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CBSE Class 12 Chemistry Term 2 Sample Paper & Solutions

Class 12 Chemistry Term 2 Sample Paper

Central Board of Secondary Education has released CBSE Class 12 Chemistry Term 2 Sample Paper 2021-22. The students must solve CBSE Class 12 Chemistry Term 2 Sample Paper to assess their preparation. Unlike the Term 1 Exam, the CBSE Term 2 examination will be subjective and the CBSE Term 2 question paper will have very short, shot, and case-study-based questions. Thus, students must consolidate their preparation by practicing as many questions as they can. In this article, we have covered CBSE Class 12 Chemistry Term 2 Sample Paper. The students appearing in the Class 12 Term 2 Examination must solve all the questions given on this page and bookmark this page to get all the latest updates.

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Class 12 Chemistry Term 2 Sample Paper and Full Revision

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CBSE Class 12 Chemistry Term 2 Sample Paper & Solutions_40.1

CBSE Term 2 Sample Paper Class 12 Chemistry: Exam Pattern

The students must know the exam patterns before starting their preparation for any exam. The exam pattern gives students a brief about the exam like how many questions will be asked. Class 12 chemistry sample paper is divided into three sections namely Section A, Section B, and Section C. Check the other details given below:

Section A: In this section, very short answer-type questions will be asked. Each question carries 2 marks.

Section B: In this section, short answer-type questions will be asked. Each question carries 3 marks.

Section C: In this section, case-based questions will be asked. Each question carries 5 marks.

Check: CBSE Class 12 History Term 2 Additional Practice Questions

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Class 12 Chemistry Term 2 Sample Paper with Solutions

1. Arrange the following in the increasing order of their property indicated (any 2):

a. Benzoic acid, Phenol, Picric acid, Salicylic acid (pka values). 

Picric acid < salicylic acid < benzoic acid <phenol 

b. Acetaldehyde, Acetone, Methyl tert butyl ketone (reactivity towards NH2OH). 

Methyl tert – butyl ketone < acetone< Acetaldehyde 

c. ethanol, ethanoic acid, benzoic acid (boiling point) 

ethanol <ethanoic acid < benzoic acid (boiling point of carboxylic acids is higher than alcohols due to extensive hydrogen bonding, the boiling point increases with increase in molar mass)

 

  1. Solutions of two electrolytes ‘A’ and ‘B’ are diluted. The Λm of ‘B’ increases 1.5 times while that of A increases 25 times. Which of the two is a strong electrolyte? Justify your answer. Graphically show the behavior of ‘A’ and ‘B’.

CBSE Class 12 Chemistry Term 2 Sample Paper & Solutions_50.1

B is a strong electrolyte. The molar conductivity increases slowly with dilution as there is no increase in number of ions on dilution as strong electrolytes are completely dissociated.  

  1. Give reasons to support the answer:

a. Presence of Alpha hydrogen in aldehydes and ketones is essential for aldol condensation. 

The alpha hydrogen atoms are acidic in nature due to presence of electron withdrawing carbonyl group. These can be easily removed by a base and the carbanion formed is resonance stabilized. 

b. 3 –Hydroxy pentan-2-one shows positive Tollen’s test. 

Tollen’s reagent is a weak oxidizing agent not capable of breaking the C-C bond in ketones . Thus ketones cannot be oxidized using Tollen’s reagent itself gets reduced to Ag.  

  1. Account for the following: 

a. Aniline cannot be prepared by the ammonolysis of chlorobenzene under normal conditions. 

In case of chlorobenzene, the C—Cl bond is quite difficult to break as it acquires a partial double bond character due to conjugation. So Under the normal conditions, ammonolysis of chlorobenzene does not yield aniline.

b. N-ethylethanamine boils at 329.3K and butanamine boils at 350.8K, although both are isomeric in nature. 

Primary and secondary amines are engaged in intermolecular association due to hydrogen bonding between nitrogen of one and hydrogen of another molecule. Due to the presence of three hydrogen atoms, the intermolecular association is more in  primary amines than in secondary amines as there are two hydrogen atoms available for hydrogen bond formation in it.  

c. Acylation of aniline is carried out in the presence of pyridine. 

During the acylation of aniline, stronger base pyridine is added. This done in order to remove the HCl so formed during the reaction and to shift the equilibrium to the right hand side. 

OR 

  1. Convert the following: 

a. Phenol to N-phenylethanamide. 

CBSE Class 12 Chemistry Term 2 Sample Paper & Solutions_60.1

b. Chloroethane to methanamine. 

CBSE Class 12 Chemistry Term 2 Sample Paper & Solutions_70.1

 

c. Propanenitrile to ethanol. 

CBSE Class 12 Chemistry Term 2 Sample Paper & Solutions_80.1

  1. Answer the following questions: 

a. [Ni(H2O)6 ] 2+ (aq) is green in colour whereas [Ni(H2O)4 (en)]2+(aq)is blue in colour , give reason in support of your answer . 

The colour of coordination compound depends upon the type of ligand and dd transition taking place . H2O is weak field ligand , which causes small splitting , leading to the d-d transition corresponding green colour , however due to the presence of ( en ) which ia strong field ligand , the splitting is increased . Due to the change in t2g -eg splitting the colouration of the compound changes from green to blue. 

b. Write the formula and hybridization of the following compound: tris(ethane-1,2–diamine) cobalt(III) sulphate 

Formula of the compound is [Co(H2NCH2CH2NH2 )3 ]2 (SO4 )3 The hybridisation of the compound is: d2sp3

OR  

  1. In a coordination entity, the electronic configuration of the central metal ion is t2g 3 eg 1 

a. Is the coordination compound a high spin or low spin complex? 

As the fourth electron enters one of the eg orbitals giving the configuration t2g 3 eg 1, which indicates ∆o < P hence forms high spin complex. 

b. Draw the crystal field splitting diagram for the above complex. 

CBSE Class 12 Chemistry Term 2 Sample Paper & Solutions_90.1

 

  1. Account for the following:

a. Ti(IV) is more stable than the Ti (II) or Ti(III). 

Ti is having electronic configuration [Ar] 3d2 4s2 . Ti (IV) is more stable as Ti4+ acquires nearest noble gas configuration on loss of 4 e-.  

b. In case of transition elements, ions of the same charge in a given series show progressive decrease in radius with increasing atomic number. 

In case of transition elements, ions of the same charge in a given series show progressive decrease in radius with increasing atomic number. As the new electron enters a d orbital each time the nuclear charge increases by unity. The shielding effect of a d electron is not that effective, hence the net electrostatic attraction between the nuclear charge and the outermost electron increases and the ionic radius decreases.  

c. Zinc is a comparatively a soft metal, iron and chromium are typically hard

Iron and Chromium are having high enthalpy of atomization due to the presence of unpaired electrons, which accounts for their hardness. However, Zinc has low enthalpy of atomization as it has no unpaired electron. Hence zinc is comparatively a soft metal.

 

  1. An alkene ‘A’ (Mol. formula C5H10) on ozonolysis gives a mixture of two compounds ‘B’ and ‘C’. Compound ‘B’ gives positive Fehling’s test and also forms iodoform on treatment with I2 and NaOH. Compound ‘C’ does not give Fehling’s test but forms iodoform. Identify the compounds A, B and C. Write the reaction for ozonolysis and formation of iodoform from B and C. 

Compound A is an alkene, on ozonolysis it will give carbonyl compounds. As both B and C have >C=O group, 

B gives positive Fehling’s test so it is an aldehyde and it gives iodoform test so it is so it has CH3C=O group. This means the aldehyde is acetaldehyde 

C does not give Fehling’s test, so it is a ketone. It gives positive iodoform test so it is a methyl ketone means it has CH3C=O group 

Compound A (C5H10) on ozonlysis gives B (CH3CHO) + C (CH3COR) So “C” is CH3COCH3 

CH3CH=C(CH3)2 (i)O3 (ii) Zn/H3O + CH3CHO + CH3COCH3 

CH3CHO + 2Cu2+ + 5OH- CH3COO- + Cu2O (red ppt) + 3H2O 

CH3COCH3 + 2Cu2+ + 5OH- No reaction

CH3CHO + 3I2 + 3 NaOH CHI3 (yellow ppt) + 3HI + HCOONa 

CH3COCH3 + 3I2 + 3 NaOH CHI3 (yellow ppt) + 3HI + CH3COONa 

A = CH3CH=C(CH3)2 

B = CH3CHO 

C = CH3COCH3

 

  1. Observe the figure given below and answer the questions that follow:

CBSE Class 12 Chemistry Term 2 Sample Paper & Solutions_100.1

a. Which process is represented in the figure? 

electrodialysis 

b. What is the application of this process? 

purification of colloidal solution

c. Can the same process occur without applying electric field? Why is the electric field applied? 

Yes. Dialysis is a very slow process to increase its speed electric field is applied 

 

  1. What happens when reactions: 

a. N-ethylethanamine reacts with benzenesulphonyl chloride. 

When N-ethylethanamine reacts with benzenesulphonyl chloride , N,N-diethylbenzenesulphonamide is formed. 

b. Benzylchloride is treated with ammonia followed by the reaction with Chloromethane. 

When benzylchloride is treated with ammonia , Benzylamine is formed which on reation with Chloromethane yields a secondary amine , N-methylbenzylamine .  

c. Aniline reacts with chloroform in the presence of alcoholic potassium hydroxide.

When aniline reacts with chloroform in the presence of alcoholic potassium hydroxide , phenyl isocyanides or phenyl isonitrile is formed . 

OR  

  1. a. Write the IUPAC name for the following organic compound: CH3 – N—CH2CH3 

CBSE Class 12 Chemistry Term 2 Sample Paper & Solutions_110.1

N-Ethyl-N-methylbenzenamine or N-Ethyl-N-ethylaniline 

b.Complete the following: 

CBSE Class 12 Chemistry Term 2 Sample Paper & Solutions_120.1

Answer

CBSE Class 12 Chemistry Term 2 Sample Paper & Solutions_130.1

  1. Represent the cell in which the following reaction takes place.The value of E˚ for the cell is 1.260 V. What is the value of Ecell ? 

2Al(s) + 3Cd2+ (0.1M)  3Cd (s) + 2Al3+ (0.01M)

Al(s) /Cd2+ (0.1M) // Al3+ (0.01M) /Cd(s) 

2Al(s) + 3Cd2+ (0.1M)  3Cd (s) + 2Al3+ (0.01M) 

Ecell = Eocell -0.059 log [Al3+] 2 n [Cd2+] 3 

Ecell= 1.26 – 0.059 log (0.01)2 6 (0.1)3  

= 1.26 – 0.059 (-1) 6 

= 1.26+0.009 

= 1.269 V

 

  1. a. Why are fluorides of transition metals more stable in their higher oxidation state as compared to the lower oxidation state? 

The ability of fluorine to stabilize the highest oxidation state is attributed to the higher lattice energy or high bond enthalpy.

b. Which one of the following would feel attraction when placed in magnetic field: Co2+ , Ag+ ,Ti4+ , Zn2+ 

Co2+ has three unpaired electrons so it would be paramagnetic in nature, hence Co2+ ion would be attracted to magnetic field.

c. It has been observed that first ionization energy of 5 d series of transition elements are higher than that of 3d and 4d series, explain why? 

The transition elements of 5d series have intervening 4f orbitals. There is greater effective nuclear charge acting on outer valence electrons due to the weak shielding by 4f electrons. Hence first ionisation energy of 5 d series of transition elements are higher than that of 3d and 4d series. 

OR  

  1. On the basis of the figure given below, answer the following questions: 

CBSE Class 12 Chemistry Term 2 Sample Paper & Solutions_140.1

a. Why Manganese has lower melting point than Chromium? 

Manganese is having lower melting point as compared to chromium , as it has highest number of unpaired electrons , strong interatomic metal bonding , hence no delocalisation of electrons

b. Why do transition metals of 3d series have lower melting points as compared to 4d series? 

There is much more frequent metal – metal bonding in compounds of the heavy transition metals i.e 4d and 5d series , whixh accounts for lower melting point of 3d series.  

c. In the third transition series, identify and name the metal with the highest melting point

Tungsten 

 

  1. Read the passage given below and answer the questions that follow. 

Are there nuclear reactions going on in our bodies? 

There are nuclear reactions constantly occurring in our bodies, but there are very few of them compared to the chemical reactions, and they do not affect our bodies much. All of the physical processes that take place to keep a human body running are chemical processes. Nuclear reactions can lead to chemical damage, which the body may notice and try to fix. The nuclear reaction occurring in our bodies is radioactive decay. This is the change of a less stable nucleus to a more stable nucleus. Every atom has either a stable nucleus or an unstable nucleus, depending on how big it is and on the ratio of protons to neutrons. The ratio of neutrons to protons in a stable nucleus is thus around 1:1 for small nuclei (Z < 20). Nuclei with too many neutrons, too few neutrons, or that are simply too big are unstable. They eventually transform to a stable form through radioactive decay. Wherever there are atoms with unstable nuclei (radioactive atoms), there are nuclear reactions occurring naturally. The interesting thing is that there are small amounts of radioactive atoms everywhere: in your chair, in the ground, in the food you eat, and yes, in your body. 

The most common natural radioactive isotopes in humans are carbon-14 and potassium-40. Chemically, these isotopes behave exactly like stable carbon and potassium. For this reason, the body uses carbon-14 and potassium-40 just like it does normal carbon and potassium; building them into the different parts of the cells, without knowing that they are radioactive. In time, carbon-14 atoms decay to stable nitrogen atoms and potassium-40 atoms decay to stable calcium atoms. Chemicals in the body that relied on having a carbon-14 atom or potassium-40 atom in a certain spot will suddenly have a nitrogen or calcium atom. Such a change damages the chemical. Normally, such changes are so rare, that the body can repair the damage or filter away the damaged chemicals. 

The natural occurrence of carbon-14 decay in the body is the core principle behind carbon dating. As long as a person is alive and still eating, every carbon-14 atom that decays into a nitrogen atom is replaced on average with a new carbon-14 atom. But once a person dies, he stops replacing the decaying carbon-14 atoms. Slowly the carbon-14 atoms decay to nitrogen without being replaced, so that there is less and less carbon-14 in a dead body. The rate at which carbon-14 decays is constant and follows first order kinetics. It has a half – life of nearly 6000 years, so by measuring the relative amount of carbon-14 in a bone, archeologists can calculate when the person died. All living organisms consume carbon, so carbon dating can be used to date any living organism, and any object made from a living organism. Bones, wood, leather, and even paper can be accurately dated, as long as they first existed within the last 60,000 years. This is all because of the fact that nuclear reactions naturally occur in living organisms. 

(source: The textbook Chemistry: The Practical Science by Paul B. Kelter, Michael D. Mosher and Andrew Scott states)

a. Why is Carbon -14 radioactive while Carbon -12 not? (Atomic number of Carbon: 6) 

Ratio of neutrons to protons is 2.3: 1 which is not the stable ratio of 1:1 

b. Researchers have uncovered the youngest known dinosaur bone, dating around 65 million years ago. How was the age of this fossil estimated? 

Age of fossils can be estimated by C-14 decay. All living organisms have C-14 which decays without being replaced back once the organism dies

c. Which are the two most common radioactive decays happening in human body? 

carbon-14 atoms decay to stable nitrogen atoms and potassium-40 atoms decay to stable calcium  

d. Suppose an organism has 20 g of Carbon -14 at its time of death. Approximately how much Carbon -14 remains after 10,320 years? (Given antilog 0.517 = 3.289) 

t = 2.303/ k log (Co/Ct) 

Co = 20 g Ct = ? 

t = 10320 years k = 0.693/6000 (half-life given in passage) 

substituting in equation: 

10320 = 2.303 / (0.693/6000) log 20/ Ct 

0.517 = log 20 / Ct anlilog (0.517) = 20/Ct 3.289 = 20/Ct 

Ct = 6.17 g  

OR 

  1. Approximately how old is a fossil with 12 g of Carbon -14 if it initially possessed 32 g of Carbon -14? (Given log 2.667 = 0.4260)  

t = 2.303/ k log (Co/Ct) 

Co = 32 g Ct = 12 

t = ? k = 0.693/6000 (half life given in passage) 

substituting in equation: 

t = 2.303 / (0.693/6000) log 32/ 12 

t = 2.303 x 60000 /0.693 log 2.667 

t = 2.303x6000x0.4260 /0.693  

= 8494 years

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