CHAPTER

SECTION

B
1 CHEMICAL REACTIONS

TOPIC-1
Chemical Reactions an Equations

WORKSHEET-1
Solutions Answering Tip
1. Correct option : (d)  Keenly observe the necessary conditions for the
Explanation : During combustion of liquefied reactions during practical, prepare a list of type of
petroleum gas (LPG), it forms CO2 and H2O. decomposition reactions, its necessary conditions
and gases released and practise it.
2. Rusting of iron. 1

3. Reaction in which a single reactant breaks down to 5. (a) (i) Physical state of reactants and products.
give simpler products. (ii) Conditions such as temperature, pressure,
Thermal decomposition : heat etc.
∆
(iii) Catalyst involved.
CaCO3  → CaO + CO2 (Or any other) (iv) Change in state. ½×4
Electrolytic decomposition : (b) Total mass of the elements present in the products
in a chemical reaction has to be equal to the total
Electric current
2 H2O → 2H2 + O2 mass of elements present in the reactants.
or
[CBSE Marking Scheme, 2016] 1 + 1 + 1
Mass can neither be created nor be destroyed in a
chemical reaction.
4. CaCO3 heat
 → CaO + CO2  [CBSE Marking Scheme, 2013] 1

2FeSO4 heat
→ Fe2O3 + SO2 + SO3  Any one 6. Process in which new substances with new
 properties are formed by the rearrangement of

2Pb(NO3)2 heat
 → 2PbO + 4NO2 + O2 atoms.
Sunlight (i) Evolution of gas : The chemical reaction between
2AgCl → 2Ag + Cl2 
 zinc and dilute H2SO4.
Sunlight  Any one Zn(s) + H2SO4(aq) → ZnSO4(aq) + H2(g) ↑
2AgBr → 2Ag + Br2 
 (ii) Change in colour : The chemical reaction between
electricity
2H2O 
→ 2H2 + O2 potassium iodide and lead nitrate.
Pb(NO3)2(aq) + 2KI(s) → 2KNO3(aq) + PbI2(s)
(or any other equation for above decomposition
reaction.) Colourless Yellow
Note : No marks to be deducted if equations are (iii) Formation of precipitate : The chemical reaction
not balanced. 1+1+1 between sulphuric acid and barium chloride.
[CBSE Marking Scheme, 2018] BaCl2(aq) + H2SO4(aq) → 2HCl(aq) + BaSO4(s)

White precipitate
Commonly Made Error (iv) Change in temperature : The chemical reaction
between quick lime and water.
 Usually students get confused in the necessary CaO(s) + H2O (l) → Ca(OH)2(aq) + Heat
conditions and liberation of gases in the reaction. 1+1+1+1+1

WORKSHEET-2
Solutions temperature copper (II) oxide is formed.
1. Correct option : (d) 2. Law of conservation of mass.
Explanation : Chemical changes involve formation Mass can neither be created nor can it be destroyed
of new compounds from one or more substances. during a chemical reaction. ½+½
On heating copper wire in presence of air at high

S OLUT I ONS P-1

 3. (i) Copper is more reactive than silver. Hence, (iii) During digestion, food containing carbohydrates
when copper wire is dipped in silver nitrate are broken down to form glucose. This glucose
solution, it displaces silver from AgNO3 solution combines with oxygen in the cells of our body
forming copper nitrate which is bluish green in and provides energy. Since energy is given, so it is
colour. exothermic. 1×3=3
(ii) Cu + 2AgNO3 → Cu(NO3)2 + 2Ag [CBSE Marking Scheme, 2012]
(Copper (II) nitrate (Silver)
: bluish green) 6. (i) CaOCl2(s) + CO2(g) → CaCO3(s) + Cl2(g)
[CBSE Marking Scheme, 2016] 2 + 1 (Bleaching (Carbon dioxide) (Calcium (Chlorine)
4. (i) Two observations are : powder) carbonate)
(a) Change in state and colour. Heat
(b) Evolution of gas (ii) CuSO4 . 5H2O  → CuSO4 + 5H2O
(ii) Decomposition reaction (Blue) (White)
(iii) 2FeSO4(s) 
Heat
→ Fe2O3(s) + SO2(g) + SO3(g) (iii) Ca(OH)2 + Cl2 → CaOCl2 + H2O
(Dry slaked lime) (Chlorine) (Bleaching powder)
1+1+1 (iv) Ca(OH)2 + CO2 → CaCO3 + H2O
5. (i) Decomposition reactions require energy either (Lime water) (Calcium
in the form of heat, light or electricity for breaking carbonate)
down the reactants. So energy is absorbed and is
(Milky)
endothermic in nature.
Heat
(ii) Iron has displaced copper from copper sulphate (v) 2NaOH + Zn  → Na2ZnO2 + H2
solution to form iron sulphate which is light green (Sodium zincate) (Hydrogen)
in colour because Fe is more reactive than copper.
[CBSE Marking Scheme, 2016] 1 × 5

WORKSHEET-3
Solutions 5. (i) Substance X-Calcium Hydroxide.
Ca(OH)2(aq) + CO2(g) ® CaCO3(s) + H2O(l)
1. (c) 2H2(g) + O2(g) ® 2H2O(l) (White ppt.)
Explanation : It is because, the standard state for (ii) Calcium hydroxide is obtained by reaction of
hydrogen and oxygen is gas and for water is liquid
at reaction temperature.
calcium oxide and water.
CaO(s) + H2O(l) ® Ca(OH)2(aq) + Heat 1 + 2
2. In a physical change, no new substance is formed.
In a chemical change, new substance is formed. 6. (i) Decomposition Reaction : Carbohydrates are
½+½ broken down to form glucose.
(ii) Oxidation Reaction : When an iron object is left
3. (i) ‘X’ is Copper (Cu), ‘Y’ is Copper oxide (CuO). 1
in moist air for a considerable time, it gets covered
(ii) Oxidation and Reduction ½+½
Heat with a red brown flaky substance called rust.
(iii) 2Cu + O2  → 2CuO
(iii) Displacement reaction : More reactive metal
Heat
CuO + H2  → Cu + H2O ½+½ displaces less reactive metal from its salt solution.
[CBSE Marking Scheme, 2012] (iv)
Displacement reaction : More reactive metal
displaces less reactive metal from its salt solution.
4. (i) Exothermic Reaction
(v) Double displacement reaction : Reaction in which
CaO + H2O → Ca(OH)2
two compounds react by an exchange of ions to
(ii) Double displacement reaction
form two new compounds. 1×5
BaCl2 + Na2SO4 → BaSO4 + 2NaCl 1½ + 1½
[CBSE Marking Scheme, 2015] [CBSE Marking Scheme, 2016]

WORKSHEET-4
Solutions 2. In the absence of air, ZnO(s) and CO2(g) are formed.
1. Quick lime reacts with water vigorously to produce Chemical Equation :
slaked lime and a large amount of heat. Heat
ZnCO3(s) Absence
 → ZnO(s) + CO2(g) ½ + ½
CaO(s) + H2O (l) → Ca(OH)2 + Heat of air

(Quick lime) (Slaked lime) 1

P-2 S C I ENC E - X
 3. (i) Na2CO3(s) + 2HCl(aq) → 2NaCl(aq) + 5. (i) Change in colour : Reaction between lead
H2O(l) + CO2(g) nitrate solution and potassium iodide solution.
(ii) CaO(s) + H2O(l) → Ca(OH)2(aq) + Heat Pb(NO3)2(aq) + 2KI(aq) →PbI2(s) + 2KNO3(aq)
(iii) Pb(NO3)2(aq) + NaCl(aq) → PbCl2(s) + In this reaction, colour changes from colourless to
2NaNO3(aq) 1+1+1 yellow.
(ii) Change in temperature : Action of dilute sulphuric
[CBSE Marking Scheme, 2014]
acid on zinc.
4. (a) (i) HCl is oxidized. ½ Zn(s) + H2SO4(aq) → ZnSO4(aq) + H2(g)
(ii) MnO2 is reduced. ½ In this reaction, heat is evolved.
(b) (i) Oxidation : Gain of Oxygen or loss of Hydro- (iii) Formation of precipitate : Action of barium
gen. 1 chloride on sodium sulphate.
(ii) Reduction : Gain of Hydrogen or loss of Oxy- BaCl2(aq) + Na2SO4(aq) → BaSO4(s) + 2NaCl(aq)
gen. [CBSE Marking Scheme, 2018] 1 (ppt.)
[CBSE Marking Scheme, 2014] 1+1+1
Commonly Made Error
6. (i) 2NaOH(aq) + Zn(s) → Na2ZnO2(aq) + H2(g)
 Usually students get confused with the oxidised
and reduced substances in the reaction. (ii) Ca(OH)2(aq) + CO2(g) → CaCO3(s) + H2O(l)
CaCO3(s) + H2O(l) + CO2(g) → Ca(HCO3)2(aq)
Answering Tip (iii) Na2CO3(s) + H2SO4(dil) → Na2SO4(aq) + H2O(l)
 Students must remember that oxidation, a process + CO2(g)
in which oxygen or an electronegative element (iv) CaCO3 + 2HCl → CaCl2 + H2O + CO2
is added, while reduction is a process in which
(v) CuO + 2HCl (dil) → CuCl2 + H2O
addition of hydrogen or an electropositive element
1+1+1+1+1
takes place.

WORKSHEET-5
Solutions 5. (i) 2Na + 2H2O → 2NaOH + H2
1. Burning of candle is both physical and chemical It is an exothermic reaction.
change. Burning of candle melts the wax and hence ∆
(ii) 2FeSO4(s)  → Fe2O3 (s) + SO2(g) + SO3(g)
physical state of wax has changed from solid to
liquid. Again the wax combines with the atmosphere It is an endothermic reaction. 1½ + 1½
oxygen and changes to carbon dioxide, heat and 6. (i) With dilute H2SO4 : H2 gas is evolved.
light. Thus both the changes are accompanied in
Zn + H2SO4 → ZnSO4 + H2 ­↑
the burning of candle. 1
(ii) With dilute HCl : H2 gas is evolved.
2. Pb(NO3)2 + 2KI → 2KNO3 + PbI2 Zn + 2HCl → ZnCl2 + H2 ­↑
(Lead (Potassium (Potassium (Lead (iii) With dilute HNO3 : Zinc nitrate, nitrous oxide and
nitrate) iodide) nitrate) iodide) 1 water are formed.
Here, HNO3, is an oxidising agent which oxidises
3. (i) CaCO3(s) + 2HCl(aq) →CaCl2(s) + H2O(l)
H2 gas to H2O.
+ CO2(g) 4Zn + 10HNO3 → 4Zn(NO3)2 + 5H2O + N2O↑
(ii) 2Al(s) + 6HCl(aq) →2AlCl3(aq) + 3H2(g) (iv) With dilute NaCl : No chemical reaction takes
(iii) MnO2 (s) + 4HCl(aq) →MnCl2 (aq) + 2H2O(l) + place.
Cl2(g) 1+1+1 (v) With dilute NaOH : Sodium Zincate is formed and
[CBSE Marking Scheme, 2014] H2 gas is evolved.
Zn + 2NaOH → Na2ZnO2 + H2 ­↑ 1×5
4. (i) 2Al + 3H2SO4 → Al2(SO4)3 + 3H2 [CBSE Marking Scheme, 2013]
(ii) Na2CO3 + 2HCl → 2NaCl + H2O + CO2
(iii) Ca(OH)2 + CO2 → CaCO3 + H2O 1×3
[CBSE Marking Scheme, 2013]

S OLUT I ONS P-3

 WORKSHEET-6
Solutions 4. The fizzing will occur strongly in test tube A, in
which hydrochloric acid (HCl) is added. This is
1. Balanced Equation : because HCl is a stronger acid than CH3COOH and
thus, produces hydrogen gas at a faster speed due
2FeSO4(s) Heat
→ Fe2O3(s) + SO2(g) + SO3(g) 1
to which fizzing occurs. 3
5. Sodium hydroxide (NaOH) is a commonly used
2. (i) In white washing, quicklime reacts with water
base and is hygroscopic, that is, it absorbs moisture
to form slaked lime.
from the atmosphere and becomes sticky. The acidic
CaO + H2O → Ca(OH)2 + Heat oxides react with base to give salt and water. The
Quick lime Slaked lime reaction between NaOH and CO2 can be given as :
(ii) Silver bromide, when exposed to light decomposes 2NaOH + CO2 → Na2CO3 + H2O 3
to silver and bromine. 6. Compound X is ethanol, Y is ethene and Z is
Sunlight
2 AgBr(s) → 2 Ag(s) + Br2(g) hydrogen gas.
(Silver bromide) (Silver) (Bromine) CH3CH2OH Conc.H
 2 SO4
Heat [443K]
→ CH2=CH2 + H2O
[CBSE Marking Scheme, 2016] 1½ + 1½ Here, sulphuric acid acts as dehydrating agent.
When ethanol (X) reacts with sodium metal, a
3. (i) Before heating : Pale green colourless gas is evolved which known as hydrogen.
After heating : Brown or reddish brown. 1 2CH3CH2OH + 2Na ® 2CH3CH2ONa + H2
(ii) SO2 and SO3 1 (Sodium ethoxide)
Heat
(iii) 2FeSO4(s) → Fe2O3(s) + SO2(g) + SO3(g) 1  1+1+1+1+1

[CBSE Marking Scheme, 2012]

TOPIC-2
Types of Chemical Reactions-Corrosion and Rancidity

WORKSHEET-7
Solutions precipitate and the reaction in which precipitate is
formed is called precipitation reaction.
1. Correct option : (c) Example : When sodium sulphate solution is mixed
Explanation : The given reaction is a redox reaction with barium chloride solution, a white precipitate
because oxidation and reduction both take place of a substance (BaSO4) is formed.
simultaneously. Also, it is a displacement reaction
because hydrogen of NH3 has been displaced by Na2SO4(aq) + BaCl2(aq) → BaSO4(↓) + 2NaCl(aq)
oxygen. [CBSE Marking Scheme, 2013] 1+1+1
2. In combination reactions, two substances combine
to form one compound and in decomposition 5. Rancidity.
reactions, a compound breaks down into two or Anti-oxidants are substances which prevents
more substances, so they are opposite to each oxidation, actually are reducing agents. When
other. 1 added to food, the fats and oils present in the
food do not get oxidized easily, hence do not turn
3. The oxidation of oils or fats in a food resulting into rancid and remain good to eat for longer time.
bad smell and bad taste is called rancidity. 1 [CBSE Marking Scheme, 2016] 1 + 1 + 1
It can be prevented by :
(i) adding anti-oxidants. 1 6. (i) Correct definition.
(ii) flushing with nitrogen gas. 1 (ii) rusting.
[CBSE Marking Scheme, 2012] (iii) Silver - black, copper - green.
(iv) Destruction of car bodies, bridges, railing, etc
4.
On mixing the clear solution of two ionic (Any two)
compounds, a substance which is insoluble in water, (v) Painting, alloying, greasing etc (Any two)
is formed. This insoluble substance is known as a [CBSE Marking Scheme, 2016] 5

P-4 S C I ENC E - X
 Detailed Answer : (iii) Silver turns black as it reacts with H2S present in air
(i) Corrosion is a process in which metals, are and form a layer of Ag2S.
deteriorated by action of air, moisture, chemicals (iv) Corrosion of iron is a serious problem because it
etc. leads to wastage of tonnes of iron every year and
(ii) Corrosion of iron is called rusting. lot of money is spent to repair or replace it.
(v) The iron articles should be painted to prevent it
from corrosion. 1+1+1+1+1

WORKSHEET-8
Solutions (iii) Iron : Colour — Reddish Brown
Chemical name — Ferric oxide. 1+1+1
1. Correct option : (c) 5. (i) 2Mg + O2 → 2MgO
Explanation : The substance which oxidises the MgO + H2O → Mg(OH)2
other substances in a chemical reaction is known as
Type of reaction : Oxidation reaction/Combination
an oxidising agent. Likewise, the substance which
reaction
reduces the other substance in a chemical reaction
is known as reducing agent. (ii) CaCO3(s) → CaO(s) + CO2(g)
2. Because heat is absorbed in this process. 1 (Lime stone) (quick lime)
3. Iron is more reactive than copper. Iron displaces Type of reaction : Decomposition reaction. 1½+1½
copper from copper sulphate solution and forms 6. (i) Corrosion is a process in which metals, are
iron sulphate, hence holes appear on the pot. deteriorated by action of air, moisture, chemicals
Chemical equation : etc.
Fe(s) + CuSO4(aq) → FeSO4(aq) + Cu(s) 1½ + 1½ Corrosion of iron is Rusting.
4. (i) Silver : Colour — Black (ii) Silver - black, copper - green.
Chemical name — Silver sulphide (iii) Car bodies, bridges, railing etc (Any two)
(ii) Copper : Colour — Green Painting, alloying, greasing etc (Any two)
Chemical name — Copper oxide

WORKSHEET-9
Solutions When a single reactant on gaining energy
decomposes to give two or more simpler products,
1. Correct option : (a) such a reaction is called a decomposition reaction.
Explanation : Exothermic reactions are the reactions
in which heat is released along with the formation Example : CaCO ∆3 → CaO + CO 2 1½

of products. When quick lime reacts with water,
a large amount of heat is released along with the [CBSE Marking Scheme, 2012]
formation of calcium hydroxide. Similarly, the
process of dissolving an acid or base in water is a 5. (i) Photochemical decomposition : A single
highly exothermic reaction. Evaporation of water reactant breaks down to give simpler products.
and sublimation of camphor are endothermic
reactions. ½+½
2. Hydrogen peroxide decomposes into H2O and O2 White silver chloride changes to grey, as it
in the presence of sunlight and hence to prevent decomposes to silver and chlorine in presence of
decomposition, they are kept in coloured bottles. sunlight. ½
2H O Sunlight 2H2O + O2 1
2 2 → (ii) 2AgCl(s) Sunlight
→ 2Ag(s) + Cl2(g) ½

3. (i) Decomposition reaction (iii) Black and white photography. 1
(ii) Displacement reaction
[CBSE Marking Scheme, 2012]
(iii) Double displacement reaction
6. (a) Corrosion. ½
(iv) Combination reaction
(v) Displacement reaction (b) Conditions are : air and moisture. ½+½
(vi) Combination reaction ½+½+½+½+½+½ Activity : (i) Take three test-tubes. Place clean iron
nails in each test-tube.
4. The chemical reaction in which a single product is (ii) Pour some water in test-tube 1, cork it.
formed from two or more reactants and energy is
(iii) Pour water (boiled/distilled) in test-tube 2, add
evolved, is known as combination reaction.
some oil and cork it.
Example : 2Mg + O2 Burn→ 2MgO 1½
(iv) Put some anhydrous calcium chloride in test-tube 3
and cork it.

S OLUT I ONS P-5

 (v) After 2-3 days, you will observe that the nails in place in the presence of air and moisture, both. 2½
test-tube 1 get rusted because they were exposed to Methods to prevent rusting : Alloying, galvanization,
air and water both, while nail in test-tube 2 and 3 painting, lubrication. (Any two) ½+½
do not get rusted. This shows rusting of iron takes

WORKSHEET-10
Solutions 4. Gold and platinum.
1. Correct option : (c) (i) Corrosion of aluminium is useful. A protective
layer of aluminium oxide is formed on the surface
Explanation : In beakers A and B, heat is given of the metal which renders the metal passive and
out, so temperature became high, hence it is an prevents its further corrosion.
exothermic reaction while in beaker C, heat is (ii) Corrosion of iron is a serious problem. Every year
absorbed from water, so temperature falls, hence it large amount of money is spent to replace damaged
is an endothermic process. iron and steel structures. So, here, corrosion is a
2. It is a combination reaction. 1 serious problem. 1+1+1
3. A reaction is considered as double displacement [CBSE Marking Scheme, 2016]
reaction if during the chemical reaction an
5. (i) At Cathode : Hydrogen gas (H2) ½
exchange of ions takes place between two ionic
substances. At Anode : Oxygen gas (O2) ½
Example : BaCl2 + Na2SO4 → BaSO4 + 2NaCl (ii) Since 2H2O Electric current
→ 2H2 + O2
In a displacement reaction, a more reactive element 2 molecules of H2 combine with 1 mol of O2 to
displaces or removes another less reactive element, form H2O, so the volume of H2, liberated is double
to form its compound. e.g. Fe + CuSO4 → FeSO4 than that of O2. 1
+ Cu; whereas in a double displacement reaction,
(iii) When a burning splinter is brought near the
the compounds react by exchanging their ions and
mouth of the liberated gases, the burning splinter
form two new compounds. 1½+1½
extinguishes near the H2 gas while the burning
[CBSE Marking Scheme, 2012]
splinter keeps burning more near the O2 gas. 1
[CBSE Marking Scheme, 2012]

WORKSHEET-11
Solutions 4. (i) Brown fumes, white residue. 1
1. Correct option : (a) (ii) Decomposition reaction 1
Explanation : A dilute ferrous sulphate solution was 2Pb(NO3)2 → 2PbO + 4NO2 + O2 1
gradually added to the beaker containing acidified [CBSE Marking Scheme, 2012]
permanganate solution. A permanganate solution
is usually purple in colour. The light purple colour 5. (i) C is getting oxidized to CO, ZnO is getting
of the solution fades and finally disappears. This reduced to Zn. ½+½
is because Potassium permanganate (KMnO4) (ii) As carbon is gaining oxygen and ZnO is losing
is relatively an unstable compound, it tends to oxygen. 1
decompose in the presence of ferrous sulphate
(iii) It is a redox reaction or oxidation and reduction
(FeSO4). This changes the colour of the solution
from purple to colourless. The FeSO4 gets oxidised reaction.
to Fe2(SO4) as KMnO4 acts as a good oxidising CuO + H2 → Cu + H2O 1
agent in an acidic medium [CBSE Marking Scheme, 2012]
2. They get tarnished by reacting with atmospheric air
to form silver sulphide. 1 6. (i) (a) Due to the decomposition of silver chloride
into silver and chlorine by sunlight.
3. (i) Yes, temperature rises as the beaker gets hot.
Sunlight
This is because large amount of heat is released or 2AgCl → 2Ag + Cl2 1
the reaction is exothermic. 1 (b) Due to the oxidation of copper powder to
(ii) Combination reaction. Reaction in which a single copper oxide, brown colour turns into black. 1
product is obtained from two or more reactants. 1 2Cu(s) + O2(g) → 2CuO(s)
(iii) CaO + H2O → Ca(OH)2 + heat 1 (ii) (a) A chemical reaction in which the more reactive
(Quick lime) (Slaked lime) element displaces the less reactive element
[CBSE Marking Scheme, 2012] from its compound is called displacement
reaction.

P-6 S C I ENC E - X
 CuSO4(aq) + Zn(s) → ZnSO4(aq) + Cu(s) 1 (c) Combination reaction is a reaction in which
(b) A chemical reaction in which hydrogen is two or more than two substances combine and
added or oxygen is removed is called reduction form a single substance.
reaction. CaO + H2O → Ca(OH)2 + Heat 1
ZnO + C → Zn + CO 1 [CBSE Marking Scheme, 2012]

WORKSHEET-12
Solutions 5. (i) Any suitable exothermic reaction like
1. Correct option : (b) CaO + H2O → Ca(OH)2 + heat 1
Explanation : Double displacement reaction is the (ii) Oxidising agent : Cl2 ½
reaction in which two different atoms or group of Reducing agent : H2S ½
atoms are mutually exchanged. Only in reaction (iii) Rancidity, keep food in airtight containers. ½+½
(Na2SO4 + BaCl2 ↓ BaSO4 + 2NaCl), sodium and
[CBSE Marking Scheme, 2012]
barium are mutually exchanged. Hence, correct
option is (b). 6. (i) (a) Calcium oxide reacts vigorously with water
2. To prevent the oil and fats of the chips from being and releases a large amount of heat. 1
oxidized or become rancid. 1 (b) Calcium hydroxide (Slaked lime) i.e., Ca(OH)2
3. Silver chloride (AgCl) and Silver bromide (AgBr) is formed. 1
½+½ (ii) (a) In small amount, solution becomes milky due
2AgCl Sunlight 2Ag + Cl ½ to the formation of calcium carbonate.
→ 2
Or
2AgBr Sunlight
→ 2Ag + Br2 ½ Ca(OH)2 + CO2 → CaCO3 + H2O. 1

(b) In excess, Milkiness disappears because cal-
Those reactions in which energy is absorbed (in cium carbonate changes to calcium hydrogen
form of heat, light or electricity) to break down the carbonate which is colourless in nature.
reactants is called decomposition reaction. 1
[CBSE Marking Scheme, 2012] CaCO3 + H2O + CO2 → Ca(HCO3)2
(Colourless) 1
4. Combination reaction : Single product is formed (iii) Paint forms a protective coating on the surface of
from two or more reactants. iron so that, oxygen and moisture present in the
CaO + H2O → Ca(OH)2 1½ air cannot have a direct contact with iron. 1
Decomposition reaction : A single reactant breaks [CBSE Marking Scheme, 2010]
down to give simpler products.
CaCO ∆3 → CaO + CO 2 1½
[CBSE Marking Scheme, 2012]

qqq

S OLUT I ONS P-7

CHAPTER
SECTION

B
2 ACIDS, BASES AND SALTS

TOPIC-1
Acids and Bases

WORKSHEET-13
Solutions 5. (i) Solution Y is a stronger acid.
1. Correct option : (c) (ii) Strong acid : An acid which dissociates completely
in water and produces a large amount of hydrogen
Explanation : Vanilla essence can be used as an
ions. e.g., HCl.
acid-base indicator by visually impaired students as
it is an olfactory indicator whose odour changes in Weak acid : An acid which dissociates partially in
acidic or basic media.] water and produces small amount of hydrogen
ions. e.g., CH3COOH.
2. Formic acid (Methanoic acid), HCOOH. 1 [CBSE Marking Scheme, 2016] 1 + 1 + 1
[CBSE Marking Scheme, 2010] 6. (i) Tomato — Malic acid and citric acid
(ii) Vinegar — Acetic acid
3. Acids which ionize completely in aqueous solution : (iii) Tamarind — Tartaric acid and citric acid. 1 + 1 + 1
strong acids. ½
7. (i) Scale for measuring [H+] concentration in a
Acids which ionize partially in aqueous
solution is called pH scale.
solution : weak acids ½
Hydrochloric acid, nitric acid : strong acid ½+½ Refer to below figure :
Acetic acid, formic acid : weak acid ½+½ Neutral
[CBSE Marking Scheme, 2012] 0 Acidic nature increasing 7 Basic nature increasing 14

4. (i) Alkali. ½
+ +
NaOH or KOH ½ Increasing in H ion Decreasing in H
(ii) Tooth decay starts when the pH of the mouth is concentration concentration
lower than 5·5. It can be prevented by using tooth
pastes which are generally basic. 1 (ii) pH of Neutral solution is 7.
(iii) Bee-sting has acid that causes pain and irritation. pH of Acidic solution is 0 to below 7.
Baking soda being alkaline, neutralizes acid and pH of Basic solution is 7 to 14.
gives relief. [CBSE Marking Scheme, 2012] 1 [CBSE Marking Scheme, 2016] 1 + 2 + 2

WORKSHEET-14
Solutions Commonly Made Error
1. Correct option : (a)  Students usually get confused with the pH values.
Explanation : Lime juice is acidic in nature as the
juice is obtained from lime, a citrus fruit. This Answering Tip
contains citric acid, and is therefore sour in taste.  Practice the different pH values mentioned with
2. When milk changes into curd, its pH will decrease. examples in the chapter.
Because curd contains lactic acid, so H+ ion
concentration increases and thus pH will decrease. 1 4 . (i) B → 7 (ii) D → 11
(iii) C → 1 (iv) A → 4
3. (i) Increase in the strength of alkali. So the nature (v) E → 9 ½ × 5 = 2½
of solution will be basic. 1 Arrangement : D, E, B, A, C or 11, 9, 7, 4, 1. ½
(ii) Hydroxide ion or hydroxyl ion or OH– ion. 1
(iii) A paper impregnated with the universal indicator 5. (i) (a) Solution of glucose will not conduct
is generally used to measure the pH of a solution. 1 electricity because it does not have ions.
[CBSE Marking Scheme, 2012] (b) Dil. HCl will conduct electricity because it
produces H+ ions in water. ½+½

P-8 S C I ENC E - X
 (ii) HCl is stronger because it gives rise to more H+ 7. (i) Universal indicator is a mixture of many different
ions than acetic acid. 1 indication (or dyes) which give different colours
(iii) The strength of the acid decreases. 1 at different pH values of the entire pH scale. The
[CBSE Marking Scheme, 2012] colour produced by universal indicator is used to
find the pH value of acid or base by matching the
6. colour with the colours on pH colour chart.
(ii) Solution A is acidic and will turn litmus solution
Colour change from blue to red.
Name of the Colour change
solution with Solution B is basic and will turn phenolphthalein
phenolphthalein with blue litmus
from colourless to pink.
1. Sodium (iii) Green colour will be obtained.
carbonate turns pink no change
[CBSE Marking Scheme, 2016] 2 + 2 + 1
2. Hydrochloric
acid no change turns red

3. Sodium
chloride no change no change

[CBSE Marking Scheme, 2012] 1 × 3 = 3

WORKSHEET-15
Solutions Commonly Made Error
1. Correct option : (c)  Usually students get confused with the action of
Explanation : The role of calcium chloride taken in wet or dry litmus paper and reason for the same.
the guard tube is to absorb moisture from the gas.
This is because calcium chloride is used as a drying Answering Tip
agent which absorbs moisture from the hydrogen
chloride (HCl) gas.  The concept of ionization of acids in aqueous
2. Hydrogen gas. 1 medium to give hydrogen or hydronium ions
3. (i) The acid must slowly be added to water. 1 should be kept in mind.
(ii) Otherwise the mixture may splash out causing
burns, as a lot of heat is generated in this process. 1 6. Sweet tooth leads to tooth decay, which is caused
(iii) Dilution of the acid. 1 by the action of bacteria on food particles remaining
in the mouth and acid is formed. As a result, the
4. Zn + 2NaOH → Na2ZnO2 + H2 pH of the mouth falls below 5.5 and the tooth
When a burning splinter is brought near the gas, it enamel dissolves resulting in cavities. Toothpastes
burns with a Pop Sound. are generally basic, they neutralize the excess acid
Gas – Hydrogen / H2 produced in the mouth and prevent tooth decay.
[CBSE Marking Scheme, 2018] 1 + 1 + 1 [CBSE Marking Scheme, 2018] 3

Commonly Made Error Ans. (i) HCl will give rise to more H+ ions and
 Usually students get confused with the liberation of CH3COOH produces less H+ ions on dissociation.
gas in the reaction of zinc with acids and bases. The colour of pH paper depends on the
concentration of H+ ion. Colour becomes red for
high H+ concentration. 1
Answering Tip (ii) Aqueous solution of acids have H+ ions which
 Keenly observe the reactions during practicals, zinc carry electric current through the solution. 1
reacts with both acids and bases to give out hydrogen
(iii) (a) Most acidic : A, Most basic : C ½+½
gas but remember the above said reactions are not
(b) C, B, D, A / C < B < D < A ½+½
possible with all metals.
(c) In C- blue, In D- green ½+½
[CBSE Marking Scheme, 2012]
5. Wet blue litmus paper ½
Reason : Hydrogen ions are produced by HCl in Answering Tip
the presence of water. 1
 Do not overlook any part of a question and avoid
It has acidic nature. ½
being in a hurry to conclude the answer.
HCl + H2O → H3O+ + Cl– 1
[CBSE Marking Scheme, 2018]

S OLUT I ONS P-9

 WORKSHEET-16
Solutions Na2CO3 + 2HCl → 2NaCl + H2O + CO2
Sodium Hydrochloric Sodium Carbon
1. Correct option : (d)
carbonate acid chloride dioxide
Explanation : Lime will not give carbon dioxide (CO2)
on treatment with dilute acid because lime is calcium
1+1+1
oxide which gives salt and water on reaction with acids Commonly Made Error
but no CO2. On the other hand, marble, limestone and
 Students usually get confused in the different
baking soda are metal carbonates and bicarbonates
reactions of acid with different substances.
which forms CO2 on treatment with acids.
2. pH = 2 (lower the pH, stronger the acid). 1 Answering Tip
3. (i) They react with metals to give out hydrogen gas.  Keenly observe the reactions during practical, make
e.g., a list and practice all the reactions of acids with
Zn + 2HCl → ZnCl2 + H2 other substances for one to be confident.
Zinc Hydrochloric Zinc chloride Hydrogen
4. (i) Solid sodium chloride and conc. sulphuric acid.1
acid
(ii) Dry litmus — no change ½
(ii) They react with bases to form salt and water. e.g.,
Wet blue litmus — changes to red ½
2NaOH + H2SO4 → Na2SO4 + 2H2O
(iii) HCl + H2O → H 3 O+ + Cl–
Sodium Sulphuric Sodium Water
hydroxide acid sulphate (Hydronium ion) (Chloride ion)
(iii) They react with metal carbonates to liberate CO2 1
gas. e.g., [CBSE Marking Scheme, 2012]

5. (i) Bulb A and B do not glow but bulb C glows. 1

(ii) Glucose and alcohol solutions do not conduct electricity as they do not have ions. 1
Dil. HCl contains ions so the flow of ions is responsible for the flow of current.
(iii) After replacement, bulb glows in B as NaOH solution contains ions (Na+ and OH– ions). 1
[CBSE Marking Scheme, 2012]
6. (a) (b) Solution A : Basic solution
Indicators Acid Base Because, [H+] is lesser than 1.0 × 10–7.
Solution B : Acidic solution
Litmus solution Blue to Red Red to Blue Because [H+] is greater than 1.0 × 10–7 m 3+2
Phenolphthalein Colourless Pink
Methyl orange Yellow Pink

WORKSHEET-17
Solutions (vi) Liquid hydrogen is used as a fuel in rockets. ½ × 6

1. Correct option : (b) 4. (i) Mg + 2HCl → MgCl2 + H2

Explanation : Citric acid is an example of organic acid Hydrogen gas is produced.
or edible acid. (ii) HCl + NaOH → NaCl + H2O
Neutralization reaction
2. 1 M HCl has a higher concentration of H+ ions
(iii) 2HCl(aq) + CaCO3(s) → CaCl2(aq) + H2O + CO2
because when HCl dissolves in water it dissociates
Calcium chloride is formed.
completely into ions while CH3COOH is a weak acid
[CBSE Marking Scheme, 2016] 1 + 1 + 1
and does not dissociate into ions completely. 1
3. (i) Hydrogen gas.
(ii) Is soluble in water. 5. (i) If H+ ion concentration is more, pH will be
less and vice versa or pH of a solution is inversely
(iii) It is lighter than air.
proportional to H+ ion concentration or pH = –log
(iv) Test for H2 Gas : Bring a burning matchstick near [H+].
the gas jar. It burns with a pop sound. (ii) This solution is neutral.
(v) Zn(s) + H2SO4(dil) → ZnSO4(aq) + H2(g)↑

P-10 S C I ENC E - X
 (iii) 1M NaOH solution. (v) As hydronium [H3O]+ ion.
(iv) When the pH in the mouth is below 5.5, bacteria [CBSE Marking Scheme, 2016] 1 + 1 + 1 + 1 + 1
present in the mouth produce acids by degradation
of sugar and corrode the tooth enamel. It can
be prevented by using toothpastes which are
generally basic.

WORKSHEET-18
Solutions 2. At lower pH below 5.5, the calcium phosphate of
enamel of tooth gets corroded, which leads to tooth
1. Correct option : (b)
decay. 1
Explanation : Citric acid is an example of organic acid
or edible acid.
3.
Stand
Delivery tube
Test-tube
Burning of hydrogen
gas with a pop sound
Dilute
sulphuric
acid Hydrogen
gas
Zinc granules bubbles

Soap bubble filled

with hydrogen

Soap solution

(i) The gas evolved is hydrogen.

(ii) Zn(s) + H2SO4(aq) → ZnSO4(aq) + H2↑
Test for H2 gas : The presence of H2 gas can be tested by passing the gas through soap solution and then bringing
a burning splinter near the soap bubbles filled with the gas. If the gas burns with a pop sound, it is hydrogen.
1+2
4. (i) Hydrogen gas will evolve with greater speed. 6. (a) (i) Acidic — Solution A ½
(ii) Almost same amount of gas is evolved. Basic — Solution B ½
(iii) If sodium hydroxide is taken, hydrogen gas will be +
(ii) Solution A because H ion concentration is
evolved only on heating.
higher in acidic solutions. 1
Zn + 2NaOH → Na2ZnO2 + H2
(b) When same concentration of HCl and acetic acid
Sodium Zincate 1×3 are taken (one molar), then these produce different
amounts of H+ ion (as HCl dissociates completely).
5. (i) Blue and white respectively. 1 HCl gives more H+ but acetic acid gives less H+
(ii) Water of crystallization is lost on heating. 1 ions (as this does not dissociate completely). 1
(iii) Yes, by adding water. 1 [CBSE Marking Scheme, 2012]
[CBSE Marking Scheme, 2012]

S OLUT I ONS P-11

 TOPIC-2
Salts, Their Properties and Uses

WORKSHEET-19
Solutions 6. Water of crystallization is the fixed number of
1. Correct option : (b) water molecules present in one formula unit of
Explanation : Water of crystallisation is the fixed salt. 1
number of water molecules presenting one formula (i) Five molecules of water. ½
unit of a salt. Potassium nitrate, barium sulphate, (ii) Formula : CuSO4.5H2O ½
potassium chloride, sodium nitrate, baking soda, etc., (iii) When heated, its colour changes from blue to
are the salts that do not contain water of crystallisation. white. [CBSE Marking Scheme, 2012] 1
2. Baking soda. 1
7. (i) Distilled water does not conduct electricity
3. (i) Salt that contains water of crystallization. 1
because it does not contain any ionic compound
(ii) Washing soda — Na2CO3.10H2O
like acids, bases or salts dissolved in it.
Gypsum — CaSO4.2H2O
(ii) When we overeat, excess of acid is produced in the
1 stomach which causes burning sensation.
Plaster of paris — CaSO4. H2O. (Any two) 1 + 1
2 (iii) Copper vessels tarnish due to formation of basic
[CBSE Marking Scheme, 2012] copper carbonate which gets neutralized when
rubbed with lemon and the copper vessel regains
4. (i) Bleaching powder : CaOCl2 its shine.
(ii) Ca(OH)2 + Cl2 → CaOCl2 + H2O (iv) Washing soda is sodium carbonate decahydrate
(iii) Two uses other than disinfection are : which when exposed to air loses 10 molecules of
(a) Paper industries (b) Chemical Industries. water and changes to white powder.
[CBSE Marking Scheme, 2016] 1 + 1 + 1 (v) Sodium chloride is a salt of strong acid HCl and
5. (i) X = Chlorine gas, Y = Calcium oxychloride strong base NaOH, so it is neutral.
(ii) Ca(OH)2 + Cl2 → CaOCl2 + H2O Sodium carbonate is a salt of weak acid H2CO3 and
(Dry slaked (Chlorine) (Calcium (Water) strong base NaOH, so it is basic. 1 + 1 + 1 + 1 +1
lime) oxy-chloride) [CBSE Marking Scheme, 2014]
(Bleaching agent) 1+1+1

WORKSHEET-20
Solutions 4. (i) Compound is NaHCO3/baking soda/sodium
hydrogen carbonate
1. Correct option : (d)
Explanation : Sodium carbonate is a basic salt of Manufacture : NH3 + NaCl + H2O + CO2 →
weak acid i.e. carbonic acid and a strong base i.e. NH4Cl + NaHCO3
sodium hydroxide. Heat
(ii) 2 NaHCO3 → Na2CO3 + H2O + CO2
2. The colour of litmus in a solution of sodium [CBSE Marking Scheme, 2016] 1 + 1 + 1
carbonate is blue. 1
3. Chemical formula : CaOCl2 ½ 5. It is a salt produced by the neutralization reaction
between a strong base (NaOH) and a weak acid
Chemical equation :
(H2CO3), hence it is a basic salt. 1
Ca(OH)2 + Cl2  → CaOCl2 + H2O 1 It is heated strongly to produce sodium carbonate.
Uses :
(i) For bleaching cotton and linen in textile industry½ 2NaHCO3 ∆ → Na2CO3 + H2O + CO2 1

(ii) As an oxidising agent in a chemical industry. ½ Sodium carbonate is recrystallized to produce
(iii) For disinfecting water. ½ washing soda.
[CBSE Marking Scheme, 2012] Na2CO3 + 10H2O  → Na2CO3.10H2O 1
[CBSE Marking Scheme, 2012]

P-12 S C I ENC E - X
 6. CaCO3 + 2HCl ¾® CaCl2 + H2O + CO2 1 Activity : (i) Heat a few crystals of copper sulphate
Marble is calcium carbonate and on reacting with in a dry boiling tube.
HCl releases CO2. (ii) Add 2-3 drops of water on the sample of copper
Ca(OH)2 + CO2 ¾® CaCO3 + H2O sulphate obtained after heating.
(Lime water) (White ppt) After heating, water is removed and salt turns
Lime water turns milky due to the formation of white.
white precipitate of CaCO3. On passing excess
If crystals are moisten again with water, blue colour
CO2, milkiness disappears because Ca(HCO3)2 is
reappears. Water of crystallization is fixed number
formed, which is soluble in water. 1
of water molecules present in one formulae unit
CaCO3 + H2O + CO2 ¾® Ca(HCO3)2 1
of a salt. Five water molecules are present in one
Soluble in water
formula unit of copper sulphate.
[CBSE Marking Scheme, 2012]
(b) Calcium sulphate hemihydrate – CaSO4.½H2O

7. (a) Hydrated copper sulphate – CuSO4.5H2O Calcium sulphate dihydrate – CaSO4.2H2O.

Anhydrous copper sulphate – CuSO4. [CBSE Marking Scheme 2015] 3+2

WORKSHEET-21
Solutions 1
4. (i) (a) CaSO4. H2O
1. Correct option : (c) 2
Explanation : The common salt obtained is an (b) CaSO4.2H2O ½+½
important raw material for various materials of (ii) By mixing with water
daily use, such as sodium hydroxide, baking soda, 1 1
CaSO4. H2O + 1 H2O → CaSO4.2H2O 1
washing soda and many more. 2 2
2. When baking soda is heated, it decomposes to (iii) For making toys, for supporting fractured bones. 1
produce sodium carbonate, water and CO2 gas. 1 [CBSE Marking Scheme, 2012]

2NaHCO3 ∆ → Na2CO3 + H2O + CO2 1 5. The three products are :

CO2 gas produced during the reaction makes the Sodium hydroxide (NaOH), Chlorine (Cl2) and
cake or bread spongy and fluffy. 1 Hydrogen (H2). 1
[CBSE Marking Scheme, 2012] NaOH — for soaps and detergents and paper
making or artificial fibers.
3. (i) It loses water molecules and becomes calcium Cl2 — PVC, pesticides, CFCs. 1

1 H2 — fuels, margarine, NH3 for fertilizers. 1
sulphate hemihydrate, CaSO4. H2O or plaster of
2 [CBSE Marking Scheme, 2012]
paris.
6. (i) Hydrochloric acid and sodium hydroxide.
(ii) Blue copper sulphate crystals on heating lose
water of crystallisation and become white. When NaOH + HCl  → NaCl + H2O ½+½
these crystals are left open, will absorb moisture NaCl, sodium chloride, Ocean water. ½+½
from air and regain blue colour. (ii) Deposits of solid salt are found in several parts of
(iii) On passing CO2 gas through lime water, insoluble the world. These large crystals are called rock salt. 2
calcium carbonate is formed. On passing excess of
Colour of rock salts is brown, due to impurities.
CO2, soluble calcium bicarbonate is formed.
Ca(OH)2(aq) + CO2(g) → CaCO3(s) + H2O(l) (iii) 2NaCl + 2H2O electricity
→ 2NaOH + Cl2 + H2 1

Insoluble
[CBSE Marking Scheme, 2012]
CaCO3(s) + CO2(g) + H2O(l) → Ca(HCO3)2(aq)
Soluble 1+1+1
[CBSE Marking Scheme, 2014]

S OLUT I ONS P-13

 WORKSHEET-22
Solutions Tartaric acid is used to avoid the bitter taste by
reacting with Na2CO3 formed.
1. Correct option : (b) Heat
2NaHCO3 + H+ From
 → Na2CO3 + H2O +
Explanation : Baking powder is a mixture of baking acid

soda (sodium hydrogen carbonate) and a mild CO2 [CBSE Marking Scheme, 2012] ½
edible acid such as tartaric acid.
2. (a) Copper sulphate crystals (CuSO4.5H2O) 5. (i) Plaster of Paris / Calcium sulphate hemihydrate
1
5 molecules of water are present in one formula CaSO4. H2O. 1
2
unit of copper sulphate. When heated, its loses its
water and salt turns white. 1 Heat 1 1
(ii) CaSO4.2H2O 373
 K
→ CaSO4. H2O + 1 H2O 1
(b) Basic Salt : 2 2
(i) Base — KOH (Potassium hydroxide) ½ (iii) Used as a plaster for supporting fractured bones.½
Acid — H2SO4 (Sulphuric acid) ½ Used for making toys / statues / decorative items.½
(ii) Base — NH4OH (Ammonium hydroxide) ½ [CBSE Marking Scheme, 2012]
Acid — HCl (Hydrochloric acid) ½
6. (a) The common name of CaOCl2 is Bleaching
[CBSE Marking Scheme, 2012]
powder. ½
3. (i) X – Sodium hydrogen carbonate or baking soda. By passing chlorine into dry slaked lime (Ca(OH)2)
Y – CO2 Ca(OH)2 + Cl2 
→ CaOCl2 + H2O 1
(ii) NaHCO3 + HCl → NaCl + H2O + CO2↑ Two uses :
1+1+1 (i) Used for bleaching cotton and linen in the
[CBSE Marking Scheme, 2013]
textile industry and wood pulp in paper
industry etc. ½
4. Baking powder. ½ (ii) It is used for disinfecting drinking water. ½
Baking soda and tartaric acid. ½+½
(b) Washing Soda : Na2CO3.10H2O ½
On heating or mixing with water, sodium
bicarbonate reacts with hydrogen ion from acid By heating baking soda, sodium carbonate is
and releases CO2 that makes the cake soft and obtained, its recrystallization gives washing soda.1
fluffy. 1 2NaHCO3 Heat
→ Na2CO3 + H2O + CO2 ½

Na2CO3 + 10H2O → Na2CO3.10H2O ½
[CBSE Marking Scheme, 2012]

qqq

P-14 S C I ENC E - X
CHAPTER
SECTION

B
3 METALS AND NON-METALS

TOPIC-1
Properties of Metals and Non-Metals

WORKSHEET-23
Solutions Answering Tip
1. Correct option : (c)  Learn and understand the different physical and
chemical properties of metals and non-metals with
Explanation : Metals in their pure state have a
unique reactions.
shining surface and thus, dullness property is not
shown by metals. 6. (i) Non-metals are electron acceptors, they cannot
2. Metal – mercury, Non-metal – bromine. ½+½ supply electrons so as to convert H+ ion to H2(g).
(ii) Like metals, hydrogen can lose an electron to form
3. Three properties of sodium are :
positive H+ ion.
(i) Sodium is so soft that it can be cut with a knife. 1 (iii) Aluminium is covered with a strong protective
(ii) It has low density. 1 layer of oxide which protects the metal from further
(iii) It has low melting points. 1 corrosion. 1+1+1
4. (a) (i) Oxides of metal are basic / amphoteric. ½ 7. (a)
Oxides of non-metal are acidic/neutral. ½ S. Metals Non-Metals
(ii) Metals are good conductors whereas non- No.
metals are poor conductors. 1
1. Lose electrons to form Gain electrons to
(b) Iodine and Mercury. ½+½ positive ions/ are elec- form negative ions/
5. (i) X — Carbon, Y — Diamond, Z — Graphite tropositive in nature. are electronegative
(ii) Non-metal in nature.
(iii) Tin. 1+1+1 2. React with dilute acids Do not react with
[CBSE Marking Scheme, 2016] to liberate hydrogen gas. dilute acids .
3. Generally metal oxides Generally non-met-
Commonly Made Error are basic in nature. al oxides are acidic
 Usually students get confused with different in nature.
physical and chemical properties of metals and non- (b) (i) Painting
metals. (ii) Oiling
(iii) Galvanization
(vi) Alloying (or any other)
 [CBSE Marking Scheme, 2018] 5

WORKSHEET-24
Solutions Commonly Made Error
1. Correct option : (a)  Usually students get confused with different
Explanation : The ability of metals to be drawn into physical properties of metals.
thin wires is called ductility.
2. Gallium. 1 Answering Tip
3. (i) Gold and silver are most malleable metal and  Learn and understand all the physical properties of
most ductile metal. metals.
(ii) Silver is the best conductor of heat and lead is the 4. (i) 4Al + 3O2 → 2Al2O3, Aluminium oxide 1
poorest conductor of heat. (ii) 3Fe + 4H2O → Fe3O4 + 4H2, Ferric oxide 1
(iii) Metal with highest melting point : Tungsten, iron.
(iii) Ca + 2H2O → Ca(OH)2 + H2, because hydrogen
Metal with lowest melting point : Gallium, Cesium.
½ + ½ + ½ + ½ +½ + ½ sticks to calcium. 1

S OLUT I ONS P-15

 Commonly Made Error 7. Different metals react with oxygen at different rates.
e.g., Sodium (Na) and potassium (K) catch fire, if left
 Usually students get confused with different in open. Hence, these are the most reactive metals.
reactions of metals showing their reactivity. To prevent accidental fires, these metals are kept
immersed in kerosene oil. Magnesium burns in air
Answering Tip only by heating. So, it is less reactive than sodium
 Learn and understand the reactivity series of metals and potassium. Copper (Cu) does not burn on
and understand reactivity of metals in different heating but blister copper burns. Hence the order
reactions. of reactivity of these metals with oxygen is :
5. (i) 2Cu + O2 → 2CuO 1 Na > Mg > Cu.
(ii) 4Al + 3O2 → 2Al2O3 1 Metals react with water to produce a metal oxide
(iii) Al2O3 + 2NaOH → 2NaAlO2 + H2O 1 and hydrogen gas. Sodium (Na) and potassium
6. (i) Take three test-tubes. Place clean iron nails in (K) react violently with cold water. So the reaction
each test-tube. is violent and exothermic. Magnesium (Mg) does
not react with cold water. It reacts with hot water.
(ii) Pour some water in test-tube-1, cork it.
Metals like lead, copper, silver do not react with
(iii) Pour water (boiled/distilled) in test-tube-2, add
water at all. The reactivity series of metals towards
some oil and cork it.
water is :
(iv) Put some anhydrous calcium chloride in test-tube-3
Na > Mg > Cu. 1½ + 1½
and cork it.
(v) After 2-3 days, we observe that the nails in test-tube Commonly Made Error
1 rust because they are exposed to air and water  Usually students get confused with the order of
both, while nail in test-tube 2 and 3 do not rust. reactivity of metals.
This shows that rusting of iron takes place in the
presence of air and moisture both. 3

TOPIC-2
Ionic compounds, Metallurgy and Corrosion
WORKSHEET-25
Solutions
1. Correct option : (a) +
Explanation : Sodium and potassium are extracted
Cathode
by electrolytic reduction. Metals obtained after
(Pure metal)
electrolytic reduction are in pure form. But, copper
and gold are in impure form after extraction. Copper
and gold are refined by electrolytic refining methods. Anode
2. Reduction of iron oxide to iron by aluminium is (Impure Solution of
called thermite reaction. 1 metal) metal salt
Fe2O3 + 2Al → 2Fe + Al2O3 + Heat. Anode mud (Electrolyte)
Electrolytic refining
3. Electrolytic Refining : This method is widely used
as purification of metals like zinc (Zn), copper (Electrolytic refining of copper with explanation)
(Cu), aluminium (Al), chromium (Cr), tin (Sn), lead
[CBSE Marking Scheme, 2014] 3
(Pb), nickel (Ni) and gold (Au).
In this process, impure metal is used as anode, a Commonly Made Error
strip of pure metal is used as cathode and soluble
salt of metal is used as electrolyte. On passing  Usually students get confused between anode,
electric current through the electrolyte, cations cathode and the electrolyte.
move towards cathode, gain electrons and pure
metal gets deposited on cathode. Answering Tip
In electrolytic refining of copper, the impurities  Carefully understand the electrolytic reductions of
left behind at anode called anode mud contains metals. Practice drawing a well-labelled diagram.
valuable metals such as gold and silver which can
be recovered in the native state.

P-16 S C I ENC E - X
 4. (i) Calcium (20) – 2, 8, 8, 2 Oxygen (8) – 2, 6 1 5. Calcination : Heating the ore strongly in limited
–2 amount of oxygen.
Ca + O (Ca2+) O Heat
ZnCO3 → ZnO + CO2
1
(ii) Constituents metals of bronze – copper, tin. 1 Reduction : Converting ZnO to Zn with the help
[CBSE Marking Scheme, 2012] of carbon.
ZnO + C → Zn + CO 1+½+1+½
[CBSE Marking Scheme, 2013]
Commonly Made Error
 Usually students get confused in writing the 6. (i) Diagrams (activity 3.5 fig. 3.1)
electronic configuration, drawing the dot structure, Procedure
cations and anions. Observation : Heat is transferred from one end of
metal wire to the free end of wire which melts the
Answering Tip wax and pin falls. Shows metals conduct heat.
(ii) Ore HgS cinnabar
 Learn to write the electronic configuration and
Roasting : 2HgS(s) + 3O2(g) → 2HgO(s) +
understand the concept of sharing in ionic
2SO2(g)
compounds. Electron donating atoms become ∆
cations and electron receiving atoms become anions. Reduction : 2HgO(s)  → 2Hg(l) + O2(g) 5
[CBSE Marking Scheme, 2018]

WORKSHEET-26
Solutions 4. (i) A large amount of energy is required to break
the strong inter-ionic attraction.
1. Correct option : (c)
Explanation : Sulphur is present in air and silver (ii) When ionic compounds is present in molten state,
gets tarnished after the reaction of sulphur. Because crystal structure deforms and they can easily
conduct electricity with the mobile ions.
of formation of layer of silver sulphide silver articles
become black on prolonged exposure to air. (iii) Due to the strong force of attraction between the
2. Paint forms a protective coating on the surface positive and negative ions. 1+1+1
of iron. So, oxygen and moisture present in the [CBSE Marking Scheme, 2012]
air cannot have a direct contact with the metal.
Therefore, surface gets protected against rusting. 1 5. Cinnabar (HgS) is an ore of mercury.
When Cinnabar, HgS is heated in air, it is first
3. (i) Self reduction converted into HgO, HgO is then reduced to Hg
Heat on further heating.
2HgO 
→ 2Hg + O2
∆
Heat
2Cu 2O + Cu 2S 
→ 6Cu + SO2 1 2HgS + 3O2 
→ 2HgO + 2SO2

∆
(ii) Reduction using carbon 2HgO 
→ 2Hg + O2
ZnO + C → Zn + CO [CBSE Marking Scheme, 2016]
Sometimes, some highly reactive metals are used
etailed Answer :
D
as reducing agents.
e.g., 3MnO2 + 4Al → 3Mn + 2Al2O3 + heat Cinnabar (HgS) is an ore of Mercury.
or Fe2O3 + 2Al → 2Fe + Al2O3 + heat 1 The metals being less reactive can be obtained by
(iii) Electrolytic reduction reducing their oxides to metals by heating alone.
e.g., Na, Mg and Ca are obtained by electrolysis of So, when Cinnabar is heated in air, it first changes
their molten chlorides. 1 into its oxide and then into mercury metal.
[CBSE Marking Scheme, 2012] Heat
2HgS(s) + 3O2(g) 
→ 2HgO(s) + 2SO2(g)
Commonly Made Error (Cinnabar) (Air) Mercury Sulphur
 Students often get confused between the methods
oxide dioxide
of reduction of metals.
∆
2HgO(s) → 2Hg(l) + O2(g)
Answering Tip
Mercuric (II) oxide Mercury metal Oxygen
 Make a list of less, moderate and highly reactive  1+1+1
metals and learn the different processes involved in
their reduction.

S OLUT I ONS P-17

 Ans. (i) (a) Calcination, (b) Reduction, (c) Purification (ii) In the extraction of copper from its Sulphide Ore
when ore is subjected to roasting some of it is
(in the given sequence only)
oxidised to Cu2O which reacts with the remaining
(ii) Sulphide ore of copper is heated in air. Cu2S to give copper metal. In this process, Cu2S
2Cu2S + 3O2 → 2Cu2O + 2SO2 behaves as reducing agent.
2Cu2O + Cu2S → 6Cu + SO2
∆
2Cu2S(s) + 3O2(g)  → 2Cu2O(s) + 2SO2 (g)
(Note : Full marks to be awarded even when only ∆
equations are written) 2Cu2O(s) + Cu2S(s)  → 6Cu(s) + SO2(g)

Labelled diagram of electrolytic refining of copper In electrolytic refining of copper, the impurities
–
Key– +e
left behind at anode called anode mud contains
– valuable metals such as gold and silver which can
be recovered in the native state.
Cathode Anode

Acidified +
copper
Cu 2+ sulphate Cathode
solution (Pure metal)
Cu
Tank
Anode
Impurities (Impure Solution of
(anode mud) metal) metal salt
 [CBSE Marking scheme, 2018] 5 Anode mud (Electrolyte)
etailed Answer :
D Electrolytic refining
(i) A metal is obtained from its carbonate ore by
converting it into its oxide by the process of Commonly Made Error
Calcination.
 Usually students get confused with the order of

In Calcination, the ore is heated to a high the steps in extraction and sulphide ore extraction
temperature in the absence of air. technique.
∆
ZnCO3(s)  → ZnO(s) + CO2(g)
Answering Tip

The metal oxides obtained by calcination are
converted into the free metal by using reducing  Understand the concept of ore extraction and the
agents like carbon, aluminium which depend upon order of the steps involved in the extraction and
the reactivity of the metal. the equation separately for sulphide and oxide ores.
Diagrams are equally important.
ZnO + C → Zn + CO.

Impure Zn obtained is refined by electrolysis
process.

WORKSHEET-27
Solutions
1. Correct option : (c) 4. Steps :
Explanation : In the process of galvanization, iron is (i) Roasting : Sulphide ores are converted into oxides
covered by a coat of zinc. This layer of zinc prevents by heating strongly in the presence of excess air. 1
Heat
iron from getting rusted. 2ZnS + 3O2 → 2ZnO + 2SO2 ½

2. Because aluminium has greater affinity for oxygen
than for carbon, therefore carbon cannot reduce (ii) Reduction : Metal oxides are then reduced to the
corresponding metal by using suitable reducing
alumina (Al2O3) to aluminium. 1
agent (carbon). 1
3. (i) Aluminium (Al) ½ ZnO + C → Zn + CO ½
(ii) Iron (Fe) ½ [CBSE Marking Scheme, 2012]
(iii) Fe2O3 + 2Al → 2Fe + Al2O3 + Heat. 1 5. The metal is copper. The metal has corroded because
(iv) Use : To join railway tracks or cracked machine of being exposed to moist air. Green compound is
parts. [CBSE Marking Scheme, 2012] 1 basic copper carbonate [CuCO3. Cu(OH)2]

P-18 S C I ENC E - X
 Two ways to prevent this process : 7. The two reactive metals are sodium (Na) and
Painting, Greasing, Oiling, Galvanizing (Any two) potassium (K). Sodium reacts with chlorine. 1 + 1
1+1+1
Na → Na+ + e–
6. (i) Metal X is obtained simply by heating their 2, 8, 1 2, 8
oxides with carbon. e.g., mercury is obtained from
(Sodium cation)
cinnabar.
Cl + e → Cl–
–
2HgS(s) + 3O2(g) Heat
 → 2HgO(s) + 2SO2(g)
2, 8, 7 2, 8, 8
2HgO(s) Heat
→ 2Hg(l) + O2(g) 1 (Chloride anion)
–
(ii) Metals in the middle of activity series can be Na + Cl [Na+] Cl
1
obtained by heating with carbon, e.g.,
Ionic compounds have high melting and boiling
ZnO(s) + C(s) Heat
 → Zn(s) + CO(g) 1
points.
(iii) Metals high in the reactivity series are obtained by They are good conductors of electricity.
electrolytic reduction of molten ores e.g., in NaCl They are generally soluble in water and insoluble
NaCl → Na+ + Cl– in organic solvents.
They are solid, hard and brittle. 4×½ = 2
Na + e– → Na (at cathode)
[CBSE Marking Scheme, 2012]
2Cl– → Cl2 + 2e– (at anode) 1
[CBSE Marking Scheme, 2014]

qqq

S OLUT I ONS P-19

SECTION
CHAPTER

B
4 CARBON COMPOUNDS

TOPIC-1
Carbon and its Properties, Homologous Series and
IUPAC Names

WORKSHEET-28
Solutions 5. Carbon cannot form C4+ cation because removal
of 4 electrons from a carbon atom would require a
1. Correct option : (b) large amount of energy. ½
Explanation : Carbon is present in the atmosphere Carbon cannot form C4– anion because it would be
in the form of carbon dioxide gas (CO2) in air (only difficult for the nucleus with 6 protons to hold on
0.03%). Carbon also occurs in the earth’s crust in to 10 electrons. ½
the form of minerals like carbonates. It also occurs Hence, carbon atoms share electrons forming
in the form of fossil, organic compounds, wood, covalent compounds 1
cotton and wool, etc. products.
Covalent compounds do not form ions/ charged
2. A homologous series is the family of organic particles and therefore do not conduct electricity. ½
compound having the same functional group, and Inter molecular forces of attraction are weak, hence
the successive (adjacent) members of which differ low melting and boiling points. ½
by CH2 unit or 14 mass unit. 1 [CBSE Marking Scheme, 2017]
[CBSE Marking Scheme, 2016] Detailed Answer :
Atomic number of carbon is six. This means that it
3. A group of organic compounds having the
has four electrons in its outermost shell and it needs
same functional group and similar structures
four more electrons to attain noble gas electronic
in which any two successive members differ by
configuration. It does not form C4+ cation, as the
– CH2. 1
removal of four valence electrons will require a
(i) All members have similar chemical properties ½
huge amount of energy. The cation formed will
(ii) There is gradation in the physical properties. ½
have six protons and two electrons. This makes it
(or any other) highly unstable. Carbon is unable to form C4– anion
Name — Ethanoic acid/Acetic acid ½ as its nucleus with six protons will not be able to
Formula — CH3COOH ½ hold ten electrons. Thus, carbon achieves noble
[CBSE Marking Scheme, 2017] gas electronic configuration by sharing its four
electrons with other elements, i.e. it forms covalent
4. The compounds that are formed due to sharing of compounds.
electrons between two atoms/compounds having (i) Covalent compounds are bad conductors of
covalent bonds. 1 electricity due to lack of free electrons.
Ionic compounds are formed due to transfer of (ii) Covalent compounds are formed by covalent bonds
electrons from one atom to another/compounds and it has been found that the intermolecular forces
having ionic bonds/compounds having attraction of attraction in covalent compounds are weak.
between oppositely charged ions 1 Thus, their melting and boiling points are quite low.
(i) They are poor conductors of electricity ½ They cannot ionize producing ions in molten state
(ii) They have low melting and boiling point. ½ which are mainly require to conduct electricity.
(or any other) 3+1+1
[CBSE Marking Scheme, 2017]

6.

P-20 S C I ENC E - X


[Topper Answer, 2017] 5

WORKSHEET-29
Solutions (i) The parent chain should have the most number of
carbon atoms.
1. Correct option : (b) (ii) The branching cannot be done from the first on
Explanation : Ethane has 7 covalent bonds. One the last atom carbon atom of the structure. 1+1+1
bond is between two carbon atoms and rest of the [CBSE Marking Scheme, 2015]
six are between hydrogen atoms.
2. The molecular formula of first two consecutive
4. (a) Carbon compounds form covalent bonds/
members of this series is :
do not dissociate into ions/do not have charged
CH3Cl (Chloromethane) particles (ions) ½
C2H5Cl (Chloroethane) ½+½ (b) Cyclohexane
3. Isomers are the compounds which have the same H H H H
molecular formula but different structural formula. C C
H H
somers of Butane :
H H H H C C
(i)
| | | | H C C H
H— C—C— C—C —H
H H H H
| | | |
H H H H Total no. of single bonds = 18
n-butane (OR any other cycloalkane with corresponding
(ii) H| number of bonds) 1 + 1 +1

H — C—H [CBSE Marking Scheme, 2018]

H
|
H— C — C—
| H
|
C—H
Commonly Made Error
 Students usually get confused in the structure
and the number of single bonds present in the
| | |
compound.
H H H
iso-butane Answering Tip
We cannot have isomers of the first three members
 Practice the structure of different and special
of the alkane series because of the following laws
compounds involved in the homologous series.
of isomers :

S OLUT I ONS P-21

 H H H 5. (i) The compounds that contain the same molecular
formula but different structures are called isomers.
| | |
The isomers of a compound have different physical
5. (i) H — C — C — C — C — OH
properties. 1
| | | ||
(ii) The two possible isomers of the compound with
H H H O
molecular formula C3H6O are :
Carboxylic group – COOH H H O H O H
H H H

—
—

—
—

—
| | | H—C—C—C—H H — C —C —C — H
(ii) H — C — C — C — Br

—
H H H H
| | |
(Propanal) (Propanone) 2
H H H
Halogen atom – Br (iii) Electron dot structure of propanal :
H H H H O
 


| |
(iii) H — C — C — C —
—C —H H C C C H






—
 
| |
H H
H H
Electron dot structure of propanone :
Triple bond – C ≡ C –
H O  H
[CBSE Marking Scheme, 2015] 1+1+1   
H C C C H
Commonly Made Error






 
 Students get confused and do mistakes while H H 2
writing the structure, name and also with the
functional group involved. Commonly Made Error
 Students get confused in writing isomeric structures
Answering Tip and make mistake while drawing electron dot
structures.
 Please understand the basic concept of alkane,
alkene and alkynes with their general formula. Then
Answering Tip
make a list of the compounds involved in the above
and also learn different functional groups. Drawing  Learn the concept of isomerism and write structures
structures and practicing them is mandatory. accordingly.

WORKSHEET-30
Solutions Commonly Made Error
1. Correct option : (c)  Students write irrelevant stories. Be specific. Read
Explanation : Pentane contains four C—C and question carefully and write only what is asked.
twelve C—H covalent bonds. Therefore, total 16
covalent bonds.
Answering Tip
2. The molecular formula for two consecutive  Do not overlook any part of a question and avoid
members of this series are : being in a hurry to conclude the answer.
CH3Br (Bromomethane)
4.
C2H5Br (Bromoethane) ½+½
(i) Ethane : C2H6
3. (i) Vegetable oil is converted into saturated fat. ½ H H
(ii) The reaction is Hydrogenation 1 x x
(iii) Vegetable oil is liquid and saturated fat is solid at H H
x
C x
x
room temperature 1 H C C H H C H
x
(iv) Nickel acts as a catalyst. ½ H H x x
[CBSE Marking Scheme, 2018]
H H
½+½

P-22 S C I ENC E - X
 (ii) Ethene : C2H4 (ii) Conc. H2SO4 acts as a dehydrating agent. 1
Conc H 2SO 4
H H

CH3CH2OH 
443K
→ C2H4 + H2O 1

H H [CBSE Marking Scheme, 2015]

C C H C C H 6. Two examples of covalent compounds are ethanol,
H H and ethanoic acid. ½+½
½+½ Difference between the properties of covalent and
(iii) Ethyne : C2H2 ionic compounds :

S. Covalent Ionic
H C C H H C C H
No. Compounds Compounds
½+½
[CBSE Marking Scheme, 2015] (i) They are readily They are not soluble in
soluble in organic organic solvent.
solvent.
Commonly Made Error
(ii) They do not ionize. They ionize in organic
 Students often make mistake while drawing electron medium.
dot structures.
(iii) They are bad They are good
Answering Tip conductor of heat conductor of heat and
and electricity. electricity.
 Understand the basic concepts involved in drawing
the dot structure. Make sure that you have dots for (iv) They have weak force They have strong force
all shared bonds. of attraction between of attraction between
the molecule. the molecule. 4
5. (i) Ethene
H H
C C

H H 1

WORKSHEET-31
Solutions H
|
1. Correct option : (b) — C— H
H—C —
Explanation : The nearest inert gas from carbon |
is Neon and it is an element which attempts to H
attain the electronic configuration of its nearest
The first member of alkynes is ethyne and its
noble gas while attaining a fully-filled outermost
structure is given below :
shell.
H—C ≡ C—H 1+1+½+½
2. The molecular formula for first two consecutive
member of this series is : Commonly Made Error
CH3OH (Methanol)  Students get confused and do mistake while writing
C2H5OH (Ethanol) ½+½ the structure and its name.
3. A homologous series is a series of organic
compounds that belongs to the same family (i.e., Answering Tip
possesses the same functional group) and show  Drawing structures and practicing them is
similar chemical properties. The members of this mandatory.
series are called homologous as they differ from 4.
A homologous series is a series of organic
each other by the number of CH2 units in the main compounds which belong to the same family (i.e.
carbon chain. possess same functional group) and show similar
General Formula chemical properties. The members of this series are
Alkenes : CnH2n, Alkynes : CnH2n-2 called homologous and differ from each other by
Structures : the number of CH2 units in the main carbon chain.

Molecular formula of two consecutive
The first member of alkenes is ethene and its
members of homologous series of aldehydes is
structure is given below.
CH3-CHO and CH3-CH2-CHO.

S OLUT I ONS P-23

 (i) Physical properties of compounds of homologous 6. (i) Pass the vapours of the given samples of saturated
series are determined by the number of -CH2 units. and unsaturated hydrocarbons into bromine water
(ii)
Chemical properties are determined by the taken in two separate test-tubes. The one which
functional groups attached to the compounds. discharges the colour of bromine water is that of
unsaturated hydrocarbon and the other represents
Thus here, additional -CH2 part determine
saturated hydrocarbon.  (Or any other test) 1
the physical properties, while the -CHO part
determines the chemical property. (ii) On burning ethane in air, the products obtained
are carbon dioxide and water, along with heat and
5. Functional group is an atom or a group of atoms light.
that is bonded to a carbon chain. It defines the
2C2H6(g) + 7O2 (g) ® 4CO2 (g) + 6 H2O (l) + Heat
chemical property of the organic compound.
+ Light 2
Compound Functional Structure (iii) It is considered a substitution reaction because the
Group hydrogen atoms of methane (CH4) are replaced by
chlorine atoms one by one. 1+1
Ethanol Hydroxyl CH3CH2OH
(–OH)
Ethanoic acid Carboxylic acid CH3COOH
(–COOH)
1+1+1

WORKSHEET-32
Solutions Carboxylic acid – COOH

1. Correct option : (c) 1+½+½+½+½

Explanation : Oxygen has a complete octet while Commonly Made Error
each atom of hydrogen has two electrons in
outermost shell.  Students confuse with names and structure of
2. Ethane (C2H6) functional groups.
Propane (C3H8) ½+½
OR Answering Tip
 Make a list of different functional groups with its
structure and learn them.
4. Name and general formula of hydrocarbons
undergoing addition reaction with hydrogen :
Name General Formula
Alkene CnH2n
[Topper Answer, 2017]
Alkyne CnH2n-2
Commonly Made Error Essential conditions required for the addition
reaction to occur :
 Usually students get confused between the first few (i) Multiple bonds (double and triple bonds) must
members of alkane, alkene and alkyne series. be present between carbon atoms in the chain of
hydrocarbon.
Answering Tip
(ii) Addition of hydrogen should be carried out in the
 Learn and practice the starting members and first presence of catalyst such as nickel or platinum.
few members of alkane, alkene and alkyne series Chemical Equation :
with their formulae.
( Ni or Pt )
3. Functional group is an atom or a group of atoms CH2=CH2 + H2 → CH3—CH3
that is bonded to a carbon chain. It defines the Ethene Ethane
chemical property of the organic compound.
( Ni or Pt )
CH ≡ CH + H2 → CH2 = CH2
Compound Functional Group Ethyne Ethene
1+1+1
Alcohol – OH
5. (a) Four characteristics of homologous series are :
Aldehyde – CHO
(i) Same functional group. ½
Ketone – C=O (ii) Similar chemical properties. ½

P-24 S C I ENC E - X
 (iii) Regular gradation in physical properties. ½ (ii) Low melting and boiling point compared to ionic
(iv) Successive member differ by – CH2. ½ compound.
(iii) Poor conductors of electricity.
(b) O C O (b) Carbon is a versatile element because it forms
1 covalent bonds with large number of elements
and has catenation capacity to form compounds
[CBSE Marking Scheme, 2012] by chain of bonds with itself. 3+2
[CBSE Marking Scheme, 2015]
6. (a) Three physical properties of carbon
compounds are :
(i) Catenation property.

WORKSHEET-33
Solutions 4. (i) Methane
H
1. Correct option : (c) H .
Explanation : Unsaturated hydrocarbons have x

—
multiple covalent bonds (double or triple bond) like H—C—H H .x C x. H
alkene and alkyne.

—
x.
2. Ethane (C2H6) H
Propane (C3H8) ½+½ H
OR 1½
(ii) Ethane
H H
H H . .
x x
—

X
H—C—C—H x. H C C x. H
X
—

x. x.
H H
H H
[Topper Answer, 2017]
1½

Commonly Made Error 5. (i) Functional group : Hetero atom or group of

atoms attached to the carbon chain, which gives
 Usually students get confused between the first few
members of alkane, alkene and alkyne series. specific properties to the compounds, is called a
functional group. 1
Answering Tip (a) Aldehyde group, (b) Carboxylic acid ½+½
 Learn and practice the starting members and first (ii)
Acetic/Ethanoic acid is formed. It is an oxidizing
few members of alkane, alkene and alkyne series agent. ½+½
with their formulae.
[CBSE Marking Scheme, 2016]
3. (i) Catenation. 6. (i) Combustion : 'This reaction involves burning of
(ii) Carbon forms large number of compounds due to
carbon compounds releasing carbon dioxide and
the following :
water with excess heat and light'
(a) Catenation : Carbon forms bond with other
atoms of carbon. CH4 + O2 → CO2 + 2H2O + Energy + Light

(b) Tetravalency : Carbon share four electrons (ii) Addition : It is the reaction in which unsaturated
with other atoms. compound reacts with hydrogen in presence of
(iii) Structure of Cyclohexane : nickel as a catalyst to form saturated compound. '
H
H H Ni
H C
CH2 = CH2 + H2 → C2H6
H
C C (iii) Substitution : This is a reaction in which an atom
H H or group of atoms of a compound are replaced by
C C other atom or group of atoms.
H C H Sunlight
H H
CH4 + Cl2 
→ CH3Cl + HCl
H 1+1+1
[CBSE Marking Scheme, 2014]

S OLUT I ONS P-25

 (iv) Esterification : Reaction between an acid and an (v) Oxidation : When there is an addition of oxygen in
alcohol, which is acid catalysed to form sweet- the reaction, it is known as oxidation reaction. This
smelling compound called esters. reaction takes place in the presence of oxidising
+ agents. ‘
CH3COOH + C2H5OH H
→ CH3COOC2H5 +
C2H5OH alk

KMnO4
→ CH3COOH.
H 2O
1+1+1+1+1
[CBSE Marking Scheme, 2016]

WORKSHEET-34
Solutions 5. (i) Propanal (aldehyde); ½+½
1. Correct option : (d) H H H

—
Explanation : A series of compounds in which the
H—C—C—C—
—O
same functional group substitutes for hydrogen in

—
a carbon chain is called as homologous series. The H H
option (d) follows the generic formula CnH2n, while
others follow the generic formula CnH2n+2. (ii)
Propanone (Ketone); ½+½
2. Propane (C3H8) H H

—
Butane (C4H10) ½+½
H—C—C—C—H
3. An atom or a group of atoms/heteroatoms which

—
—

—
determines the chemical properties of an organic H O H
compound is called functional group. 1
(iii)
Isomers (same molecular formula but different
Name Structural Functional structural formula/different functional group) 1
Formula Group [CBSE Marking Scheme, 2016]
Ethanol H H – OH
6. (i) Hydrogenation reaction : It is the process
H—C—C—O—H in which unsaturated compound reacts with
hydrogen in presence of nickel as a catalyst to
H H form saturated compound.
R R R R
Ethanoic H O – COOH | |
acid Nickel
H — C — C — OH O C =C + H2 H — C — C —H
| |
H — C — OH R R R R
[CBSE Marking Scheme, 2015] 1 + 1 Vegetable Oil Vegetable Ghee

4. (i) C3H4, C2H4 will undergo addition reactions 1

because they are unsaturated compounds. 1 (ii) Oxidation reaction : When there is an addition of
(ii) C3H4 – Alkyne 1 oxygen in the reaction, it is known as oxidation
C2H4 – Alkene [CBSE Marking Scheme, 2012] 1 reaction.

CH3CH2OH + [O] Alkaline

 → CH3COOH + H2O
Commonly Made Error KMnO4
Ethanol Ethanoic acid 1
 Students write vague answers. It seems they are
unaware of the concept of addition reaction and (iii) Substitution reaction : This is a reaction in which
homologous series.
an atom or group of atoms of a compound are
Answering Tips replaced by other atom or group of atoms.
Sunlight
 First understand the homologous series (saturated CH4(g) + CH3Cl(g) Cl2(g) 
→
and unsaturated compounds), then learn their Methane ChloromethaneChlorine
general formula, first few compounds, their + HCl(g)
structures and names.
Hydrogen
 Always remember unsaturated compounds
chloride 1
undergo addition reactions.

P-26 S C I ENC E - X
 (iv) Combustion reaction : Ethanol is highly (v) Saponification reaction : It is the reaction in which
inflammable liquid, i.e., it catches fire very easily. an ester reacts with sodium hydroxide to form
It burns with blue flame in presence of oxygen to sodium salts of acid and alcohol.
form carbon dioxide and water. CH3COOC2H5 + NaOH →
C2H5OH + 3O2 → Ethyl ethanoate Sodium hydroxide
Ethanol Oxygen CH3COONa + C2H5OH
2CO2 + 3H2O Sodium ethanoate Ethanol 1
Carbon dioxide Water 1 [CBSE Marking Scheme, 2014]

WORKSHEET-35
Solutions
1. Correct option : (d) Answering Tip
Explanation : Any atom apart from carbon and  Keenly observe experimental tests done during
hydrogen is called heteroatom. practicals, understand and then learn the tests.
Make a list of experimentally done tests used for
2. (i) C3H6, (ii) C5H8. ½+½ distinguishing organic compounds with different
[CBSE Marking Scheme, 2016] functional groups and practice them.
3. (i) Test 1 (Litmus Test) : Take two strips of blue 4. (i) C2H4O4 : CH3–COOH - ethanoic acid
litmus paper. Place a drop each of the alcohol and (ii) C2H5OH : ethanol
carboxylic acid on these strips separately. The blue (iii) X is CH3COOC2H5 - ethyl acetate [ethyl ethanoate
litmus paper turns red in the case of carboxylic acid -ester] 1+1+1
and remains unaffected in the case of alcohol. 1
(ii) Test 2 (Sodium hydrogen carbonate test/sodium 5. (i) All the members of a series have the same
carbonate test) : A pinch of sodium hydrogen functional group, similar structure and same
carbonate or sodium carbonate is added to both general formula. (Any two) ½ + ½
test tubes, separately. (ii) CH3OH, C2H5OH ½+½
If brisk effervescence with the evolution of a (iii) The physical properties are determined by alkyl
colorless gas is observed, it indicates the presence group/hydrocarbon part/part other than the
of carboxylic acid. ½ functional group. ½
If no change is observed then it confirms the (iv) The chemical properties are determined by
presence of the alcohol. ½ functional group such as –OH group, or any other
(iii) Test 3 (Ester test or any other suitable test) example from any other homologous series. ½
(Any two) ½ + ½ [CBSE Marking Scheme, 2015]
[CBSE Marking Scheme, 2015] 6. (a) (i) CH3 — CH2 — CH2 — CH — CH3 1
Detailed Answer : |
(ii) We can distinguish between an alcohol and a Br
carboxylic acid on the basis of their reaction with (ii) CH3 — CH — CH3
sodium carbonate and sodium hydrogen carbonate. |
Carboxylic acids reacts with sodium carbonate and
sodium hydrogen carbonate to evolve CO2 gas that CH3 1
turns lime water milky. (iii) CH3CH2CH2CHO 1
Alcohol, on the other hand, do not react with (iv) HC ≡ C – CH2 – CH2 – CH2 – CH3 1
sodium carbonate and sodium hydrogen carbonate. (b) Electrons dot structure of ethanoic acid
(iii) Alcohol reacts with sodium metal to produce H
hydrogen gas with rapid effervescence. On the
other hand, carboxylic acid does not show this type H C C O H
of chemical reaction with sodium metal. 1½+1½
Commonly Made Error
H O 1
 Students usually get confused between various [CBSE Marking Scheme, 2012]
distinguishing tests for functional groups.

S OLUT I ONS P-27

 WORKSHEET-36
Solutions R R R R
| |
1. Correct option : (a) Ni catalyst
C =C + H2 H — C — C —H
Explanation : The hydrocarbons containing one H2 | |
or more triple bonds are called alkynes. The first
R R R R 1
member is ethyne.
[CBSE Marking Scheme, 2012]
2. Name : Ethane
Formula : C2H6 ½+½ 6. (i) Carbon has 4 electrons in its outermost shell,
[CBSE Marking Scheme, 2013] and needs to gain or lose 4 electrons to attain noble
gas configuration. 1
3. (i) Butyne 1 (ii) Losing or gaining 4 electrons is not possible due to
(ii) Ethanol 1 energy considerations; hence it shares electrons to
(iii) Propanone [CBSE Marking Scheme, 2012] 1 form covalent bonds. 1
Two reasons for large number of carbon
4. Hydrogenation : The process in which compounds :
unsaturated compounds reacts with hydrogen (i) Catenation : The unique ability of carbon to form
in the presence of nickel (as a catalyst) to form bonds with other atoms of carbon, giving rise to
saturated compounds are called hydrogenation. 1 long chains of different types of compounds. 1

This reaction is commonly used in the (ii) Tetravalency : Since carbon has a valency of 4,
hydrogenation of vegetable oils. Vegetable oils it is capable of bonding with four other atoms
have long unsaturated carbon chains, which are of carbon or atoms of elements like oxygen,
converted into vegetable ghee i.e., saturated fatty hydrogen, nitrogen, sulphur, chlorine, etc. 1
acids. 1 The reason for the formation of strong bonds by
carbon is its small size which enables the nucleus to
hold on to the shared pairs of electrons strongly. 1
[CBSE Marking Scheme, 2015]

TOPIC-2
Carbon Compounds, Soap and Detergents

WORKSHEET-37
Solutions (iii) Propanol is oxidised to Propanoic acid. 1
[CBSE Marking Scheme, 2016]
1. Correct option : (a)
Explanation : Hydrogenation reaction means 4. In hard water, soap reacts with calcium and
addition of hydrogen to double bonds of magnesium salts which are present in hard water
and form insoluble substances called scum. 1
unsaturated compounds found in oil in the
Detergents are ammonium or sulphonate salts
presence of catalysts such as palladium or nickel to of long chain carboxylic acids. The charged
give saturated hydrocarbons. end of these compounds do not form insoluble
precipitates with calcium and magnesium ions in
2. Acetic acid or vinegar. hard water. [CBSE Marking Scheme, 2012] 2
[CBSE Marking Scheme, 2012] 1
5. Molecular formula of ethyne = C2H2
3. (i) It is a substance which can give oxygen to other
Electronic formula = H C C H
substances. 1

(ii) CH3 —CH2 —CH2 —OH 
Alkaline KMnO4 + Heat
 → Structural formula = H — C —
— C — H
OR Acidified K Cr O + Heat
2 2 7
[CBSE Marking Scheme, 2015] 1+2
Propanol
CH 3 -- CH 2 -- C -- OH 6. (i) Soaps are the sodium or potassium salts of
|| long chain carboxylic acids while detergents are
O 1 the ammonium of sulphonate salts of long chain
Propanoic acid carboxylic acids.

P-28 S C I ENC E - X
 (ii) The dirt is oily in nature and when soap is added Cleansing Action of Soap : When soap is dissolved
to water, its molecules form structures called in water, it forms a colloidal suspension. In this
micelles in which carbon chain of the molecules colloidal suspension, the soap molecules cluster
dissolves in together to form micelles and remain radially
the oil while the ionic end dissolves in water and suspended in water with the hydrocarbon end
faces outside. The micelles thus help in dissolving towards the centre and the ionic end directed
the dirt in water. (Note : 1 mark to be awarded if outward. The dirt particles always adhere to the
only labelled diagram of micelle is given) oily or greasy layer present on the skin or clothes.
Ca2+ and Mg2+ present in hard water form When a dirty cloth is dipped into a soap solution,
insoluble substance (scum) with soap. its non-polar hydrocarbon end of micelles gets
Two problems : attached to the grease or oil present in dirt and
(a) Non-biodegradable polar end remains in water layer.
(b) Water pollution / soil pollution The mechanical action of rubbing subsequently
(Note : 1 mark to be awarded for any one of dislodges the oily layer from the dirty surface
the problems.) shaping it into small globules. A stable emulsion of
[CBSE Marking Scheme, 2017] oil in water is formed. The emulsified oil or grease
Detailed Answer : globules bearing the dirt can now be readily washed
with water.
S.No. Soap Detergents Soaps do not form lather when the water is hard.
(i) They are sodium These are sodium When soap is added to hard water, calcium and
salts of long chain or potassium magnesium salts present in water displace sodium
fatty acids. salts of sulphonic or potassium ions from the soap molecules forming
acids of hydro- an insoluble substance called scum.
carbons. Problems that aries due to use of detergents instead
(ii) Soaps cannot be Detergents work of soap :
used with hard well with hard (i) Detergents being non-biodegradable, they
water. and soft water accumulate in the environment causing pollution.
both. (ii) In soil, the presence of detergents leads to pH
(iii) They are fully They are non-bi- changes making soil infertile.
biodegradable. odegradable. (iii) The entry of detergents into food chain leads to
bioaccumulation in living beings and leads to
(iv) They take time to They dissolve serious health issues. (Any two) 5
dissolve in water. faster in water.

WORKSHEET-38
Solutions 2. Correct option : (c)
1. Correct option : (b) Explanation : In esterification, RCOOH, – H is
Explanation : Compounds with –OH functional replaced by – R’ of R’OH in the presence of acid to
group are ended with suffix –ol. form RCOOR’.

3.

[Topper Answer, 2017] 3

S OLUT I ONS P-29

 4.

[Topper Answer, 2017] 3
OR
In esterification, a carboxylic acid and alcohol react to form ester in presence of acid. It is reverse of saponification.
+
H
RCOOH + R’OH → RCOOR’ + H2O
Ester
In saponification, an ester reacts with a strong base or an acid to give soap and alcohol.
RCOOR’ + NaOH → RCOONa + R’OH 2
Soap
(i) Use of esters :
As esters have fragrant odours, they are used as a constituent of perfumes, essential oils, food flavourings etc.
(ii) Use of saponification process :
In the manufacturing of soap used as cleansing agent. 1
5. CH3COOH + C4H9OH ¾® CH3COOC4H9 + H2O 6. (i) Soap molecules have two ends– one end is
Acid Alcohol Ester the hydrocarbon chain which is water repellent,
H O H H H H where as the other end is the ionic part which it
—

—
—

water soluble end. When soap is dissolved in water

H — C — C — OH + H — C — C — C — C — O — H it forms a group of many molecules, known as
—

micelle. 1
H H H H H (ii) These micelles are formed because their
Ethanoic acid Butanol hydrocarbon chains come together and the polar
Process is Esterification. 1+ 1 + 1 ends are projected outwards. 1
(iii) Micelle formation in ethanol will not occur because
Commonly Made Error the hydrocarbon chain end of the soap will dissolve
 Students often write incorrect formula. in ethanol. 1
(iv) Soaps in the form of micelle are able to clean
Answering Tip dirty clothes having oily spots, as the oily dirt is
collected in the centre of the micelle, which forms
 Be clear with concepts and how the reaction occurs. an emulsion in water and on rinsing, the water
In the above reaction a water is removed from washes away the micelles with dirt attached to
reactants (acid and alcohol) to form ester. Cautiously them. 2
count the number of carbon atoms present and then [CBSE Marking Scheme, 2016]
write the name.

WORKSHEET-39
Solutions of soap molecules in water in which hydrocarbon
ends are directed towards the centre and ionic ends
1. Correct option : (a) are directed outwards.
Explanation : A micelle is a spherical aggregation

P-30 S C I ENC E - X
 2. It is unsaturated compound because they have 6. (i) The force of attraction between the molecules
more carbon. It burns with sooty or smoky flame. of covalent compounds are not strong as ionic
compounds. So, they have low melting and low
[CBSE Marking Scheme, 2012] 1
boiling points. 1
(ii) Esters are organic compounds which have sweet
3. CH3COOH + Na2CO3 ¾® smell.
Sodium carbonate Activity :
2CH3COONa + H2O + CO2 Aim : To demonstrate esterification process using
Water Carbondioxide ethanol and acetic acid. 1
CH3COOH + NaHCO3 ¾® Materials Required : Beakers, water, test-tube,
Sodium hydrogen CH3COONa + H2O + CO2 ethanol, ethanoic acid, conc. H2SO4.
carbonate Water Carbon
Test-tube
cantaining
dioxide
ethanol
CH3COOH+ NaOH ¾® CH3COONa + H2O
ethanoic acid Beaker
Sodium hydroxide Water
conc. H2SO4
2CH3COOH + 2 Na ¾® CH3COONa + H2
Wire
Sodium Hydrogen gas
gauge
(Any three reaction) 1+1 +1 Water
[CBSE Marking Scheme, 2016]

4. (i) Esters. ½
Chemical Equation :
O Tripod
Burner
stand
—
—

CH3 — C — OH + CH3CH2OH
Conc. H2SO4 Procedure :
O
(a) Take 2 ml of ethanol in a test-tube. 1
—
—

CH3 — C — O — CH2 — CH3 +H2O 1 (b) Add 2 ml of ethanoic acid into it.
(c) Add few drops of conc. H2SO4.
Product’s chemical name—Ethyl ethanoate ½
(d) Warm it in a beaker containing water.
(ii) Conc. H2SO4 acts as a dehydrating agent. (Helps
in the removal of water formed in the reaction.) 1 (e) Observe the smell of the products formed.
[CBSE Marking Scheme, 2016] Observation : Pleasant fruity smelling compound
(called ester) is formed. 1
5. (i) Ethene ½ Chemical Reaction :
Conc. H2 SO4 + Heat
CH3COOH(l) + C2H5OH(l) Conc.H
 2 SO 4
→
(ii) C2H5OH  443 K
 → H2C = CH2 + H2O ½
Ethanoic acid Ethanol

Ethene CH3COOC2H5 + H 2O
(iii) Conc. H2SO4 acts as a dehydrating agent/removes Ethyl ethanoate Water 1
water from the reactant. 1
Conclusion : Carboxylic acid reacts with alcohol in
(iv) Ethane/C2H6 will be formed. 1 presence of conc. H2SO4 which act as a dehydrating
[CBSE Marking Scheme, 2016] agent to from esters.

WORKSHEET-40
Solutions 3. Ethanol, C2H5OH 1
Conc.H SO
1. Correct option : (c) C2H5OH 
2 4
→ CH2 = CH2 + H2O 1
Explanation : Ethanol (C2H5OH) reacts with Ethene
sodium to form sodium ethoxide (C2H5ONa) along

Role of conc. H2SO4 : It act as a dehydrating
with liberation of hydrogen gas.
agent.
2C2H5OH + 2Na → 2C2H5ONa + H2 ↑
[CBSE Marking Scheme, 2016] 1
2. N N
[CBSE Marking Scheme, 2012] 1

S OLUT I ONS P-31

 4. (i) CH3COOC2H5 + NaOH → CH3 COONa
CH2 = CH2 + H2 → C2H6
Ni
+ C2H5OH
(ii) CH3COOH + NaOH → CH3COONa + H2O Use : Hydrogenation of vegetable oils in the
Conc. H 2 SO4 presence of Ni as catalyst. 2
(iii) C2H5OH + CH3COOH → Catalyst : Increases the rate of reaction. 1
CH3COOC2H5 + H2O 1×3 Hydrogenation reaction : It is the process in which
[CBSE Marking Scheme, 2017]
unsaturated compound reacts with hydrogen in
presence of nickel as a catalyst to form saturated
5. P-Ethanol, Q-Ethanoic acid, R-Hydrogen
compound.
½+½+½ R R R R
Acidified K 2 Cr2 O7 | |
CH3CH2OH  → CH3COOH ½ Nickel
C =C + H2 H — C — C —H
2CH3COOH + 2Na ¾® 2CH3COONa + H2 1 | |
[CBSE Marking Scheme, 2016] R R R R

6. When any molecule like H2 adds to unsaturated Vegetable Oil Vegetable Ghee
hydrocarbon because of double and triple bond in [CBSE Marking Scheme, 2015] 2
the presence of Ni as catalyst, it is called addition
reaction.

WORKSHEET-41
Solutions 5.
O
1. Correct option : (c)
—
—

Conc. H2SO4 + heat

Explanation : Vinegar is a 5%–8% aqueous solution CH3 — C — OH + H — OCH2CH3
(Esterification)
of acetic acid. Ethanoic acid Ethanol

2.
O
—
—
CH3 — C — OCH2CH3 + H2O
[Topper Answer, 2017] Ethyl acetate Water
Reverse reaction = Saponification 2+1
Commonly Made Error [CBSE Marking Scheme, 2015]
 Students write vague answers.
Commonly Made Error
 Generally students get confused and write the
Answering Tip
reaction and their structures incorrectly.
 Learn the reactions, how they react, step by step,
changes involved, product formed. Carefully Answering Tip
observe reactions done during practicals.
 Be clear with concepts and how the reaction occurs.
In the above reaction a water is removed from
3. (i) 2CH3COOH + 2Na ¾® 2CH3COONa + H2 reactants (acid and alcohol) to form ester. Cautiously
Sodium ethanoate/Sodium acetate ½ + ½ count the number of carbon atoms present and then
write the name.
(ii) CH 3 COOH + NaOH ¾® CH3COONa + H2O
Conc.H SO
Sodium ethanoate/Sodium acetate ½ + ½ 6. (a) (i) CH3CH2OH  2
443 K
4
→

(iii) CH3COOH + C2H5OH ¾® CH3COOC2H5 + H2O CH2 = CH2 + H2O 1
Ethyl ethanoate/Ester ½ + ½ Alkaline KMnO
(ii) CH3CH2OH 4 → CH3COOH
1
[CBSE Marking Scheme, 2016]
Sunlight
(b) (i) CH4 + Cl2 
→ CH3Cl + HCl 1
4. X : C2H5OH, Y : H2 gas ½+½ NaOH
2C2H5OH + 2Na ¾® 2C2H5ONa + H2­­­­­↑ 1 (ii)CH3COOC2H5 
H2 O
→ C2H5OH+ CH3COONa


Ethene, C2H4. ½+½
1
[CBSE Marking Scheme, 2016]
(iii) CH4 + 2O2 → CO2 + 2H2O + light + heat energy
1

P-32 S C I ENC E - X
 WORKSHEET-42
Solutions R R H H
| |
1. Correct option : (a) Ni catalyst
C =C + H2 R — C — C —R
Explanation : Mineral acids like nitric acid, sulphuric Heat | |
acid are stronger than carboxylic acid as they can R R R R
ionize 100% in their solution.
Vegetable oil Vanaspati ghee
2. (Liquid state) (Solid state)
The essential condition for the reaction is presence
of nickel as a catalyst and temperature.
[Topper Answer, 2017] Change observed in the physical property during
hydrogenation is the change of the unsaturated
Commonly Made Error compound from the liquid state to the
 Students write irrelevant stories. Be specific. Read corresponding saturated compound in the solid
question carefully and write only what is asked. state. Its boiling and melting point will increase.
[CBSE Marking Scheme, 2015] 1+2
Answering Tip
5. The name of the compound formed is Ethene and
 Remember that salts of calcium and magnesium are
responsible for hardness of water, which result in its structural formula is :
scum. H H
| |
H—C=C—H
3. (i) 2CH3COOH + Na2CO3 → 2CH3COONa + H2O
Chemical equation :
+ CO2 1 Conc H SO
(ii) CH4 + 2O2 → CO2 + 2H2O 1 CH3CH2OH  2
443 K
4
→ CH2=CH2+H2O

(iii) 2C2H5OH + 2Na → 2 C2H5ONa + H2 1 Ethanol Ethene

[CBSE Marking Scheme, 2017] Role of H2SO4 : It works as a catalyst to initiate the
4. The addition of hydrogen to an unsaturated reaction to lose water molecule to form alkene.
hydrocarbon to get a saturated hydrocarbon is [CBSE Marking Scheme, 2015] 1+1+1
called hydrogenation.
6. (i) CH4 + 2O2 → CO2 + 2H2O + heat + light 1
Example : Vegetable oils such as groundnut oils,
(ii) 2CH3CH2OH + 2Na → 2CH3CH2ONa + H2 1
cotton seed oils and mustard oils are unsaturated
(iii) CH3COOH + NaOH → CH3COONa + H2O 1
and also contain double bonds (C=C). They are in
(iv) 2CH3COOH + Na2CO3 → 2CH3COONa + H2O
the liquid state.
+ CO2 1
On hydrogenation (addition of hydrogen) in Acid
(v) C H
2 5 OH + CH 3 COOH  → CH 3 COOC 2H 5
the presence of nickel as catalyst. Vegetable oil
+
produces vanaspati ghee. This is solid at room
H 2O 1
temperature.
[CBSE Marking Scheme, 2012]

WORKSHEET-43
Solutions is replaced by —O— alkyl or aryl group to form
an ester. Esters are represented by the formula
1. Correct option : (b) R1COOR2 where R1 and R2 represent an alkyl or
Explanation : While cooking, if the bottom of the aryl group.
vessel is getting blackened on the outside, then it Preparation of esters : When carboxylic acids are
means that the fuel is not burning completely. heated with alcohols in the presence of an acid
2. CH3COOH + NaHCO3 → CH3COONa catalyst, esters are produced.
O O
+ H2O+CO2
R– C + R'OH R– C + H2 O
[CBSE Marking Scheme, 2012] 1
3. Esters are the derivatives of carboxylic acids which O –H O–R'
contain —COOR group. The –OH group in an acid

S OLUT I ONS P-33

 Uses of esters : a soap solution, its non-polar hydrocarbon end of
(i) Esters are used in food as flavours and fragrances. micelles gets attached to the grease or oil present
(ii) In making of soaps. in dirt and polar end remains in water layer.
(iii) Esters are used as solvents. The mechanical action of rubbing subsequently,
(iv) Esters are used in medicines. dislodges the oily layer from the dirty surface
(v) Esters are used as emulsifying agents. (Any two) shaping it into small globules. A stable emulsion
1+1+1 of oil in water is formed. The emulsified oil or
grease globules bearing the dirt can now be
4. Detergents are sodium salt of long chain benzene readily washed with water. 3
sulphonic acids.
[CBSE Marking Scheme, 2015]
Soaps are sodium or potassium salt of long chain
fatty acids that have cleaning action in water. e.g., Conc.H SO
5. (a) (i) CH3CH2OH 2 4
→ CH2 = CH2
1
Sodium stearate. Heat

Cleansing Action of Soap : A soap molecule (ii) CH3COOH + KHCO3 → CH3COOK + H2O +
consists of two dissimilar parts : CO2 1
(i) A short ionic part comprising the carboxylate salt- (iii) CH4 + Cl2 → CH3Cl + HCl 1
- COO–Na+ which is water soluble. (b) (i) CH3CH2COOH : propanoic acid ½
(ii) A long hydrocarbon chain which is hydrophobic. (ii) CH3CH2CH2Br : 1-bromo propane ½
When soap is dissolved in water, it forms a colloidal
(c) Electron dot structure of ethene (C2H4) :
suspension. In this colloidal suspension, the soap
molecules cluster together to form micelles and H H
. . . .
remain radially suspended in water with the
hydrocarbon end towards the centre and the ionic
. .

. .
. .

. .
H C C H
end directed outward. The dirt particles always 1
adhere to the oily or greasy layer present on the
skin or clothes. When a dirty cloth is dipped into [CBSE Marking Scheme, 2012]

WORKSHEET-44
Solutions In pure oxygen, ethyne undergoes complete
combustion and high temperature suitable for
1. Correct option : (c) welding is attained. 1
Explanation : The beaker becomes hot because it is Whereas air contains less percentage/amount of
an exothermic reaction. oxygen which results in incomplete combustion of
ethyne and the temperature required for welding
2. is not attained. 1
[CBSE Marking Scheme, 2015]
etailed Answer :
D
Electron dot structure of ethyne is :
[Topper Answer, 2017] H C C H
No, the mixture of ethyne and air cannot be used
NaOH for welding because when ethyne is burnt in air,
3. (i) CH3COOC2H5 
→ C2H5OH it gives a sooty flame. This is due to incomplete
+ CH3COONa combustion caused by limited supply of air.
(ii) CH3COOH + NaHCO3 → CH3COONa + H2O However, if ethyne is burnt with oxygen, it gives
a clean flame with temperature 3000°C because of
+ CO2 complete combustion. This oxy-acetylene flame is
Sunlight
used for welding. It is not possible to attain such a
(iii) CH4 + Cl2  → CH3Cl + HCl 1+1+1
high temperature without mixing oxygen. 1 + 2
4.
.. .. 5. (i) Ethanol is oxidised to ethanoic acid with the
. . . .
H . C .. .. C . H / H . C .. .. C . H help of acidified K2Cr2O7.
CH3CH2OH  +  2[O] —→ CH3COOH + H2O
1

P-34 S C I ENC E - X
 (ii) Ethanol reacts with sodium metal to form sodium 6. 2CH3COOH + 2Na → 2CH3COONa + H2O.
ethoxide and hydrogen gas. (A) (B)
2C2H5OH + 2Na —→ 2C2H5ONa + H2 CH3COOH + C2H5OH → CH3COOC2H5 + H2O
(iii) Ethene is formed when ethanol is heated at 443 (C) (D)
K with concentrated sulphuric acid. Concentrated CH3COOC2H5 + NaOH → CH3COONa
sulphuric acid acts as a dehydrating agent in this + C2H5OH
reaction and removes a molecule of water. (A) CH3COOH
H SO (B) CH3COONa
CH CH2OH 
2 4
→ CH2 = CH2 + H2O 1+1+1
(C) C2H5OH
3 443 K

[CBSE Marking Scheme, 2014] (D) CH3COOC2H5 [CBSE Marking Scheme, 2018] 5

Commonly Made Error Commonly Made Error

 Students often write incorrect reactions.  Students usually make mistakes in writing the
name of the structures and also writing products in
Answering Tip reactions.
 Understand the reactions and how it takes place.
Make a list of important reactions (reactants, Answering Tip
conditions and products) in this chapter and  Please understanding basic concept of naming
practice them. in carbon compounds (involving saturated,
unsaturated hydrocarbons, functional groups).
Understand the reactions and how the reactions
take place. Then make a list of every important
reactions and practice them.

WORKSHEET-45
Solutions 3. The oxidizing agents used for the conversion of
ethanol to ethanoic acid are alkaline potassium
permanganate (KMnO4) and acidified potassium
1. CH3CH2OH KMnO
 4 [ O]
→ CH3COOH dichromate (K2Cr2O7). 1
Ethanol Ethanoic acid
S. Ethanoic
[CBSE Marking Scheme, 2014] 1 Test Ethanol
No. Acid
2. (i) The carboxylic acid involved in the reaction is
acetic acid (CH3COOH). (i) Litmus Test No change in Blue litmus
H O colour of litmus solution turns
| || solution. red.
H — C — C — OH
(ii) Reaction with No brisk Brisk
|
sodium effervescence. effervescence
H
carbonate due to
(ii) The alcohol involved in the reaction is ethanol evolution of
(CH3CH2OH). CO2.
H H
1+1
| |
4. (i) Ethanol reacts with ethanoic acid in the presence
H — C — C — OH of acid catalyst to give ethyl ethanoate.
| |
CH3COOH + CH3CH2OH  Acid
→
H H cataltyst

(iii) X is the ester formed by the condensation of acetic CH3COOCH2CH3 + H2O

acid and ethanol is ethyl acetate (CH3COOC2H5). (ii) When ethanol reacts with sodium, sodium
H O H H ethoxide is produced along with the liberation of
| || | | hydrogen.
H— C — C — O— C— C— H 2CH3CH2OH + 2Na → 2CH3CH2ONa + H2
| | | 1½+1½
H H H 1+1+1

S OLUT I ONS P-35

 5. (i) A : CH2 = CH2 Ethene 6. (i)
B : CH3 – CH3 Ethane Blue Lit- Red Lit-
Solution Sodium Metal
C : CH3 – CH2 – Cl Chloroethane 1½ mus Paper mus Paper
(ii) CH2 = CH2 hydrogen
 → CH3 – CH3 1 Ethanol No No change Hydrogen 1
nickel
change gas
(A) (B)
Ethanoic Turns red No change Hydrogen 1
Type of reaction : Addition reaction. ½ acid gas
[CBSE Marking Scheme, 2012] Soap No Turns blue Hydrogen 1
change gas
(full credit may be given to the candidate with the
answer showing test only with litmus paper)
(ii)
Hard water contains calcium ions or magnesium
ions or both. These ions on reacting with soap
solution form insoluble substance called scum.
[CBSE Marking Scheme, 2016] 1 + 1

qqq

P-36 S C I ENC E - X
CHAPTER
SECTION

B
5 PERIODIC CLASSIFICATION OF ELEMENTS

TOPIC-1
Periodic Laws and their Limitations
WORKSHEET-46
Solutions (b X has 6 valence electrons it belongs to group 16. 1
(c) Valency will be 2. To acquire noble gas
1. Correct option : (b) configuration it will gain 2 electrons. 1
Explanation : It was found that the Law of Octaves [CBSE Marking Scheme, 2018]
was applicable only upto calcium, as after calcium
every eighth element did not possess properties Commonly Made Error
similar to that of the first.
 Usually students get confused with the group
2. It states that “Properties of elements are periodic
and period to which they belong to. They also get
function of their atomic number.” 1
confuse with the valency.
3. (i) Henry Moseley
(ii) Atomic number Answering Tip
(iii) Modern Periodic Law : Properties of elements are  Understand the basic concepts involved in the
the periodic functions of their atomic numbers. separation of periods and groups in the modern
[CBSE Marking Scheme, 2014] 1 + 1 + 1 periodic table. After writing the electronic
configuration, check the number of energy shells
4. (i) Aim of Classification : For systematic and
simplified study of elements and their com- and this will be equal to the period number. Then
pounds. ½ check for how many outermost (valence) electrons
(ii) Basic property : Atomic Number. ½ are there and then write the Group to which it
(iii) Modern periodic Law : The properties of elements belongs. From the valence electrons we can write
are a periodic function of their atomic number. ½ the valency of that element.
(iv) Metals are found on the left side and centre of the
Modern Periodic Table. ½ 6. (i) To study the properties of elements and to keep
(v) Metalloids are found in a zig-zag manner between the elements with similar properties together. 1
the metals and the non-metals. ½
(ii) Chemical properties of elements and atomic
(vi) Non-metals are found on the right side of the
number. 1
Modern Periodic Table. ½
[CBSE Marking Scheme, 2015] (iii) Metals lies on extreme left, metalloids lie in the
middle and non-metals lie on the right side. 2
5. X : 2, 8, 6 (iv) They should be placed in the same slot. Since they
(a) Since ‘X’ has three energy shells and period have same numbers of electrons.
number of an element is equal to the number of [CBSE Marking Scheme, 2011] 1
energy shells. X belongs to 3rd period. 1

WORKSHEET-47
Solutions in properties between metals and non-metals are
called metalloids.
1. Correct option : (c) (ii) Boron, silicon, germanium and arsenic.
Explanation : Mendeleev arranged the known
1+½+½+½+½
elements according to increasing order of their atomic
masses because according to him, fundamental 4. (i) Electronic Configuration — 2, 8, 2
property of an element was atomic mass. Valency — 2.
2. There are seven horizontal rows in the modern (ii) Metal
periodic table. These rows are called periods. ½ + ½ There are two electrons in its outermost shell and it
3. (i) Borderline elements which are intermediate easily loses them to form a positive ion.

S OLUT I ONS P-37

 (iii) M O check for how many outermost (valence) electrons
2 2 are present, and use them to write the Group to
M2O2 = MO which it belongs. From the valence electrons we can
write the valency of that element.
It is a basic oxide. 1+1+1
5. Electronic Configuration of ‘P’ — 2, 8, 7 6. (i) X is a metal. Nature of it is Basic.
Group number — 17 (ii) X(NO3)2, XSO4
Period number — 3rd Period These compounds are Ionic/electrovalent. 1+1+1
Electronic Configuration of ‘Q‘ — 2, 8, 8, 1 7. (a) Gaps were left for undiscovered elements in the
Group number — 1 periodic table.
(b) (i) Position of hydrogen was not justified.
Period number — 4th Period
(ii) Increasing order of atomic mass could not be
[CBSE Marking Scheme, 2017] 1 + 1 + 1
maintained.
Commonly Made Error (iii) Isotopes have atomic number but different
 Students usually get confused with the group and atomic masses, they cannot be given separate
period to which an element belong to. They also get places.
confused with the valency. (c) Number of shells remains the same, number of
valence electrons goes on increasing from left to
Answering Tip right in a period till octet is complete. e.g.,
 Understand the basic concepts involved in the Li Be B C N O F Ne
making of periods and groups in the modern 2, 1 2, 2 2, 3 2, 4 2, 5 2, 6 2, 7 2, 8
periodic table. After writing the electronic 1+3+1
configuration, check the number of energy shells
and this will be equal to the period number. Then

WORKSHEET-48
Solutions 6. (i) In the periodic table, elements are placed
according to their electronic configuration. If
1. Correct option : (a)
an element has only one shell in its electronic
Explanation : Many gaps for the undiscovered
configuration, it is placed in the first period.
elements were left in the periodic table by
Mendeleev. For instance, the elements scandium If the element has two shells then it is placed in the
(Sc), gallium (Ga), and germanium (Ge) were not second period, and so on. Vertical columns in the
known at Mendeleev’s time, but he had predicted periodic table are called groups.
their existence in advance of their discovery. There are eighteen groups and in a group, all
2. There are 18 vertical columns in the modern periodic the elements have same number of valence
table and these are known as groups. ½+½ electrons. 1+1
3. (a) (i) K (Potassium — 2, 8, 8, 1)
(ii) They will be placed at the same slot as their atomic
(ii) Be and Ca in same group because both have same
number is same, valence electron and valency are
number of valence electrons in their outermost
shell. The number of this group is 2. same and have same chemical properties. 1
(b) Ca X [CBSE Marking Scheme, 2012]
Valency 2 1
7. (a) Atomic mass 1
Thus, the formula of the compound is Ca1X2 =
CaX2 1+1+1 (b) (i) He could classify all the 63 elements known at

that time.
4. (i) Cations are Na and Al (Sodium and Aluminium) (ii) He left gaps for the yet to be discovered
Anions are Cl and O (Chlorine and Oxygen) elements.
(ii) Inert elements are He and Ne (Helium and Neon) (iii) He predicted the properties of such elements.
[CBSE Marking Scheme, 2015] 1 + 1 + 1 (any two) 1 × 2
(c) (i) Position of isotopes
5. (i) Mercury (ii) Irregular increase in atomic masses in going
(ii) Bromine from one element to the next, making the
(iii) Helium or Neon or Argon. (Any one) 1 + 1 + 1 prediction of undiscovered elements difficult.
(iii) Position of Hydrogen. (any two) 1 × 2
[CBSE Marking Scheme, 2015]
[CBSE Marking Scheme, 2018]

P-38 S C I ENC E - X
 Commonly Made Error Answering Tip
 Students write irrelevant stories. Be specific. Read  Do not overlook any part of a question and avoid
question carefully and write only what is asked. being in a hurry to conclude the answer.

TOPIC-2
Periodic Elements and Periodic Properties
WORKSHEET-49
Solutions Detailed Answer :
In the modern periodic table, there are 18 vertical
1. Correct option : (c)
columns known as Groups and 7 horizontal rows
Explanation : There are 18 vertical columns, known
known as Periods.
as groups and 7 horizontal rows in Modern Periodic
Table. Metallic character increases on moving down
a group in the Modern Periodic table. As we
2. Due to the presence of 2 electrons in the valence move down the group, the electrostatic attraction
shell and similar chemical properties. between the nucleus and the outermost electron
[CBSE Marking Scheme, 2015] 1 decreases due to increase in the distance between
them. This happens because; on moving down the
group a new shell is added. So, the valence electron
3. The distance from centre of nucleus to outermost
can be easily lost by the element, thereby metallic
shell of an atom is atomic radius.
character increases on moving down a group.
Atomic radius decreases across a period because
electron is added in the same shell. So attraction The size of atomic radius decreases on moving
between nucleus and valence shell increases due left to right in a horizontal row. When, we move
to which outermost shell is pulled in closer to the across a period, the number of electrons in the same
nucleus. [CBSE Marking scheme, 2015] 1 + 2 shell increases. This leads to greater electrostatic
attraction between the nucleus and the outermost
4. (i) Atomic radius decreases ½ electron. This increased attraction pulls the
Reason : Nuclear charge increases which tends to outermost electron closer to the nucleus, thereby
pull the electrons closer to the nucleus. 1 decreasing the atomic size. 3
Ans. (i) A and B belong to group 1 and 2 because they
(ii) Atomic radius increases ½
form basic oxides. C belongs to group 13 as it has 3
Reason : Number of shells increases on going
down the group. 1 valence electrons. D belongs to group 14 as it forms
[CBSE Marking Scheme, 2017] almost neutral oxide. E and F belong to group 15 and
16 as they form acidic oxides, G belongs to group 17
Ans. Vertical Column — Groups ½ as it has 7 valence electrons, and H to group 18. They
belong to 3rd period of the periodic table.
Horizontal Rows — Periods ½
(ii) H belongs to noble gas
(i) Metallic character increases. (iii) A has largest atomic radius
Reason : Ability to lose electrons increases on (iv) E and F are likely to be non–metals
moving down the group due to increase in distance
(v) D is likely to be metalloid or semi–metal.
between the nucleus and the valence electrons/
1+1+1+1+1
decrease in the attraction between the nucleus and
the valence electrons. 1 Commonly Made Error
(ii) Atomic radius decreases.  Students write irrelevant stories. Be specific. Read
Reason : The nuclear charge increases on moving question carefully and write only what is asked.
from left to right across a period resulting in
increase in the attraction between the nucleus and Answering Tip
the valence electrons. 1  Do not overlook any part of a question and avoid
[CBSE Marking Scheme, 2017] being in a hurry to conclude the answer.

S OLUT I ONS P-39

 WORKSHEET-50
Solutions (a) Valency is the combining capacity of an element.
Valence electrons are the number of electrons in
1. Correct option : (c) the outermost shell. Valency and valence electrons
Explanation : Element with electronic configuration are same till the outermost electron is 4, but when
2, 8 has octet configuration, so must be placed in it goes beyond 4, then the outermost electrons are
group 18. subtracted from 8 and valency is determined. Thus
2. A and C will show similar properties because they valency first increases along the period and then
have same number of valence electrons.  1 decreases.
(b) Along the period, from left to right effective nuclear
3. (i) Chlorine ½
charge increases as the number of protons increase,
(ii) 3rd period 1 due to which force of attraction between nucleus
(iii) 17th group 1 and the valence electron increases therefore, atomic
(iv) 2, 8, 7. [CBSE Marking Scheme, 2016] ½ radius decreases.
(c) Along the period, from left to right effective nuclear
4. There are 18 groups and 7 periods in the modern charge increases as the number of protons increase,
periodic table. due to which force of attraction between nucleus
(i) Atomic size generally increases down a group due and the valence electron increases thus, it becomes
to the addition of new shell and metallic character difficult to lose electron across the period and
also increases down a group due to the increase in metallic to non-metallic character increases.
tendency to lose electrons. (d) Along the period, from left to right effective nuclear
(ii) Atomic size goes on decreasing along a period charge increases as the number of protons increase,
from left to right. due to which force of attraction between nucleus and
the valence electron increases. Hence the electron
Metallic character decreases along a period due to
from the outermost orbit is difficult to remove,
decrease in tendency to lose electrons. 1 + 1 + 1
atomic size decreases therefore electronegativity
[CBSE Marking Scheme, 2015]
increases.
5. Atomic number is more appropriate parameter (e) Along the period, from left to right effective nuclear
than atomic mass as atomic number determines charge increases as the number of protons increase,
the number of valence electrons which decide the due to which force of attraction between nucleus and
chemical properties of an atom of an element. 1 the valence electron increases, hence the electron
Metallic character decreases from left to right in from the outermost orbit is difficult to remove.
a period, because the tendency to lose electrons Therefore,across the period the metallic to non-metallic
increase in size and decreases due to increased character increases. So nature of oxide formation across
attraction between nucleus and valence electrons. the period changes from basic to acidic.
½+½
Metallic character increases down the group, as Commonly Made Error
the tendency to lose electrons increases, due to  Usually students get confused with variation in
decreased attraction between nucleus and valence periodic properties across the period and down the
electrons because outermost electrons are farther group.
away. ½+½
Ans. (a) Valency first increases, then decreases Answering Tip
(b) Decreases  Students should understand the basic concept of
(c) Increase how the elements are arranged in the periodic
table, across the period and down the group, their
(d)
Increases
electronic configurations etc. Then they should
(e)
Change from basic to acidic
learn the variation of periodic properties and how
[CBSE Marking scheme, 2018] 5
it varies.
Detailed Answer :

WORKSHEET-51
Solutions 2. 2nd period has 8 elements, 5th period has 18
elements. ½ + ½
1. Correct option : (b) 3. (a) (i) Valency : The combining power or the com-
Explanation : Carbon is an essential element for all bining capacity of an atom is called its valency.
organic compounds and belongs to group 14.

P-40 S C I ENC E - X
 (ii) Atomic size: Atomic size or atomic radius is 4. Vertical columns of the periodic table are known as
the distance between the centre of the nucleus groups.
and the outermost shell of an isolated atom. (i) The number of valence electrons remains constant
(b) On moving from left to right in the periodic table, when we move down the group.
valency increases up to 4 and then decreases. (ii) The number of occupied shells increases down the
The electrons present in the last shell determine group.
the valency of a particular element. (iii) The size of atom increases down the group.
If the number of valence electrons is less than or (iv) The metallic character of elements increases down
equal to 4, valency = number of valence electrons. the group.
If the number of valence electrons is more than 4, (v) The effective nuclear charge decreases down the
valency = 8 - number of valence electrons group. ½×6=3
Atomic size decreases along a period. This is because 5. No. of periods : 7. ½
on moving across a period, the number of valence
Valency across a period increases from 1 to 4, then
shells remains the same and the electrons increase
decreases from 4 to zero. 1
by one unit. As a result, the nuclear charge increases
and thus, the atomic radius decreases. Metallic character of elements across a period
decreases. ½
1 + 1 + ½ + ½
Valency remains the same down a group ½
Atomic size of elements increases down a group. ½

6.

[Topper Answer, 2017]

WORKSHEET-52
Solutions Since the atomic radius decreases along a period,
the atomic radius of calcium is smaller than that of
1. Correct option : (b) potassium. ½
Explanation : The elements of 2nd period contain (iii) The formula of oxide of calcium is CaO, because
two shells, K and L shell. the valency of calcium as well as that of oxygen
2. (i) Magnesium : 2, 8, 2 ½ is 2. [CBSE Marking Scheme, 2016] ½+½
(ii) Calcium : 2, 8, 8, 2. ½
4. (i) The electronic configuration (2, 8, 2) of the
3. (i) It is a metal. ½
element ‘M’ suggests that it belongs to group 2
Since it has two electrons in its outermost shell/ and period 3 of the modern periodic table and its
two valence electrons, which it can lose easily. ½ valency is 2. ½+½
(ii) K (19) is placed before Ca (20) in the same period/
fourth period. ½

S OLUT I ONS P-41

 (ii) The chemical formula of the compounds are : 5. Sr. No. Characteristics A B
M (NO3)2 / Mg (NO3)2; MSO4 / MgSO4; M3 (PO4)2 /
(i) Number of elec- 12 13
Mg3 (PO4)2. 3×½ = 1½
trons in their atoms
(iii) ‘M’ will form ionic compounds by losing two
electrons. [CBSE Marking Scheme, 2016] ½
(ii) Size of their atoms Bigger Smaller
Commonly Made Error (iii) Their tendencies to More Less
 Usually students get confused with the position lose electrons
of elements arranged in groups and periods of (iv) The formula of AO B2O3
the periodic table. They also get confuse with the their oxides
electronic configuration, valence electrons, formula
of compound these elements form. (v) Their metallic More Less
character metallic metal-
Answering Tip lic

 Basically students should understand the basic (vi) The formula of ACl2 BCl3
their chlorides
concept of how the elements are arranged in the
periodic table, across the period and down the [CBSE Marking Scheme, 2016] 6 × ½ = 3
group, their position, electronic configurations,
6. (i) Two elements of group 1 are Na, K / Sodium,
no. of shells, valence electrons. Valence electrons potassium. 2×½
are the electrons of outermost orbit. Correctly find Electronic configurations Na = 2,8,1; K = 2,8,8,1
the number of valence electrons. While writing 2×½
the formula of the compound formed, first find (ii) Similarity : Both have one valence electron / One
the valence electrons of both elements and then electron in outermost shell. ½
interchange their valency as subscript of the (iii) Oxide – Na2O / K2O. ½
compound’s formula. [CBSE Marking Scheme, 2016]

WORKSHEET-53
Solutions (ii) Formula with hydrogen- H2X or H2S
xx
1. Correct option : (d)
Explanation : In the Modern Periodic Table, the
atomic radii decrease with increasing the atomic
H . x
X x .
H
number from left to right. The atomic number of xx
½+½
F, O, and N are 9, 8, and 7, respectively so atomic
(iii) Sulphur; non-metal. ½+½
radius will decrease from N to F. [CBSE Marking Scheme, 2016]
2. Atomic number of X = 2 + 8 + 7 = 17
Atomic number of Y = 2 + 8 + 8 + 3 = 21 ½ + ½ 6. Atomic number of X = Mass number of X – No. of
neutrons ½
3. (i) Electronic Configuration − 2,8,2. ½ = 35 – 18 = 17 ½
(ii) Metal, as it can easily lose electrons (from outer Therefore, Electronic configuration of
most orbit) ½+½ X = 2, 8, 7 ½
(iii) (a) X Y (b) X Z Group number =17 ½
2 2 2 1 ½ Period = 3
Valency = 8 – 7 = 1 ½+½
Compound – XY XZ2 ½+½
[CBSE Marking Scheme, 2016]
[CBSE Marking Scheme, 2016]
OR
4. (i) Electronic Configuration of X(19) − 2,8, 8, 1. ½
(ii) Fourth Period, Valency 1. ½+½
(iii) Basic oxide (X2O). ½
(iv) X2O + H2O ¾® 2 XOH. ½+½
[CBSE Marking Scheme, 2016]

5. (i) Electronic configuration of X = 2,8,6.

Valence electrons = 6.
Valency = 8-6= 2. ½+½ [Topper Answer, 2016]

P-42 S C I ENC E - X
 WORKSHEET-54
Solutions (vi) B
e < Mg < Ca < Rb
[Topper Answer, 2016]
1. Correct option : (c)
Explanation : Na and K are in the same group whereas
5. (i) Na/Sodium.
Na and Mg belong to same period. In a group from
Reason : The atomic size decreases from left to
top to bottom, atomic radius increase whereas it
right due to the increase in the nuclear charge.
decreases in period from left to right. Hence, the
order would be K> Na >Mg and K> Ca> Mg. (ii) Al/Aluminium.
Reason : The tendency to lose electrons decreases
2. Group 1 elements have 1 valence electron while
from left to right.
Group 2 elements have 2 valence electrons. ½+½
[CBSE Marking Scheme, 2015] 1½ + 1½
3. (i) X (7) = 2,5. Group 15; Period 2 ½ Detailed Answer :
Y(8 ) = 2,6. Group 16; Period 2 ½ (i) Na will have the largest atomic radius as atomic size
Z(9) = 2,7. Group 17; Period 2 ½ goes on decreasing along a period from left to right.
(ii) X> Y>Z ½
It is due to increase in nuclear charge (number
(iii) XZ3 1
[CBSE Marking Scheme, 2016] of protons in nucleus) which pulls the electrons
towards it, i.e., force of attraction between nucleus
4. (i) Ca = 2, 8, 8, 2 ½ and valence electrons increases, therefore atomic
(ii) Valence electrons in Rb = 1 ½ size decreases.
(iii) Five ½ (ii) Al is least reactive because reactivity of an element
(iv) Metal ½ depends upon the ability of its atoms to donate or
(v) Rb is biggest in size ½ accept electrons. Tendency to lose electrons along a
(vi) Be < Mg < Ca < Rb. ½ period generally decreases with decrease in atomic
[CBSE Marking Scheme, 2016] size, i.e., the force of attraction between the valence
OR electrons and the nucleus increases, therefore
electrons cannot be removed easily. 1+2
6. (i) Valence electrons in ‘D’ = 5 and Valency of
‘D’ = 3. 1
(ii) ‘A’ will have largest atomic radii because atomic
radius decreases across a period from left to right. 1
(iii) ‘A’ will form the most basic oxide as it is most
metallic. [CBSE Marking Scheme, 2016] 1

WORKSHEET-55
Solutions 4.
1. Correct option : (c) Sr. No. Property P Q
Explanation : Mg and Ca are alkaline earth metals (i) No. of elec- 3 4
whereas Na and K are alkali metals. Alkali metals trons in the 11 12
are more electropositive compared to alkaline earth atom 19 20
metals and can easily lose electrons. Out of Na and (any one pair) ½
K, K will easily lose electrons because of bigger size
(ii) Size of the Bigger Smaller
and less electronic attraction. ½
atom
2. Atomic number of X = 2 + 8 + 2 = 12
Atomic number of Y = 2 + 8 + 6 = 16 ½+½ (iii) Metallic More Less metallic
½
character metallic
3. (a) Non-metals
(iv) Tendency to More Less
(b) (i) Increases lose electrons
(ii) Decreases
[CBSE Marking Scheme, 2015] 1 + 1 + 1 (v) Formula of P2 O QO
½
oxides
(vi) Formula of PCl QCl2
½
chlorides

S OLUT I ONS P-43

 Note : For parts (v) and (vi) examples using 5. (i) The elements which have one electron in the
symbols of elements may also be accepted. ½ outermost shell are
[CBSE Marking Scheme, 2015] 19K = 2, 8, 8, 1
(ii) Two elements of the same group are Beryllium (Be)
etailed Answer :
D
and Calcium (Ca).
(i) P has 1 valence electron, Q has 2 valence electrons. 4Be = 2, 2
(ii) ‘P‘ is bigger than ‘Q‘. 20Ca = 2, 8, 8, 2
(iii) ‘P‘ is more metallic than ‘Q‘. Formula of element K and element X :
(iv) ‘P‘ has more tendency to lose electrons than ‘Q‘. Both elements K and X have valency one.
(v) P2O and QO are formula of their oxides. So, K X
1 1
(vi) P Cl and Q Cl2 are formula of their chlorides.
K1X1 = KX/KCl
½+ ½+ ½+ ½+ ½+ ½ It is an electrovalent compound. 1+1+1
Commonly Made Error [CBSE Marking Scheme, 2015]
 Usually students get confused with the position 6. Electronic configuration = 2, 8, 8, 2. ½
of elements arranged in groups and periods of (i) ‘X’ is present in the 2nd group and 4th period of the
the periodic table and also with their periodic periodic table. ½+½
properties. They also confuse with the valence (ii) XY. ½
electrons, formula of compound these elements (iii) Basic because X is a metal and the oxides of
form. metals are basic in nature (Y, Atomic number = 8,
Oxygen). ½+½

WORKSHEET-56
Solutions Answering Tip
1. Correct option : (b)  Basically students should understand the basic
Explanation : As we move from left to right in concept of how the elements are arranged in the
periodic table, metallic characters’ decreases and periodic table, across the period and down the
non-metallic characters’ increases. group, their electronic configurations, stability,
reactivity, family and other properties. Then they
2. X and Y will show similar chemical properties as
should recollect the element in the modern periodic
these have same valence electrons.
table and answer the questions.
X = 2, (1)
5. Electronic configuration of Q = 2, 8, 3
Y = 2, 8, (1) ½ + ½
Valency of Q = 3 1
3. (i) E ½
(ii) B ½ Electronic Configuration of R = 2, 8, 5
(iii) C ½ Valency of R = 8 – 5 = 3 1
(iv) B, because atomic radius decreases from left to right Electronic configuration of P = 2, 8, 1
due to increase in the nuclear charge. 1 Electronic configuration of S = 2, 8, 7 1
(v) Noble gases. ½ Formula : PS/NaCl
6. (i)Valency of group 1 elements = 1
4. (i) They all belong to group 2 because all three
Valency of oxygen = 2
have 2 electrons in their outermost shell.
Oxides of group 1 elements :
(ii) Be is least reactive because it has 2 shells and due to Formula of the oxides of group 1 is M2O, where M
more nuclear change it is not easy to take electrons is the group 1 element and O is oxygen.
from it. (ii) Valency of group 13 elements = 3
(iii) Ca is the element having largest atomic radius Valency of halogens = 1
because it has 4 shells. 1+1+1 Halides of group 13 elements :
Formula of the halides of group 13 is MX3,
[CBSE Marking Scheme, 2015]
where M is the group 13 element and X is
halogen.
Commonly Made Error
(iii) Valency of group 2 elements = 2
 Usually students get confused with element Valency of group 16 elements = 2
arranged in groups and periods in the periodic table
Compounds of group 2 and group 16 elements :
and their periodic properties.
Formula of the compounds of group 2 and 16 is
MN, where M is the group 2 element and
N is the group 16 element. 1+1+1

P-44 S C I ENC E - X
 Commonly Made Error group, their position, electronic configurations, no.
of shells, valence electrons. Valence electrons are
 Usually students get confused with the position
the electrons of outer most orbit. Correctly find
of elements arranged in groups and periods of the
the number of valence electrons. While writing
periodic table. They also confuse with the valence
the formula of the compound formed, first find
electrons, formula of compound these elements
the valence electrons of both elements and then
form.
interchange their valency as subscript of the
compound’s formula.
Answering Tip
 Basically students should understand the basic
concept of how the elements are arranged in the
periodic table, across the period and down the

WORKSHEET-57
Solutions 4.

1. Correct option : (b) Elements Valence Period

Explanation : These all elements belong to same electrons
period in which non-metallic characters increases A 1 3
from left to right.
B 3 3
2. Hydrogen and Helium. ½+½
C 5 3
[CBSE Marking Scheme, 2012]
3. (i) Position of the elements in the periodic table : D 7 3
Electronic configuration of A : 2, 8, 1
Element Period Group
Electronic configuration of D : 2, 8, 7
A 3 17 Atomic number of A = 11
B 4 1 Electronic configuration A = 2, 8, 1
Number of valence electrons of A = 1
(ii) Atomic number of A = 17
Valency of A = 1
Electronic configuration A = 2, 8, 7
Atomic number of D = 17
Number of valence electrons of A = 7
Electronic configuration D = 2, 8, 7
Valency of A = 8 – 7 = 1
Number of valence electrons of D = 7
Atomic number of B = 19
Valency of D = 8 – 7 = 1
Electronic configuration B = 2, 8, 8, 1
Element D A
Number of valence electrons of B = 1
Valency of A = 1
Element B A

Valency 1 1
So, the formula of the compound formed when
Valency elements A and D combine is AD. 1+1+1
1 1
5. (i) Element E will form covalent compounds by
So, the formula of the compound formed when
sharing its four valence electrons.
elements A and B combine is BA.
(ii) Element D is a metal with valency three.
(iii) B. + A B A
(iii) Element B is a non-metal with valency three.
+ – (iv) We know that the size of elements decreases on
Lewis Structure : B A moving left to right in a period. Therefore, element
1+1+1 D is greater than element E.
(v) The name of the family to which elements C and F
belong is the noble gas family (group 18).
½+½+½+1+½

S OLUT I ONS P-45

 WORKSHEET-58
Solutions (ii) Elements P and Q are metals as they have 2
electrons in their valence shell and they are
1. Correct option : (a) positively charged ions whereas elements R and S
Explanation : Formation of acidic oxides is are non-metals as they gain electrons to complete
characteristic of non-metals. Here, element with their octet.
atomic number 7 is a non-metal that is nitrogen. (iii) P and Q will form basic oxides as they are metals.
Rest three elements are metals and hence from basic 1+1+1
oxide. 5. (i) A and B have same period(second period). C and
2. Metals are placed on the left side while non-metals D have same period (third period).
are placed on the right side of the periodic table. (ii) A and C have one valence electron, so they belong
[CBSE Marking Scheme, 2012] ½+½ to same group.
(iii) C is more reactive because C is placed below
3. (i)They are in group 1 because all the elements in A in the periodic table and reactivity increases
group have 1 valence electron. They form positively down the group. 1+1+1
charged ions by losing one electron.
(ii) Lithium is least reactive because reactivity of an 6. (i) Number of electrons in the outermost orbit of
element depends upon the ability of its atoms to X = 1 and Y = 2. ½
donate or accept electrons. (ii) Valency of X = 1 and Y =2. ½
(iii) Potassium will have largest atomic radius due to (iii) Metal X is more metallic than Y. ½
the addition of a new shell. Atomic size generally (iv) Atomic size of X is bigger than that of Y. ½
increases from top to bottom in a group. 1+1+1 (v) Chloride XCl ; YCl2 ½
4. (i) The valency of Q is 3 as its valence shell has three (vi) Sulphate X2SO4 ; YSO4 ½
electrons in it.
[CBSE Marking Scheme, 2012]

WORKSHEET-59
Solutions 4. (i) 12. ½
1. Correct option : (b) (ii) 2. ½
(iii) Be. ½
Explanation : Atomic number 14 is for Silicon
element which is a metalloid and exhibit properties Because Be has two electrons in its valence shell.
of both metals and non-metals. It has 4 electrons Given element also has two electrons in its valence
in its valence shell so tends to form covalent acidic shell.
oxide like non-metals. Since valence electrons determine the chemical
property, hence the given element has same
2. Valency is same i.e. 1. 1 chemical properties as that of Be. 1½
[CBSE Marking Scheme, 2012] [CBSE Marking Scheme, 2012]
3. (i) Z. ½
5. (i) The position of elements X in the modern periodic
(ii) Halogens. ½ table is group number 17 and period number 3. The
(iii) Magnesium and nitrogen. ½+½ position of element Y in the modern periodic table
(iv) Silicon. ½ is group number 2 and period number 4. Electronic
(v) X has bigger size than P because X has less effective configuration of element X = 2, 8, 7.
nuclear charge. ½ It has 3 shells so period number 3.
Commonly Made Error Halogens are kept in group 17.
 Students usually get confused with elements Electronic configuration of element Y = 2, 8, 8, 2
arranged in groups and periods in the given table. It has 4 shells so period number is 4.
The valence shell has 2 electrons so the group
Answering Tip number is 2. 1½
 Students should understand the basic concept of (ii) Formula of the compound :
how the elements are arranged in the periodic Valency of element Y Valency of element X
table, across the period and down the group, their 2 1
electronic configurations, stability reactivity, family \ YX2 1½
and other properties.

P-46 S C I ENC E - X
 Commonly Made Error (ii) 4Be (Beryllium) and 20Ca (Calcium) belong to the
same group, because they have same number of
 Students usually get confused with the position valence electrons.
of elements arranged in groups and periods of
Be = 2, 2 and Ca = 2, 8, 8, 2
the periodic table. They also get confused with the
formula of compound. (iii) Beryllium (Be) and Fluorine (F) have same period
i.e., second period which has 2 shells (K and L)
6. (i) 19K (Potassium) is the element that has one whereas Potassium (K) and Calcium (Ca) also
electron in the outermost shell. belong to same period i.e., fourth period which
Electronic configuration = 2, 8, 8, 1 have 4 shells (K, L, M and N). 1+1+1

WORKSHEET-60
Solutions 4. There are 7 horizontal rows known as periods, that
the modern periodic table has.
1. Correct option : (c) In periods, the number of valence electrons
Explanation : Elements of the same group have increases from left to right and due to decrease
same number of valence electrons hence valency in atomic size, the force of attraction between
will remain same. the valence electrons and the nucleus increases.
2. Electronegativity increases across a period. 1 Metallic character decreases along the period due
[CBSE Marking scheme, 2012] to decrease in tendency to lose electrons.
Atomic size generally increases from top to bottom
3. (i) Element ‘A’ is a metal because it is present in in a group due to the addition of a new shell i.e.,
group I (1 valence electron) and can lose electron number of shells goes on increasing down the
easily. 1 group. 3
(ii) Element B has larger size than ‘C’ because it has 5. (i) Element — E 1
more number of shells than ‘C’. (ii) K L M 1
Also, B lies in the third period and has three shells 2 8 6
whereas C lies in second period and has two (iii) X and Z. 1
shells. 1 [CBSE Marking Scheme, 2012]
(iii) ‘C’ being an element of 3rd group has three valence
electrons, therefore its valency is 3. 1
[CBSE Marking Scheme, 2014]
qqq

S OLUT I ONS P-47

SECTION
CHAPTER

B
6 LIFE PROCESSES

TOPIC-1
Nutrition
WORKSHEET-61
Solutions 5. Three event which occur during photosynthesis :
1. Correct option : (c) (i) Absorption of light energy by chlorophyll.
Explanation : Autotrophs take in food from the (ii) Conversion of light energy to chemical energy and
outside world and convert them into stored forms splitting of water molecules into hydrogen and
of energy. This material is taken in the form of oxygen : Photochemical reaction.
carbon dioxide and water which is converted into
carbohydrates in the presence of sunlight and (iii) Reduction of carbon dioxide to carbohydrates :
chlorophyll. Dark reaction. 1+1+1
[CBSE Marking Scheme 2016]
2. Glucose/Amino acid 1
6. Photosynthesis takes place in the grana and stroma
[CBSE Marking Scheme, 2018] of the chloroplast (Plastid) in green parts of plants.
3. Conditions necessary for autotrophic nutrition are The raw materials required for this process are
: Sunlight, chlorophyll, carbon dioxide and water. carbon dioxide and water in the presence of
The by-products are carbohydrates, in the form sunlight and chlorophyll.
of starch and oxygen. The source of by-product Carbon dioxide enters the leaves through stomata
(oxygen) is water. 1+1+1 and cells of the roots absorbs water from the soil.
4. It is the mode of nutrition in which an organism Balanced equation for photosynthesis :
cannot make its own food and depends on other Chlorophyll
organisms for food. All the animals including man, 6CO2 + 6H2O 
Sunlight
→ C6H12O6 + 6O2
most bacteria and some fungi have heterotrophic
mode of nutrition and these organisms are called The by-products in this process is the evolution of
heterotrophs. oxygen gas. 1+1+1+1+1
Holozoic, Saprophytic and Parasitic. 3
[CBSE Marking Scheme, 2014]

WORKSHEET-62
Solutions are generally basic, they neutralize the excess acid
produced in the mouth and prevent tooth decay.3
1. Correct option : (d) [CBSE Marking Scheme 2018-19]
Explanation : The autotrophic mode of nutrition
requires carbon dioxide, water, chlorophyll and
sunlight. 5.
2. Saliva moistens the ingested food with mucus, S. Name of the Name of Secretions
sterilises it with lysozyme and partially digests No. glands
starch part of food into sugar with the help of
salivary amylase or ptyalin. 1 (i) Salivary glands Saliva contains enzyme,
3. Digestion starts in the buccal cavity in the mouth. ptyalin
The enzyme secreted in buccal cavity is salivary (ii) Gastric glands Secretes gastric juice,
amylase. Salivary amylase helps in breakdown of HCl, mucus, pepsin.
starch into maltose and dextrin. 1+1+1
(iii) Liver Bile juice.
4. Sweet tooth leads to tooth decay, which is
(iv) Intestinal glands Intestinal juice.
caused by the action of bacteria on food particles
remaining in the mouth and acid is formed. (v) Pancreas Pancreatic juice which
The pH of the mouth falls below 5.5 and the tooth contains trypsin, lipase
and amylase.
enamel dissolves resulting in cavity. Toothpastes
(Any three) 1 + 1 + 1
[CBSE Marking Scheme, 2012]

P-48 S C I ENC E - X
 6. (a) (i) Starch (ii) ATP
Ingested food Food
(b) Food particle Fresh food
vacuole
Contractive particle Lysosome vacuole
vacuole Old food Undigested
vacuole particle

A B C D E
Holozoic nutrition in Amoeba
Protozoans like Amoeba capture food with the help of temporary finger-like processes called pseudopodia.
As soon as Amoeba comes in contact with a food particle or prey, it throws pseudopodia all around the food
particle. The tips of encircling pseudopodia fuse and the prey comes to lie in a vesicle or phagosome. This
method of intake of food is called circumvallation. Amoeba can also ingest food by other methods like import
circumfluence and invagination.
(c) In Paramecium, the food is taken in at a specific spot and is moved to this spot by the movement of cilia, which
cover the entire surface of the cell. [CBSE Marking Scheme, 2014] 1 + 3 + 1

WORKSHEET-63
Solutions 4. Small intestine. ½
1. Correct option : (b) Secretions of liver and pancreas mixes with food.
Pancreatic enzymes make it alkaline. Bile juice
Explanation : These are saprotrophs and digestion
from liver too helps in it. Bile salts break the fat
in saprotrophs takes place before ingestion. They
present in the form of large globules into smaller
break down and convert complex organic molecules
ones, increasing the efficiency of enzyme action.
present in dead and decaying matter into simpler
Enzymes like trypsin digests proteins and lipase
substances outside their body.
breaks down fats. Intestinal juice convert proteins
2. Digestion of fats takes place in small intestine with to amino acids, complex carbohydrate to glucose
the help of bile juice which acts as emulsifiers and and fat into fatty acids and glycerol. 2½
breaks down the larger fat droplets into smaller [CBSE Marking Scheme, 2012]
ones. They are then acted upon by lipase enzyme
found in pancreatic juice, which converts them into
5. They are called as Villi.
monoglycerides and fatty acids. 1
(i) Villi are richly supplied with blood vessels which
3. (i) Small intestine. take the absorbed food to each and every cell of the
(ii) Herbivores have longer, small intestine for body.
digestion of cellulose while carnivores have a (ii) It also absorbs water.
shorter small intestine for early digestion of meat.
(iii) They increase the surface area for the absorption of
1+2
food. [CBSE Marking Scheme, 2013] 1 + 1 + 1
[CBSE Marking Scheme, 2013]

6. (a)

Tongue Mouth (Buccal Cavity)

Oesophagus

Diaphragm
Gall bladder
(stores bile)
Stomach
Bile duct
Liver Small
Pancreas intestine

Large intestine (Colon)

Appendix
Anus

S OLUT I ONS P-49

 2 marks for diagram + 2 marks for labelling = 4
(b) Peristaltic movement : The lining of alimentary canal has muscles that contract rhythmically in order to push
the food forward. This is called peristaltic movements. [CBSE Marking Scheme, 2012] 1

Commonly Made Error

 Students fail to draw a neat diagram. Make sure the lines are properly marked. Lines which are not properly
marked will deduct your marks. Many students forget the next part of the question.

WORKSHEET-64
Solutions (v) HCl and Enzymes–pepsin, mucus.
[CBSE Marking Scheme, 2012] 1 + 1 + 1
1. Correct option : (d)
Explanation : The sequence of organs in human 4. (a) (i) Dilute HCl makes the medium acidic. 1
alimentary canal are mouth, oesophagus, stomach,
(ii) It activates the enzyme pepsin. 1
small intestine, large intestine and anus. (b) CO2 and water. [CBSE Marking Scheme, 2012] 1
2. Gastric glands are present in the wall of stomach
which releases hydrochloric acid, mucus and 5. (i) Helps in mixing the food thoroughly with more
protein digesting enzyme pepsin. 1 digestive juice by peristalsis. 1
[CBSE Marking Scheme, 2014] (ii) Creates an acidic medium, which facilitates the
action of the enzyme pepsin. 1
3. (i) Small intestine (iii) Protects the inner lining of the stomach from the
(ii) Pancreatic juice action of the acid. 1
(iii) Liver [CBSE Marking Scheme, 2011]
(iv) Large intestine

6. (i) Process : Photosynthesis ½

Type of Autotrophic Nutrition
Explanation : Autotrophic nutrition is the process where plants prepare their own food, using inorganic mate-
rial such as CO2 and H2O in presence of Sunlight and Chlorophyll.½+½
Chlorophyll, Sunlight
6CO2 + 12H2O 
→ C H O + 6O + 6H O
6 12 6 2 2 1
(Glucose)
Raw material : CO2, H2O½+½
(ii) (a) Absorption of light energy by chlorophyll.
(b) Conversion of light energy to chemical energy and splitting of water molecules into hydrogen and oxygen.
(c) Reduction of Carbon dioxide to Carbohydrates.
 [CBSE Marking Scheme, 2018] 1½

WORKSHEET-65
Solutions (ii) The purpose of making urine is to filter out waste
products (urea or uric acid) from the blood. 1
1. Correct option : (b) [CBSE Marking Scheme, 2012]
Explanation : If salivary amylase is lacking in the
saliva, the process of starch digestion will get disturb 4. (i) Trypsin + Proteins → Amino acids
as salivary amylase helps in digestion of starch. (ii) Amylase + Carbohydrates → Simple sugars
2. HCl of gastric juice disinfects the food and acidifies
(iii) Lipase + Fats → Fatty acids + Glycerol.
it for proper functioning of proteolytic enzyme
pepsin. 1 [CBSE Marking Scheme, 2012] 1 + 1 + 1
Detailed Answer :
3. (i) The food coming from the stomach is acidic and
Trypsin, chymotrypsin, pancreatic amylase
has to be made alkaline for the pancreatic enzyme
and lipase are the main enzymes present in the
to act. Bile juice accomplishes this. Bile salts break-
pancreatic juice. They help in the break down of
down larger globules of fats into smaller globules
food components as follows :
increasing the efficiency of enzyme action. 2

P-50 S C I ENC E - X
 Trypsin acts upon proteins and converts it into 6. Just before Starch test — Pale yellow
peptids and amino acids, pancreatic amylase
Just after Starch test – Blue black
converts starch into disaccharide and other simple
Chlorophyll
sugars while pancreatic lipase acts upon fats and 6CO2 + 6H2O 
Sunlight
→ C6H12O6 + 6O2
converts it into fatty acids and glycerol.
O2 is obtained from water (H2O), as splitting of
5. (i) The cake will have a bitter taste because of the water results in formation of hydrogen (used for
formation of Na2CO3/sodium carbonate while making glucose) and oxygen (by-product).
baking/heating ½+½  [CBSE Marking Scheme, 2018-19] 3
(ii) By adding tartaric acid 1
(iii) The liberated CO2 gas 1
[CBSE Marking Scheme, 2018]

TOPIC-2
Respiration
WORKSHEET-66
Solutions Detailed Answer :
A physical process by which oxygen is taken in and
1. Correct option : (b)
carbon dioxide is given out is called breathing.
Explanation : The breakdown of pyruvate to give
carbon dioxide, water and energy takes place in Breathing in humans involves three steps :
mitochondria. (i) Inspiration : When we breathe in, ribs move up and
2. The respiratory pigment in human being is flatten the diaphragm due to which the chest cavity
haemoglobin. Haemoglobin is present in RBC’s of becomes larger. As a result air is sucked into the
blood in humans. ½+½ lungs and fills the expanded alveoli.
3. Breathing is the process of letting in oxygen from (ii) Gaseous exchange : Haemoglobin binds with
air into the lungs and CO2 out of the lungs. the oxygen and carries it along the blood in the
Mechanism : Involuntary, rate controlled by body. As blood passes through the tissues of the
brain. Outward and inward movement of ribs body, oxygen from the blood diffuses into the cell,
whereas carbon dioxide which is produced during
increases or decreases the space of thoracic
respiration diffuses into the blood and is carried to
cavity, action assisted by diaphragm continuous
the lungs for expiration.
inhalation and exhalation of the air. (iii) Expiration : Ribs move down and diaphragm
[CBSE Marking Scheme, 2016] becomes dome-shaped decreasing the chest cavity.
Thus, pushing the air out from lungs. 3
4. (i) Take two healthy potted plants which are nearly the same size.
(ii) Keep them in a dark room for three days.
(iii) Now place each plant on separate glass plates. Place a watch glass containing potassium hydroxide by the side of
one of the plants. The potassium hydroxide is used to absorb carbon dioxide. When the leaves of both the plants
were tested for starch, it was found that the leaves of the plant kept in bell jar (b), which is without potassium
hydroxide gave the positive test of starch. This shows that CO2 is essential for photosynthesis.

Bell Jar

Watch glass
containing potassium
hydroxide
(a) (b)
Experimental set up (a) With potassium hydroxide (b) Without potassium hydroxide. 3

S OLUT I ONS P-51

 Answering Tip
 Write answers point-wise rather than in the form of an essay.
 Steps should be written in the correct sequence.

5. (i) It has large surface area and are branched. 1

(ii) Contain an extensive network of blood vessels. 1
(iii) It is thin, delicate and fine. 1
[CBSE Marking Scheme, 2012]

6. (i)
Nasal passage
Trachea
Mouth cavity
Pharynx
Larynx Rings of cartilage

Lung

Bronchi Ribs
Bronchioles
Alveolar sac
Diaphragm 4
(ii) So that there is sufficient time for oxygen to be absorbed and for CO2 to be released. 1
[CBSE Marking Scheme, 2012]

WORKSHEET-67
Solutions Answering Tip
1. Correct option : (d)  Always be specific and give clear and complete
Explanation : Yeast is unicellular eukaryote which answers. Incomplete and vague answers must be
carries out ethanol fermentation. In the first phase, pointed out.
glucose is converted into pyruvate (glycolysis) in
4.
The exchange of gases takes place between the
the cytoplasm of the cell. Due to limited oxygen
blood capillaries that surround the alveoli and the
availability, pyruvate remains in cytoplasm where
pyruvate decarboxylase and alcohol dehydrogenase gases present in the alveoli. Thus, alveoli are the
enzymes carry out the second phase of anaerobic site for exchange of gases. The lungs get filled up
respiration and produce ethanol and carbon with air during the process of inhalation as ribs are
dioxide. lifted up and diaphragm is flattened. The air that is
2. Aerobic respiration occurs in mitochondria of the rushed inside the lungs fills the numerous alveoli
cell. 1 present in the lungs. Each lung contains 300–350
3. A terrestrial organism can obtain oxygen directly million alveoli. These numerous alveoli increase
from the air and have slow breathing rate but; the surface area for gaseous exchange making the
aquatic organisms have to obtain oxygen for process of respiration more efficient. 3
respiration which is dissolved in water. Since, the 5.
Plants perform two processes all together,
amount of oxygen dissolved in water is fairly low as
photosynthesis and respiration. Photosynthesis
compared to the amount of oxygen in air, the rate of
takes place during day time when there is enough
breathing in aquatic organisms is much faster. 3
sunlight. During this process, the green tissue of the
Commonly Made Error plants exposed to light makes glucose in presence
 Students often write vague answer. It seems they of chlorophyll pigment, water and carbon dioxide.
are unaware of the concept of rate of breathing. During this process oxygen is released and it is
possible only in presence of sunlight. 2

P-52 S C I ENC E - X
 Respiration is another process which is carried out by (ii) Oxygen ½
all animals and plants. This is a continuous process By splitting of water (photolysis)
where glucose is broken down to CO2 and H2O and
energy is released in the form of ATP molecules. This 2H2O splits
→ H2 + O2 1
is continued regardless of daylight or darkness. 1
(iii)
They take up CO2 at night through stomata, which
6. (i) 2 open during night and produce an intermediate
(b) Cuticle organic acid which is acted upon by the energy
absorbed by chlorophyll during the day and
breaks up to release of CO2. The CO2 so produced
internally is used in photosynthesis during day
(a) Chloroplast when stomata are closed. 1½

Desert plants. [Xerophytic CAM plants] 1

WORKSHEET-68
Solutions
4. (a) Glucose In Cytoplasm
 → Pyruvate
1. Aquatic organisms like fishes obtain oxygen from
In absence of oxygen
water present in dissolved state through their gills. 
→ Ethanol + CO2 + Energy 1½
Since the amount of dissolved oxygen in water is (b) Fishes take in water through the mouth and force
fairly low compared to the amount of oxygen in the it past the gills where the dissolved oxygen is taken
air. Hence, the rate of breathing is much faster in up by the blood. 1
aquatic organisms than in terrestrial organisms. 1
(c) Alveoli ½
2. Adenosine triphosphate (ATP). It is produced during
respiration in living organisms in mitochondria. 1 Functions : They contain an extensive network of
3. blood vessels which exchange gases. ½
They increase surface area of absorption of gases. ½
S. No. Aerobic respiration Anaerobic
(d) Haemoglobin ½
respiration
Role : Due to high affinity for O2, it helps in its
(i) Oxygen is utilized Oxygen is not transport from alveoli to the tissue ½
for the breakdown required. [CBSE Marking Scheme, 2018]
of respiratory sub-
strate. 5. Breathing : Breathing is the process of taking in
(ii) It takes place in It takes place in oxygen-rich air and giving out carbon dioxide-
cytoplasm (glycoly- cytoplasm only. rich air. Breathing in is ‘inhalation’ and breathing
sis) and inside mi- out is ‘exhalation’. The breathing mechanism of
tochondria (Krebs lungs is controlled by the diaphragm and the inter-
cycle). costal muscles. Diaphragm is a membrane which
separates the thoracic chamber from the abdominal
(iii) End products are End products are
cavity. When diaphragm moves down, the lungs
carbon dioxide and lactic acid or etha-
expand and air is inhaled. When diaphragm moves
water. nol and carbon di-
up, the lungs contract and air is exhaled. Inhalation
oxide.
involves bringing in air from outside the body into
(iv) More energy is re- Less energy is re- the lungs. When we breathe in, the size of our chest
leased. leased. increases. This happens because when air enters,
(v) Complete oxidation There is incomplete the lungs expand and the ribs move outwards.
of glucose happens. oxidation of glu- Simultaneously, the diaphragm contracts and
cose. becomes flat. Thus, the size of our chest increases.
Exhalation involves removal of CO2 from the
 (Any Three) 1 + 1 + 1
body. When we breathe out, the size of our chest

S OLUT I ONS P-53

 decreases. This happens because when air moves Breathing in Breathing out
out of our lungs, lungs contract, the ribs move back,
and the diaphragm curves upwards into the chest
decreasing the size of the chest. Chest Chest
expands contracts

Ribs Lung

Diaphragm

Diaphragm
contracts Diaphragm
relaxes

Breathing in Man 5

TOPIC-3
Circulation and Transportation
WORKSHEET-69
Solutions plants by physical forces such as transpiration pull.
Transport in phloem : Phloem tissue helps in
1. Correct option : (a)
transport of food. Here, food is transported in both
Explanation : Valves ensure that blood does
not flow backwards when the atria or ventricles upward and downward directions. Transport of
contract. Semilunar valves, the valves present food in phloem requires energy in the form of ATP.
between ventricles and their attached vessels, (b)
The transport of soluble products of photosynthesis
serve to prevent the backflow of blood to ventricles to other parts of plant through phloem is known as
from their respective attached vessels. Likewise,
translocation. 1+1+1
atrioventricular (AV) valve between atrium and
ventricle directs the flow of blood and prevents any 5.
backflow into atria.
S. No. Artery Vein
2. (i) Phloem (ii) Xylem. ½+½
(i) Wall is thick. Wall is thin.
3. The loss of water in the form of vapour from
the aerial parts of the plants is known as (ii) Valves absent. Valves present.
transpiration. 1
(iii) Blood flows from Blood flows from
Functions : heart to different different organs to
(i) It helps in the absorption and upward movement organs. heart.
of water and minerals dissolved in it from roots to
(iv) The flow of The flow of blood
the leaves. 1
blood is fast, is slow, steady and
(ii) It also helps in temperature regulation. 1 jerky and with with less pressure.
[CBSE Marking Scheme, 2012] great pressure.
4. (a) Transport in xylem : Xylem tissue helps in (Any three) [CBSE Marking Scheme, 2016]
transport of water and minerals. Here, water is 1+1+1
transported upwards from roots to aerial parts of

P-54 S C I ENC E - X
 Aorta
6. (a)
Vena Cava Pulmonary arteries
from upper
body Pulmonary veins

Right atrium Left atrium

Vena Cava
from lower
body
Right
ventricle Left ventricle
Septum
(Dividing wall)
4
(b) (i) As ventricle has to pump blood into various organs. Therefore it has wall thicker them that of atria.
(ii) Since the blood emerges from the heart under high pressure. 1

WORKSHEET-70
Solutions Detailed Answer :
(i) Capillaries.
1. Correct option : (b) (ii)
Explanation : Hippocampus, Exocoetus, Anabas
belong to class pisces. Fishes have two chambered
heart and exhibit single circulation while three
chambered heart of amphibian and reptiles and Pulmonary
Lung Pulmonary vein
four chambered heart of birds and mammals exhibit artery to lungs
capillaries from lungs
double circulation.
2. It is the loss of water in the vapour form from the
aerial exposed parts of a plant. 1

3. (a) Lymph is a colourless fluid containing white

blood cells. 1 Vena cava Aorta to
body
(b)
Lymph is colourless and contains less protein than from body
plasma. 1
Capillaries in
(c) Functions : body organs apart
(i) Carry digested fat/absorption of fat. ½ from the lungs
(ii) Drains excess fluid from extracellular space
back into the blood. ½
[CBSE Marking Scheme, 2012]
Schematic representation of transport of exchange of
4. (i) Because fish has two chambered heart . The
oxygen & carbon dioxide
blood is pumped to the gills for oxygenation from 1+2
where it is passed directly to the rest of the body 6. (a) Two advantages :
and de-oxygenated blood returns to the heart to be
(i) Helps in absorption and upward movement of
pumped into gills again. 1 water and minerals from roots to the leaves. 1
(ii) Because plant bodies have a large proportion of (ii) Temperature regulation. 1
dead cells. 1
(b) (i)
(iii) Capillaries are smallest vessels which have one cell
thick wall. The exchange of material between the S. No. Transpiration Translocation
blood and the surroundings take place through (i) Plays role in It is a transport of soluble
capillaries. 1 transport of water products of photosynthesis.
[CBSE Marking Scheme, 2012] and minerals.
(ii) Occurs through Occurs through phloem
5. (i) Capillaries xylem by simple in the form of sucrose by
physical forces. utilising energy.
(ii) Diagram. [CBSE Marking Scheme, 2016] 3
2

S OLUT I ONS P-55

 (ii) Because plants have a large proportion of dead cells Answering Tip
in many tissues. So, their energy needs are low and
they can afford to have slow transport system. 1  Differences when asked should be compatible.
 Answers should be specific and precise. Answers
Commonly Made Error like present and not present are not acceptable.
 Sometimes students write irrelevant stories. Be
specific. Read question carefully and write only
what is asked.

WORKSHEET-71
Solutions 4. (i) Since ventricles have to pump blood into
various organs, they have thicker muscular walls
1. Correct option : (a)
than atria do. 1
Explanation : The anterior vena cava collects
deoxygenated blood from the head, chest, and arms (ii) The lining of alimentary canal has muscles that
and enters the right atrium while the inferior vena contract rhythmically in order to push the food
cava collects blood from the lower body regions. forward. These are peristaltic movements. 1
Both venae cavae pass the deoxygenated blood to (iii) In desert plant, stomata open at night and absorb
the right atrium. Therefore, blood from tissues is CO2 and store it as an intermediate compound,
rich in carbon dioxide. that is converted into carbohydrate during the day
2. (i) Plasma time. [CBSE Marking Scheme, 2012] 1
(ii) Haemoglobin present in RBCs. ½+½
3. 5. Xylem (vessels) of roots, stems and leaves are inter-
connected to form a continuous column. Roots
S. No. Blood Vessels Function also take up mineral salts actively, water moves
in and as a result by creating pressure pushes the
(i) Arteries They carry blood away from water up. Transpiration pull creates a suction force
the heart to various organs of for pulling water up. 3
the body. [CBSE Marking Scheme, 2014]
(ii) Veins They collect blood from
different organs and bring it 6. (a) Plasma, red blood cells, white blood cells,
back to the heart. platelets (Any two) 2
(iii) Capillaries Exchange of material between (b) Lungs → Left side of the heart → aorta → body
organs 1
the blood and surrounding
(c) Prevent back flow of blood. 1
cells takes place across the thin
(d) Artery has thick elastic wall and vein is thin walled/
walls of capillaries.
valves are present in the veins and not in arteries.
[CBSE Marking Scheme, 2012] 1 + 1 + 1 [CBSE Marking Scheme, 2018] 1

WORKSHEET-72
Solutions 4.
The components of xylem tissue (tracheids
and vessels) of roots, stems and leaves are
1. Correct option : (a)
interconnected to form a continuous system of
Explanation : In a plant, the xylem is responsible for
water conducting channels that reaches all parts of
transport of water.
the plant. Transpiration creates a suction pressure,
2. Transpiration [CBSE Marking Scheme, 2016] 1 as a result of which water is forced into the xylem
cells of the roots. Then there is a steady movement
3. (i) This allows a highly efficient supply of oxygen, of water from the root xylem to all the plant parts
and meet their high energy needs (as they are through the interconnected water conducting
warm–blooded animals). 1 channels. 3
(ii) Because they have to pump blood throughout the
body or into various organs. 1 5. Differences between Xylem and Phloem :
(iii) To allow cellulose to be digested. 1 S. No. Xylem Phloem
[CBSE Marking Scheme, 2012]
(i) Xylem tissue helps Phloem tissue helps
in the transport of in the transport of
water and minerals. food.

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 (ii) Water is transported Food is transported Pulmonary Vein
upwards from roots in both upward and ¯
to all other plant downward direc- Left Auricle
parts. tions. ¯
Left Ventricle
(iii) Transport in xylem Transport of food ¯
occurs with the help in phloem requires Truncus Arteriosus
of simple physical energy in the form ¯
forces such as tran- of ATP. Systemic Circulation
spiration pull. The blood from the systemic circulation comes to
 1+1+1 the right auricle. From the right auricle, the blood is
pumped into the right ventricle. The right ventricle
Ans. The heart is the key organ of cardiovascular system,
pumps the blood to lungs, through pulmonary
which acts as a muscular pump. It is organized with
a set of arteries and blood vessels which regulates artery. Oxygenation of blood takes place in alveoli
the blood flow systematically. in lungs. The oxygenated blood, from the lungs,
comes to the left auricle through pulmonary vein.
The following flow chart shows the circulation of
From left auricle, the blood is pumped into the left
blood through the heart :
ventricle. Finally, the left ventricle pumps the blood
Systemic Vein into the systemic circulation.
¯ In our body two types of circulatory system
Sinus Venosus completes together the whole cardiovascular
¯ system.
Right Auricle (i) First is the transport of blood for the exchange of
¯ gases (Oxygen and Carbon dioxide) which includes
Right Ventricle the connection of heart to liver.
¯ (ii) Second, includes the transport and distribution
Pulmonary Artery
of oxygenated blood from heart to all the other
¯
body parts.
Lungs
¯

TOPIC-4
Excretion
WORKSHEET-69
Solutions The four substances reabsorbed from initial filtrate
1. Correct option : (d) are :
(i) Amino acid (ii) glucose (iii) salts (iv) major
Explanation : Nephrons are the structural and amount of water. 1+2
functional filtration unit of kidney that serve in 4. Nitrogenous waste present in urine is uric acid or
filtration, reabsorption and secretion. Ureters are urea.
small muscular tubes that extend from the kidney
The basic filtration unit of kidney is nephron.
and carry urine into the urinary bladder. The
urethra is a canal that carries urine from bladder Urine production is regulated by :
and expels it out of body. Neurons are structural (i) amount of excess water in the body.
and functional unit of nervous system. (ii) amount of dissolved wastes that need to be excreted.
2. Respiratory unit of lungs — Alveoli 1+1+1
Excretory unit of kidneys — Nephrons ½+½ 5. (i) Many plants store waste materials in the vacuoles
of mesophyll cells and epidermal cells. When old
3. Kidney → Ureters → Urinary bladder → Urethra. leaves fall, the waste materials are excreted along
Glucose, amino acids, salts and major amount of with the leaves.
water. [CBSE Marking Scheme 2016] 1 + 2 (ii) Gaseous waste is removed through stomata in
leaves.
Detailed Answer :
The pathway of urine starting from the organ of its (iii) Excess of water is also excreted from the plant
formation is : body through the stomatal pores. The process of
Kidney → Ureters → Urinary bladder → Urethra. elimination of water is called transpiration. 3

S OLUT I ONS P-57

 6. (a)

Right renal artery

Left kidney
Right kidney

Vena cava Ureter

Urinary bladder

Urethra
3
(b) Vital functions of kidney :
(i) To regulate right amount of water in body.
(ii) Helps in filtering out nitrogenous waste like urea from blood. [CBSE Marking Scheme, 2016] 2

WORKSHEET-74
Solutions (ii) Function of nephron is filtration, reabsorption and
1. Correct option : (c) secretion. 1
Explanation : Kidneys are the paired organ where (iii) Function of artificial kidney : Helps to remove
urine formation takes place. Small muscular tube, harmful wastes, extra salts and water, control
called as ureter, extend from kidneys and carry blood pressure. Maintain the balance of sodium
blood to urinary bladder. The urethra is a small tube potassium salts in a patient whose kidneys have
that extends from the urinary bladder to an external failed. (Any one) 1
opening.
[CBSE Marking Scheme, 2014]
2. The amount of urine produced depends on the
amount of excess water and dissolved wastes
4. (a) Process involved in removal of nitrogenous
present in the body. Some other factors such as
harmful metabolic waste from the body. 1
habitat of an organism and hormone such as
Antidiuretic hormone (ADH) also regulates the (b) Nephron. 1
amount of urine produced. 1 (c) Diagram of Human Excretory System.
3. (i) Labelling of the following parts :
Bowman's Capillaries (i) Kidney
capsule Glomerulus (ii) Ureter
(iii) Urinary bladder

Left kidney
Renal Artery

Ureter
Collecting duct
Venule

Urinary bladder

[CBSE Marking Scheme, 2018] 3

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3
Uriniferous tubule with its blood vessels

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