Solved question paper for Chemistry Mar-2017 (PSEB 12th)

Question paper 1

1. 1. Under what conditions the van't Hoff factor is greater than one ?

   Answer:

   i = observed value of CP / calculated the value of CP    (when i  > 1)

   when the solute undergoes dissociation in the solⁿ

2. 2. Define order of reaction. 

   Answer:

   Rate = K[A]^P  [B]^q  

   Order of R x n = p + q

3. 3. Write down IUPAC name of  CH3-NH-CH3

   Answer:

   Amine N-methyl methen - amine 

4. 4. Complete the following reaction:- 

   

   Answer:

   

6. 5. Write down an isomer of C2H3OH.

   Answer:

   The isomer of ethanol (CH3-CH2-OH) is diethyl ether. 
   (CH3-O-CH3) this is the functional isomer of ethanol.

7. 6. What are food preservatives

   Answer:

   Food preservatives are chemicals that prevent food from going bad by preventing microbial growth.

    

8. 7. What are antacids ?

   Answer:

   Antacids are the drugs used to prevent the overproduction acid in the stomach. 

9. 8. What are disaccharides?

   Answer:

   Disaccharides are sugars that from when two simple sugars. ie. monasaccharides combine to from a disaccharides.

11. 9. The radius of Na+ ion is 95pmand that of Cl- ion is 181 pm pridict whether the Co-ordination number of Na+ ion is 6 or 4

    Answer:

    Radius ratio = radius of cation / radius of anion      =        95 / 181 

    Radius of cation = 95 pm,  Radius of anion = 181 pm   =  0.524

    we will check the value of co-ordination no = 0.524 lies b/w 0.732 - 0.414 

    Therefore the coordination number is 6.

12. 10. A first order reaction is 20% complete in 20 minutes. Calculate the time it will take the reaction to complete 80%

    Answer:

    

13. 11. what is gravity separation method for concentration of or eg ?

    Answer:

    This method of concentration of the ore is based upon the difference in specific gravities of the metallic ore and gangue particles. Generally, metal ores are heavier than the gangue associated with them.

14. 12. Write down difference between terylence fiberes and Buna-s-rubber (elastromes).

    Answer:

     

    Terylene	Buna-s rubber
    Terylene is a fibre	Buna-s is an elastomere
    Terylene has stronger intermolecular forces of attraction.	Buna-s rubber has weakest intermolecular forces.
    Thread forming solids which possess high tensile strength and high modulus.	Weak binding forces permit the polymer to be stretched.

     

16. 13. Express linkage isomerism in [Co (NH3) NO2] CL2

    Answer:

    Linkage isomerism:- This type of isomerism is found in complexes that contain ambidentate ligands.

    

17. 14. Write down any two Difference between nucleoside and necleotide

    Answer:

    A nucleoside is formed is formed by attachment of a base to the 1st position of a sugar ie Nucleoside = sugar + Base
    Nucleotides on the other hand, all the three basic components of nucleic acids are present in nucleotide.
    Nucleotide = Sugar + Base + Phosphoric acid

18. 15. Write down carbylamine reation 

    Answer:

    A primary amine (both aliphatic and aromatic) when warmed with chloroform and alcoholic potassium hydroxide gives isocyanides. This is called a Carbylamine reaction.
    R-NH2 + CHCl3 + 3KOH → RNC (Carbylamine) + 3KCl + 3H20.

19. 16. Write down reactions involved in preparation of Potassium dichromate from chromite ore?

    Answer:

    I.    2FeCr2O4  + 8NaOH + 7/2O2→ 4Na2Cr2O4  + Fe2O3 + 4H2O.
                               (Chromite)                    (Sodium Chromate)
    II.    2FeCr2O4 + H3SO4 → Na2Cr2O4 + Na2SO4 + H2O
              (Sodium chromate)      (Sodium dichromate)
    III.    Na2Cr2O7 + 2KCl → K2Cr2O7 + 2NaCl
          (Sodium dichromate)       (potassium dichromate).

21. 17. Determine the type of cubic lattice to which the crystal of the element indicated here belongs It has an edge length of 290 pm and a density 80 g cm-3 Atomic mass of element = 56amu

    Answer:

    

    It belongs to crystal lattice.

22. 18 (i) Prove that osmotic pressure is colligative property.

    (ii) Calculate the molar concentration of urea solution if it exerts an osmotic pressure of  2.45 atmosphere at 300K. [R= 0.0821 Latm.mol-1 K-1 ]

    Answer:

    (i) Collative property depends only on the number of dissolve parties in solⁿ& not on their identity. Both solⁿ have the same freezing point, boiling point, vapour pressure & osmotic pressure because that collagative property of a solⁿ only depend on the number of dissolved particles 

    => Colligative property is proportional to the molarity & temperature 

    \(\pi v = nRT\) the basic equation for ideal solⁿ    or 

    \(\pi = CRT\)  where  \({n\over v } = v\)

    \(\pi\) is osmotic pressure,

    C is concentration, V is volume, T is absolute Temperature, n is the number of males of the solute of the solution.

    (ii) \(\Pi = {nrt \over v}\) 

    Osmotic pressure = 2.45 atm

    T = 300 K

    R = gas constant = 0. 0821L

    so n/v is molar concentration that is \({n \over v} = { \pi \over rt}\)

    molar concentration = 2.45 / (300 * 0.0821)

      = 2.45 / 24. 63 

      = 0. 099 mol 

23. 19. What is electro chemical theory of rusting of iron and give two methods of prevention of rusting of iron?

    Answer:

    Electrochemical theory of rusting of iron and two methods of prevention of rusting of iron.

    At  Anode: Fe_(s) undergoes oxidation to release electrons.

    Fe_(s) → Fe²⁺_(aq) + 2e^-

    At Cathode: O_2(g) + 4H_(aq) + 4e^-1  →  2H₂O_(l)

    Electrons released at the anode move to another metal and reduce oxygen in presence of H⁺. It is available from H₂CO₃ formed from the dissolution of CO₂ from air into water. H⁺ in water may also be available through dissolution of other acidic oxides from the atmosphere.

    RXN Fe_(s) + 2H⁺_(aq)  + 1/2O_2(g) → Fe²⁺_(aq) + H₂O_(l)

    Fe²⁺ is oxidised again to form rust.

    2Fe²⁺_(s)  + 1/2O_2(g) + 2H₂O_(l) → Fe₂O_3(s) + 4H⁺_(aq)

    (rust)

24. OR

    19. Write the Nernst equation and calculate the emf of following cell at 298K:-
    Ni/Ni²⁺ (0.01M) || Cu²⁺ / Cu (0.01M)
    Given E⁰ (Cu²⁺/Cu) = + 0.34V, E⁰(Ni²⁺/Ni)= -0.22V

    Answer:

    E⁰cell = (E_C – E_a) =     

    E_c – E_a = 0.34 – (0.22) = 0.56V

    E⁰cell = 0.56 –      

    = 0.56 –     

    = 0.56 – 0.03 = 0.52  

    E_cell = 0.53V

26. 20. Define Tyndall effect. Differentiate between electrophoresis and electroosmosis

    Answer:

    Tyndall effect is the phenomenon of scattering of light of colloidal particle and hence the path of the beam becomes visible is called tyndall effect.

    Electrophoresis	Electroosmosis
    Refers to the movement of colloidal particles under an applied electric field	Refers to the movement of molecules of the dispersion medium under the influence of electric field where colloidal particles are not allowed to move

27. 21. (i) Why are interhalogen compounds more reactive than halogens.
          (ii) All the five bonds in PCl₅ are not equivalent. Justify.

    Answer:

    (i) Interhalogens are generally more reactive than halogens except fluorine. This is because A-X bonds in interhalogens are weaker than the X-X in dihalogen molecules. Reaction of interhalogens is similar to halogens. Hydrolysis of interhalogen compounds yields halogen acid and oxy-acid.

    (ii) PCl₅ has a trigonal bipyramidal structure due to SP³ hybridisation. As a result of this, two Cl atoms lie along axial line along equatorial plane. Hence all atoms which lie along the axis have different bond length than that of Cl atoms lying on the equatorial plane.

    Structure of PCl₅

    

     

     

28. 22. (i) Phenol has a higher boiling point than toluene. Why?
          (ii) why alcohol easily protonated but phenGtrs'.are not protonated?

    Answer:

    (i) Phenol has a higher boiling point then toluene due to the presence of intermolecular hydrogen bonding in phenols. The formation of hydrogen bonds increases intermolecular forces of attraction among phenol molecules hence the increase in boiling point.

    (ii) In phenol, the lone pair of electrons on oxygen involved in delocalization is not freely available for protonation whereas in alcohols, the electrons on oxygen atom are not delocalized so they are available for protonation.

29. 23. (i) Write HeIl-Volhard-Zelinslry reaction.  

          (ii) Write cross aldol condensation.

          (iii) Ethanoic acid is weaker acid than benzoic acid. Why ?

    Answer:

    (i) Equation for the reaction Hell-Volhard-Zelinsky reaction:-

    

    (ii) Equation Goss-Aldol condensation reaction:-

    

    (iii) Stronger acids have a lower PKa value. Ethanoic acid is a weaker acid than benzoic acid due to presence of a benzene ring in benzoic acid, delocalization of electrons makes it more acidic than ethanoic acid but such stabilisation is not present in ethanoic acid.

31. 24. (i)Why is H ₃PO₃ diprotic in nature ? Draw structure.  

    (ii) Why is H₂S less acidic than H₂Te ?

    (iii) Give hybridization and draw structure of XeF₆.

    Answer:

    (i) H₃PO₃ is dipole in nature because H atoms are attached to oxygen. As a result phosphorus acid is dibasic as it has two P-OH bonds. Similarly, phosphoric acid, H₃PO₄ is tribasic as it has three P-OH.

    

    (ii) Acidic character increases, H₂Te has less bond dissociation enthalpy than H₂S, hence less energy is required to break H₂Te bond easily releasing [H⁺] hence acidity of H₂Te is higher whereas H₂S has high bond dissociation energy and hence its acidity is less.

    (iii) Hybridisation of this molecule is SP³d³. According to VSEPR theory, this molecule has seven electron pairs. It has octahedral structure.

    

32. OR

    24. (i) How is nitric acid manufactured by Qstwald process ?

    (ii) Write down there action of Ozone with blacklead sulphide.

    (iii) Draw structure of IF .

    Answer:

33. 25. (i) Scandium (z = 21) is a transition element but zinc (z = 30) is not. Explain.

    (ii) Calculate equivalent weight of KMnO₄ in acidic medium.

    (iii) What do you mean by Lanthanoid contraction? 

    Answer:

    (i) Transition metals are elements that form at least stable where the compound has an incomplete d subshell. Zinc is not a transition metal because it forms only Zn²⁺ ions with all the 3d-electrons present. The valence configuration should have at least unpaired electrons.

    (ii) Eq.wt =  = 31.6g per equivalent.

    So equivalent weight =

    = = 29gm/equivalent.

    As Mn7+in KMNO₄ changes to Mn²⁺ in acidic medium, equivalent weight becomes

    (iii) Lanthanoid contractions is greater than expected decrease in ionic radii of the elements in the lanthanoid series, from atomic number 57, lanthanum to 71, IU resulting in smaller atomic radii.

34. OR

    25. (i) Write down any three differences between Lanthanoids and Actinoids

    (ii) The melting and boiling points of Zy, Cd and Hg are low. Why? 

    (iii) Draw the structure of manganate ion.

    Answer:

    (i)

    Lanthanoids	Actinoids
    They are generally non-radioactive	They are radioactive
    Most of their ions are colourless.	They have coloured ions
    Less tendency	High tendency

     

    (ii) Zinc has relatively low melting and boiling point. Cadmium (CD) is similar to Zinc in many aspects but forms complex compound. Mercury (Hg) has a low melting point for d-block element.

    (iii) Structure of magnetic ion:-

    

36. 26. Write the following reactions :

    (i) Williamson' s synthesis

    (ii) Mendius reaction

    (iii) Friedel Craft' s Alkylation

    (iv) Haloform reaction

    (v) Gutterman reaction

    Answer:

    (i) William synthesis

    R―X + R  ―  Ö Na     ―>      R―Ö―R  +  NaX

    (alkythalide)(sodium alkoxide)       (ether)

    (ii) Mendius Rxn

    RCN              +           2H₂       →    RCH₂NH₂

    (alkylcyanide)    (hydrogen) 

    (iii) Fredal Craft Alkylation:- 

     

    (iv) Halo form Rxn:- 

    

    (v) Carbylamine Rxn:-

    RNH₂    +   CHCl₃  + 3KOH(alc)  ―>  RNC  +  3KCl  + 3H₂O

     

    (vi) Gatterman Rxn:- 

    

     

37. OR

    26. (i) Hydrogen atom of chloroform is acidic. Explain.

    (ii) Why is dehydrohalogenation reaction in haloalkanes termed as Beta-elimination?

    Answer:

    (i) Hydrogen atom of chloroform is acidic because it creates a positive charge, partial deficiency of electron with carbon, more electronegativity than H atom, hence starts to pull electrons from H.

    The presence of 3Cl aggravates the pull resulting into loss of H⁺. It also accepts the lone pair of electrons left by H termed back donation into its empty 3d shell thereby stabilising the negative charge on carbon.

    Structure of CHCl₃

    

    (ii) Alkenes are generally prepared through B-elimination reaction, in which two atoms on the adjacent carbon atoms are removed in formation of a double bond. Preparations include dehydration of alcohols, dehydrohalogenation of alkyl halides and dehalogenation of alkanes.

Question paper 2

1. 1. Under what conditions the van't Hoff factor is less than one ?

   Answer:

   i =    (when i > 1)

   when the solute undergoes dissociation in the solution.

2. 2. Define molecularity of a reaction.

   Answer:

   It’s defined as the number of molecules or ions that participate in the rate determining step.

3. 3. Write down IUPAC name of

    

   Answer:

   N-Methylmethanamine.

4. 4. Complete the following reaction : -
      

   Answer:

   

6. 5. Write down the chain Isomer of 

   

   Answer:

   CH₃ CH₂ COCH₂ CH₃

      Pentan - 3 - One

7. 6. Write down name of one antiseptic.

   Answer:

   Hydrogen Peroxide.

8. 7. What are artificial sweetners?

   Answer:

   Artificial sweetners agents are chemicals that sweeten food. Unlike natural sweetners they do not add calories to our body. Some artificial sweetners are saccharin, sucrolose.

9. 8. What are polysaccharides ?

   Answer:

   These are polymers of monosaccharides.

   Eg. Starch, Glycogen etc.

11. 9. The two ions A+ and B- have ready 88 pm and 200 pm respectively. In the close-packed crystal of compound AB, predict the coordination number of A+.

    Answer:

       =      =  0.44 

    It lies in the range of 0.414 - 0.723

    The co-ordination number of A⁺  = 6

12. 10. A first-order reaction is 20% complete in 10 minutes. Calculate the time for 75% completion of the reaction.

    Answer:

    Initial conc. A = 100, final conc. B = 80, T = 10 min, So

    K = -2.303 log(BA)/T

    = -2.303 log(80100)/10

    K = 0.0223  => final conc. C = 25 (because 75 of reaction finished)

    New Reaction T = -2.303 log(CA)/K

     = -2.303 log(25100)/0.0223 

     T  = 62.17 min.

13. 11. What is 'Froth flotation process' for concentration of ore ?

    Answer:

    Ore contains rocky impurities which are needed to be separated before processing. This process is called concentration of ore. The principle of froth floatation process is that sulptids ores are preferentially wetted by the pine oil, the particles are wetted by water. In the process a suspension of powdered ore is made with water. The frothis formed which is lighter and skimmed off. The frothis dried for recovery of the ore particles.

14. 12. Write down difference between additional and condensation Polymers 

    Answer:

    Addition Polymer is made when the monomers lose an atom or group of atoms while forming the polymers. A condensation polymer is formed when monomers bond to each other without the loss of atoms.

16. 13. Express coordination isomerism in 

    

    Answer:

    This type of isomerism is common in hetroleptic complex. It arises due to the different possible geometrical arrangement of ligands.  

    

17. 14. What is mutarotation?
     

    Answer:

    Mutarotation is the change in the optical rotation because of the change in equilibrium between two anomers, when the corresponding stereocenter interconvert. Cyclic sugars show mutarotation as α and β anomeric forms interconvert.

18. 15. Write down coupling reaction of amines.

    Answer:

    Palladium – catalysed synthes of aryl amines. Starting materials are aryl halides or pseudohalides and primary or secondary amines.

19. 16. Explain how the colour of K₂Cr2O₇, solution depends sn PH of the solution 

    Answer:

    K₂Cr₂O₇ contains dischromate anion.

    This Cr₂O₇ is responsible for the change in colour of K₂Cr₂O₇ altogether.

    => When the solⁿ is acidic (pH < 7) = Orange colour.

    => When the solⁿ is alkaline (pH > 7) = Yellow colour.

21. 17. Unit cell of an element (atomic mass = 108 amu and density = 10.5 g cm-3) has an edge length of 409 pm. Deduce the type of crystal lattice

    Answer:

    Molar Mass (M) = 108g/mol

    Density (d) = 10.5g/cm³

    Edge length (a) 409pm

    Z =

    Z =

    =  4

    Number of atoms = 4

    Element packed in FCC structure.

22. 18.  (i) Prove that depression in freezing point is a colligative property

           (ii)45g of ethylene glycol (C₂H₆O₂) is mixed with 600g of water. Calculate the freezing point depression (K_f for water = 1.86 K Kg mol^-1

    Answer:

    (i) Freezing point of depression is a colligative property observed in solⁿ that result from the introduction solute molecule to the solvents. The freezing point of solⁿ are lower that that of pure solvent and is directly proportional to the molality of the solute.

    ΔT_f = T_f (solvent) – T_f (solution) = K_f x m

    Where ΔT_f is freezing point depression, T_f (solution) is the freezing point of solⁿ, T_f (solvent), K_f is the freezing point of depression constant and mis the molarity.

    (ii) Required -> molality

    For molality, you need to know the moles of ethylene glycol.

    No. Of moles  =    =    =  0.72 moles

    Molality =     =   

                   =      =  1.2 molality

    ΔT_f   =   m x K_f

             =  1.2 x 1.86            [ i=1 since α = 0]

             =  2.232⁰

23. 19. Explain the variation of molar conductivity of strong and weak electrolytes with dilution.

    Answer:

    Variation of motet conductivity with strong electrolytes

    For strong electrolytes the motet conductivity increases slowly with dilution:- The plot between the motet conductivity& is  a straight line having y-intercept equal to E^Om can be determined from the graph or with the help of kohlrausch law

    mc=m∞- bc

     

    Where A is constant equal to slope of line, the value of ‘A’ depend on type of electrolytes at particular temp.

    Variation of molar conductivity with concentration for weak electrolyte:-

    For weak electrolyte the graph plotted b/w molar conductivity & c1/2 (where c is concentration) is not straight line weak electrolytes have lower molar conductivites& lower degree of dissociation at higher concentration which increase steepy at lower concentration Emcnnot be molar conductivity to zero concentrations kohlrausch law of independent migration of ions for limiting molar conductivity Em of weak electrolytes.

    

    Conductivity decrease with decrease in concentration as the number of ions per unit volume tha carry in a sol decrease on dilution…… Variation of molaconcentratin is different for strong & weak electrolytes.

24. OR

    19. Write the Nernst equation and calculate the emf of following cell at 298K:-
    Mg(s)/Mg²⁺ (0.001M) || Cu²⁺ (0.0001M)/ Cu(s)
    Given E⁰ Mg²⁺/Mg = -2.37V, E⁰Cu²⁺/Cu= 0.34V

    Answer:

    E_cell = E⁰_cell- =   log       --------(i)

    E⁰_cell = E⁰(cu2+/cu) - E⁰- E⁰(Mg2+/Mg)

    E⁰(cu2+/cu) = 0.34V  ,   E⁰(Mg2+/Mg)  =  -2.37V

    E⁰_cell = 0.34V  - (-2.37) = 2.71

    Substitute equation in 1

    E_cell= 2.71 – 0.0295 = 2.6805 V

26. 20. Define coagulation. Differentiate between physical adsorption and chemical adsorption.

    Answer:

    Coagulation :- The phenomenon of precipitation of the colloidal practice by the addition of excess of an electrolyte is called coagulation for eg:- Milk

    Difference:-

    Physical Adsorption It is reversible in nature. It is not specific in nature. Decrease of pressure cause desorption.   Low Enthalpy of adsorption in order of 80 to 240 km/mol They does not require activation energy. It forms multi molecular layers.

    Chemical Adsorption It is a irreversible in nature It is specific nature Decrease of pressure does not cause desorption High Enthalpy of adsorption in order of 80 to 240 km/mol They does not requires activation energy It forms molecular layers

27. 21. (i) Among noble gases, only Xe is known to form chemical compounds. Why?

    (ii) Sulphur is a solid but oxygen is a gas. Why?

    Answer:

    (i) Among the noble gases only Xenon is well known to form chemical compounds-only xenon is known to form chemical compounds because Xenon is large in size & having higher atomic Mass. Due to having large atomic radius the force of attraction b/w the outer electron & the protons in the nucleus is weaker

    (ii) Oxygen is smaller in size as compared to sulphur. Also the inter molecular force in Oxygen as weak van der Walls, which cause it to exist gas . On the other hand, sulphur does not form M2 molecular but exist as a puckered structure held tighter by strong covalent bonds. Hence it is solid.

28. 22. (i)Alcohols have a higher boiling point than alkanes. Why?

    (ii) Discuss oxidation of primary, secondary and tertiary alcohols.

    Answer:

    (i) In alkanes, the only intermolecular forces are van der Walls dispersion forces – Hydrogen bonds are much stronger than these & therefore it takes more energy to separate alcohol molecule than it does a separate alkane molecule. That’s the main reason that the boiling points are higher.

    (ii) Discuss oxidation of Aldehyde, then Carboxylic Acid. Secondary Alcohol to ketone . Tertiary alcohols?

    Primary Alcohalto Aldehyde, then Carboxylic Acid. Secondary Alcohol to ketone. Tertiary Alcohol No reaction. Oxidation is usually with potassium dichromate sol. Which turns from orange to green.

    Oxidation of Alcohol:-

    

    

29. 23. (i) Write Cannizzaro reaction.

    (ii) Write aldol condensation.

    (iii) Why aliphatic carboxylic acids are stronger than phenols?

    Answer:

    (i) Cannizzaro reaction:-

    

    This redox disproportional of non-enolizable aldehyde to carboxylic acids & alcohol

    L-Keto aldehyde gives the product of intramolecular disproportional.

    

     

    (ii) Write Adol Condensation:-

    

    (iii) On the other hand in case of phenol are –ve charge is less effectively delocalized over one oxygen atom & less electronegative carbon atom in phenoxide ion

    The carboxylate ions exhibit higher stability in compared to Phenoxide ion. Hence the carboxylic acid is more acidic than phenols.

31. OR

    23. (i) Carboxylic acids do not give characteristic reactions of carbonyl group. Explain.

    (ii) Why do aldehydes and ketones have high dipole moment?

    Answer:

    (i) The carbonyl carbon in ketons& aldehydes is more electrophilic than carboxylic acid. Because the lone pair on oxygen atom attached to hydrogen atom in the - COOH group are involved in resonance thereby making the carbon atom less electrophilic

    

    (ii) Aldehydes &Ketons have high dipole moment due to the presence of oxygen atom in term that is highly electronegative.  The bond in the carbonyl group is lesser than carbon-oxygen single bond in alcohols etc leading to more polarity in the carbonyl group.

32. 24. (i) PbCl₂ is known but PbCl₄ not known. Explain with inert pair effect.

    (ii) Why is SF₆ is much less reactive than SF₄?

    (iii) Give Hybridization and draw structure of XeF₂.

    Answer:

    (i) This is due to inner pair effect. Pb has four electrons in its outermost shell, two are in s- orbital & two in p-orbital. Due to d-block contraction the s-orbital are more strongly held than s-electron in upper periods in the same group. The s-electron are inert & are not that easily removed to give the group of valency 4. Therefore, Pb tends to forms 2+ ion instead.

    (ii) SF₄ is assymmeterical& has a lone pair of electrons on the sulphur atom, which can react further. In SF₆ all of the electrons are paired, giving great stability to the molecule & reducing its reactivity

    (iii) The bond angle will be 90⁰& 180 in the plane of molecules. Acc to lewis structure, XeF₂ has three lone pairs & two bonds to the central Xe atoms. Five valence atomic orbitals on Xe must hyberdised to form five sp³d hybrid orbitals

    Structure of XeF₂

    

    These are arranged in trigonal bipyramid geometry 3 non – bonded pair of electrons prefer the equatorial position.

33. OR

    24. (i) Draw flow chart for Haber's process for the manufacture of ammonia.

    (ii) Write down the reaction of ozone with potassium nitrite.

    (iii) Draw structure of IF₅.

    Answer:

    (i) The Haber Process combines nitrogens form the air with hydrogen derived mainly from natural gas into ammonia Hxn is reversible & product of ammonia is exothermic

    N₂(g) + 3H₂ (g) ßà 2NH₃(g) AH = - 92KJ mol^-1

    

    (ii) 3KNO₂ + O₃ → 3KNO₃

                                   (Potassium Nitrate)

    Potassium Nitrate itself strong Oxidising Agent So, it will give potassium Nitrate, or we can say reduce Nitrate. Ozone usually reacts with nitrite to give Nitrate which is less toxis as compared to nitrites.

     

    (iii) Iodine pentafluoride is an interhalogen compound

    

    I₅ contains five bonded & one nonbounded electrons & square pyramidal molecular geometry.

34. 25. (i) Why do transition elements exhibit higher enthalpies of atomization?

    (ii) Calculate equivalent weight of KMnO₄ in alkaline medium.

    (iii) What are the consequences of Lanthanoid contraction?

    Answer:

    (i) Transition elements have high effective nuclear charge & number of valence electrons they from very strong metallic bonds. As a result, the enthalpy of atomization of transition netal is high.

    (ii) The Mn is KMnO₄ exist in + 7 state

    In acidic medium, this Mn+7 goes to Mn+2 state & gain of % electrons

    Equivalent weight  = Molar mass / No of electrons gain or lost

    Equ. weight = 158 / 5  = 31.6g

    For alkaline medium, there are two possibilities

    (1) Alkaline neutral

    Mn+7 changes into Mn+4 gain of 3 electrons

    Eq.wt = 158/3 = 52.67g

    (2) Highly alkaline

    Mn+7 changes into Mn+6 , gain of 1 electrons

    Eq.wt = 158/1 = 158g

    In many cases through alkine medium, it mostly means the neutral one, for the highly alkaline thing.

    (iii)  Consequences of Lanthanoid contraction:-

    • Separation of Lanthanoid is possible due to Lanthanoid Contraction.
    • it is due to Lanthanoid contraction that there is variation in the basic strength of Lanthanoid contraction.
    • due to Lanthanoid Contraction, size of mions decrease & increase in covalent character in M- OH & basic character decrease.
    • The atomic radii of second row transition elements are almost similar the third row transition elements because increase in size on moving down the group

36. OR

    25. (i) Write down general electronic configuration and any two uses of block elements.

    (ii) Copper is regarded as transition metal though it has completely filled d-orbitals (3d¹⁰4s¹). Explain. 

    (iii) Draw the structure of chromate ion.

    Answer:

    (i) General Configuration of elements:-

    S - block elements is = ns^1-2

    p - block element is = ns²

    d - block element = (n-1)d^1-10 ns^0-2

    f – block element = (n-2)^f1-4 (n-1)d^0-1 , ns²

    uses of d-block elements & s-block elements , p elements

    d – block

    1. iron& amalgam, stell are utilized broadly development industry

    2. Tungsten comes in use in making electrical fibres

    3. Magnese dioxide comes in used part of dry battery cells.

    4. Titanium is part of manufacture of airstrip &spacestup

    S – block

    5. lithium is used in making electrochemical cells.

    6. lithium in combination with magnesium. It is used to make armour plates.

    P – block

    7. p – elements are commonly used as mutagenic agents.

    8. The p – block elements encodes for the protein P transposase & terminal inverted repeat which is important for mobility.

    F – block

    9. f-block elements are Lanthanide alloys utilized for the creation of instrumental steels and heat resistance

    10. Carbides, Borides & nitrides of Lanthanoids in use as refractories.

    (ii) Although copper has 3d¹⁰ 4s¹ configuration, it can lose one electron from this arrangement. Cu²⁺has  3d⁹configuration. So, according to the transition metal that cations have partially filled (n-1)d subshell, copper can be regarded as transition metal.

    (iii) Structure of chromate ion:-

    

37. 26. Write.the following reactions :

    (i) Wurtz reaction

    (ii) Sandmeyer'sreaction

    (iii) Hunsdiecker reaction

    (iv) Reimer-Tiemann reaction

    (v) Friedel Craft's acylation

    (vi) Ullman reaction

    Answer:

    (i) Wurtz reaction:-

    

    (ii) Sandmeyer's reaction:-

    

    (iii) Hunsdiecker reaction:-

    CH₃ CooAg + Br₂  CH₃Br + AgBr + CO₂

    (iv) Reimer-Tiemann reaction:-

    

    (v) Friedel Craft's acylation :-

    

    (vi) Ullman reaction:-

    

38. OR

    26. (i) Why are haloarenes more stable than haloalkanes? 
    (ii) Alkyl halides react with AgNO₂ to give R-NO₂ or R-ONO. Explain.

    Answer:

    (i) Haloarenes are more stable because they can donate there lone pair of electrons inside the ring for resonance. Due to resonance, the electron density increase more at ortho & para position the halogen atom I effect & having tendency to withdraw electrons from the benzene ring. As a result the ring gets activated as compared to benzene &electrophillic substitution rxn occur

    

    (ii) Alkyl halides react with AgNo₂ to give R-NO₂ or R-ono Explain?

     On treating ethanolic sol of nitroalkane with silver nitrate (Ag-O-N=0 , Ag NO₂₎ , nitroalkane is formed because since the bond b/w Ag- O is covalent, the lone pair on nitrogen act as attacking site for nucleophilic rxn

            R – x + Ag No₂à R - No₂+ Agx

    But on other hand, if haloalkane is treated with potassium nitrite (KNo₂), alkyl nitrite is formed as major product because the bond b/w K-O ionic nature, -ve charge on Oxygen serve attacking site

    R-X + KNo₂à R – O – N = O + Kx

Question paper 3

1. 1. Under what conditions the van't Hoff factor is equal to one?

   Answer:

   [i = 1] when the solute doesn’t undergo any dissociation or association in a solution that is a non-electrolyte solute.

2. 2. Define half life period of a reaction.

   Answer:

   It’s defined as the time taken for half of the reaction to be completed ie the time in which the concentration of the reactant is reduced to half of its original value is called half period of a reaction.

3. 3. write down IUPAC name of

   

   Answer:

   N-phenylbenzenamine or Diphenylamine.

4. 4. Complete the following reaction

   

   Answer:

   

6. 5. Write down the chain Isomer of CH₃ - CH₂ - CH₂  -CH₃ 

   Answer:

   

7. 6. Write down name of one antibiotic.

   Answer:

   Cephalexin, Amoxicillin.

8. 7. What are analgesics?

   Answer:

   Analgesics are medicines that help to control pain and reduce fever. Analgesics that are available over counter include Aspirin, Ketoprofen.

9. 8. What are monosaccharides ?

   Answer:

   Monosaccharides are also called simple sugars. They are the simplest form of sugars and the most basic unit of carbohydrates. General formula is C_nH_2nO_n.

   They are usually colourless, water soluble.

11. 9. A solid has NaCl structure.If the radius of cation A is l00pm,what is the radius of anior B?

    Answer:

    Radius of Na⁺ = 100pm

    Radius of Cl^- = ?

    Radius ratio

    Radius ratio

    Hence radius of anion B is 241pm.

12. 10. calculate the time required for the completion of 90% of a reaction of first order kinetics t_1/2=44.1 minutes

    Answer:

    K =

    t =

    Given:  t^1/2 = 44.1min

                  Ao = 100

                  At = (100 - 90) = 10

                   K =  = 0.0157

     

    Substitute all values in formula of first order.

    t =

     

    t =  * 1

    A = 146.7min.

13. 11. what is magnetic separation method for concentration of ore ?

    Answer:

    Magnetic separation method is used when either the ore particles or the gangue associated with it possesses magnetic properties eg chromite Fe(CrO₂)₂ being magnetic can be separated from non magnetic silicon gangue by finely ground ore magnetic separation.

    

     

14. 12. Write down differences between thermosetting polymes and thermoplastic polymers

    Answer:

    Thermo setting polymers	Thermo plastic polymers
    These are formed by condensation polymerization	They are formed by additional polymerization
    They have higher molecular weight	They have low molecular weight
    They are hard, strong and more brittle	They are soft, weak and less brittle.
    They can’t be reshaped or remodelled	They can be reshaped and remodelled
    Examples include Phenol, Nylon etc.	Examples include PVC, PVA etc.

16. 13. Express coordination isomerism in [CO(NH₃)₆] [Cr (CN)₆]

    Answer:

    This type of isomerism occurs in compounds containing both cation and anion entities. Isomers differ in the distribution of ligands in the co-ordination of anion and cation parts. The complex [Co(NH₃)₆]  [Cr(CN)₆] are the examples of co-ordination isomerism. It occurs due to exchange of ligands between cation and anion.

17. 14. What do you mean by inversion of cane sugar ?

    Answer:

    Monday’s Molecule #46 was the chemical reaction shown above. Sucrose is an optically active compound, which causes polarized light to rotate when you shine it through a solution of sucrose. The rotation is measured by a polarimeter.

    

     

18. 15. Write down Hinsberg's test for primary amines.

    Answer:

    Primary amines can be identified by Hinsberg test. In this test, the amines are allowed to react with Hinsberg reagent,(C₆H₅SO₂Cl) benzene sulfonyl chloride. Three types of amines react with Hinesberg reagent. Primary amines react with benzene sulfonyl chloride to form N-alkyl benzene sulfonyl amide which is soluble in alkali.

    

19. 16. Write down reaction involved in the preparation of potassium permanganate from pyrolusitev  ore.

    Answer:

    Potassium permanganate (KMnO₄) is prepared from pyrolusite ore (MnO₂). Pyrolusite ore is fused with alkali metal hydroxide like potassium hydroxide in the presence of air or oxidizing agent like potassium nitrate to give a dark green potassium manganate (K₂MnO₄). Potassium manganate in a disproportional acidic or neutral solution gives potassium permanganate.

    2MnO₂ + 4KOH + O₂→ 2K₂MnO₄ + 2H₂O

    3MnO₄ + 4H → 2MnO₄ + MnO₂ + 2H₂O

    Potassium permanganate is prepared by alkaline oxidative fusion of pyrolusite ore and followed by electrolytic oxidation of manganate (4) ion.

    2MnO₂ + 4KOH + O₂ → 2K₂MnO₄ + 2H₂O

    Mn⁴⁺ (electrolytic oxidation) → MnO₄ + e^-

21. 17. The density ofchromium metal is 7.2g cm-3,If the unit cell is cubic with an edge length of 289 pm, determine the type of unit cell. (Atomic mass of chromiu m= 529 amu)

    Answer:

    Gram atomic mass of Cr(m) = 52.0gmol^-1

    Edge length of unit cell(a) = 289pm.

    Density of unit cell(p) = 7.2gcm^-3

    Avogadro No.(No) = 6.022 * 10²³ mol^-1

    P = OR    Z =     ----(1)

    Put values in equation (1);

     

    Z =

    Z = 2

    Hence the unit cell has 2 atoms. It is body centre in nature.

22. 18. (i) Prove that elevation in boiling point is a colligative property.

    (ii) The boiling point of benzene is 353.23K. When 1.8Og of a non.volatile solute was dissolved in 90'g of benzene, the boiling pointis raised to 354.11 K. Calculate the rnolar mass of solute. (\ for benzene = 2.53KKgmel-t).

    Answer:

    (i) Both boiling point elevation and freezing point depression are proportional to the lowering of vapour pressure in a dilute solution. These properties are colligatives in systems where the solute is essentially confined to the liquid phase. Since boiling point is the temperature at which vapour pressure of any liquid becomes equal to that of the atmosphere. Less molecules will leave the surface because of the solute dissolution, it will take more temperature to get pressure and no matter which solute eg glucose. Same elevation if taken in same concentration.

    (ii) T_b⁰ = 353.23k

    W₂ = 1.80g

    M₂ = ?

    W₁ = 90g

    T_b = 354.11k

    T_b = 354.11 – 353.23

        = 0.88k

    K_b = 2.53 kgmol^-1

    T_b  =

    M₂ =

    

    => 57.5g

    Molar mass of solute is = 57.5g

23. 19. What is corrosion? what are the factors affecting corrosion?

    OR

    Write the Nernst equation and calculate the emf of following cell at 298K:-Cu(s)/Cu²⁺ (0.130M) || Ag⁺ (1.0 x 10^-4M) / Ag(s)
    Given E⁰ (Cu²⁺/Cu) = + 0.34V, E⁰(Ag²⁺/Ag)= +0.80V

    Answer:

    Corrosion refers to the formation of undesirable compounds such as oxides, sulphides or carbonates at the surface of metal by reaction with moisture and other atmospheric gases.

    Factors affecting corrosion:-

    • A rise in temperature increases the rate of reaction
    • Presence of natural water increases rusting of iron
    • Impurities help in setting up voltaic cells which increase speed of corrosion.
    • When iron surface is coated with a layer of metal such as alloys, rate of corrosion is retarded.
    • The more the reactivity of the metal, the more will be the possibility of the metal getting corroded.

    OR

    Cu_(s)→ Cu²⁺_(aq) + 2e^-

    2Ag⁺_(aq) + 2e^-→ 2Ag_(s)

    Cu_(s) + 2Ag⁺_(aq)→ Cu²⁺_(aq) + 2Ag_(s)

    Nernst equation

    E_cell = E⁰_cell -

     = - (Cu²⁺/Cu)

    = 0.80 – 0.34 = 0.46V

    E_cell = 0.46 - 

    =  0.46 - 

    = 0.46 – 0.21 = 0.25V

     

24. 20. Define colloidal solution. Differentiate between lyophillic colloids and lyophobic colloids.

    Answer:

    A colloidal solution, occasionally identified as a colloidal suspension, is a mixture in which the substances are regularly suspended in a fluid. Colloidal systems can occur in any of the three key states of matter gas, liquid or solid. In other words, a colloidal is a microscopically small substance that is equally dispersed throughout in another material.

    Lyophilic colloid	Lyophobic colloid
    They are reversible in nature	They are irreversible in nature
    The particles move in any direction	The particles move in specific direction
    They do not show tyndall effect	They show tyndall effect
    They are easily formed by direct mixing	They are easily formed by a special method

26. 21. (i) SO₂ act as both oxidising and reducing agent but H₂S acts as only reducing agent. Why?

    (ii) Why halogens are coloured?

    Answer:

    (i) So₂ can act as both oxidising as well as reducing agent since it has +4 oxidation state which is exactly b/w its highest state +6 & lowest oxidation state +2 it can change its oxidation number either from +4 to -2 (reduction) & +4 to +6 (oxidation)

    Where asH₂S the oxidation state of sulphur is -2 so it can lose electrons to +4 & +6 oxidation state but it cannot gain electron H₂s act as only reducing agent.

    (ii) Halogens have an unpaired electron that is present in the outermost shell of atom. When photons of suitable energy hit atom, the electron gets excited & moves higher energy state in the atom, they absorb energy from visible region & show the colour.

27. 22. (i) Why do ethers possess dipole moment ?

          (ii) Boiling points of ethers are lower than their corresponding isomeric dlcohols. Why ?

    Answer:

    (i) Because ethers contain O atom that fuses two carbons, the overall shape of the molecule somewhat similar to H₂o it is V shaped molecule because there is angle between oxygen bonds.

    (ii) Because hydrogen bonds can’t form b/w the molecule in the ether, the boiling point of this compound is more than so, lower than the corresponding alcohol. As a result, ethers are less likely to be soluble in water than the alcohol with same molecular weight.

28. 23. (i) Write Clemmensen reduction reaction.

    (ii) Write Rosenmund reaction.

    (iii) Formaldehyde gives Cannizzaro reaction whereas acetaldehyde does not. Why?

    Answer:

    (i) Clemmensen reduction reaction:-

    

    (ii) Rosenmund reaction:-

    

    (iii) The Cannizzaro reaction is a chemical rxn that involves the base-induced disproportional of an aldehyde lacking of hydrogen atom in alpha position…. Hydrogen present. Hence formaldehyde undergoes Cannizzaro reaction whereas acetaldehyde does not.

29. OR

    (i) Aldehydes and ketones undergo a number of nucleophilic addition reactions. Why?

    (ii) Acetic acid is liquid while aromatic acids are solids. Give reasons.

    Answer:

    (i) The Carbonyl group on aldehyde &ketons is particularly to nucleophilic addition due to high electronegativity of oxygen atom double bonded to carbon atom. Hence oxygen acquires partial +ve change.

    (ii) But the aromatic carboxylic acid carry atleast 7 carbons atom. It belongs to higher molecules weight carboxylic acid. They are solid at room temperature because of vander walls force in addition to hydrogen bonding. More force of attraction means molecular are more closer to form solid

31. 24. (i) Unlike Phosphorus, nitrogenshows,little tehdency for catenation. Why ?

    (ii) SF₆ is known but SH₆ is not known. Explain

     (iii) Give hybridization and draw structure of XeF₄

    Answer:

    (i) Phosphorus shows marked tendency for catenation but nitrogen shows little tendency for catenation. Nitrogen has very little tendency to show catenation as N-N bond is very weak because of the repulsion in the electron pair on the nitrogen atom.

    (ii) SF₆ exists but SH₆ does not because fluorine is the most electronegative element in the periodic table, the size is extremely small so it has greater polarising power & has small h bonding, but in SH₆ then the electronegativity of sulphur is much more than hydrogen, hydrogen doesn't have sufficient Nuclear charge.

    (iii)

    

    In XeF₄, central atom Xe is Sp₃d₂hyberdised

    Having 2 lone pair on it so shape of molecule will be square planer. The valense shell of Xe contains 2 electrons in 5s orbital & six electrons in 5p orbital. In the 5p orbital there is place for d orbital f-orbital electrons in outermost orbit of Xe two of the 5s & 5p electrons get excited to the vaccent 5d orbitals. 2 in 5p & 2 in 5d orbitals four fluorine atom bond with these four electrons. The remaining two points pairs hybridised electrons are free on either side of central atom.

32. Or 

    (i) Explain the steps involved in manufacture of sulphuric acid by contact process.

    (ii) Write down the reaction of ozone with potassium iodide. 

    (iii) Draw structure of ClF₃

    Answer:

    (i) Steps involved in contact process:-

    SO₂ is produced by roasting metallic sulphides in air

    4FeS₂  +  11O₂à 2Fe₂O₃ + 8SO₂

    Purification of Gases

    The efficiency of catalyst, various impurities present in mixture of sulphur dioxide & air are first removed

    Catalytic Oxidation of Sulpher Dioxide:-

    2SO₂+O₂   >  2SO₃

    Absorption of Sulpher trioxide in Sulphuric Acid:-

    SO₃  + H₂ SO₄ à  H₂S₂O₇   (oleum)

    Dilution of Oleum to Obtain Sulphuric Acid:-

    A calculated amount of water is added to obtain sulphuric acid of desired strength

    H₂S₂O₇  + H₂Oà 2H₂SO₄ 

    (ii)   KI + 3O₃ àKIO₃  + 3O₂

    Potassium iodide react with ozone to produce potassium iodate &oxygee. This reaction take place in heat, concentrated solution potassium hydroxide.

    (iii) The structure of Clf₃ is trigonal bipyramidal with a 175⁰ F-Cl-F bond angle. There are 2 equatorial lone pairs making the final structure T shaped

    

33. 25. (i) Why do transition elements show catalytic properties ?

         (ii) Calculate equivalent weight of KMnO4 in neutral medium.

         (iii) What is the cause of Lanthanoid contraction ?

    Answer:

    (i)

    • Their partically empty d-orbitals provide surface area.
    • They show multiple oxidation state & by giving electrons to reactants they from complexes & lower their energies.

    (ii) MnO₄  +  2H₂O  + 3e  à  MnO₂ (s)  + 4OH (gained 3 electrons)

    In this case

    3 equivalent per mole 

    equivalent mass of KMNO₄  =  M/3

                                                     = 158.04 /3

                                                     = 52.68 gram/equivalent

     

    (iii) The Lanthanoid contraction is caused by poor shielding effect of the 4f electrons. Gd because as atomic number increases, the decrease the atomic number radii. Yb because it has large atomic number. Because the elements in row 3 have 4f electrons.

34. OR 

    (i) Write down any three similarities between lanthanoids and actinoids.

    (ii) Out of Co²⁺ and Zn²⁺ which will be paramagnetic and why?

    (iii) Draw the structure of permanganate ion:

    Answer:

    (i)

    • Both have prominent oxidation state +3
    • They are involved in filling of (n-2) f orbitals
    • They are highly electropositive & very reactive in nature.
    • They are two additional rows below the periodic table
    • Due to similar outermost electronic configuration
    • They are two big families of iso-structural natural & synthesized chemical elements.

    (ii) Co²⁺  will be paramagnetic because it has unpaired electrons.

    (iii) Structure of permanganate ion:-

    

36. 26. Write the following reactions:-

    (i) Balz-Schiemann reaction

    (ii) Fitting's reaction

    (iii) Gattermann reaction 

    (iv) Finkelstein reaction 

    (v) Diazotisation reaction

    (vi) Grooves process

    Answer:

    (i) Balz-Schiemann reaction:-

    

    (ii) Fitting's reaction:-

    

    (iii) Gattermann reaction  :-

    

    (iv) Finkelstein reaction:-

    CH₃CH₂Cl + NaI     CH₃CH₂I + NaCl

    (ethyl chloride)

    (v) Diazotisation reaction:-

    

37. OR

    26. (i) The treatment of alkylhalide with aqueous KOH leads to the formation of alcohols while in the presence of alcoholic KOH, alkenes are formed as the major product Explain.

    (ii) How aryl halides react with sodium metal? Explain why alkyl halides show nucleophilic substitution reaction?

    Answer:

    (i) Aqueous Kott is alkaline in nature, These hydroxide ion act as strong nucleophile & replace the halogen atom in an alkyl halide

    Rcl + KOH (aq) à ROH + Kcl

    This result in formation of alcohol & the reaction is known as nucleophile substitution reaction

    Alcohlic KOH, especially in ethanol produces C₂H₅O^- ions. The C₂H₅O^-ions is stronger base than OH^-ions the former abstract the B-hydrogen of an alkylhalide to produce alkenes.

    CH₃ CH₂ Br + KOH(alc)  à  H₂C = CH₂ + KBr + H₂O

     

    (ii) The courtz-fitting rxn is the chemical rxn of aryl halides with alkyl halides & sodium metal in the presence of dry ether to give substituted aromatic compounds. The rxn works best for forming assymeterical products if the halide reactant are separate in their relative chemical reactivities.

    

    Alkyl halides are generally used example for nucleophilic substation reaction because they have good tendency leaving group the halide group Regardless of whether you’re looking at an SN₁& SN ₂ reaction, You will need a good leaving group. Halides are relatively weak base, which basically means that they are energetically stable. –ve charge stabilized, It is less reactive It can just hang out there in Sol as a counter ion & long as you have a stronger nucleophile in solⁿ.