorganic Chemistry, Fifth edition, John Wily and Sons, Singapore, 1995.
• Day, C.M., Selbin, J., Theoritical inorganic Chemistry, second edition, Affiliated East-West Press Pvt. Ltd., New Delhi, 2002.
• https://www.sizes.com/units/dobson_unit.htm#:~:text=A%20unit%20expressing%20the%20concentration,It%20is%20named%20for%20G.%20M.%20B.
• https://www.epa.gov/ozone-pollution-and-your-patients-health/what-ozone#:~:text=Ozone%20(O3)%20is%20a%20highly,either%20good%20or%20bad%20ways.
• https://www.epa.gov/ground-level-ozone-pollution/ground-level-ozone-basics

The post Ozone – Preparation, Uses, Test, Ozone layer depletion appeared first on Online Chemistry notes.

The two or more forms of same element having similar chemical properties but different physical properties are called allotropes and the phenomenon is called allotropy. For example- diamond and graphite are two allotropes of carbon.

Allotropes of carbon, sulphur and phosphorus are described below.

Allotropes of carbon

Carbon exists on following allotropic forms:

.

Crystalline form :

1. Diamond :

Diamond is the purest form of carbon. In diamond 1 carbon atom is bonded with 4 other carbon atoms by covalent bond forming tetrahedral structure. The bond angle between carbon atoms in diamond is found to be 109.5⁰ and its bond length is 1.54 Å.

• It is transparent and sparks when light falls on it.
• It is the hardest substance known.
• It is purest form of carbon ( 100% carbon)
• It’s density is very high (3.5 at 15⁰C)
• It has very high refractive index ( 2.417)
• Its melting point is very high (3750⁰C).
• It is bad conductor of heat and electricity.

Note : Diamond is bad conductor because its electrons are not free to move due to the formation of sp³ hybrid covalent bonds between carbon atoms.

Uses of diamond :

• It is used in jewellery.
• It is used for cutting glass.

Note : Diamond is weighed in carat, 1 carat = 0.2 gm

2. Graphite :

In graphite one atom is attached with other three carbon atoms by covalent bond which forms hexagonal ring. Out of four valence electrons only three valence electrons are used to form three covalent bonds leaving one electron free. This free electron per atoms is responsible for making graphite a good conductor. The hexagonal rings of graphite form a layer structure. The two adjacent layers of graphite are held with each other by Vander Waal’s force of attraction. Distance between two layers is 3.4 Å. The bond length between two carbon atoms is 1.42 Å and bond angle is 120⁰.

• It is good conductor of heat and electricity.
• It’s melting point is 3500⁰C.
• It is resistant towards many chemical substances but it burns on strong heating.

Uses of graphite :

• It can mark the paper. So, it is used to prepare lead of pencil.
• It is used to prepare electrodes.
• It is used for the preparation of artificial diamond.

Note : Graphite can be converted into diamond at about 1600⁰C and 50,000 to 60,000 atm pressure. Because of high cost and poor quality, diamond is seldom (rarely) made artificially.

3. Fullerene :

It is latest discovered allotropic form of carbon (in 1985). The most common form of fullerene is C₆₀. It is named as Buckminster fullerene after the name of Rechard Buckminster. It contains 60 carbon atoms arranged in hexagonal and pentagonal ring which forms spherical bucky ball or soccer like structure. So, C₆₀ molecules are also called bucky balls.

• It is soluble in organic solvent.
• It is good conductor of electricity.

Note : Other forms of fullerene are C₃₂ , C₅₀ , C₇₀ , C₇₆ , C₈₄ , etc.

Uses of fullerene :

• It is used to make superconductor (due to its reasonable electric conductivity).
• It is used to make carbon nanotubes.
• It is used as molecular sieve.

Amorphous forms :

1. Coal : It is black or almost black solid combustible substance. It is naturally formed by the partial decomposition of wood or vegetable matter in presence of moisture at high pressure and temperature. It is used as fuel.

2. Coke : It is the residue of coal obtained after destructive distillation of coal. During this process volatile organic substances escape out. It is pure form of carbon, used as fuel as it burns without smoke. It is also used as a reducing agent in metallurgy.

3. Charcoal : It is porous form of carbon produced by the destructive distillation of organic materials like wood, sugar, bone, etc. Organic materials are heated strongly in limited supply of air. Charcoals are highly porous and used as adsorbent for absorbing toxic gases and purifying liquids.

4. Gas carbon : Gas carbon is dense form of deposited carbon on the interior part of gas retort during the manufacture of coal gas. It is good conductor of electricity and is used for making electrode in dry cell.

5. Lamp black : Lamp black is a finely divided black powdered soot obtained by burning natural gas and other carbon rich compounds in limited supply of air. It is used in manufacture of ink, black paints, boot polishes, carbon paper, etc.

Allotropes of sulphur

Sulphur exists on following allotropic forms :

Crystalline Allotropes :

1. Rhombic sulphur ( α – sulphur ) :

• It is most stable form of sulphur at ordinary temperature.
• The crystals are octahedral (eight sided).
• It has pale yellow colour.
• It is insoluble in water but soluble in carbon disulphide (CS₂).
• It’s melting point is 114⁰C.
• It is bad conductor of heat and electricity.
• Above 96⁰C, the rhombic sulphur changes to monoclinic sulphur.

2. Monoclinic sulphur ( β – sulphur) :

• It is needle shaped crystalline sulphur.
• It is stable allotrope of sulphur above 96⁰C.
• It is insoluble in water but soluble in carbon disulphide (CS₂).
• It’s melting point is 119⁰C.

Note : 96⁰C is the transition temperature at which both α and β forms can co-exist. Below 96⁰C rhombic sulphur is stable and above 96⁰C monoclinic sulphur is stable.

Amorphous allotropes :

1. Plastic sulphur ( γ – sulphur) :

• When boiling sulphur is poured into cold water, a rubber like substance is formed which is plastic sulphur.
• It is insoluble in water and CS₂.
• It is super cooled liquid and does not have sharp melting point.
• On long standing, plastic sulphur changes to rhombic sulphur.

2. Colloidal sulphur ( δ – sulphur) :

It is prepared by passing H₂S gas to the solution of oxidizing agents like SO₂, HNO₃, KMnO₄, K₂Cr₂O₇, etc.

• It is insoluble in water but soluble in CS₂.
• It has no sharp melting point.
• On long standing or heating, it changes into rhombic sulphur.

3. Milk of sulphur :

• When sulphur is boiled with milk of lime [Ca(OH)₂] then water soluble mixture of calcium pentasulphide (CaS₅) and calcium thiosulphate (CaS₂O₃) is obtained. This mixture is then treated with dil. HCl and amorphous milk of sulphur is precipitated.

• It is white solid insoluble in water but soluble in CS₂.
• It has no sharp melting point.
• It is bad conductor of heat and electricity.
• On heating it changes to rhombic sulphur.

Allotropes of phosphorus

Phosphorus exists on following allotropic forms :

Among them white and red phosphorus are more common.

White Phosphorus :

White phosphorus is manufactured by heating calcium phosphate i.e. Ca₃(PO₄)₂ with sand (SiO₂) and coke (C) .

Calcium phosphate reacts with sand to form calcium silicate and phosphorus pentoxide.

Phosphorus pentoxide in vapour form is reduced by coke and liberates phosphorus at 1500⁰C.

• It is soft, waxy white solid with garlic odour.
• Its melting point is 44⁰C and boiling point is 287⁰C.
• It is extremely poisonous.
• Its ignition temperature is 35⁰C. It burns (oxidize) in air giving yellow-green flame forming phosphorus pentoxide and trioxide. This phenomenon is called phosphorescence.

Hence, it is stored under water.

Uses of white phosphorus :

• It is to prepare poison for killing rats.
• It is used to prepare phosphine gas, phosphoric acid, etc.
• Hypophosphite prepared from white phosphorus is used in medicines as tonic.

Red Phosphorus :

• When white phosphorus is heated at about 250⁰C in an inert atmosphere( of nitrogen or carbondioxide or coal gas) for several hours then it is converted into red phosphorus.
• It is dark red powder.
• It is colourless and non-poisonous.
• Its melting point is 550⁰C and sublimes at 290⁰C in absence of air.
• It is less reactive than white phosphorus.
• It does not show the phenomenon of phosphorescence.

Uses of red phosphorus :

• It is used in the manufacture of phosphate fertilizer.
• It is used for making match stick in match industry.

Note: The tip of match stick contains combustible material Sb₂S₃ along with oxidizing agent like KClO₃, PbO₂ or K₂CrO₄ and the two sides of the match box is coated with a mixture of powdered glass (abrasive) and red phosphorus. Friction applied to the side of the match box vaporizes red phosphorus and ignites. Thus produce fire to the head of match stick.

References

• Lee, j. D., Concise Inorganic Chemistry, Fifth Edition, Joh, Wiley and Sons, Inc., 2007.
• Sarkar, R., General and Inorganic Chemistry, Second Edition, New Central Book Agency(P) Ltd., India, 2007.
• Shriver, D. F., Atkins, P. W., Inorganic Chemistry, Fifth Edition, Oxford university Press, 2010.
• Agrawal, S. K., Lal, K., Advanced Inorganic Chemistry, Fifth Revised Edition, Pragati Prakashan, Meerut, 2001.
• Cotton, F. A., and Wilkinson, G., Advanced Inorganic Chemistry, Fifth edition, John Wily and Sons, Singapore, 1995.
• https://www.simply.science/images/content/chemistry/metals_and_non_metals/oxygen_family/conceptmap/Monoclinic_sulphur.html#:~:text=Monoclinic%20sulfur%20is%20a%20crystalline,ring%20molecules%20in%20crystalline%20structure.
• https://geology.com/minerals/graphite.shtml
• https://www.sciencedirect.com/topics/engineering/red-phosphorus

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