XII CHEMISTRY Comprehensive Study Paper 2023


Short Answer Questions

Q1. Give the general valence shell electronic configuration of the following:

(i) Representative elements                           

(ii) Inner transition elements

(iii) Nobel gases                                                

(iv) Outer transition elements (except group IB,VIB)

(v) Alkali metals                                                

(vi) Alkaline earths                           

(vii) Halogens                                    

(viii) Pnictogens                                

(ix) Chalcogens

Q2. Give the general valence shell electronic configuration of the following:

(i) Group IIA and group IIB            

(ii) Group IVA and group VIB        

Q3. Using electronic configuration, identify the position (block, period & group) of the elements with the following configuration. 11, 15, 18, 29, 24, 35, 36, 37, 42, 47, 53 and 54.        

Q4. Identify the group, period and block of each element in the periodic table.

(i)  3d10,4s2                          

(ii)  3s2 3p5                          

(iii)  2s2 2p6                         

(iv)  5s2 4d5                                              

(v)  4s1 3d10  

Q5. Justify the statements: 

i)    The elements of a group in the Periodic table have the same valance shell configuration.

ii)   A period starts with s1 configuration and ends at s2 p6 configuration.

Q6. Define periodicity and its cause. State modern periodic law in the light of Electronic configuration.

Q7. Differentiate between Mendeleev’s periodic law and Modern Periodic Law. Why Mendeleev’s law modified? Give at least two reasons.

Descriptive Answer Questions

Q8. What are long and short periods in the Periodic table. Give a brief account of them

Q9. Describe classification of elements based on electronic configuration.



Short Answer Questions

Q1. Explain the preparation and properties of Ionic and Covalent.

Q2. Briefly explain:

(i) Hydrogen exhibits +1 and -1 oxidation states in its compounds.

(ii) Why atomic hydrogen is more reactive than the ordinary molecular hydrogen?                     

(ii) Hydrogen can be placed with the Alkali metals of group IA and halogens. OR give the similarities and dissimilarities of hydrogen with the elements of group IA and VIIA.  

(iii) Write down Electronic configuration of simplest ions of hydrogen. Show their reactions with water. OR Give the reaction of H+ and H¯ ion with water.                

(iv) D2O is heavier than H2O.

Descriptive Answer Questions

Q3. What are binary compounds and hydrides?  Define six kinds of hydrides. Explain its first three types.

Q4. Differentiate between atomic and nascent hydrogen with their method of preparation. Give action of Nascent Hydrogen on phosphorus, oxygen, tungsten oxide, arsenic, FeCl3 and sodium.



Short Answer Questions

Q1. Describe the reaction of s-block elements with air, nitrogen, sulphur and water.

Q2. Give the balanced equations to show that how NaOH react with following?

(i)  Al            (ii) boric acid             (iii) CO2/CO     (iv) bauxite              (v) lead chromate       

Q3. What do you mean by diagonal relationship? Explain diagonal relationship between Li and Mg.

Q4. Give reasons of the following:

(i) Zinc hydroxide is soluble in excess of sodium hydroxide solution.

(ii) s-block elements (or Alkali metals) are powerful reducing agent.

(iii) Elements of group IA and IIA readily lose their valence electrons.

(iv) Li and Be differ to great extent from other members of their respective group.

(v) Alkali metals cannot be used in voltaic cells.       

(vi) I.P. or E.N. decreases down the group.

(vii) The salts of alkaline earths are more strongly hydrated than alkali metals.

(viii) Na+  ion is smaller than Na atom

 (ix) M.P. and B.P. of IA group elements are very low.              

(x) Atomic radii of alkali metals are larger than alkaline earth metal.

Descriptive Answer Questions

Q3. Discuss the trends of the following properties in representative elements:

Hydration energy, ionization potential, electrode potential, m.p, b.p and hardness.

Q4. Describe electrolytic method for large scale production of Caustic soda. Give merits and Demerits of process.

Q5. Describe with flow sheet diagram, preparation of soda ash by Solvay’s Process.

Q6. Describe the extraction of sodium metal from halite ore.



Short Answer Questions

Q1. Define Auto-Redox reactions. Explain auto-redox reactions of chlorine with examples.

Q2. Give the reactions of the following compounds with Aluminium metal;

(i) Sulfuric Acid          (ii) Potassium Hydroxide      (iii)   ferric oxide      (iv)   manganese dioxide.(v) Caustic soda

Q3. Give chemical reactions of nitric acid with copper, Mg, Zn, S, P, methane and nitrobenzene.

Q4. By the help of chemical equations for each show that Nitric acid is a good oxidant.

Q5. Give the formulae of the following compounds:

Blue Vitriol, Borax/Suhaga, Washing soda, Photographic hypo, Plaster of Paris, Gypsum, Colemanite, Epsomite, Cryolite, Lunar caustic, Magnesite, Phosgene, Marsh gas, Silane, Oil of Vitriol, Oleum, Bleaching powder, Barium peroxide, Quick lime, Yellow pigment, Read lead pigment, Chromite ore, Alunite , Potash alum, Chrome alum

Descriptive Answer Questions

Q6. Describe the extraction of Aluminium from its oxide-ore containing both Silica & Ferric oxide as major impurity.

Q7. Describe the manufacture of Nitric acid by Ostwald’s method.

Q8. Describe the manufacture of chlorine gas by electrolytic methods.

Q9. Give short answers of following:

i)             How Aqua Regia dissolves noble metals? (Give equation).

ii)            Bromine is displaced from its salts by chlorine, not by iodine. Explain.

iii)          Aluminium utensils must not be washed with water containing NaOH or Na2CO3.

iv)           Write down names and formulae of ores of aluminium.

v)         What is thermite, ammonal, duralumin and aluminium bronze? Explain thermite process?

Q10. Complete and balance the following equations.

CaC2        +              2H2O      →

Al4C3      +              6H2O      →

WO3       +              3H2         →

2Al          +              3H2SO4  →

Fe2O3     +              2Al          →

Cl2           +              H2O        →

K2Cr2O7 +              2KOH     →

2K2MnO4+            Cl2           →

H2O2       +              H2S         →

Ca3(PO4)2+            2H2SO4  →

Na2CO3  +              SiO2        →

Al2O3.nH2O+        2NaOH   →

4NaBO2 +              CO2         →

CO           +              Cl2           →

SO2         +              Cl2           →

Ca3(PO4)2+            4H3PO4  →

CO           +              2H2         →

Al2O3.nH2O+        Na2CO3  →

2CuSO4  +              4KI          →

2K2CrO4 +              H2SO4     →

2PbCrO4+              2NaOH   →

2Pb(NO3)2                            →



Q1           Give the formulae of the following complexes.

(a)           dichlorotetraamminechromium(III) chloride                                           

(b)          tetracarbonylnickel(0)                                                     

(c)           tetranitrochromate(III) ion                                                                           

(d)          potassium hexacyanoferrate(III)                                   

(e)           Tollen’s reagent                                                                                                                     

(f)         Nessler’s Reagent                                                                                

Q2           Name the following compounds by IUPAC System:

(1)          K3 [Fe(CN)6]                                                       

(2)          Na3 [Co(NO2)6]                                   

(3)          NH4 [Cr(SCN)4(NH3)2]                                      

(4)          [Co(NH3)5 Cl] Cl2                                                                

(5)          [Cu (NH3)4] SO4                                                   

(6)          [Ni(CO)4]                                                                                                             

(7)          [MnO4] ¯                                                

(8)          [Cr (en)2 Br2] SO4                                                                                      

(9)          [Pt (NH3)2Br4]                                                                    

(10)        [Co(NH3)6]2+                                                           

(11)        [Cr (NH3)2(NSC)4]¯                                                            

(12)        [Ni(en)3](NO3)3                                                 

(13)        [Fe(CN)6]4¯                                                                

 (14)       [Cu(CN)4]                                          

(15)        Na2[Fe(CN)5NO]                                

(16)        (NH4)3[Fe(CN)6]                                                                                

(17)        [Cr(NH3)4Cl2]Cl                                                                                  

(18)      Sodiumcobaltinitrite                           

Q3         What are transition elements? How are they classified?

Q4. Write short notes on:

(i)     Chemistry of Copper sulphate           

(ii)   Chemistry of potassium chromate   

(iii)   Corrosion and its prevention            

Q5. What are Transition elements? Explain their following characteristics.

(i)  Complex formation   (ii) Color formation (iii) Magnetic behavior  (iv) Catalytic property (v) Variable oxidation states

Q6. How is copper extracted from its sulphide ore?

Q7. Give brief answers of following:

(i). NH3 and H2O can act acts as ligands but NH4+ and H3O+ cannot. Explain.   

(ii). Hydrated copper sulphate is blue but anhydrite is white.

(iii).What are complexes? Why do transition metals form complex?

(iv). Write down name and formulae of 5 neutral, 2 bidentate and 2 polydentate ligands.

(v).   How is blue vitriol prepared?



Q1. Define the following with examples:

homologous series, functional group, catenation, Polymerization (polymers), isomerism (isomers), metamersim, knocking, heterocylics.    

Q2.Write down difference between following with examples and give two chemical tests to identify them:

(i) Aliphatic and aromatic compounds (Soot Test)               

(ii) Saturated and unsaturated compounds (BUT

Q3. Write short notes on each of the following:

(i) Classification of organic compounds        (ii)  Polymerization

(iii)  Isomerism & its types                             (iv) Natural sources of Organic Compounds     

Q4.Give structural formulae of following compounds:


Short Answer Questions on Aliphatic Hydrocarbons

Q1. Define & write the general formulae of the Alkanes (CnH2n+2), Alkenes (CnH2n), Alkynes (CnH2n-2), alcohols (CnH2n+2O or CnH2n+1-OH) and carbonyl compounds (CnH2nO or CnH2n+1CHO for aldehydes and CnH2n+1COR for ketones).

Q2. Distinguish between s & p bond. Identify each of them in ethane, ethene & ethyne, by drawing their orbital structures.

Q3. Give a simple test to distinguish between the following:

(i).Alkanes & Alkyl halides (Ethane & ethyl chloride)  (Silver Nitrate Test)       

(ii) Alkanes & alkenes (ethane and ethene) (Baeyer’s unsaturation test; BUT)

(iii) Alkenes & Alkynes (Ethene and ethyne)(Substitution reaction of ethyne)

Q4.Which alkane may be prepared by the reduction of? [Learn Wurtz’s reaction]

(i) 2–bromopropane                        (2,4-dimethylbutane)   

(ii) 2- bromo-2-methylpropane   (2,2,4,4,-tetramethylbutane)

(iii) secondarybutyl chloride      (3,4-dimethylhexane)   

(iv) iso-butyl bromide (2,4-dimethylhexane)

Q5. What alkyl halide would yield following alkenes upon dehydrohalogentaion by strong base?

(i) 2-butene (2-bromobutane)        

(ii)  1-butene (1-bromobutane)     

(iii) isobutylene (1-bromo-2-methylpropane)

Q6. Starting from Acetylene how would you obtain Formic acid, Acetaldehyde, Benzene, oxalic acid & vinyl cyanide?

Q7.Give three methods of preparations for ethyne and ethylene.

Q8. Acetylene shows Acidic properties. Give two reactions to justify it. Or What are white & red solids obtained from Acetylene.

Q9. State Markonokoff’s rule. Give its two applications in alkenes and alkynes.

Q10. Complete and balance the following reactions:



Q11. Write down the names of compounds according to IUPAC System:



Short Answer Questions on Aromatic Hydrocarbons (Benzene)

Q1. Define Aromaticity. How will you explain the stability of benzene?

Q2. What is meant by Orientation & orienting effects? Name 3 different types of directing groups with examples.

Q3. Define Resonance. Draw all the possible structures of benzene.                                                                  

Q4. How will you prepare a sample of benzene from petroleum (coal tar or n-hexane or n-heptane)?       

Q5. How will you obtain benzene from the following?                                                           

(i).  Aliphatic sources (n-hexane or n-heptane or ethyne)

(ii) Aromatic sources (phenol or sodium benzoate or toluene)

Q6. Give two reactions in which benzene ring is not retained. (Hydrogenation and halogenation in sunlight)

Q7. Under what conditions benzene reacts with chlorine to form

(i) monochlorobenzene (ESR)                       

(ii) hexachlorocyclohexane (Halogenation in sunlight)

Q8. Give 3 evidences in support of Kekule’s structures of benzene. What objection was raised to it and how was it defended?

Short Answer Questions on Aromatic Hydrocarbons (Phenol)

Q1. What are phenols and naphthols? Define their types with two examples each. Give the uses of Phenol.

Q2. How does Phenol react with nitric acid and sulphuric acid?

Q3. How will you convert phenol into following?

(i)   Cyclohexanol               (Catalytic hydrogenation)               

(ii)  Benzene                       (Reduction)                         

(iii) picric acid                    (Successive nitration)                      

(iv) Phenoxide                   (Neutralization)

Descriptive Answer Questions (Aliphatic Hydrocarbons)

Q1. How will you prepare methane or ethane from following?

Alkyl halide (reduction), Grignard’s reagent (Hydrolysis), sodium carboxylate (decarboxylation), and alkynes

Q2. Explain Halogenation in Ethane or Methane with mechanism.                    

Q3. Write the preparation of the following:

(i)  Preparation of ethene from alcohol or vicinal dihalide   (Dehydration, dehalogenation)      

(ii) Preparation of ethyne from vicinal dihalide or ethene  (Double De-HX, Halogenation+ Double De-HX)

(ii) Preparation of formic and oxalic acid from ethyne (Oxidation by cold and hot KMnO4)   

(iv)  Preparation of n-butane by Wurtz reaction. (Reduction or Coupling reaction)                 

(v) Preparation of mustard gas from ethene.   (Reaction with S2Cl2)                       

(vi)  Preparation of Acetaldehyde from acetylene (Catalytic hydration + tautomeric rearrangement)

(vii) Preparation of Acetone from propyne  (Catalytic hydration + tautomeric rearrangement)  

Q4. What happens when?

(i) Ethene or propene reacts with acidified water.   (Catalytic hydration into alcohol)                

(ii) Ethene reacts with alkaline KMnO4. (Oxidation/hydroxylation into diol)

(iii)  Ethyne reacts with ammonical AgNO3 or CuCl.    (Substitution reaction)                    

(iv) Ethyne reacts with water or hydrogen cyanide.        (Hydration and hydronitrilation)

(v) Ethene reacts with acidified Cl2 water or bromine water (Hypobromous acid).[Halohydrination)

(vi)  Propene is treated with Hydrogen Iodide.     (Hydrohalogenation)                        

(vii) Acetylene is treated with Iodine          (Partial halogenation)

Q5. How will you obtain the following?

(i) Ethane from sodium salt of carboxylic acid     [Decarboxylation]

(ii) 2,3–dimethylbutane by Wurtz’s synthesis    [Reduction or Coupling reaction]

(iii) Ethane from methyl Iodide                      [Reduction or Coupling reaction]   

(iv) Isobutylene from alcohol                            [Dehydration]

(v) Acetylene from Calcium dicarbide            [Hydrolysis]

(vi) Ethane from Grignard’s reagent           [Hydrolysis]  

(vii) Ethyne from 1,2-dibromoethane      [Double dehydrohalogenation]

Descriptive Answer Questions (Aromatic Hydrocarbons i.e. Benzene and Phenol)

Q1. Starting from benzene how will you prepare the following?             

(i) Acetophenone                                [acylation]                                          

(ii) Benzenesulphonic acid              [sulphonation]                       

(iii) Cyclohexane                                 [Hydrogenation]                               

(iv) Carbolic acid or Phenol             [Chlorination, hydrolysis, acidification]

(v) Bromobenzene/Halobenzene   [Bromination or halogenation in dark]                               

(vi) Cumene (isopropylbenzene) [alkylation or isopropylation with isopropyl chloride]

(vii) p-nitro benzoic acid  [Alkylation (methylation) + Nitration + oxidation]  

(viii) meta-nitrotoluene                     [Nitration + Alkylation(methylation)]

(ix) Ortho & para nitrotoluene        [Alkylation + Nitration(methylation)]   

Q2. Attempt the following conversions:

(i) Benzene into o-nitro benzoic acid & p-nitro benzoic acid  (Alkylation followed by nitration and then oxidation)

(ii) Benzene into m-nitro benzoic acid   (Alkylation followed by oxidation and then nitration)

(iii) Toluene into T.N.T.          (Successive nitration of toluene thrice)    

(iv)  Benzene into benzoic acid (Alkylation followed by oxidation)

(v) Toluene into benzene     (Hydro-demethylation of toluene with H2, CH4 is also obtained)  

Q3. Benzene acts as Saturated as well as unsaturated compound. Support your answer by two chemical reactions for each type of  behavior.              

Q4. Discuss the Molecular Orbital Treatment of benzene. Also give reason for hyper reactivity of benzene.                         

Q5. Define Orientation in benzene & orienting effect in benzene. Give the examples of ortho & para directors.  

Q6. What is ESR? Why benzene undergoes ESR? Explain Nitration & Acylation in Benzene with mechanism.



Short Answer Questions on Alkyl Halides

Q1. What are Alkyl Halides? How are they classified? Give their general structures and examples.                

Q2. Draw all the possible isomers of an alkyl halide with composition C5H11Cl (8 isomers).

Q3. How would you prepare alkyl halide from alcohol and alkenes?        

Q4.Outline the step-wise reaction mechanism of the following:

  (i).  SN2 reaction between bromomethane and NaOH

  (ii)  SN1 reaction between 2-chloro-2-methylpropane and NaCN

Descriptive Answer Questions

Q5. What is Nucleophilic substitution reaction? Write down difference between SN1 and SN2. Explain SN1 mechanism.

Q6. What is elimination reaction? Write down difference between E1 and E2. Explain E1 mechanism.

Q7.  How will you obtain the following?

(i) Ethane from methyl magnesium chloride                             [Alkylation]

(ii) Ethanoic acid from methyl magnesium chloride                [Carbonation]

(iii) Tertiarybutyl alcohol from ketone [Nucleophilic carbonyl addition of GR]

(iv) Ethyl alcohol from methyl magnesium iodide  [Nucleophilic carbonyl addition of GR]

(v) Secondary alcohol from Grignard’s reagent [Nucleophilic carbonyl addition of GR]



Short Answer Questions on Alcohols

Q1. Define primary, secondary & tertiary alcohols. Write the structures & the names of primary, secondary & tertiary alcohols of the molecular formula C5H11OH (6 isomers).      

Q2.Give two examples of each of monohydric, dihydric and trihydric alcohols with their IUPAC names.

Descriptive Answer Questions

Q1.          How do you prepare alcohol from carbonyl compounds, Grignard’s reagent, alkenes & esters?

Q2.          How is methyl alcohol prepared from water gas and wood? (Petret process/Catalytic Hydrogenation), Destructive distillation of wood)

Q3.         Give industrial preparation of ethyl alcohol from starch. How does different types of alcohol prepare?       

Q4.          Briefly explain the following reactions.

(i)           Condensation of alcohols into ether             

(ii)          Reaction of alcohols with halogen acids (Lucas test)

(iii).        Oxidation of alcohols                                       

(iv)         Catalytic hydrogenation of carbon monoxide

Short Answer Questions on Aldehydes and Ketones

Q1.          What are Aldehydes & Ketones? Write down two simple tests to distinguish between aldehydes & ketones.      

Q2.          How would you obtain acetaldehyde from ethyl alcohol and ethyne?

Q3.          How would you obtain methanal from methanol and calcium formate?

Q4.          How would you prepare acetone from pyrolysis of acetic acid, propyne, calcium acetate and alcohol?

Q5.          Give two methods of preparation of dimethyl ketone with its two chemical properties

General Questions on Aldehydes and Ketones

Q1. Briefly explain the following: Haloform reaction, Cannizaro’s reaction,

Q2. Briefly explain the following reactions.

(i) Self addition of aldehydes                                               

(ii)          Reaction of methanal with Na-C2H5OH.

(iii)         Catalytic hydrogenation of acetone.                             

(iv)         Reaction of Phenyl Hydrazine with methanal

Q3. Explain how you will prepare the following?

(i)Acetal from formaldehyde.

(ii)An oxime from ethanal.

(iii)Metaformaldehyde from formaldehyde.

(iv)Paraldehyde from acetaldehyde

(v)Phenyl Hydrazone from methanal and acetone.  

(vi)Acetaldol and acetketol from carbonyl compounds.        



Q1.Define Amino acids. Give their classification with examples.               

Q2. What are carbohydrates? Write their biochemical importance. Give their classification with examples.               

Q3. Define the following terms:

Reducing & non reducing sugars, Glycosidic linkage, Peptide linkage, Zwetterion, , life chemistry (biochemistry),enzymes 

Q4.What are proteins? Give their classification with examples. 

Q5. Write down differences between fats and oils and drying and non-drying oils.       

     


Q1.What are fertilizers? Give their importance. Give a brief account of the fertilizers industry in Pakistan.

Q2. What is glass? Give an account of the composition of glass. How is soft glass prepared?  

Q3. What are plastic? Define their two Plastics. Write down preparation and uses of Teflon, polythene, Perspex, PVC, PVA and Bakelite.

Q4.  Write down brief note synthetic fibers. 


Solution of Important Questions of Study Paper

Chapter # 1 ……Periodic Classification



Chapter # 3 ……s-Block Elements



Q5. Give reasons of the following:

Answer

(i) Zinc hydroxide is soluble in excess of sodium hydroxide solution.

The amphoteric insoluble hydroxides of certain amphoteric metals like Zn, Be, Al, Cr, Sn (both stannous and stannic), and lead (both plumbous and plumbic) etc. redissolves in excess of sodium hydroxide solution forming either soluble complex anions or soluble simple oxyanions. Zinc hydroxide is an amphoteric  hydroxide and thus soluble in excess of NaOH solution due to formation of soluble complex called tetrahydroxozincate (II) ion.

(ii) s-block elements (or Alkali metals) are powerful reducing agent.

The elements with high negative electrode (reduction) potentials are easily oxidized and behave as powerful reducing agents. The s-block elements are powerful reducing agent due to their high negative aqueous electrode (reduction) potential.

(iii) Elements of group IA and IIA readily lose their valence electrons.

Elements of group IA and IIA readily lose their valence electrons to form M1+ and M2+ cations respectively, this is due to their large atomic size, low ionization enthalpy and also by forming cations, they acquire stable configuration of respective inert gases.

(iv) Li and Be differ to great extent from other members of their respective group.

Lithium and Beryllium, the first members of groups IA and IIA differ markedly from other members of their respective groups. This is because of their small atomic sizes, which results in high charge densities on Li+ and Be2+ ions, which produces high heat of hydration. Thus Li and Be have more tendency to form covalent compounds.

(v) Alkali metals cannot be used in voltaic cells. 

Alkali (and alkaline earth) metals have high negative values of aqueous electrode (reduction) potential due to which they cannot be used in voltaic cells based on water as solvent because of their rapid oxidation by that solvent.

(vi) I.P. or E.N. decreases down the group.

Electronegativity is the relative tendency of an atom to attract the shared pair of electron towards itself. It varies directly with nuclear charge and inversely with atomic size and shielding effect. Since down the group atomic size and shielding effect increases, therefore, electronegativity decreases down each group (except IIIA)

I.P. decreases down each group.

Ionization Potential is defined as the amount of energy required to remove most loosely bonded electron from an atom of an element in gaseous state. I.P. values vary inversely with atomic size and shielding effect and vary directly with nuclear charge.

I.P. decreases down each group in all s and p-block elements (except IIIA). This is because I.P. is inversely proportional to atomic size as well as shielding effect of inner energy levels, i.e. the greater the atomic size and shielding effect, the less is the hold of nucleus on outer electrons, so less is the I.P. Atomic size and shielding increase down the group, hence I.P. decreases down the group.                  

(vii) The salts of alkaline earths are more strongly hydrated than alkali metals.

The alkaline earth metals ions (M2+) are more strongly hydrated than alkali metal ions (M1+). This is because M2+ ions have greater charge density due to greater charge and smaller size of ions thereby increasing its    interaction with water molecule which extensively increases hydration energy. The alkali and alkaline earth metals require considerable amount of energy to form M1+ and M2+ ions respectively. But when these ions are formed in solution, the energy required in the formation of gaseous ions is off set by the high negative value of hydration enthalpies of ions which evolve during hydration of ions.

For example:

Ionization potential of Na+ = +500 kJ/mole

But,

Hydration energy of Na+ = – 390 kJ/mole.

(viii) Na+  ion is smaller than Na atom

The size or ionic radii of cations are much smaller than the parent atom [i.e. Na1+ is   smaller than Na atom; Mg2+ ion is smaller than Mg atom]. This is because by loss of valence electrons.

(a)   The number of protons exceeds the number of electrons, due to which nuclear charge increases exerting greater force on valence electrons.

(b)  During the formation of cation, the outer electronic level empty, thereby decreasing total number of shells. Due to above factors, shrinkage of shells occur. Thus ionic radii or size of cation becomes smaller than parent atom.

(ix) M.P. and B.P. of IA group elements are very low.        

Down the group IA and IIA, hardness, M.P. and B.P. decreases (Na is softer than Li), because due to increase  in atomic radii, there are less number of atoms per unit volume, so less number of electrons are available for interaction, so hardness, M.P, B.P decreases.

Alkaline earth metals are more harder having high M.P. and B.P. than alkali metals. This is because of greater   interaction among A.E.M. due to following 2 reasons:

(i) Their smaller atomic size permitting close packing.

(ii) Presence of an extra (two) valence electron for interaction, producing greater binding force which keeps the atoms held together.

(x) Atomic radii of alkali metals are larger than alkaline earth metal.

Atomic radii of group IA elements are larger than those of respective group IIA elements. This difference is due to greater nuclear charge of group IIA elements. High nuclear charge tends to draw valence electrons more strongly towards the nucleus thereby decreasing electronic clouds. Consequently the atomic size decreases. Hence elements of IIA group show smaller atomic radii than elements of group IA.


Chapter # 4 ……p-Block Elements



Q6. Give short answers of following:

i) How Aqua Regia dissolves noble metals? (Give equation).

Answer

A mixture of concentrated nitric acid (HNO3) and concentrated hydrochloric acid (HCl) in the ratio of 1:3 by volume is known as Aqua regia or Royal water. It is called aqua regia because it can dissolve noble metals like gold and platinum due to formation of chloro-complexes of metals due to liberation of nascent chlorine.



In aqua regia, HNO3 acts as an oxidizing agent and oxidizes HCl to nascent chlorine and itself reduces to Nitrosyl Chloride (NOCl). Formation of ionic chloro-complexes of noble metals favours the dissolution of noble metals. Thus formation of [Cl] makes aqua regia a stronger solvent.



ii) Bromine is displaced from its salts by chlorine, not by iodine. Explain.

Answer

The order of electronegativity of Halogens is Cl2 > Br2 > I2. Thus chlorine is chemically more reactive than bromine and iodine, so chlorine displaces both bromine and iodine from their salts.

Bromine is more electronegative (reactive) than iodine but less than chlorine. Therefore bromine can displace iodine from its salts but not chlorine.

Iodine is the least reactive of the three halogens. Thus iodine cannot displace bromine or chlorine from their salts.


iii) Aluminium utensils must not be washed with water containing NaOH or Na2CO3.

Answer

Aluminium dissolves in hot dilute solution of NaOH or KOH, forming hydrogen gas and aluminate (tetrahydroxoaluminate) salt. Owing to the reaction of aluminium with alkali, aluminium pans should never be washed water-containing alkali or with washing soda which forms an alkaline solution.



iv) Write down names and formulae of ores of aluminium.

Answer



v) What is thermite, ammonal, duralumin and aluminium bronze? Explain thermite process?

Answer

A mixture of aluminium powder and ferric oxide is called thermite.

A mixture of aluminium powder and aluminium nitrate is called ammonal.

Duralumin is a high-strength aluminium alloy having tensile strength as high as that of steel and light weight showing resistant to corrosion by acids and sea water. It is used in aircrafts, boats and machines. It comprises of 95% Al, 4% Cu  and 1% Mg.

Aluminium bronze is used in imitation jewelry due to golden colour and high tensile strength containing 10% Al and 90% Cu.

Aluminothermic or Thermite Process is highly exothermic process of reduction of metal oxides into molten metal (along with molten Al2O3) by finely divided powdered aluminium as reducing agent. The process is also called Goldschmidt’s Process after the name of the discoverer of the process. The mixture of iron oxide (Fe2O3) and aluminium powder is called Thermite, that is why the process is termed as thermite process.    

The formation of aluminium oxide is strongly and highly exothermic reaction releasing large amount of heat (– 798 kcal or  – 3341 kJ). Because of high heat of formation of aluminium oxide, aluminium has great affinity for oxygen. The high exothermicity of the reaction makes powdered aluminium an excellent and powerful Reducing Agent and can be used very effectively to extract metals by the reduction of their metallic oxides which cannot be reduced by ordinary reducing agents like coke (carbon) and hydrogen.

4Al + 3O2 → 2Al2O3                            ΔH  =  –798 kcal (–3341/3230 kcal)

Q7. Complete and balance the following equations.

Answer


Chapter # 5 ……Transition Elements

Q2. Name the following compounds by IUPAC System:

Answer

Q7. Give brief answers of following:

Answer

(i). NH3 and H2O can act acts as ligands but NH4+ and H3O+ cannot. Explain.

Ligands are the atom or molecules or ions which have lone pair and have ability to donate it to transition metal forming co-ordinate bond.  Thus ligands are electron pair donor. The structure of NH3 and H2O show that NH3 has 1 and water has 2 lone pairs, they can donate these lone pairs to central atom.  Thus NH3 and H2O act as ligands. The structure of NH4+ (ammonium ion) and H3O+ (oxonium ion) show that as NH4+ and H3O+ have already donated their lone pair, they do not act as ligand.  Although H3O+ has 1 lone pair but this pair is unlike the ordinary lone pair because this lone pair experiences more inward force due to increased nuclear charge.

(ii). Hydrated copper sulphate is blue but anhydrite is white.

Answer

Generally, for any transition metal compound to be coloured, the degeneracy of its d-orbitals must be lifted, as this allows, the so-called d-d transitions which occur in visible region imparting colour to the compound. Lifting of the degeneracy of d-orbitals is caused when these orbitals experience unequal electrostatic force by crystal (ligand) fields

In hydrated CuSO4 the water molecules surrounding the central metal (Cu) function as ligands which bring d-d transition and hence emits blue colour in visible region due which hydrated CuSO4 appears blue and as anhydrous CuSO4 doesn't have any  water of crystallization hence remains white in colour.

This can be explained on the basis of Crystal field theory.

Hydrated CuSO4.5H2O is made up of [Cu(H2O)4]2+ square planar complex cations, the water is covalently bound to the Cu2+ ion and these water molecules act as ligand which splits the d-orbitals into two sets of different energy one of which contain an unpaired electron which is used for d-d electronic transition absorbing orange light which appear blue to our eyes.

(iii). What are complexes? Why do transition metals form complex?

Answer

All transition metals exhibit a characteristic property of complex compound formation. A compound in which a metal atom or ion mostly transition metal (central atom) is surrounded or coordinated by a definite number of oppositely charged ions (Cl-, CN-, OH- etc) or neutral molecules (NH3, H2O, CO, NO, -en etc) called Ligands which donate their lone pairs of electrons to central metal cation or atom (center of co-ordination) to form Metal-Ligand (M←L) co-ordinate bond and which retains its identity in the solid as well as in solution is called a complex compound or co-ordination compounds. Since metal-ligand bond is co-ordinate, such compounds are termed as co-ordination compounds.

The complex formation tendency of the transition metals is attributed by two reasons; firstly transition metal cations due to their small size and highly effective nuclear charge with high positive charge density which makes it easy for their cations to accept the lone pair of electrons from the ligands. Secondly, availability of vacant inner (n–1)d-orbitals of appropriate energy in their cations facilitates the acceptance of lone pair of electrons from ligands.

 iv. Write down name and formulae of 5 neutral, 2 bidentate and 2 polydentate ligands.

Answer

1. Monodentate or Unidentate Ligands (Non-chelated Ligands)

The ligands which have only one donor atom i.e. the ligands which are co-ordinated to the central metal ion by donating one lone pair of electrons are called monodentate or unidentate ligands. They are co-ordinated to the central metal ion at one site or by one metal-ligand bond only or one position on central metal ion. They can attach to the central metal or ion though one point and occupy one position on central atom. They are capable of donating one lone pair of electrons. The monodentate ligands are called Non-chelated Ligands and the complex containing monodentate ligands are called Non-chelated complex.

They may be negative, neutral or positive. e.g. X, H, O2–, O22–, O2, HS, OH, NH2, NH2–, CN, NC, SCN, NCS, NO2, ONO, CH3COO, OCN, etc. are the examples of monodentate anionic ligands. NH3, PH3, H2O, N2, PR3, NO, CO, R2S, R3As, C6H5N, P(C6H5)3, (NH2)2CS, -en or NH2–CH2–CH2–NH2, etc. are the examples of monodentate neutral ligands. 

2. Polydentate or Multidentate Ligands or Chelating Agents

The ligands having two or more donor atoms i.e. the ligands which are coordinated to the central metal ion by donating two or more lone pairs of electrons are called polydentate or multidentate ligands. They are coordinated to the central metal ion at more than two sites or by two or more metal – ligand bonds or occupy two or more coordination positions on the central atom or ion within the complex molecules.

They can attach to the central metal or ion through more than one point by donating two or more lone pair of electrons and occupy two or more positions on the central atom. These Ligands capable of donating two or more lone pairs of electrons to central atom. The polydentate ligands are also called chelates or chelating ligands or Chelating Agents because they form one or more ring structures called chelates (having claw). e.g. Nickel dimethyl glyoximate.


(v). How is blue vitriol prepared?

Preparation by Action of Sulphuric acid on Cu in presence of air or Its compounds

Copper sulphate is prepared by the action of sulphuric acid on copper in a current of air, copper, cupric hydroxide, cupric oxide or cupric carbonate followed by evaporation and crystallization.



Important Formulae of Compounds, Hydrates, Ores


Important Formulae of Organic Compounds



Important Equations of Inorganic Chemistry

Hydrogen, Hydrides, Atomic Hydrogen

Sodium Hydroxide

Sodium Carbonate

s-Block Elements

s-Block Compounds

Chlorine

Hydrogen Sulphide

Aluminium

Nitric Acid

Sulphuric Acid

Miscellaneous Topics

d-Block Compounds

Distinguishing Laboratory Tests for Organic Compounds

Important Differences


IUPAC NOMENCLATURE OF COMPLEXES



Q1. Complete and balance the following reactions:


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