SAMPLE QUESTION PAPER (2021-22)
CHEMISTRY (043)
TERM II
CLASS 12
Time: 2 Hrs Max. Marks: 35
GENERAL INSTRUCTIONS
1. There are 12 questions in this question paper with internal choice.
2. SECTION – A, Q. No. 1 – 3 are very short answer questions carrying 2 Marks each.
3. SECTION – B, Q. No. 4 – 11 are short answer questions carrying 3 Marks each.
4. SECTION – C, Q. No. 12 is case based question carrying 5 Marks.
5. All questions are compulsory.
6. Use of log tables and Calculator is not allowed.
SECTION A
Attempt all questions: –
Q1. For the reaction:
(i) What is the order and molecularity of the reaction?
Zero order, Bimolecular
(ii) Write unit of k.
mol L-1s-1
Q2. Give reasons:
(i) The -hydrogen atoms of aldehydes and ketones are acidic in nature.
Oxygen being more electronegative than carbon draws the electrons towards it making carbon positive. Carbon, in turns, draws the electrons.
(ii) Propanone is less reactive than Ethanal towards Nucleophilic addition reactions.
Because the nucleophile CN– faces steric hindrance while attaching to Propanone.
Q3. Draw structures of the following derivatives:
(i) Cyanohydrin of cyclobutanone
(ii) Hemiacetal of ethanol
Draw their structures of the following derivatives.
SECTION B
Q4. A first order reaction takes 20 minutes for 25% decomposition. Calculate the time when
75% of the reaction will be completed. [Given log2 = 0.3010, log3 = 0.4771, log4 = 0.6021]
t = 2.303/k log [R]0 /[R], t = 96.3 minutes.
OR
The decomposition of N2O5 (g) is a first order reaction with rate constant of 5 x 10-4 s-1 at 450 C
2N2O5 (g) → 4NO2 (g) + O2 (g)
If the initial concentration of N2O5 is 0.25 M. calculate its initial concentration after 2 minutes.
Also calculate the half-life for decomposition of N2O5 (g).
t = 2.303/k log [R]0/[R], [R] = 0.23 M, t1/2 = 0.693/5 x 10-4s-1, 1386 s.
Q5. (a) Adsorption of gas on the surface of solid is generally accompanied by decrease in
entropy. Still it is a spontaneous process. Explain.
In adsorption, ∆S is –ve, ∆H is –ve (Exothermic), ∆G is –ve (Spontaneous)
(b) How does an increase in temperature affect both physical as well as chemical adsorption?
Physical adsorption decreases with increase of temperature while Physical adsorption increases with increase of temperature.
(c) Identify the functional groups in the molecular of non-ionic detergent.
(i) ether, (ii) primary alcoholic group.
Q6. Write the difference in each of the following:
(i) Multimolecular colloids and Associated colloids
On dissolution, a large number of atoms or smaller molecules of a substance aggregate to form species having size in colloidal range. These species are called multimolecular colloids.
There are some substances which at lower concentration behave as normal electrolytes but at higher concentration exhibit colloidal behaviour. Such substances are called associated colloids
(ii) Coagulation and Peptisation
The process of settling of colloidal particles is called coagulation. Peptisation is the process of conversion of precipitate into colloidal solution.
(iii) Lyophilic sol and lyophobic sol.
Lyophilic sols are reversible and lyophobic sols are irreversible.
Q7. The magnetic moments of few transition metal ions are given below:
Which of the given metal ions?
(i) Has maximum number of unpaired electrons?
Cr2+
(ii) Forms colourless aqueous solution?
Sc3+
(iii) Exhibits the most stable +3 oxidation state?
Sc3+
OR
Account for the following:
(i) CuCl2 is more stable than Cu2Cl2.
CuCl2 is more stable than Cu2Cl2. Because in CuCl2, Cu is in +2 oxidation states is more stable due to high hydration enthalpy compared to Cu2Cl2.
(ii) Transition metals form complex compounds.
Transition metals form complex compounds due to comparatively smaller sizes of the metal ions, high ionic charges and the availability of d-orbitals for bond formation.
(iii) Atomic radii of 4d and 5d series elements are nearly same.
Because of lanthanoids contraction.
Q8. (a) What type of isomerism is shown by the complex [Co (NH3)5 (SCN)]2+?
Linkage isomer
(b) Why is [NiCl4]2- paramagnetic while [Ni (CN)4]2- is diamagnetic?
[NiCl4] 2- paramagnetic (sp3), 2 unpaired electrons while [Ni (CN) 4] 2- is diamagnetic (dsp2), 2 no unpaired electrons.
(c) Why are low spin tetrahedral complexes rarely observed?
For tetrahedral complexes, ∆t = (4/9) ∆0. Crystal field splitting energies are not large; pairing of electrons does not take place.
OR
(a) What type of isomerism is shown by the complex [Cr(H2O)6]Cl3?
[Cr (H2O) 6] Cl3 shows hydrate isomerism.
(b) On the basis of crystal field theory, write the electronic configuration for d4ion if ∆0 >P.
t2g4eg0
(c) Write the hybridization and shape of [CoF6]3- . (Atomic number of Co = 27)
sp3d2, octahedral
Q9. Give reasons:
(i) Benzoic acid does not give Friedel-Craft reaction.
Benzoic acid does not give Friedel-Craft reaction because –COOH group is deactivating and the aluminium chloride (Lewis acid) gets bonded to the carbonyl group.
(ii) O2N-CH2-COOH has lower pKa value than CH3COOH.
–NO2 Exerts electron withdrawing effect which increases the acidic strength or decreases the pKa value.
(iii) (CH3)2CH-CHO undergoes Aldol condensation whereas (CH3)3C-CHO does not.
(CH3)3C-CHO does not have -hydrogen atom.
Q10. (a) Write the structure of main products when aniline reacts with the following reagents:
(i) HCl
Aniline hydrochloride (Anilinium chloride)
(ii) Bromine water
2, 4, 6-tribromoaniline (white ppt.)
(iii) (CH3CO)2O/pyridine
Acetanilide
Q11. (a) Arrange the following in the increasing order of their basic character in an aqueous
solution: C2H5NH2, (C2H5)2NH, (C2H5)3N
C2H5NH2 < (C2H5)2NH < (C2H5)3N
(b) Arrange the following in the increasing order of their boiling points:
C2H5NH2, C2H5OH, (CH3)3N
(CH3)3N < C2H5NH2 < C2H5OH
(c) Give a simple chemical test to distinguish between the following pair of compounds:
(CH3)2NH, and (CH3)3N
(CH3)2NH reacts with Hinsberg’s reagent but (CH3)3N does not react with Hinsberg’s reagent.
SECTION C
Q12. Case Based Question:
Read the passage given below and answer the following questions:
Weak electrolyte like acetic acid has lower degree of dissociation at higher concentration
and hence for such electrolytes, the change in Ʌm with dilution is due to increase the degree of
dissociation and consequently the number of ions in total volume of solution that contains one
mol of electrolyte. In such cases Ʌm increases steeply on dilution, especially near lower
concentrations. Therefore, Ʌ0m cannot be obtained by extrapolation of Ʌm to zero concentration.
At infinite dilution (i.e. concentration c –> zero) electrolyte dissociates completely (α = 1), but at
such low concentration the conductivity of the solution is so low that it cannot be measures
accurately. Therefore, Ʌ0m for weak electrolytes is obtained by using Kohlrausch’s law of
independent migration of ions.
(i) Which of the following is not a weak electrolyte?
(a) Ammonium acetate
(b) Acetic acid
(c) Ammonium chloride
(d) Ammonium hydroxide
(ii) The value of Ʌm in case of a weak electrolyte on dilution.
(a) Increases steeply
(b) Decreases steeply
(c) Increases gradually
(d) Decreases gradually
(iii) Ʌ0m for weak electrolytes is obtained by using
(a) Kohlrausch’s law
(b) Arrhenius law
(c) Avogadro’s law
(d) Maxwell’s law
(iv) For strong electrolytes, Ʌm increases slowly with dilution and can be represented by the equation:
(a) Ʌ0m = Ʌm – AC1/2
(b) Ʌm = Ʌ0m – AC1/2
(c) Ʌ0m = Ʌm – AC
(d) Ʌm = Ʌ0m – AC
(v) Limiting molar conductivity (S cm mol-1) of K+ and SO42-are respectively 73.5 and 160.0. The Limiting molar conductivity of K2SO4 will be
(a) 393.5
(b) 233.5
(c) 467
(d) 307