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Class 10 Chemistry Chapter 21 Organic Chemistry Test 5
Grade 10 Chemistry chapter 21 Organic Chemistry Test 5
Prepare for the Punjab Board Class 10 Chemistry Examination 2027 with Chapter 21: Organic Chemistry – Class Test No. 5. This test covers textbook topics 21.1 to 21.3 and includes the complete chapter exercise for comprehensive exam preparation. Designed to support students following the new Chemistry syllabus issued by PECTAA Punjab, this test is part of our Punjab Board 2027 Model Papers, Full Unit Model Papers, and Chapter-Wise Test Series. Practice important questions, revise key concepts, assess your understanding, and improve your preparation for the Class 10 annual Chemistry examination. Use this full chapter test to strengthen your knowledge of Organic Chemistry and prepare confidently for board examinations.
Chapter Wise Class Test
T-5
Chapter-21
Organic Chemistry
1.
Organic chemistry is defined as the study of compounds of carbon and hydrogen, which are collectively known as:
(a) Halogens(b) Hydrocarbons(c) Carbonates(d) Oxides
Answer: (b) Hydrocarbons
2.
Open chain or acyclic compounds are also commonly known as:
(a) Aromatic compounds(b) Aliphatic compounds(c) Alicyclic compounds(d) Heterocyclic compounds
Answer: (b) Aliphatic compounds
3.
A benzene ring is characterized by having six carbon atoms with:
(a) Only single bonds(d) Only double bond(b) Three alternate double and single bonds(c) Two alternate double and single bonds
Answer: (b) Three alternate double and single bonds
4.
In a condensed structural formula, which of the following bonds are typically omitted (not drawn)?
(a) Carbon-to-Carbon double bonds (C=C)(b) Carbon-to-Hydrogen bonds (C-H)(c) Carbon-to-Carbon triple bonds (C≡C)(d) All of the above are shown
Answer: (b) Carbon-to-Hydrogen bonds (C-H)
5.
What is the correct condensed structural formula for Ethane?
(a) CH4(b) CH2 = CH2(c) CH3 – CH3(d) HC ≡ CH
Answer: (c) CH3 – CH3
6.
If a molecule of Propane is drawn as a Displayed Formula, how many C–H bonds would be visible?
(a) 3(b) 6(c) 8(d) 10
Answer: (c) 8 (Propane C3H8 has 8 hydrogen atoms, each joined to carbon by one C–H bond.)
7.
Successive members of a homologous series differ in molecular mass by:
(a) 12 units(b) 14 units(c) 16 units(d) 18 units
Answer: (b) 14 units (difference of one CH2 group = 12 + 2)
8.
Which property shows a regular change with the increase in molecular mass?
(a) Chemical properties(b) Preparation methods(c) Physical properties(d) Functional groups
Answer: (c) Physical properties
(P.T.O) ➦
Chapter Wise Class Test
T-5
Chapter-21
Organic Chemistry
2. Write short answers to any six (6) questions.[6 × 2 = 12]
(i)
What is an organic compound?
Organic compounds are compounds of carbon (mostly with hydrogen, and often oxygen, nitrogen, halogens, sulphur) in which carbon atoms are joined by covalent bonds. Examples: methane, ethanol, glucose.
(ii)
Which carbon compounds are studied as inorganic compounds?
Oxides of carbon (CO, CO2), carbonates (e.g. CaCO3), bicarbonates (e.g. NaHCO3), carbides (e.g. CaC2), cyanides and cyanates are studied as inorganic compounds.
(iii)
Why is it necessary to classify the organic compound?
There are millions of organic compounds. Classification arranges them into groups with similar structure and properties, which makes their study easier, systematic and helps predict their behaviour.
(iv)
How organic compounds are broadly classified?
On the basis of carbon chain they are classified into: (1) Acyclic (open chain / aliphatic) compounds and (2) Cyclic (closed chain) compounds. Cyclic compounds are further divided into homocyclic (alicyclic and aromatic) and heterocyclic compounds.
(v)
Define Alicyclic compounds and give their general formula.
Alicyclic compounds are cyclic compounds containing a closed ring of carbon atoms only, with properties similar to aliphatic compounds. General formula (for saturated ones): CnH2n. Example: cyclohexane, C6H12.
(vi)
Why do members of a homologous series have similar chemical properties?
Because all members contain the same functional group, so they react in a similar way.
(vii)
Write any two characteristics of Homologous Series.
(1) All members have the same general formula (e.g. alkanes CnH2n+2).
(2) Successive members differ by a CH2 group (14 units of mass).
(Also: physical properties change regularly; similar chemical properties.)
(2) Successive members differ by a CH2 group (14 units of mass).
(Also: physical properties change regularly; similar chemical properties.)
(viii)
Give two examples of branced chain compounds.
(1) Isobutane (2-methylpropane): (CH3)2CH–CH3
(2) Neopentane (2,2-dimethylpropane): C(CH3)4
(2) Neopentane (2,2-dimethylpropane): C(CH3)4
(ix)
What are Aromatic compounds? Give an example.
Aromatic compounds are cyclic compounds containing a benzene ring (six carbon atoms with three alternate double and single bonds). Example: benzene, C6H6.
Note: Attempt any one question.[5 × 1 = 5]
3.
What is an organic compound? How organic compounds are classified based on the type of carbon chain present in them?
Organic compound: a compound of carbon (usually with hydrogen and other elements) held by covalent bonds.
Classification:
1. Acyclic (open chain/aliphatic): straight or branched chains, e.g. CH3–CH2–CH3.
2. Cyclic (closed chain):
(a) Homocyclic (ring of carbon atoms only): alicyclic (e.g. cyclohexane) and aromatic (e.g. benzene).
(b) Heterocyclic: ring contains carbon and another atom like N, O or S (e.g. pyridine).
Classification:
1. Acyclic (open chain/aliphatic): straight or branched chains, e.g. CH3–CH2–CH3.
2. Cyclic (closed chain):
(a) Homocyclic (ring of carbon atoms only): alicyclic (e.g. cyclohexane) and aromatic (e.g. benzene).
(b) Heterocyclic: ring contains carbon and another atom like N, O or S (e.g. pyridine).
4.
Explain structural Formulas.
A structural formula shows how atoms are arranged and bonded in a molecule. Types:
1. Displayed (full) formula: all atoms and all bonds are drawn (e.g. ethane, H–C–C–H with three H on each carbon).
2. Condensed formula: C–H bonds are omitted, e.g. ethane CH3–CH3, ethene CH2=CH2, ethyne HC≡CH.
3. Skeletal formula: only the carbon skeleton is shown as lines.
1. Displayed (full) formula: all atoms and all bonds are drawn (e.g. ethane, H–C–C–H with three H on each carbon).
2. Condensed formula: C–H bonds are omitted, e.g. ethane CH3–CH3, ethene CH2=CH2, ethyne HC≡CH.
3. Skeletal formula: only the carbon skeleton is shown as lines.
H H
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H – C – C – H = CH3 – CH3
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H H