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

Solved Question Paper

Chemistry Mar-2017

PSEB • Class 12th • 1st • Mar-2017

Chemistry previous year question papers on BRpaper help students looking for PSEB 12th Class Question Papers to browse Chemistry papers for board exam practice and revision. This subject section is useful for finding PSEB 12th Chemistry old question papers, PYQ-style papers, and related exam-preparation material where available, so students can study the question format, recurring topics, and marks pattern used in the Punjab board exam. If you are preparing for the Punjab School Education Board Class 12 Chemistry paper, this page can help you compare past question styles with sample paper and model paper trends, and focus on important questions more confidently. BRpaper is not the official website of Punjab School Education Board or any institution, and it does not publish official notices or academic updates.

Solved Questions

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

    Very Short Answer Mar-2017 • PSEB Class 12th

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

  2. Define half life period of a reaction.

    Very Short Answer Mar-2017 • PSEB Class 12th

    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. write down IUPAC name of

    Very Short Answer Mar-2017 • PSEB Class 12th

    N-phenylbenzenamine or Diphenylamine.

  4. Complete the following reaction

    Very Short Answer Mar-2017 • PSEB Class 12th

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

    Very Short Answer Mar-2017 • PSEB Class 12th

  6. Write down name of one antibiotic.

    Very Short Answer Mar-2017 • PSEB Class 12th

    Cephalexin, Amoxicillin.

  7. What are analgesics?

    Short Answer Mar-2017 • PSEB Class 12th

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

  8. What are monosaccharides ?

    Short Answer Mar-2017 • PSEB Class 12th

    Monosaccharides are also called simple sugars. They are the simplest form of sugars and the most basic unit of carbohydrates. General formula is CnH2nOn.

    They are usually colourless, water soluble.

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

    Short Answer Mar-2017 • PSEB Class 12th

    Radius of Na+ = 100pm

    Radius of Cl- = ?

    Radius ratio

    Radius ratio

    Hence radius of anion B is 241pm.

  10. calculate the time required for the completion of 90% of a reaction of first order kinetics t1/2=44.1 minutes

    Short Answer Mar-2017 • PSEB Class 12th

    K =

    t =

    Given:  t1/2 = 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.

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

    Short Answer Mar-2017 • PSEB Class 12th

    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.

     

  12. Write down differences between thermosetting polymes and thermoplastic polymers

    Short Answer Mar-2017 • PSEB Class 12th

    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.

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

    Short Answer Mar-2017 • PSEB Class 12th

    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.

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

    Short Answer Mar-2017 • PSEB Class 12th

    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.

     

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

    Long Answer Mar-2017 • PSEB Class 12th

    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.

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

    Long Answer Mar-2017 • PSEB Class 12th

    Potassium permanganate (KMnOâ‚„) 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-

  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)

    Long Answer Mar-2017 • PSEB Class 12th

    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.

  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).

    Long Answer Mar-2017 • PSEB Class 12th

    (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) Tb0 = 353.23k

    W2 = 1.80g

    M2 = ?

    W1 = 90g

    Tb = 354.11k

    Tb = 354.11 – 353.23

        = 0.88k

    Kb = 2.53 kgmol-1

    Tb  =

    M2 =

    => 57.5g

    Molar mass of solute is = 57.5g

  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

    Long Answer Mar-2017 • PSEB Class 12th

    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

    Ecell = E0cell -

     = - (Cu2+/Cu)

    = 0.80 – 0.34 = 0.46V

    Ecell = 0.46 - 

    =  0.46 - 

    = 0.46 – 0.21 = 0.25V

     

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

    Long Answer Mar-2017 • PSEB Class 12th

    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

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

    (ii) Why halogens are coloured?

    Long Answer Mar-2017 • PSEB Class 12th

    (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.

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

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

    Long Answer Mar-2017 • PSEB Class 12th

    (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.

  23. (i) Write Clemmensen reduction reaction.

    (ii) Write Rosenmund reaction.

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

    Long Answer Mar-2017 • PSEB Class 12th

    (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.

  24. 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.

    Long Answer Mar-2017 • PSEB Class 12th

    (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

  25. (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

    Long Answer Mar-2017 • PSEB Class 12th

    (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.

  26. 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

    Long Answer Mar-2017 • PSEB Class 12th

    (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 2H2SO

    (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

  27. (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 ?

    Long Answer Mar-2017 • PSEB Class 12th

    (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 KMNO=  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.

  28. 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:

    Long Answer Mar-2017 • PSEB Class 12th

    (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:-

  29. Write the following reactions:-

    (i) Balz-Schiemann reaction

    (ii) Fitting's reaction

    (iii) Gattermann reaction 

    (iv) Finkelstein reaction 

    (v) Diazotisation reaction

    (vi) Grooves process

    Long Answer Mar-2017 • PSEB Class 12th

    (i) Balz-Schiemann reaction:-

    (ii) Fitting's reaction:-

    (iii) Gattermann reaction  :-

    (iv) Finkelstein reaction:-

    CH3CH2Cl + NaI     CH3CH2I + NaCl

    (ethyl chloride)

    (v) Diazotisation reaction:-

  30. 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?

    Long Answer Mar-2017 • PSEB Class 12th

    (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 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.

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