Article

CBSE Class 12 Chemistry Question Paper 2016 Comptt (Delhi) with Solutions

Students can use CBSE Previous Year Question Papers Class 12 Chemistry with Solutions and CBSE Class 12 Chemistry Question Paper 2016 Comptt (Delhi) to familiarize themselves with the exam format and marking scheme.

CBSE Class 12 Chemistry Question Paper 2016 Comptt (Delhi) with Solutions

Time allowed : 3 hours

General Instructions:

  1. All questions are compulsory.
  2. Q.no. 1 to 5 are very short answer questions and carry 1 mark each.
  3. Q.no. 6 to 10 are short answer questions and carry 2 marks each.
  4. Q.no. 11 to 22 are also short answer questions and carry 3 marks each.
  5. Q.no. 23 is a value based question and carries 4 marks.[*]
  6. Q.no. 24 to 26 are long answer questions and carry 5 marks each.
  7. Use log tables if necessary, use of calculators is not allowed.
    + Deleted from Syllabus.

Question 1.
Define rate of reaction. [1]
Answer:
The change in concentration of reactant or product per unit time is called rate of reaction.

Question 2.
Why is adsorption always exothermic? [1]
Answer:
Due to the force of attraction/bond formation between adsorbate and adsorbent.

Question 3.
Write the IUPAC name of the following complex : [1]
[Co(NH3)6]3+
Answer:
Hexaamminecobalt (III) ion.

Question 4.
Write equation of the nitration of anisole. [1]
Answer:

Question 5.
Draw the structure of 2-methylbutanal. [1]
Answer:

Question 6.
Define the following terms: [2]
(i) n-type semiconductor (ii) Ferrimagnetism
Answer:
(i) n-type semiconductor : When Si/Ge is doped with group 15 element.
(ii) Ferrimagnetism : When magnetic domains are aligned in parallel and anti-parallel directions in unequal numbers.

Question 7.
What is osmotic pressure? Why it is a colligative property? [2]
Answer:
The excess pressure applied on solution side to stop the process of osmosis. It is a colligative property because it depends upon the number of solute particles but not on their nature.

Question 8.
Draw the structures of the following compounds
(i) BrF3
(ii) XeF4
Answer:

Question 9.
Give reasons : [2]
(i) Transition metals show variable oxidation states.
(ii) Actinoids show wide range of oxidation states.
Answer:
(i) Because in transitional elements ns and (n – 1) d electrons have approximate equal energies.
(ii) Due to comparable energies of 5f, 6d and 7s orbitals.

Question 10.
Write the definitions of the following : [2]
(i) Racemic mixture
(ii) Chiral compound
Answer:
(i) The mixture of equimolar quantities of d and l isomers.
(ii) The compounds in which at least one of the carbon atom is linked with four different atoms or groups of atoms.

Or

Write the ambident nucleophiles? Given an example.
Answer:
A group containing two nucleophilic centres.
Example: CN (Cyanide) and NC (Isocyanide).

Question 11.
If NaCl is doped with 10-3 mol % of SrCl2, what is the concentration of cation vacancies? [3]
Answer:
Concentration of SrCl2 = 10-3 mol% = 10-3/100 mol = 10-5 mol
1 mol of NaCl on doping procuces = 6.022 × 1023 cation vacancies
Therefore, 10-5 mol of NaCl on doping produces = 6.022 × 1023 × 10-5 = 6.022 × 1018 cation vacancies

Question 12.
Calculate the mass of a non-volatile solute (molecular mass 40) which should be dissolved in 114 g octane to reduce the vapour pressure to 80%. [3]
Answer:

Question 13.
If the half-life period of a first order reaction in A is 2 minutes, how long will it take [A] to reach 25%; of its initial concentration?
Answer:

Question 14.
What are emulsions? What are their different types? Give an example of each type. [3]
Answer:
A liquid-liquid colloidal systems is called emulsions. They are of two types :
(i) Oil in water
(ii) Water in oil.
Examples :
o/w—Vanishing cream, milk
w/o—Butter, cold cream, cod-liver oil

Question 15.
Describe the following : [3]
(i) Role of depressant in froth floatation process
(ii) Role of silica in the metallurgy of copper
(iii) Role of cryolite in the metallurgy of aluminium
Answer:
(i) To selectively prevent the formation of froth by one of the sulphide ore present in a mixture of sulphide ores.
(ii) To remove impurity FeO in the form of slag (FeSiO3)
FeO + SiO2 → FeSiO3
(iii) Lowering down the melting point and increasing the conductivity of the melt.

Question 16.
In the 3d series (Sc = 21 to Zn = 30) : [3]
(i) Which element shows maximum number of oxidation states?
(ii) Which element shows only +3 oxidation state?
(iii) Which element has the lowest enthalpy of atomization?
Answer:
(i) Mn
(ii) Sc
(iii) Zn

Question 17.
Write down the hybridization and magnetic character of the following complexes : [3]
(i) [Ni(CO)4]
(if) [CoF6]3
(Atomic number : Ni = 28, Co = 27)
Answer:
(i) Hybridisation : sp3
Magnetic character: Diamagnetic [Ni(CO)4]
Tetrahedral shape,

(ii) Hybridisation : sp3d2
Magnetic character: Paramagnetic
Co3+ orbitals =

Question 18.
In the following pairs of the halogen compounds which compound undergoes faster SN1 reaction. [3]

Answer:

(ii) CH3CH2-I

Question 19.
How are the following conversions carried out? [3]
(i) Propene → Propan-2-ol
(ii) Benzyl chloride → Benzyl alcohol
(iii) Ethyl magnesium chloride → Propan-1-ol
Answer:

Question 20.
Write major product(s) in the following reactions :

Answer:

Or

Describe a method for the identification of primary, secondary and tertiary amines. Also write chemical equations of the reaction involved.
Answer:
Benzenesulphonyl chloride (C6H5S02C1), which is also known as Hinsberg’s reagent, reacts with primary and secondary amines to form sulphonamides and tertiary amine does not reacts.

The product will precipitate after addition of HC1.

It does not react with tertiary amines.

Question 21.
Write two uses of each of the following polymers. [3]
(i) Polypropylene
(ii) PVC
(iii) Nylon-66
Answer:
(i) Polypropylene : Manufacture of ropes, toys, pipes etc.
(ii) PVC : Manufacture of raincoats, handbags, vinyl flooring, water pipes etc.
(iii) Nylon-66 : Manufacture of sheets, bristles from brushes, textiles etc.

Question 22.
Define the following as related to proteins : [3]
(i) Peptide linkage
(ii) Primary structure
(iii) Denaturation
Answer:
(i) Peptide linkage : Two amino acids of same type or different types combine together by the elimination of H20 molecule to form -CONH- linkage.
(ii) Primary structure : It refers to the sequence in which amino acids are joined.
(iii) Denaturation : When a native protein is subjected to change in temperature or pH, hydrogen bonds get disturbed and globules get uncoiled and proteins lose their biological activity.

Question 23.
Ms. Kirti was pursuing her studies in medicine at Bangalore. When she visited her family during holidays, she noticed that their maid was always complaining of stomach ache and some burning sensation. Kirti took her maid to the doctor where she was diagnosed for early stage of ulcers due to excessive use of Antacids.
Answer the following: [4]
*(i) What are the values displayed by Kirti?
(ii) What are antacids? Give two examples.
(iii) Why prolong use of antacids is harmful?
Answer:
(i) As per latest CBSE curriculum, Value Based Questions will not be asked in the examination.
(ii) Antacids are substances which neutralise excess acid in the stomach.
Example : NaHCO3, milk of magnesia
(iii) It triggers the overproduction of acid by making stomach alkaline.

Question 24.
Define the following :
(a) (i) Molar conductivity
(ii) Fuel cell

(b) The molar conductivity of a 1.5 M solution of an electrolyte is found to be 138.9 S cm2 mol-1 Calculate the conductivity of the solution. [5]
Answer:
(i) Molar conductivity :
(i) Molar conductivity (Λm) : Molar conductivity can be defined as the conductance of the volume V of electrolytic solution kept between two electrodes of a conducting cell at distance of unit length but having area of cross section large enough to accomodate sufficient volume of solution that contains one mole of the electrolyte.
Λm = KV

(ii) Fuel cell : Galvanic cells that are designed to convert the chemical energy of combustion of fuels like hydrogen, methane etc. into electrical energy are called fuel cells, e.g. H2 – O2 fuel cell

(b) Λm = \(\frac{k \times 1000}{c}\) S cm2 mol-1
138.9 = \(\frac{k \times 1000}{1.5}\)
k = 0.208 S cm-1

Or

(a) Define the following terms :
(i) Primary batteries (it) Corrosion
(b) The resistance of a conductivity cell containing 0.001 M KCl solution at 298 K is 1500 Ω. What is the cell constant if conductivity of 0.001 M KCl solution at 298 K is 0.146 × 10-3 S cm-1?
Answer:
(a) (i) Primary batteries : Batteries which can’t be rechared/reused.
(ii) Corrosion : Loss of useful metals due to oxidation on exposure to atmospheric gases and moisture.

(b) k = \(\frac{1}{R}\) (\(\frac{l}{A}\))
\(\frac{l}{A}\) = k × R = 0.146 × 10-3 S cm-1 × 1500 Ω
\(\frac{l}{A}\) = 0.219 cm-1

Question 25.
(a) What happens when :
(i) HCl reacts with finely powdered iron.
(ii) Cl2 reacts with hot concentrated solution of NaOH.

(b) Give appropriate reason for each of the following :
(i) Sulphur vapour exhibits some paramagnetic character.
(ii) NH3 is more basic than PH3
(iii) Dioxygen is a gas but sulphur a solid.
Answer:
(a) (i) It forms FeCl2 and H2 gas
Chemical Equation : Fe + 2HCl > FeCl2 + H2

(ii) It forms NaClO3
Chemical Equation:

(b) (i) Sulphur Above 1000 K exists partially as S2 polecule which has two unpaired electrons in its Anti Bonding orbitals.
(ii) Due to small size of N, NH3 molecule has small surface area and high electron density in comparison to PH3. Hence it can donate lone pair of electrons more readily.
(iii) In dioxygen there is Pn – Pn multiple bonding and no cation hence it is a gas.

Or

(a) Complete the following equations :
(i) PCl5 \(\rightarrow{\text { heat }}\)
(ii) C + Conc.H2SO4

(b) Explain giving reason in each case :
(i) Halogens are strong oxidising agents
(ii) Fluorine form only one oxoacid, HOF
(iii) Helium is used in diving apparatus
Answer:
(a)

(b) (i) Due to their high electron affinity. As they have seven electrons, they accept one electron readily.
(ii) Due to high electronegativity and small size of fluorine.
(iii) Due to low solubility in blood.

Question 26.
(a) Write the product(s) in the following : dil NaOH

(b) Give simple tests to distinguish the following pairs of compounds :
(i) Ethanal and Propanal
(ii) Benzaldehyde and Acetophenone
(iii) Benzoic acid and Ethyl benzoate

Or

(a)

(b)
(i) On heating with NaOH and I2, ethanal forms yellow ppt of CHI3 whereas propanal can not.

CH3CHO + 3I2 + 4NaOH → CHI3 + 3NaI + HCOONa + 3H2O

(ii) On heating with NaOH and I2, aceptophenone forms yellow ppt of CHI3 whereas benzaldehyde does not.

C6H5COCH3 + 3NaOI → C6H5COONa + CHI3 ↓ + 2NaOH

(iii) On adding NaHCO,, benzoic acid produces brisk effervescence of CO2 gas whereas ethylbenzoate does not.

C6H5COOH + NaHCO3 → C6H5COONa + CO2 ↑ + H2O

Or

(a) Give reasons :
(i) CH3-CHO is more reactive than CH3COCH3 towards HCN.
(ii) 4-nitrobenzoic acid is more acidic than benzoic acid.

(b) Describe the following :
(i) Acetylation
(ii) Cannizzaro reaction
(iii) Cross aldol condensation
Answer:
(a) (i) Because carbonyl carbon of CH3-CHO is more electrophilic than CH3COCH3 due to only one electron donating CH3– group.
(ii) Because of electron withdrawing nature of -NO2 group.

(b) (i) Acetylation : Introduction of an acetyl group/CH3CO- by heating an organic compound with acetyl chloride/acetic anhydride.

(ii) Aldehydes having no a-hydrogen atom when treated with cone. NaOH, undergoes self-oxidation and self-reduction simultaneously.

(iii) When aldol condensation is carried out between two different aldehydes or ketones, it is called cross aldol condenstation


Show More
यौगिक किसे कहते हैं? परिभाषा, प्रकार और विशेषताएं | Yogik Kise Kahate Hain Circuit Breaker Kya Hai Ohm ka Niyam Power Factor Kya hai Basic Electrical in Hindi Interview Questions In Hindi