Interesting Careers 4: How Do They Do It - Mining Trucks, IKEA, Air Traffic - Video

Interesting Careers 4: How Do They Do It - Mining Trucks, IKEA, Air Traffic - Video

Interesting Careers 2- How Do They Do It - Expensive perfumes, Genius recycling solution

Interesting careers 4 - high tech manufacturing

http://ashimaggrwl.blogspot.in/2014/12/interesting-careers-1.html

Interesting careers 3 - Food

http://ashimaggrwl.blogspot.in/2014/12/interesting-careers-3.html

Interesting Science Careers - How Do They Do It - Air to Air, Ski Slopes, Car Production - Video

Electrochemistry Class XII (chapter3)

d & f block elements (Chapter 8) class XII

d- and f- block elements

d-block elements represent change in properties from most electropositive s-block elements to least electropositive p-block elements. So, these are called transition metals.

Transition elements are those elements which have partially filled d-subshells in their elementary state or in their commonly occurring oxidation states.

General configuration = (n-1) d^1-10 ns^1-2

There are four transition series:

3d – Sc        Ti         V        Cr         Mn           Fe         Co        Ni      Cu       Zn

4d – Y          Zr        Nb     Mo        Tc            Ru          Rh       Pd      Ag      Cd

5d – La        Hf        Ta       W         Re           Os            Ir        Pt       Au      Hg

6d – Ac        Ku       Ha

Cu = 3d¹⁰ 4s¹. But in its most common oxidation state (+2), it contains one unpaired electron in its d-orbital. So, it is a transition element. Cu²⁺ = 3d⁹ 4s⁰. Similarly, Ag and Au are also transition elements.

Physical Properties –

1) Atomic radii – Across the period (series), atomic radii of elements of a particular series decrease with increase in atomic number and then  become constant in mid-way and at last, there is an increase in atomic radius.

Reason – With increase in atomic number, nuclear charge goes on increasing and force of attraction between nucleus and valence electrons increases due to which atomic radius decreases with increase in atomic number in the beginning.

But as the electrons are increased in d-orbitals. These electrons screen the outermost s-electrons. So, as the d-electrons increase, screening effect increases which neutralises the effect of increased nuclear charge and hence, atomic radius remains almost unchanged after chromium.

At the end of the series, there is slight increase in the atomic radii due to increased electron-electron repulsions between electrons in same orbitals due to which electron cloud expands and size increases.

Down the group, atomic radii increase due to increase in number of shells. But atomic radii of elements of 4d-series and 5d-series are nearly same due to Lanthanide Contraction.

2) Metallic character - All transition elements are metals. They have high density, hardness, high melting point and boiling point, malleable, ductile, conduct electricity and heat.

Reason – Metallic character is due to their low ionization energies and number of vacant orbitals in outermost shell. Greater the number of unpaired d-electrons, greater is the number of bonds and greater is the strength of metallic bonds. E.g. Cr, Mo and W (tungsten) have maximum number of unpaired electrons and are very hard metals.

Zn, Cd and Hg do not have any unpaired electrons. So, they are not hard (nd¹⁰). They are also known as non-transition metals.

3) Density – All metals have high density. Osmium has the highest density.

4) Melting point and boiling point – They have high melting point and boiling point due to strong metallic bonds between atoms of elements.

Metallic bond is formed due to the interaction of electrons in outermost orbitals. Greater the number of valence electrons, stronger is the metallic bonding and higher is the melting point. In a particular series, metallic strength increases upto middle with increasing availability of unpaired electrons upto d⁵ configuration (Sc =1, Ti =2, V =3, Cr =4, Mn =5) and then decreases with decreasing availability of unpaired electrons in d-orbital. Therefore, melting point decreases.

5) Ionization enthalpy – Energy required to remove an electron from valence shell of an isolated gaseous atom. First ionization enthalpy is higher than those of s-block elements and lesser than p-block elements.

Reason – Due to increase in nuclear charge and force of attraction between nucleus and valence electrons. Hence more energy is required to remove an electron from outermost shell.

Also, the effect of increasing nuclear charge is opposed by additional screening effect of nucleus and hence, ionization enthalpy increases but quite slowly among d-block series.

First ionization enthalpy of 5-d series is higher than those of 3d- and 4d- series due to poor shielding effect of 4f-orbitals. As a result, outer electrons have greater nuclear charge acting on outer valence electrons.

Consequences: a) Ni (II) compounds are thermodynamically more stable than Pt(II) compounds. Pt(IV) compounds are more stable than Ni(IV) compounds. E.g. K₂PtCl₆ exist but not K₂NiCl₆.

6) Oxidation state – Transition metals exhibit variable oxidation states.

Reason- This is due to the participation of inner (n-1) d-electrons in addition to outer n electrons because energies of ns- and (n-1) d subshells are almost equal. They contain many unpaired electrons.

Highest oxidation state is shown by Osmium(Os) and Rhuthenium(Ru) of +8.

Few transition metals also form compounds in low oxidation states like +1, 0, -1. E.g. [Ni(CO)₄], [Fe(CO)₅] in which Ni and Fe have zero oxidation states.

7) Formation of coloured ions – s and p-block elements are white in colour while d-block elements are coloured in solid or solution form.

Reason- It occurs due to the presence of unpaired electrons. The energies of five d-orbitals in the same subshell do not remain equal during the formation of complexes or compounds of transition metals. Under the influence of approaching ions towards the central metal ion, d-orbitals split into different energy levels. This phenomenon is known as crystal field splitting (CFS).

In case of transition metal ions, electrons are excited from one energy level to another in same d-subshell. These are known as d-d transitions. The amount of energy required to excite unpaired electrons to higher excited states within the same d-subshell corresponds to energy of certain colours of visible light and the observed colour of substance is always complementary colour of colour absorbed by the substance.

Zn, Cd and Hg compounds are generally white in colour because of completely filled d-orbitals i.e. they do not have any unpaired electron (d¹⁰).

8) Magnetic properties – Most of the compounds of transition elements are paramagnetic due to the presence of unpaired electrons in the d-orbitals of metal ion. Unit of magnetic moment (μ) = Bohr Magneton.              μ =    n ( n + 2 )        where n = number of unpaired electrons

Iron oxide = Ferromagnetic substance (highly magnetic).

Zn²⁺, Cd²⁺, Hg²⁺, Cu⁺ compounds = Diamagnetic (no unpaired electron).

9) Tendency to form complexes – Transition elements form large number of coordination compounds. The transition metal ions bind to number of anions or neutral molecules (ligands) in the complexes like [Ni(NH₃)₆]²⁺, [Co(NH₃)₆]³⁺, etc.

Reason- due to small size of transition metal atoms or ions, high nuclear charge, presence of vacant d-orbitals of suitable energy are available to accept lone pair of electrons donated by ligands.

10) Formation of interstitial compounds – Transition metals form interstitial compounds with elements like B, C and N i.e. they get trapped in vacant spaces of lattices of these metals.

Reason-  Transition metals can easily accommodate the vacant spaces of these non-metals due to their small size, defects in structures and their variable oxidation states.

11) Catalytic properties – Many transition metals and their compounds act as good catalyst for many reactions like Ni, Pt, Pd, V₂O₅, MnO₂, etc.

Reason- (i) due to the presence of vacant orbitals and their tendency to show variable oxidation states, (ii) Transition metals provide large surface area on which reactants may be absorbed for the reaction to occur. E.g. in Contact process, solid V_2­O₅ adsorbs SO₂ for reaction.

12) Alloy formation - Transition metals form large number of alloys.

Reason- Transition metals are quite similar in size and atoms of one element in its crystal lattice substitute atoms of other metal in its crystal lattice. E.g. Brass = Cu + Zn, Bronze = Cu + Sn, Stainless steel = Fe + Cr + Ni (high melting point, hard, more resistant to corrosion).

13) First row transition metals – Oxides- General formulae: MO, M₂O₃, M₃O₄, MO₂, M₂O₅, MO₃ .

Acidic character increases as oxidation state increases.

Oxide :                   MnO        Mn₂O₃          Mn₃O₄          MnO₂       Mn₂O₇

Oxidation:               +2               +3               +8/3               +4             +7

Reducing character of oxides –Across the period, reducing character first increases then decreases as oxidation state. Lower oxidation state is more stable in 3d-series i.e. +2 state. V₂O₅ and CrO are strong reducing agents.

 As we move down the group, higher oxidation states become more stable. E.g. Osmium(VIII) oxide (OsO₄) is known but iron (VIII) oxide is not known. Similarly, hexafluorides of 4d- and 5d-series exist.

K₂Cr₂O₇  (potassium dichromate):

Preparation: from chromite ore

(i) 4FeCr_2­O₄ + 16NaOH + 7O₂ → 8Na₂CrO₄ + 2 Fe₂O₃ + 8H₂O

(ii) 2Na₂CrO₄ + H₂SO₄ → Na₂Cr₂O₇  + Na₂SO₄ + H₂O.  Na₂SO₄ can be removed by crystallization.

(iii) Na₂Cr₂O₇  + 2KCl → K₂Cr₂O₇  + 2NaCl. So, orange crystals of K₂Cr₂O₇  are obtained.

Properties: (i) It is orange red crystalline solid, soluble in water.

(ii) Action of heat: 4 K₂Cr₂O₇     →  4 K₂Cr₂O₄ + 2 Cr₂O₃ + 3O₂

(iii) K₂Cr₂O₇  + 2 KOH → 2 K₂Cr₂O₄ + H₂O

2 K₂CrO₄ + H₂SO₄ → K₂Cr₂O₇  + K₂SO₄ + H₂O

(iv) Chromyl chloride test – used to test the presence of chloride ions.

K₂Cr₂O₇ + 4NaCl + 6H₂SO₄ → 2KHSO₄ + 4NaHSO₄ + 3H₂O + 2Cr₂O₂Cl₂ (chromyl chloride red vapours)

(v) K₂Cr₂O₇ + 4HCl → 2KCl + 2CrCl₃ + 7H₂O + 3Cl₂

(vi) Oxidising character- K₂Cr₂O₇ is a powerful oxidizing agent.

K₂Cr₂O₇ + 4H₂SO₄ → K₂SO₄ + Cr₂(SO₄)_­3­ + 4H₂O + 3[O]

Strong oxidizing agent in acidic medium:

Cr₂O₇^2- + 14 H⁺ + 6e^- → 2Cr³⁺ (green) + 7 H₂O

Uses: a) in volumetric analysis of Fe²⁺, I^-.    b) in photography for hardening gelatin film.  c) in dyeing.

Note: a) 2I^- → I₂ + 2e^-           b) Fe²⁺ → Fe³⁺  + e^-                 c) H₂S → S + 2H⁺ + 2e^-

d) H₂C₂O₄ → 2CO₂ + 2H⁺ + 2e^-                e) SO₃^2- + H₂O  → SO₄^2- + 2e^- + 2H⁺ 

KMnO₄ = Potassium permanganate

Preparation: 1) a) 2MnO₂ + 4KOH + O₂ →2K₂MnO₄ + 2H₂O

b) 3MnO₄^2- + 4H⁺ → 2MnO₄^- + MnO₂ + 2H₂O   [Disproportionation reaction]

2) Commercial process:  KMnO₄ is prepared by the alkaline oxidative fusion of MnO₂ followed by electrolytic oxidation of manganite (VI)

Properties: a) It is dark violet crystalline solid.

b) It is fairly soluble in water.

c) 2KMnO₄                K₂MnO₄ + MnO₂ + O₂

d) 4KMnO₄ + 4KOH → 4K₂MnO₄ + 2H₂O + O₂

e) It is powerful oxidising agent in neutral, alkaline or acidic solutions. MnO₄^-  + e^-  → MnO₄^2-_,

MnO₄^- + 4H⁺ +3e^-  → MnO₂ + 2H₂O,                MnO₄^- + 8H⁺ +5e^-  → Mn²⁺ + 4H₂O.

Uses: as an oxidizing agent in labs and industries, used as disinfectant for water, in quantitative and qualitative analysis.

E⁰ values (standard electrode potential) become less negative across the series is related to the general increase in sum of the first and second ionization enthalpies.

More the ionization enthalpy is, more is the reduction potential and lesser is the oxidation potential. So, E⁰ value (reduction potential) becomes more negative.

Exception: a) E⁰ for Mn, Ni and more (-) than expected from the trend. This is due to the stability of the half-filled d subshell in Mn²⁺ and completely filled d¹⁰ configuration in Zn²⁺, whereas E⁰ for Ni is related to the highest negative Δ_hydH⁰ (hydration energy).

b) Cu²⁺ = 3d⁹ 4s⁰,   Cu⁺ ₌ 3d¹⁰ 4s⁰

Cu²⁺ is exceptionally more stable than Cu⁺ in aqueous solutions because of more negative Δ_hydH⁰ of Cu²⁺ (aq) than Cu⁺ , which is more than compensates for the second ionization enthalpy of Cu. So, Cu⁺ undergoes disproportionation in aqueous solution. 2 Cu⁺ (aq) → Cu²⁺ + Cu. E⁰ = + 0.34V (favourable E⁰ value).

c) All Cu²⁺ halides are known except the iodide. In this case, Cu²⁺ oxidises I^-  to I₂.

2 Cu²⁺ + 4 I^- → Cu₂I₂ (s) + I₂

Stable d-configurations – d³ = stable half-filled t_2g configuration, d⁵ = half-filled d-orbital, d⁶ = fully-filled t_2g level, d¹⁰ = fully-filled d-orbital.

f-block elements: Elements in which last electron enters the f-orbital of their atoms are called f-block elements. They are known as inner transition elements.        

 General Configuration: (n-2)f^1-14 (n-1)d^0-10 ns^1-2      (n-2) = Anti-penultimate shell,   (n-1) = Penultimate shell,   n = main ( valence ) shell.

They consist of two series of elements placed at the bottom of periodic table by filling of electrons in 4f- and 5f-orbitals.

(i) Lanthanoids or lanthanides-  The series involving the filling of 4f-orbitals following Lanthanum (Z=57), Cerium (Z=58) to Lutetium (Z=71).

They occur very rarely and are known as rare earth metals.

Z=57 = La (Lanthanum) = 5d¹ 6s²,    Z=58 = Ce (Cerium) = 4f¹  5d¹ 6s²,     Z=59 = Pr (Praseodymium) = 4f²  6s²,      Z=63 = Eu (Europium) = 4f⁷ 6s²     ,        Z=64 = Gd  (Gadolinium) = 4f⁷  5d¹ 6s²,                            Z=71 = Lu (Lutetium) = 4f¹⁴  5d¹ 6s².

(ii) Actinides - The series involving the filling of 5f-orbitals following Actinium (Z=89), Thorium (Z=90) to Lawrencium (Z=103).

Actinium (Z=89) = Ac = 6d¹ 7s²,  Thorium (Z=90) = Th = 6d² 7s², Curium (Z=96) = Cm = 5f⁷ 6d¹ 7s², Lawrencium (Z=103) = Lr = 5f¹⁴ 6d¹ 7s²

The Lanthanoids- All Lanthanoids have common oxidation state of +3 with +2 and +4 also by attaining stable f⁰, f⁷, f¹⁴ configurations.

Physical Properties: All lanthanides are soft, malleable, and ductile with low tensile strength, bad conductors of heat and electricity.

General Charcteristics:  1) Trivalent ions of lanthanides are coloured in solid state as well as in solution due to the absorption in visible light of spectrum by unpaired electrons resulting in f-f transitions.

2) They are paramagnetic. All lanthanides are highly electropositive metals and have almost similar reactivity due to the fact that lanthanides differ only in number of 4f electrons. Since these electrons are very effectively shielded from interaction by 6s, 5p electrons, hence they show little differences in their chemical reactivity. So, they are difficult to separate.

f⁰ and f¹⁴ are colourless and diamagnetic [La³⁺ and Lu³⁺]

3) Atomic radii- The steady decrease in atomic sizes and ionic sizes of lanthanides with increasing atomic number is known as lanthanide contraction.

Reason: As atomic number increases across the series, nuclear charge increases by one unit. The new electrons are added to same inner 4f-subshells. But 4f electrons shield each other from nuclear charge quite poorly because of very diffused shapes of f-orbitals.

Consequences of lanthanide contraction:  

a) Elements of 4f- and 5f- transition series resemble each other due to similar sizes.

b) It is difficult to separate these elements in pure state. Fractional crystallization method is done repeatedly to separate them.

c) Due to lanthanide contraction, size of lanthanide ions decrease with increase in atomic number due to which the covalent character between Lanthanide ion and OH^- ions increase from La³⁺ to Lu³⁺. Therefore, basic strength of hydroxides decreases with increase in atomic number across the series. Thus, La(OH)₃ is most basic and Lu(OH)₃ is least basic.

The actinoids-

General Characteristics: 1) These metals are all silvery in appearance.

2)Antinoid contraction- The steady decrease in atomic sizes and ionic sizes of antinoids with increasing atomic number is known as actinide contraction.

Reason: As atomic number increases across the series, nuclear charge increases by one unit. The new electrons are added to same inner 5f-subshells. But 5f electrons shield each other from nuclear charge quite poorly.

3) They are radioactive elements and earlier members have relatively long half lives, the latter ones have half lives ranging from a day to 3 minutes for Lawrencium (Z=103). The latter members are prepared only in nanogram quantities. These facts render their study more difficult.

Lanthanoids	Actinoids
1) Tendency to form complexes is less.	1) More tendency to form complexes.
2) Non-radioactive except promethium.	2) All are radioactive.
3) All show +3 oxidation state and few show +2 and +4 also.	3) All show +3, +4, +5, +6 and +7 oxidation state.
4) They are less basic.	4) They are more basic.

Misch Metall- An alloy of a lanthanoid metal (95%) and iron (5%) with traces of S, C, Ca and Al. It is used in Mg-based alloy to produce bullets, shell and lighter flint.

Uses of Lanthanoids: (i) For production of alloy steels for plates and pipes. (ii) Mixed oxides of lanthanoids are used as catalysts in petroleum cracking.

Reactions of Lanthanoids:

2Ln + 3X₂ → 2LnX₃ ,        2Ln + 6H₂O →  2Ln(OH)₃ + 3H₂ ,

Ln + O₂ → Ln₂O₃ ,              Ln + S →Ln₂S₃ ,                Ln + N₂ → LnN,                         Ln + C  →  LnC₂  

  

Chemistry in everyday life (Chapter 16) Class XII

Chemistry in everyday life

Drugs – are the chemicals of low molecular mass (100-500u) which interact with macromolecular targets and produce a biological response i.e. can act as medicines or as poisons.

Chemotherapy - Branch in medicine where we use chemicals to cure the diseases or reduce suffering from pains (for therapeutic effect).

Classification of drugs: a) On the basis of pharmacological effect – analgesics for pain killing effect, antiseptics for killing or arresting the growth of micro-organisms.

b) On the basis of drug action – It is based on the action of a drug on a particular biochemical process. E.g. all antihistamines inhibit the action of the compound, histamine which causes inflammation in the body.

c) On the basis of chemical structure – on the basis of common structural features and similar pharmacological activity. E.g. sulphonamides have common structure:

d) On the basis of molecular targets – Drugs possessing some common structural features may have the same mechanism of action on targets. (most useful classification).

Consideration of drug designing – Two considerations are essential: (i) Choice of drug target – Drugs usually interact with biological macromolecules known as targets, (ii) Drug metabolism – A drug travels through the body in order to reach the target. So, its design should be such that it reaches the target without being metabolised in between and also, after its action, it should be excreted without causing harm to the body.

Drugs are designed by lead compounds (Sources = plants, trees, bushes, metabolites of microorganisms)

Q. How do drugs interact with targets?

Drugs interact with biological macromolecules like proteins, carbohydrates, lipids and nucleic acids. E.g. Enzymes are the proteins which perform the role of biological catalysts. They hold the substrate for a chemical reaction and they provide functional group (-OH, -NH₂, -COOH) that will attack the substrate and carry out chemical reaction (by H-bonding, ionic bonds, Vander Waal’s forces, dipole-dipole interactions). Drugs are designed to inhibit any of the above activities of enzymes and are known as enzymes inhibitors which can block the binding site and prevent the binding of substrate with enzyme.

Drugs compete with the natural substrate for active site. Such drugs are called as competitive inhibitors. Some drugs do not bind to the active site but bind to the different site of enzyme known as allosteric site. This changes the shape of active site such that substrate cannot recognize it.

Receptors – are the proteins that are crucial to the body’s communication process which are embedded in cell membrane (made up of phospholipid bilayer). Two chemical messengers which help the receptor to transfer message into the cell are:

a) Hormones – are released from endocrine glands. These enter into the blood stream and travel in the body, activating all the receptors which recognize them for message transfer (that helps in the growth and development of the body). E.g. Adrenaline (epinephrine) is a hormone released from adrenal medulla in the situation of stress or danger.

b) Neurotransmitters – like acetylcholine, dopamine and serotonin. Nerves transfer message through them.

Drugs are categorised as – a) Antagonists : drugs that bind to receptors site and inhibits its natural function by blocking the active site.

b) Agonists : Drugs that mimic the natural messenger by switching on the receptor. These are useful when there is lack of natural chemical messenger in the body.

Side effects are caused when drugs bind to more than one type of receptor or are taken in excess.

Therapeutic action of different classes of drugs:

 a) Antacids – Chemical substances which can reduce or neutralise the acidity of stomach which occurs due to release of excess of gastric acid (HCl) in stomach. E.g. Ranitidine (Zantac), Cimetidine. They are better than NaHCO₃ and Mg(OH)₂ as they prevent interaction of histamine with receptors (or anti-allergic action) in stomach and helps in quick healing of ulcers as they  are mild bases.

b) Antihistamines (or anti-allergic drugs)- Drugs which combat or prevent the effects of histamine, a chemical released by certain cells of body (mast cells) during an allergic reaction. E.g. Dimetapp and seldane.

Histamine is a potent vasodilator. It contracts smooth muscles in bronchi and gut and relaxes other muscles of blood vessels. It is also responsible for nasal congestion associated with common colds and allergic response to pollen grains. 

Antihistamines do not affect the secretion of acid in stomach because antiallergic and antacid drugs work on different receptors.

c) Neurologically active drugs –

(i) Tranquilizers or sedatives – drugs given to patients suffering from anxiety and tension. These relieve anxiety, stress, irritability or excitement. Antipsychotic tranquilizers are the major tranquilizers while sedative tranquilizers are used during intense anxiety or panic. These are used for making sleeping pills like Barbituric acid, lyminal and seconal. Non-adrenaline is a neurotransmitter that plays role in mood changes. If level of non-adrenaline is low, the person suffers from depression. So, anti-depressant drugs are required like Iproniazid, Nardil (phenelzine).

(ii) Analgesics – Chemical substances which are used for relieving pains in the body. Two types of analgesics are: Non-narcotic drugs (non-addictive) – Salicylates (for e.g. aspirin and paracetamol) belong to this class. Aspirin inhibits the synthesis of chemicals known as prostaglandins which stimulate inflammation in tissue and cause pain. These drugs are effective in reducing fever (antipyretic) and preventing platelet coagulation. Because of its anti-blood clotting action, aspirin finds its use in prevention of heart attacks.(Aspirin = 2-Acetoxy benzoic acid).                                                                  Narcotic analgesic drugs – such as morphine and many of its homologous, when administered in medicinal doses, relieve pain and produce sleep and cause addiction. Morphine narcotics are sometimes referred to as opiates, since they are obtained from the opium poppy.  

d) Antibiotics – Chemical substances produced by micro-organisms (bacteria, fungi, moulds) that inhibit the growth or even destroy other micro-organisms. First antibiotic was discovered by Alexander Fleming and it was penicillin (for malaria). The complete range of micro-organisms which can be killed by a particular antibiotic is called its spectrum. There are two types of antibiotics-: Broad – spectrum antibiotics: antibiotics which can kill different types of micro-organisms. They can cause several infections. E.g. Tetracyclines, chloramphenicol. Narrow – spectrum antibiotics: antibiotics which can kill only few micro-organisms. E.g. Penicillin, streptomycin, chloromycetin.

Two major classes of antibiotics: Bactericidal is an antibiotic which kills the micro-organisms in the body. E.g. Penicillin, ofloxacin. Bacteriostatic is an antibiotic which inhibit or control the growth of micro-organisms. E.g. tetracycline, chloramphenicol.

e) Antiseptics – can kill or prevent the growth of micro-organisms. They do not harm the living tissue and can be applied to the skin. They are used for dressing wounds, ulcers and in treatment of diseased skin. E.g. Dettol (a mixture of chloroxylenol and terpineol), Tincture of iodine (2-3% iodine solution of alcohol-water)

Disinfectants – can kill micro-organisms. They are toxic to living tissues. So, it cannot be applied to the skin. These are used for disinfecting floors, toilet drains, instruments, etc. E.g. DDT (Dichloro Diphenyl Trichloroethane), BHC (Benzene hexachloride).

Phenol 1-2% solution is used as disinfectant while 0.2% solution of phenol is used as antiseptic.

f) Antifertility drugs – are hormonal contraceptives and are available for females only. Birth control pills are derivatives of synthetic oestrogen and progesterone. E.g. Norethindrone (progesterone derivative), novestrol (oestrogen derivative).

Chemicals in food: a) Artificial sweetners – Natural sweetners sucrose adds to calorie intake and therefore many people prefer to use artificial sweetners. Its use is of great value to diabetic persons. E.g. Aspartame (unstable at cooking temperature as it contains peptide linkages), Saccharine(low quality), Sucrulose, Alitame (high potency sweetner).  

b) Preservatives – They prevent the spoilage of food due to microbial growth. They are classified into two groups: Class 1 preservatives contain table salts, sugars and vegetable oils and Class 2 preservatives contain sodium benzoate (C_6­H₅COONa).

c) Antioxidants – Substances which when added to the fats and fat containing food prevents their  oxidation (rancidity) and thus prolong their life. Example : Butylated  Hydroxy Anisole (BHA), Butylated Hydroxy Toluene (BHT), Vitamin C.

 Chemistry of Cleansing agents :

SOAPS  ::  are sodium and potassium salts of long chain fatty acids. Soaps are formed  by treating NaOH with fatty acids. This process is known as saponification.

The soap obtained is in colloidal form which is precipitated by adding NaCl and is obtained by fractional distillation. Cleansing action of soaps –

Hydrophobic tail and Hydrophilic head forming micilles.

Soaps do not work in hard water (disadvantage): Hard water contains calcium and magnesium ions which form insoluble salts with soaps as scum which are useless as cleansing agent.

2C₁₇H₃₅COONa  +  CaCl₂ (in hard water)     è     2NaCl  +  (C₁₇H₃₅COO)₂Ca    (insoluble)

Soaps can be biodegraded.

Advantage of detergents :  They do not form scum with hard water.

 Disadvantage of detergent: They are non-biodegradable.

STRUCTURE :

Sodium palmitate :  C₁₅H₃₁COONa

Sodium oleate : C₁₇H₃₁COONa

Types of synthetic detergents:

a) Anionic detergents – are sodium salts of sulphonated long chain alcohols. They are effective in acidic solutions. Their larger part of molecule is anion. E.g. ABS (alkyl benzene sulphonate) CH₃(CH₂)₁₆OSO₃Na and Sodium lauryl sulphonate  CH₃(CH₂)₁₀CH₂OSO₃Na.

b) Cationic detergents – are chlorides or bromides of quarternary amines. Cationic part possesses a long hydrocarbon chain and a positive charge. E.g. Cetylmethylammonium bromide(non-biodegradable). They have germicidal properties but have high cost.

c) Non-ionic detergents – Liquid dishwashing detergents are of non-ionic type. They have same mechanism of cleansing action as of soaps.

HO-CH₂-CH₂-OH   +   n  CH₂-CH₂     →     HO(CH₂CH₂O)_nCH₂CH₂OH (polyethylene glycol)

Q. Why medicines or sleeping pills should not be taken without consulting doctors?

Ans. Because most drugs are potential poisons if taken in doses higher than the dose recommended.

Q. What is biothionol? Give its use.

Ans. Biothionol is an aromatic compound containing sulphur and it is added to soaps to impart antiseptic  properties.

Q. What are hypnotics? Give an e.g.

Ans. Hypnotics are drugs which produce sleep. They are habit forming. E.g. Luminal, Seconal.

Q. What kind of medicines are given to agitated and violent patient?

Ans. Tranquilizers or sedatives or anti-depressant drugs.

Q. Low level of non-adrenaline is the cause of depression. What type of drugs are needed to cure this problem? Name two drugs.

Ans. Psychotherapeutic drugs or tranquilizers. E.g. Luminal, Equanil.

Q. Why aspirin should not be taken empty stomach?

Ans. It decreases pH and causes ulcer. 

Chemistry Class XII syllabus

http://www.slideshare.net/ashima1986/20142015-syllabus-12chemistry

Name Reactions (Organic Chemistry) Class XII

CBSE Chemistry Sample Paper 2014-15 Class XII

http://www.slideshare.net/ashima1986/chemistry-sample-paper-2014-15

Chapter 1&2 test Class XII

Time:  1 hr.                                                                                        Max. Marks: 30

Q1. What is the value of Van’t Hoff factor for: (i) Al₂(SO₄)₃      

(ii) Dimerization of benzoic acid?                                     (1)

Q2. Which magnetic substances can behave as permanent magnets and why?                                                                   (1)

Q3. Which point defect lowers the density of a crystal?                                                                                                                  (1)

Q4. Which colligative property is preferred for the calculation of molecular mass of macromolecules?                     (1)

Q5. Which type of semi-conductor is formed when: (i) Si is doped with P, (ii) Si is doped with In.                                                (1)

Q6. Calculate the fraction of M²⁺ and M³⁺ ions present in M_0.98O_1.00 .                                                                                        (2)

Q7. Determine amount of CaCl₂ dissolved in 2.5 L of water such that its osmotic pressure is 0.75 atm at 27⁰C, assuming that it is completely dissociated.                                                                                                                                         (2)

Q8. Why LiCl acquires pink colour when heated in Li vapours?                                                                                                     (2)

Q9. (i) Atoms of element B form hcp lattice and those of the element A occupy 2/3 rd of tetrahedral voids. What is the formula of the compound formed by the elements A and B?                                                                                                      (2)

(ii) State the condition resulting in reverse osmosis.

Q10. Gold (atomic mass = 197 u, atomic radius = 0.144nm) crystallises in fcc lattice. Determine the density of the gold.                                                                                                                               (2)

Q11. Determine the molarity of an antifreeze solution containing 250g water mixed with 222g ethylene glycol (C₂H₆O₂). The density of this solution is 1.07g/mL.                                                                                                                         (2)

Q12. Derive the relationship between relative lowering of vapour pressure and molecular mass of solute.            (2)

Q13. Vapour pressure of chloroform (CHCl₃ ) and dichloromethane (CH₂Cl₂) at 298 K are 200 mm Hg and 415 mm Hg respectively.(i) Calculate the vapour pressure of the solution prepared by mixing 25.5 g of CHCl₃ and 40 g of CH₂Cl₂ at 298 K and,(ii) mole fraction of each component in vapour phase.                                                                                                          (3)

Q14. Write main difference between the following with an example of each:                        

(i) Ideal and Non-ideal solution

(ii) Schottky and Frenkel Defect.                                                                                                                     (3)

Q15. a) Draw the graph showing depression of the freezing point of a solvent in a solution. Also, define Cryoscopic constant.

b) Two elements A and B form compounds having formula AB₂ and AB₄. When dissolved in 20g of benzene, 1g of  AB₂ lowers the freezing point by 2.3K whereas 1g of AB₄ lowers the freezing point by 1.3K. The molal depression constant for benzene is 5.1 KKg/mol. Calculate atomic messes of A and B.                                                                                                      (5)   

Previous Year Paper Class XII

Class – XII
Subject – Chemistry

Time:3hrs                                                                                                                                               MM:70

General instructions:

All questions are compulsory.

Questions number 1 to 8 are very short –answer questions, carrying 1 mark each.

Questions number 9 to18 are short –answer questions, carrying 2 marks each.

Questions number19 to27 are short –answer questions, carrying 3 marks each.

Questions number28 to30 are long-answer questions of 5 marks each.

Q1. Define F-centres.

Q2. Why is enthalpy of chemisorption higher than that of physisorption?

Q3. An ore sample of PbS is contaminated with ZnS. Name one chemical which can be used to concentrate galena selectively by froth floatation method.

Q4. A and B liquids on mixing produce a warm solution. Which type deviation from  Raoult’s law is there?

Q5. Give the IUPAC name of the following compound: CH₃COCH₂CH₂CHO. 

Q6. Except for vitamin B₁₂, all other vitamins of group B, should be taken regularly in diet. Why?

Q7. The decomposition of ammonia gas on platinum surface has a rate constant k = 2.5 X 10^-4 mol/L/s. What is the order of a reaction?

Q8. What is the molecularity of the reaction: Cl → ½ Cl₂?

Q9. How does molar conductance vary with concentration in strong and weak electrolytes? Explain with the help of graph.

Q10. An element E crystallises in bcc structure. If the edge length of the cell is 1.469 X 10^-10 m and the density is 19.3 g/cm³, calculate the atomic mass of this element.

Q11. Define emulsions. Classify them giving an example of each.

Q12. Classify colloids on the basis of interaction between dispersed phase and dispersion medium. Write two differences between them.

Q13. (a) What is the covalence of nitrogen in N₂O₅?

(b) Explain why both N and Bi do not form pentahalides.

Q14. (a)Write the reaction involved in the preparation of tert-butyl ethyl ether by Williamson’s synthesis.

(b) Give reason: Propanol has higher boiling point than propane.

Q15. (a) How can you convert an amide into an amine having one carbon less than the starting compound?

(b) Give the IUPAC name and structure of the amine obtained by the above method if the amide is 3-Chlorobutanamide.

Q16. (a) Why does chlorine water lose its yellow colour on standing?

(b) What happens when Cl₂ reacts with cold and dilute solution of sodium hydroxide? Write the reaction involved.

Q17. How would you distinguish between 1⁰, 2⁰ and 3⁰ amines by a chemical test. Write the reactions to justify your answer.

Q18. Give the mechanism for the preparation of ethanol from ethene by acidic hydration.

Q19. Geeta and her sister Reeta called refrigeration mechanic for the repair of their refrigerator which was not lowering the temperature and food particles were getting destroyed. The mechanic came told the sisters that it requires filling of the gas. “But where has the earlier gas gone”, asked the sisters. The mechanic told that gas leaked out perhaps due to the poor repairs done by the earlier mechanic. They knew the ill effects of Freon gas and told the mechanic to fill and seal the gas carefully.

(a) What are the ill effects of Freon as discussed by Geeta and Reeta?

(b) How do you evaluate the consciousness of the sisters towards the environment?    

Q20. The decomposition of N₂O₅ (g) is a first order reaction with a rate constant of 5 X 10^-4 s^-1 at 45^oC, i.e. 2 N₂O₅ (g) → 4 NO₂ (g) + O₂ (g). If initial concentration of N₂O₅ is 0.25M, calculate its concentration after 2 min. Also calculate half life for decomposition of N₂O₅ (g).

Q21. (a) Name the method used for refining of (i) nickel, (ii) zirconium.

(b) The extraction of Au by leaching with NaCN involves both oxidation and reduction. Justify giving equations.

                                    

Q22. Write down the equations of hydrolysis of XeF₄ and XeF₆. Which of these two reactions is a redox reaction?

Q23. (a) Explain bondings in metal carbonyls with orital diagram.

(b) Write the configuration of Ni in the complex [NiCl₄]^2- in terms of t_2g and e_g.

Q24.

Q25. i) Explain the mechanism of acid catalysed dehydration of ethanol forming ethene.

ii) Convert methanal to Ethanol by using Grignard’s reagent.

iii) Write the reaction between phenol and Br₂ (aq).

Q26. Giving an example for each describe the following reaction:

i) HVZ reaction

ii) Reimer – Tiemer reaction

iii) Aldol condensation

Q27. i)Write the difference between antiseptic and disinfectants with one example in each.

ii) Give one example each of artificial sweetener and narrow spectrum antibiotic.

                Q28. (i) Account for the following : (i) Out of the ions Co²⁺, Sc³⁺ and Cr³⁺ ,which one would

 give coloured aqueous solutions  and why ?

(ii) Explain why chromium is a typical hard metal while mercury is a liquid.

(iii) Why do the transition elements form coloured compounds? Explain.

(iv)Why do the transition elements form interstitial compound?          

 (v) Complete the given reaction :                                                                                         

                  Cr₂O₇^2-  + H⁺  +Fe⁺²     →     ?

OR

 i) Transition metal compounds generally act as catalyst.(give reason)

ii) E⁰Mn³⁺/Mn²⁺ has higher positive value than E⁰ Cr³⁺/Cr^2+.(Atomic number  Cr=24,Mn=25)

iii) How does KMnO₄ can be prepared from pyrolusite ore?

iv) Why do the transition elements form coloured compounds? Explain.

Q29. (i) State Kohlrausch’s law.

(ii) Write down the reactions involved in the working of a H₂––O₂ fuel cell.

(iii) A solution of Ni(NO₃)₂ is electrolysed between platinum electrodes using a current of     5.0 amperes for 20.0 minutes. What mass of Ni is deposited at the cathode? [ Atomic mass of Ni = 58.7u ]

OR

i)Write the anode and cathode reaction of lead storage battery.

ii) Define the molar conductivity .

iii) Calculate the equilibrium constant for the reaction

2Cr(s) +3Cd⁺²→2Cr⁺³(s) +3Cd

[ E⁰Cr³⁺/Cr   =  – 0.74 V and   E⁰Cd²⁺/Cd  =  + 0.40V ]

           

Q30. i) An organic compound with the molecular formula C₉H₁₀O forms 2,4-DNP derivative, reduces Tollen’s reagent and undergoes Cannizzaro reaction .On vigorous oxidation it gives 1,2-benzenedicarboxylic acid. Identify the organic compound.

ii) Arrange the following acid in increasing order of acidity:

CH₃CH₂CH(Br)COOH, CH₃CHBrCH₂COOH, (CH₃)₂CHCOOH, CH₃CH₂CH₂COOH

(iii) Convert toluene  to benzaldehyde.

OR

Complete the following reaction.

i) Benzenamide to Chlorobenzene

ii)     Benzoyl chloride to benzaldehde                 

iii)Distinguish between the following by suitable chemical test and write chemical reaction:

a)Phenol and Benzoic acid

b)Benzylamine and Aniline.