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. Calculate the degree of dissociation (α) of acetic acid if its molar conductivity (Ʌm) is 39.05 S cm2 mol-1 given: λ0 (H+) = 349.6 S cm2 mol-1 and λ0 (CH3COO–) = 40.9 S cm2 mol-1
α = 0.1
Q2. Write a chemical equation each to represent
(a) Gatterman reaction
Gatterman reaction: Benzene Diazonium halides on treatment with Cu/HX liberate N2 and aryl halide is formed.
(b) Carbylamine reaction.
Carbylamine reaction: Aliphatic/aromatic primary amines on heating with chloroform and alkali produce alkyl/aryl isocyanide.
Q3. Show the mechanism of acylation of Ethanamine and write IUPAC name of the product
formed.
Give mechanism, The product N-ethylethanamide is formed.
SECTION B
Q4. Calculate the standard cell potential of the galvanic cell in which the following reaction
takes place:
2Cr (s) + 3Cd2+ (aq) —> 2Cr3+ (aq) + 3Cd (s)
Also calculate the ∆rG0 value of the reaction.
[Given: E0Cr3+/Cr) = – 0.74 V, E0Cd2+/Cd) = – 0.40 V and F = 96500 C mol-1]E0cell = -0.40 V – (-0.76 V) = 0.34 V, ∆rG0 = – nF E0 cell = -196.86 kJ mol-1
OR
The following chemical reaction is occurring in an electrochemical cell:
Mg (s) + 2Ag+(0.0001M) → Mg2+ (0.10M) + 2Ag (s)
The E0 electrode values are Mg2+/ Mg = -2.36 V, Ag+/Ag = 0.81 V
For this cell calculate:
(a) Standard cell potential (E0 cell)
E0cell = 0.81 V – (-2.36 V) = 3.17 V
(b) Cell potential (Ecell)
Ecell = 2.9635 V
(c) (i) Symbolic representation of the above cell
Mg (s)|Mg2+ (0.10M) || Ag+(0.0001M)|Ag (s)
(ii) Will the above cell reaction be spontaneous?
Yes, E0cell is +ve.
Q5. (a) For the following reaction: 2NO (g) + O2 (g) → 2NO2 (g),
Write the rate law and order of reaction.
Rate = k[NO]2[O2]1 and order of reaction 2 +1 = 3
(b) After 24 hours, only 0.125 g out of the initial quantity of 1 g of a radioactive isotope remains
behind. What is the half-life period?
k = 2.303/t log (a/a-x) = (2.303/24) log (1/0.125) = 0.0866 h-1
t1/2 = 0.693/ 0.0866 =8 hours.
OR
Following data are obtained for the reaction: N2O5 → 2NO2 + 1/2O2
(a) Show that it follows first order reaction.
k = 2.303/t log [R]0 /[R] = 0.00231 s-1
(b) Calculate half-life period.
t1/2 = 0.693/ 0.00231 = 300 s
Q6. For the reaction:
Write:
(i) Rate reaction expression
Rate = -d/dt[C12H22O11] = -d/dt[H2O] = d/dt[C6H12O6] = d/dt[C6H12O6]
(ii) Rate law equation.
Rate = k[C12H22O11]
(iii) Order and molecularity of the reaction.
Two and first order reaction.
Q7. Explain the following terms:
(i) Electrophoresis
The movement of colloidal particles under an applied electric potential is called electrophoresis.
(ii) Tyndall effect
The path of light is clearly visible due to scattering of light by colloidal particles.
(iii) Zeta potential
The potential difference between the fixed layer and the diffused layer of opposite charges.
Q8. Explain the following observations:
(a) Ferric hydroxide sol gets coagulated on addition of sodium chloride solution.
Fe(OH)3 is positively charged sol. It gets coagulated by the Clions from NaCl.
(b) Cottrell’s smoke precipitator is fitted at the mouth of the chimney used in factories.
The charged colloidal particles of carbon, after coming into contact with oppositely charged electrode in Cottrell precipitator lose their charge and settle down at the bottom.
(c) Physical adsorption is multilayered, while chemisorption is monolayer.
In physical adsorption, there are weak Vander waals forces. Therefore, it forms multilayers. In chemisorption, adsorbate is attached by chemical bond. There is a strong force of attraction. Therefore, only one layer is obtained.
Q9.
(a) For the complex [Fe (CN) 6]3-, write the hybridization type, magnetic character and spin
nature of the complex.
Hybridization d2sp3, paramagnetic due to presence of one unpaired electron and low spin.
(b) Draw one of the geometrical isomers of the complex [Pt (en) 2Cl2]2+ which is optically
active.
cis- form is optically active.
OR
(a) Name the following compound according to IUPAC system of nomenclature:
[Pt (NH3)4 Cl (NO2)]SO4Tetraaminechloridonitrito-N-platinum (IV) sulphate
(b) On the basis of crystal field theory, write the electronic configuration for d4ion if ∆0 < P.
t2g3eg1
(c) Write the hybridization and magnetic behaviour of the complex [Ni (CN)4]2-
Hybridization dsp2, Diamagnetic
Q10. Write the products in the following reactions:
(i) Cyclohexane cyanohydrin
(ii) Benzene
(iii) But-2-en-1-al
Q11. Complete the following sequence of reactions:
(i) Identify (A) to (D)
A = 4-Hydroxy-4-methylpentan-2-one
B = 4-methylpent-3-en-2-one
C = Iodoform
D = Sodium 3-methylbut-2-enoate
(ii) Give the IUPAC name of (A)
4-Hydroxy-4-methylpentan-2-one
SECTION C
Q12. Case Based Question:
Read the passage given below and answer the following questions:
The d-block of the periodic table contains the elements of the groups 3–12 and is known
as transition elements. In general, the electronic configuration of the elements is (n-1) d1-10ns1-2. The d-orbitals of the penultimate energy level in their atoms receive electrons giving rise to the three rows of the transition metals i.e. 3d, 4d and 5d series. However, Zn, Cd and Hg are not regarded as transition elements. Transition elements exhibit certain characteristic properties like variable oxidation states, complex formation,
formation of coloured ions and alloys, catalytic activity, etc. Transition metals are hard
(except Zn, Cd and Hg) and have a high melting point.
The following questions ate multiple choice questions. Choose the most appropriate answer.
(i) The electronic configuration of d-block elements is
(a) (n-2) d1-10 (n-1)s1-2
(b) (n-1) d10ns1-2
(c) (n-2) d1-10ns1-2
(d) (n-1) d1-10ns1-2
(ii) Which transition element of 3d series does not show variable oxidation state?
(a) Chromium
(b) Titanium
(c) Scandium
(d) Mercury
(iii) Why are melting points of transition metals high?
(a) It is because of vacant d-orbitals.
(b) It is because of high ionization enthalpy.
(c) It is due to metallic bonding.
(d) It is because they occupy positions between s- and p-block.
(iv) 4d series of elements consists of the elements
(a) Sc – Zn
(b) Rf – Cn
(c) La – Hg
(d) Y – Cd
(v) Why are Zn, Cd and Hg not regarded as transition elements?
(a) They have low melting points.
(b) They have low boiling points.
(c) (n-1) d orbitals are completely filled.
(d) They do not show variable oxidation state.