ICSE Board Class 10 2026 Chemistry – Science Paper 2 Question Paper with Solutions : Download PDF

Step 1: Understanding the Chemical Reaction:

Concentrated sulphuric acid (\(H_2SO_4\)) acts as a powerful oxidizing agent. When it reacts with elemental carbon, it oxidizes the carbon to carbon dioxide while being reduced to sulphur dioxide.




Step 2: Identifying the Product Gases:

The balanced chemical equation for the reaction is: \[ C + 2H_2SO_4 (conc.) \rightarrow CO_2 + 2SO_2 + 2H_2O \]
The two gaseous products are Carbon dioxide (\(CO_2\)) and Sulphur dioxide (\(SO_2\)).


Step 3: Lime Water Test:

Both \(CO_2\) and \(SO_2\) have the property of reacting with lime water [calcium hydroxide, \(Ca(OH)_2\)] to form a white precipitate (\(CaCO_3\) and \(CaSO_3\) respectively), which makes the solution appear milky.


Step 4: Final Answer:

Carbon is the unique non-metal in this list that produces both of these gases simultaneously upon reaction with concentrated acid. Quick Tip: While both gases turn lime water milky, \(SO_2\) can be distinguished because it has a pungent smell of burning sulphur and turns acidified potassium dichromate paper green!

Step 1: Understanding Ionic Bonding:

An ionic bond is formed by the complete transfer of electrons from a metallic element (electropositive) to a non-metallic element (electronegative).




Step 2: Analyzing the Pairs:


Pair P: Both Group 1 and Group 2 are metals. Two metals do not form ionic bonds with each other.
Pair Q: Elements of Group 14 (like Carbon) and Group 16 (like Oxygen) are both non-metals and typically form covalent bonds.
Pair R: Group 2 (Metals) and Group 17 (Non-metals/Halogens) have a high difference in electronegativity, leading to ionic bond formation (e.g., \(MgCl_2\)).
Pair S: Group 18 consists of noble gases which are chemically inert and rarely form bonds.



Step 3: Conclusion:

The combination of a metal from Group 2 and a non-metal from Group 17 results in a stable ionic compound.


Step 4: Final Answer:

The correct pair is R. Quick Tip: Ionic bonds usually form between elements on opposite sides of the periodic table (excluding noble gases)!

Step 1: Identifying the Element:

The total number of electrons is \(2 + 8 + 2 = 12\). The element with atomic number 12 is Magnesium (\(Mg\)).


Step 2: Determining Valency:

Magnesium has 2 electrons in its outermost shell. To achieve stability, it loses these 2 electrons to form a divalent cation, \(Mg^{2+}\).


Step 3: Writing the Hydroxide Formula:

Hydroxide is a radical with a valency of 1 (\(OH^-\)). To balance the \(Mg^{2+}\) ion, two \(OH^-\) ions are required. The chemical formula is \(Mg(OH)_2\).




Step 4: Final Answer:

Upon dissociation, one molecule of \(Mg(OH)_2\) produces 2 hydroxyl ions. Quick Tip: The number of ionizable \(OH^-\) ions in a base is known as its acidity. Therefore, Magnesium hydroxide is a diacidic base.

Step 1: Understanding the Electrolysis Setup:

When using active copper electrodes, the anode (positive) dissolves and the cathode (negative) receives deposits.


Step 2: Reaction at the Cathode (Negative):

Copper ions (\(Cu^{2+}\)) from the solution move toward the cathode, gain electrons, and turn into solid copper: \[ Cu^{2+} + 2e^- \rightarrow Cu(s) \]
This results in an increase in the mass of the negative electrode.


Step 3: Reaction at the Anode (Positive):

The copper anode itself loses electrons and dissolves into the electrolyte: \[ Cu(s) \rightarrow Cu^{2+} + 2e^- \]
This results in a decrease in the mass of the positive electrode.


Step 4: Final Answer:

Statement (a) is the only correct observation for this specific electrolytic process. Quick Tip: In this process, the blue color of the \(CuSO_4\) solution does not fade because the number of \(Cu^{2+}\) ions leaving the solution at the cathode is replenished by the anode!

Step 1: Understanding the Reaction Type:

Ethane is a saturated hydrocarbon (alkane). Alkanes undergo substitution reactions with halogens where hydrogen atoms are replaced one by one.




Step 2: The Role of UV Light:

UV light provides the energy required to break the \(Cl-Cl\) bond to start the free-radical chain reaction.


Step 3: Writing the Products:

One hydrogen atom from ethane (\(C_2H_6\)) is replaced by one chlorine atom to form chloroethane (\(C_2H_5Cl\)). The displaced hydrogen atom bonds with the second chlorine atom to form hydrogen chloride (\(HCl\)).


Step 4: Final Answer:

The balanced equation is \(C_2H_6 + Cl_2 \rightarrow C_2H_5Cl + HCl\). Quick Tip: Substitution reactions are characteristic of saturated compounds, while addition reactions are characteristic of unsaturated compounds (alkenes/alkynes)!

Step 1: Understanding the Reaction (Cracking):

The reaction described is thermal cracking, where a long-chain saturated hydrocarbon (alkane) is broken down into a smaller alkane and an unsaturated hydrocarbon (alkene).




Step 2: Identifying the Compounds:


X (Butane): \(C_4H_{10}\). This is an alkane. All carbon-carbon bonds are single bonds (Saturated).
Y (Propene): \(C_3H_6\). This is an alkene. It contains at least one carbon-carbon double bond (Unsaturated).
Z (Methane): \(CH_4\). This is the simplest alkane (Saturated).



Step 3: Defining Unsaturated:

Unsaturated hydrocarbons are those that contain double or triple covalent bonds between carbon atoms. In this reaction, only propene (Y) fits this definition.


Step 4: Final Answer:

Compound Y only is unsaturated. Quick Tip: A quick way to spot an unsaturated hydrocarbon is its formula: Alkenes follow \(C_nH_{2n}\), while Alkanes follow \(C_nH_{2n+2}\).

Step 1: Understanding Metal-Acid Reactivity:

Metals that are more reactive than hydrogen (positioned above hydrogen in the reactivity series) displace hydrogen from dilute acids to produce hydrogen gas.




Step 2: Evaluating the Observations:


Copper (Cu): Below hydrogen in the reactivity series. It does not react with dilute HCl. (Observation 1 is Incorrect).
Iron (Fe): Above hydrogen. It reacts to give \(H_2\) gas. (Observation 2 is Correct).
Magnesium (Mg): Highly reactive, above hydrogen. It reacts vigorously to give \(H_2\) gas. (Observation 3 is Incorrect).
Zinc (Zn): Above hydrogen. It reacts to give \(H_2\) gas. (Observation 4 is Correct).



Step 3: Identifying Correct Observations:

Only observations 2 (Iron) and 4 (Zinc) accurately reflect the chemical reality.


Step 4: Final Answer:

The correct choice is Both 2 and 4. Quick Tip: The "gas given off" in these reactions is Hydrogen (\(H_2\)), which can be tested using a burning splinter; it extinguished with a characteristic 'pop' sound!

Step 1: Understanding Ion Formation:

The equation \(X \rightarrow X^{2-}\) implies that the atom X accepts 2 electrons to form a negatively charged ion (anion).


Step 2: Applying the Octet Rule:

Non-metals gain electrons to complete their valence shell (usually reaching 8 electrons). An atom that gains 2 electrons must have had 6 valence electrons in its neutral state.




Step 3: Checking the Atomic Numbers:


16 (Sulphur): 2, 8, 6. (Needs 2 electrons to become 2, 8, 8). Matches.
10 (Neon): 2, 8. (Noble gas, stable).
12 (Magnesium): 2, 8, 2. (Loses 2 electrons to form \(Mg^{2+}\)).
14 (Silicon): 2, 8, 4. (Usually forms covalent bonds).



Step 4: Final Answer:

The atomic number is 16. Quick Tip: Atoms in Group 16 (Chalcogens) always form \(2-\) ions, while atoms in Group 2 (Alkaline Earth Metals) always form \(2+\) ions.

Step 1: Understanding Salt Preparation:

Soluble salts can be made by reacting an acid with a metal, base, or carbonate. However, the "Metal + Acid" method only works for metals more reactive than hydrogen.


Step 2: Checking Copper's Reactivity:

As established in question (vii), Copper is below hydrogen in the reactivity series. This means it cannot displace hydrogen from dilute acids like \(HCl\) or \(H_2SO_4\).


Step 3: Evaluating Alternatives:

Copper salts \textit{can be prepared by reacting Copper oxide (a base) or Copper carbonate with dilute acids, as these are neutralization/decomposition reactions that do not require hydrogen displacement.


Step 4: Final Answer:

Method 3 (Action of dilute acid on metals) cannot be used for copper. Quick Tip: To make Copper(II) Sulphate in a lab, we typically react Copper(II) Oxide with hot dilute Sulphuric acid.

Step 1: Comparing Bond Types:

The melting point of a substance depends on the strength of the forces holding its particles together.


Step 2: Analyzing the Options:


Methane, Ammonia, Ethanol: These are Covalent compounds. They have weak intermolecular forces (Van der Waals or Hydrogen bonding). They are usually gases or liquids at room temperature with very low melting points.
Sodium Chloride (\(NaCl\)): This is an Ionic compound. It forms a giant 3D crystal lattice held together by strong electrostatic forces of attraction between oppositely charged ions.



Step 3: Comparing Strengths:

The energy required to break ionic bonds is significantly higher than that required to overcome intermolecular forces in covalent molecules.


Step 4: Final Answer:

Sodium chloride has the highest melting point. Quick Tip: The melting point of \(NaCl\) is approximately 801°C, whereas Methane melts at -182°C. That is a massive difference!

Step 1: Understanding Electrolytes:

An electrolyte's strength is determined by its ability to conduct electricity, which depends on the presence of free, mobile ions.


Step 2: Comparing Dilute and Concentrated Acid:

Concentrated sulphuric acid is largely composed of molecules (\(H_2SO_4\)). Dilution with water causes the acid to ionize into \(H^+\) and \(SO_4^{2-}\) ions.



Step 3: Linking Ionization to Conductivity:

Because dilute acid contains a significantly higher number of dissociated, mobile ions compared to the molecular form of concentrated acid, it conducts electricity much more effectively.


Step 4: Final Answer:

Both statements are true; the higher concentration of ions in the dilute form is exactly why it is a stronger electrolyte. Quick Tip: Pure 100% sulphuric acid is a poor conductor because it contains almost no ions. It’s the water that "unlocks" the ions through the process of electrolytic dissociation!

Step 1: Understanding Gay-Lussac's Law:

According to Gay-Lussac's Law of Combining Volumes, gases react in a simple whole-number ratio by volume, provided temperature and pressure remain constant.


Step 2: Analyzing the Stoichiometry:

From the balanced equation: \[ \underbrace{1 vol.}_{C_3H_8} + 5 vol. O_2 \rightarrow \underbrace{3 vol.}_{CO_2} + 4 vol. H_2O \]
This means 1 litre of propane produces 3 litres of carbon dioxide.


Step 3: Calculating for 5 Litres:

Volume of \(CO_2 = 5 litres (propane) \times 3 = 15 litres\).


Step 4: Final Answer:

The volume of \(CO_2\) produced is 15 litres. Quick Tip: As long as all reactants and products are gases at the same temperature and pressure, you can use the coefficients of the balanced equation directly as volume ratios!

Step 1: Analyzing the Molecular Formula:

The given information states the compound has 3 Carbon atoms and 6 Hydrogen atoms. The molecular formula is \(C_3H_6\).


Step 2: Matching with General Formulas:


Alkanes: \(C_nH_{2n+2}\) (For \(n=3\), \(H=8\)) \(\rightarrow\) Propane
Alkenes: \(C_nH_{2n}\) (For \(n=3\), \(H=6\)) \(\rightarrow\) Propene
Alkynes: \(C_nH_{2n-2}\) (For \(n=3\), \(H=4\)) \(\rightarrow\) Propyne





Step 3: Verifying Unsaturation:

Unsaturated hydrocarbons contain double or triple bonds. Propene (\(C_3H_6\)) contains one carbon-carbon double bond, making it unsaturated.


Step 4: Final Answer:

The compound is propene. Quick Tip: Remember the suffixes: \textbf{-ane} (single bond), \textbf{-ene} (double bond), and \textbf{-yne} (triple bond). Propene belongs to the alkene family.

Step 1: Understanding Water Electrolysis:

The decomposition of water into its constituent gases is represented by the equation: \[ 2H_2O (l) \rightarrow 2H_2 (g) + O_2 (g) \]




Step 2: Evaluating the Assertion:

From the balanced equation, 2 volumes of Hydrogen are produced for every 1 volume of Oxygen. Thus, Assertion (A) is True.


Step 3: Evaluating the Reason:

Water (\(H_2O\)) consists of Hydrogen and Oxygen in a ratio of 2:1 by volume (and 1:8 by mass). Reason (R) states the ratio is 1:2, which is False.


Step 4: Final Answer:

Assertion (A) is true but Reason (R) is false. Quick Tip: In a Hoffmann voltameter, the level of water in the cathode limb (where \(H_2\) collects) will be twice as low as the level in the anode limb (where \(O_2\) collects)!

Step 1: Understanding Electrode Polarity:

The negative electrode is the Cathode. Cations (positively charged ions) are attracted to the cathode to undergo reduction.




Step 2: Identifying the Cation:

In molten aluminium oxide (\(Al_2O_3\)), the cation is \(Al^{3+}\).


Step 3: Describing the Cathode Reaction:

At the cathode, aluminium ions gain three electrons each to form neutral aluminium atoms: \[ Al^{3+} + 3e^- \rightarrow Al \]
This is a reduction reaction.


Step 4: Final Answer:

Reaction 1 is the correct cathode reaction. Quick Tip: An easy way to remember: \textbf{RED CAT} (Reduction at Cathode) and \textbf{AN OX} (Anode Oxidation).

(a) Name the process:

The industrial manufacture of ammonia is known as the Haber Process.


(b) Which catalyst is used in the above process?

The catalyst used is Finely divided Iron (Fe).

\textit{Note: Molybdenum is often used as a promoter to enhance the efficiency of this catalyst.

(c) In the above diagrammatic setup, how is ammonia gas separated from the unreacted gases to obtain liquid ammonia?

Ammonia gas is separated by the process of liquefaction. Because ammonia has a higher boiling point and is more easily liquefied under pressure than nitrogen or hydrogen, it is condensed into a liquid in a cooling chamber while the unreacted gases are pumped back into the catalytic chamber.

(d) Which two properties of ammonia gas can be demonstrated by the Fountain Experiment?

The two properties demonstrated are:

High solubility of ammonia gas in water.
Basic (alkaline) nature of the gas. Quick Tip: In the Fountain Experiment, the vacuum created by the rapid dissolution of ammonia in the first drop of water causes the water to rush up like a fountain. If phenolphthalein is used, the fountain turns pink!

(a) A compound which reacts with water to give acetylene gas:

Calcium Carbide (\(CaC_2\)).

The reaction is: \(CaC_2 + 2H_2O \rightarrow Ca(OH)_2 + C_2H_2\uparrow\)


(b) The gas evolved when sodium propanoate is heated with soda lime:

Ethane (\(C_2H_6\)).

This is a decarboxylation reaction: \(C_2H_5COONa + NaOH \xrightarrow{CaO, \Delta} C_2H_6 + Na_2CO_3\).

(c) A reddish-brown precipitate formed when ferric chloride solution reacts with an alkali:

Ferric Hydroxide [\(Fe(OH)_3\)].

The reaction with sodium hydroxide is: \(FeCl_3 + 3NaOH \rightarrow Fe(OH)_3\downarrow + 3NaCl\).

(d) A pair of electrons present in an atom which is not shared with any other atom during bond formation:

Lone pair (or unshared pair).


(e) The relative molecular mass of a substance expressed in grams:

Gram Molecular Mass (or one mole). Quick Tip: When using soda lime for decarboxylation, the resulting alkane always has one carbon atom less than the starting carboxylic acid salt. This is why sodium propanoate (\(C_3\)) yields ethane (\(C_2\)).

(a) — 4. Redox: In this reaction, Copper Oxide is being reduced (loses oxygen) and Carbon is being oxidized (gains oxygen).
(b) — 3. Reduction: Gain of electrons is defined as reduction. Aluminum ions gain three electrons to form aluminum atoms.
(c) — 5. Electrolytic dissociation: The separation of ions of a pre-existing ionic compound (molten or aqueous) is called electrolytic dissociation.
(d) — 1. Oxidation: Loss of electrons is defined as oxidation. Oxide ions lose electrons at the anode to form oxygen gas.
(e) — 2. Ionization: The process by which a polar covalent compound (like gaseous HCl) forms ions when dissolved in water. Quick Tip: Remember the acronym \textbf{OIL RIG}: \textbf{O}xidation \textbf{I}s \textbf{L}oss (of electrons), \textbf{R}eduction \textbf{I}s \textbf{G}ain (of electrons).

(a) ZnO: Zinc Oxide is an amphoteric oxide, meaning it reacts with both acids and strong bases (alkalis) like KOH to form salt and water. CuO is basic and does not dissolve in KOH.


(b) 2Br\(^-\) \(\rightarrow\) Br\(_2\) + 2e\(^-\): At the anode (positive electrode), oxidation occurs. Bromide ions lose electrons to form neutral bromine atoms, which then form bromine gas molecules.



(c) 5.6 (Note: Provided options may be incorrect; 5.6 is the calculation):

Molecular weight of \(O_2 = 16 \times 2 = 32\) g.
32 g of \(O_2\) occupies 22.4 L at STP.
8 g of \(O_2\) occupies \(\frac{22.4}{32} \times 8 = \mathbf{5.6}\) litres.

Correction: If restricted to brackets, 8.96 is mathematically closest but incorrect based on standard 22.4 L molar volume.


(d) Cu: Copper is below hydrogen in the reactivity series and cannot displace hydrogen from acids. Manganese (Mn) and Magnesium (Mg) are the only two metals that react with very dilute \((1%)\) \(HNO_3\) to evolve hydrogen.


(e) HCl: Hydrogen Chloride has a large difference in electronegativity between H and Cl, creating a partial charge (dipole). Carbon tetrachloride (\(CCl_4\)) is non-polar because its symmetrical tetrahedral shape causes individual bond polarities to cancel out. Quick Tip: To remember which oxides are amphoteric, think of \textbf{ZAP: \textbf{Z}inc, \textbf{A}luminum, and \textbf{P}b (Lead) oxides!

1. Propan-1-ol: The longest chain contains 3 carbon atoms (propane). The functional group is an alcohol (\(-OH\)), which gets the suffix "-ol". Since it is attached to the first carbon, it is propan-1-ol.


2. But-1-ene: The parent chain has 4 carbon atoms (but-). The presence of a double bond indicates an alkene (suffix "-ene"). The double bond starts at the first carbon, hence but-1-ene. Quick Tip: When numbering the carbon chain, always give the functional group (like the \(-OH\) or the double bond) the lowest possible number!

1. Butanal:
Butanal is an aldehyde with 4 carbon atoms. The functional group is \(-CHO\). \[ CH_3-CH_2-CH_2-CHO \]


2. Pent-2-yne:
Pent-2-yne is an alkyne with 5 carbon atoms and a triple bond starting at the second carbon atom. \[ CH_3-C\equiv C-CH_2-CH_3 \]


3. Isomer of n-butane (Iso-butane / 2-Methylpropane):
n-butane is a straight chain of 4 carbons. Its only structural isomer is a branched chain where 3 carbons form the main chain and 1 carbon is a methyl branch. \[ CH_3-CH(CH_3)-CH_3 \] Quick Tip: Carbon always forms 4 bonds. When drawing structures, double-check that every carbon atom has exactly 4 lines (bonds) connected to it!

(a) Anion: The anion is the Chloride ion (\(Cl^-\)).
When silver nitrate is added to a chloride solution, it forms a white precipitate of Silver Chloride (\(AgCl\)) which dissolves in excess ammonium hydroxide to form a clear complex solution.



(b) Cation: The cation is the Copper(II) ion (\(Cu^{2+}\)).
The formation of a pale blue precipitate (Copper(II) hydroxide) with sodium hydroxide is a characteristic test for copper ions. Quick Tip: Salt Y is Copper(II) Chloride (\(CuCl_2\)). Remember that while \(AgCl\) is soluble in \(NH_4OH\), it is insoluble in dilute nitric acid!

(a) Reason for temperature \(< 200^\circC\):
Maintaining the temperature below 200°C is important for several reasons:

It prevents the formation of a hard crust of sodium sulphate (\(Na_2SO_4\)), which is difficult to remove from the glass retort.
It prevents the thermal decomposition of the product, Nitric acid (\(HNO_3\)).
It saves fuel and prevents the glass apparatus from cracking.



(b) Use of concentrated Sulphuric acid:
Concentrated sulphuric acid is used because it is a non-volatile acid. It can displace the more volatile nitric acid from its salt (sodium nitrate) when heated. Quick Tip: Nitric acid prepared in the lab often appears yellow. This is due to the presence of dissolved nitrogen dioxide (\(NO_2\)) produced by the slight decomposition of the acid!

(a) Identification:
An element in Period 2 has 2 shells. Group 15 indicates 5 valence electrons. The element is Nitrogen (\(N\)) (Atomic number 7, configuration 2, 5).


(b) Formula:
Nitrogen requires 3 electrons to complete its octet, and Hydrogen has 1 electron. The compound formed is Ammonia (\(NH_3\)).


(c) Electron Dot Structure:
The Nitrogen atom (center) shares three of its five valence electrons with three Hydrogen atoms, leaving one lone pair. Quick Tip: The presence of the lone pair on the nitrogen atom is what gives ammonia its basic properties and its ability to form coordinate bonds!

(a) Dehydrohalogenation of ethylene dibromide: \[ C_2H_4Br_2 + 2KOH (alc.) \xrightarrow{\Delta} C_2H_2 + 2KBr + 2H_2O \]
The product is Ethyne (Acetylene).


(b) Reaction of excess ammonia with chlorine: \[ 8NH_3 + 3Cl_2 \rightarrow 6NH_4Cl + N_2 \]
The products are Ammonium chloride (white fumes) and Nitrogen gas.

(c) Dehydration of Cane Sugar: \[ C_{12}H_{22}O_{11} \xrightarrow{conc. H_2SO_4} 12C + 11H_2O \]
The concentrated sulphuric acid removes water from the sugar, leaving behind a black spongy mass of Sugar Charcoal. Quick Tip: In equation (b), if Chlorine were in excess instead of Ammonia, the product would be the highly explosive liquid, Nitrogen Trichloride (\(NCl_3\))!

(a) Acetic acid vs. Sulphuric acid:

Acetic acid (\(CH_3COOH\)): It is monobasic, meaning it has one replaceable hydrogen ion per molecule.
Sulphuric acid (\(H_2SO_4\)): It is dibasic, meaning it has two replaceable hydrogen ions per molecule.




(b) Electrolyte vs. Metallic conductor:

Electrolyte: The particles conducting electricity are mobile ions (cations and anions).
Metallic conductor: The particles conducting electricity are free electrons. Quick Tip: In acetic acid, although there are four hydrogen atoms in total, only the one attached to the oxygen atom in the carboxyl group (\(-COOH\)) is ionizable!

Step 1: Apply Avogadro's Law:

Avogadro's Law states that equal volumes of all gases, under the same conditions of temperature and pressure, contain an equal number of molecules (and thus an equal number of moles). \[ Moles of NH_3 = Moles of Cl_2 \]

Step 2: Calculate moles of Ammonia (\(NH_3\)):

Molecular weight of \(NH_3 = 14 + (1 \times 3) = 17\) g/mol. \[ Moles = \frac{Mass}{Molar Mass} = \frac{34 g}{17 g/mol} = 2 moles \]

Step 3: Determine molecules in Chlorine (\(Cl_2\)):

Since the volumes are equal, Chlorine also has 2 moles. \[ Number of molecules = Moles \times Avogadro's Number (N_A) \] \[ Number of molecules = 2 \times 6.022 \times 10^{23} = 1.2044 \times 10^{24} molecules \]

Final Answer: The chlorine gas contains \(1.2044 \times 10^{24}\) molecules (or simply \(2N_A\) molecules). Quick Tip: You don't actually need the atomic weight of Chlorine for this specific problem because Avogadro's Law links volume directly to the number of molecules, regardless of the gas's identity!

Based on the descriptions: P is an Alkali Metal (Group 1), Q is likely a Group 14 element (like Carbon or Silicon), and R is a Halogen (Group 17).


(a) Which element would be most difficult to reduce?

Element P. P is highly electropositive and wants to lose electrons (oxidize). Forcing it to gain electrons (reduction) is extremely difficult.


(b) Smallest atomic radius?

Element R. Atomic radius decreases across a period from left to right due to increased nuclear charge. R is furthest to the right.


(c) Decreasing order of ionization potential:

Ionization potential increases across a period. Therefore, the order of decreasing IP is: \[ \mathbf{R > Q > P} \] Quick Tip: This specific problem because Avogadro's Law links volume directly to the number of molecules, regardless of the gas's identity!

(a) Lead (II) carbonate: (Prepared via double decomposition/precipitation) \[ Pb(NO_3)_2 + Na_2CO_3 \rightarrow PbCO_3\downarrow + 2NaNO_3 \]


(b) Copper (II) chloride: (Prepared via action of acid on a carbonate) \[ CuCO_3 + 2HCl \rightarrow CuCl_2 + H_2O + CO_2\uparrow \]

(c) Iron (II) chloride: (Prepared via direct synthesis/combination) \[ Fe + 2HCl \rightarrow FeCl_2 + H_2\uparrow \]
\textit{Note: Iron reacting with dry Chlorine would produce Iron (III) chloride, so dilute HCl is used for the (II) state. Quick Tip: To prepare an insoluble salt like Lead Carbonate, you must always start with two soluble reactants!

(a) Addition: Since the hydrocarbon is unsaturated (contains multiple bonds) and reacts with hydrogen to become saturated without losing any atoms, it is an addition reaction.


(b) Catalyst: Suitable catalysts include Nickel (Ni), Platinum (Pt), or Palladium (Pd).



(c) Balanced Chemical Equation:
Since the hydrocarbon has 2 carbon atoms and reacts with two moles of \(H_2\), the starting material must be Ethyne (\(C_2H_2\)). \[ C_2H_2 + 2H_2 \xrightarrow{Ni, 200^\circC} C_2H_6 \]
(Ethyne reacts with two moles of Hydrogen to form Ethane). Quick Tip: If the hydrocarbon had only reacted with \textbf{one mole} of hydrogen, the starting material would have been Ethene (\(C_2H_4\))!

(a) Sulphur trioxide (\(SO_3\)): The catalytic oxidation of \(SO_2\) in the Contact Process yields \(SO_3\).


(b) Sulphuric acid (\(H_2SO_4\)): Dissolving \(SO_3\) in water directly forms \(H_2SO_4\).


(c) Reason for not dissolving directly: The reaction between \(SO_3\) and water is highly exothermic. It produces a dense fog or mist of tiny sulphuric acid droplets that is very difficult to condense and can damage the apparatus. To avoid this, \(SO_3\) is first dissolved in concentrated sulphuric acid to form Oleum (\(H_2S_2O_7\)), which is then safely diluted. Quick Tip: Hydrocarbon had only reacted with \textbf{one mole} of hydrogen, the starting material would have been Ethene (\(C_2H_4\))!

(a) Cathode Reaction: The silver ions migrate to the spoon (cathode) and gain electrons to deposit as metal. \[ Ag^+ + e^- \rightarrow Ag(s) \]

(b) Reason for avoiding \(AgNO_3\): Silver nitrate is a strong electrolyte and causes very rapid migration of ions. This leads to an uneven, loose, and non-uniform coating. Sodium argentocyanide (\(Na[Ag(CN)_2]\)) provides a slow and steady supply of silver ions, ensuring a smooth and firm plate.


(c) Why AC is not used: Alternating current (AC) reverses polarity periodically. This would cause the metal to deposit during one half-cycle and dissolve back into the solution during the next, resulting in no net electroplating.


(d) Observation at Anode: The silver block (anode) gradually decreases in size/mass as silver atoms lose electrons and enter the solution as ions (\(Ag \rightarrow Ag^+ + e^-\)). Quick Tip: To ensure a perfect finish during electroplating, the article to be plated must always be placed at the \textbf{Cathode} and the electrolyte must contain ions of the metal to be deposited!

(a) Basicity of HCl: Hydrochloric acid (\(HCl\)) is a monobasic acid. It contains only one ionizable hydrogen atom per molecule (\(HCl \rightarrow H^+ + Cl^-\)). Acid salts are only formed by polybasic acids (like \(H_2SO_4\)) where there is at least one replaceable hydrogen left after partial neutralization. Since \(HCl\) has only one hydrogen, it can only form normal salts.


(b) Nuclear Charge: As we move across a period from left to right, the nuclear charge (number of protons) increases while the number of shells remains the same. This increases the force of attraction exerted by the nucleus on the shared pair of electrons, thereby increasing electronegativity. Quick Tip: Remember: \textbf{Basicity} refers to the number of ionizable \(H^+\) ions in an acid. \(HCl\) is monobasic, \(H_2SO_4\) is dibasic, and \(H_3PO_4\) is tribasic!

(a) False: Isomers are compounds that have the same molecular formula but different structural formulas.


(b) True: This is the chemical definition of a salt. For example, replacing \(H^+\) in \(H_2SO_4\) with \(Na^+\) gives \(NaHSO_4\) (partial replacement) or \(Na_2SO_4\) (complete replacement). Quick Tip: Isomerism is like having the same set of Lego bricks (molecular formula) but building two different shapes (structural formula) with them!

(a) Storage Error: Anil should have stored the nitric acid in an amber-colored (dark) bottle and kept it in a cool, dark place away from sunlight.


(b) Thermal Decomposition: Nitric acid is unstable in the presence of heat or sunlight. It decomposes to give nitrogen dioxide (\(NO_2\)), which appears as brown fumes and dissolves in the remaining acid to give it a yellow tint. \[ 4HNO_3 \xrightarrow{sunlight} 4NO_2 \uparrow + O_2 \uparrow + 2H_2O \]

(c) Removal of Yellow Tinge: The yellow color can be removed by bubbling dry air or carbon dioxide through the acid (to drive out the dissolved \(NO_2\)) or by adding a small amount of water. Quick Tip: Nitric acid is a powerful oxidizing agent. Its yellow color is purely due to dissolved \(NO_2\) impurity; pure nitric acid is actually colorless!

(a) Moles of Zinc: \[ Moles of Zn = \frac{Given Mass}{Atomic Mass} = \frac{32.5}{65} = \mathbf{0.5 moles} \]

(b) Mass of Nitric Acid:
From the equation, 1 mole of \(Zn\) reacts with 4 moles of \(HNO_3\).
Therefore, 0.5 moles of \(Zn\) will react with: \[ 0.5 \times 4 = 2 moles of HNO_3 \]
Molecular weight of \(HNO_3 = 1 + 14 + (16 \times 3) = 63\) g/mol. \[ Mass = Moles \times Molar Mass = 2 \times 63 = \mathbf{126 g} \]

(c) Volume of \(NO_2\) at STP:
From the equation, 1 mole of \(Zn\) produces 2 moles of \(NO_2\).
Therefore, 0.5 moles of \(Zn\) will produce: \[ 0.5 \times 2 = 1 mole of NO_2 \]
Since 1 mole of any gas occupies 22.4 L at STP: \[ Volume = \mathbf{22.4 litres} \] Quick Tip: Always check the coefficients in the balanced chemical equation; they tell you the "recipe" or the molar ratio required for the reaction to occur!

(a) Acid Name: The acid used is Concentrated Nitric Acid (\(HNO_3\)). Concentrated nitric acid is a powerful oxidizing agent that oxidizes non-metals like carbon to their respective oxides or oxyacids.


(b) Balanced Equation: \[ C + 4HNO_3 (conc.) \rightarrow CO_2 + 2H_2O + 4NO_2 \] Quick Tip: Concentrated nitric acid usually acts as an oxidizing agent because it decomposes to give nascent oxygen: \(2HNO_3 \rightarrow 2NO_2 + H_2O + [O]\).

(a) Reason for Caustic Soda: Caustic soda (Sodium Hydroxide) is added because aluminium oxide is amphoteric. It reacts with the hot concentrated alkali to form soluble Sodium Meta-aluminate, while the main impurities in bauxite (like Ferric Oxide) are basic and do not dissolve, allowing them to be filtered out as "red mud."


(b) Balanced Equation: \[ Al_2O_3 \cdot 2H_2O + 2NaOH \rightarrow 2NaAlO_2 + 3H_2O \]
(Bauxite reacts with sodium hydroxide to produce sodium meta-aluminate and water). Quick Tip: The Bayer’s process is specifically a \textbf{chemical} method of refining, whereas the subsequent Hall-Héroult process is an \textbf{electrolytic} method of extraction.

(a) Observation: A white precipitate of Calcium hydroxide [\(Ca(OH)_2\)] is formed, which is sparingly soluble in excess of sodium hydroxide.


(b) Observation: A gas is evolved which has a characteristic foul smell of rotten eggs (Hydrogen Sulphide, \(H_2S\)).


(c) Observation: The metal dissolves with brisk effervescence as a colorless, odorless gas (Hydrogen) is evolved, which burns with a 'pop' sound. Quick Tip: To distinguish between Calcium and Magnesium salts using NaOH: Calcium forms a white precipitate that is sparingly soluble, whereas Magnesium forms a white precipitate that remains strictly insoluble in excess NaOH!

Reactant A: Concentrated Sulphuric Acid (\(H_2SO_4\)) at 170°C. It acts as a dehydrating agent to convert ethanol into ethene.


Reactant B: Acetic Acid (\(CH_3COOH\)). Ethanol is oxidized to acetic acid, which then reacts with more ethanol (in the presence of an acid catalyst) to form Ethyl acetate (esterification).


Reactant C: Oxygen (\(O_2\)) or an Oxidizing agent. This facilitates the conversion of Ethanol to B (Acetic acid) or the combustion of Ethanol/Ethene to \(CO_2\) and water. Quick Tip: The conversion of Ethanol to Ethene is a \textbf{Dehydration} reaction, while the conversion of Ethanol to Ethyl Acetate is called \textbf{Esterification}!

Step 1: Calculate Molecular Weight:

According to the molar volume concept, 22.4 litres of any gas at STP is equal to its molecular weight in grams.
\[ Weight of 5.6 L of gas Z = 32 g \] \[ Weight of 1 L of gas Z = \frac{32}{5.6} \] \[ Weight of 22.4 L of gas Z = \frac{32}{5.6} \times 22.4 = \mathbf{128 g} \]
Therefore, the Molecular Weight of Z is 128.

Step 2: Calculate Vapour Density:

The relationship between molecular weight and vapour density is given by the formula: \[ Molecular Weight = 2 \times Vapour Density \] \[ Vapour Density = \frac{Molecular Weight}{2} = \frac{128}{2} = \mathbf{64} \] Quick Tip: Vapour density is a ratio and has no units, whereas molecular weight is usually expressed in atomic mass units (amu) or grams per mole!

(a) Fluorine (F): Electronegativity increases across a period from left to right. Fluorine is the furthest to the right in Period 2 (excluding noble gases) and is the most electronegative element in the entire periodic table.


(b) Sodium (Na): Atomic size decreases across a period due to an increase in nuclear charge. Therefore, the element at the beginning of Period 3, Sodium, has the largest atomic radius in that period. Quick Tip: Remember that Noble Gases (like Neon or Argon) are often excluded from general electronegativity trends because they do not readily form chemical bonds!

(a) Fluorine (F): Fluorine is in Period 2 and Group 17. Compared to the alkali metals listed (Li, K, Cs), it has more protons in the same/fewer shells, pulling the electrons closer.


(b) None of these: Li, K, and Cs are Group 1 elements (1 valence electron). F is a Group 17 element (7 valence electrons).


(c) Cesium (Cs): Electropositivity (metallic character) increases down a group. Since Cesium is at the bottom of Group 1 among the choices, it loses its valence electron most easily. Quick Tip: Cesium is so electropositive and reactive that it will explode upon contact with even cold water or ice!

(a) Beaker L: Sodium chloride (\(NaCl\)) is a strong electrolyte. In aqueous solution, it dissociates completely into \(Na^+\) and \(Cl^-\) ions. (Technically it also contains water molecules, but among the choices, it is the one characterized by full ionic dissociation).


(b) Beaker M: Distilled water is a non-electrolyte. While it undergoes self-ionization to a negligible extent, it is considered to consist almost entirely of \(H_2O\) molecules.


(c) Beaker K: Acetic acid (\(CH_3COOH\)) is an organic acid. All acids have a pH value less than 7. Quick Tip: Aqueous acetic acid is a weak electrolyte, meaning it contains both ions and molecules, as it only partially dissociates in water!