Chemistry Answer Key for Class 12  Term 2 

Chemistry Answer Key 2022 for Class 12 Term 2 Exam: CBSE Chemistry Class 12 Term 2 Exam has been concluded on 07th May 2022 (Saturday). Undoubtedly, each student after attempting the exam wants to know which questions were correctly attempted by him/her and in which question he/she missed the chance. Here have discussed the detailed Class 12 Chemistry Answer Key for the Term 2 Exam in brief to give you an idea of how many marks you are going to score in the Chemistry Term 2 Exam. No need to go anywhere else and there just stay with us and we have covered all the questions one by one with their correct answers here after the exam is over. 

Chemistry Answer Key Class 12

CBSE conducted Class 12 Chemistry Board Term 2 Exam in offline mode. In Term 2 the exam is divided into three sections with 12 questions for which the exam pattern that has been followed is as follows-  

1. Section A consists of 1 to 3 questions carrying 2 marks each [Very Short Answer Questions]. 

2. Section B consists of 4 to 11 questions carrying 3 marks each [Short Answer Questions]. 

3. Section C consists of 1 question carrying 5 marks [Case Based Question].

Our faculty will be discussing all the questions asked in the Term 2 Chemistry Exam, you just have to tally with those you have attempted & calculate your approximate marks. 

CBSE Class 12 Chemistry Answer Key Term 2

Term 2 Chemistry Answer Key has been discussed in this article after the conduction of Class 12 Chemistry Term 2 Exam on 07th May 2022. 

CBSE Class 12 Chemistry Answer Key Term2
Exam Conducting BodyCentral Board of Secondary Education
ClassCBSE Class 12
Exam NameChemistry
ExaminationTerm 2
Post CategoryAnswer Key
Exam Date07th May 2022 (Saturday)
Chemistry Answer Key Class 12 (Unofficial )07th May 2022
Official Websitehttps://www.cbse.nic.in/

Class 12 Chemistry Term 2 Answer Key 2022

As per CBSE Class 12 Chemistry Syllabus, In Term II, questions from different formats i.e case-based, situation-based, open-ended- short answer, long answer types will be asked. Term-II comprises of remaining units i.e. Electrochemistry, Chemical Kinetics, Surface Chemistry, d -and f -Block Elements, Coordination Compounds, Aldehydes, Ketones and Carboxylic Acids, and Amines. Cross-check your Chemistry Term 2 Solution for each question from below section. 

Question 1- Arrange the following compounds in the increasing order of their property indicated.

a) Acetaldehyde, Benzaldehyde, Acetophenone, Acetone

Solution- Acetaldehyde, Acetone, Benzaldehyde, Acetophenone

(b) (CH3)2CHCOOH, CH3CH2CH(Br)COOH, CH3CH(Br)CH2COOH (Acidic Strength)

Solution- (CH3)2CHCOOH, CH3CH(Br)CH2COOH ,  CH3CH2CH(Br)COOH, 

(c) CH3CH2OH, CH3CHO, CH3COOH (Boiling point)

Solution- CH3COOH, CH3CH2OH, CH3CHO

Question 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. 

Answer- Yes, we can do it by Kohlrausch's law which states that the equivalent conductivity of an electrolyte at infinite dilution is equal to the sum of the conductances of the anions. 

Question 3- Write reasons for the following statements : 

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

Solution- 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.

Solution- The difference between an aldehyde and a ketone is the presence of a hydrogen atom attached to the carbon-oxygen double bond in the aldehyde. Ketones don't have that hydrogen. The presence of that hydrogen atom makes aldehydes very easy to oxidize (i.e., they are strong reducing agents)

Question 4- Write reasons for the following: 

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

Solution- 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. 

Solution- Nitration is carried out in an acidic medium. In an acidic medium, aniline is protonated to give anilinium ion (which is meta-directing). For this reason, aniline on nitration gives a substantial amount of m-nitroaniline.

(iii) Amines behave as nucleophiles.

Solution- 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.

OR 

Question 4- How will you carry out the following conversions:

i) Nitrobenzene to Aniline 

Solution- 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.

ii) Ethanamide to Methanamine

Solution- Methenamine can be prepared from Ethanamine, firstly by using HNO2HNO2 which gives ethyl alcohol which is oxidized to acetaldehyde and then to ethanoic acid. Now ethanoic acid is heated with ammonia to form ethanamide. And finally, ethanamide is treated with B{r_2}/KOH to form methenamine.

iii) Ethanenitrile to Ethanamine 

Solution- Nitroethane can be converted into corresponding ethylamine by doing the reduction reaction of nitroethane. By passing hydrogen gas in the presence of finely divided nickel, platinum or palladium, nitroethane can be reduced to ethylamine.

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

Solution- Sp³d² → t2g³ eg²

(ii) [Fe(CN)6]³- is weakly paramagnetic whereas [Fe(CN)6]4- is diamagnetic. Give reason to support this statement [Atomic number- 26].

Solution [Fe(CN)6]³- is weakly paramagnetic due to the presence of one unpaired electron as there will be d5 configuration. 

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

Solution- 3 ions 

OR

Question 5- (i) Calculate the spin only magnetic moment of the complex [CoF613-. (Atomic no. of Co = 27)

Solution- 4.9

(ii) Write the IUPAC name of the given complex- [CrCl2(H20)4]Cl

Solution- dichloro tetra aqua chromium(III) chloride

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

[Fe(H20)6]3+, [Fe(C204)3]3–

Solution- [Fe(C204)3]3– due to chelation effect. 

Question 6- (i) Which ion amongst the following is colourless and why?

Ti4+, Cr3+, V3+ (Atomic number of Ti = 22, Cr = 24, V = 23)

Solution- Ti4+ will be colorless because all eletcrons have been extracted from it. 

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

Solution- 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.

Solution- 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.

Question 7- A compound ‘A’ (C2H4O) on oxidation gives 'B' (C2H4O2). ‘Aundergoes 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. 

Chemistry Class 12 Answer Key

Question 9- (a) Write equations involved in the following reactions :

(i) Ethanamine reacts with acetyl chloride.

Chemistry Class 12 Answer Key

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

Chemistry Class 12 Answer Key

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

Chemistry Class 12 Answer Key

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

Solution- N-Ethyl-N-Methyl Ethanamine

(ii) Write the equations for the following: 

(1) Gabriel phthalimide synthesis

Chemistry Class 12 Answer Key Term 2

 

(2) Hoffmann bromamide degradation 

Chemistry Class 12 Answer Key 2022

Chemistry Class 12 Term 2 Question Paper [QP Code- 56/4/1]

Chemistry Class 12 Term 2 Question Paper [QP Code- 56/4/3]

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Chemistry Answer key Class 12: FAQs

Ans. At Careerpower, our faculty has discussed the complete Chemistry Answer Key term 2 exam.

Ans. In Term 2 the exam is divided into three sections with 12 questions

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