Q: Question 2(a)Draw the electron dot structure of:(i) Nitrogen molecule [N = 7](ii) Sodium chloride [Na = 11, Cl = 17](iii) Ammonium ion [N = 7, H = 1]

(i) Electron dot structure of Nitrogen molecule is shown below:(ii) Electron dot structure of Sodium chloride is shown below:(iii) Electron dot structure of Ammonium ion is shown below:

Q: Question 2(c)Study the extract of the Periodic Table given below and answer the questions that follow.Give the alphabet corresponding to the element in question. DO NOT repeat an element.A C DE B GF (i) Which element forms electrovalent compound with G?(ii) The ion of which element will migrate towards the cathode during electrolysis?(iii) Which non-metallic element has the valency of 2?(iv) Which is an inert gas?

(i) B forms an electrovalent compound with G.(ii) A has positive ions which migrate towards cathode.(iii) E is non-metallic and has a valency of 2 as it belongs to group 16.(iv) F is an inert gas as it belongs to the zero group.

Q: Question 3(b)Distinguish between the following pairs of compounds using the reagent given in the bracket.(i) Manganese dioxide and copper (II) oxide. (using concentrated HCl)(ii) Ferrous sulphate solution and ferric sulphate solution. (using sodium hydroxide solution)(iii) Dilute hydrochloric acid and dilute sulphuric acid. (using lead nitrate solution)

(i) When each of the compound is heated with conc. hydrochloric acid, greenish yellow (chlorine) gas is evolved in case of manganese dioxide and filtrate is brownish in colour whereas, no chlorine gas is evolved in case of copper (II) oxide and filtrate is bluish in colour.MnO2 + 4HCl ⟶ MnCl2 + 2H2O + Cl2CuO + 2HCl ⟶ CuCl2 + H2O(ii) When sodium hydroxide is added to the two solns., ferrous sulphate solution gives a dirty green ppt. of Fe(OH)2 whereas, ferric sulphate solution forms a reddish brown ppt. of Fe(OH)3. Hence, the two compounds can be distinguished.(iii) Sulphuric acid forms a white precipitate with lead nitrate solution. This precipitate does not dissolve on warming the reaction mixture.H2SO4 (dil.) + Pb(NO3)2 ⟶ 2HNO3 + PbSO4 ↓ [white ppt. formed which does not dissolve on warming the mixture].Hydrochloric acid forms a white precipitate with lead nitrate solution. This precipitate dissolves on warming the reaction mixture so as to form clear solution.2HCl (dil.) + Pb(NO3)2 ⟶ 2HNO3 + PbCl2 ↓ [white ppt. formed which dissolves on warming the mixture].Hence, dilute hydrochloric acid and dilute sulphuric acid can be distinguished using lead nitrate solution.

Q: Question 4(b)Write a balanced chemical equation for the preparation of:(i) Ethene from bromoethane(ii) Ethyne using calcium carbide(iii) Methane from sodium acetate.

(i) Ethene from bromoethaneC2H5-Br Bromoethane [ethyl bromide]+KOH alcoholic KOH→boilH2C=CH2 ethene [ethylene]+KBr+H2O\underset{\text{ Bromoethane [ethyl bromide]}}{\text{C}_2\text{H}_5\text{-Br}} + \underset{\text{ alcoholic KOH}}{\text{KOH}} \xrightarrow{\text{boil}} \underset{ \text{ ethene [ethylene]} } {\text{H}_2\text{C}=\text{C}\text{H}_2} + \text{KBr} + \text{H}_2\text{O} Bromoethane [ethyl bromide]C2H5-Br+ alcoholic KOHKOHboil ethene [ethylene]H2C=CH2+KBr+H2O(ii) Ethyne from calcium carbideCaC2calcium carbide+2H2Ocold water⟶C2H2ethyne [acetylene]+Ca(OH)2\underset{\text{calcium carbide}}{\text{CaC}_2} + \underset{\text{cold water}}{2\text{H}_2\text{O}} \longrightarrow \underset{\text{ethyne [acetylene]}}{\text{C}_2\text{H}_2} + \text{Ca(OH)}_2calcium carbideCaC2+cold water2H2O⟶ethyne [acetylene]C2H2+Ca(OH)2(iii)CH3.COONa sodium acetate+NaOHsodalime→CaOΔCH4methane+Na2CO3\underset{\text{ sodium acetate}}{\text{CH}_3\text{.COONa}} + \underset{\text{sodalime}}{\text{NaOH}} \underset{\Delta}{\xrightarrow{\text{CaO}}} \underset{\text{methane}}{\text{C}\text{H}_4} + \text{Na}_2\text{CO}_3 sodium acetateCH3.COONa+sodalimeNaOHΔCaOmethaneCH4+Na2CO3

Question 1

Question 1(ii)The most electronegative element from the following elements is:MagnesiumChlorineAluminiumSulphur

Accepted Answer

ChlorineReason — Chlorine is the most electronegative element among the given options because electronegativity increases left to right in a period.

Question 2

Question 1(iv)The drying agent used to dry HCl gas is:Conc. H2SO4ZnOAl2O3CaO

Accepted Answer

Conc. H2SO4Reason — Drying agent used for drying should only remove the moisture and not react with it, hence, conc. sulphuric acid is used as the drying agent.

Question 3

Question 1(b)Fill in the blanks with the choices given in brackets:(i) Conversion of ethanol to ethene by the action of concentrated sulphuric acid is an example of ............... (dehydration/dehydrogenation/dehydrohalogenation)(ii) When sodium chloride is heated with concentrated sulphuric acid below 200°C, one of the products formed is ............... (sodium hydrogen sulphate / sodium sulphate / chlorine)(iii) Ammonia reacts with excess chlorine to form ............... (nitrogen / nitrogen trichloride / ammonium chloride)(iv) Substitution reactions are characteristic reactions of ............... (alkynes / alkenes / alkanes)(v) In Period 3, the most metallic element is ............... (sodium / magnesium / aluminium)

Accepted Answer

(i) Conversion of ethanol to ethene by the action of concentrated sulphuric acid is an example of dehydration(ii) When sodium chloride is heated with concentrated sulphuric acid below 200°C, one of the products formed is sodium hydrogen sulphate (sodium hydrogen sulphate / sodium sulphate / chlorine)(iii) Ammonia reacts with excess chlorine to form nitrogen trichloride(iv) Substitution reactions are characteristic reactions of alkanes(v) In Period 3, the most metallic element is sodium.

Question 4

Question 1(c)Write a balanced chemical equation for each of the following reactions:(i) Reduction of copper (II) oxide by hydrogen.(ii) Action of dilute sulphuric acid on sodium hydroxide.(iii) Action of dilute sulphuric acid on zinc sulphide.(iv) Ammonium hydroxide is added to ferrous sulphate solution.(v) Chlorine gas is reacted with ethene.

Accepted Answer

(i) Reduction of copper (II) oxide by hydrogen.CuO+H2→ΔCu+H2O[g]	ext{CuO} + 	ext{H}_2 \xrightarrow{\Delta} 	ext{Cu} + 	ext{H}_2	ext{O} [	ext{g}]CuO+H2​Δ​Cu+H2​O[g](ii) Action of dilute sulphuric acid on sodium hydroxide.2NaOH + H2SO4 ⟶ Na2SO4 + 2H2O(iii) Action of dilute sulphuric acid on zinc sulphide.ZnS + H2SO4 (dil.) ⟶ ZnSO4 + H2S(iv) Ammonium hydroxide is added to ferrous sulphate solution.2NH4OH + FeSO4 ⟶ (NH4)2SO4 + Fe(OH)2(v) Chlorine gas is reacted with ethene.

Question 5

Question 1(f)(i) Give the IUPAC name of the following organic compounds:(ii) What is the special feature of the structure of ethyne(iii) Name the saturated hydrocarbon containing two carbon atoms.(iv) Give the structural formula of acetic acid.

Accepted Answer

(i) The IUPAC names are:     A: Propyne     B: Ethanal(ii) The special feature of the structure of ethyne is that the two carbon atoms are linked by triple covalent bond formed by sharing three pairs of electrons between the two carbon atoms. The availability of electrons on the triple bond makes ethyne more reactive and hence it undergoes characteristics addition reactions only.(iii) Ethane [C2H6](iv) Structural formula of acetic acid is shown below:

Question 6

Question 1(h)Arrange the following according to the instructions given in brackets:(i) K, Pb, Ca, Zn. (In the increasing order of the reactivity)(ii) Mg2+, Cu2+, Na1+, H1+ (In the order of preferential discharge at the cathode)(iii) Li, K, Na, H (In the decreasing order of their ionization potential)(iv) F, B, N, O (In the increasing order of electron affinity)(v) Ethane, methane, ethene, ethyne. (In the increasing order of the molecular weight) [H = 1, C = 12]

Accepted Answer

(i) Pb < Zn < Ca < K(ii) Cu2+, H1+, Mg2+, Na1+Lower the position of the ion, greater the tendency to be liberated at the cathode (or respective electrode).(iii) H > Li > Na > KIonization Potential decreases as we move down the group.(iv) B < N < O < FElectron Affinity increases from left to right across a period.(v) Methane < Ethyne < Ethene < EthaneEthane [C2H6] : M.W. = 2[12] + 6[1] = 30Methane [CH4] : M.W. = 12 + 4[1] = 16Ethene [C2H4] : M.W. = 2[12] + 4[1] = 28Ethyne [C2H2] : M.W. = 2[12] + 2[1] = 26Hence, increasing order of molecular weight : Methane < Ethyne < Ethene < Ethane

Question 7

Question 2(a)Draw the electron dot structure of:(i) Nitrogen molecule [N = 7](ii) Sodium chloride [Na = 11, Cl = 17](iii) Ammonium ion [N = 7, H = 1]

Accepted Answer

(i) Electron dot structure of Nitrogen molecule is shown below:(ii) Electron dot structure of Sodium chloride is shown below:(iii) Electron dot structure of Ammonium ion is shown below:

Question 8

Question 2(c)Study the extract of the Periodic Table given below and answer the questions that follow.Give the alphabet corresponding to the element in question. DO NOT repeat an element.A          C DE B     GF         (i) Which element forms electrovalent compound with G?(ii) The ion of which element will migrate towards the cathode during electrolysis?(iii) Which non-metallic element has the valency of 2?(iv) Which is an inert gas?

Accepted Answer

(i) B forms an electrovalent compound with G.(ii) A has positive ions which migrate towards cathode.(iii) E is non-metallic and has a valency of 2 as it belongs to group 16.(iv) F is an inert gas as it belongs to the zero group.

Question 9

Question 3(b)Distinguish between the following pairs of compounds using the reagent given in the bracket.(i) Manganese dioxide and copper (II) oxide. (using concentrated HCl)(ii) Ferrous sulphate solution and ferric sulphate solution. (using sodium hydroxide solution)(iii) Dilute hydrochloric acid and dilute sulphuric acid. (using lead nitrate solution)

Accepted Answer

(i) When each of the compound is heated with conc. hydrochloric acid, greenish yellow (chlorine) gas is evolved in case of manganese dioxide and filtrate is brownish in colour whereas, no chlorine gas is evolved in case of copper (II) oxide and filtrate is bluish in colour.MnO2 + 4HCl ⟶ MnCl2 + 2H2O + Cl2CuO + 2HCl ⟶ CuCl2 + H2O(ii) When sodium hydroxide is added to the two solns., ferrous sulphate solution gives a dirty green ppt. of Fe(OH)2 whereas, ferric sulphate solution forms a reddish brown ppt. of Fe(OH)3. Hence, the two compounds can be distinguished.(iii) Sulphuric acid forms a white precipitate with lead nitrate solution. This precipitate does not dissolve on warming the reaction mixture.H2SO4 (dil.) + Pb(NO3)2 ⟶ 2HNO3 + PbSO4 ↓ [white ppt. formed which does not dissolve on warming the mixture].Hydrochloric acid forms a white precipitate with lead nitrate solution. This precipitate dissolves on warming the reaction mixture so as to form clear solution.2HCl (dil.) + Pb(NO3)2 ⟶ 2HNO3 + PbCl2 ↓ [white ppt. formed which dissolves on warming the mixture].Hence, dilute hydrochloric acid and dilute sulphuric acid can be distinguished using lead nitrate solution.

Question 10

Question 4(b)Write a balanced chemical equation for the preparation of:(i) Ethene from bromoethane(ii) Ethyne using calcium carbide(iii) Methane from sodium acetate.

Accepted Answer

(i) Ethene from bromoethaneC2H5-Br Bromoethane [ethyl bromide]+KOH alcoholic KOH→boilH2C=CH2 ethene [ethylene]+KBr+H2O\underset{	ext{ Bromoethane [ethyl bromide]}}{	ext{C}_2	ext{H}_5	ext{-Br}} + \underset{	ext{ alcoholic KOH}}{	ext{KOH}} \xrightarrow{	ext{boil}} \underset{ 	ext{ ethene [ethylene]} } {	ext{H}_2	ext{C}=	ext{C}	ext{H}_2} + 	ext{KBr} + 	ext{H}_2	ext{O} Bromoethane [ethyl bromide]C2​H5​-Br​+ alcoholic KOHKOH​boil​ ethene [ethylene]H2​C=CH2​​+KBr+H2​O(ii) Ethyne from calcium carbideCaC2calcium carbide+2H2Ocold water⟶C2H2ethyne [acetylene]+Ca(OH)2\underset{	ext{calcium carbide}}{	ext{CaC}_2} + \underset{	ext{cold water}}{2	ext{H}_2	ext{O}} \longrightarrow \underset{	ext{ethyne [acetylene]}}{	ext{C}_2	ext{H}_2} + 	ext{Ca(OH)}_2calcium carbideCaC2​​+cold water2H2​O​⟶ethyne [acetylene]C2​H2​​+Ca(OH)2​(iii)CH3.COONa sodium acetate+NaOHsodalime→CaOΔCH4methane+Na2CO3\underset{	ext{ sodium acetate}}{	ext{CH}_3	ext{.COONa}} + \underset{	ext{sodalime}}{	ext{NaOH}} \underset{\Delta}{\xrightarrow{	ext{CaO}}} \underset{	ext{methane}}{	ext{C}	ext{H}_4} + 	ext{Na}_2	ext{CO}_3 sodium acetateCH3​.COONa​+sodalimeNaOH​ΔCaO​​methaneCH4​​+Na2​CO3​

Question 11

Question 4(c)Name the following organic compounds:(i) The compound with 3 carbon atoms whose functional group is a carboxyl.(ii) The first homologue whose general formula is CnH2n(iii) The compound that reacts with acetic acid to form ethyl ethanoate.(iv) The compound formed by complete chlorination of ethyne.

Accepted Answer

(i) Propanoic acid [C3H7CHO](ii) Ethene [C2H4](iii) Ethyl alcoholC2H5OHethyl alcohol+CH3COOHacetic acid→Conc. H2SO4CH3−COO−C2H5ethyl ethanoate+H2O\underset{	ext{ethyl alcohol}}{	ext{C}_2	ext{H}_5	ext{OH}} + \underset{	ext{acetic acid}}{	ext{CH}_3	ext{COOH}} \xrightarrow{	ext{Conc. H}_2	ext{SO}_4} \underset{	ext{ethyl ethanoate}}{	ext{CH}_3-	ext{COO}-	ext{C}_2	ext{H}_5} + 	ext{H}_2	ext{O}ethyl alcoholC2​H5​OH​+acetic acidCH3​COOH​Conc. H2​SO4​​ethyl ethanoateCH3​−COO−C2​H5​​+H2​O(iv) 1,1,2,2 tetrachloroethane

Question 12

Question 5(b)Answer the following questions based on the extraction of aluminium from alumina by Hall-Heroult's Process.:(i) What is the function of cryolite used along with alumina as the electrolyte?(ii) Why is powdered coke sprinkled on top of the electrolyte?(iii) Name the electrode, from which aluminium is collected.

Accepted Answer

(i) Addition of cryolite —lowers the fusion point of the mixture i.e., mixture fuses around 950°C instead of 2050°C.enhances the mobility of the fused mixture by acting as a solvent for the electrolytic mixture.enhances the conductivity of the mixture since, pure alumina is almost a non-conductor of electricity.(ii) The layer of powdered coke is sprinkled over the electrolytic mixture as :it prevents burning of carbon electrodes in air at the emergence point from the bath.it minimizes or prevents heat loss by radiation.(iii) Cathode

Question 13

Question 6(a)Identify the substances underlined :(i) The catalyst used to oxidise ammonia.(ii) The organic compound which when solidified, forms an ice like mass.(iii) The dilute acid which is an oxidizing agent.

Accepted Answer

(i) Platinum4NH3+5O2→800°CPt.4NO+6H2O+Δ4	ext{NH}_3 + 5	ext{O}_2 \xrightarrow[800 \degree 	ext{C}]{	ext{Pt.}} 	ext{4NO} + 	ext{6H}_	ext{2}	ext{O} + \Delta4NH3​+5O2​Pt.800°C​4NO+6H2​O+Δ(ii) Glacial acetic acid(iii) Nitric acid

Question 14

Question 6(b)Copper sulphate solution reacts with sodium hydroxide solution to form a precipitate of copper hydroxide according to the equation:2NaOH + CuSO4 ⟶ Na2SO4 + Cu(OH)2↓(i) What mass of copper hydroxide is precipitated by using 200 g of sodium hydroxide.[H = 1, O = 16, Na = 23, S = 32, Cu = 64](ii) What is the colour of the precipitate formed.

Accepted Answer

(i)2NaOH+CuSO4⟶Na2SO4+Cu(OH)2↓2[23+1664+2[40]+1]=98g=80 g\begin{matrix} 2	ext{NaOH} & + & 	ext{CuSO}_4 & \longrightarrow & 	ext{Na}_2	ext{SO}_4 & + & 	ext{Cu(OH)}_2\downarrow \ 2[23 + 16 & & & & & & 64 + 2[40] \ + 1] & & & & & & = 98 	ext{g} \ = 80 	ext{ g} & & & & & & \ \end{matrix}2NaOH2[23+16+1]=80 g​+​CuSO4​​⟶​Na2​SO4​​+​Cu(OH)2​↓64+2[40]=98g​80 g of sodium hydroxide precipitates 98 g of copper hydroxide.∴ 200 g of sodium hydroxide will precipitate 9880\dfrac{98}{80}8098​ x 200 = 245 g of copper hydroxide.Hence, 245 g. of copper hydroxide is precipitated.(ii) Pale blue precipitate is formed.

Question 15

Question 6(c)Find the empirical formula and the molecular formula of an organic compound from the data given below:C = 75.92%, H = 6.32% and N = 17.76%The vapour density of the compound is 39.5.[C = 12, H = 1, N = 14]

Accepted Answer

Element% compositionAt. wt.Relative no. of atomsSimplest ratioCarbon75.921275.9212\dfrac{75.92}{12}1275.92​ = 6.326.321.26\dfrac{ 6.32 }{ 1.26}1.266.32​ = 5Hydrogen6.3216.321\dfrac{6.32}{1}16.32​ = 6.326.321.26\dfrac{6.32 }{ 1.26}1.266.32​ = 5Nitrogen17.761417.7614\dfrac{ 17.76}{14}1417.76​ = 1.261.261.26\dfrac{ 1.26}{ 1.26}1.261.26​ = 1Simplest ratio of whole numbers = C : H : N = 5 : 5 : 1Hence, empirical formula is C5H5NEmpirical formula weight = 5(12) + 5(1) + 14 = 79V.D. = 39.5Molecular weight = 2 x V.D. = 2 x 39.5 = 79n=Molecular weightEmpirical formula weight=7979=1\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \[0.5em] = \dfrac{79}{79} = 1n=Empirical formula weightMolecular weight​=7979​=1∴ Molecular formula = n[E.F.] = 1[C5H5N] = C5H5N

Question 16

Question 7(a)Name the gas evolved in each of the following cases:(i) Alumina undergoes electrolytic reduction.(ii) Ethene undergoes hydrogenation reaction.(iii) Ammonia reacts with heated copper oxide.

Accepted Answer

(i) Oxygen gas(ii) Ethane gas(iii) Nitrogen gas

Question 17

Question 7(b)Study the flow chart given and give balanced equations to represent the reactions A, B and C:Mg3N2→ANH3⇄CBNH4Cl	ext{Mg}_3	ext{N}_2 \xrightarrow{	ext{A}} 	ext{NH}_3 \xtofrom[	ext{C}]{	ext{B}} 	ext{NH}_4	ext{Cl}Mg3​N2​A​NH3​BC​NH4​Cl

Accepted Answer

A : Mg3N2 + 6H2O ⟶ 3Mg(OH)2 + 2NH3 [g]B : NH3 + HCl ⟶ NH4ClC: NH4Cl + NaOH ⟶ NaCl + H2O + NH3 [g]

Question 18

Question 7(c)Copy and complete the following table which refers to the industrial method for preparation of ammonia and sulphuric acid.Name of the compoundName of the processCatalytic equation[with the catalyst]Ammonia(i)(ii)Sulphuric acid(iii)(iv)

Accepted Answer

(i) Haber's process(ii) N2+3H2⇌450-500 °CFe2NH3+Δ	ext{N}_2 + 3	ext{H}_2 \xrightleftharpoons[	ext{450-500 \degree C}]{	ext{Fe}} 	ext{2NH}_3 + \DeltaN2​+3H2​Fe450-500 °C​2NH3​+Δ(iii) Contact Process(iv) 2SO2+O2⇌450−500°C /1−2 atmosV2O52SO3+45 K cals2	ext{SO}_2 + 	ext{O}_2 \xrightleftharpoons [450-500	ext{\degree C }/1 - 2 	ext{ atmos}]{	ext{V}_2	ext{O}_5} 2	ext{SO}_3 + 	ext{45 K cals}2SO2​+O2​V2​O5​450−500°C /1−2 atmos​2SO3​+45 K cals

Solved 2019 Question Paper ICSE Class 10 Chemistry

Section I 40 Marks

Section Ii 40 Marks