Solved 2016 Question Paper ICSE Class 10 Chemistry

(i) Metals are good reducing agents because they are electron donors.

(ii) Electrovalent compounds have high melting points.

(iii) Higher the pH value of a solution, the more alkaline it is.

(iv) AgCl, a white precipitate is soluble in excess NH₄OH.

(v) Conversion of ethene to ethane is an example of hydrogenation.

17

Reason — The element with atomic number 19 has electronic configuration of (2, 8, 8, 1) and that with atomic number 17 has electronic configuration of (2, 8, 7). This element with atomic number 17 needs 1 electron to complete its octet whereas element with atomic number 19 has an extra electron. Hence, an element with atomic number 19 is most likely to combine with the element with atomic number 17.

1:1

Reason — 1 mole of any substance contains same (Avogardo No.) of molecules. Hence, ratio is same.

Copper and tin

Reason — The two main metals in Bronze are Copper and tin .

mainly ions

Reason — The particles present in strong electrolytes are mainly ions.

HCl is highly soluble in water

Reason — The aim of the Fountain Experiment is to prove that HCl is highly soluble in water.

(i) AlN + 3H₂O ⟶ Al(OH)₃ + NH₃ [g]

(ii) Cu + 4HNO₃ ⟶ Cu(NO₃)₂ + 2H₂O + 2NO₂

(iii) NaHCO₃ + HCl ⟶ NaCl + H₂O + CO₂ ↑

(iv) Na₂SO₃ + H₂SO₄ (dil) ⟶ Na₂SO₄ + H₂O + SO₂

(v) $\underset{\text{ Chloroethane [ethyl chloride]}}{\text{C}_2\text{H}_5\text{-Cl}} + \text{NaOH [aq.]} \xrightarrow{\text{boil}} \underset{\text{ Ethanol [ethyl alcohol]}}{\text{C}_2\text{H}_5\text{-OH}} + \text{NaCl}$

(i) White precipitate of PbCl₂ is formed which is soluble in hot water.
Pb(NO₃)₂ + 2HCl ⟶ PbCl₂ ↓ + 2HNO₃

(ii) White coloured precipitate of barium sulphate is formed when sodium sulphate is mixed with barium chloride.
Na₂SO₄ (dil.) + BaCl₂ (aq.) ⟶ 2NaCl + BaSO₄ ↓ [white ppt. formed]

(iii) Black spongy charred mass of carbon is formed. Steam is seen to be evolved.

$\text{C}_{12}\text{H}_{22}\text{O}_{11}\xrightarrow{\text{Conc. H}_2\text{SO}_4} \text{12C} \small{\space[sugar\space charcoal]} + \text{11H}_2\text{O}$

(iv) Effervescence of CO₂ seen which turns lime water milky.
CuCO₃ + 2HCl ⟶ CuCl₂ + H₂O + CO₂

(v) Yellow particles of S precipitate.
Na₂S₂O₃ + 2HCl ⟶ 2NaCl + SO₂ + S + H₂O

(i) Electronegativity

(ii) Froth flotation process

(iii) Nickel or platinum or palladium.

(iv) Addition reaction

(v) Valence Electrons

(i) 20 lit. of gas weighs = 32 g.

1 mole = 22.4 lit. of gas will weigh = $\dfrac{32}{20}$ x 22.4 = 35.84 g.

Hence, gram mol. weight of the gas = 35.84 g.

(ii)

$\begin{matrix} 2\text{Ca}(\text{NO}_3)_2 & \longrightarrow & 2\text{CaO} & + & 4\text{NO}_2 & + & \text{O}_2 \\ 2[(40) + & & 2(40 & & 4[22.4] \\ 2(14 +3(16))] & & + 16) & & \text{ lit.} \\ = 328 \text{ g} & & 112 \text{g} \\ \end{matrix}$

(a) 328 g of Ca(NO₃)₂ produces 112 g of calcium oxide.

∴ 82 g of Ca(NO₃)₂ will produce = $\dfrac{112}{328}$ x 82 = 27.99 g = 28 g

Hence, 28 g of calcium oxide is produced.

(b) 328 g of Ca(NO₃)₂ produces 4(22.4) lit of nitrogen dioxide.

∴ 82 g of Ca(NO₃)₂ will produce = $\dfrac{4 \times 22.4}{328}$ x 82 = 22.4 lit.

Hence, vol of nitrogen dioxide evolved = 22.4 lit.

(i) 1. Propene

2. 2-butyne

3. ethanal

(ii) 1. The ionization potential of potassium is < (less than) that of Sodium.

2. The electronegativity of iodine is < (less than) that of chlorine.

(i) Atom J has 5 valence electrons as it belongs to the fifth group in the periodic table.

(ii) M forms ions with a single negative charge as it has 7 electrons in the valence shell and obtaining one more electron completes it's octet.

(iii) Metal T is more reactive than R because elements at the bottom of the group are more reactive.

(iv) T has it's electrons arranged in 4 shells as it belongs to 4^th period.

(i) If an element has a low ionization energy then it is likely to be metallic.

(ii) If an element has seven electrons in it's outermost shell then it is likely to have the smallest atomic size among all the elements in the same period.

(i) Type of bond formed

(a) W (2,8,1) and X (2,8,7) — Electrovalent bond is formed as W donates 1 electron and X takes up that electron and both attain a stable octet.

(b) Y (2, 5) and Z (1) — Covalent bond as electron sharing takes place between the two elements. The valency of Z element is 1 and that of Y is 3. Z needs one electrons to attain stable duplet structure of nearest noble gas - He [2] and Y needs three electrons to attain stable octet structure of nearest noble gas - Ne [2,8]
One Y atom shares three electron pairs one with each of the three atoms of Z such that Z acquires a duplet configuration and Y attain a stable octet configuration resulting in the formation of three single covalent bonds.

(ii) Formula of the compound formed

(a) X (2,8,7) and Z (1) — XZ as X needs one 1 electron to complete it's octet and Z needs needs one electron to complete it's duplet hence, they both share one electron pair and forms ZX compound.

(b) W (2,8,1) and X (2,8,7) — WX as W has the tendency to donate one electron and gain stability whereas X has the tendency to gain one electron to be stable. Hence, W donates one electron and X takes it and form WX compound.

(i) Carbon dioxide and water are produced:

$2\text{C}_2\text{H}_6 + 7\text{O}_2[\text{excess}] \longrightarrow 4\text{CO}_2 + 6\text{H}_2\text{O} + \Delta$

(ii) Calcium hydroxide and ethyne are produced

$\underset{\text{calcium carbide}}{\text{CaC}_2} + \underset{\text{water}}{2\text{H}_2\text{O}} \longrightarrow \underset{\text{ethyne [acetylene]}}{\text{C}_2\text{H}_2} + \text{Ca(OH)}_2$

(iii) Ethene and water are produced

$\underset{\text{ ethyl alcohol}}{\text{C}_2\text{H}_5\text{OH}} \xrightarrow[170\degree\text{C}]{\text{Conc. H}_2\text{SO}_4\text{[excess]}} \underset{ \text{ethene}}{\text{C}_2\text{H}_4} + \text{H}_2\text{O}\$

(i) 2-methyl propane :

(ii) Ethanoic acid :

(iii) Butan – 2 – ol :

(i) Compound A is Ethene. Its structure is shown below:

(ii) Brown colour of bromine is discharged.

(i) Cold, dilute nitric acid reacts with copper to given nitric oxide.

(ii) Hot, concentrated nitric acid reacts with sulphur to form nitrogen dioxide.

(i) SO₂ is evolved when dilute hydrochloric acid reacts with sodium sulphite.

Na₂SO₃ + 2HCl ⟶ 2NaCl + H₂O + SO₂ [g]

(ii) H₂S gas is evolved when dilute hydrochloric acid reacts with Iron [II] sulphide.

FeS + 2HCl ⟶ FeCl₂ + H₂S

(i) Pale blue precipitate of Cu(OH)₂ is formed, which becomes deep blue when excess of ammonium hydroxide is added.

CuSO₄ + 2NH₄OH ⟶ (NH₄)₂SO₄ + Cu(OH)₂ ↓

Cu(OH)₂ + (NH₄)₂SO₄ + 2NH₄OH [in excess] ⟶ 4H₂O + [Cu(NH₃)₄]SO₄

(ii) Gelatinous white precipitate of zinc hydroxide is formed which becomes colourless solution on adding excess of ammonium hydroxide.

ZnSO₄ + 2NH₄OH ⟶ (NH₄)₂SO₄ + Zn(OH)₂ ↓

Zn(OH)₂ +(NH₄)₂SO₄ + 2NH₄OH [in excess] ⟶ 4H₂O + [Zn(NH₃)₄]SO₄

(i) Electrode — Cathode : Copper, Anode : Copper

Dissociation of aq. CuSO₄ :

CuSO₄ ⇌ Cu²⁺ + SO₄^2-

H₂O ⇌ H¹⁺ + OH^1-

H₂SO₄ ⇌ 2H¹⁺ + SO₄^2-

Reaction at cathode [pure thin sheet of Cu]:
Cu²⁺ + 2e^- ⟶ Cu [Cu deposited on thin sheet]

Reaction at anode [impure block of active Cu]:
Cu - 2e^- ⟶ Cu²⁺ [product nil - Cu²⁺ ions]

(ii) Electrode — Cathode : Iron or Graphite, Anode : Graphite

Dissociation of molten PbBr₂ : PbBr₂ ⇌ Pb²⁺ (cathode) + 2Br^1- (anode)

Reaction at cathode:
Pb²⁺ + 2e^- ⟶ Pb [product lead metal]

Reaction at anode:
Br^1- - 1e^-⟶ Br
Br + Br ⟶ Br₂

(i) Oxygen

(ii) Silver ions

Gram molecular mass of oxygen = 32 g = 1 mole and contains 6 x 10²³ molecules i.e., 6 x 10²³ molecules weigh 32 g

∴ 12 x 10²⁴ molecules will weigh = $\dfrac{32}{ 6\times 10^{23}}$ x 12 x 10²⁴ = 640 g

(ii) 6 x 10²³ molecules occupy 22.4 lit. of vol.

∴ 12 x 10²⁴ molecules will occupy = $\dfrac{22.4}{ 6\times 10^{23}}$ x 12 x 10²⁴ = 448 lit.

Hence, mass of O₂ present in the cylinder = 640 g and volume of O₂ at S.T.P. present in the cylinder = 448 lit.

Simplest ratio of whole numbers = C : H = 1 : 2.5 = 2 : 5

Hence, empirical formula is C₂H₅

Empirical formula weight = 2(12) + 5(1) = 29

V.D. = 29

Molecular weight = 2 x V.D. = 2 x 29

$\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \dfrac{2 \times 29}{29} =2$

∴ Molecular formula = n[E.F.] = 2[C₂H₅] = C₄H₁₀

[By Lussac's law]

$\begin{matrix} 4\text{NH}_3 & + & 5\text{O}_2 &\longrightarrow & 4\text{NO} & + & 6\text{H}_2\text{O} \\ 4 \text{ vol.} & : & 5 \text{ vol.} & \longrightarrow & 4\text{ vol.} \end{matrix}$

To calculate the volume of oxygen required :

$\begin{matrix}\text{NH}_3 & : & \text{O}_2 \\ 4 \text{ vol.} & : & 5 \text{ vol.} \\ 100 \text{ cm.}^3 & : & \text{x} \end{matrix}$

$\therefore x = \dfrac{5}{4} \times 100 = 125 \text{ cm.}^3$

Hence, volume of oxygen required = 125 cm³

Electron dot diagram showing the formation of ammonium ion is given below:

(i) Ammonia gas

2NH₄Cl + Ca(OH)₂ ⟶ CaCl₂ + 2H₂O + 2NH₃

(ii) Nitrogen gas

NH₄Cl + NaNO₂ ⟶ NaCl + NH₄NO₂

NH₄NO₂ ⟶ 2H₂O + N₂

(i) 8NH₃ [excess] + 3Cl₂ ⟶ 6NH₄Cl + N₂

(ii) 2NH₃ + 3CuO ⟶ 3Cu + 3H₂O + N₂ [g]

(i) B: Non-volatile acid

$\text{NaCl} + \text{H}_2\text{SO}_4 [\text{conc.}]\xrightarrow{\lt 200 \degree\text{C}} \text{NaHSO}_\text{4} + \text{HCl}$

(ii) A: Typical acid property

CuO + H₂SO₄ (conc.) ⟶ CuSO₄ + H₂O

(iii) C: Oxidizing agent

S + 2H₂SO₄ (conc.) ⟶ 3SO₂ + 2H₂O

(i) The ions Na⁺ and Cl^– are not free but held together by strong electrostatic force of attraction. In fused or molten state the ions break free and move. Hence, NaCl will conduct electricity only in fused state or aq. soln. state.

(ii) Migration of Ag¹⁺ ions from sodium argentocyanide solution is slow compared to that from silver nitrate. Hence, an even deposition of metal silver is obtained on the article. Therefore, the electrolyte sodium argentocyanide soln. is preferred over silver nitrate solution.

(iii) Copper does not undergo chemical decomposition due to flow of electric current through it. Hence, copper is a good conductor of electricity but it is a non-electrolyte.

(i) Sodium hydroxide

(ii) Conversion of Al(OH)₃ - to pure alumina - heat on Al(OH)₃

$\underset{\text{[aluminium hydroxide]}}{\text{2Al(OH)}_3} \xrightarrow{1100\degree\text{C}} \underset{\text{[pure alumina]}}{\text{Al}_2\text{O}_3} + \text{3H}_2\text{O}$

(iii) Cryolite (Na₃AlF₆)

(iv) 2Al³⁺ + 6e^- ⟶ 2Al

(v) The oxygen evolved at the anode escapes as a gas or reacts with the carbon anode. The carbon anode is thus oxidized to carbon monoxide which either burns giving carbon dioxide or escapes out through an outlet. [2C + O₂ ⟶ 2CO; 2CO + O₂ ⟶ 2CO₂] The carbon anode is hence consumed and renewed periodically after a certain period of usage. It is therefore preferable to use a number of graphite electrodes as anode instead of a single electrode.

(i) When crystals of washing soda are exposed to air, it loses its water of crystallisation due to efflorescence.

(ii) When ferric chloride is exposed to the atmosphere, it absorbs moisture from the atmospheric air to become moist and ultimately dissolves in the absorbed water, forming a saturated solution.

(i) Ferric (Fe³⁺) ion

(ii) Lead (Pb²⁺) ion

(iii) Calcium (Ca²⁺) ion