Chemistry SCERT Questions

 

Unit 1: Nomenclature of Organic Compounds and Isomerism

The embedded exercises in this unit primarily involve naming structures or drawing formulae based on IUPAC rules.

I. Activity: Functional Group Identification
Complete the table:

Functional group	Name of functional group	Common name
−OH	Hydroxyl	Alcohol
−COOH	Carboxyl	Carboxylic acid
−CHO	Aldehydic	Aldehyde
C=O	Keto	Ketone
−O–R	Alkoxy	Ether
−F,−Cl,−Br,−I	Halo	Halo compounds

II. Let us assess...
Question 5: i) CH₃–CH₂–O–CH₂–CH₃ ii) CH₃–CH₂–CH₂–CH₂–OH
a) What type of isomerism do these compounds exhibit?
b) Write the structural formula of the metamer of compound (i).
Answer:
a) They have the same molecular formula (C₄H₁₀O) but different functional groups (Ether and Alcohol). They exhibit Functional isomerism.
b) Metamers have the same molecular formula but different alkyl groups attached to either side of the functional group (O or C=O). Structural formula of the metamer of (i) (Ethoxyethane): CH₃–O–CH₂–CH₂–CH₃ (Methoxypropane).

Question 6: i) CH₃–CH₂–CH₂–CH₂–CHO ii) CH₃–CH₂–CH₂–CO–CH₃
a) What is the IUPAC name of the first compound?
b) These two compounds are isomers. Why?
c) What type of isomerism do these compounds exhibit?
d) Write the structural formula of the position isomer of the second compound.
Answer:
a) The main chain has 5 carbon atoms including the aldehyde group (CHO). IUPAC name: Pentanal.
b) They are isomers because they have the same molecular formula (C₅H₁₀O) but different structural formulae and hence different chemical and physical properties.
c) They have the same molecular formula but different functional groups (Aldehyde in (i), Ketone in (ii)). They exhibit Functional isomerism.
d) Compound (ii) is Pentan-2-one (Keto group at C₂). A position isomer would be Pentan-3-one (Keto group at C₃). Structural formula: CH₃–CH₂–CO–CH₂–CH₃.

Question 7: Examine the compounds given below and identify the isomeric pairs. What type of isomerism is shown by each pair?
a) Methoxypropane
b) 2,3–Dimethylbutane
c) Propan–1–ol
d) Ethoxyethane
e) Propan–2–ol
f) Hexane
Answer:

1. Chain Isomers: Hexane (C₆H₁₄) and 2,3–Dimethylbutane (C₆H₁₄). (Pair b and f).
2. Position Isomers: Propan–1–ol (C₃H₈O) and Propan–2–ol (C₃H₈O). They have the same functional group but differ in its position. (Pair c and e).
3. Functional Isomers: (Based on common molecular formulae for functional isomers: C₃H₈O corresponds to Propan-1-ol, Propan-2-ol, and Methoxyethane (CH₃−O−CH₂−CH₃).)
4. Metamers: Methoxypropane (CH₃−O−CH₂CH₂CH₃, C₄H₁₀O) and Ethoxyethane (CH₃CH₂−O−CH₂CH₃, C₄H₁₀O). (Pair a and d).

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Unit 2: Chemical Reactions of Organic Compounds

I. Let us assess...
Question 1:
a) In which of the following situation is methane converted to chloromethane? (i) Chlorine + sunlight
b) Write the name of such types of reactions.
Answer:
a) (i) Chlorine + sunlight.
b) Substitution reactions.

Question 2:
a) How many hydrogen molecules are required to convert CH≡CH (ethyne) into C₂H₆ (ethane)?
b) Write the chemical equation of the reaction.
c) To which category does this chemical reaction belong?
Answer:
a) Two hydrogen molecules are required (C₂H₂ + 2H₂ → C₂H₆).
b) H–C≡C–H + 2H₂ → CH₃–CH₃.
c) Addition reactions.

Question 4:
a) Which of the given polymers is used to coat the inner surface of cookware? (Polythene, polyvinyl chloride, teflon).
b) What is the monomer of this polymer?
Answer:
a) Teflon (Polytetrafluoroethene).
b) Monomer: Tetrafluoroethene (CF₂=CF₂).

Question 5:
a) Which among the following is a condensation polymer? (Polyvinyl chloride, nylon 66, teflon).
b) What are the monomers of nylon 66?
Answer:
a) Nylon 66.
b) Monomers: Adipic acid (HOOC–(CH₂)₄–COOH) and Hexamethylenediamine (NH₂–(CH₂)₆–NH₂).

Question 6:
C₁₂H₂₂O₁₁ + H₂O --(A)--> C₆H₁₂O₆(B) + C₆H₁₂O₆(C)
C₆H₁₂O₆ --(D)--> 2C₂H₅OH + 2CO₂
a) Identify A,B,C and D in the given chemical reactions.
b) What is wash?
c) How is rectified spirit obtained from wash?
d) What is the purpose of denaturing rectified spirit?
Answer:
a) A = Invertase (enzyme); B = Glucose; C = Fructose; D = Zymase (enzyme).
b) Wash is the 8%–10% ethanol solution obtained after the fermentation of molasses.
c) Rectified spirit (95.6% ethanol) is obtained when wash is subjected to fractional distillation.
d) Denaturing rectified spirit involves adding toxic substances (like methanol or pyridine) to ethanol to prevent its misuse as a beverage.

Question 7:
a) How is ethanoic acid prepared industrially?
b) 5%–8% ethanoic acid is called ...................
Answer:
a) Ethanoic acid is prepared industrially by treating methanol (CH₃OH) with carbon monoxide (CO) in the presence of a catalyst.
b) 5%–8% ethanoic acid is called vinegar.

Question 8:
CH₃–COOH + CH₃–OH --(conc. H₂SO₄)--> CH₃–COO–CH₃ + H₂O
a) What is the name of this chemical reaction?
b) What is the name of the ester formed?
c) Write any two uses of esters.
Answer:
a) Esterification.
b) The ester formed is Methyl ethanoate (from ethanoic acid and methanol).
c) Esters are used to make artificial perfumes and juices because they possess the fragrance of flowers and fruits.

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Unit 3: Periodic Table and Electron Configuration

I. Let us assess...
Question 1: The element X having 3 shells belongs to group 17.
a) Write the subshell electron configuration of this element.
b) To which block does this element belong?
c) What is its period number?
d) Write the molecular formula of the compound formed when X reacts with an atom of element Y which belongs to the third period and has one electron in its p subshell.
Answer:
a) Element X is in period 3 (3 shells) and group 17 (p-block). It must have 7 outermost electrons (2 in s and 5 in p subshells). Subshell electron configuration: 1s²2s²2p⁶3s²3p⁵. (Cl)
b) p-block (since the last electron is added to the p subshell).
c) 3 (The highest shell number is 3).
d) Element Y is in period 3 and has 1 electron in its p subshell. Outermost configuration: 3s²3p¹. This element is in Group 13 (2 + 10 + 1 = 13). (Aluminium, Al). Element X (Cl) has oxidation state −1. Element Y (Al) has oxidation state +3. Molecular formula: YX₃ (e.g., AlCl₃).

Question 2: A few subshells are given. 3p, 4d, 3f, 2d, 2p.
a) Among these, which subshells are not possible?
b) Explain the reason.
Answer:
a) The subshells that are not possible are 3f and 2d.
b) The Azimuthal quantum number (l) defines subshells and its value ranges from 0 to (n−1).

• For 3f: n=3. The maximum possible value for l is (3−1)=2 (representing s,p,d). Since l=3 corresponds to the f subshell, 3f is not possible.
• For 2d: n=2. The maximum possible value for l is (2−1)=1 (representing s,p). Since l=2 corresponds to the d subshell, 2d is not possible.

Question 7: Which subshell is represented by each pair of the quantum number values given below?
a) n=1, l=0
b) n=2, l=1
Answer:
a) n=1, l=0: 1s subshell (where l=0 denotes the s subshell).
b) n=2, l=1: 2p subshell (where l=1 denotes the p subshell).

Question 10: Iron (Fe) takes part in chemical reactions and becomes Fe³⁺ ion. (Atomic number of Fe=26).
a) Write the electron configuration of this ion.
b) Write the chemical formula of the compound formed when this ion combines with sulphate ion (SO₄²⁻).
c) Which is the other oxidation state of this element? Write the electron configuration of the ion thus formed.
d) Iron shows variable oxidation states. Why?
Answer:
a) Neutral Fe configuration: 1s²2s²2p⁶3s²3p⁶3d⁶4s². To form Fe³⁺, three electrons are lost: two from the outermost 4s subshell and one from the 3d subshell. Fe³⁺ electron configuration: 1s²2s²2p⁶3s²3p⁶3d⁵.
b) The Fe³⁺ ion combines with SO₄²⁻ ion. Chemical formula: Fe₂(SO₄)₃.
c) The other common oxidation state is +2 (Ferrous). The ion formed is Fe²⁺ (loses two 4s electrons). Electron configuration of Fe²⁺: 1s²2s²2p⁶3s²3p⁶3d⁶.
d) Iron (a transition element) shows variable oxidation states because there is only a slight energy difference between the outermost s subshell and the penultimate d subshell. As a result, electrons from the d subshell can also participate in chemical reactions under favourable conditions.

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Unit 4: Gas Laws and Mole Concept

I. Activity: Calculation of Volume using Combined Gas Equation
Question: The volume of a gas at 27°C and 1 atm pressure is 100 mL. What will be its volume at 273 K temperature and 2 atm pressure?
Answer: Initial conditions: T₁ = 27°C = (27+273) K = 300 K; P₁ = 1 atm; V₁ = 100 mL. Final conditions: T₂ = 273 K; P₂ = 2 atm; V₂ = ?. Using the Combined Gas Equation: P₁V₁/T₁ = P₂V₂/T₂.
(1 × 100) / 300 = (2 × V₂) / 273
V₂ = (100 × 273) / (300 × 2) = 273 / 6 = 45.5 mL.
Volume at 273 K and 2 atm: 45.5 mL.

II. Let us assess
Question 2: Complete the table. (Atomic Mass - H=1, C=12, O=16, N=14).

Substance	Molar Mass (g)	Given mass (g)	Number of moles	Volume at STP (L)
H₂	2	10	5	112
CO₂	44	440	10	224
NH₃	17	340	20	448
(Derivations based on Number of moles = Given mass / Molar mass, and Volume at STP = Number of moles × 22.4 L):

• H₂: Molar Mass = 2×1=2. Moles = 112/22.4=5. Given Mass = 5×2=10.
• CO₂: Molar Mass = 12+2(16)=44. Moles = 440/44=10. Volume = 10×22.4=224.
• NH₃: Molar Mass = 14+3(1)=17. Moles = 340/17=20. Volume = 20×22.4=448.

Question 5: 400 L of gas is stored in a cylinder at 27°C and constant pressure.
a) What will be the temperature if the volume of this gas is reduced to 200 L at the same pressure?
b) Which gas law is relevant to this context?
c) The boiling point of a substance is 3°C. Above what temperature in Kelvin does this substance obey the gas laws?
Answer:
a) This is governed by Charles’s Law (constant pressure). T₁ = 27°C = 300 K; V₁ = 400 L. T₂ = ?; V₂ = 200 L. V₁/T₁ = V₂/T₂.
400 / 300 = 200 / T₂
T₂ = (200 × 300) / 400 = 150 K.
b) Charles’s Law (At constant pressure, volume is directly proportional to temperature in Kelvin).
c) Gas laws describe the properties of gases. Since the boiling point is 3°C, the substance must be in the gaseous state above this temperature. 3°C = (3+273) K = 276 K. The substance obeys the gas laws above 276 K.

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Unit 5: Electrochemistry

I. Conceptual Questions
Question: Why can silver nitrate solution not be stored in a copper vessel?
Answer: Copper is a more reactive metal than silver. The more reactive metal (Copper) will displace the less reactive metal (Ag⁺ ion) from its salt solution. The reaction is Cu + 2AgNO₃ → Cu(NO₃)₂ + 2Ag, meaning the copper vessel would dissolve.

Question: Which of the following metals can displace hydrogen from hydrochloric acid? (Sodium, gold, silver, aluminium)
Answer: Sodium and Aluminium. The metals placed above hydrogen in the reactivity series can displace hydrogen from dilute acids.

Question: Define fuel cells and write its advantage.
Answer: Fuel cells are galvanic cells designed to convert the chemical energy produced by the combustion of fuels (such as hydrogen, methane, and methanol) directly into electrical energy. An advantage is that they produce electricity with high efficiency.

II. Let us assess...
Question 1: Examine the diagram of a galvanic cell (Zn/Cu) and find out whether the given statements are true or false.
a) As the cell operates, the mass of the zinc rod, Zn(s) decreases.
b) The copper electrode is the anode.
c) Electrons flow from the zinc electrode to the copper electrode through the external circuit.
d) During cell reaction, reduction takes place at the copper electrode.
e) During cell reaction, the concentration of Cu²⁺ decreases.
Answer:
a) True (Oxidation occurs: Zn(s) → Zn²⁺(aq) + 2e⁻).
b) False (Zinc is the anode (oxidation), Copper is the cathode (reduction)).
c) True (Flow is from anode (Zn) to cathode (Cu)).
d) True (Reduction occurs at the cathode: Cu²⁺(aq) + 2e⁻ → Cu(s)).
e) True (Cu²⁺ ions are consumed to form Cu metal at the cathode).

Question 4: When an aqueous sodium chloride solution is electrolysed, the reaction that takes place at the anode is:
(i) Na⁺(aq) + e⁻ → Na(s)
(ii) 2H₂O(l) → 4H⁺(aq) + O₂(g) + 4e⁻
(iii) H⁺(aq) + e⁻ → ½H₂(g)
(iv) Cl⁻(aq) → ½Cl₂(g) + e⁻
Answer: (iv) Cl⁻(aq) → ½Cl₂(g) + e⁻ (Chloride ions are oxidised to chlorine gas at the anode).

Question 5: Write the products obtained when the following solutions undergo electrolysis.
(i) Aqueous solution of NaCl
(ii) Aqueous solution of CuSO₄ (using copper electrodes)
Answer:
(i) Products: H₂ gas (at cathode), Cl₂ gas (at anode), and an aqueous solution of NaOH.
(ii) The process is electrolytic refining. Products: The impure copper anode dissolves (Cu → Cu²⁺), and pure copper is deposited on the cathode (Cu²⁺ → Cu).

Question 6: Brass is an alloy of zinc and copper. When brass comes into contact with saline water, corrosion of metal takes place and zinc gradually dissolves in the solution, leaving copper behind. Explain why zinc dissolves in comparison to copper.
Answer: Zinc is more reactive than copper; this is indicated by its higher position in the reactivity series. When two metals are in contact in the presence of an electrolyte (like saline water), the more reactive metal acts as the anode and is destroyed (corrodes), thus protecting the less reactive metal (copper).

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Unit 6: Metals

I. Conceptual Questions
Question: What are the qualities of a mineral which is used to extract a metal?
Answer: The mineral should have: Abundance, Ease of extraction, High metal content, and Low production cost.

Question: The nature of ores is given. Complete the table using appropriate concentration methods.

Nature of ore	Concentration method
1. Ores are less dense than the impurities.	Froth floatation
2. Ores have magnetic properties but impurities are non magnetic.	Magnetic separation
3. A solvent which dissolves the ore is used.	Leaching
4. Ores are denser than the impurities.	Levigation (Hydraulic washing)

Question: What type of flux should be used if the gangue in the ore is basic in nature?
Answer: An acidic flux should be used, as the flux combines with the gangue to form molten slag.

II. Let us assess...
Question 1: Define the following and answer the given questions.
(1) Ore - What is the ore of aluminium?
(2) Roasting - Which type of ores are subjected to roasting?
(3) Reducing agent - What is the reducing agent used in the manufacture of aluminium?
(4) Flux - Which flux is used in the manufacture of copper? Why?
(5) Leaching - Which metal ore is leached with sodium cyanide?
Answer:
(1) Ore: Any mineral from which a metal can be extracted easily and economically. Ore of aluminium: Bauxite (Al₂O₃·2H₂O).
(2) Roasting: Heating the concentrated ore at a temperature below the melting point in the presence of air. Ores subjected to roasting: Sulphide ores.
(3) Reducing agent for aluminium: Electricity (or e⁻).
(4) Flux used in copper manufacture: Sand (SiO₂). This is an acidic flux, and it is used because the ore contains basic impurities (such as iron oxide).
(5) Metal ores leached with sodium cyanide (NaCN): Gold and Silver ores.

Question 2: Find the relation and write the answer.
Zinc sulphide : Roasting :: Calcium carbonate : ....................
Magnetite : Magnetic separation :: Bauxite : ....................
Answer:

• Zinc sulphide : Roasting :: Calcium carbonate : Calcination.
• Magnetite : Magnetic separation :: Bauxite : Leaching.

Question 4: Following are two facts related to the manufacture of an industrially important metal.

• The ore is treated with hot NaOH solution.
• Electricity is used as the reducing agent to extract the metal.
  (i) These facts are related to the production of which metal?
  (ii) What is the reason for using electricity as the reducing agent?
  (iii) Which substance is used as the electrolyte here?
  (iv) Which gas is liberated at the anode?
  Answer:
  (i) The metal is Aluminium.
  (ii) Aluminium is a highly reactive metal (and forms stable compounds). Strong reducing agents (like electrolysis) are required to extract highly reactive metals.
  (iii) The electrolyte is anhydrous Alumina (Al₂O₃) dissolved in molten Cryolite (Na₃AlF₆).
  (iv) Oxygen gas (O₂) is liberated at the anode.

Question 6: Corrosion is a process by which iron is converted into its oxide. This is an oxidation reaction.
(i) What is oxidation?
(ii) Complete the following chemical equation. ....... Fe + ....... O₂ → ....... Fe₂O₃
(iii) Suggest two methods for preventing the corrosion of iron.
Answer:
(i) Corrosion (as an oxidation process): A process in which the metal reacts with its surrounding medium and undergoes chemical change.
(ii) 4Fe + 3O₂ → 2Fe₂O₃.
(iii) Two methods: Electroplating and Cathodic protection (e.g., attaching zinc or magnesium blocks to iron).

Question 7: Alloys containing iron are given. Find a,b,c,d.

Alloys	Constituent elements	Uses
Alnico	(a) Fe, Al, Ni, Co	(b) Making of permanent magnets
Stainless steel	(c) Fe, Cr, Ni, C	Resist the corrosion of iron. Making of utensils
Silicon steel	Fe, Si, C	(d) Used in the core of electromagnetic instruments (e.g., motors, generators, transformers) as it reduces electric loss

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Unit 7: Some Compounds of Industrial Importance

I. Conceptual Questions
Question: Why is the jar used for collecting ammonia kept inverted?
Answer: Ammonia gas is less dense than atmospheric air.

Question: Why does the water entering the flask change its colour (to pink) in the ammonia fountain experiment?
Answer: Ammonia dissolves in water to form ammonium hydroxide (NH₄OH), which is an alkaline substance (base). Phenolphthalein is an indicator that gives a pink colour with alkaline substances.

Question: Concentrated H₂SO₄ is not used as a drying agent in the preparation of NH₃. Why?
Answer: Concentrated H₂SO₄ and NH₃ react chemically.

Question: What are the essential qualities of fertilizers?
Answer:

1. The compounds should be soluble in water.
2. The elements present in the compounds must be easily available to plants.
3. The compounds must be stable and remain in the soil for a long time.
4. They should not cause significant variations to the pH of the soil.
5. They should not be toxic to plants.

II. Let us assess...
Question 1: Chemical formula of certain salts are given below: KCl, (NH₄)₂SO₄, AlCl₃, CH₃COONa. Choose the correct statements.
(i) KCl does not undergo hydrolysis.
(ii) (NH₄)₂SO₄ is an acidic salt.
(iii) AlCl₃ is a basic salt.
(iv) CH₃COONa is used to increase the basicity of an aqueous solution.
Answer:

• (i) Correct (KCl is formed from Strong Acid (HCl) + Strong Base (KOH), hence neutral).
• (ii) Correct (Formed from Strong Acid (H₂SO₄) + Weak Base (NH₄OH), hence acidic).
• (iii) Incorrect (AlCl₃ is formed from Strong Acid (HCl) + Weak Base (Al(OH)₃), hence acidic).
• (iv) Correct (Formed from Weak Acid (CH₃COOH) + Strong Base (NaOH), hence basic, increasing basicity).
  The correct statements are (i), (ii), and (iv).

Question 4: Explain with chemical equations of the chemical reactions that take place in the left and right chambers of the membrane cell during the chlor-alkali process.
Answer: This process involves the electrolysis of concentrated aqueous sodium chloride solution (brine).

Chamber	Electrode	Reaction Type	Chemical Equation	Product(s) Formed
Left	Anode (+)	Oxidation	2Cl⁻ → Cl₂ + 2e⁻	Chlorine gas (Cl₂)
Right	Cathode (–)	Reduction	2H₂O + 2e⁻ → H₂ + 2OH⁻	Hydrogen gas (H₂) and Sodium Hydroxide (NaOH)

Question 6: Explain the effect of pressure on the following reversible reactions.
(i) H₂(g) + I₂(g) ⇌ 2HI(g)
(ii) N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
Answer: Pressure affects equilibrium only if the number of gaseous molecules changes in the reaction.
(i) No effect: Total moles of reactants (H₂ + I₂) is 2; total moles of product (2HI) is 2. Since the number of moles of gaseous molecules does not change, pressure will have no effect on the chemical equilibrium.
(ii) Pressure increases product yield: Total moles of reactants (N₂ + 3H₂) is 4; total moles of product (2NH₃) is 2. According to Le Chatelier’s principle, increasing pressure shifts the equilibrium in the direction where the number of moles decreases. Thus, increasing pressure favors the forward reaction, increasing the production of NH₃.

Question 7: 2NO(g) + O₂(g) ⇌ 2NO₂(g) + heat (Forward reaction is exothermic, 3 moles → 2 moles). Complete the table by indicating the possible measures to increase the amount of NO₂ (product).
Answer:

Change Applied	Effect on NO₂ Production	Reason (Le Chatelier’s Principle)
Temperature	Decrease	Favours the exothermic reaction (forward reaction).
Pressure	Increase	Favours the direction with fewer moles (forward reaction, 3 → 2).
Amount of oxygen	Increase	Increases reactant concentration, shifts equilibrium forward.
Amount of NO	Increase	Increases reactant concentration, shifts equilibrium forward.

Question 11:
(i) Which substance is used as the drying agent in the laboratory preparation of ammonia?
(ii) Can concentrated sulphuric acid be used as a drying agent in this process? Write the chemical equation that occurs if it is used.
Answer:
(i) The drying agent used is Quick lime (CaO).
(ii) Concentrated sulphuric acid cannot be used. This is because ammonia (NH₃) reacts with concentrated sulphuric acid (H₂SO₄). The chemical equation (implied reaction): 2NH₃ + H₂SO₄ → (NH₄)₂SO₄.

Question 12: Complete the table.

Function in plants	Fertilizers that can be used
To accelerate plant growth.	Urea, sodium nitrate (Nitrogenous fertilizers)
To increase productivity.	Ammonium phosphate, potassium chloride (Phosphate and Potash fertilizers)
To boost immunity of plants.	Potassium chloride (Potash fertilizers)