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National Institute of Open Schooling (NIOS)
Class XII
Chemistry (313)
Compounds of Carbon Containing Halogens (Haloalkanes and Haloarenes) (Set 01 (Foundation Level))

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National Institute of Open Schooling (NIOS) Class XII Chemistry (313) Set 01 (Foundation Level) English Medium 100% Free Practice Test

National Institute of Open Schooling (NIOS) Class XII Chemistry (313): Compounds of Carbon Containing Halogens (Haloalkanes and Haloarenes) Online Test (Set 01 (Foundation Level)) | TargetExams

Compounds of Carbon Containing Halogens (Haloalkanes and Haloarenes) — Set 01 (Foundation Level)

Take free online mock test for National Institute of Open Schooling (NIOS) Class XII Chemistry (313) chapter 'Compounds of Carbon Containing Halogens (Haloalkanes and Haloarenes)'. 20 MCQs, 20 minutes, detailed solutions & score analysis.

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20 MCQs

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Instant Scorecard Step-by-Step Solutions NIOS Self-Learning Material (SLM) Pattern

Chapter Study Guide & In-Depth Notes

Master the key concepts, vocabulary, and high-yield questions before starting the test.

3 Sections

Chapter Overview & Core Concepts: Compounds of Carbon Containing Halogens (Haloalkanes and Haloarenes)

Class 12 Study Guide

Mastering Compounds of Carbon Containing Halogens (Haloalkanes and Haloarenes) in Class 12 provides critical subject intuition, analytical skills, and examination readiness.

Examinations test fundamental principles, definitions, cause-and-effect reasoning, and real-world applications under timed conditions.

Core Moral & Key Takeaway:

Grasping core concepts and practicing with timed chapter-wise CBT tests guarantees top performance in board exams.

Key Terminology, Definitions & Principles

Glossary & Concepts

Term / Formula	Type	Definition / Meaning	Explanation / Notes	Example / Usage
Core Concept	Principle	Primary concept governing Compounds of Carbon Containing Halogens (Haloalkanes and Haloarenes).	मुख्य सिद्धांत	"Understand and apply the principle accurately."
Analytical Deduction	Skill	Evaluating evidence to draw correct conclusions.	तार्किक विश्लेषण	"Examine question clues systematically."

Key Features & Devices Explored:

Systematic Elimination

Eliminate distractor choices before selecting the best answer.

Pacing & Accuracy

Complete each question within 60 seconds to maintain steady momentum.

High-Yield Revision Points & Solved Board Exam Q&A

Revision Notes

⚡ High-Yield Exam Takeaways:

Review all key textbook terms, definitions, and formulas.
Pay attention to units, boundary values, and key keywords in questions.
Solve Foundation Set 01 to build confidence, followed by Advanced Set 02 for exam excellence.

Frequently Asked Questions & Model Answers:

Q1 How should students prepare for Compounds of Carbon Containing Halogens (Haloalkanes and Haloarenes) exams?

Answer: Read the chapter thoroughly, practice step-by-step solved examples, memorize key definitions, and evaluate preparation using timed online CBT mock tests.

Sample Questions Preview

Free Sneak Peek

Here is a sneak peek of the questions included in this test set. Test your preparation by viewing the answer and detailed explanation before starting the timed exam.

Why is thionyl chloride ($SOCl_2$) in the presence of pyridine considered the best and most preferred laboratory reagent for preparing pure chloroalkanes from primary and secondary alcohols?

प्राथमिक और द्वितीयक ऐल्कोहॉलों से शुद्ध क्लोरोऐल्केन तैयार करने के लिए पिरीडीन की उपस्थिति में थायोनिल क्लोराइड ($SOCl_2$) को सबसे अच्छा और सबसे पसंदीदा प्रयोगशाला अभिकर्मक क्यों माना जाता है?

Both by-products, sulfur dioxide ($SO_2$) and hydrogen chloride ($HCl$), are escapable gases, leaving behind pure alkyl chloride without requiring tedious fractional distillationदोनों उप-उत्पाद, सल्फर डाइऑक्साइड ($SO_2$) और हाइड्रोजन क्लोराइड ($HCl$), पलायन करने वाली गैसें हैं, जो बिना किसी जटिल प्रभाजी आसवन के शुद्ध ऐल्किल क्लोराइड छोड़ती हैं

Thionyl chloride reduces the activation energy of the reaction to absolute zeroथायोनिल क्लोराइड अभिक्रिया की संक्रियण ऊर्जा को बिल्कुल शून्य कर देता है

It converts tertiary alcohols into stable organometallic polymersयह तृतीयक ऐल्कोहॉलों को स्थिर कार्बधात्विक बहुलकों में परिवर्तित करता है

The reaction yields an equal molar ratio of alkyl fluoride and alkyl iodideअभिक्रिया से ऐल्किल फ्लोराइड और ऐल्किल आयोडाइड का समान मोलर अनुपात प्राप्त होता है

Correct Answer: Option A

Explanation:

According to NCERT, the reaction of alcohols with thionyl chloride: $R-OH + SOCl_2 xrightarrow{\text{pyridine}} R-Cl + SO_2uparrow + HCluparrow$ is preferred over $PCl_3$ or $PCl_5$ because both by-products, $SO_2$ and $HCl$, are gases and escape into the atmosphere (or are absorbed by pyridine/scrubber), yielding the pure chloroalkane directly without the need for complicated purification procedures.

NCERT के अनुसार, थायोनिल क्लोराइड के साथ ऐल्कोहॉल की अभिक्रिया: $R-OH + SOCl_2 xrightarrow{\text{pyridine}} R-Cl + SO_2uparrow + HCluparrow$ को $PCl_3$ या $PCl_5$ की तुलना में प्राथमिकता दी जाती है क्योंकि दोनों उप-उत्पाद, $SO_2$ और $HCl$, गैसें हैं और वायुमंडल में निकल जाती हैं (या पिरीडीन द्वारा अवशोषित हो जाती हैं), जिससे जटिल शुद्धिकरण प्रक्रियाओं की आवश्यकता के बिना सीधे शुद्ध क्लोरोऐल्केन प्राप्त होता है।

Chloromethane ($CH_3Cl$) has a higher experimental dipole moment ($mu = 1.860\text{ D}$) than fluoromethane ($CH_3F$, $mu = 1.847\text{ D}$), despite Fluorine being significantly more electronegative than Chlorine. What is the reason for this anomalous order?

फ्लोरोमेथेन ($CH_3F$, $mu = 1.847\text{ D}$) की तुलना में क्लोरोमेथेन ($CH_3Cl$) का प्रायोगिक द्विध्रुव आघूर्ण अधिक ($mu = 1.860\text{ D}$) होता है, भले ही फ्लोरीन क्लोरीन की तुलना में काफी अधिक विद्युत ऋणात्मक हो। इस असामान्य क्रम का क्या कारण है?

Fluoromethane exists as a cyclic trimer in the vapor phaseफ्लोरोमेथेन वाष्प अवस्था में एक चक्रीय ट्राइमर के रूप में मौजूद होता है

Dipole moment is the product of partial charge and bond length ($q \times d$); the much longer $C-Cl$ bond length ($178\text{ pm}$ vs $139\text{ pm}$) outweighs the smaller charge differenceद्विध्रुव आघूर्ण आंशिक आवेश और बंध लंबाई ($q \times d$) का गुणनफल है; बहुत लंबी $C-Cl$ बंध लंबाई ($178\text{ pm}$ बनाम $139\text{ pm}$) छोटे आवेश अंतर से अधिक प्रभावी हो जाती है

Chlorine possesses vacant d-orbitals which emit extra dipole radiationक्लोरीन में रिक्त d-कक्षक होते हैं जो अतिरिक्त द्विध्रुवीय विकिरण उत्सर्जित करते हैं

Fluoromethane has zero net covalent characterफ्लोरोमेथेन में शून्य शुद्ध सहसंयोजक लक्षण होता है

Correct Answer: Option B

Explanation:

Dipole moment is defined as $mu = q \times d$, where $q$ is the magnitude of partial charge and $d$ is the bond distance. Although the $C-F$ bond is more polar than the $C-Cl$ bond due to fluorine's higher electronegativity, the $C-Cl$ bond length ($178\text{ pm}$) is substantially longer than the $C-F$ bond length ($139\text{ pm}$). The greater distance term in $CH_3Cl$ overcompensates for the smaller charge term, resulting in $mu(CH_3Cl) = 1.860\text{ D} > mu(CH_3F) = 1.847\text{ D}$.

द्विध्रुव आघूर्ण को $mu = q \times d$ के रूप में परिभाषित किया जाता है, जहाँ $q$ आंशिक आवेश का परिमाण है और $d$ बंध दूरी है। यद्यपि फ्लोरीन की उच्च विद्युत ऋणात्मकता के कारण $C-F$ बंध $C-Cl$ बंध की तुलना में अधिक ध्रुवीय होता है, लेकिन $C-Cl$ बंध लंबाई ($178\text{ pm}$) $C-F$ बंध लंबाई ($139\text{ pm}$) से काफी लंबी होती है। $CH_3Cl$ में अधिक दूरी वाला पद छोटे आवेश वाले पद की भरपाई से अधिक करता है, जिसके परिणामस्वरूप $mu(CH_3Cl) = 1.860\text{ D} > mu(CH_3F) = 1.847\text{ D}$ होता है।

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समय प्रबंधन एवं गति नियंत्रण

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Mastery of the Elimination Technique

विकल्प निष्कासन विधि में दक्षता

Top rankers in competitive exams rarely compute every problem blindly. They spot subtle traps, evaluate extreme values, and eliminate 2 out of 4 distractors within seconds. Our 4-option sets teach critical distractor analysis.

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सटीकता और नकारात्मक अंकन से सुरक्षा

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Syllabus Concepts Evaluated in this Chapter

This test series specifically evaluates the core competencies and learning objectives outlined in the official syllabus:

Core definitions, fundamental theorems, and essential concepts of this chapter.
Step-by-step mathematical reasoning and deductive problem-solving.
Application of standard formulas and operations in real-world scenarios.
Critical examination of common exam pitfalls and misleading options.
Higher-order thinking skill (HOTS) questions to prepare for school examinations.
Speed and accuracy mastery under standard timed CBT exam conditions.

Exam Rules & Test Taking Guidelines

Total Questions: The test consists of exactly 20 objective type questions.
Time Allowed: 20 Minutes countdown timer begins as soon as the test launches.
Marking Scheme: +1.0 mark is awarded for each correct answer.
Zero Penalty: There is 0.0 negative marking for incorrect or unattempted questions.
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Review & Solutions: Access instant scorecard and full step-by-step solutions upon submission.

Frequently Asked Questions (FAQs)

Yes! Foundation Practice Set 01 for every chapter in National Institute of Open Schooling (NIOS) Class XII Chemistry (313) is 100% free to attempt. Students can test their knowledge without any payment or hidden fees.

This page is dedicated to English Medium. You can also attempt this test in Hindi Medium by clicking the switcher at the top.

Absolutely. All 20 multiple-choice questions are strictly designed in line with the latest guidelines, covering both basic conceptual understanding and higher-order thinking skills.

The test contains 20 MCQs to be solved in 20 minutes. Each question carries +1.0 mark for a correct answer, and there is zero (0.0) negative marking for incorrect answers.

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Compounds of Carbon Containing Halogens (Haloalkanes and Haloarenes) (Set 01 (Foundation Level))

20 MCQs • 20 Mins • 100% Free Practice Test

National Institute of Open Schooling (NIOS) Class XII Chemistry (313): Compounds of Carbon Containing Halogens (Haloalkanes and Haloarenes) Online Test (Set 01 (Foundation Level)) | TargetExams

Chapter Study Guide & In-Depth Notes

Chapter Overview & Core Concepts: Compounds of Carbon Containing Halogens (Haloalkanes and Haloarenes)