CBSE Class 12 Chemistry The p Block Elements Assignment Set 03

Question. What happens when (i) PCl₅ is heated? (ii) H3PO₃ is heated? Write the reactions involved.
Answer:
(i) When \( \text{PCl}_5 \) is warmed, it decomposes into phosphorus trichloride and chlorine gas:
\[ \text{PCl}_5 \xrightarrow{\Delta} \text{PCl}_3 + \text{Cl}_2 \]
(ii) Heating orthophosphorous acid causes it to undergo a disproportionation reaction, yielding orthophosphoric acid and phosphine gas:
\[ 4\text{H}_3\text{PO}_3 \xrightarrow{\Delta} 3\text{H}_3\text{PO}_4 + \text{PH}_3 \]
In simple words: Heating phosphorus pentachloride splits it into phosphorus trichloride and chlorine gas. Heating phosphorous acid turns it into two different acids and a gas.

Exam Tip: Make sure to balance the disproportionation reaction of phosphorous acid correctly, as it is frequently asked in examinations.

Question. How are interhalogen compounds formed? What general compositions can be assigned to them?
Answer: Interhalogen molecules are created when distinct halogen elements directly combine or react under specific conditions. They have the general formula \( \text{XX}'_n \), where \( \text{X} \) represents the larger, less electronegative halogen and \( \text{X}' \) stands for the smaller, more electronegative halogen. The value of \( n \) can be 1, 3, 5, or 7. These compounds belong to four main categories:
(i) \( \text{XX}' \) (e.g., \( \text{ClF}, \text{BrF}, \text{IF}, \text{BrCl}, \text{ICl}, \text{IBr} \))
(ii) \( \text{XX}'_3 \) (e.g., \( \text{ClF}_3, \text{BrF}_3, \text{IF}_3, \text{ICl}_3 \))
(iii) \( \text{XX}'_5 \) (e.g., \( \text{ClF}_5, \text{BrF}_5, \text{IF}_5 \))
(iv) \( \text{XX}'_7 \) (e.g., \( \text{IF}_7 \))
In simple words: Interhalogen compounds are made when two different halogens join together. The larger one is always written first, and they can pair up in four different ratios.

Exam Tip: Remember that the central halogen (X) is always larger and less electronegative, while the surrounding halogens (X') are smaller and more electronegative.

Question. Explain the following: (a) NO₂ readily forms a dimer. (b) BiCl₃ is more stable than BiCl₅.
Answer:
(a) Nitrogen dioxide (\( \text{NO}_2 \)) has an odd number of valence electrons (23 electrons), possessing a single unpaired electron. To become stable, two \( \text{NO}_2 \) molecules easily combine to form a stable dimer, dinitrogen tetroxide (\( \text{N}_2\text{O}_4 \)), which has paired electrons.
(b) Bismuth trichloride (\( \text{BiCl}_3 \)) is far more stable than bismuth pentachloride (\( \text{BiCl}_5 \)) because of the inert pair effect. As we descend Group 15, the outer s-electrons become increasingly reluctant to participate in bonding, making the +3 state highly stable and the +5 oxidation state unstable.
In simple words: NO2 has an odd electron that wants to pair up, so two molecules join to form stable N2O4. For bismuth, the inert pair effect makes the +3 state much more stable than +5.

Exam Tip: Mentioning the "inert pair effect" is mandatory to get full marks for bismuth stability questions.

Question. Complete the following chemical equations: (i) Ca₃P₂ + H₂O -> (ii) Cu + H₂SO₄ (conc.) ->
Answer:
(i) Calcium phosphide reacts with water to produce phosphine gas and calcium hydroxide:
\[ \text{Ca}_3\text{P}_2 + 6\text{H}_2\text{O} \rightarrow 2\text{PH}_3 + 3\text{Ca(OH)}_2 \]
(ii) Hot, concentrated sulfuric acid oxidizes copper metal to form copper sulfate, sulfur dioxide gas, and water:
\[ \text{Cu} + 2\text{H}_2\text{SO}_4 \, (\text{conc.}) \rightarrow \text{CuSO}_4 + \text{SO}_2 + 2\text{H}_2\text{O} \]
In simple words: Mixing calcium phosphide with water makes smelly phosphine gas and calcium hydroxide. Copper and concentrated sulfuric acid react to give copper sulfate, choking sulfur dioxide gas, and water.

Exam Tip: Ensure all equations are balanced, especially the stoichiometric coefficients of water and phosphine.

Question. Arrange the following in the order of property indicated against each set: (i) HF, HCl, HBr, HI - increasing bond dissociation enthalpy. (ii) H₂O, H₂S, H₂Se, H₂Te - increasing acidic character.
Answer:
(i) Increasing bond dissociation enthalpy order:
\[ \text{HI} < \text{HBr} < \text{HCl} < \text{HF} \]
As the halogen atom size increases, the H-X bond length increases, which lowers the energy required to break the bond.
(ii) Increasing acidic character order:
\[ \text{H}_2\text{O} < \text{H}_2\text{S} < \text{H}_2\text{Se} < \text{H}_2\text{Te} \]
Descending the group, the bond dissociation enthalpy of the H-E bond decreases, allowing hydrogen ions to release more readily.
In simple words: HI has a weak, long bond that is easy to break compared to HF, which has a very short and strong bond. Acid strength increases down the hydrides because the bond holding the hydrogen gets weaker.

Exam Tip: Keep in mind that bond dissociation energy of hydrogen halides has the exact opposite trend of their acidic strength.

Question. Complete the following equations: (i) P₄ + H₂O -> (ii) XeF₄ + O₂F₂ ->
Answer:
(i) Phosphorus reacts with water at high temperatures or in alkaline media to form phosphine gas and phosphorous acid:
\[ \text{P}_4 + 6\text{H}_2\text{O} \rightarrow 2\text{PH}_3 + 2\text{H}_3\text{PO}_3 \]
(ii) Xenon tetrafluoride reacts with dioxygen difluoride at \( 143\text{ K} \) to yield xenon hexafluoride and oxygen:
\[ \text{XeF}_4 + \text{O}_2\text{F}_2 \xrightarrow{143\text{ K}} \text{XeF}_6 + \text{O}_2 \]
In simple words: Phosphorus reacts with water to give phosphine gas and phosphorous acid. Xenon tetrafluoride absorbs fluorine from O2F2 to form xenon hexafluoride.

Exam Tip: Specify the reaction temperature of \( 143\text{ K} \) above the reaction arrow to demonstrate complete understanding.

Question. Complete the following equations: (i) Ag + PCl₅ -> (ii) CaF₂ + H₂SO₄ ->
Answer:
(i) Silver metal reacts with phosphorus pentachloride upon heating to yield silver chloride and phosphorus trichloride:
\[ 2\text{Ag} + \text{PCl}_5 \rightarrow 2\text{AgCl} + \text{PCl}_3 \]
(ii) Calcium fluoride reacts with concentrated sulfuric acid to generate calcium sulfate and hydrogen fluoride gas:
\[ \text{CaF}_2 + \text{H}_2\text{SO}_4 \rightarrow \text{CaSO}_4 + 2\text{HF} \]
In simple words: Heated silver reacts with PCl5 to make solid silver chloride and phosphorus trichloride. Calcium fluoride and acid make calcium sulfate and hydrofluoric acid gas.

Exam Tip: Be sure to balance the silver chloride equation correctly, as it involves a two-electron redox transfer.

Question. Complete the following equations: (i) C + conc. H₂SO₄ -> (ii) XeF₂ + H₂O ->
Answer:
(i) Carbon reacts with hot, concentrated sulfuric acid to yield carbon dioxide, sulfur dioxide, and water:
\[ \text{C} + 2\text{H}_2\text{SO}_4 \, (\text{conc.}) \rightarrow \text{CO}_2 + 2\text{SO}_2 + 2\text{H}_2\text{O} \]
(ii) Xenon difluoride undergoes rapid hydrolysis with water, decomposing to form xenon gas, hydrogen fluoride, and oxygen gas:
\[ 2\text{XeF}_2 + 2\text{H}_2\text{O} \rightarrow 2\text{Xe} + 4\text{HF} + \text{O}_2 \]
In simple words: Carbon gets oxidized by heavy acid to form carbon dioxide, sulfur dioxide, and water. XeF2 breaks down in water into xenon gas, HF acid, and oxygen gas.

Exam Tip: Pay special attention to the stoichiometry of water and HF in the hydrolysis of xenon difluoride.

Question. Draw the structure of each of the following: (i) H₂SO₄ (ii) Solid PCl₅
Answer:
(i) Sulfuric acid (\( \text{H}_2\text{SO}_4 \)) structure consists of a tetrahedral layout around sulfur with two double-bonded oxygen atoms and two hydroxyl groups.
(ii) In the solid state, \( \text{PCl}_5 \) exists as an ionic crystal lattice composed of tetrahedral tetrachlorophosphonium cations, \( [\text{PCl}_4]^+ \), and octahedral hexachlorophosphate anions, \( [\text{PCl}_6]^- \).
In simple words: H2SO4 has a tetrahedral shape with two O groups and two OH groups. Solid PCl5 is not a single molecule; it exists as positive tetrahedral ions and negative octahedral ions packed together.

Exam Tip: Clearly state that solid PCl5 is ionic and describe the respective geometries of both the cation and the anion to score full marks..

Question. Complete the following chemical equations: (i) PCl₅ --Heat--> (ii) NaHCO₃ + HCl ->
Answer:
(i) Heating phosphorus pentachloride results in thermal decomposition to form phosphorus trichloride and chlorine gas:
\[ \text{PCl}_5 \xrightarrow{\Delta} \text{PCl}_3 + \text{Cl}_2 \]
(ii) Sodium bicarbonate reacts with hydrochloric acid to produce sodium chloride, water, and carbon dioxide gas:
\[ \text{NaHCO}_3 + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} + \text{CO}_2 \]
In simple words: Heating PCl5 breaks it into phosphorus trichloride and chlorine gas. Sodium bicarbonate (baking soda) plus acid fizzes to make salt, water, and carbon dioxide.

Exam Tip: Remember that bicarbonate-acid reactions always liberate carbon dioxide gas with brisk effervescence.

Question. Complete the following chemical equations: (i) SO₂ + MnO₄^- + H₂O -> (ii) F₂ (g) + H₂O (l) ->
Answer:
(i) Sulfur dioxide reduces the purple permanganate ion in aqueous conditions to colorless manganese(II) ions, while being oxidized to sulfate ions:
\[ 5\text{SO}_2 + 2\text{MnO}_4^- + 2\text{H}_2\text{O} \rightarrow 5\text{SO}_4^{2-} + 2\text{Mn}^{2+} + 4\text{H}^+ \]
(ii) Fluorine gas oxidizes water to form either oxygen gas or ozone, depending on the reaction conditions:
- Formation of oxygen:
\[ 2\text{F}_2 \, (g) + 2\text{H}_2\text{O} \, (l) \rightarrow 4\text{H}^+ \, (aq) + 4\text{F}^- \, (aq) + \text{O}_2 \, (g) \]
- Formation of ozone:
\[ 3\text{F}_2 \, (g) + 3\text{H}_2\text{O} \, (l) \rightarrow 6\text{H}^+ \, (aq) + 6\text{F}^- \, (aq) + \text{O}_3 \, (g) \]
In simple words: SO2 decolorizes purple permanganate solution, turning it into manganese ions and sulfate. Fluorine is a super strong oxidizer that breaks water into acid, fluoride, and oxygen or ozone.

Exam Tip: The decolorization of \( \text{MnO}_4^- \) by \( \text{SO}_2 \) is a primary chemical test used to detect sulfur dioxide gas.

Question. Complete the following chemical reaction equations: (i) KClO₃ --Heat / MnO₂--> (ii) XeF₄ + H₂O ->
Answer:
(i) Potassium chlorate decomposes upon heating in the presence of manganese dioxide catalyst to yield potassium chloride and oxygen gas:
\[ 2\text{KClO}_3 \xrightarrow[\text{MnO}_2]{\Delta} 2\text{KCl} + 3\text{O}_2 \]
(ii) Xenon tetrafluoride reacts violently with water to undergo disproportionation, forming xenon gas, xenon trioxide, hydrogen fluoride, and oxygen gas:
\[ 6\text{XeF}_4 + 12\text{H}_2\text{O} \rightarrow 4\text{Xe} + 2\text{XeO}_3 + 24\text{HF} + 3\text{O}_2 \]
In simple words: Heating potassium chlorate with a catalyst releases lots of oxygen. Xenon tetrafluoride in water splits into plain xenon gas, solid xenon trioxide, acid, and oxygen gas.

Exam Tip: Note that the reaction of XeF4 with water is a redox disproportionation reaction, which is extremely important for chemical examinations.

Question. Complete the following chemical equations: (i) P4 + SOCl2 -> (ii) F2 (Excess) + Cl2 --300 C-->
Answer:
(i) White phosphorus reacts with thionyl chloride to form phosphorus trichloride, sulfur dioxide, and disulfur dichloride:
\[ \text{P}_4 + 8\text{SOCl}_2 \rightarrow 4\text{PCl}_3 + 4\text{SO}_2 + 2\text{S}_2\text{Cl}_2 \]
(ii) When chlorine reacts with excess fluorine at high temperatures (\( 300^\circ\text{C} \)), chlorine trifluoride is formed:
\[ \text{Cl}_2 + 3\text{F}_2 \, (\text{excess}) \xrightarrow{300^\circ\text{C}} 2\text{ClF}_3 \]
In simple words: Phosphorus reacts with thionyl chloride to make phosphorus trichloride, sulfur dioxide gas, and sulfur monochloride. Excess fluorine gas mixed with chlorine gas at high heat yields chlorine trifluoride.

Exam Tip: Notice that thionyl chloride reacts with phosphorus to give \( \text{PCl}_3 \), whereas sulfuryl chloride (\( \text{SO}_2\text{Cl}_2 \)) yields \( \text{PCl}_5 \).

Question. Explain the following: (i) Nitrogen is much less reactive than phosphorus. (ii) NF3 is an exothermic compound but NCl3 is an endothermic compound.
Answer:
(i) Nitrogen exists as a diatomic molecule (\( \text{N}_2 \)) with an exceptionally strong triple bond between the nitrogen atoms (\( \text{N}\equiv\text{N} \)). This gives it a very high bond dissociation enthalpy (\( 941.4\text{ kJ/mol} \)), making it highly inert at room temperature. Conversely, phosphorus exists as tetrahedral \( \text{P}_4 \) molecules with single \( \text{P-P} \) bonds which are much weaker and highly strained, making it extremely reactive.
(ii) Fluorine has a smaller atomic size than chlorine, so the \( \text{N-F} \) bond is much shorter and stronger than the \( \text{N-Cl} \) bond. Also, the bond dissociation energy of \( \text{F}_2 \) is extremely low compared to \( \text{Cl}_2 \). Hence, a significant amount of energy is released when \( \text{NF}_3 \) forms (exothermic), while the formation of \( \text{NCl}_3 \) requires a net input of energy (endothermic).
In simple words: Nitrogen gas has a super strong triple bond that is very hard to break, while phosphorus has strained single bonds that pop easily. NF3 is exothermic because the N-F bonds are incredibly strong, but NCl3 is unstable and takes energy to form because the N-Cl bonds are too weak.

Exam Tip: Use terms like "bond dissociation enthalpy" and "interelectronic repulsion" to provide a rigorous, complete explanation.

Question. What happens when: (i) SO2 gas is passed through an aqueous solution of Fe3+ salt? (ii) XeF4 reacts with SbF5?
Answer:
(i) Sulfur dioxide gas acts as a reducing agent in water, reducing yellow \( \text{Fe}^{3+} \) ions to pale green \( \text{Fe}^{2+} \) ions, while itself getting oxidized to sulfate ions:
\[ 2\text{Fe}^{3+} + \text{SO}_2 + 2\text{H}_2\text{O} \rightarrow 2\text{Fe}^{2+} + \text{SO}_4^{2-} + 4\text{H}^+ \]
(ii) Xenon tetrafluoride reacts with antimony pentafluoride (a strong fluoride acceptor) to form an ionic adduct consisting of a cationic xenon species and an anionic antimony species:
\[ \text{XeF}_4 + \text{SbF}_5 \rightarrow [\text{XeF}_3]^+ [\text{SbF}_6]^- \]
In simple words: Passing sulfur dioxide gas through an iron(III) solution changes its color by reducing the iron from +3 to +2. XeF4 donates a fluoride ion to SbF5 to make a salt.

Exam Tip: Write the ionic charges correctly for the Xe-Sb adduct: \( [\text{XeF}_3]^+ \) and \( [\text{SbF}_6]^- \).

Question. What happens when: (i) Concentrated H₂SO₄ is added to calcium fluoride? (ii) SO₃ is passed through water?
Answer:
(i) Concentrated sulfuric acid reacts with solid calcium fluoride to release hydrogen fluoride gas:
\[ \text{CaF}_2 + \text{H}_2\text{SO}_4 \rightarrow \text{CaSO}_4 + 2\text{HF} \]
(ii) When sulfur trioxide gas is dissolved in water, it reacts vigorously to yield sulfuric acid:
\[ \text{SO}_3 + \text{H}_2\text{O} \rightarrow \text{H}_2\text{SO}_4 \]
In simple words: Adding sulfuric acid to calcium fluoride produces hydrogen fluoride gas. Passing sulfur trioxide through water creates sulfuric acid.

Exam Tip: Both chemical reactions represent essential processes in the laboratory preparation of inorganic acids.

Question. Complete the following reactions: (i) NH₃ + 3Cl₂ (excess) -> (ii) XeF₆ + 2H₂O ->
Answer:
(i) Ammonia reacts with excess chlorine gas to form nitrogen trichloride (an explosive liquid) and hydrochloric acid:
\[ \text{NH}_3 + 3\text{Cl}_2 \, (\text{excess}) \rightarrow \text{NCl}_3 + 3\text{HCl} \]
(ii) Xenon hexafluoride undergoes partial hydrolysis with water to yield xenon dioxydifluoride and hydrogen fluoride:
\[ \text{XeF}_6 + 2\text{H}_2\text{O} \rightarrow \text{XeO}_2\text{F}_2 + 4\text{HF} \]
In simple words: Ammonia and excess chlorine yield explosive nitrogen trichloride. Xenon hexafluoride reacts with a limited amount of water to give xenon dioxydifluoride and acid.

Exam Tip: Always pay attention to whether chlorine or ammonia is in excess, as the product changes completely.

Question. What happens when H₃PO₃ is heated?
Answer: Heating orthophosphorous acid (\( \text{H}_3\text{PO}_3 \)) causes it to disproportionate, producing orthophosphoric acid and phosphine gas:
\[ 4\text{H}_3\text{PO}_3 \xrightarrow{\Delta} 3\text{H}_3\text{PO}_4 + \text{PH}_3 \]
In simple words: Heating phosphorous acid breaks it down into phosphine gas and a different, more oxidized phosphorus acid.

Exam Tip: Label this as a disproportionation reaction to secure maximum grade points.

Question. Draw the structures of the following: (i) H3PO2 (ii) XeF4
Answer:
(i) Hypophosphorous acid (\( \text{H}_3\text{PO}_2 \)) is monobasic and tetrahedral, featuring one double-bonded oxygen, one OH group, and two directly attached hydrogen atoms.

(ii) Xenon tetrafluoride (\( \text{XeF}_4 \)) has a square planar geometry with 4 Xe-F bonds and 2 lone pairs.

In simple words: H3PO2 is a tetrahedron with two hydrogens linked straight to phosphorus. XeF4 is a flat square of fluorines surrounding xenon.

Exam Tip: Highlighting that H3PO2 is monobasic because of only one OH group is crucial in exams.

Question. Complete the following reactions: (i) Cl₂ + H₂O -> (ii) XeF₆ + 3H₂O ->
Answer:
(i) Chlorine dissolves in water to form a mixture of hydrochloric acid and hypochlorous acid, which eventually releases nascent oxygen:
\[ \text{Cl}_2 + \text{H}_2\text{O} \rightarrow \text{HCl} + \text{HOCl} \rightarrow 2\text{HCl} + [\text{O}] \]
(ii) Xenon hexafluoride undergoes complete hydrolysis with water to yield xenon trioxide and hydrogen fluoride:
\[ \text{XeF}_6 + 3\text{H}_2\text{O} \rightarrow \text{XeO}_3 + 6\text{HF} \]
In simple words: Chlorine reacts with water to form a bleaching mixture of hydrochloric and hypochlorous acids. Xenon hexafluoride reacts with three water molecules to form xenon trioxide and hydrofluoric acid.

Exam Tip: Complete hydrolysis of XeF6 yields the highly explosive compound XeO_3.

Question. What happens when (i) conc. H₂SO₄ is added to Cu? (ii) SO₃ is passed through water? Write the equations.
Answer:
(i) Hot, concentrated sulfuric acid oxidizes copper metal to copper(II) sulfate while releasing sulfur dioxide gas:
\[ \text{Cu} + 2\text{H}_2\text{SO}_4 \, (\text{conc.}) \rightarrow \text{CuSO}_4 + \text{SO}_2 + 2\text{H}_2\text{O} \]
(ii) Passing sulfur trioxide through water results in the energetic formation of sulfuric acid:
\[ \text{SO}_3 + \text{H}_2\text{O} \rightarrow \text{H}_2\text{SO}_4 \]
In simple words: Hot sulfuric acid reacts with copper to make copper sulfate and release sulfur dioxide gas. Sulfur trioxide mixed with water makes sulfuric acid.

Exam Tip: Remember that hot, concentrated H2SO4 behaves as a strong oxidizing agent capable of oxidizing metals like copper.

Question. Complete the following chemical equations: (i) F₂ + 2Cl- -> (ii) 2XeF₂ + 2H₂O ->
Answer:
(i) Fluorine gas, being the strongest oxidizing agent among halogens, displaces and oxidizes chloride ions to chlorine gas:
\[ \text{F}_2 + 2\text{Cl}^- \rightarrow \text{Cl}_2 + 2\text{F}^- \]
(ii) Xenon difluoride undergoes a redox reaction with water, yielding xenon gas, oxygen, and hydrofluoric acid:
\[ 2\text{XeF}_2 + 2\text{H}_2\text{O} \rightarrow 2\text{Xe} + 4\text{HF} + \text{O}_2 \]
In simple words: Fluorine gas kicks out chlorine from its salts because it is more reactive. XeF2 breaks down in water to yield xenon gas, oxygen gas, and hydrofluoric acid.

Exam Tip: Be mindful of redox oxidation states: fluorine is reduced from 0 to -1, and oxygen in water is oxidized to 0.

Question. What happens when: (i) HCl is added to MnO₂? (ii) PCl₅ is heated? Write the equations involved.
Answer:
(i) Adding hydrochloric acid to manganese dioxide leads to oxidation of chloride ions, releasing chlorine gas along with manganese chloride and water:
\[ \text{MnO}_2 + 4\text{HCl} \rightarrow \text{MnCl}_2 + 2\text{H}_2\text{O} + \text{Cl}_2 \]
(ii) When heated, phosphorus pentachloride undergoes thermal dissociation to form phosphorus trichloride and chlorine gas:
\[ \text{PCl}_5 \xrightarrow{\Delta} \text{PCl}_3 + \text{Cl}_2 \]
In simple words: MnO2 reacts with HCl to liberate green chlorine gas. Heating PCl5 breaks it down into PCl3 and chlorine.

Exam Tip: The reaction of MnO2 with HCl is the standard laboratory method for preparing chlorine gas.

Question. "Orthophosphoric acid (H₃PO₄) is non-reducing whereas hypophosphorous acid (H₃PO₂) is a strong reducing agent." Explain and justify the above statement with suitable example.
Answer: Orthophosphoric acid (\( \text{H}_3\text{PO}_4 \)) contains no direct \( \text{P-H} \) bonds, meaning all three hydrogen atoms are connected to oxygen as \( \text{P-OH} \) bonds. Hypophosphorous acid (\( \text{H}_3\text{PO}_2 \)) has two direct \( \text{P-H} \) bonds that readily act as hydride donors to reduce other elements. For example, \( \text{H}_3\text{PO}_2 \) easily reduces silver nitrate to metallic silver, while \( \text{H}_3\text{PO}_4 \) is incapable of doing so:
\[ 4\text{AgNO}_3 + \text{H}_3\text{PO}_2 + 2\text{H}_2\text{O} \rightarrow 4\text{Ag} \downarrow + \text{H}_3\text{PO}_4 + 4\text{HNO}_3 \]
In simple words: H3PO4 has no P-H bonds, so it can't reduce anything. H₃PO₂ has two P-H bonds which easily reduce chemicals like silver nitrate into metallic silver.

Exam Tip: Always write the complete balanced equation with silver nitrate to illustrate the reducing property of H₃PO₂.

Question. (a) What is the covalence of nitrogen in N₂O₅? (b) BiH₃ is a stronger reducing agent than SbH₃, why?
Answer:
(a) The covalency of nitrogen in \( \text{N}_2\text{O}_5 \) is 4. This is because each nitrogen atom forms four covalent bonds (consisting of single, double, and dative coordinate bonds) and cannot expand its octet to 5 since d-orbitals are absent.
(b) Bismuth hydride (\( \text{BiH}_3 \)) is a much stronger reducing agent than antimony hydride (\( \text{SbH}_3 \)) because the \( \text{Bi-H} \) bond length is longer and weaker compared to the \( \text{Sb-H} \) bond. This allows \( \text{BiH}_3 \) to release hydrogen ions much more easily.
In simple words: Nitrogen can only share 4 pairs of electrons because it doesn't have d-orbitals. BiH3 is a better reducing agent because its bond is longer and weaker, making it easy to release hydrogen.

Exam Tip: Be sure to cite "absence of d-orbitals" for nitrogen's limited covalency of 4.

Question. Account for the following: (i) The two oxygen-oxygen bond lengths in ozone molecule are identical. (ii) Most of the reactions of fluorine are exothermic.
Answer:
(i) Due to resonance, the ozone molecule is a hybrid of two equivalent structures, which distributes the double bond evenly. Consequently, the bond orders are identical, giving equal oxygen-oxygen bond lengths of \( 128\text{ pm} \).
(ii) Fluorine reacts very vigorously because of its low bond dissociation energy of \( \text{F}_2 \) molecules (due to interelectronic repulsions) and the extremely high electronegativity of fluorine atoms, which forms very strong bonds with other elements.
In simple words: Ozone resonance averages the double and single bonds, making both oxygen bonds identical in length. Fluorine's own bond is very weak and easy to split, but it forms extremely strong bonds with other elements, releasing lots of heat.

Exam Tip: Explicitly sketch the ozone resonance structures to support your answer visually.

Question. Account for the following: (i) Bond angle is NH₄⁺ is higher than that in NH₃. (ii) ICl is more reactive than I2.
Answer:
(i) In the ammonium ion (\( \text{NH}_4^+ \)), nitrogen has four bonding pairs and no lone pairs of electrons, giving it a symmetric tetrahedral geometry with a bond angle of \( 109^\circ28' \). In ammonia (\( \text{NH}_3 \)), the lone pair of electrons causes stronger lone pair-bond pair repulsion, compressing the bond angle down to \( 107.3^\circ \).
(ii) The heteronuclear \( \text{I-Cl} \) bond is weaker and more polar than the homonuclear \( \text{I-I} \) bond because of the electronegativity difference between iodine and chlorine. This allows \( \text{ICl} \) to break much more easily to initiate chemical reactions.
In simple words: NH4+ is perfectly symmetrical with no lone pair, so it has a larger angle than NH3, where a lone pair squeezes the bonds together. ICl is polar and has a weaker bond than pure I2, making it react faster.

Exam Tip: Discuss both molecular symmetry and electron-pair repulsion when comparing bond angles.

Question. "Orthophosphoric acid (H₃PO₄) is not a reducing agent whereas hypophosphorus acid (H₃PO₂) is a strong reducing agent." Explain and justify the above statement with the help of a suitable example.
Answer: Orthophosphoric acid (\( \text{H}_3\text{PO}_4 \)) has three hydroxyl groups connected to phosphorus with no direct \( \text{P-H} \) bonds, so it lacks reducing properties. In contrast, hypophosphorous acid (\( \text{H}_3\text{PO}_2 \)) contains two direct \( \text{P-H} \) bonds that break easily to release hydrides. For instance, \( \text{H}_3\text{PO}_2 \) readily reduces silver nitrate into metallic silver:
\[ 4\text{AgNO}_3 + \text{H}_3\text{PO}_2 + 2\text{H}_2\text{O} \rightarrow 4\text{Ag} \downarrow + \text{H}_3\text{PO}_4 + 4\text{HNO}_3 \]
In simple words: H3PO4 lacks the P-H bonds needed to reduce anything. H3PO2 has two P-H bonds that easily reduce chemicals like silver nitrate into metallic silver.

Exam Tip: Mentioning the silver nitrate reduction reaction is standard proof of the reducing power of phosphorous acids.

Question. Account for the following: (i) NH₃ is a stronger base than PH₃. (ii) Sulphur has a greater tendency for catenation than oxygen. (iii) Bond dissociation energy of F₂ is less than that of Cl₂.
Answer:
(i) Nitrogen has a smaller atomic size and higher electronegativity than phosphorus, giving a much higher electron density on nitrogen's lone pair. Consequently, \( \text{NH}_3 \) acts as a much stronger Lewis base and donates its lone pair more readily than \( \text{PH}_3 \).
(ii) The \( \text{O-O} \) single bond is exceptionally weak due to intense interelectronic repulsion between the lone pairs on the small oxygen atoms. Sulfur is larger, which reduces lone pair repulsion and makes the \( \text{S-S} \) single bond much stronger, leading to higher catenation.
(iii) Fluorine atoms are very small, which forces their lone pairs incredibly close together. This causes strong interelectronic repulsion that weakens the \( \text{F-F} \) single bond, giving it a lower dissociation energy than the \( \text{Cl-Cl} \) bond.
In simple words: NH3 has a high electron density because nitrogen is small, making it a stronger base. Sulfur has a larger size so its lone pairs don't repel each other, giving it strong chain-forming bonds. F2 has a weak bond because its tiny size squeezes lone electron pairs too close together, creating a lot of repulsion.

Exam Tip: The key term "interelectronic repulsion" should always be used to explain the anomalies in fluorine and oxygen single-bond chemistry.

Question 108. Explain the following situations: (i) In the structure of HNO₃ molecule, the N-O bond (121 pm) is shorter than N-OH bond (140 pm). (ii) SF₄ is easily hydrolysed whereas SF₆ is not easily hydrolysed. (iii) XeF₂ has a straight linear structure and not a bent angular structure.
Answer:
(i) In nitric acid (\( \text{HNO}_3 \)), the \( \text{N-O} \) bonds exhibit partial double bond character due to resonance and coordinate bonding. The \( \text{N-OH} \) bond is a pure single covalent bond, which makes it significantly longer.
(ii) Sulfur in \( \text{SF}_4 \) is less sterically hindered and has empty d-orbitals accessible to attack by water molecules. In \( \text{SF}_6 \), the sulfur atom is completely shielded and protected by six fluorine atoms, preventing water from reaching it.
(iii) In \( \text{XeF}_2 \), the central xenon has two bonding pairs and three lone pairs of electrons. To minimize electron repulsions, the three lone pairs occupy equatorial positions while the two fluorine atoms occupy axial positions, yielding a linear geometry.
In simple words: The N-O bond has double bond character, while the N-OH bond is a single bond and therefore longer. SF4 can be attacked by water easily, but SF6 has six fluorine atoms guarding the sulfur atom from water. XeF2 is linear because three pairs of lone electrons push the two fluorines to opposite ends.

Exam Tip: Use the term "steric hindrance" to explain the extreme chemical inertness of SF6 towards hydrolysis.

Question 109. Explain the following observations: (i) Fluorine does not exhibit any positive oxidation state. (ii) The majority of known noble gas compounds are those of Xenon. (iii) Phosphorus is much more reactive than nitrogen.
Answer:
(i) Fluorine is the most electronegative element in the periodic table and lacks any vacant d-orbitals in its valence shell. Thus, it cannot expand its octet to lose or share electrons in a way that yields positive oxidation states.
(ii) Xenon has a larger atomic size and therefore the lowest ionization enthalpy among the non-radioactive noble gases. This allows strong oxidizing agents like oxygen and fluorine to easily share or polarize its electrons, forming stable compounds.
(iii) Nitrogen exists as a diatomic gas with an exceptionally strong triple bond (\( \text{N}\equiv\text{N} \)) that requires high energy to break. Phosphorus is a solid consisting of tetrahedral \( \text{P}_4 \) units with weak single bonds that suffer from high angular strain, making it very reactive. forms chemical compounds particularly with oxygen and fluorine.
(iii) Because P – P single bond is much weaker than N ≡ N triple bond and the bond length of nitrogen is small and bond dissociation energy is very large which makes it inert and unreactive and thus phosphorus becomes more reactive.
In simple words: Fluorine is so electronegative and small that it can only show a negative oxidation state. Xenon has a large size and low ionization energy, making it easier to form bonds than other noble gases. Nitrogen is inert because of its tight triple bond, while phosphorus has single bonds with lots of angle strain.

Exam Tip: Be sure to emphasize "low ionization energy" when describing xenon reactivity, and "bond dissociation enthalpy" when discussing nitrogen's inertness.