Class 12 Chemistry Chapter 7 P-Block Elements

Key Concept: Properties and Structures of Nitrogen Oxides

d) Nitric oxide ($NO$)
[Solution Description]
Nitric oxide ($NO$) is paramagnetic because it has an unpaired electron in its molecular orbital configuration. The electronic structure of $NO$ shows one unpaired electron, making it paramagnetic. Other oxides like $N_2O$, $N_2O_3$, and $N_2O_4$ are diamagnetic as they do not have unpaired electrons.

Key Concept: Properties, Dehydration

d) Carbon and water
[Solution Description]
Concentrated sulfuric acid has a strong affinity for water and acts as a dehydrating agent. When it reacts with sucrose, it removes water molecules from sucrose, leaving behind carbon. The reaction is:
$C_{12}H_{22}O_{11} \rightarrow 12C + 11H_2O$
Thus, the main product is carbon.

Key Concept: Manufacture of Nitric Acid

b) Platinum
[Solution Description]
The first step of Ostwald's process involves the oxidation of ammonia gas ($NH_3$) to nitric oxide ($NO$) using a platinum ($Pt$) catalyst. The reaction occurs at 1155 K and is given by:
$4NH_3 + 5O_2 \xrightarrow{Pt} 4NO + 6H_2O$
Thus, the correct catalyst is platinum.

Key Concept: Occurrence of Halogens

b) Chlorine
[Solution Description]
Among halogens, chlorine ($Cl$) is the most abundant and found primarily in the form of sodium chloride ($NaCl$).

Key Concept: Physical State and Molecular Structure, Hydrogen Bonding

d) Nitrogen forms $p\pi-p\pi$ multiple bonds while phosphorus does not
[Solution Description]
Nitrogen forms strong $p\pi-p\pi$ bonds due to its small size and high electronegativity, resulting in stable $N_2$ molecules with a triple bond. Phosphorus cannot form such strong multiple bonds due to larger atomic size and lower electronegativity, so it forms weaker single bonds in $P_4$ molecules which are solids at room temperature.
The correct explanation involves:
1) Nitrogen's ability to form strong triple bonds ($N \equiv N$) due to effective $p\pi-p\pi$ overlap
2) Small size allowing close approach for $p\pi-p\pi$ bonding
3) These factors make $N_2$ gaseous at room temperature
4) Phosphorus forms $P_4$ with weaker single bonds and exists as solid

Key Concept: Reactivity of Nitric Acid

d) Copper (Cu)
[Solution Description]
Concentrated nitric acid (\$HNO_3\$) reacts with certain metals (e.g., Zn, Fe, Pb, Cu, Ag) to produce nitrogen dioxide (\$NO_2\$) along with the respective metal nitrate. For example:
$Cu + 4HNO_3 \rightarrow Cu(NO_3)_2 + 2NO_2 + 2H_2O$

Key Concept: Sulphur compounds, hydrolysis

c) Sulphur hexafluoride ($SF_6$)
[Solution Description]
The question tests knowledge of sulphur hexafluoride ($SF_6$) which is resistant to hydrolysis because its six fluorine atoms create steric hindrance, preventing water molecules from attacking the sulphur atom. This property makes it useful as a gaseous electrical insulator.

Key Concept: Chemical Properties of Hydrogen Chloride

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Dry hydrogen chloride gas ($HCl$) does not dissociate into $H^+$ and $Cl^-$ ions in the absence of water. The acidic property of $HCl$ is due to the presence of $H^+$ ions, which are only produced when $HCl$ dissolves in water ($HCl + H_2O \rightarrow H_3O^+ + Cl^-$). Since dry $HCl$ does not have free $H^+$ ions, it cannot turn dry blue litmus red. Thus, both the Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Applications and Compounds, Interhalogen Compounds

c) $ClF_3$
[Solution Description]
Among interhalogen compounds, $ClF_3$ reacts vigorously with water to form hydrofluoric acid ($HF$) and oxygen gas ($O_2$). The reaction proceeds as follows:
$2ClF_3 + 3H_2O \rightarrow 6HF + O_2 + Cl_2O$

Key Concept: Occurrence of Sulphur and Chlorine

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that sulphur occurs in both native (free) and combined states, which is true as per the syllabus. Chlorine, being highly reactive, does not occur freely in nature and is always found in combined forms such as metal halides. The reason correctly explains why chlorine is not found in the free state and also provides the correct explanation for sulphur's dual occurrence, linking it to reactivity. Thus, both assertions are true, and the reason correctly explains them.

Key Concept: Structure of Interhalogen Compounds

c) $sp^3d^2$
[Solution Description]
The general formula for interhalogen compounds with a square pyramidal structure is $XY_5$, which involves $sp^3d^2$ hybridization.

Key Concept: Structure of Ozone

c) 116.8°
[Solution Description]
The ozone molecule ($O_3$) has a bent or angular structure due to its lone pair on the central oxygen atom. The bond angle is 116.8°, which is less than the ideal tetrahedral angle (109.5°) due to lone pair repulsion.

Key Concept: Manufacture of Nitric Acid

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
In Ostwald's process, ammonia ($NH_3$) is oxidized to nitric oxide ($NO$) using platinum ($Pt$) as the catalyst. The platinum gauze provides a large surface area for the reaction to occur efficiently. Additionally, platinum remains chemically unchanged at high temperatures (1155 K), which is why it is suitable as a catalyst. Therefore, both the Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Atomic radius, Van der Waals' radii

c) Noble gases exhibit van der Waals' radii, while halogens exhibit covalent radii.
[Solution Description]
The atomic radii of noble gases correspond to van der Waals' radii, which are larger than covalent or ionic radii of halogens due to increased electronic repulsion in their completely filled valence shells. Halogens have smaller radii because they form covalent bonds, leading to shorter interatomic distances.

Key Concept: Strong acid strength comparison

d) HClO$_4$
[Solution Description]
The acid strength of oxoacids of a halogen increases with an increase in the oxidation number of the halogen. The oxidation states for the given oxoacids are:
- HClO (chloric(I) acid, oxidation state +1),
- HClO$_2$ (chloric(III) acid, oxidation state +3),
- HClO$_3$ (chloric(V) acid, oxidation state +5),
- HClO$_4$ (chloric(VII) acid, oxidation state +7).
Since HClO$_4$ has the highest oxidation state (+7), it is the strongest acid among the given options.

Key Concept: Reactivity of Halogens

a) Fluorine
[Solution Description]
Fluorine is the most reactive halogen due to its low dissociation energy and high negative electron gain enthalpy. It can displace other halogens from their salt solutions.

Key Concept: Colour of Halogens

d) Violet
[Solution Description]
The colours of halogens darken as we move down the group. Fluorine is pale yellow, chlorine is greenish-yellow, bromine is reddish-brown, and iodine is violet. Therefore, iodine is the correct answer.

Key Concept: Oxidizing Action

c) Copper(II) sulfate, sulfur dioxide, and water are formed.
[Solution Description]
Concentrated sulfuric acid oxidizes copper to form copper(II) sulfate ($CuSO_4$), sulfur dioxide ($SO_2$), and water ($H_2O$). The balanced chemical equation is:
$Cu + 2H_2SO_4 \rightarrow CuSO_4 + SO_2 + 2H_2O$
Here, $H_2SO_4$ supplies oxygen for oxidation, reducing itself to $SO_2$.

Key Concept: Bleaching Powder

c) $Ca(OCl)_2 \cdot CaCl_2 \cdot Ca(OH)_2 \cdot 2H_2O$
[Solution Description]
Bleaching powder has the chemical formula $Ca(OCl)_2 \cdot CaCl_2 \cdot Ca(OH)_2 \cdot 2H_2O$. It is prepared by treating slaked lime with chlorine gas.

Key Concept: Catalyst in Contact Process

c) Vanadium pentoxide ($V_2O_5$)
[Solution Description]
The catalyst used in the contact process for the oxidation of $SO_2$ to $SO_3$ is vanadium pentoxide ($V_2O_5$). It is preferred over platinum-based catalysts due to its cost-effectiveness and resistance to poisoning.

Key Concept: Oxidation States of Nitrogen

c) $N_2O$
[Solution Description]
To determine the oxidation state of nitrogen in each compound:
1. For $N_2O$, let the oxidation state of N be x. Since O has an oxidation state of -2:
$2x + (-2) = 0 \implies 2x = 2 \implies x = +1$.
2. For $NO$, equation would be: $x + (-2) = 0 \implies x = +2$.
3. For $HNO_2$, $1 + x + 2(-2) = 0 \implies x = +3$.
4. For $NH_4^+$, $x + 4(+1) = +1 \implies x = -3$.
Thus, only $N_2O$ has nitrogen in the +1 oxidation state.

Key Concept: Oxidation States, Group 16 Elements

c) $OF_2$
[Solution Description]
Oxygen usually shows $-2$ oxidation state, but in $OF_2$, it exhibits $+2$ because fluorine is more electronegative and shared electrons are counted with fluorine.

Key Concept: Manufacture of dioxygen

c) Electrolysis of acidulated water
[Solution Description]
The electrolysis of acidulated water is used for the commercial preparation of dioxygen ($O_2$) at places where electricity is cheap. During this process, hydrogen ($H_2$) is produced at the cathode and dioxygen ($O_2$) at the anode.

Key Concept: Acidic and Basic Oxides

c) Assertion is true, but Reason is false.
[Solution Description]
The Assertion is true as $CO_2$ is indeed an acidic oxide since it reacts with water to form carbonic acid ($H_2CO_3$). However, the Reason is false because not all non-metal oxides are acidic; some like $N_2O$ are neutral. Hence, the correct option is (c).

Key Concept: Uses of Nitric Acid

d) Preparation of aqua regia
[Solution Description]
Nitric acid is used in the preparation of aqua regia when mixed with hydrochloric acid ($HCl$) in a 1:3 ratio. Aqua regia can dissolve noble metals like gold and platinum due to its strong oxidizing ability. The reaction for gold is:
$Au + HNO_3 + 4HCl \rightarrow HAuCl_4 + NO + 2H_2O$
Hence, the correct answer is "Preparation of aqua regia."

Key Concept: Amphoteric oxides

c) $Al_2O_3$
[Solution Description]
Amphoteric oxides exhibit both acidic and basic character. They react with acids as well as bases to form salts. Examples include $Al_2O_3$ and $ZnO$. Here, $Al_2O_3$ reacts with HCl to form $AlCl_3$ and with NaOH to form sodium aluminate:
$Al_2O_3(s) + 6HCl(aq) \rightarrow 2AlCl_3(aq) + 3H_2O(l)$
$Al_2O_3(s) + 6NaOH(aq) + 3H_2O(l) \rightarrow 2Na_3[Al(OH)_6](aq)$

Key Concept: Chemical Properties of Sulphur Dioxide

d) Assertion is false, but Reason is true.
[Solution Description]
When sulphur dioxide is passed through lime water initially, it forms insoluble calcium sulphite which causes the solution to turn milky:
$Ca(OH)_2 + SO_2 \rightarrow CaSO_3 + H_2O$
On passing excess sulphur dioxide, the milkiness disappears due to the formation of soluble calcium bisulphite:
$CaSO_3 + H_2O + SO_2 \rightarrow Ca(HSO_3)_2$
Here, Assertion states that SO$_2$ turns lime water milky in excess, which is false because milkiness disappears in excess SO$_2$. The Reason correctly explains the disappearance of milkiness but does not justify the Assertion. Thus, Assertion is false while Reason is true.

Key Concept: Haber's Process Conditions

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that high pressure (200-900 atm) is used in Haber’s process for ammonia production, which is true as per the syllabus. The reason explains that the forward reaction $N_2 + 3H_2 \rightleftharpoons 2NH_3$ leads to a decrease in volume (4 moles of reactants $\rightarrow$ 2 moles of products), and Le Chatelier’s principle confirms that high pressure favors reactions with a decrease in volume. Thus, both the assertion and reason are correct, and the reason logically explains the assertion.

Key Concept: Occurrence in Nature, Extraction of Sulphur

a) Driving a 30 cm diameter pipe to the sulphur bed
[Solution Description]
The Frasch process involves specific steps for extracting sulphur from underground deposits. According to the syllabus, the first step is driving a 30 cm diameter pipe to the sulphur bed. This allows access to the sulphur deposit before introducing superheated water or air.

Key Concept: Bonding in Nitrogen

c) Triple bond
[Solution Description]
Nitrogen forms a triple bond ($N \equiv N$) due to $p\pi - p\pi$ overlapping, making it highly stable. This consists of one sigma and two pi bonds.

Key Concept: Acid Strength of Oxoacids

c) $HClO < HClO_2 < HClO_3 < HClO_4$
[Solution Description]
The acid strength of oxoacids of halogens increases with an increase in the oxidation number of the halogen. For chlorine, the order is $HClO < HClO_2 < HClO_3 < HClO_4$.

Key Concept: Ammonia Manufacture

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
According to the syllabus, ammonia is manufactured by Haber's process under specific conditions: low temperature (720-770 K) and high pressure (200-900 atm). This is because the given reaction is exothermic ($\Delta H = -93.6 \, \text{kJ mol}^{-1}$) and proceeds with a decrease in volume (4 moles of reactants form 2 moles of product). As per Le Chatelier's principle, lowering the temperature favors the forward reaction in an exothermic process, and increasing the pressure favors the side with fewer gas moles. Therefore, both the Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Halates as Oxidizing Agents

d) In fireworks and safety matches
[Solution Description]
Potassium chlorate ($KClO_3$) is a halate and a strong oxidizing agent. One of its primary uses is in fireworks and safety matches due to its ability to release oxygen upon decomposition, which supports combustion.

Key Concept: Physical Properties of Ammonia

a) Gas
[Solution Description]
Ammonia is a colorless gas with a characteristic pungent odor at room temperature and pressure. Its boiling point is 239.7 K, which is below room temperature, meaning it remains gaseous under standard conditions.

Key Concept: Oxidizing Nature of Ozone

b) $KNO_2 + O_3 \rightarrow KNO_3 + O_2$
[Solution Description]
Ozone acts as a strong oxidizing agent and oxidizes potassium nitrite ($KNO_2$) to potassium nitrate ($KNO_3$). The reaction is:
$KNO_2 + O_3 \rightarrow KNO_3 + O_2$
Here, ozone oxidizes nitrite to nitrate while itself getting reduced to oxygen.

Key Concept: Oxidation States of Nitrogen

c) Nitrogen dioxide (NO$_2$)
[Solution Description]
The oxidation states of nitrogen in its compounds are:
- $-3$ in $NH_3$
- $+1$ in $N_2O$
- $+2$ in $NO$
- $+3$ in $HNO_2$
- $+4$ in $NO_2$
- $+5$ in $HNO_3$
Thus, nitrogen dioxide ($NO_2$) has nitrogen in the +4 oxidation state.

Key Concept: Electronic Configuration, Group Variation

c) It is a non-metal belonging to Group 16.
[Solution Description]
The given electronic configuration is $3s^2 3p^4$, which means it has 6 valence electrons (2 from $s$ and 4 from $p$). This places it in Group 16 (Oxygen family). As we move down Group 16, metallic character increases, but since this is a third-period element (n=3), it will exhibit non-metallic properties.
Step-by-step reasoning:
1. Valence electrons = 2 ($s$) + 4 ($p$) = 6 → Group 16.
2. Period number = Principal quantum number (n) = 3 → Third period.
3. Group 16 elements are non-metals in periods 2 and 3, so X is non-metallic.

Key Concept: Thermal decomposition of oxygen-rich salts

c) Potassium chloride and dioxygen
[Solution Description]
The thermal decomposition of potassium chlorate ($KClO_3$) in the presence of manganese dioxide ($MnO_2$) as a catalyst produces potassium chloride ($KCl$) and dioxygen ($O_2$). The reaction is given by:
$2KClO_3 \xrightarrow{\text{Heat, } MnO_2} 2KCl + 3O_2$.
Therefore, the products are potassium chloride and dioxygen.

Key Concept: Fertilizer Composition

c) Urea (CO(NH$_2$)$_2$)
[Solution Description]
Calculate nitrogen \% for each option:
- **Urea (CO(NH$_2$)$_2$)**: Molar mass = 60 g/mol; N mass = 28 g/mol. \% N = $(28/60) \times 100 = 46.67\%$.
- **Ammonium nitrate (NH$_4$NO$_3$)**: Molar mass = 80 g/mol; N mass = 28 g/mol. \% N = $(28/80) \times 100 = 35\%$.
- **Ammonium sulphate ((NH$_4$)$_2$SO$_4$)**: Molar mass = 132 g/mol; N mass = 28 g/mol. \% N = $(28/132) \times 100 = 21.21\%$.
- **Ammonium phosphate ((NH$_4$)$_3$PO$_4$)**: Molar mass = 149 g/mol; N mass = 42 g/mol. \% N = $(42/149) \times 100 = 28.19\%$.
Urea has the highest \% N.

Key Concept: Interhalogen Compounds

d) $BrF_3$
[Solution Description]
Interhalogen compounds of type $XY_3$ include examples like $ClF_3$, $BrF_3$, and $ICl_3$. These compounds have one central halogen atom bonded to three other halogen atoms.

Key Concept: Contact Process, Le-Chatelier's Principle

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The reaction $2SO_2 + O_2 \rightleftharpoons 2SO_3$ involves a decrease in the number of moles of gas (from 3 moles to 2 moles). According to Le-Chatelier’s principle, increasing the pressure will shift the equilibrium towards the side with fewer moles of gas, which is the forward reaction producing $SO_3$. Therefore, maintaining a higher pressure (2 atm) increases the yield of $SO_3$. Both Assertion and Reason are true, and Reason correctly explains Assertion.

Key Concept: Affinity for water

b) A decrease in volume and evolution of heat occur
[Solution Description]
When concentrated sulphuric acid (\$H_2SO_4\$) is added to water, a large amount of heat is released due to its strong affinity for water. This exothermic reaction can cause accidents if water is added to the acid instead of vice versa. Therefore, concentrated acid should always be added slowly to water while stirring.

Key Concept: Oxidising Power of Halogens

a) Fluorine
[Solution Description]
The oxidising power of halogens decreases from fluorine to iodine. Thus, fluorine has the highest oxidising power among the halogens.

Key Concept: Ostwald process, Catalyst efficiency, Reaction conditions

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description] The assertion is true as the Ostwald process does not require continuous external heating after the initial phase due to its exothermic nature. The reason is also correct because the reaction $4NH_3 + 5O_2 \xrightarrow{Pt} 4NO + 6H_2O$ has $\Delta H = -1275\ \text{kJ}$, indicating significant heat release, which maintains the high temperature necessary for the catalytic process. The reason accurately explains the assertion by linking the reaction's exothermic nature to the sustained temperature.

Key Concept: Catalysts, Promoters

c) It increases the surface area of iron by preventing sintering (correct promoter role)
[Solution Description]
In Haber’s process, the primary catalyst is finely divided iron. Molybdenum acts as a promoter, enhancing the catalyst's efficiency by preventing sintering (loss of surface area) and maintaining activity over time. Promoters do not directly participate in the reaction but stabilize the catalyst.

Key Concept: Halates Reaction

b) Iodine ($I_2$), potassium chloride ($KCl$), and water ($H_2O$)
[Solution Description]
Potassium chlorate ($KClO_3$) acts as an oxidizing agent and reacts with potassium iodide ($KI$) in an acidic medium to produce iodine ($I_2$), potassium chloride ($KCl$), and water ($H_2O$). The balanced reaction is:
$KClO_3 + 6KI + 6H^+ \rightarrow 3I_2 + 7KCl + 3H_2O$
Here, $I_2$ is liberated as a reddish-brown solution.

Key Concept: Melting point, Physical state of halogens

d) Larger atomic size and stronger van der Waals forces in iodine
[Solution Description]
The strength of van der Waals forces increases with increasing atomic size from bromine to iodine. Iodine being larger has stronger intermolecular forces, requiring more energy to overcome these forces, thus existing as a solid at room temperature compared to bromine which exists as a liquid.

Key Concept: Assertion-Reason

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The Assertion is correct because $XeF_6$ indeed reacts with water as per the reaction:
$XeF_6 + H_2O \rightarrow XeOF_4 + 2HF$
The Reason is also correct because the high electronegativity of fluorine makes $XeF_6$ susceptible to hydrolysis. However, the reason does not fully explain why $XeOF_4$ forms instead of other products like $XeO_3$. Therefore, both statements are true, but the Reason is not the correct explanation of the Assertion.

Key Concept: Oxidation States of Group 16 Elements

d) $+2$
[Solution Description]
In $OF_2$, fluorine is more electronegative than oxygen. Each fluorine atom has an oxidation state of $-1$. Since there are two fluorine atoms, the total oxidation contribution from fluorine is $-2$. Letting $x$ be the oxidation state of oxygen, we solve for $x + (-2) = 0$, giving $x = +2$.

Key Concept: Hybridisation and Structure

b) 93.6°
[Solution Description]
The bond angles for group 15 hydrides decrease down the group due to decreasing repulsion between bonding pairs as the size of the central atom increases. The given bond angles are:
$NH_3$ (107.8°),
$PH_3$ (93.6°),
$AsH_3$ (91.8°),
$SbH_3$ (91.3°),
$BiH_3$ (~90°).
Thus, the bond angle in $PH_3$ is 93.6°.

Key Concept: Acidic Oxides, Reactions with Water

d) $CaO$
[Solution Description]
To determine which oxide does not produce an acid when reacted with water, we analyze each option:
- $CO_2$ reacts with water to form carbonic acid ($H_2CO_3$).
- $SO_3$ reacts with water to form sulfuric acid ($H_2SO_4$).
- $N_2O_5$ reacts with water to form nitric acid ($HNO_3$).
- $CaO$ is a basic oxide and reacts with water to form calcium hydroxide ($Ca(OH)_2$), a base, not an acid.
Thus, the correct answer is $CaO$.

Key Concept: Assertion-Reason

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Nitrous oxide ($N_2O$) is classified as a neutral oxide because it does not react with acids or bases to form salts, which matches the assertion. The reason correctly explains why $N_2O$ is considered a neutral oxide, as its chemical behavior confirms its neutrality.

Key Concept: Physical Properties of HCl

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that dry hydrogen chloride gas does not turn dry blue litmus paper red, which is correct because it lacks free $H^+$ ions in the absence of water. The reason provided is also correct as dry HCl does not dissociate into $H^+$ and $Cl^-$ ions without water. Furthermore, the reason correctly explains the assertion since the absence of $H^+$ ions is why dry HCl cannot exhibit acidic properties like turning litmus paper red.

Key Concept: Distinctive Properties of Phosphorus

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that white phosphorus is stored under water to prevent its spontaneous combustion in air. This is true because white phosphorus is highly reactive and catches fire when exposed to air, forming phosphorus pentoxide ($P_4 + 5O_2 \rightarrow 2P_2O_5$). The reason explains that white phosphorus reacts with oxygen in the air to form phosphorus pentoxide, which is also true. Moreover, the reason correctly explains why white phosphorus needs to be stored under water—to cut off its contact with air and prevent this exothermic reaction. Hence, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Halogens occurrence

c) They are highly reactive.
[Solution Description]
Halogens are highly reactive due to their high electronegativity, so they form compounds (such as metal halides) rather than existing freely in nature.

Key Concept: Sulphur in Minerals

c) Gypsum
[Solution Description]
The question asks for the mineral containing sulfur with the formula $CaSO_4 \cdot 2H_2O$. From the syllabus, gypsum is mentioned to have this formula.

Key Concept: Oxidation States, Inert Pair Effect

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The inert pair effect refers to the tendency of the outermost s-electrons to remain non-bonding due to their poor shielding and high effective nuclear charge. In bismuth ($Bi$), the $6s^2$ electrons are less likely to be involved in bonding, making the +3 oxidation state more stable than the +5 state. Thus, both Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Physical Properties, Chemical Properties

c) 221 pm
[Solution Description]
The ionic radius increases down the group as the size of the atom increases. From the data provided, the ionic radii are: $O^{2-} = 140$ pm, $S^{2-} = 184$ pm, $Se^{2-} = 198$ pm, $Te^{2-} = 221$ pm, and $Po^{2-} = 230$ pm (approx.). Thus, the ionic radius of $Te^{2-}$ is 221 pm.

Key Concept: Oleum Formation, Concentration Control

c) To avoid hazardous mist formation and control dilution
[Solution Description]
Directly reacting $SO_3$ with water produces a fine mist of sulfuric acid, which is hazardous and difficult to handle. Dissolving $SO_3$ in conc. $H_2SO_4$ forms oleum ($H_2S_2O_7$), which can then be safely diluted to obtain sulfuric acid of desired concentration.

Key Concept: Laboratory preparation, stoichiometry

b) 4.48 L (correct stoichiometric volume)
[Solution Description]
The reaction is: $2NH_4Cl + Ca(OH)_2 \rightarrow 2NH_3 + CaCl_2 + 2H_2O$.
Moles of $NH_4Cl$ = mass/molar mass = $10.7 \, \text{g} / 53.5 \, \text{g/mol} = 0.2 \, \text{mol}$.
From stoichiometry, 2 mol $NH_4Cl$ produces 2 mol $NH_3$, so 0.2 mol $NH_4Cl$ yields 0.2 mol $NH_3$.
At STP, 1 mol gas occupies 22.4 L; thus, 0.2 mol occupies $0.2 \times 22.4 = 4.48 \, \text{L}$.

Key Concept: Uses of Chlorine

a) Sterilization of municipal water supply
[Solution Description]
Chlorine is widely used for sterilizing municipal water supplies, bleaching in the paper and textile industries, and manufacturing plastics like PVC, dyestuffs, and pesticides. Among the options provided, sterilization of water is one of its primary industrial applications.

Key Concept: Group 15 Elements

c) Arsenic
[Solution Description]
Arsenic ($As$) and antimony ($Sb$) are metalloids in Group 15. Among the options provided, only $As$ is listed as a metalloid.

Key Concept: Reaction of HCl with ammonia

c) Ammonium chloride
[Solution Description]
When hydrogen chloride gas reacts with ammonia gas, they combine to form dense white fumes of ammonium chloride. The chemical equation for this reaction is: $HCl(g) + NH_3(g) \rightarrow NH_4Cl(s)$. This is an example of a gas-phase reaction forming a solid compound.

Key Concept: Oxidation States, Halogens

c) $+9$
[Solution Description]
Chlorine exhibits $-1$, $+1$, $+3$, $+5$, and $+7$ oxidation states due to the availability of vacant d-orbitals that allow electron unpairing and promotion. However, chlorine cannot exhibit a $+9$ oxidation state because it lacks sufficient orbitals to accommodate such a high oxidation state.

Key Concept: Bond Energy, Stability of N$_2$

b) 313.8 kJ mol$^{-1}$
[Solution Description]
The N≡N bond consists of one sigma bond and two pi bonds. The total bond energy is 941.4 kJ mol$^{-1}$. Breaking just one bond in N$_2$ implies breaking one of the three bonds. Since all three bonds are not equivalent (one sigma and two pi bonds), the bond energy is not simply divided by 3. However, for approximation, we can consider the average bond energy per bond as $\frac{941.4}{3} = 313.8$ kJ mol$^{-1}$.

Key Concept: Oxygen isotopes, Molecular properties

d) Assertion is false, but Reason is true.
[Solution Description]
The Assertion states that $^{18}_{8}O$ is radioactive, which is false because $^{18}_{8}O$ is a stable isotope. The Reason states the natural abundance of $^{18}_{8}O$ as 0.204\%, which is true. Thus, the Assertion is false, but the Reason is true.

Key Concept: Oxidation State in Halogens

a) It has no vacant d-orbitals and high electronegativity
[Solution Description]
Fluorine is the most electronegative element and lacks vacant d-orbitals, preventing it from unpairing its electrons to show higher positive oxidation states. Hence, it always gains one electron to attain noble gas configuration, exhibiting only -1 oxidation state.

Key Concept: Haber’s Process

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that a low temperature is maintained in the Haber’s process to increase the yield of ammonia, which is true because the reaction $N_2 + 3H_2 \rightleftharpoons 2NH_3$ is exothermic (\$ \Delta H = -93.6 \text{ kJ mol}^{-1} \$). According to Le Chatelier’s principle, lowering the temperature favors the exothermic forward reaction, increasing ammonia production. The reason correctly explains why low temperature increases ammonia yield as it identifies the exothermic nature of the reaction. Hence, both Assertion and Reason are true, and Reason correctly explains Assertion.

Key Concept: Electronic Configuration, Catenation

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that oxygen shows limited catenation compared to sulphur because of its smaller atomic size and higher electronegativity. This is true because oxygen's small size leads to strong lone pair-lone pair repulsion, making $O—O$ bonds weaker and less stable for long chains.
The reason states that the $O—O$ bond energy is lower than the $S—S$ bond energy, which is also true. The bond energy of $O—O$ ($138 \, \text{kJ mol}^{-1}$) is indeed lower than that of $S—S$ ($266 \, \text{kJ mol}^{-1}$).
Moreover, the reason correctly explains the assertion since the lower bond energy of $O—O$ makes it less favorable for forming long chains (catenation), while the stronger $S—S$ bond allows sulphur to form extended structures like $S_8$ rings and polysulfides. Thus, both statements are correct, and the reason justifies the assertion.

Key Concept: Ammonia Manufacture and Uses, Nitric Acid Manufacture

c) Assertion is true, but Reason is false.
[Solution Description]
The assertion states that the manufacture of ammonia via Haber's process is carried out at high pressure to increase the yield of ammonia based on Le Chatelier's principle. This is true because the reaction
$N_2 + 3H_2 \rightleftharpoons 2NH_3$
involves a decrease in volume (4 moles of reactant gases produce 2 moles of product gas). Therefore, increasing the pressure shifts the equilibrium towards the side with fewer moles of gas (ammonia), increasing its yield.
However, the reason states that increasing the pressure shifts the equilibrium towards the side with fewer moles of gas for exothermic reactions. While the first part of the reason is correct (shifting towards fewer moles of gas), the reason incorrectly links this behavior only to exothermic reactions. Pressure changes affect all equilibrium reactions involving gases, regardless of whether they are exothermic or endothermic.
Thus, the Assertion is true, but the Reason is false because it inaccurately associates pressure effects exclusively with exothermic reactions.

Key Concept: Oxidation States

c) $+2$
[Solution Description]
Chlorine can exhibit various oxidation states like $-1$, $+1$, $+3$, $+5$, and $+7$. However, it does not show a $+2$ oxidation state in any of its compounds.

Key Concept: Occurrence of Sulphur

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Assertion states that sulphur occurs in both free and combined states, which is correct as per the syllabus. The Reason mentions that sulphur is found as sulphides and sulphates in the combined state, which is also accurate. Moreover, the Reason correctly explains why sulphur occurs in the combined state. Thus, both Assertion and Reason are true, and the Reason is the correct explanation of the Assertion.

Key Concept: Electron Gain Enthalpy, Atomic Radius

a) Fluorine’s smaller size causes higher electron-electron repulsion upon gaining an electron.
[Solution Description]
Despite fluorine's higher electronegativity, chlorine has a more negative electron gain enthalpy because the addition of an electron to the small fluorine atom leads to significant electron-electron repulsion in its compact 2p subshell. Chlorine, with a larger 3p subshell, accommodates the extra electron with less repulsion, resulting in a more favorable (more negative) electron gain enthalpy.

Key Concept: Isotopes of Oxygen

c) $^{18}_8O$
[Solution Description]
The syllabus states that naturally occurring oxygen has three isotopes: $^{16}_8O$ (99.76\%), $^{17}_8O$ (0.037\%), and $^{18}_8O$ (0.204\%). Among these, $^{18}_8O$ is radioactive. Thus, the correct answer is $^{18}_8O$.

Key Concept: Xenon Compounds, Hybridization

b) $sp^3d^2$, Square planar
[Solution Description]
In $XeF_4$, xenon has 8 valence electrons. It forms 4 bonds with fluorine atoms and retains 2 lone pairs. To accommodate these, xenon undergoes $sp^3d^2$ hybridization, resulting in an octahedral electron pair geometry. However, due to the presence of 2 lone pairs, the molecular geometry is square planar.

Key Concept: Physical Properties

a) Helium (He)
[Solution Description]
Ionization enthalpy decreases down the group for noble gases. From Table 7.25, Helium (He) has the highest ionization enthalpy of 2372 kJ mol$^{-1}$, followed by Neon (2080 kJ mol$^{-1}$), and so on. Thus, Helium has the highest ionization enthalpy.

Key Concept: General Characteristics of Interhalogen Compounds

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that interhalogen compounds are more reactive than individual halogens, which is true because their $X-Y$ bonds are weaker compared to $X-X$ or $Y-Y$ bonds in pure halogens. The reason correctly explains the assertion since weaker bonds make interhalogen compounds more reactive.

Key Concept: Nitric acid reactions

b) Nitrobenzene ($C_6H_5NO_2$)
[Solution Description]
Nitration of benzene in the presence of conc. $H_2SO_4$ at 323 K substitutes a hydrogen atom with a nitro ($-NO_2$) group, forming nitrobenzene ($C_6H_5NO_2$).

Key Concept:

d) $H_2O$
[Solution Description]
Thermal stability decreases down the group as the bond energy ($E_{M—H}$) decreases with increasing size of the central atom. Among the given options, $H_2O$ has the highest bond energy (463 kJ mol^{-1}) and thus the highest thermal stability.

Key Concept: Acidic and Basic Oxides

c) Assertion is true, but Reason is false.
[Solution Description]
Assertion (A) is true because $CO_2$ reacts with water to form carbonic acid ($H_2CO_3$), which confirms it as an acidic oxide. The Reason (R) states that all non-metal oxides are acidic, but this is not entirely accurate. While many non-metal oxides are acidic, there are exceptions like neutral oxides (e.g., $CO$, $N_2O$) which do not form acids or bases in water. Therefore, Assertion is true, but Reason is false.

Key Concept: Reducing nature, Acidic character

c) Decolourization of the solution
[Solution Description]
Sulphur dioxide acts as a reducing agent in acidic medium due to liberation of nascent hydrogen ($[H]$). It reduces $KMnO_4$ (purple) to $Mn^{2+}$ (colorless), decolorizing the solution. The reaction involved is:
$5SO_2 + 2KMnO_4 + 2H_2O \rightarrow K_2SO_4 + 2MnSO_4 + 2H_2SO_4$

Key Concept: Oxidation States in Group 16

b) Sulfur (S)
[Solution Description]
Among group 16 elements, sulfur ($S$) shows the most stable $+6$ oxidation state because it can promote its $ns^2$ electrons to vacant $d$-orbitals, forming compounds like $SF_6$. The stability decreases down the group due to the inert pair effect, making heavier elements like polonium ($Po$) prefer lower oxidation states.

Key Concept: Reactivity of Interhalogen Compounds

b) They have weaker bonds
[Solution Description]
Interhalogen compounds are more reactive than halogens because the bond between two different halogen atoms ($X-Y$) is weaker than the bond between two identical halogen atoms ($X-X$ or $Y-Y$).

Key Concept: Classification of Interhalogen Compounds

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that $ClF_3$ is an interhalogen compound of the type $XY_3$, which is true as per the classification given in the syllabus. The reason states that in $ClF_3$, chlorine is the central atom and it is larger in size than fluorine, which is also true. Moreover, the reason correctly explains why $ClF_3$ is classified as $XY_3$ because the larger halogen (chlorine) is the central atom and the smaller halogens (fluorine) are attached to it. Thus, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Group 16 Elements

d) Chlorine
[Solution Description]
The Group 16 elements, also known as chalcogens, include oxygen (O), sulfur (S), selenium (Se), tellurium (Te), and polonium (Po). Chlorine (Cl) is not a member of Group 16; it belongs to Group 17 (Halogens).

Key Concept: Chalcogens

c) First four elements of Group 16
[Solution Description]
Chalcogens are the first four elements of Group 16: oxygen ($O$), sulphur ($S$), selenium ($Se$), and tellurium ($Te$), named for their ore-forming properties.

Key Concept: Haber's Process, Le Chatelier's Principle

c) The yield of ammonia increases because higher pressure favors the product side.
[Solution Description]
According to Le Chatelier's principle, increasing the pressure favors the side with fewer moles of gas. The reaction is $N_2 + 3H_2 \rightleftharpoons 2NH_3$. On the left side, there are 4 moles of gas (1 mole of $N_2$ and 3 moles of $H_2$), while on the right side, there are 2 moles of $NH_3$. Increasing the pressure from 200 atm to 900 atm will shift the equilibrium towards the side with fewer moles of gas, thus increasing the yield of ammonia.

Key Concept: Hydrogen halides, Thermal stability

d) Due to decreasing hydrogen-halogen bond energy down the group.
[Solution Description]
The thermal stability of hydrogen halides depends on bond dissociation energy and bond length. From HF to HI:
1. Bond length increases as atomic size increases (H-F < H-Cl < H-Br HCl (432 \, \text{kJ/mol}) > HBr (366 \, \text{kJ/mol}) > HI (298 \, \text{kJ/mol})$.
3. Higher bond dissociation energy means higher thermal stability, so HF is the most stable and HI the least.

Key Concept: Uses of Chlorine

a) Sterilization of municipal supply water
[Solution Description]
Chlorine is commonly used for the sterilization of municipal water supplies to kill harmful bacteria and pathogens. It ensures safe drinking water.

Key Concept: Chemical Properties of Ammonia

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that ammonia reacts with copper oxide to produce nitrogen gas and copper, which is true based on the chemical reaction: $3CuO + 2NH_3 \rightarrow 3Cu + N_2 + 3H_2O$.
The reason states that ammonia acts as a reducing agent in this reaction, which is correct because ammonia donates hydrogen to copper oxide, reducing it to copper.
Both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Electronic Configuration and Allotropy

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Oxygen has the electronic configuration $1s^2 2s^2 2p^4$, with two unpaired electrons in the p-subshell. These unpaired electrons allow oxygen to form multiple bonds ($p\pi-p\pi$ bonding), resulting in different allotropic forms like dioxygen ($O_2$) and ozone ($O_3$). Thus, the assertion is true. The reason correctly explains the assertion because the unpaired electrons in the p-subshell facilitate the formation of multiple bonds, which is the underlying cause for allotropy in oxygen. Therefore, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Thermodynamics, Preparation of Hydrides

c) Yield of $NH_3$ increases
[Solution Description]
According to Le Chatelier's principle, increasing the pressure shifts the equilibrium in the direction that reduces the number of moles of gas. The reaction for ammonia synthesis is:
$N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$
Here, 4 moles of reactant gases produce 2 moles of product gas. Thus, increasing the pressure favors the forward reaction, increasing the yield of $NH_3$.

Key Concept: Sulphuric Acid Applications

b) It dissolves sulfur trioxide to form oleum
[Solution Description]
Oleum ($H_2S_2O_7$) is formed when sulphur trioxide ($SO_3$) dissolves in sulphuric acid ($H_2SO_4$), represented by:
$H_2SO_4 + SO_3 \rightarrow H_2S_2O_7$
This reaction enhances the sulfur trioxide concentration for industrial applications.

Key Concept: Electronic Configuration, Physical Properties of Noble Gases

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the boiling point of noble gases increases down the group, which is true as seen in Table 7.25 (boiling points: He = 4.2 K, Ne = 27.1 K, Ar = 87.2 K, Kr = 119.7 K, Xe = 165.0 K, Rn = 211 K). The reason explains this trend by stating that van der Waals' forces increase with atomic size, making it harder to separate the atoms and thus increasing the boiling point. This explanation is correct because larger atoms have more delocalized electron clouds, leading to stronger instantaneous dipole-induced dipole interactions (van der Waals' forces), which require more energy to overcome.

Key Concept: Interhalogen Compounds

d) Iodine monochloride ($ICl$)
[Solution Description]
Interhalogen compounds have different geometries based on their hybridization states. For example:
- $ClF_3$ ($sp^3d$) has a T-shaped geometry.
- $BrF_5$ ($sp^3d^2$) has a square pyramidal geometry.
- $IF_7$ ($sp^3d^3$) has a pentagonal bipyramidal geometry.
- $ICl$ ($sp^3d$) forms a linear molecule.

Key Concept: Catalytic Oxidation of Ammonia

c) Assertion is true, but Reason is false.
[Solution Description]
The Assertion states that the oxidation of ammonia in the Ostwald process requires a platinum catalyst, which is true as per the syllabus. The Reason claims that platinum increases the activation energy, which is false. In reality, catalysts like platinum lower the activation energy, not increase it, to make the reaction feasible at lower temperatures. Thus, Assertion is true, but Reason is false.

Key Concept: Ionisation Energy

a) Fluorine
[Solution Description]
Ionisation energy decreases down the halogen group as atomic size increases, reducing the effective nuclear charge on valence electrons. Fluorine (\$F\$) has the smallest atomic radius and highest ionisation energy among halogens.

Key Concept: Acidity and Stability of Oxoacids of Phosphorus

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Hypophosphorus acid ($H_3PO_2$) has only one ionizable $P-OH$ group, making it monobasic. The reason correctly explains that the basicity of oxoacids of phosphorus depends on the number of ionizable hydrogen atoms attached to oxygen ($P-OH$ groups). Hence, both Assertion and Reason are true, and Reason is the correct explanation of the Assertion.

Key Concept: Molecular Properties

d) By the action of ultraviolet rays on $O_2$
[Solution Description]
Ozone is formed in the upper atmosphere when ultraviolet rays act on dioxygen molecules ($O_2$), splitting them into oxygen atoms, which then react with another $O_2$ molecule to form $O_3$.

Key Concept: Oxidation State

a) +1
[Solution Description]
The formula for nitrous oxide is $N_2O$. Let the oxidation state of nitrogen be $x$. Oxygen typically has an oxidation state of $-2$. Therefore, solving $2x + (-2) = 0$ gives $x = +1$.

Key Concept: Acidic Oxides, Basic Oxides

c) $CaO$
[Solution Description]
The oxide that reacts with water to form a basic solution (turns red litmus blue) must be a basic oxide. Basic oxides are typically formed by highly electropositive metals. Among the given options, $CaO$ (calcium oxide) is a basic oxide and reacts with water to form $Ca(OH)_2$, which is a base.

Key Concept: Ionisation Energy, Electronegativity

c) Fluorine has a smaller atomic radius and higher effective nuclear charge.
[Solution Description]
Ionisation energy decreases down the halogen group due to increasing atomic size. Fluorine, being the smallest atom in the group, has its valence electrons closest to the nucleus, experiencing the strongest effective nuclear charge. This makes it difficult to remove an electron, resulting in high ionisation energy. Iodine, being larger, has valence electrons farther from the nucleus, experiencing weaker attraction, leading to lower ionisation energy.

Key Concept: Density at STP

b) He < Ne < Ar < Xe
[Solution Description]
The densities of noble gases at STP increase down the group. Given data: He ($1.8 \times 10^{-4} \, \text{g cm}^{-3}$), Ne ($9.0 \times 10^{-4} \, \text{g cm}^{-3}$), Ar ($1.8 \times 10^{-3} \, \text{g cm}^{-3}$), Xe ($5.9 \times 10^{-3} \, \text{g cm}^{-3}$). The correct order is He < Ne < Ar < Xe.

Key Concept: Nitrogen Forms in Ammonium Nitrate

c) Half as ammonium and half as nitrate nitrogen
[Solution Description]
The syllabus explains that ammonium nitrate provides a balanced supply of nitrogen, with approximately half in the ammonium form and half in the nitrate form. Thus, each form contributes roughly equal amounts.

Key Concept: Contact Process - Oxidation of SO$_2$

c) Vanadium pentoxide ($V_2O_5$)
[Solution Description]
The Contact process involves the oxidation of sulfur dioxide ($SO_2$) to sulfur trioxide ($SO_3$) using vanadium pentoxide ($V_2O_5$) as a catalyst. This reaction occurs in the contact tower at around 725 K.

Key Concept: Occurrence of Oxygen

c) 88.8\%
[Solution Description]
The question asks for the percentage of oxygen by weight in water. According to the syllabus, water consists of 88.8\% oxygen by weight. Therefore, the correct answer is 88.8\%.

Key Concept: Bleaching Powder Preparation

b) $2Ca(OH)_2 + 2Cl_2 \rightarrow Ca(OCl)_2 + CaCl_2 + 2H_2O$
[Solution Description]
The reaction for the preparation of bleaching powder involves chlorine gas reacting with slaked lime (calcium hydroxide). The correct balanced equation is:
$2Ca(OH)_2 + 2Cl_2 \rightarrow Ca(OCl)_2 + CaCl_2 + 2H_2O$
This shows the formation of calcium hypochlorite (bleaching powder), calcium chloride, and water.

Key Concept: Combustion, Oxidation

c) Nitric oxide ($NO$)
[Solution Description] When a mixture of ammonia and oxygen is passed over a platinum catalyst at 1073 K, ammonia is oxidized to nitric oxide (NO) and water. The balanced chemical equation for this reaction is:
$4NH_3 + 5O_2 \xrightarrow{Pt, 1073 K} 4NO + 6H_2O$

Key Concept: Nitrogen Hydrides

c) Haber's process.
[Solution Description]
Ammonia is prepared on an industrial scale using the Haber's process. In this method, a mixture of nitrogen and hydrogen gases is passed over a finely divided iron catalyst mixed with molybdenum at high temperature (720-770 K) and pressure (200 atmospheres). The reaction occurs as follows:
$N_2(g) + 3H_2(g) \xrightarrow{\text{Catalyst}} 2NH_3(g)$

Key Concept: Catalyst in Haber’s Process

b) Potassium oxide ($K_2O$)
[Solution Description]
In the Haber’s process, finely divided iron is the main catalyst, while molybdenum or $K_2O$ and $Al_2O_3$ act as promoters to increase its efficiency.
Hence, potassium oxide ($K_2O$) is one such promoter.

Key Concept: Haber's Process

a) High pressure
[Solution Description]
The Haber’s process involves the reaction $N_2 + 3H_2 \rightleftharpoons 2NH_3$, which is reversible and exothermic. To maximize yield:
1) Low temperature favors the forward reaction (Le Chatelier’s principle for exothermic reactions).
2) High pressure promotes the forward reaction as fewer moles of gas are formed.
3) A catalyst speeds up the reaction without affecting equilibrium.
Thus, high pressure is a favorable condition.

Key Concept: Assertion-Reason

c) Assertion is true, but Reason is false.
[Solution Description]
The assertion states that sulphur is found in the native state in volcanic regions, which is true as per the syllabus (e.g., Sicily, Italy, Japan). However, the reason claims sulphur is highly reactive and does not occur freely, which contradicts the syllabus as sulphur does occur in the native state. Thus, the assertion is true, but the reason is false.

Key Concept: Superphosphate Production, Phosphorus Extraction

c) 340 kg
[Solution Description]
First reaction (extraction): 2Ca$_3$(PO$_4$)$_2$ + 6SiO$_2$ + 10C → 6CaSiO$_3$ + P$_4$ + 10CO. Second reaction (superphosphate): Ca$_3$(PO$_4$)$_2$ + 2H$_2$SO$_4$ → Ca(H$_2$PO$_4$)$_2$ + 2CaSO$_4$. Pure Ca$_3$(PO$_4$)$_2$ = 620 × 0.85 = 527 kg. Molar mass Ca$_3$(PO$_4$)$_2$ = 310 g/mol. Moles = 527,000/310 = 1700 moles. According to second equation, each mole requires 2 moles H$_2$SO$_4$. So need 3400 moles H$_2$SO$_4$. Molar mass H$_2$SO$_4$ = 98 g/mol. Pure H$_2$SO$_4$ needed = 3400 × 98 = 333,200 g = 333.2 kg. With 98\% purity, actual needed = 333.2/0.98 = 340 kg.

Key Concept: Hydrolysis of Interhalogen Compounds

c) $ICl$, $HIO_2$, and $HCl$
[Solution Description]
The hydrolysis reaction of $ICl_3$ is given by:
$2ICl_3 + 3H_2O \rightarrow ICl + HIO_2 + 5HCl$
Therefore, the products are $ICl$, $HIO_2$, and $HCl$.

Key Concept: Oxidizing property, Perhalates

b) Perbromate ($BrO_4^-$) due to the weakest Br—O bond
[Solution Description]
The oxidizing power of perhalates follows the order:
$ClO_4^- IO_4^-$
Perbromates ($BrO_4^-$) are the strongest oxidizing agents among all perhalates due to the weaker Br—O bond compared to Cl—O and I—O bonds, making it easier for $BrO_4^-$ to accept electrons.

Key Concept: Reactivity and hydrolysis of interhalogen compounds

c) $ICl$, $HIO_3$, and $HCl$.
[Solution Description]
$ICl_3$ undergoes hydrolysis with water to form $ICl$, $HIO_3$, and $HCl$. The reaction can be represented as:
$2ICl_3 + 3H_2O \rightarrow ICl + HIO_3 + 5HCl$

Key Concept: Reducing nature of SO_2

c) Liberates nascent hydrogen
[Solution Description]
In the presence of moisture, sulphur dioxide liberates nascent hydrogen which enables it to act as a reducing agent. The reaction involved is: $SO_2 + 2H_2O \rightarrow H_2SO_4 + 2[H]$. The liberated nascent hydrogen is responsible for the reducing property of sulphur dioxide.

Key Concept: Oxidation States of Nitrogen

c) $+1$
[Solution Description]
In $N_2O$, the oxidation state of oxygen is $-2$. Let the oxidation state of nitrogen be $x$. The equation becomes: $2x + (-2) = 0$. Solving gives $x = +1$.

Key Concept: Sulfur in Sulfuric Acid Production - Contact Process

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that vanadium pentoxide is preferred over platinised asbestos as a catalyst in the Contact process. This is true because vanadium pentoxide is more cost-effective and efficient in industrial settings. The reason provided explains why vanadium pentoxide is preferred: it is less prone to poisoning (which can deactivate the catalyst) and operates effectively within the optimal temperature range (670–720 K). Since both statements are correct and the reason logically justifies the assertion, they are directly related.

Key Concept: Properties of Nitric Acid, Oxidizing Nature

d) Nitrogen dioxide ($NO_2$)
[Solution Description]
Concentrated nitric acid ($HNO_3$) oxidizes copper ($Cu$) to copper(II) nitrate ($Cu(NO_3)_2$) and produces nitrogen dioxide ($NO_2$) gas. The reaction is highly vigorous and exothermic.
Step-by-step explanation:
1. The balanced chemical equation is: $Cu + 4HNO_3 \rightarrow Cu(NO_3)_2 + 2NO_2 + 2H_2O$.
2. Nitrogen dioxide ($NO_2$) is a reddish-brown gas with a pungent odor.
3. This reaction demonstrates the strong oxidizing nature of concentrated nitric acid.

Key Concept: Occurrence in Nature, Combined State Forms

c) Calcium carbonate ($CaCO_3$)
[Solution Description]
From the syllabus, sulphur occurs in combined states as sulphides (e.g., copper pyrites, iron pyrites), sulphates (e.g., gypsum, barytes), and organic substances (e.g., proteins, garlic). However, $CaCO_3$ (calcium carbonate) is not listed as a compound containing sulphur in any form.

Key Concept: Chlorine preparation, stoichiometry

b) 3.36 L
[Solution Description]
The balanced equation is:
$2KMnO_4 + 16HCl \rightarrow 2KCl + 2MnCl_2 + 8H_2O + 5Cl_2$.
Moles of HCl = $6 \, \text{M} \times 0.1 \, \text{L}$ = 0.6 mol.
From the equation, 16 moles of HCl produce 5 moles of $Cl_2$.
Moles of $Cl_2$ expected = $\frac{5}{16} \times 0.6$ = 0.1875 mol.
Considering 80\% yield: Actual moles = $0.1875 \times 0.8$ = 0.15 mol.
Volume at STP = $0.15 \, \text{mol} \times 22.4 \, \text{L/mol}$ = 3.36 L.

Key Concept: Preparation

c) $2Pb(NO_3)_2 \xrightarrow{673 K} 2PbO + 4NO_2 + O_2$
[Solution Description]
From the preparation methods listed, nitrogen dioxide ($NO_2$) is prepared by heating lead nitrate ($Pb(NO_3)_2$) at 673 K, which decomposes to form $PbO$, $NO_2$, and $O_2$. The other reactions produce different nitrogen oxides like nitrous oxide ($N_2O$) or nitric oxide ($NO$).

Key Concept: Uses of Noble Gases

c) Helium (He)
[Solution Description]
Helium is used in deep-sea diving gas mixtures because it is less soluble in blood than nitrogen, reducing the risk of decompression sickness (the bends). This property makes it safer for divers at high pressures.

Key Concept: Uses of Noble Gases

d) Mixing with oxygen for respiration
[Solution Description]
Helium is mixed with oxygen for respiration in deep-sea diving because it is less soluble in blood than nitrogen under high pressure.

Key Concept: Oxidation States in Group 16 Elements

c) $OF_2$
[Solution Description]
Oxygen usually shows -2 oxidation state, but in $OF_2$, since fluorine is more electronegative, the shared electrons are counted with fluorine, resulting in oxygen exhibiting a +2 oxidation state.

Key Concept: Chemical Properties

c) Magnesium (Mg)
[Solution Description]
Magnesium is a highly reactive metal that continues to burn in sulphur dioxide gas. The reaction produces magnesium oxide and sulphur:
$2Mg + SO_2 \rightarrow 2MgO + S$
This demonstrates that magnesium can reduce sulphur dioxide to sulphur while itself getting oxidized to magnesium oxide. Other metals like potassium also show similar behavior.

Key Concept: Amphoteric Oxides

b) $Al_2O_3$
[Solution Description]
Amphoteric oxides exhibit both acidic and basic properties and react with both acids and bases. $Al_2O_3$ is amphoteric as it reacts with $HCl$ to form $AlCl_3$ and with $NaOH$ to form $Na_3[Al(OH)_6]$.

Key Concept: Contact Process - Conditions for Maximum Yield

c) Assertion is true, but Reason is false.
[Solution Description]
The assertion states that the Contact Process favors a low temperature (670–720 K), which is correct as per Le-Chatelier’s principle for this exothermic reaction. However, the reason incorrectly claims that lower temperatures increase the reaction rate. In reality, while lower temperatures favor equilibrium yield (due to exothermicity), they decrease the reaction rate. Thus, the assertion is true, but the reason is false.

Key Concept: Reducing properties, Reaction with acids

b) Phosphoric acid and sulphur dioxide.
[Solution Description]
Phosphorus acts as a reducing agent when reacting with concentrated sulphuric acid, leading to the formation of phosphoric acid and sulphur dioxide:
$P_4 + 10H_2SO_4 \rightarrow 4H_3PO_4 + 10SO_2 + 4H_2O$
The key products are phosphoric acid ($H_3PO_4$) and sulphur dioxide ($SO_2$).

Key Concept: Acidic Character

d) $H_2Te$
[Solution Description]
The acidity of the hydrides increases down the group due to weaker $M-H$ bonds and easier dissociation. The order is: $H_2O < H_2S < H_2Se < H_2Te$. Therefore, $H_2Te$ is the most acidic among them.

Key Concept: Uses of Chlorine

c) Extraction of aluminum from bauxite
[Solution Description]
Chlorine is widely used for sterilizing water, bleaching textiles, and manufacturing PVC, dyes, and explosives. It is also used to produce refrigerants like freon ($CCl_2F_2$) and pesticides such as DDT. However, chlorine is not used in the extraction of aluminum, which is primarily extracted from bauxite via the Hall-Héroult process.

Key Concept: Nitric Acid, Chemical Properties

c) Oxalic acid
[Solution Description]
Concentrated nitric acid acts as an oxidizing agent. When cane sugar ($C_{12}H_{22}O_{11}$) reacts with it, it oxidizes to oxalic acid ($(COOH)_2$). The balanced reaction is:
$C_{12}H_{22}O_{11} + 36HNO_3 \rightarrow 6(COOH)_2 + 36NO_2 + 23H_2O$
Here, the carbon in cane sugar is oxidized to oxalic acid, while nitrogen in $HNO_3$ is reduced to $NO_2$.

Key Concept: Oxidation State, Bonding

b) +7, $sp^3d^3$
[Solution Description]
In $IF_7$ (an $XY_7$ type interhalogen compound), iodine is the central atom bonded to 7 fluorine atoms. Each F has an oxidation state of $-1$, so the oxidation state of I is $+7$. The hybridization of iodine is $sp^3d^3$ to accommodate 7 bond pairs, giving a pentagonal bipyramidal structure.

Key Concept: Occurrence of Nitrogen

c) Nitrogen
[Solution Description]
The syllabus mentions that nitrogen occurs as a diatomic gas ($N_2$) and constitutes about 78\% by volume of the atmospheric air.

Key Concept: Reaction with Ammonia

b) Ammonium chloride
[Solution Description]
Hydrogen chloride gas (\$HCl\$) reacts with ammonia gas (\$NH_3\$) to form dense white fumes of ammonium chloride (\$NH_4Cl\$). The reaction is:
$HCl(g) + NH_3(g) \rightarrow NH_4Cl(s)$

Key Concept: Assertion-Reason

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Assertion states that the atomic radii of noble gases increase down the group from Helium (He) to Xenon (Xe). This is true as seen in Table 7.25, where the atomic radius increases from He (120 pm) to Xe (220 pm). The Reason explains that this increase is due to the addition of a new electronic shell at each succeeding element, which is also correct and provides the correct explanation for the Assertion. Thus, both Assertion and Reason are true, and Reason correctly explains the Assertion.

Key Concept: Assertion-Reason

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that sulphuric acid is used to manufacture fertilisers such as ammonium sulphate and super phosphate of lime, which is true as per the syllabus. The reason states that sulphuric acid reacts with ammonia to form ammonium sulphate, which is also correct and directly explains why sulphuric acid is used in fertiliser production. Hence, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Combustibility of HCl gas, Thermal dissociation

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The Assertion states that dry HCl gas does not support combustion because it is non-combustible. This is true as per the syllabus which mentions HCl extinguishes a burning splint. The Reason states that dry HCl dissociates into hydrogen and chlorine gases at high temperatures, which is also true as given in the syllabus ($2HCl \xrightarrow{>500^\circ C} H_2 + Cl_2$). However, the thermal dissociation property is unrelated to its non-combustibility or inability to support combustion. Hence, while both statements are true, the Reason does not explain the Assertion.

Key Concept: Uses of Nitric Acid

c) Galvanizing iron sheets
[Solution Description]
Nitric acid is widely used in manufacturing explosives (\textit{e.g., T.N.T., nitroglycerine}), fertilizers (\textit{e.g., ammonium nitrate}), artificial silk, dyes, drugs, and as an oxidizer in rocket fuels. However, it is not used for galvanizing metals, which involves coating them with zinc.

Key Concept: Electronic Configuration of Noble Gases

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that noble gases have a stable closed-shell electronic configuration, which is true as they have completely filled outer orbitals (either $1s^2$ for helium or $ns^2np^6$ for other noble gases). The reason correctly explains the assertion by stating that the completely filled outermost shell makes them chemically inert due to no tendency to gain or lose electrons. Hence, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Reaction with ammonia

b) Dense white fumes are produced
[Solution Description]
When hydrogen chloride gas ($HCl(g)$) reacts with ammonia gas ($NH_3(g)$), they combine to form dense white fumes of ammonium chloride ($NH_4Cl(s)$). The reaction is given by:
$HCl(g) + NH_3(g) \rightarrow NH_4Cl(s)$
This phenomenon is due to the formation of a solid product from two gases.

Key Concept: Occurrence of Sulphur

c) Both in free state and combined state (as sulphides and sulphates)
[Solution Description]
Sulphur occurs both in the native (free) state and combined state. In the combined state, it is found as sulphides (e.g., $Cu_2S$, $FeS_2$), sulphates (e.g., $CaSO_4 \cdot 2H_2O$, $BaSO_4$), and traces of hydrogen sulphide in volcanic gases.

Key Concept: Thermal Dissociation

b) It decomposes into hydrogen and chlorine gas.
[Solution Description]
When hydrogen chloride gas ($HCl$) is heated above 500$^\circ$C, it undergoes thermal dissociation to form hydrogen gas ($H_2$) and chlorine gas ($Cl_2$). The reaction can be represented as:
$2HCl(g) \xrightarrow{>500^\circ C} H_2(g) + Cl_2(g)$
This shows that two molecules of $HCl$ decompose into one molecule of $H_2$ and one molecule of $Cl_2$.

Key Concept: Ammonia Manufacture

c) 720–770 K
[Solution Description]
The Haber process optimally operates between 720–770 K as per Le Chatelier's principle, favoring lower temperatures for the exothermic reaction $N_2 + 3H_2 \rightleftharpoons 2NH_3$.

Key Concept: Chalcogens

d) They form large quantities of metal ores.
[Solution Description]
Chalcogens are called ore-forming elements since many metal ores like oxides and sulphides occur naturally. This is the reason behind their name.

Key Concept: Physical Properties

c) It is very soluble in water, forming sulphurous acid ($H_2SO_3$).
[Solution Description]
Sulphur dioxide has several physical properties. It is a colourless gas with a pungent and suffocating smell, 2.2 times heavier than air, very soluble in water forming sulphurous acid ($H_2SO_3$), and can be liquefied at room temperature under pressure.

Key Concept: Thermal Stability and Reducing Character

a) HF
[Solution Description]
The thermal stability of hydrogen halides decreases from $HF$ to $HI$ ($HF > HCl > HBr > HI$) due to the increasing bond length and decreasing bond strength. $HF$ has the highest bond dissociation energy, making it the most thermally stable.

Key Concept: Laboratory preparation, Ammonium salts, Alkali reactions

c) Hydrolysis of $Mg_3N_2$
[Solution Description]
The hydrolysis of magnesium nitride ($Mg_3N_2$) produces ammonia, represented by the equation:
$Mg_3N_2 + 6H_2O \rightarrow 3Mg(OH)_2 + 2NH_3$.
Magnesium nitride is formed by heating magnesium metal in nitrogen gas, making this method unique as it involves the direct combination of nitrogen with magnesium. Other methods involve ammonium salts reacting with alkalis.