Solved Sample Paper 4

Lone pair electrons

Reason — Lone pair of electrons are valence electrons that in a covalent bond are not shared with another atom.

have the same number of protons

Reason — An atom forms its respective ion by accepting or donating electrons. Hence, number of electrons between an atom and its ion will differ but both will have the same number of protons.

Solid

Reason — Electrovalent compounds consists of ions but the strong electrostatic force keeps ions in fixed position in the solid state. The force is weakened in the molten state and disappears in aqueous solution state, hence free ions formed help in the conduction of electricity.

electron affinity

Reason — Electron affinity is the amount of energy released when an atom in the gaseous state accepts an electron to form an anion.

addition reaction

Reason — The availability of electrons in the double or triple bonded unsaturated hydrocarbons makes them more reactive and they therefore undergo characteristic addition reactions only.

C₃H₄

Reason — C₃H₄ is an Alkyne as it follows the general formula C_nH_2n-2 and Alkynes contain a triple carbon-carbon covalent bond.

Only P

Reason — The reaction is only a combination reaction as magnesium and oxygen combine to form magnesium oxide.

2Mg + O₂ ⟶ 2MgO

Cathode is impure copper block and anode is pure copper rod.

Reason — In electro-refining of copper, cathode is thin strip of pure copper and anode is impure copper block.

$\dfrac{1}{12}$

Reason — Atomic mass unit is defined as $\dfrac{1}{12}$th the mass of an carbon atom C-12.

Both P and Q

Reason — Hydrogen gas is evolved which is lighter than air hence both are correct.

S

Reason — In beaker S, zinc will displace iron from iron sulphate solution as zinc is above iron in the metal activity series.

8

Reason — Electronic configuration of the element is 2, 8, 8. Hence, 8 electrons are present in the valence shell.

X

Reason — Compound B₂C is wrongly formed. Valency of B is 2 and C is 1. Hence, correct formula of the compound will be BC₂

H₃PO₄

Reason — H₃PO₄ is a weak electrolyte.

NO + H₂O

Reason — Ammonia gets catalytically oxidised in presence of platinum at 800°C to give nitric oxide and water vapours.
4NH₃ + 5O₂ $\xrightarrow[800 \degree \text{C}]{\text{Pt}.}$ 4NO↑ + 6H₂O + Δ

(a) Structural formulae are shown below:

(1) Isobutane

(2) Propanal

(3) But-2-yne

(b) IUPAC names of the organic compounds are:

(1) But-1-ene

(2) 1-bromo 2, 3-dichloropropane

(a) Electrorefining

(b) Hydrogen chloride gas

(c) Ethanol

(d) Addition reactions

(e) Copper sulphate solution

(a) An acid is a compound which when dissolved in water forms hydronium ions as the positively charged ions.

(b) Al metal is used in making alloys such as duralumin and magnalium.

(c) Cold and dilute nitric acid reacts with sulphur to form nitrogen dioxide.

(d) Pure water consists almost entirely of molecules.

(e) One carbon atom in a hydrocarbon contain the root word as meth.

(a) Dil. sulphuric acid is preferred to acidify water because it is a non-volatile acid, while dil. HCl or HNO₃ are volatile acids and may decompose and interfere with the electrolytic reaction.

(b) H₂O ⟶ H⁺ + OH^-

H₂SO₄ ⟶ 2H⁺ + SO₄^2-

H⁺ ions migrate to cathode, SO₄^2- and OH^- both ions migrate to anode.

(c) The ratio of hydrogen and oxygen liberated at cathode and anode is 2 : 1 by volume.
Overall reaction is:

2H₂O $\xrightarrow[\text{Electrical Energy}]{\text{H}_2\text{SO}_4}$ 2H₂ + O₂

(a) SO₄^2- (sulphate)
Reason — White precipitate formed is of barium sulphate [BaSO₄], hence anion present is SO₄^2-.

(b) CO₃^2- (carbonate)
Reason — Sodium carbonate will react with sulphuric acid liberate carbon dioxide along with sodium sulphate and water.
Na₂CO₃ + H₂SO₄ ⟶ CO₂ + Na₂SO₄ + H₂O

Ca(NO₃)₂ + 2NaOH ⟶ Ca(OH)₂ + 2NaNO₃

Hence, the reactants are calcium nitrate and sodium hydroxide

(a) (1) True
Reason — Fluorine is placed in period 2 group 17 (VIIA) in the modern periodic table i.e., it occupies the upper right hand corner of the periodic table. Along a period from left to right electronegativity increases and down a group from top to bottom it decreases. Hence, Fluorine being the topmost and rightmost element in its group and period is the most electronegative element of the periodic table.

(2) True
Reason — Atomic size increases on moving down the group. As potassium lies below sodium in its group, hence, size of potassium atom is greater than the size of sodium atom.

(b) Molar mass of water [H₂O] = 2(1) + 16 = 18

Number of moles = $\dfrac{\text{Given mass}}{\text{Molar mass}}$ = $\dfrac{9}{18}$ = 0.5 moles

Hence, 0.5 moles are present in 9 gm of water.

(a) Formation of NH₄⁺ involves ammonia molecule and hydrogen ion.

(b) Hydroxyl ion or hydroxide ion is formed when one H⁺ ion is removed from water molecule.

(c) Hydrogen needs one electron to complete its duplet.

Simplest ratio of whole numbers = C : H : O = 3 : 5 : 2

Hence, empirical formula is C₃H₅O₂

Empirical formula weight = 3(12)+ 5(1) + 2(16) = 36 + 5 + 32 = 73

Vapour density (V.D.) = 73

Molecular weight = 2 x V.D. = 2 x 73 = 146

$\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \dfrac{146}{73} = 2$

∴ Molecular formula = n[E.F.] = 2[C₃H₅O₂] = C₆H₁₀O₄

(a) C₂H₅CHO — aldehyde group [-CHO]

(b) H₃C-CH₂-CH=CH-CH₃ — carbon-carbon double bond [-C=C-]

(a) Cryolite [Na₃AlF₆]

(b) 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.

(c) The element belongs to group 16 [VI A].
Reason — The element Z has atomic number 16 and so the electronic configuration will be 2, 8, 6. The number of valence electrons determine the group of the element. Hence, Z will belong to group 16 [VI A] as it has 6 electrons in the valence shell.

(a) C — reaction of a dilute acid with an insoluble base
Cu(OH)₂ + H₂SO₄ (dil.) ⟶ CuSO₄ + 2H₂O

(b) E — reaction of two solutions of salts to form a precipitate
CaCl₂ + Na₂CO₃ ⟶ CaCO₃ ↓ + 2NaCl

(c) B — reaction of a dilute acid with a metal
Zn + H₂SO₄ (dil.) ⟶ ZnSO₄ + H₂

(a) Hydronium ion

(b) Methane

(a) On strong heating, ammonium chloride sublimes to form dense white fumes which condense to white powdery substance on cooler parts of the test tube whereas no white fumes are seen on heating sodium chloride. Hence, the two can be distinguished.
NH₄Cl $\xrightarrow{\Delta}$ NH₃ + HCl

(b) On reaction with ammonium hydroxide, ferrous salt forms a dirty green ppt whereas ferric salt forms a reddish brown ppt. Hence, the two can be distinguished easily.

(a) NaCl — Na⁺, Cl^-

(b) CH₃COOH — CH₃COO^-, H⁺

(c) H₂O — H⁺, OH^-

(a) All elements in group VII A have same number of valence electrons (i.e., 7), hence they have similar chemical properties.

(b) Helium (He) has highest ionisation potential, as it has fully-filled electronic configuration due to which it requires maximum energy to remove an electron from its valence shell.

(c) The element is chlorine (Cl) with atomic number 17.

(a) The high solubility of HCl gas in water is demonstrated by the Fountain experiment.

(b) Hydrogen chloride gas is not collected over water since it is highly soluble in water.

(c) Nitrogen is obtained by fractional distillation of liquid air.

(a) Aluminium is placed above zinc in reactivity series. Hence, aluminium is more reactive than zinc and its oxide is stable and cannot be reduced by a reducing agent. Zinc being less reactive metal, its oxide is less stable and so it can be easily reduced by carbon monoxide.

(b) As graphite is unaffected by the reactive bromine vapours released at the anode hence, a graphite anode is preferred during the electrolysis of molten lead bromide

(c) Electrolysis of molten lead bromide involves oxidation and reduction reactions and hence is a redox reaction.

Reduction reaction at cathode:

Pb²⁺ + 2e^- ⟶ Pb

Oxidation reaction at the anode:

Br^1- - 1e^2- ⟶ Br

Br + Br ⟶ Br₂

$\begin{matrix} \text{MgCO}_3 & \longrightarrow & \text{MgO} & + & \text{CO}_2 \\ 24 + 12 + 3(16) \text{ g} & & = 24 + 16 & & [ 12 + 2(16)] \\ = 84 & & = 40 & & 12 + 32 \\ & & & & 44 \text{ g} \\ \end{matrix}$

(a) Number of moles of MgCO₃ = $\dfrac{\text{Given weight}}{\text{molecular weight}}$ = $\dfrac{4.2}{84}$ = 0.05 moles

1 mole of MgCO₃ gives 1 mole of CO₂

Hence, 0.005 mole of MgCO₃ gives 0.05 mole of CO₂

(a) Volume of carbon dioxide at STP = no. of moles of carbon dioxide x volume in litres
= 0.05 x 22.4 = 1.12 L

Hence, volume of carbon dioxide obtained at STP = 1.12 L

(b) 84 grams of MgCO₃ produce 40 grams of MgO

∴ 4.2 g will produce = $\dfrac{40}{84}$ x 4.2 = 2 g

Hence, mass of MgO formed = 2g

(a) As sulphuric acid reacts chemically with ammonia to form ammonium sulphate hence, it is not used as a drying agent for drying ammonia.

2NH₃ + H₂SO₄ ⟶ (NH₄)₂SO₄

(b) AlCl₃ + 3NH₄OH ⟶ 3NH₄Cl + Al(OH)₃

(a) Hydrogen sulphide [H₂S] gas

(b) Sulphur dioxide gas [SO₂]

(a) Ethyne from (ethylene dibromide) 1, 2-di bromoethane.

$\underset{\text{ ethylene dibromide}}{\text{C}_2\text{H}_4\text{Br}_2} + \underset{\text{ alcoholic KOH}}{\text{2KOH [aq.]}} \xrightarrow{\text{boil}} \underset{\text{ Ethene [ethylene]} }{\text{C}_2\text{H}_2} + \text{2KBr} +\text{2H}_2\text{O}$

(b) Ethanoic acid from ethanol.

$\underset{ \text{ethanol}}{\text{C}_2\text{H}_5\text{OH}} \xrightarrow[\underset{\text{[acidified]}}{\text{K}_2\text{Cr}_2\text{O}_7}]{\text{[O]}} \underset{\text{ethanal}}{\text{CH}_3\text{CHO}} \xrightarrow[\underset{\text{[acidified]}}{\text{K}_2\text{Cr}_2\text{O}_7}]{\text{[O]}} \underset{ \text{ethanoic acid} }{\text{CH}_3\text{COOH}}$

(c) Sodium acetate from acetic acid.

CH₃COOH + NaOH ⟶ CH₃COONa + H₂O

(a) On heating, zinc carbonate, which is a light amorphous white solid, changes to pale yellow and gives off a colourless and odourless gas that turns lime water milky. The milkiness disappears on passing excess of gas. This proves that the gas evolved is carbon dioxide.

ZnCO₃ $\xrightarrow{\Delta}$ ZnO + CO₂ ↑

(b) Chalky white precipitate of lead hydroxide is formed which is insoluble in excess of ammonium.

Pb(NO₃)₂ + 2NH₄OH ⟶ 2NH₄NO₃ + Pb(OH)₂ ↓

(c) When sodium hydrogen carbonate is added to nitric acid, a colourless and odourless gas is evolved that turns lime water milky. The milkiness disappears on passing excess of gas. Hence, the gas evolved is carbon dioxide.

NaHCO₃ + HNO₃ ⟶ NaNO₃ + H₂O + CO₂ ↑

[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

(a) Electronaffinity — The amount of energy released while converting a neutral gaseous isolated atom into a negatively charged gaseous ion (anion) by the addition of electron is called Electron Affinity (E.A.)

(b) Electronegativity — The tendency of an atom in a molecule to attract the shared pair of electrons towards itself is called its electronegativity.

(a) Y is the metallic element.

(b) Ionisation energy of X is greater than Y.

(c) Y is the reducing agent.

Explanation

(a) Y is the metallic element as it has one valence electron.

(b) Atomic size of X is less than Y as X has 2 shells while Y has 3 shells. As, ionisation energy decreases with increase in atomic size, hence ionisation energy of X is greater than Y.

(c) Y has the tendency to lose its one valence electron and get oxidised. Hence, it acts as a reducing agent.

(1) C : An electrolyte

(2) A : Oxidation

(3) B : Cathode