Long Answer Questions

(a) 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.

(b) The graphite anodes are continuously replaced during the electrolysis because:

1. The oxygen evolved at the anode escapes as a gas or reacts with the carbon anode.
2. 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

(c) Cryolite and fluorspar are added to the electrolytic mixture because their addition:

1. lowers the fusion point of the mixture i.e., mixture fuses around 950°C instead of 2050°C.
2. enhances the mobility of the fused mixture by acting as a solvent for the electrolytic mixture.
3. enhances the conductivity of the mixture since, pure alumina is almost a non-conductor of electricity.

(a) NH₃ + 3PbO ⟶ 3Pb + 2H₂O + 2NO₂

(b) Zn + 4HNO₃ ⟶ Zn(NO₃)₂ + 2H₂O + 2NO₂

(c) 2H₂SO₄ + S ⟶ 3SO₂ + 2H₂O

(A) Zn + 2HCl ⟶ ZnCl₂ + 2HCl

(B) ZnCl₂ + 2NaOH ⟶ Zn(OH)₂ + 2NaOH

(C) Zn(OH)₂ + H₂SO₄ ⟶ ZnSO₄ + 2H₂O

(a) The article to be electroplated is placed is made as cathode. This is because metal cations from the electrolyte will move toward the cathode and get reduced and deposited on the spoon.

(b) Ni²⁺ + 2e^- ⟶ Ni

(c) The anode should be made of pure nickel metal. During electroplating, nickel atoms from the anode dissolve into the solution as Ni²⁺ ions, maintaining a steady concentration of nickel ions in the electrolyte.

Given,
Mass of gas = 30g
Volume = 1000 c.c = 1dm³ = 1 liter

We know one mole of gas at STP occupies 22.4 l

1 litres of gas at S.T.P. has mass = 30 g

∴ 22.4 litre (molar volume) has mass

= $\frac{30}{1}$ x 22.4

= 672 g/mol

Vapour density = $\frac{\text{Molecular mass}}{2}$

= $\frac{672}{2}$

= 336

(b) 22.4 litres of the gas contains 6.023 x 10²³ molecules

So, 44.8 litres of the gas contain 2 x 6.023 x 10²³

= 12.046 x 10 ²³ molecules

Element X is a metal, and its oxide is X₂O₃ so X must have a valency of 3 (X³⁺). The oxide is a good conductor of electricity. Since it requires electrolysis to extract X, it must be a very reactive metal.

(a) The oxide (X₂O₃) is reacting with hydrochloric acid, so we can say the oxide is ionic and basic in nature. The reaction will give salt and water.

X₂O₃ + 6HCl ⟶ 2XCl₃ + 3H₂O

(b) Since X forms X³⁺, it must lose 3 electrons, meaning it has 3 valence electrons.

(c) To form X³⁺, element X must lose 3 electrons and loss of electron is oxidation.

(a) NaOH

(b) C₂H₅OH

(c) 2NaOH + H₂SO₄ ⟶ Na₂SO₄ + 2H₂O

(d) Sodium plumbite

(a) Element X has the highest ionisation energy because it has the smallest atomic size, fewer electron shells, and its valence electrons experience a stronger attraction to the nucleus.

(b) Z has the configuration of 2,8,8,2, electrons occupying K,L,M and N shells. Hence, Z belongs to Period 4.

(c) The electron dot structure of the compound formed between Z and oxygen is shown below:

(a) When the student mixed potassium sulphite solution (K₂SO₃) with barium nitrate (Ba(NO₃)₂), the white precipitate of Barium sulphite (BaSO₃) is formed.

K₂SO₃ + Ba(NO₃)₂ ⟶ BaSO₃ + 2KNO₃

(b) When white precipitate of BaSO₃ reacts with dilute HCl, it dissolves, forming barium chloride (BaCl₂), water, and sulphur dioxide (SO₂) gas.

BaSO₃ + 2HCl ⟶ BaCl₂ + H₂O + SO₂ ↑

(c) From the above reaction, the gas produced is sulphur dioxide (SO₂).

(a) Empirical formula = C₂H₃

Relative molecular mass = 54

$$\begin{aligned}\text{n} = \frac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \frac{54}{27} = 2\end{aligned}$$

∴ Molecular formula = n[E.F.] = 2[C₂H₃] = C₄H₆

(b) The structural formula of butyne is shown below:

(c) The hydrocarbon C₄H₆ is an alkyne, and the general formula for alkyne is C_nH_2n-2.

(a) Electroplating

(b) Iron

(c) Iron tools and instruments are often electroplated with zinc to protect against rusting.

(a) Calcium carbide (CaC₂)

(b) Acetylene/Ethyne.

(c)

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

Atomic number of element M = 12

Electronic configuration of element M = 2, 8, 2

∴ Element M is a metal with valency 2, it will form M²⁺ ion

(a)

Atomic Number	Configuration	Evaluation
14	2, 8, 4	Tends to share electrons and forms covalent bonds
10	2, 8	Noble gas – inert, does not form ions
8	2, 6	Non-metal, gains 2 electrons to form ionic compounds

The correct match for L will be (iii), because M can form an ionic compound with L (M²⁺ + L^2- = ML)

(b) As element M contains 2 valenced electrons, it belongs to Group 2 of the Periodic Table and group 2 elements are called Alkaline Earth Metal.

(c) The electron dot structure of the compound formed between M and L is shown below:

(a) The gas X is Ammonia (NH₃).

(b) 4NH₃ + 5O₂ $\xrightarrow[800 ^{\circ} \text{C}]{\text{Pt.}}$ 4NO↑ + 6H₂O + Δ
     2NO + O₂ ⟶ 2NO₂ [brown gas]