CBSE Sample Papers for Class 12 Chemistry Term 2 Set 2 with Solutions

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CBSE Sample Papers for Class 12 Chemistry Standard Term 2 Set 2 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.
Arrange the following in the order of their property indicated (Any two):
(a) Formic acid is heated with Tollen’s reagent?
(b) Acetaldehyde is treated with concentrated NaHSO₃ solution?
(c) 2-Pentanone is reacted with iodine in presence of sodium hydroxide?
Answer:
(a) Formic acid contains both aldehyde (-CHO) and acidic group (-COOH). Due to presence of aldehydic group, it acts as reducing ageñt and reduces Tollen’s reagent to metallic silver (silver mirror test) according to the following reaction:
HCOOH + 2[Ag(NH₃)₂]^{2 +}+ 2OH^– → 2Ag + CO₂ + 2H₂O + 4NH₃
(b) Sodium hydrogen sulphite adds to aldehydes and ketones to form the addition products.

(c) 2-Pentanone would undergo iodoform reaction with I₂ and NaOH as it is a methyl ketone, to give
CHI₃ (iodoform) and corresponding sodium carboxylate.

Question 2.
Give reason to support your answer:
(a) Why does sky looks blue?
(b) Colloidal medicines are more effective. Why?
Answer:
(a) The dust particles along with water suspended in the air reflect the sunlight coming from the sun and this is called scattering of light. The blue colour is scattered the most by the dust and gas particles present in the atmosphere hence, sky appears to be blue.

(b) Medicines are more effective in colloidal state because colloids have a larger surface area which enhances their absorption in the body and therefore increases the effectiveness of medicines given in colloidal form. Thus, colloidal medicines easily gets assimilated, absorbed and digested in our body.

Question 3.
Answer the following:
(a) In a first-order reaction the concentration of reactant is reduced from 0.6 mol L^-1 to 0.2 mol L^-1 in 5 minutes. Calculate the rate constant for the reaction.
(b) What would be the unit of rate equation in case of gaseous reaction?
Answer:
(a) For a first-order reaction:
Rate constant k = \(\frac{2.303}{\left(t_{2}-t_{1}\right)} \log \left(\frac{\left[R_{1}\right]}{\left[R_{2}\right]}\right)\)
Putting the values given: k = \(\frac{2.303}{5 \mathrm{~min}} \times \log \left(\frac{0.6 \mathrm{~mol} \mathrm{~L}^{-1}}{0.2 \mathrm{~mol} \mathrm{~L}^{-1}}\right)\)
= 0.4606 x 0.477
= 0.22 min^-1
(b) Unit of rate equation in case of gaseous reaction wouki be atm s^-1.

Section – B

Question 4.
Account for the following:
(a) Why zinc, cadmium and mercury are not regarded as transition metals?
(b) Why Zr and Hf occur in nature together?
(c) In which industry AgBr is used?
OR
(a) Why Cu²⁺ ion is stable in aqueous solution
(b) Trivalent Lanthanoid ions are coloured. Why?
(c) Cul₂ is not known. Why?
Answer:
(a) The elements such as Zn, Cd, and Hg are not transitioning elements because of their electronic configuration. The orbitals of these elements are completely filled both in their ground state as well as in their general oxidation state. Therefore, these elements do not transition elements.

(b) As a consequence of lanthanoid contraction, the atomic radii of (160 pm) and Hf (159 pm) is almost similar. That is why Zr and Hf occur in nature together.

(c) AgBr is widely used in photography industry.
OR
(a) In an aqueous medium, Cu²⁺ is more stable because although energy is required to remove one electron from Cu⁺ to Cu²⁺, high hydration energy of Cu²⁺ compensates for it.
Therefore, Cu⁺ ion in an aqueous solution is unstable. It disproportionates to give Cu²⁺ and Cu.
2Cu⁺ (aq) → Cu²⁺(aq) + Cu(s)

(b) The lanthanide metals are silvery-white but the trivalent lanthanide ions show different colours. Colour of the ions depend on the number of unpaired electrons because the elements with (x)f electrons often have a similar colour to those of (14-x)f electrons.

The colour of lanthanide ions is due to the presence of partially filled f orbitals. As a result, it is possible to absorb certain wavelength from the visible region of the spectrum. This results in transitions from one 4f orbital to another 4f orbital known asf-f transition.
(c) CuI₂ is not known because Cu²⁺ oxidises I^– to I₂ as follows:

Question 5.
(a) Name the isomerism shown by the complex [Co(Pn)₂Cl₂]+ and [Co(tn)₂Cl₂]+.
(b) What type of bond is present between metal and carbon in metal carbonyls?
(c) [FeF₆]^3- has five unpaired electrons while [Fe(CN)₆]^3- has only one unpaired electron. Explain.
OR
Answer the following questions:
(a) Which type of compounds show linkage isomerism? What would be the structural formula for linkage isomer of [CO(NH₃)5(NO₂)]Cl₂?
(b) How are the inner orbital complexes formed?
(c) What is a chelate ligand? Give one example.
Answer:
(a) Ligand isomerism.
(b) It deals with both σ and π bonds. Here M-C σ’ bond is formed by the donation of lone pair of electrons from carbonyl carbon to the vacant orbital of metal. M- C π bond is formed by the back donation of electrons from a filled d-orbital of metal into the vacant antibonding π* orbital of carbon monoxide.
(c) In [FeF₆]^3-, Fe(III) has got 3d⁵ system. As is a weak field ligand, according to crystal field theory (CFT) electronic arrangement will be as follows.

Hence, it has 5 unpaired electrons.
In [Fe(CH)₆], Fe (Ill) has got 3d⁵ system. As is a strong field ligand, according to CFT electronic
arrangement will be as follows.

Hence, it has only one unpaired ligand.
OR
(a) Linkage isomerism arises in a coordination compound containing ambidentate ligand. The structural
formula for linkage isomer of [Co(NH₃)₅(NO₂) ]Cl₂ would be [Co(NH₃)₅(ONO)]Cl₂.

(b) Inner orbital complexes are coordination compounds composed of a central metal atom having hybridization of the atomic orbitals including d orbitals of inner shell and s, p orbitals from the outer shell. In other words, the central metal atom of these complexes uses inner shell d orbitals for the hybridization of atomic orbitals which means these complexes are formed due to the participation of (n – 1)d orbitals. Therefore, these d orbitals are in a lower energy level than s and p orbitals. The most common hybridisation of the metal atom in inner orbital complexes is d²sp³ and their shape is octahedral.

(c) Chelating ligand is a ligand which is mostly attached to a central metal ion by bonds that are from two or more donor atoms. For example, ethane-1, 2-diamine.

Question 6.
Consider a first-row transition metal M. M²⁺ ion has a 3d⁶ configuration. Its M(II) chloride is dissolved in water to give hexahydrated green coloured complex A. On progressive addition of bidentate ligand, ethane-1, 2-diamine(en), three different compounds are formed:

S.No.	en: M	Compound	Colour of compound
1.	1:1	B	Pale blue
2.	2:1	C	Blue/Purple
3.	3:1	D	Violet

What is the formula for compound A to D? Provide the sequence of reactions for the formation of compounds B to D.
Answer:
The M is Ni here (3d⁸ configuration), the compound A is [Ni(H₂O)₆]²⁺ obtained after dissolving NiCl₂ in water.
The complete set of reactions are:
[Ni(H₂O)₆]²⁺ (aq) + en (aq) → [Ni(H₂O)₄(en)]²⁺ (aq) + 2H₂O
(Ethane-1-2-diamine) (Pale blue)

Question 7.
Account for the following:
(a) What happens to Δ H and ΔS during adsorption of a gas on solid?
(b) Gelatin is what type of sol?
(c) Give one example of peptising agents.
Answer:
(a) During the adsorption of a gas on the surface of a solid there is always a decrease in residual forces of the surface i.e., there is decrease in surface energy which appears as heat. Adsorption therefore is an exothermic process. Thus, ΔH for adsorption is always negative. Again when a gas is adsorbed, the freedom of movement of its molecules also becomes restricted. This leads to decrease in the entropy of the gas after adsorption. Therefore, AS is negative.
(b) Lyophilic type of sol.
(c) Sodium chloride (NaCl) is an example of a peptising agent.

Question 8.
Account for the following:
(a) What is electrochemical series? The cell potential of mercury cell remains constant during its lifetime. Why?
(b) Define cell constant, what is its unit?
(c) Limiting the molar conductivity of an electrolyte cannot be determined experimentally. Why?
Answer:
(a) Electrochemical series can be defined as the arrangement of various elements in the order of increasing value of their standard reduction potential values. The cell potential of a Mercury cell remains constant throughout it’s life because the overall reaction does not involve any ion in the solution whose concentration may change.

(b) Cell constant of a conductivity cell is defined as the ratio of the distance between the electrodes divided by the area of cross-section of the electrodes. It is denoted by b and can be expressed as:
∴ Cell constant = b = \(\frac{l \mathrm{~cm}}{a \mathrm{~cm}^{2}}\)
Thus, it is expressed in unit cm^-1 or m^-1.
(c) In weak electrolytes the conductivity of the solution increases very slowly with a dilution of solution and goes on increasing upto infinity. Hence, it can not be measured experimentally.

Question 9.
(a) Complete the following reactions:
(i) CH₃CH₂CHO + Cl₂ →
(ii) C₆H₅ – COCH₃ + NH₂OH →
(b) Which type of aldeydes undergo Cannizzaro’s reaction?
(c) Explain why the boiling points of aldehydes and ketones are lower than the corresponding alcohols?
OR
(a) How will you convert:
(i) Benzoyl chloride to benzaldehyde.
(ii) Propanone to 2- propanol.
(b) What happens when CH₃CHO is treated with K₂Cr₂O₇ in presence of H₂SO₄?
(c) Ester of which acid is used in perfumery?
Answer:
(a)
(i) CH₃CH₂CHO + Cl₂→ CH₃CH₂COCl +HCl

(b) Aromatic and aliphatic aldehydes which do not contain hydrogen undergoes Cannizzaro’s reaction e.g.,- HCHO, C₆H₅CHO etc.
(c) The presence of intermolecular hydrogen bonding increases the molecular association of a particular molecule and thereby increases the effective mass of that molecule which in turn is directly proportional to the boiling point. Thus, the absence of intermolecular H-bonding in aldehydes and ketones causes the lowering in the boiling points of aldehydes and ketones than the corresponding alcohols (which involve intermolecular hydrogen bonding due to the presence of -OH group).
OR

(b) When CH₃CHO is treated with K₂Cr₂O7 in presence of H₂SO₄, methanoic acid is formed. The
reaction can be represented as follows:

(c) Esters of benzoic acids are used in the perfumery industry. For example, ethyl benzoate is a component of some fragrances and artificial flavours.

Question 10.
Complete the following equation:
(a) CH₃CHO + C₆H₅NHNH₂ →


Answer:

Question 11.
(a) Determine the order of reaction and also determine the units of rate constant:

(b) The activation energy of a reaction is zero. Will the rate constant of the reaction depend upon temperature? Give reason.
(c) Write rate law expression and order of the following reaction:
Step 1. H₂O₂ +I^– → H₂O + IO^– (slow)
Step 2. H₂O₂ +I^– → H₂O + IO^– + O₂ (fast)
OR
Write three points of differences between rate of reaction and rate constant.
Answer:
(a) Order of Reaction: First order; Unit of rate constant: s^-1.
(b) According to Arrhenius equation:
k = Ae^-E_a/RT
If E_a = O, then wf= A. Frequency factor (a) does not depend upon temperature, therefore, rate constant
and rate does not depend on temperature.
(c) Rate K[H₂O₂] [I^–] because step 1 is rate-determining step and,
Order =1+1=2
OR

Rate of Reaction	Rate Constant
1. It depends upon the concentration of the reactant.	It is independent of the concentration of the reactant.
2. It is expressed in terms of consumption of reactants or formation of product per unit time.	It is proportionality constant in differential form in rate law or rate equation.
3. It generally decreases with the progress of the reaction,	It does not depend on the progress of the reaction.

Section – C

Question 12.
Read the passage given below and answer the questions that follow?
Properties of Amines
Due to presence of lone pair of electrons in ‘N’ atoms the amines are basic in nature because these lone pair of electrons can be donated to electron-deficient compound. Aliphatic amines are stronger bases then ammonia. The basicity increases with the increase in number of alkyl groups attached to the ‘N’ atom. However, the observed basic nature of amines is 2°>1°>3°. Primary and secondary amines are soluble in water due to hydrogen bonding. Solubility decreases with increase in number of ‘C’ atoms.
(a) Boiling points of amines are less than the alcohols of similar molecular mass. Why?
(b) Amines behave as nucleophiles. Why?
(c) Which reaction is used for the test of primary aromatic amines?
(d) Lower aliphatic amines are soluble in water. Why?
OR
(d) Amines are less soluble in water than alcohols. Why?
Answer:
(a) Amines are less polar than alcohols because electronegativity difference between N-H is less than that of O-H. Intermolecular H-bond in amines is weaker than that in alcohols. Hence, boiling point of amines are less than the alcohols of similar molar mass.

(b) Amines contain a lone pair of electrons on nitrogen atom. Therefore they behave as nucleophiles.
(c) Diazo test is used to detect primary aromatic amines.
(d) Lower aliphatic amines are soluble in water because they can form hydrogen bonds with water molecules.
OR
(d) Alcohol molecules form H-bond with water molecules more strongly than amines and therefore solubility of amine is less than that of alcohols.