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CBSE Class 12 Chemistry Term 2 Answer key 2022

Class 12 Chemistry Term 2 Answer key

Class 12 Chemistry Term 2 Answer Key: The Central Board of Secondary Education is all set to conduct CBSE Class 12 Chemistry Term 2 Exam 2022 today on 7th May 2022. We have covered the CBSE Class 12 Chemistry Term 2 Answer Key on this page in detail. The students will get 2 hours to solve the CBSE Class 12 Term 2 question paper. The chemistry exam will conclude at 12:30 pm, after completion of the examination we will update the CBSE Answer Key of Chemistry Term 2 on this page. The students appearing in the examination can match their answers with the unofficial CBSE Class 12 Chemistry Term 2 Answer Key 2022 prepared by the expert facilities of Adda247. 

CBSE Class 12 Term 2 Chemistry Answer Key

CBSE Class 12 Answer Key Chemistry Term 2 given on this page is error-free and all the answers to CBSE Class 12 Chemistry Term 2 Answer Key 2022 are framed by the faculties who have years of experience in teaching Class 12 Chemistry. Below we have given the highlights of CBSE Class 12 Answer Key Chemistry Term 2. The students must check brief information of Term 2 CBSE Class 12 Chemistry Answer Key 2022 provided in the table below: 

CBSE Class 12 Answer Key Chemistry Term 2
Exam Conducting Body Central Board of Secondary Education
Exam & Subject Name CBSE Class 12 Chemistry
Category Answer Key
Exam Date 7th May 2022, Saturday 
Unofficial Answer Key 7th May 2022, Saturday 
Official Answer Key To be notified
Official Website


CBSE Class 12 Chemistry Term 2 Answer Key 2022: Exam Pattern 

As per the CBSE Class 12 Term 2 sample paper, the CBSE Class 12 Chemistry Term 2 question paper consists of three sections. All the sections hold different marks and different types of questions. The board will ask the very short answer type, short answer type, and case-based questions in the CBSE Class 12 Term 2 Chemistry Examination. Check out the detailed section-wise pattern listed below: 

Section A: This section consists of very short answer-type questions of 2 marks each.

Section B: This section consists of short answer-type questions of 3 marks each.

Section C: This section consists of case-based questions of 5 marks each.

CBSE Class 12 Chemistry Answer Key- CBSE Paper Solution 2022


CBSE Class 12 Term 2 Chemistry Exam Analysis 2022

Here we have given the complete CBSE Class 12 Term 2 Chemistry Exam Analysis 2022. In CBSE Class 12 Term 2 Chemistry Exam Analysis 2022, we will cover the difficulty level of the exam, out of syllabus questions, and mistakes in the question paper. After completion of the CBSE Term 2 Chemistry Exam, the students must match their responses with the Term 2 Chemistry Answer Key provided on this page and must check the CBSE Class 12 Term 2 Chemistry Exam Analysis 2022 given below: 

CBSE Class 12 Term 2 Chemistry Answer Key 2022: Paper Code-56/4/1

Section A

Q.1: Arrange the following compounds in the increasing order of their property indicated.

i) Acetaldehyde, Benzaldehyde, Acetophenone, Acetone(Reactivity towards HCN)

Answer. Acetaldehyde<Acetone<Benzaldehyde<Acetophenone

iii) CH3CH2OH, CH3CHO, CH3COOH (Boiling point)


Q.2: In a plot of m against the square root of concentration (C12) for strong and weak electrolytes, the value of limiting molar conductivity of a weak electrolyte cannot be obtained graphically. Suggest a way to obtain this value. Also state the related law, if any.

Yes, we can do it by Kohlrausch’s law.

Kohlrausch’s law:  a statement in physical chemistry: the migration of an ion at infinite dilution is dependent on the nature of the solvent and on the potential gradient but not on the other ions present.

Q.3: Write reasons for the following statements : 

(i) Benzoic acid does not undergo Friedel-Crafts reaction.

No, benzoic acid does not undergo Friedel Craft reaction because the carboxylic group is deactivating and the Lewis acid catalyst and carboxylic group are bonded.

(ii) Oxidation of aldehydes is easier than that of ketones.

oxidation of aldenyde is easier than ketone due to presence of H-atoms linked to carbonyl group carbon which is absent in ketones.

Section B

Q.4: Write reasons for the following: 

(i) Ethylamine is soluble in water whereas aniline is insoluble. 

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.

(ii) Amino group is o- and p-directing in aromatic electrophilic substitution reactions, but aniline on nitration gives a substantial amount of m-nitroaniline. 

Nitration is carried out in an acidic medium. In an acidic medium, aniline is protonated to give anilinium ion (which is meta-directing).

CBSE Class 12 Chemistry Term 2 Answer key 2022_40.1

For this reason, aniline on nitration gives a substantial amount of m-nitroaniline.

(iii) Amines behave as nucleophiles.

A nucleophile is a substance that is attracted to, and then attacks, a positive or slightly positive part of another molecule or ion. All amines contain an active lone pair of electrons on the very electronegative nitrogen atom. It is these electrons that are attracted to positive parts of other molecules or ions.


b) How will you carry out the following conversions:

i) Nitrobenzene to Aniline 

Nitrobenzene is reduced to aniline by Sn and concentrated HCl. Instead of Sn, Zn or Fe also can be used. Aniline salt is given from this reaction. Then aqueous NaOH is added to the aniline salt to get released aniline. This reaction is called nitrobenzene reduction.

CBSE Class 12 Chemistry Term 2 Answer key 2022_50.1
  • Benzene is a clear, colourless, highly flammable and volatile, liquid compound.
  • Aniline is a yellowish to brownish oily liquid with a musty fishy odour organic compound.

ii) Ethanamide to Methanamine

CBSE Class 12 Chemistry Term 2 Answer key 2022_60.1

React ethanamine with nitrous acid to form an azo compound, which further reacts with water to form ethanol, which on oxidising gives ethanoic acid. After treating with an excess of ammonia the ethanoic acid becomes ethanamide, which on further reacting with Bomine and a strong base (Hoffmann bromamide degradation reaction) to form methenamine.

iii) Ethanenitrile to Ethanamine 

it’s simple
reduction of nitriles with lithium aluminium hydride or catalytic hydrogenation produce primary amines. the reaction is

CH3C≡N + H2/Ni OR LiAlH4 →  CH3CH2NH2

Q.5(a)(i) Write the electronic configuration of d5 on the basis of crystal field splitting theory if Δ0 < P.

When Δo < P, it is weak field and high spin situation’. As a result one electron entered in eg orbital and 3 electrons in t2g.

CBSE Class 12 Chemistry Term 2 Answer key 2022_70.1

(ii) [Fe(CN)6]³- is weakly paramagnetic whereas [Fe(CN)6]4- is diamagnetic. Give reason to support this statement [Atomic no. Fe = 26].

In both CBSE Class 12 Chemistry Term 2 Answer key 2022_80.1and CBSE Class 12 Chemistry Term 2 Answer key 2022_90.1, Fe exists in the +3 oxidation state i.e., in d5 configuration.

CBSE Class 12 Chemistry Term 2 Answer key 2022_100.1

Since CN is a strong field ligand, it causes the pairing of unpaired electrons. Therefore, there is only one unpaired electron left in the d-orbital.

CBSE Class 12 Chemistry Term 2 Answer key 2022_110.1


CBSE Class 12 Chemistry Term 2 Answer key 2022_120.1

On the other hand, H2O is a weak field ligand. Therefore, it cannot cause the pairing of electrons. This means that the number of unpaired electrons is 5.


CBSE Class 12 Chemistry Term 2 Answer key 2022_130.1

Thus, it is evident that CBSE Class 12 Chemistry Term 2 Answer key 2022_80.1is strongly paramagnetic, while CBSE Class 12 Chemistry Term 2 Answer key 2022_90.1is weakly paramagnetic.

(iii) Write the number of ions produced from the complex [Co(NH3)6]C12 in solution. 

The given complex can be written as [Co(NH3)6]Cl2. Thus, [Co(NH3)6]+ along with two Cl− ions are produced.


Q.(b)(i) Calculate the spin only magnetic moment of the complex [CoF613-. (Atomic no. of Co = 27)

Given ion is  with the atomic no. 27.

So, electronic configuration of  is  [Ar] 
              3d                               4s
No. of unpaired  s =3
So, the spin only magnetic moment is given by,
=       =  = = BM
=3.87 BM
(ii) Write the IUPAC name of the given complex: [CrCl2(H20)4]Cl
The correct I.U.P.A.C. name for  complex is tetraaquadichlorochromium(III) nitrate.

(iii) Which out of the two complexes is more stable and why

[Fe(H20)6]3+, [Fe(C204)3]3–

[Fe(C204)3]3– is more stable out of two complexes. The central metal ion is Fe3+ and C2O4 2– is negative bi-dentate ligand which forms more stable complex than neutral or monodentate ligand.

Q.6 (i) Which ion amongst the following is colourless and why?

Ti4+, Cr3+, V3+

(Atomic number of Ti = 22, Cr = 24, V = 23)

Ti is colourless because it is having no unpaired ee- for excitation to higher energy level and it is colourless.

(ii) Why is Mn²+ much more resistant than Fe²+ towards oxidation?

Mn2+ is much more resistant than Fe2+ towards oxidationAs Mn2+ has stable configuration hence it is resistant towards oxidation. while in Fe2+ electronic configuration is 3d6 so it can lose one electron to give stable configuration 3d5.

(iii) Highest oxidation state of a metal is shown in its oxide or fluoride only. Justify the statement.

The highest oxidation state of a metal is exhibited in its oxide or fluoride only. This is because fluorine (F) and oxygen (O)are the most electronegative elements and the highest oxidation state shown by any transition element is +8.

Q.7 A compound ‘A’ (C2H4O) on oxidation gives ‘B’ (C2H4O2). ‘A’ undergoes an Iodoform reaction to give yellow precipitate and reacts with HCN to form the compound ‘C’. ‘C’ on hydrolysis gives 2-hydroxypropanoic acid. Identify the compounds ‘A’, ‘B’ and ‘C’. Write down equations for the reactions involved. 

CBSE Class 12 Chemistry Term 2 Answer key 2022_160.1

Q.9(a) Write equations involved in the following reactions :

(i) Ethanamine reacts with acetyl chloride.

CBSE Class 12 Chemistry Term 2 Answer key 2022_170.1

(ii) Aniline reacts with bromine water at room temperature. 

CBSE Class 12 Chemistry Term 2 Answer key 2022_180.1

(iii) Aniline reacts with chloroform and ethanolic potassium hydroxide.

When aniline reacts with chloroform and alcoholic  it gives an offensive smelling liquid i.e., phenyl isocyanide as product. It is called as isocyanide test. It is given by aliphatic and aromatic primary amines.

CBSE Class 12 Chemistry Term 2 Answer key 2022_190.1

Q.9 (b)  (i) Write the IUPAC name for the following organic compound:  (CH3CH2)2NCH3

it is a tertiary amine, and the parent chain is ethanamine. IUPAC name is N – Ethyl – N- methyl ethanamine.

(ii) Write the equations for the following: 

(I) Gabriel phthalimide synthesis

Gabriel Phthalimide Synthesis Mechanism has 3 steps. The Synthesis is used to get primary amines from primary alkyl halides and is named after the German scientist Siegmund Gabriel.

The reaction has been generalized for applications in the alkylation of sulfonamides and imides & their deprotection in order to obtain amines. Alkylation of ammonia is quite inefficient, therefore it is substituted with phthalimide anion in the Gabriel synthesis.

(II) Hoffmann bromamide degradation 

When an amide is treated with bromine in an aqueous or ethanolic solution of sodium hydroxide, degradation of amide takes place leading to the formation of primary amine. This reaction involving degradation of amide and is popularly known as Hoffmann bromamide degradation reaction. The primary amine thus formed contains one carbon less than the number of carbon atoms in that amide.

RCONH2 +Br2 + 4NaOH

R-NH2 + Na2CO3 + 2NaBr + 2H2O

Q.11(a) i) Silver atom has completely filled d-orbitals in its ground state, it is still considered to be a transition element. Justify the statement. 

Silver (Ag) belongs to group 11 of d-block and its ground state electronic configuration is 4d10 5s1. It shows an oxidation state of +2 in its compounds like AgO and AgF2 in which its electronic configuration is d9 so it is a transition element.

ii) Why are E° values of Mn and Zn more negative than expected? 

Negative E° values of Mn2+ and Zn2+ are because of the stabilities of half-filled (3d5 : Mn2+)and fullyfilled (3d10 : Zn2+) configuration respectively. Ni2+ ion has higher E°  value due to highest negative enthalpy of hydration.

iii) Why do transition metals form alloys?

Transition metals have very similar atomic sizes. One metal can easily replace the other metal from its lattice to form solid solution (alloy). Transition metals are miscible with one another in the molten state. The molten state solution of two or more transition metals on cooling forms alloy.

CBSE Class 12 Chemistry Term 2 Answer Key With Marking Scheme

1. (a) Picric acid < salicylic acid < benzoic acid <phenol
(b) Methyl tert – butyl ketone < acetone< Acetaldehyde
(c) ethanol <ethanoic acid < benzoic acid (boiling point of carboxylic acids is higher than alcohols due to extensive hydrogen bonding , boiling point increases with increase in molar mass)

2. 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.

CBSE Class 12 Chemistry Term 2 Answer key 2022_200.1

(a) 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) 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.

4 a) 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) 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) 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.


CBSE Class 12 Chemistry Term 2 Answer key 2022_210.1

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

a)As the fourth electron enters one of the eg orbitals giving the configuration t2g3 eg,which indicates ∆o < P hence forms high spin complex.

b) CBSE Class 12 Chemistry Term 2 Answer key 2022_220.1

6 (a) 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. 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) 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.

7. 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

8 (a)electrodialysis
(b)purification of colloidal solution
(c)Yes. Dialysis is a very slow process to increase its speed electric field is applied

9 (a)When N-ethylethanamine reacts with benzenesulphonyl chloride , N,N-diethylbenzenesulphonamide is formed.

b)When benzylchloride is treated with ammonia , Benzylamine is formed which on creation with Chloromethane yields a secondary amine, N-methylbenzylamine .
c)When aniline reacts with chloroform in the presence of alcoholic potassium hydroxide , phenyl isocyanides or phenyl isonitrile is formed .

(I) N-Ethyl-N-methylbenzenamine or N-Ethyl-N-ethylaniline

ii) CBSE Class 12 Chemistry Term 2 Answer key 2022_230.1

10. 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 – 0.059/6 log (0.01)2/(0.1)3

= 1.26 – 0.059 (-1)/6

= 1.26+0.009

= 1.269 V

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

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

(c) 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.


a)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) 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) Tungsten

12. (a) Ratio of neutrons to protons is 2.3: 1 which is not the stable ratio of 1:1
(b)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) carbon-14 atoms decay to stable nitrogen atoms and potassium-40 atoms decay to stable calcium
(d)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
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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CBSE Class 12 Chemistry Term 2 Answer Key 2022: FAQs

Q. Where can I get the Class 12 Term 2 Chemistry Answer Key 2022?

On this page, you will get the Class 12 Term 2 Chemistry Answer Key 2022, check your responses with the answer key, and calculate your marks.

Q. Does the CBSE publish an official Answer Key for Class 12 Term 2 Chemistry?

The board has not been informed yet. When the board will publish the CBSE Class 12 Term 2 Chemistry official answer key, we will update the same here.

Q. What are the units covered in CBSE Class 12 Term 2 Chemistry Exam 2022?

In CBSE Class 12 Term 2 Chemistry Exam 2022, the questions will be asked from the remaining 7 units. The units are Electrochemistry, Chemical Kinetics, Surface Chemistry, d -and f -Block Elements, Coordination Compounds, Aldehydes, Ketones and Carboxylic Acids, and Amines.

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