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Assertion & Reason Test: Organic Chemistry - Some basic Principles & Techniques


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6 Questions MCQ Test Chemistry Class 11 | Assertion & Reason Test: Organic Chemistry - Some basic Principles & Techniques

Assertion & Reason Test: Organic Chemistry - Some basic Principles & Techniques for NEET 2023 is part of Chemistry Class 11 preparation. The Assertion & Reason Test: Organic Chemistry - Some basic Principles & Techniques questions and answers have been prepared according to the NEET exam syllabus.The Assertion & Reason Test: Organic Chemistry - Some basic Principles & Techniques MCQs are made for NEET 2023 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Assertion & Reason Test: Organic Chemistry - Some basic Principles & Techniques below.
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Assertion & Reason Test: Organic Chemistry - Some basic Principles & Techniques - Question 1

Direction: In the following questions a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.

Assertion (A) : Simple distillation can help in separating a mixture of propan-1-ol (boiling point 97°C) and propanone (boiling point 56°C).

Reason (R) : Liquids with a difference of more than 20°C in their boiling points can be separated by simple distillation.

Detailed Solution for Assertion & Reason Test: Organic Chemistry - Some basic Principles & Techniques - Question 1 Simple distillation technique is used for separating components that have a difference of more than 20C in their boiling points. Thus, the statement 'simple distillation can help in separating a mixture of propan-1-ol (boiling point 97C) and propanone (boiling point 56C)' is correct. Thus, the assertion is correct.
Assertion & Reason Test: Organic Chemistry - Some basic Principles & Techniques - Question 2

Direction: In the following questions a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.

Assertion (A): Energy of resonance hybrid is equal to the average of energies of all canonical forms.

Reason (R): Resonance hybrid cannot be presented by a single structure.

Detailed Solution for Assertion & Reason Test: Organic Chemistry - Some basic Principles & Techniques - Question 2 Canonical structures always have more energy than resonance hybrid. Resonance hybrids are always more stable than any of the canonical structures. The delocalization of electrons lowers the orbitals energy and gives stability.
Assertion & Reason Test: Organic Chemistry - Some basic Principles & Techniques - Question 3

Direction: In the following questions a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.

Assertion (A): Pent-1-ene and pent-2-ene are position isomers.

Reason (R): Position isomers differ in the position of functional group or substituent.

Detailed Solution for Assertion & Reason Test: Organic Chemistry - Some basic Principles & Techniques - Question 3 Both A and R are correct and R is the correct explanation of A. When two or more compounds differ in the position of substituent atom or functional group on the carbon skeleton then it is position isomerism. Double bond is a functional group whose position varies.
Assertion & Reason Test: Organic Chemistry - Some basic Principles & Techniques - Question 4

Direction: In the following questions a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.

Assertion (A): All the carbon atoms in H2C=C=CH2 are sp2 hybridized

Reason (R): In this molecule, all the carbon atoms are attached to each other by double bonds.

Detailed Solution for Assertion & Reason Test: Organic Chemistry - Some basic Principles & Techniques - Question 4 Hybridization of C can be calculated by counting p bonds and σ bonds present on C atom. If C has 3σ bonds, it is sp2 hybridized. If C has 2σ bonds, it is sp hybridized.
Assertion & Reason Test: Organic Chemistry - Some basic Principles & Techniques - Question 5

Direction: In the following questions a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.

Assertion (A): Sulphur present in an organic compound can be estimated quantitatively by Carious method.

Reason (R): Sulphur is separated easily from other atoms in the molecule and gets precipitated as light yellow solid.

Detailed Solution for Assertion & Reason Test: Organic Chemistry - Some basic Principles & Techniques - Question 5 Sulphur is estimated by Carius method in the form of white precipitate of BaSO4 on heating with fuming and BaCl2. If light yellow solid is obtained means impurities are present. It is filtered, washed and then dried to get pure BaSO4.
Assertion & Reason Test: Organic Chemistry - Some basic Principles & Techniques - Question 6

Direction: In the following questions a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.

Assertion (A): Components of a mixture of red and blue inks can be separated by distributing the components between stationary and mobile phases in paper chromatography.

Reason (R): The colored components of inks migrate at different rates because paper selectively retains different components according to the difference in their partition between the two phases.

Detailed Solution for Assertion & Reason Test: Organic Chemistry - Some basic Principles & Techniques - Question 6 In paper chromatography, a chromatography paper is used. It contains water in it, which acts as the stationary phase. A strip of chromatography paper spotted at the base with ink is suspended in a suitable solvent. Solvent acts as the mobile phase.

The solvent rises up the paper by capillary action and flows over the spot. The paper selectively retains different components according to their differing partition in two phases.

Hence, components of ink will megrate at different rates and are separated.

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