ICSE Class 10 Chemistry — 2017 Paper

(i) The energy required to remove an electron from a neutral isolated gaseous atom and convert it into a positively charged gaseous ion is called Ionization Potential

(ii) The compound that does not have a lone pair of electrons is carbon tetrachloride.

(iii) When a metallic oxide is dissolved in water, the solution formed has a high concentration of OH^- ions.

(iv) Potassium sulphite on reacting with hydrochloric acid releases SO₂ gas.

(v) The compound formed when ethene reacts with Hydrogen is C₂H₆.

Copper chloride.

Reason — Copper chloride forms a precipitate that is soluble in excess of ammonium hydroxide.

Alkyne

Reason — As the general formula for alkyne is C_nH_2n-2. Hence, when C is 10 then H is 18.

Consists of molecules

Reason — In covalent compound, valence shell electrons are mutually shared by atom of each element to achieve a stable electronic configuration. As there is no transfer of electrons between the atoms so the atoms remain electrically neutral. Hence they don't form ions and covalent compounds consists of molecules.

An alkali

Reason — The pH of alkali solutions is more than 7 therefore in order to increase the pH value of a neutral solution, an alkali should be added.

direct combination

Reason — Anhydrous iron [III] chloride is prepared by direct combination of iron and chloride as follows:
2Fe + 3Cl₂ ⟶ 2FeCl₃

(i) Calcium [Ca²⁺] ions

(ii) sodium argentocyanide solution

(iii) Molecules only

(iv) Carboxylic acids

(v) Ammonium chloride.

(i) 3Cu + 8HNO₃ [dil.] ⟶ 3Cu(NO₃)₂ + 4H₂O + 2NO [g]

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

(iii)
$\underset{\text{methyl iodide}}{\text{CH}_3\text{I}} + \underset{\text{nascent hydrogen}}{2\text{[H]}} \xrightarrow[\text{alcohol}]{\text{Zn/Cu couple}} \underset{\text{methane}}{\text{CH}_4} +\text{HI}$

(iv) S + 2H₂SO₄ (conc.) ⟶ 3SO₂ + 2H₂O

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

(i) On warming, the mixture gives fruity smell. Acetic acid on heating with an alcohol and a dehydrating agent [conc. H₂SO₄] forms an ester — ethyl acetate.

$\underset{\text{ethyl alcohol}}{\text{C}_2\text{H}_5\text{OH}} + \underset{\text{acetic acid}}{\text{CH}_3\text{COOH}} \xrightarrow{\text{Conc. H}_2\text{SO}_4} \underset{\text{ethyl ethanoate}}{\text{CH}_3-\text{COO}-\text{C}_2\text{H}_5} + \text{H}_2\text{O}$

(ii) H₂S gas is evolved. It has a foul smell like rotten eggs.
FeS + 2HCl [g] ⟶ FeCl₂ + H₂S

(iii) A dirty green precipitate is formed which is insoluble in excess of sodium hydroxide solution.
FeSO₄ + 2NaOH ⟶ Fe(OH)₂↓ (dirty green ppt.) + Na₂SO₄

(iv) Ammonia burns in the atmosphere of excess oxygen with a green or greenish yellow flame, forming nitrogen and water vapour.
4NH₃ + 3O₂ ⟶ 2N₂ + 6H₂O

(v) The blue coloured hydrous copper sulphate changes to white anhydrous copper sulphate as the water of crystallization is removed.

$\text{CuSO}_4\text{.5H}_2\text{O} \xrightarrow{\text{Conc. H}_2\text{SO}_4} \text{CuSO}_4 \text{ [White Anhydrous]} + 5\text{H}_2\text{O}$

1. 2, 3 – dimethyl butane :

2. Diethyl ether :

3. Propanoic acid :

(ii) The matching terms are:

1. Crushing of the ore into a fine powder — Pulverisation
2. Heating of the ore in the absence of air to a high temperature — Calcination

(i) 23 g of Na = 1 g. atom

∴ 4.6 g. of Na = $\frac{1}{23}$ x 4.6 = 0.2 g

Hence, 0.2 g. atoms

(ii) Molecular weight of hydrated copper sulphate CuSO₄.5H₂O = 64 + 32 + 4(16) + 5[2(1) + 16]

= 64 + 32 + 64 + 5[18]

= 64 + 32 + 64 + 90

= 250 g

250 g of CuSO₄.5H₂O contains 90 g of water of crystallization.

∴ Percentage of water of crystallization = $\frac{90}{250}$ x 100 = 36%

Hence, percentage of water of crystallization is 36%

(iii) Empirical formula = XY₂

Empirical formula weight = V.D.

Molecular weight = 2 x V.D.

$$\begin{aligned}\text{n} = \frac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] \text{n} = \frac{\text{2 x V.D.}}{\text{V.D.}} \\[0.5em] \text{n} = 2\end{aligned}$$

∴ Molecular formula = n[E.F.] = 2[XY₂] = X₂Y₄

(i) 15

(ii) 19

(iii) 8

(iv) 4

(v) 2

(i) He < Ne < Ar
Reason — Number of electron shells increases as we move down the group.

(ii) K < Na < Li
Reason — Ionization Potential decreases as we move down the group.

(iii) Br < Cl < F
Reason — Electronegativity decreases as we move down the group.

(iv) Li < Na < K
Reason — Atomic Size increases as we move down the group.

(i) Covalent bonding
Reason — The valency of hydrogen element is 1 and that of oxygen is 2. Hydrogen needs one electrons to attain stable duplet structure of nearest noble gas - He [2] and oxygen needs two electrons to attain stable octet structure of nearest noble gas - Ne [2,8]
Each of the two hydrogen atoms shares an electron pair with the oxygen atom such that hydrogen acquires a duplet configuration and oxygen an octet configuration resulting in the formation of two single covalent bonds [H-O-H] in the molecule of water.

(ii) Electrovalent bonding
Reason — The valency of calcium (2,8,8,2) element and that of oxygen (2,6) is 2. Calcium donates two electrons and oxygen takes up those two electrons and electrovalent bond is formed.

(i) With Sodium hydroxide, copper sulphate (CuSO₄) solution forms a pale blue precipitate which is insoluble in excess of sodium hydroxide and with ammonium hydroxide it forms a pale blue precipitate which dissolves in excess of ammonium hydroxide and forms a deep/inky blue solution.

(ii) 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.

(i) P is potassium chloride.

Reason — K⁺ ions produces lilac coloured flame and Cl^- ions react with silver nitrate to form silver chloride ppt. (soluble in excess of ammonium hydroxide).

(ii) Q is zinc carbonate.

Reason — CO₃^2- ions produces carbon dioxide with HCl. Zinc chloride forms white ppt. with ammonium hydroxide ( soluble in excess of ammonium hydroxide).

(i) Electron dot diagram showing the formation of Methane is given below:

(ii) Electron dot diagram showing the formation of Magnesium chloride is given below:

(i) The observations at the anode & at the cathode are:
At cathode — silvery grey deposit of lead metal.
At anode — reddish brown fumes of bromine vapours.

(ii) The observations at the anode & at the cathode are:
At cathode — Brownish pink copper metal is deposited at cathode during electrolysis of copper sulphate.
At anode — Copper ions are formed. Copper anode diminishes in mass.

Blue colour of CuSO₄ remains unchanged.

(i) OH^- ion will get discharged in preference to SO₄^2- or NO₃^- ions.

(ii) Ag⁺ ions will get discharged in preference to Pb²⁺ or Cu²⁺ ions.

A : Sodium hydrogen sulphate (NaHSO₄) and HCl gas.

B : Upward displacement of air.

C : Magnesium nitride (Mg₃N₂) and water (H₂O).

D : Quick lime [CaO]

E : Downward displacement of air.

(i) C + 2H₂SO₄ (conc.) ⟶ CO₂ + 2SO₂ + 2H₂O

(ii) Mg + H₂SO₄ (dil.) ⟶ MgSO₄ + H₂ (g)

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

SO₃ + H₂SO₄ (conc.) ⟶ H₂S₂O₇ (oleum)
H₂S₂O₇ + H₂O ⟶ 2H₂SO₄ (conc.)

(i) Given, 20% of air contains oxygen

∴ 20% of 1000 cm³ = $\frac{20}{100}$ x 1000 = 200 cm³

[By Lussac's law]

$$\begin{matrix} \text{C}_3\text{H}_8 & + & 5\text{O}_2 &\longrightarrow & 3\text{CO}_2 & + & 4\text{H}_2\text{O} \\ 1 \text{ vol.} & : & 5 \text{ vol.} & \longrightarrow & 3\text{ vol.} & : & 4\text{ vol.} \end{matrix}$$

To calculate the volume of propane consumed :

$$\begin{matrix}\text{O}_2 & : & \text{C}_3\text{H}_8 \\ 5 \text{ vol.} & : & 1 \text{ vol.} \\ 200 \text{ cm.}^3 & : & \text{x} \end{matrix}$$

$\therefore x = \frac{1}{5} \times 200 = 40 \text{ cm.}^3$

Hence, volume of propane is consumed is 40 cm³

(ii) 11.2 lit. weighs 24 g

∴ 22.4 lit. will weigh = $\frac{24}{11.2}$ x 22.4 = 48 g

Hence, gram molecular mass of the gas = 48 g.

(i) 2 g of hydrogen gas = 1 mole

∴ 1000 g of hydrogen gas = $\frac{1000}{2}$ = 500 moles.

(ii) Molar mass of CO₂ = C + 2(O) = 12 + 2(16) = 44 g

1 mole of carbon dioxide = 44 g

∴ 500 moles of carbon dioxide = 44 x 500 = 22,000 g = 22 kg.

Hence, Weight of carbon dioxide in cylinder = 22 kg

(iii) According to Avogadro's law — Equal volumes of all gases under similar conditions of temperature and pressure contain equal number of molecules.

∴ molecules in the cylinder of carbon dioxide = X.

(iv) Avogadro's law — Equal volume of all gases under similar conditions of temperature and pressure contain equal number of molecules.

(i) CuSO₄ + Na₂CO₃ ⟶ Na₂SO₄ + CuCO₃

(ii) 2NH₄OH + H₂SO₄ ⟶ (NH₄)₂SO₄ + 2H₂O

(i) S + 6HNO₃ [conc.] ⟶ H₂SO₄ + 2H₂O + 6NO₂

(ii) $4\text{NH}_3 + 5\text{O}_2 \xrightarrow[800 ^{\circ} \text{C}]{\text{Pt.}} \text{4NO} + \text{6H}_\text{2}\text{O} + \Delta$

(iii) KNO₃ + H₂SO₄ [conc.] $\xrightarrow{\lt 200 ^{\circ}\text{C}}$ KHSO₄ + HNO₃

(iv) NH₃ + HNO₃ ⟶ NH₄NO₃

(i) Glacial acetic acid

(ii) Ethyne or acetylene

(iii) Isomerism

(iv) Ketone or Alkanone

(i) Preparation of ethane from Sodium propionate:

$\underset{\text{sodium propionate}}{\text{C}_2\text{H}_5\text{COONa}} + \underset{\text{sodalime}}{\text{NaOH}} \xrightarrow[\Delta]{\text{CaO}} \underset{\text{ethane}}{\text{C}_2\text{H}_6} + \text{Na}_2\text{CO}_3$

(ii) Action of alcoholic KOH on bromoethane:

$\underset{\text{ Bromoethane [ethyl bromide]}}{\text{C}_2\text{H}_5\text{-Br}} + \underset{\text{ alcoholic KOH}}{\text{KOH [aq.]}} \xrightarrow{\text{boil}} \underset{ \text{ ethyl alcohol}} {\text{C}_2\text{H}_5\text{OH}} + \text{KBr}$

(i) Galvanization

(ii) Solder [fuse metal]

(iii) Zinc blende [ZnS]

(iv) Copper oxide [CuO]

(i) Cryolite and fluorspar are components of the electrolyte. Addition of cryolite enhances the conductivity of the mixture since, pure alumina is almost a non-conductor of electricity. Addition of both cryolite & fluorspar lowers the fusion point of the mixture i.e., mixture fuses around 950°C instead of 2050°C.

(ii) The layer of powdered coke is sprinkled over the electrolytic mixture as :

1. it prevents burning of carbon electrodes in air at the emergence point from the bath.
2. it minimizes or prevents heat loss by radiation.

(i) The metal which does not react with water or dilute H₂SO₄ but reacts with concentrated H₂SO₄ is Cu.

(ii) The metal whose oxide, which is amphoteric, is reduced to metal by carbon reduction — Pb.

(iii) The divalent metal whose oxide is reduced to metal by electrolysis of it's fused salt is — Mg.