Question 1

Question 5Give examples of eight metals which show variable valency. State the valency of sulphur in :(a) SO2(b) SO3

Accepted Answer

Eight metals which show variable valency are :1+2+3+4+CuCuHgHgAgAgAu  Au  FeFe  Pb  Pb  Sn  Sn  Mn  Mn(a) Valency of sulphur in SO2 is 4By interchanging subscript and writing as superscript:S11 ↗↖ O2⇒S22 ↗↖ O1\underset{\phantom{1}{1}}{\text{S}} \space {\nearrow}\mathllap{\nwarrow} \space \underset{2}{\text{O}} \Rightarrow \overset{\phantom{2}{2}}{\text{S}} \space {\nearrow}\mathllap{\nwarrow} \space \overset{1}{\text{O}} \[0.5em]11S​ ↗↖ 2O​⇒S22 ↗↖ O1But valency of O is 2. Multiplying by 2, we get:S2×2 ↗↖ O2×1⇒S44 ↗↖ O2\overset{{2 \times 2}}{\text{S}} \space {\nearrow}\mathllap{\nwarrow} \space \overset{2 \times 1}{\text{O}} \Rightarrow \overset{\phantom{4}{4}}{\text{S}} \space {\nearrow}\mathllap{\nwarrow} \space \overset{2}{\text{O}} \[0.5em]S2×2 ↗↖ O2×1⇒S44 ↗↖ O2Therefore, valency of Sulphur is 4.(b) Valency of sulphur in SO3 is 6S11 ↗↖ O3⇒S33 ↗↖ O1\underset{\phantom{1}{1}}{\text{S}} \space {\nearrow}\mathllap{\nwarrow} \space \underset{3}{\text{O}} \Rightarrow \overset{\phantom{3}{3}}{\text{S}} \space {\nearrow}\mathllap{\nwarrow} \space \overset{1}{\text{O}} \[0.5em]11S​ ↗↖ 3O​⇒S33 ↗↖ O1But valency of O is 2. Multiplying by 2, we get:S2×3 ↗↖ O2×1⇒S66 ↗↖ O2\overset{{2 \times 3}}{\text{S}} \space {\nearrow}\mathllap{\nwarrow} \space \overset{2 \times 1}{\text{O}} \Rightarrow \overset{\phantom{6}{6}}{\text{S}} \space {\nearrow}\mathllap{\nwarrow} \space \overset{2}{\text{O}} \[0.5em]S2×3 ↗↖ O2×1⇒S66 ↗↖ O2Therefore, valency of Sulphur is 6.

Question 2

Question 12(1)Write the formula of the following compounds :Potassium(a)chloride(b)nitrate(c)carbonate(d)bisulphate(e)sulphite(f)dichromate(g)permangnate(h)zincate(i)plumbite(j)sulphate(k)bicarbonate(l)aluminate(m)hydroxide(n)iodide(o)nitrite(p)bisulphite

Accepted Answer

PotassiumSl.No.CompoundFormula(a)chlorideKCl(b)nitrateKNO3(c)carbonateK2CO3(d)bisulphateKHSO4(e)sulphiteK2SO3(f)dichromateK2Cr2O7(g)permangnateKMnO4(h)zincateK2ZnO2(i)plumbiteK2PbO2(j)sulphateK2SO4(k)bicarbonateKHCO3(l)aluminateKAIO2(m)hydroxideKOH(n)iodideKI(o)nitriteKNO2(p)bisulphiteKHSO3

Question 3

Question 12(4)Write the formula of the following compounds :Magnesium(a)chloride(b)nitrate(c)carbonate(d)sulphate(e)bicarbonate(f)hydroxide(g)oxide

Accepted Answer

Sl.No.CompoundFormula(a)chlorideMgCl2(b)nitrateMg(NO3)2(c)carbonateMgCO3(d)sulphateMgSO4(e)bicarbonateMg(HCO3)2(f)hydroxideMg(OH)2(g)oxideMgO

Question 4

Question 12(5)Write the formula of the following compounds :Zinc(a)chloride(b)nitrate(c)carbonate(d)sulphate(e)hydroxide(f)oxide

Accepted Answer

Sl.No.CompoundFormula(a)chlorideZnCl2(b)nitrateZn(NO3)2(c)carbonateZnCO3(d)sulphateZnSO4(e)hydroxideZn(OH)2(f)oxideZnO

Question 5

Question 12(6)Write the formula of the following compounds :Aluminium(a)chloride(b)nitrate(c)carbonate(d)sulphate(e)hydroxide(f)oxide

Accepted Answer

Sl.No.CompoundFormula(a)chlorideAlCl3(b)nitrateAl(NO3)3(c)carbonateAl2(CO3)3(d)sulphateAl2(SO4)3(e)hydroxideAl(OH)3(f)oxideAl2O3

Question 6

Question 12(7)Write the formula of the following compounds :CopperCopper [I] chlorideCopper [II] chlorideCopper [I] oxideCopper [II] oxideCopper [I] sulphideCopper [II] sulphideCopper [II] nitrateCopper [II] sulphateTetra amine copper [II] sulphate

Accepted Answer

CompoundFormulaCopper [I] chlorideCuClCopper [II] chlorideCuCl2Copper [I] oxideCu2OCopper [II] oxideCuOCopper [I] sulphideCu2SCopper [II] sulphideCuSCopper [II] nitrateCu(NO3)2Copper [II] sulphideCuSTetra amine copper [II] sulphate[Cu(NH3)4]SO4

Question 7

Question 12(8)Write the formula of the following compounds :IronIron [II] chlorideIron [III] chlorideIron [II] oxideIron [III] oxideIron [II] sulphateIron [III] sulphateIron [II] sulphideIron [III] sulphideIron [II] hydroxideIron [III] hydroxide

Accepted Answer

CompoundFormulaIron [II] chlorideFeCl2Iron [III] chlorideFeCl3Iron [II] oxideFeOIron [III] oxideFe2O3Iron [II] sulphateFeSO4Iron [III] sulphateFe2(SO4)3Iron [II] sulphideFeSIron [III] sulphideFe2S3Iron [II] hydroxideFe(OH)2Iron [III] hydroxideFe(OH)3

Question 8

Question 19Balance the following simple equations :C + O2 ⟶ CON2 + O2 ⇌ NOZnS + O2 ⟶ ZnO + SO2Al + O2 ⟶ Al2O3Mg + N2 ⟶ Mg3N2Al + N2 ⟶ AlNNO + O2 ⟶ NO2SO2 + O2 ⇌ SO3H2 + Cl2 ⟶ HClFe + Cl2 ⟶ FeCl3H2S + Cl2 ⟶ S + HClFeCl2 + Cl2 ⟶ FeCl3CO2 + C ⟶ COKHCO3 ⟶ K2CO3 + H2O + CO2K + CO2 ⟶ K2O + CCa(OH)2 + HNO3 ⟶ Ca(NO3)2 + H2OK + H2O ⟶ KOH + H2Ca + H2O ⟶ Ca(OH)2 + H2Al + H2O ⟶ Al2O3 + H2Fe + H2O ⇌ Fe3O4 + H2Zn + NaOH ⟶ Na2ZnO2 + H2Zn + HCl ⟶ ZnCl2 + H2Al + H2SO4 ⟶ Al2(SO4)3 + H2H2 + O2 ⟶ H2ON2 + H2 ⟶ NH3Fe2O3 + H2O ⟶ Fe + H2OKBr + Cl2 ⟶ KCl + Br2NaOH + Cl2 ⟶ NaCl + NaClO + H2ONaHCO3 + H2SO4 ⟶ Na2SO4 + H2O + CO2Mg + CO2 ⟶ MgO + CFe2O3 + CO ⟶ Fe + CO2CaO + HCl ⟶ CaCl2 + H2O

Accepted Answer

2C + O2 ⟶ 2CON2 + O2 ⇌ 2NO2ZnS + 3O2 ⟶ 2ZnO + 2SO24Al + 3O2 ⟶ 2Al2O33Mg + N2 ⟶ Mg3N22Al + N2 ⟶ 2AlN2NO + O2 ⟶ 2NO22SO2 + O2 ⇌ 2SO3H2 + Cl2 ⟶ 2HCl2Fe + 3Cl2 ⟶ 2FeCl3H2S + Cl2 ⟶ S + 2HCl2FeCl2 + Cl2 ⟶ 2FeCl3CO2 + C ⟶ 2CO2KHCO3 ⟶ K2CO3 + H2O + CO24K + CO2 ⟶ 2K2O + CCa(OH)2 + 2HNO3 ⟶ Ca(NO3)2 + 2H2O2K + 2H2O ⟶ 2KOH + H2Ca + 2H2O ⟶ Ca(OH)2 + H22Al + 3H2O ⟶ Al2O3 + 3H23Fe + 4H2O ⟶ Fe3O4 + 4H2Zn + 2NaOH ⟶ Na2ZnO2 + H2Zn + 2HCl ⟶ ZnCl2 + H22Al + 3H2SO4 ⟶ Al2(SO4)3 + 3H22H2 + O2 ⟶ 2H2ON2 + 3H2 ⟶ 2NH3Fe2O3 + 3H2 ⟶ 2Fe + 3H2O2KBr + Cl2 ⟶ 2KCl + Br22NaOH + Cl2 ⟶ NaCl + NaClO + H2O2NaHCO3 + H2SO4 ⟶ Na2SO4 + 2H2O + 2CO22Mg + CO2 ⟶ 2MgO + CFe2O3 + 3CO ⟶ 2Fe + 3CO2CaO + 2HCl ⟶ CaCl2 + H2O

Question 9

Question 20Write balanced equations for the following word equations :Potassium nitrate ⟶ Potassium nitrite + OxygenCalcium + Water ⟶ Calcium hydroxide + HydrogenIron + Hydrochloric acid ⟶ Iron [II] chloride + HydrogenNitrogen dioxide + Water + Oxygen ⟶ Nitric acidLead dioxide [lead (IV) oxide] ⟶ Lead monoxide + OxygenAluminium + Oxygen ⟶ Aluminium oxideIron + Chlorine ⟶ Iron [III] chloridePotassium bromide + Chlorine ⟶ Potassium chloride + BrominePotassium bicarbonate ⟶ Potassium carbonate + Water + Carbon dioxideCalcium hydroxide + Ammonium chloride ⟶ Calcium chloride + Water + Ammonia

Accepted Answer

2KNO3 ⟶ 2KNO2 + O2Ca + 2H2O ⟶ Ca(OH)2 + H2Fe + 2HCl ⟶ FeCl2 + H24NO2 + 2H2O + O2 ⟶ 4HNO32PbO2 ⟶ 2PbO + O24Al + 3O2 ⟶ 2Al2O32Fe + 3Cl2 ⟶ 2FeCl32KBr + Cl2 ⟶ 2KCl + Br22KHCO3 ⟶ K2CO3 + H2O + CO2Ca(OH)2 + 2NH4Cl ⟶ CaCl2 + 2H2O + 2NH3

Question 10

Question 21Balance the following important equations :NaHCO3 + H2SO4 ⟶ Na2SO4 + H2O + CO2NaOH + H2SO4 ⟶ Na2SO4 + H2OPb(NO3)2 + NaCl ⟶ NaNO3 + PbCl2FeSO4 + NaOH ⟶ Na2SO4 + Fe(OH)2FeCl3 + NaOH ⟶ NaCl + Fe(OH)3CuSO4 + NaOH ⟶ Na2SO4 + Cu(OH)2FeCl3 + NH4OH ⟶ NH4Cl + Fe(OH)3ZnO + NaOH ⟶ Na2ZnO2 + H2OPb(OH)2 + NaOH ⟶ Na2PbO2 + H2OAl2O3.2H2O + NaOH ⟶ NaAlO2 + H2ONaAlO2 + H2O ⟶ NaOH + Al(OH)3Al(OH)3 ⟶ Al2O3 + H2OZnS + O2 ⟶ ZnO + SO2Fe2O3 + Al ⟶ Al2O3 + FeAl + Cl2 ⟶ AlCl3NaCl + H2SO4 ⟶ Na2SO4 + HClFe + HCl ⟶ FeCl2 + H2Na2CO3 + HCl ⟶ NaCl + H2O + CO2Pb(NO3)2 + HCl ⟶ PbCl2 + HNO3AgCl + NH4OH ⟶ Ag(NH3)2Cl + H2OMnO2 + HCl ⟶ MnCl2 + H2O + Cl2Pb3O4 + HCl ⟶ PbCl2 + H2O + Cl2KMnO4 + HCl ⟶ KCl + MnCl2 + H2O + Cl2K2Cr2O7 + HCl ⟶ KCl + CrCl3 + H2O + Cl2NH4Cl + Ca(OH)2 ⟶ CaCl2 + H2O + NH3(NH4)2SO4 + NaOH ⟶ Na2SO4 + H2O + NH3Mg3N2 + H2O ⟶ Mg(OH)2 + NH3AlN + H2O ⟶ Al(OH)3 + NH3NH3 + O2 ⟶ N2 + H2O [burning of NH3]NH3 + O2 ⟶ NO + H2O [Catalytic oxidation of NH3]NH4OH + H2SO4 ⟶ (NH4)2SO4 + H2ONH3 + CuO ⟶ Cu + H2O + N2NH3 + Cl2 ⟶ HCl + NCl3 [nitrogen trichloride]HNO3 ⟶ H2O + NO2 + O2Ca(HCO3)2 + HNO3 ⟶ Ca(NO3)2 + H2O + CO2C + HNO3 [conc.] ⟶ CO2 + H2O + NO2S + HNO3 [conc.] ⟶ H2SO4 + H2O + NO2Cu + HNO3 [conc.] ⟶ Cu(NO3)2 + H2O + NO2C + H2SO4 [conc.] ⟶ CO2 + H2O + SO2S + H2SO4 [conc.] ⟶ SO2 + H2OCu + H2SO4 [conc.] ⟶ CuSO4 + H2O + SO2

Accepted Answer

2NaHCO3 + H2SO4 ⟶ Na2SO4 + 2H2O + 2CO22NaOH + H2SO4 ⟶ Na2SO4 + 2H2OPb(NO3)2 + 2NaCl ⟶ 2NaNO3 + PbCl2FeSO4 + 2NaOH ⟶ Na2SO4 + Fe(OH)2FeCl3 + 3NaOH ⟶ 3NaCl + Fe(OH)3CuSO4 + 2NaOH ⟶ Na2SO4 + Cu(OH)2FeCl3 + 3NH4OH ⟶ 3NH4Cl + Fe(OH)3ZnO + 2NaOH ⟶ Na2ZnO2 + H2OPb(OH)2 + 2NaOH ⟶ Na2PbO2 + 2H2OAl2O3.2H2O + 2NaOH ⟶ 2NaAlO2 + 3H2ONaAlO2 + 2H2O ⟶ NaOH + Al(OH)32Al(OH)3 ⟶ Al2O3 + 3H2O2ZnS + 3O2 ⟶ 2ZnO + 2SO2Fe2O3 + 2Al ⟶ Al2O3 + 2Fe2Al + 3Cl2 ⟶ 2AlCl32NaCl + H2SO4 ⟶ Na2SO4 + 2HClFe + 2HCl ⟶ FeCl2 + H2Na2CO3 + 2HCl ⟶ 2NaCl + H2O + CO2Pb(NO3)2 + 2HCl ⟶ PbCl2 + 2HNO3AgCl + 2NH4OH ⟶ Ag(NH3)2Cl + 2H2OMnO2 + 4HCl ⟶ MnCl2 + 2H2O + Cl2Pb3O4 + 8HCl ⟶ 3PbCl2 + 4H2O + Cl22KMnO4 + 8HCl ⟶ 2KCl + 2MnCl2 + 4H2O + Cl2K2Cr2O7 + 14HCl ⟶ 2KCl + 2CrCl3 + 7H2O + 3Cl22NH4Cl + Ca(OH)2 ⟶ CaCl2 + 2H2O + 2NH3(NH4)2SO4 + 2NaOH ⟶ Na2SO4 + 2H2O + 2NH3Mg3N2 + 6H2O ⟶ 3Mg(OH)2 + 2NH3AlN + 3H2O ⟶ Al(OH)3 + NH34NH3 + 3O2 ⟶ 2N2 + 6H2O [burning of NH3]4NH3 + 5O2 ⟶ 4NO + 6H2O [Catalytic oxidation of NH3]2NH4OH + H2SO4 ⟶ (NH4)2SO4 + 2H2O2NH3 + 3CuO ⟶ 3Cu + 3H2O + N2NH3 + 3Cl2 ⟶ 3HCl + NCl3 [nitrogen trichloride]4HNO3 ⟶ 2H2O + 4NO2 + O2Ca(HCO3)2 + 2HNO3 ⟶ Ca(NO3)2 + 2H2O + 2CO2C + 4HNO3 [conc.] ⟶ CO2 + 2H2O + 4NO2S + 6HNO3 [conc.] ⟶ H2SO4 + 2H2O + 6NO2Cu + 4HNO3 [conc.] ⟶ Cu(NO3)2 + 2H2O + 2NO2C + 2H2SO4 [conc.] ⟶ CO2 + 2H2O + 2SO2S + 2H2SO4 [conc.] ⟶ 3SO2 + 2H2OCu + 2H2SO4 [conc.] ⟶ CuSO4 + 2H2O + SO2

Question 11

Question 22Give balanced equations for (1) & (2) by partial equation method, [steps are given below]Reaction of excess ammonia with chlorine – Ammonia as a reducing agent(a) Ammonia first reacts with chlorine to give hydrogen chloride and nitrogen.(b) Hydrogen chloride then further reacts with excess ammonia to give ammonium chloride.Oxidation of Lead [II] Sulphide by Ozone(a) Ozone first decomposes to give molecular oxygen & nascent oxygen.(b) Nascent oxygen then oxidizes lead [II] sulphide to lead [II] sulphate.

Accepted Answer

1. Reaction of excess ammonia with chlorine – Ammonia as a reducing agent2NH3+3Cl2⟶6HCl+N26NH3+6HCl⟶6NH4Cl8NH3+3Cl2⟶6NH4Cl+N2\begin{array}{l} 2	ext{NH}_3 + 3	ext{Cl}_2 \longrightarrow \cancel{6	ext{HCl}} + 	ext{N}_2 \ 6	ext{NH}_3 + \cancel{6	ext{HCl}} \longrightarrow 6	ext{NH}_4	ext{Cl} \ \hline 8	ext{NH}_3 + 3	ext{Cl}_2 \longrightarrow 6	ext{NH}_4	ext{Cl} + 	ext{N}_2 \end{array}2NH3​+3Cl2​⟶6HCl+N2​6NH3​+6HCl⟶6NH4​Cl8NH3​+3Cl2​⟶6NH4​Cl+N2​​​2. Oxidation of Lead [II] Sulphide by OzoneO3⟶O2+[O]×4PbS+4[O]⟶PbSO4PbS+4O3⟶PbSO4+4O2\begin{array}{l} 	ext{O}_3 \longrightarrow 	ext{O}_2 + \cancel{	ext{[O]}} 	imes 4 \ 	ext{PbS} + \cancel{4	ext{[O]}} \longrightarrow 	ext{PbSO}_4 \ \hline 	ext{PbS} + 4	ext{O}_3 \longrightarrow 	ext{PbSO}_4 + 4	ext{O}_2 \end{array}O3​⟶O2​+[O]​×4PbS+4[O]​⟶PbSO4​PbS+4O3​⟶PbSO4​+4O2​​​

Question 12

Question 23Define the terms – (a) Relative atomic mass (b) Relative molecular mass. State why indirect methods are utilized to determine the absolute mass of an atom. Explain in brief the indirect method used.

Accepted Answer

(a) The number of times one atom of an element is heavier than 1⁄12th the mass of an atom of carbon [C12] is known as the Relative Atomic Mass [RAM] of the element.Mass of one atomRAM=of the element(112) Mass of one atom‾of carbon [C12]\begin{matrix} & 	ext{Mass of one atom} \ 	ext{RAM} = & 	ext{of the element} \ & \overline{(\frac{1}{12}) \space 	ext{Mass of one atom}} \ & 	ext{of carbon [C}^{12}] \end{matrix}RAM=​Mass of one atomof the element(121​) Mass of one atom​of carbon [C12]​(b) The number of times one molecule of the substance is heavier than 1⁄12th the mass of an atom of carbon [C12] is known as Relative Molecular Mass [RMM] of the element.Mass of one moleculeRMM=of the substance(112) Mass of one atom‾of carbon [C12]\begin{matrix} & 	ext{Mass of one molecule} \ 	ext{RMM} = & 	ext{of the substance} \ & \overline{(\frac{1}{12}) \space 	ext{Mass of one atom}} \ & 	ext{of carbon [C}^{12}] \end{matrix}RMM=​Mass of one moleculeof the substance(121​) Mass of one atom​of carbon [C12]​As atoms are extremely small and very light, hence cannot be weighed directly. So indirect methods are utilized to determine the absolute mass of an atom.The relative mass of an atom or molecule is hence considered by considering a mass of a light atom and relating the mass of other atoms or molecules to it.

Question 13

Question 24(1)Calculate relative molecular mass of(a) ZnCO3(b) CaSO4[Zn = 65, S = 32, O = 16, Ca = 40, C = 12]

Accepted Answer

(a) Relative Molecular Mass of ZnCO3:Molecular wt. of ZnCO3 = At. wt. of Zn + At. wt. of C + 3(At. wt. of O)= 65 + 12 + 3(16)= 65 + 12 + (48)= 65 + 60= 125Hence, Relative Molecular Mass of ZnCO3 = 125(b) Relative Molecular Mass of CaSO4:Molecular wt. of CaSO4 = At. wt. of Ca + At. wt. of S + 4(At. wt. of O)= 40 + 32 + 4(16)= 40 + 32 + 64= 136Hence, Relative Molecular Mass of CaSO4 = 136

Question 14

Question 24(2)Calculate the percentage composition of(a) calcium chloride(b) calcium nitrate[Ca = 40 , Cl = 35.5 , N = 14 , O = 16]

Accepted Answer

(a) Percentage composition of Calcium Chloride:Chemical Formula of Calcium Chloride = CaCl2Mol. wt. of calcium chloride = At. wt. of Ca + 2(At. wt. of Cl)= 40 + 2(35.5)= 40 + 70= 110110 g of CaCl2 contains 40 g of Calcium,∴ 100 g of CaCl2 will contain 40110\dfrac{40}{110}11040​ x 100 % = 36.36% of Calcium [percentage].Similarly, 110 g of CaCl2 will contain 70 g of Chlorine,∴ 100 g of CaCl2 will contain 70110\dfrac{70}{110}11070​ x 100 % = 63.64% of Chlorine.Hence, Calcium Chloride contains 36.36% of Calcium and 63.64% of Chlorine.(b) Percentage composition of Calcium Nitrate:Chemical Formula of Calcium Nitrate = Ca(NO3)2Mol. wt. of Calcium Nitrate = At. wt. of Ca + 2[At. wt. of N] + 6[At. wt. of O]= 40 + 2(14) + 6(16)= 40 + 28 + 96= 164164 g of Ca(NO3)2 contains 40 g of Calcium,∴ 100 g of Ca(NO3)2 will contain 40164\dfrac{40}{164}16440​ x 100 % = 24.39% of Calcium.Similarly, 164 g of Ca(NO3)2 will contain 28 g of Nitrogen,∴ 100 g of Ca(NO3)2 will contain 28164\dfrac{28}{164}16428​ x 100 % = 17.07% of Nitrogen.Similarly, 164 g of Ca(NO3)2 will contain 96 g of Oxygen,∴ 100 g of Ca(NO3)2 will contain 96164\dfrac{96}{164}16496​ x 100 % = 58.54% of Oxygen.Hence, Calcium Nitrate contains 24.39% of Calcium, 17.07% of Nitrogen and 58.54% of Oxygen.

Question 15

Question 7Write chemical formulae of the sulphates of Aluminium, Ammonium and Zinc.

Accepted Answer

(a) Chemical formula of Aluminium Sulphate:Al3+ SO42−Al33 ↙↘ SO42⇒Al22 ↙↘SO433Al2(SO4)3\text{Al}^{3+} \space \text{SO}_4^{2-} \[1em] \overset{\phantom{3}{3}}{\text{Al}} \space {\swarrow}\mathllap{\searrow} \space \overset{2}{\text{SO}_4} \Rightarrow \underset{\phantom{2}{2}}{\text{Al}} \space {\swarrow}\mathllap{\searrow} \underset{\phantom{3}{3}}{\text{SO}_4} \[1em] \text{Al}_2\text{(SO}_4)_3Al3+ SO42−​Al33 ↙↘ SO4​2​⇒22Al​ ↙↘33SO4​​Al2​(SO4​)3​∴ Chemical formula of Aluminium Sulphate is Al2(SO4)3(b) Chemical formula of Ammonium Sulphate:NH4+ SO42−NH431 ↙↘ SO42⇒NH422 ↙↘SO431(NH4)2SO4\text{NH}_4^{+} \space \text{SO}_4^{2-} \[1em] \overset{\phantom{3}{1}}{\text{NH}_4} \space {\swarrow}\mathllap{\searrow} \space \overset{2}{\text{SO}_4} \Rightarrow \underset{\phantom{2}{2}}{\text{NH}_4} \space {\swarrow}\mathllap{\searrow} \underset{\phantom{3}{1}}{\text{SO}_4} \[1em] \text{(NH}_4)_2\text{SO}_4NH4+​ SO42−​NH4​31​ ↙↘ SO4​2​⇒22NH4​​ ↙↘31SO4​​(NH4​)2​SO4​∴ Chemical formula of Ammonium Sulphate is (NH4)2SO4(c) Chemical formula of Zinc Sulphate:Zn2+ SO42−Dividing by H.C.F. it becomesZn1+ SO41−Zn221 ↙↘ SO41⇒Zn21 ↙↘SO421ZnSO4\text{Zn}^{2+} \space \text{SO}_4^{2-} \[1em] \text{Dividing by H.C.F. it becomes} \[1em] \text{Zn}^{1+} \space \text{SO}_4^{1-}\[1em] \overset{\phantom{22}{1}}{\text{Zn}} \space {\swarrow}\mathllap{\searrow} \space \overset{1}{\text{SO}_4} \Rightarrow \underset{\phantom{2}{1}}{\text{Zn}} \space {\swarrow}\mathllap{\searrow} \underset{\phantom{2}{1}}{\text{SO}_4} \[1em] \text{ZnSO}_4Zn2+ SO42−​Dividing by H.C.F. it becomesZn1+ SO41−​Zn221 ↙↘ SO4​1​⇒21Zn​ ↙↘21SO4​​ZnSO4​∴ Chemical formula of Zinc Sulphate is ZnSO4

Question 16

Question 10The formula of the sulphate of an element M is M2(SO4)3. Write the formula of its(a) Chloride(b) Oxide(c) Phosphate(d) Acetate

Accepted Answer

Given, formula of sulphate is M2(SO4)3.M2(SO4)3=M22 ↗↖ SO43⇒M33 ↗↖ SO42\text{M}_2(\text{SO}_4)_3 = \underset{\phantom{2}{2}}{\text{M}} \space {\nearrow}\mathllap{\nwarrow} \space \underset{3}{\text{SO}_4} \Rightarrow \overset{\phantom{3}{3}}{\text{M}} \space {\nearrow}\mathllap{\nwarrow} \space \overset{2}{\text{SO}_4} \[0.5em]M2​(SO4​)3​=22M​ ↗↖ 3SO4​​⇒M33 ↗↖ SO4​2​∴ Valency of M is 3.(a) Formula of Chloride:M3+ Cl1−M33 ↙↘ Cl1⇒M11 ↙↘Cl33MCl3\text{M}^{3+} \space \text{Cl}^{1-} \[1em] \overset{\phantom{3}{3}}{\text{M}} \space {\swarrow}\mathllap{\searrow} \space \overset{1}{\text{Cl}} \Rightarrow \underset{\phantom{1}{1}}{\text{M}} \space {\swarrow}\mathllap{\searrow} \underset{\phantom{3}{3}}{\text{Cl}} \[1em] \text{M}\text{Cl}_3M3+ Cl1−M33 ↙↘ Cl1⇒11M​ ↙↘33Cl​MCl3​∴ Formula of chloride of element M is MCl3(b) Formula of Oxide:M3+ O2−M33 ↙↘ O2⇒M22 ↙↘O33M2O3\text{M}^{3+} \space \text{O}^{2-} \[1em] \overset{\phantom{3}{3}}{\text{M}} \space {\swarrow}\mathllap{\searrow} \space \overset{2}{\text{O}} \Rightarrow \underset{\phantom{2}{2}}{\text{M}} \space {\swarrow}\mathllap{\searrow} \underset{\phantom{3}{3}}{\text{O}} \[1em] \text{M}_2\text{O}_3M3+ O2−M33 ↙↘ O2⇒22M​ ↙↘33O​M2​O3​∴ Formula of oxide of element M is M2O3(c) Formula of Phosphate:M3+ PO43−Dividing by H.C.F. it becomesM1+ PO41−M31 ↙↘ PO41⇒M21 ↙↘PO431MPO4\text{M}^{3+} \space \text{PO}_4^{3-} \[1em] \text{Dividing by H.C.F. it becomes} \[1em] \text{M}^{1+} \space \text{PO}_4^{1-} \[1em] \overset{\phantom{3}{1}}{\text{M}} \space {\swarrow}\mathllap{\searrow} \space \overset{1}{\text{PO}_4} \Rightarrow \underset{\phantom{2}{1}}{\text{M}} \space {\swarrow}\mathllap{\searrow} \underset{\phantom{3}{1}}{\text{PO}_4} \[1em] \text{MPO}_4M3+ PO43−​Dividing by H.C.F. it becomesM1+ PO41−​M31 ↙↘ PO4​1​⇒21M​ ↙↘31PO4​​MPO4​∴ Formula of phosphate of element M is MPO4(d) Formula of Acetate:M3+ CH3COO1−M33 ↙↘ CH3COOCHCOOO1⇒M11 ↙↘CH3COO33M(CH3COO)3\text{M}^{3+} \space \text{CH}_3\text{COO}^{1-} \[1em] \overset{\phantom{3}{3}}{\text{M}} \space {\swarrow}\mathllap{\searrow} \space \overset{\phantom{CHCOOO}1}{\text{CH}_3\text{COO}} \Rightarrow \underset{\phantom{1}{1}}{\text{M}} \space {\swarrow}\mathllap{\searrow} \underset{\phantom{3}{3}}{\text{CH}_3\text{COO}} \[1em] \text{M}\text{(CH}_3\text{COO})_3M3+ CH3​COO1−M33 ↙↘ CH3​COOCHCOOO1​⇒11M​ ↙↘33CH3​COO​M(CH3​COO)3​∴ Formula of acetate of element M is M(CH3COO)3

Question 17

Question 16How are the elements with variable valency named ? Explain with an example.

Accepted Answer

If an element exhibits two different positive valencies, then the suffix "ous" is used for the lower valency and the suffix "ic" is used for the higher valency. Modern chemists use Roman numerals in place of these trivial names.For example, Iron can exhibit a valency of +2 or +3. The ion with valency +2 is named as Ferrous ion (Fe2+) and with valency +3 is named as Ferric ion (Fe3+). Its oxides will be named as Ferrous oxide or Iron (II) oxide [FeO] and Ferric oxide or Iron (III) oxide [Fe2O3].

Question 18

Question 17(a) What is a chemical formula ?(b) What is the significance of a formula ? Give an example to illustrate.

Accepted Answer

(a) A chemical formula also known as molecular formula employs symbols to denote the molecule of an element or of a compound.(b) The chemical/molecular formula of a compound has quantitative significance. It represents:both the molecule and the molecular mass of the compound.the respective numbers of different atoms present in one molecule of the compound.the ratios of the respective masses of the elements present in the compound.For example, the formula CO2 means that:the molecular formula of carbon dioxide is CO2.each molecule contains one carbon atom joined by chemical bonds with two oxygen atoms.the molecular mass of carbon dioxide is 44, given that the atomic mass of carbon is 12 and that of oxygen is 16.

Question 19

Question 19Write the basic radicals and acidic radicals of the following and then write the chemical formulae of these compounds.(a) Barium sulphate(b) Bismuth nitrate(c) Calcium bromide(d) Ferrous sulphide(e) Chromium sulphate(f) Calcium silicate(g) Stannic oxide(h) Sodium zincate(i) Magnesium phosphate(j) Sodium thiosulphate(k) Stannic phosphate(l) Nickel bisulphate(m) Potassium manganate(n) Potassium ferrocyanide

Accepted Answer

S.No.CompoundBasicradicalAcidicradicalFormulaaBariumsulphateBa2+SO42-BaSO4bBismuthnitrateBi3+NO3-Bi(NO3)3cCalciumbromideCa2+Br-CaBr2dFerroussulphideFe2+S2-FeSeChromiumsulphateCr3+SO42-Cr2(SO4)3fCalciumsilicateCa2+SiO32-CaSiO3gStannicoxideSn4+O2-SnO2hSodiumzincateNa1+ZnO22-Na2ZnO2iMagnesiumphosphateMg2+PO43-Mg3(PO4)2jSodiumthiosulphateNa1+S2O32-Na2S2O3kStannicphosphateSn4+PO43-Sn3(PO4)4lNickelbisulphateNi2+HSO4-Ni(HSO4)2mPotassiummanganateK1+MnO42-K2MnO4nPotassiumferrocyanideK1+Fe(CN)64-K4[Fe(CN)6]

Question 20

Question 20Give the names of the elements and number of atoms of those elements, present in the following compounds.(a) Sodium sulphate(b) Quick lime(c) Baking soda(d) Ammonia(e) Ammonium dichromate

Accepted Answer

(a) Sodium sulphate     Formula: Na2SO4   ∴ It contains 2 atoms of Sodium (Na), 1 atom of Sulphur (S) and 4 atoms of Oxygen (O).(b) Quick lime     Formula: CaO   ∴ It contains 1 atom of Calcium (Ca) and 1 atom of Oxygen (O).(c) Baking soda     Formula: NaHCO3   ∴ It contains 1 atom of Sodium (Na), 1 atom of Hydrogen (H), 1 atom of Carbon (C) and 3 atoms of Oxygen (O).(d) Ammonia     Formula: NH3   ∴ It contains 1 atom of Nitrogen (N) and 3 atoms of Hydrogen (H).(e) Ammonium dichromate     Formula: (NH4)2Cr2O7   ∴ It contains 2 atoms of Nitrogen (N), 8 atoms of Hydrogen (H), 2 atoms of Chromium (Cr) and 7 atoms of Oxygen (O).

The Language of Chemistry

Additional Questions

Exercise 1A

Exercise 1C Descriptive Type

Exercise 1C Multiple Choice Type

Exercise 1C Numericals

Exercise 1C Short Answer Type

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