Students can access the CBSE Sample Papers for Class 12 Chemistry with Solutions and marking scheme Term 2 Set 7 will help students in understanding the difficulty level of the exam.
CBSE Sample Papers for Class 12 Chemistry Standard Term 2 Set 7 with Solutions
Max. Marks: 35
Time: 2 Hours
General Instructions:
Read the following instructions carefully:
- There are 12 questions in this question paper with internal choice.
- Section A – Q. No. 1 to 3 are very short answer questions carrying 2 marks each.
- Section B – Q. No. 4 to 11 are short answer questions carrying 3 marks each.
- Section C- Q. No. 12 is case-based question carrying 5 marks.
- All questions are compulsory.
- Use of log tables and calculators is not allowed.
Section – A
Question 1.
Answer the following:
(a) What is Fehling’s solution?
(b) To what oxidation state ethanal converts Fehling’s solution.
Answer:
(a) Fehling’s solution is an alkaline solution of CuSO4 along with some Rochelle salt.
(b) Ethanol converts Cu(II) of Fehlmg’s solution to Cu(l) i.e., + 1 oxidation state.
Question 2.
Given the standard electrode potentials,
K+/K = – 2.93 V,
Ag+/Ag = 0.80 V,
Hg2+/Hg = 0.79 V,
Mg2+Mg=- 2.37 V,
Cr3+/Cr = -0.74 V
Arrange these metals in their increasing order of reducing power.
Answer:
The lower the reduction potential more the tendency to get oxidised, the higher is the reducing power.
The given standard electrode potentials increase in the order of:
K+/K <Mg2+Mg < Cr3+/Cr < Hg2+/Hg <Ag+/Ag.
Hence, the reducing power of the given metals increases in the following order:
Ag < Hg < Cr < Mg < K.
Question 3.
Explain what is observed
(a) When a beam of light is passed through a colloidal sol.
(b) An electrolytic NaCl is added to hydrated ferric oxide sol.
Answer:
(a) When a beam of light is passed through a colloidal solution, then scattering of light is observed. This is known as the Tyndall effect. This scattering of light illuminates the path of the beam in the
colloidal solution.
(b) When NaCI is added to ferric oxide sol, it dissociates to give Na+ and Cl– ions. Particles of ferric
oxide sol are positively charged. Thus, they get coagulated in the presence of negatively charged Cl– ions.
Section – B
Question 4.
(a) On the basis of crystal field theory, write the electronic configuration for d4 ion if Δ0 > P.
(b) [NiCl4]2- is paramagnetic, while [Ni(CO)4‘] is diamagnetic, though both are tetrahedral. Why? (Atomic number of Ni = 28).
(c) What is meant by the chelate effect? Give an example.
OR
(a) What are the oxidation state of Ni and Fe in Ni(CO)4 and Fe(CO)5.
(b) NH3 acts as a ligand but NH4+ does not.
(c) CN– is an ambidentate ligand.
Answer:
(a) If Δ0> P, it becomes more energetically favourable for the fourth electron to occupy a t2g orbital with
configuration t42g eg.
(b) Though both [NiCl4]2- and [Ni(CO)4] are tetrahedral, their magnetic characters are different. This is due to a difference in the nature of ligands. Cl– is a weak field ligand and it does not cause the pairing of unpaired 3d electrons. Hence, [NiCl4]2- is paramagnetic. On the other hand, CO is a strong field ligand and it causes the pairing of unpaired 3d electrons. Hence, [Ni(CO)4] is diamagnetic.
(c) When a polydentate ligand attaches to the metal ion in a manner that forms a ring, then the metal-ligand association is found to be more stable. In other words, we can say that complexes containing chelate rings are more stable than complexes without rings. This is known as the chelate effect.
For example:
Ni2+(aq) + 6NH3(aq) → [Ni(NH)3]62+(aq)
log b =8.61
Ni2+(aq) + 3en(aq) → [Ni(en3)]2+(aq)
log b 18.28 (more stable)
OR
(a) In Ni(CO)4 and Fe(CO)5 the oxidation state of Ni and Fe is zero oxidation state.
(b) NH3 acts as a ligand because it has lone pair of electrons, whereas NH4+ does not have lone pair of
electrons.
(c) CN– is an ambidentate ligand because it can coordinate through either the nitrogen or the carbon
atom to central metal ion.
Question 5.
(a) Co (iv) is a good analytical reagent. Explain.
(b) Unlike 4f orbitals, 5f -orbitals can take part in bond formation. Explain.
OR
(a) State with the help of equations how solution of potassium dichromate can be used for strength determination of aqueous iron solutions.
(b) Why all the actinoid metals are attacked by hydrochloric acid but slightly affected by nitric acid?
Answer:
(a) The E° value for Ce+4/Ce+3 is 1.74 V (This value shows that Ce+4 has sufficient kinetic stability due to
which the reaction rate is very slow) which refers that it can oxidise water. Hence, Ce (iv) is a good
analytical reagant.
(b) Though the 5f-subshell resemble 4f-subshell in their angular part of the wave functions (because both 5f& 4f have / = 3) but 5f are not buried as 4f (thus electrons present is 5f orbitals experience less nuclear attraction than electrons present in 4f orbitals) hence, 5f orbitals can take part in bonding to a far greater extent.
OR
(a) Potassium dichromate is a strong oxidising agent and is used as a primary standard in a volumetric analysis done through titration. In acidic solution, its oxidising action can be represented as follow:
Cr2O7– + 14H+ + 6e– → 2Cr3+ + 7H2O
Thus, acidified potassium dichromate will oxidise iodides to iodine, sulphides to sulphur, iron(II) salts to iron(III) etc., and hence can be used to determine their concentrations in solutions of unknown strength.
Cr2O7– + 14H+ + 6Fe2+ → 2Cr3++ 6Fe3+ + 7H2O
(b) Actinoid metals are only slightly affected by nitric acid due to the formation of protective oxide layers on coming in contact with nitric acid, which prevents any further reaction.b
Question 6.
Explain the bonding in coordination compounds in terms of Werner’s postulates.
Answer:
Werner’s postulates explain bonding in coordination compounds as follows:
(a) In a coordination compound a metal exhibits two types of valency namely, primary and secondary valency. Primary valency is satisfied by negative ions while secondary valency is satisfied by both negative and neutral ions. Primary valency corresponds to the oxidation number of the metal ion, whereas the secondary valency refers to the coordination number of the metal ion and is satisfied by ligands.
(b) A metal ion has a definite number of secondary valency around the central atom. Also, this valency project in a specific direction in the space assigned to the definite geometry of the coordination compound.
(c) Primary valency is usually ionisable, while secondary valency is non-ionisable.
(d) The negative ligands satisfy both primary arid secondary valency.
Question 7.
(a) What is the use of Hinsberg test?
(b) How would you achieve aniline to benzonitrile?
(c) Why ethylamine is soluble in water but aniline is insoluble?
Answer:
(a) Hinsberg test is used to distinguish between 10, 2° and 30 amines. Here amines are treated with benzene sulphonyl chloride and excess of KOH.
(b)
(c) Ethylamine when added to water forms intermolecular H-bonds with water. Hence it is soluble in water. But aniline can form H-bonding with water to a very small extent due to the presence of a large hydrophobic -C6H5 group. Hence aniline is insoluble in water.
Question 8.
For a reaction A + B → P, the rate is given by Rate = k[A][B]2.
(a) How is the rate of reaction affected if the concentration of B is doubled?
(b) What is the overall order of reaction if A is present in large excess?
(c) Define order of reaction.
Answer:
(a) For a reaction, A + B → P
R1 = k[A][B]2 ………………………….. (i)
If the concentration of B is doubled,
R2 = k[A][2B]2 ……………………………………..(ii)
On dividing (i) and (ii)
\(\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{k[\mathrm{~A}][\mathrm{B}]^{2}}{k[\mathrm{~A}][2 \mathrm{~B}]^{2}}\)
\(\frac{R_{1}}{R_{2}}=\frac{B^{2}}{4 B^{2}}\)
R2 = 4R1
The rate of reaction will be four times the initial rate.
(b) If the concentration of B is doubled i.e., [B]2 the overall reaction will be two, because if A is present in
large excess, then the reaction will be independent of the concentration of A and will be dependent only on the concentration of B. Order of reaction 2.
(c) If A is present in large excess, then the rate of the reaction will be independent of A and will depend
only on the concentration of B. The overall rate of the reaction will be 2.
Question 9.
(a) Explain what is observed when electric current is passed through a colloidal sol.
(b) Why is the ester hydrolysis slow in the begining and becomes faster after some time?
(c) Why does physisorption decreases with rise in temperature?
OR
Explain what is observed when:
(a) Silver nitrate solution is added to potassium iodide solution.
(b) The size of the finest gold particles increases in a gold sol.
(c) Two oppositely charged sols are mixed in almost equal proportions.
Answer:
(a) On passing electric current through a sol, colloidal particles start moving towards oppositely charged electrode where they lose their charge and get coagulated. This process is known as electrophoresis.
(b) In simple ester hydrolysis reaction that can be written as:
Ester + Water → Acid + Alcohol
The acid i.e. product of the hydrolysis will release W ions in solution which acts as a catalyst (auto- catalysis) for the reaction. Therefore, the slow process of hydrolysis will become faster.
(c) Physissorption is exothermic in nature (ΔH is negative) which means that the rise in temperature favours the reverse process i.e., desorption. Therefore, physisorption decreases with rise in temperature.
OR
A yellow precipitate or coagulated silver iodide is formed.
(b) On dissolution, a large number of gold atoms agitate. Thus, the size of the finest gold sol particles increases in the gold. A large number of atoms gold aggregate together to form species having size in the colloidal range (1-1000 nm).
(c) When two oppositely charged sols are mixed in almost equal proportions, their charges are neutralized resulting in coagulation. This type of coagulation is called mutual coagulation or material coagulation.
Question 10.
(a) Give a method to distinguish between formaldehyde and acetaldehyde.
(b) What happens when:
(a) Acetaldehyde reacts with hydrazine
(b) Acetone reacts with hydroxylamine
Answer:
(a) Formaldehyde and acetaldehyde can be distinguished with the help of iodoform reaction Acetaldehyde bearing methyl carbonyl group when heated with I2 and NaOH, gives a yellow precipitate of iodoform (iodoform test) whereas formaldehyde does not give a yellow precipitate of iodoform.
(b) (i) Acetaldehyde reacts with hydrazine to give hydrazone.
(ii) Acetone reacts with hydroxylamine to give oxime.
Question 11.
The change in concentration of a reactant or product in unit time is called rate of reaction. The reaction 2NO(g) + O2(g) →2NO2(g) was studied by the initial rate method. The following kinetic data was obtained.
(a) What do you understand by rate law?
(b) What is the order of the reaction with respect to NO and O2?
(c) Find out the rate of formation of NO2 when [NO] is 0.1 and [O2] is 0.2 mol L-1.
Or
Calculate the half-life of a first-order reaction from their rate constants given below :
(a) 200 s-1,
(b) 2 min-1,
(c) 4 years-1.
Answer:
(a) Rate law is an experimentally determined expression which relates the rate of reaction with a concentration of the reactants.
(b) Suppose the order of reaction w.rt. NO is x and w.r.t. O2’ it is y, then,
Rate = k NO]x[O2]y
0.096 = k (0.03)x (0.30)y …………………………… (i)
0.384 = k (0.60)x(0.60)y ……………………………… (ii)
0.192 = k (0.30)x (0.60)y’ …………………………………. (iii)
0.768 = k (0.60)x (0.60)y’ ……………………………………… (iv)
Dividing equation (ii), by (i), we get
\(\frac{0.384}{0.096}=\frac{k(0.60)^{x}(0.30)^{y}}{k(0.30)^{x}(0.30)^{y}}\)
4 =2x’or 22=2x or x=2
Dividing (iv) by (ii), we get
\(\frac{0.768}{0.384}=\frac{k(0.60)^{x}(0.60)^{y}}{k(0.60)^{x}(0.30)^{y}}\)
2 = 2y or y = 1
Hence, order or reaction w.r.t. NO = 2
order or reaction w.r.t. O2= 1
(c) Rate k [NO]2[O2]
Rate 11.85 (0.1)2 (02) or rate = 2.37 x 10-2 mol-2Ls-1
Or
Calculate the half-life of a first order reaction from their rate constants given below:
(a) Half life, t1/2 = \(\frac{0.693}{k}\)
= \(\frac{0.693}{200 \mathrm{~s}^{-1}}\) = 3.47 s x 10-3 s
(b) Half life, t1/2 = \(\frac{0.693}{k}\)
= \(\frac{0.693}{2 \min ^{-1}}\)
= 0.35 min (approximately) or 3.5 x 10-1 min
(c) Half life, t1/2 = \(\frac{0.693}{k}\)
= \(\frac{0.693}{4 \text { years }^{-1}}\)
= 0.173 years (approximately) or 1.73 x 10-1 year.
Section – C
Question 12.
Read the passage given below and answer the questions that follow:
The recently discovered photochemical reductive animation and the enantioselective amine synthesis by combined photoredox/enzymatic catalysis is reviewed.
In this mini-review, we discuss recent advances in the area of photo-driven multi-electron transfer with a particular focus on our own work on reductive animation and the enantioselective synthesis of amines by combined photoredox and enzyme catalysis. Polarity-matched hydrogen atom transfer (HAT) between photochemically-generated a-amino alkyl radicals and thiols is a key step in these reactions. A cyclic reaction network comprised of light-driven imine reduction by an Ir-photocatalyst and enantioselective amine oxidation by the enzyme monoamine oxidase (MAO-N-9) was used to obtain enantioenriched amines from imines.
(a) Give the IUPAC name of C6H5NH2.
(b) What is the final product obtained upon reaction of benzylamine with 2 moles of methyl chloride?
(c) Hoffmann bromamide degradation reaction is used for preparation of which type of amine?
(d) Predict the product: 
Or
(e) Predict the product:
Answer:
(a) Benzenamine
(b) N, N-Dimethylphenylmethanamine
(c) It is used to prepare Primary aliphatic amines or aromatic amines.
