SOLUTIONS
2024 BOARD EXAM
CBSE CLASS 12-PCB CHEMISTRY Board Paper 2024 — Set 4
2024
CHEMISTRY
CLASS 12-PCB
CBSE EXAMINATION PAPER-2024
CHEMISTRY
(Solved)
General Instructions :
Read the following instructions carefully and follow them :
- This question paper contains 16 questions. All questions are compulsory.
- This question paper is divided into 3 sections.
- Section A – questions number 1 to 3 are case based questions
- Section B – questions number 4 to 15 are multiple choice questions
- Section C – questions number 16 to 16 are very short answer
- There is no overall choice given in the question paper. However, an internal choice has been provided in few questions.
- Use of calculator is NOT allowed.
Section A
20. The nature of bonding, structure of the coordination compound can be explained to some extent by valence bond theory. The central metal atom/ion makes available a number of vacant orbitals equal to its coordination number. The appropriate atomic orbitals (s, p and d) of the metal hybridise to give a set of equivalent orbitals of definite geometry such as square planar, tetrahedral, octahedral and so on. A strong covalent bond is formed only when the orbitals overlap to the maximum extent. The d-orbitals involved in the hybridisation may be either inner d-orbitals i.e. (n—l) d or outer d-orbitals i.e. nd. The complexes formed are called inner orbital complex (low spin complex) and outer orbital complex (high spin complex) respectively. Further, the complexes can be paramagnetic or diamagnetic in nature. The drawbacks of this theory are that this involves number of assumptions and also does not explain the colour of the complex.
Answer the following questions :
Section B
Which of the following acids represents Vitamin C?
The correct answer is Ascorbic acid, as it is the chemical name for Vitamin C, which is vital for various bodily functions, including the maintenance of healthy skin, blood vessels, bones, and cartilage.
Rosenmund reduction is used for the preparation of Aldehydes. The catalyst used in this reaction is.
The correct answer is Pd-BaSO₄. In the Rosenmund reduction, palladium on barium sulfate (Pd-BaSO₄) is used as a catalyst to selectively reduce acyl chlorides to aldehydes in the presence of hydrogen.
Which alkyl halide from the given options will undergo SN1 reaction faster?
The correct option is (CH₃)₃C-Br, which is a tertiary alkyl halide. SN1 reactions generally proceed more quickly with tertiary substrates due to the stability of the carbocation formed during the reaction. In contrast, primary and secondary alkyl halides do not stabilize the carbocation as effectively, making (CH₃)₃C-Br the fastest to undergo an SN1 reaction.
From the elements of 3d series given below, which element shows the maximum number of oxidation states?
Manganese shows the maximum number of oxidation states in the 3d series, with a total of seven oxidation states ranging from -3 to +7, due to its electronic configuration and the ability to lose or share its 3d and 4s electrons.
Identify the tertiary amine from the following:
The correct answer is CH₃–N(CH₃)–CH₃, which is a tertiary amine. Tertiary amines have the structure R₃N, where R represents alkyl groups. In this case, the nitrogen atom is bonded to three methyl groups (CH₃), which classifies it as tertiary. The other options represent primary or secondary amines based on their structures.
Nucleophilic addition of Grignard reagent to ketones followed by hydrolysis with dilute acids forms
The correct answer is 'Tertiary alcohol' because when a Grignard reagent adds to a ketone, it forms a tertiary alcohol after hydrolysis. According to the context provided, nucleophilic addition to ketones yields a tertiary alcohol, while primary alcohols are produced from aldehydes.
In a given graph of zero order reaction, the slope and intercept are:
The correct option is 'Slope = -k, Intercept = [R]₀'. In a zero-order reaction, the concentration of the reactant [R] decreases linearly over time, so the graph of [R] vs. time has a slope equal to -k (where k is the rate constant) and the y-intercept is the initial concentration [R]₀.
Assertion (A): p-nitrophenol is less acidic than phenol. Reason
(R): Nitro group is electron withdrawing and helps in the stabilization of p-nitrophenoxide ion.
Assertion (A) is false, but Reason (R) is true. p-Nitrophenol is actually more acidic than phenol due to the strong electron-withdrawing effect of the nitro group, which stabilizes the p-nitrophenoxide ion. This contradicts the assertion.
Assertion (A): Benzoic acid does not undergo Friedel–Crafts reaction. Reason
(R): Carboxyl group is deactivating and the catalyst aluminium chloride gets bonded to the carboxyl group.
Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). While it's true that the carboxyl group is a deactivating group that prevents Friedel-Crafts reactions, the explanation that aluminium chloride bonds to the carboxyl group is inaccurate. The main reason for the lack of reaction is the electron-withdrawing nature of the carboxyl group, which decreases the electrophilicity of the aromatic ring.
Assertion (A): Fructose is a reducing sugar.
Reason (R): Fructose does not reduce Fehling solution and Tollen's reagent.
Assertion (A) is true, but Reason (R) is false. Fructose is indeed a reducing sugar as it can reduce Fehling's solution and Tollen's reagent under certain conditions, despite it being a ketose, which may not react as easily as aldoses.
The general electronic configuration of d-block elements is :
The correct option is (n-1)d¹⁰ ns¹⁻² because d-block elements, which include transition metals, typically have their outermost electrons filling the d subshell. The general configuration indicates that the d-subshell can hold a maximum of 10 electrons while the s-subshell can hold 2, leading to the configuration of (n-1)d¹⁰ ns¹⁻².
Consider the following reaction :
Section C
Step 1: Calculate freezing point depression, ΔTf = 273.15 K - 272.07 K = 1.08 K
Step 2: Use formula for molecular mass M = (Kf * w2 * 1000) / (ΔTf * w1)
Step 3: Substitute values: M = (1.86 * 18 * 1000) / (1.08 * 200) = (33480) / (216) = 155 g/mol
Answer: Molecular mass of the solute is 155 g/mol.
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