CBSE Class 12 Chemistry The p Block Elements Questions

Consequences :

1. The first element in p-block element has four valence orbitals i.e. one 2s and three 2p, Hence maximum covalency of the first element in limited to four. The other elements of the p-block elements have vacant d-orbitals in their valence shell, e.g. elements of the third period have nine (9) one 3s, three 3p and five three 3d orbitals. Hence these show maximum covalence greater than four. Following questions can be answered -

(i) Nitrogen (N) does not from pentahalide while P froms PCl₅, PF₅, and PF₆ –. Why?

(ii) Sulphur (S) forms SF6 but oxygen does not form OF₆. Why?

(iii) Though nitrogen forms pentoxide but it does not form pentachloride. Explain. Why?

(iv) Fluorine forms only one oxoacid while other halogens form a number of oxoacids. Why?

(2) The first member of p-block elements displays greater ability to from p_Π- p_Π bond (s) with itself, (e.g., C = C, C ≡ C, N = N, N ≡ N) and with the other elements of second period (e.g., C = O, C ≡ N, N = O) compared to the subsequent members of the group.

This is because p-orbitals of the heavier members are so large and diffuse that they cannot have effective sideways overlapping. Heavier members can form p_Π – d_Π bonds with oxygen.

Question. When Cl₂ gas reacts with hot and concentrated sodium hydroxide solution, the oxidation number of chlorine changes from
(a) zero to +1 and zero to –5
(b) zero to –1 and zero to +5
(c) zero to –1 and zero to +3
(d) zero to +1 and zero to –3 

Question. Which of the following is used in the preparation of chlorine?
(a) Both MnO₂ and KMnO₄
(b) Only KMnO₄
(c) Only MnO₂
(d) Either MnO₂ or KMnO₄ 

Question. Which of the following elements is extracted commercially by the electrolysis of an aqueous solution of its compound?
(a) Cl
(b) Br
(c) Al
(d) Na 

Question. When chlorine is passed over dry slaked lime at room temperature, the main reaction product is
(a) Ca(ClO₂)₂
(b) CaCl₂
(c) CaOCl₂
(d) Ca(OCl)₂ 

Question. In the manufacture of bromine from sea water, the mother liquor containing bromides is treated with
(a) carbon dioxide
(b) chlorine
(c) iodine
(d) sulphur dioxide.

Question. The bleaching action of chlorine is due to
(a) reduction
(b) hydrogenation
(c) chlorination
(d) oxidation. 7.20 Hydrogen Chloride

Question. Bleaching powder reacts with a few drops of conc. HCl to give
(a) chlorine
(b) hypochlorous acid
(c) calcium oxide
(d) oxygen.

Question. Among the following, the correct order of acidity is
(a) HClO₂ < HClO < HClO₃ < HClO₄
(b) HClO₄ < HClO₂ < HClO < HClO₃
(c) HClO₃ < HClO₄ < HClO₂ < HClO
(d) HClO < HClO₂ < HClO₃ < HClO₄

Question. Which of the statements given below is incorrect?
(a) O3 molecule is bent.
(b) ONF is isoelectronic with O₂N–.
(c) OF₂ is an oxide of fluorine.
(d) Cl₂O₇ is an anhydride of perchloric acid.

Question. The correct order of increasing bond angles in the following species is
(a) Cl₂O < ClO₂ < ClO₂–
(b) ClO₂ < Cl₂O < ClO₂–
(c) Cl₂O < ClO₂– < ClO₂
(d) ClO₂– < Cl₂O < ClO₂ 

Question. Which one of the following oxides is expected to exhibit paramagnetic behaviour?
(a) CO₂
(b) SiO₂
(c) SO₂
(d) ClO₂

Question. Identify the incorrect statement, regarding the molecule XeO₄.
(a) XeO₄ molecule is square planar.
(b) There are four pp - dp bonds.
(c) There are four sp₃ - p, s bonds.
(d) XeO₄ molecule is tetrahedral.

Question. Which compound has planar structure?
(a) XeF₄
(b) XeOF₂
(c) XeO₂F₂
(d) XeO₄

Question. The ion or group detected by K₂[HgI₄] is :
(a) NO
(b) Cl^– 
(c) NH₂–
(d) NH⁴⁺

Question. Which gas cannot be collected over water ?
(a) O₂
(b) PH₃ 
(c) N₂
(d) SO₂

Question. H₂S does not produce metallic sulphide with :
(a) CuCl₂
(b) COCl₂ 
(c) CdCl₂
(d) ZnCl₂

VSA QUESTIONS (1 - MARK QUESTIONS)

Question 1. In group 15 elements, there is considerable increase in covalent radius from N to P but small increase from As to Bi. Why?
Answer: The considerable increase in covalent radius from N to P is due to an increase in the principal quantum number and addition of a new main shell. From As to Bi, the small increase is due to the presence of completely filled \(d\)- and/or \(f\)-orbitals in heavier members (As, Sb, Bi), which have poor shielding effect, leading to a higher effective nuclear charge that pulls the electron cloud inward.
In simple words: Moving down from nitrogen to phosphorus adds a whole new electron shell, but moving from arsenic to bismuth brings in crowded \(d\)- and \(f\)-orbitals that shield poorly, causing the size to stay relatively steady.

Exam Tip: Mention both factors clearly: addition of a new shell for the top elements and poor shielding by \(d\)/\(f\)-electrons for the heavier elements.

Question 2. The tendency to exhibit -3 oxidation state, decreases down the group in group 15 elements. Explain.
Answer: The tendency to exhibit a \(-3\) oxidation state depends upon the electronegativity and size of the element. As we move down group 15 from N to Bi, the atomic size increases significantly and the electronegativity decreases. Consequently, the tendency to accept or share electrons to attain a \(-3\) state decreases.
In simple words: Larger atoms with lower electronegativity find it harder to attract or hold onto extra electrons, so their tendency to show a \(-3\) oxidation state drops as you go down the group.

Exam Tip: Always link oxidation state trends directly to atomic size and electronegativity changes.

Question 3. Maximum covalence of Nitrogen is '4' but the heavier elements of group 15 show covalence greater than '4'. Why?
Answer: Nitrogen has only four valence orbitals (\(2s\) and three \(2p\)) in its valence shell, limiting its maximum covalence to 4. The heavier elements of group 15 possess vacant \(d\)-orbitals in their valence shell, which can be utilized for bonding, allowing them to expand their octet and exhibit covalence greater than 4 (e.g., \(PCl_5\), \(PF_6^-\)).
In simple words: Nitrogen has no \(d\)-orbitals so it can form a maximum of 4 bonds, whereas heavier elements use empty \(d\)-orbitals to form 5 or 6 bonds.

Exam Tip: Keywords to include: absence of \(d\)-orbitals in nitrogen and presence of vacant \(d\)-orbitals in heavier elements.

Question 4. Nitrogen exists as a diatomic molecule with a triple bond between the two atoms, whereas the heavier elements of the group do not exist as \(E_2\) at room temperature. Assign a reason.
Answer: Nitrogen has a small size and high electronegativity, which enables it to form stable \(p\pi - p\pi\) multiple bonds with itself (\(N \equiv N\)). The heavier elements of group 15 have larger atoms with diffuse orbitals that cannot undergo effective sideways overlapping, making \(p\pi - p\pi\) bonding ineffective. Instead, they prefer single bonds and exist as tetraatomic molecules (\(P_4\), \(As_4\), etc.).
In simple words: Small nitrogen atoms overlap sideways easily to form strong triple bonds, but bigger atoms can only form single bonds.

Exam Tip: Emphasize effective sideways overlapping of \(p\)-orbitals for nitrogen versus diffuse orbitals for heavier elements.

Question 5. The ionization enthalpies of group 15 elements are higher than those of corresponding members of group 14 and 16 elements. Assign the reason.
Answer: Group 15 elements have an extra stable electronic configuration with exactly half-filled \(p\)-orbitals (\(ns^2 np^3\)). Because half-filled subshells possess extra exchange energy and spherical symmetry, removing an electron requires more energy than removing an electron from group 14 (\(ns^2 np^2\)) or group 16 (\(ns^2 np^4\)) elements.
In simple words: Group 15 elements have half-filled orbitals which are very stable, so it takes extra energy to pull an electron away from them.

Exam Tip: Explicitly mention the half-filled \(ns^2 np^3\) configuration stability as the primary reason.

Question 6. The boiling point of \(PH_3\) is lesser than \(NH_3\). Why?
Answer: Nitrogen is highly electronegative and small in size, allowing \(NH_3\) molecules to form intermolecular hydrogen bonds. \(PH_3\) does not form hydrogen bonds because phosphorus has lower electronegativity. Consequently, \(NH_3\) requires more thermal energy to break these strong intermolecular forces, giving it a higher boiling point than \(PH_3\).
In simple words: Ammonia molecules stick together strongly using hydrogen bonds, whereas phosphine molecules only have weak van der Waals forces.

Exam Tip: Mention intermolecular hydrogen bonding in \(NH_3\) and its absence in \(PH_3\).

Question 7. \(NO_2\) dimerises to form \(N_2O_4\). Why?
Answer: Nitrogen dioxide (\(NO_2\)) is an odd-electron molecule containing an unpaired electron on the nitrogen atom, making it a reactive radical. To achieve a more stable paired electron configuration with a complete octet, two \(NO_2\) molecules combine (dimerize) to form \(N_2O_4\).
In simple words: \(NO_2\) has an unpaired electron making it unstable, so two molecules join together to pair up their electrons.

Exam Tip: Key phrase to include: "presence of odd electron on nitrogen makes \(NO_2\) prone to dimerization."

Question 8. Draw the structure of \(N_2O_5\) molecule.
Answer: In \(N_2O_5\) (dinitrogen pentoxide), each nitrogen atom is bonded to three oxygen atoms, with a bridging oxygen atom connecting the two nitrogen atoms (\(O_2N-O-NO_2\)). The structure features resonance with terminal \(N=O\) double bonds and single bonds involving coordinate/dative overlap.
In simple words: Two \(NO_2\) groups are bridged by a single oxygen atom in the middle.

Question 9. How does ammonia solution react with \(Ag^+\text{ (aq)}\)? Write the balanced chemical equation.
Answer: When aqueous ammonia is added to a solution containing silver ions (\(Ag^+\)), it first precipitates silver oxide, which then dissolves in excess ammonia to form a soluble coordination complex, diamminesilver(I) ion.
\(Ag^+ (aq) + OH^- (aq) \rightarrow AgOH\)
\(2AgOH \rightarrow Ag_2O_{(s)} + H_2O_{(l)}\)
\(Ag_2O_{(s)} + 4NH_3 (aq) + H_2O_{(l)} \rightarrow 2[Ag(NH_3)_2]^+ (aq) + 2OH^- (aq)\)
In simple words: Ammonia reacts with silver ions to form a clear soluble complex salt.

Exam Tip: Write the net ionic equation for the complex formation: \(Ag^+ + 2NH_3 \rightarrow [Ag(NH_3)_2]^+\).

Question 10. Why does \(NH_3\) forms intermolecular hydrogen bonds whereas \(PH_3\) does not?
Answer: Nitrogen has a high electronegativity (3.0) and a small atomic radius, allowing it to polarize the N-H bond strongly and form hydrogen bonds. Phosphorus has a lower electronegativity (2.1) and a larger size, which is insufficient to polarize the P-H bond for hydrogen bonding.
In simple words: Nitrogen is small and very electronegative, pulling electrons hard enough to form hydrogen bonds, while phosphorus cannot.

Exam Tip: Highlight the high electronegativity and small size of nitrogen as the key prerequisites for hydrogen bonding.

Question 11. Write disproportionation reaction of \(H_3PO_3\)?
Answer: When orthophosphorous acid (\(H_3PO_3\)) is heated, it undergoes a disproportionation reaction to form phosphine (\(PH_3\)) and phosphoric acid (\(H_3PO_4\)).
\(4H_3PO_3 \xrightarrow{\text{heat}} PH_3 + 3H_3PO_4\)
In simple words: Heating \(H_3PO_3\) causes it to simultaneously oxidize into phosphoric acid and reduce into phosphine.

Exam Tip: Check oxidation states: P goes from +3 in \(H_3PO_3\) to -3 in \(PH_3\) and +5 in \(H_3PO_4\).

Question 12. How does \(NH_3\) acts as a complexing agent?
Answer: Ammonia possesses a lone pair of electrons on the nitrogen atom, which it can donate to metal ions with vacant orbitals. This allows \(NH_3\) to act as a Lewis base and form stable coordination complexes with transition metal cations like \(Cu^{2+}\), \(Ag^+\), and \(Zn^{2+}\).
In simple words: Ammonia donates its lone pair of electrons to metal ions, forming coordination complexes.

Exam Tip: Define ammonia's role as a Lewis base providing a lone pair to metal cations.

Question 13. Why HF is the weakest acid and HI is the strongest?
Answer: Acid strength depends on the bond dissociation enthalpy of the H-X bond. Fluorine is small, making the H-F bond very short and exceptionally strong (high bond dissociation enthalpy), so it releases \(H^+\) ions with difficulty. Conversely, iodine is large, making the H-I bond long and weak, so it dissociates easily to release protons.
In simple words: The H-F bond is extremely tight and hard to break, making it a weak acid, whereas the H-I bond is loose and breaks easily, making it a strong acid.

Exam Tip: Connect acid strength inversely to bond dissociation enthalpy down the group 17 hydrides.

Question 14. Explain why halogens are strong oxidising agents.
Answer: Halogens have high electronegativities and high negative electron gain enthalpies, meaning they have a strong tendency to accept an electron to attain a stable noble gas configuration (\(X + e^- \rightarrow X^-\)). Consequently, they readily oxidize other substances while getting reduced themselves.
In simple words: Halogens desperately want one more electron to complete their octet, so they snatch electrons from other substances and act as strong oxidants.

Exam Tip: Mention high negative electron gain enthalpy and low bond dissociation energy as key driving factors.

Question 15. Why is Bi(V) a stronger oxidant than Sb(V)?
Answer: Due to the inert pair effect, the +3 oxidation state of bismuth is much more stable than its +5 oxidation state. Therefore, Bi(V) readily accepts two electrons to convert into the more stable Bi(III) state, making it a powerful oxidizing agent.
In simple words: Bismuth prefers the +3 oxidation state due to the inert pair effect, so Bi(V) quickly grabs electrons to drop down to +3.

Question 16. Why \(SF_4\) is easily hydrolysed, whereas \(SF_6\) is resistant to hydrolysis?
Answer: In \(SF_6\), the central sulfur atom is sterically protected by six surrounding fluorine atoms, and sulfur is coordinatively saturated, preventing water molecules from attacking. In contrast, in \(SF_4\), the sulfur atom has a lone pair and is less crowded, allowing water molecules to easily attack and initiate hydrolysis.
In simple words: Sulfur in \(SF_6\) is completely shielded by six fluorines so water cannot reach it, but in \(SF_4\) water can easily attack.

Exam Tip: Use the terms steric hindrance and coordinative saturation in your explanation.

Question 17. Bond dissociation enthalpy of \(F_2\) is less than that of \(Cl_2\). Why?
Answer: Due to the extremely small size of the fluorine atom, its valence lone pairs are very close to each other. This causes severe interelectronic repulsion between the lone pairs of the two fluorine atoms, weakening the F-F bond and resulting in an unexpectedly low bond dissociation enthalpy compared to \(Cl_2\).
In simple words: Fluorine atoms are so tiny that their electrons crowd each other, pushing the atoms apart and making the bond easy to break.

Exam Tip: Highlight lone pair-lone pair repulsions caused by the small size of fluorine.

Question 18. Write the reaction of \(PCl_5\) with heavy water.
Answer: Phosphorus pentachloride reacts with heavy water (\(D_2O\)) to form phosphorus oxychloride and deuterium chloride.
\(PCl_5 + D_2O \rightarrow POCl_3 + 2DCl\)
In simple words: Heavy water hydrolyses phosphorus pentachloride to produce phosphoryl chloride and deuterated hydrochloric acid.

Exam Tip: Ensure balanced stoichiometry with 2 moles of DCl produced.

Question 19. How many \(P - O - P\) bonds are there in cyclotrimetaphosphoric acid?
Answer: Cyclotrimetaphosphoric acid (\(H_3P_3O_9\)) has a cyclic structure consisting of a six-membered ring with alternating phosphorus and oxygen atoms (\(-P-O-P-O-\)). There are exactly 3 \(P - O - P\) bonds in the ring.
In simple words: There are 3 alternating phosphorus-oxygen-phosphorus bridges in the ring structure.

Question 20. In group 16, the stability of +6 oxidation state decreases and that of +4 oxidation state increases down the group. Why?
Answer: This is due to the inert pair effect. As we move down group 16, the energy required to promote \(ns^2\) electrons into bonding decreases, but the reluctance of the \(ns^2\) electron pair to participate in bonding increases due to poor shielding by intervening \(d\)- and \(f\)-orbitals, making the lower oxidation state (+4) more stable for heavier elements.
In simple words: Down the group, inner \(d\) and \(f\) electrons shield poorly, holding onto the outer \(ns^2\) electrons and making lower oxidation states more stable.

Exam Tip: Always invoke the inert pair effect for stability shifts in heavier p-block elements.

Question 21. Why we can not prepare HBr by heating KBr with sulphuric acid?
Answer: Concentrated sulfuric acid is a strong oxidizing agent. When KBr is heated with conc. \(H_2SO_4\), the initially formed HBr is oxidized by \(H_2SO_4\) into bromine gas (\(Br_2\)) instead of being released as free HBr.
\(2KBr + H_2SO_4 \rightarrow K_2SO_4 + 2HBr\)
\(2HBr + H_2SO_4 \rightarrow Br_2 + SO_2 + 2H_2O\)
In simple words: Sulfuric acid is too strong an oxidizer and destroys the HBr by turning it into bromine gas.

Exam Tip: State that non-oxidizing acids like phosphoric acid (\(H_3PO_4\)) must be used instead of sulfuric acid to prepare HBr.

Question 24. Fluorine exhibit only -1 oxidation state whereas other halogens exhibit +ve oxidation states also. Explain.
Answer: Fluorine is the most electronegative element in the periodic table and has no vacant \(d\)-orbitals in its valence shell (\(2s^2 2p^5\)) to promote electrons for higher oxidation states. Other halogens possess vacant \(d\)-orbitals, allowing them to unpair their electrons and exhibit positive oxidation states like +3, +5, and +7.
In simple words: Fluorine is too electronegative and lacks empty \(d\)-orbitals, so it can only ever show a -1 oxidation state.

Exam Tip: Keywords: highest electronegativity and absence of \(d\)-orbitals.

Question 25. Arrange the following oxoacids of chlorine in increasing order of acidic strength. \(HOCl, HOClO, HOClO_2, HOClO_3\).
Answer: The acidic strength of oxoacids increases with an increase in the oxidation state of the halogen and the number of oxygen atoms attached to it. More oxygen atoms withdraw electron density more effectively from the O-H bond, making proton release easier.
Increasing order: \(HOCl < HOClO < HOClO_2 < HOClO_3\)
In simple words: More oxygen atoms attached to chlorine mean a stronger acid because electrons are pulled away from hydrogen faster.

Exam Tip: Acid strength is directly proportional to the number of terminal oxygen atoms.

Question 26. The majority of known noble gas compounds are those of Xenon. Why?
Answer: Xenon has a relatively large atomic size and a low ionization enthalpy compared to other lighter noble gases, making its valence electrons easier to remove. Furthermore, fluorine and oxygen are extremely electronegative elements capable of oxidizing xenon and stabilizing its compounds.
In simple words: Xenon is large with a low ionization energy, allowing strong oxidizers like fluorine to react with it and form compounds.

Question 27. "Hypophosphorus acid is a good reducing agent." Justify with an example.
Answer: Hypophosphorous acid (\(H_3PO_2\)) contains two P-H bonds which are strong reducing centers. It reduces silver nitrate to metallic silver, demonstrating its strong reducing property:
\(4AgNO_3 + H_3PO_2 + 2H_2O \rightarrow 4Ag + HNO_3 + H_3PO_4\)
In simple words: Because it has two hydrogen atoms directly bonded to phosphorus, \(H_3PO_2\) easily reduces metal ions like silver.

Question 28. Draw the structure of \(H_4P_2O_7\) and find out its basicity?
Answer: Pyrophosphoric acid (\(H_4P_2O_7\)) features two phosphorus atoms linked by a bridging oxygen atom (\(HO)_2P(=O)-O-P(=O)(OH)_2\)). Each phosphorus is bonded to one terminal oxygen (\(=O\)) and two hydroxyl groups (\(-OH\)). Since it contains four ionizable hydroxyl groups, its basicity is 4 (tetrabasic).
In simple words: It has four replaceable hydrogen atoms attached to oxygen, so its basicity is 4.

Question 29. Arrange the following triatomic species in the order of increasing bond angle. \(NO_2, NO_2^+, NO_2^-\)
Answer: \(NO_2^+\) has a linear structure with no lone pairs on nitrogen (bond angle \(180^\circ\)). \(NO_2\) has one non-bonding odd electron causing slight repulsion (bond angle \(\approx 134^\circ\)). \(NO_2^-\) has two non-bonding electrons on nitrogen causing strong lone pair-bond pair repulsion, compressing the bond angle (\(\approx 115^\circ\)).
Increasing order of bond angle: \(NO_2^- < NO_2 < NO_2^+\)
In simple words: More lone or odd electrons on the central nitrogen push the oxygen atoms closer, reducing the bond angle.

Question 30. With what neutral molecule \(ClO^-\) is isoelectronic?
Answer: The hypochlorite ion (\(ClO^-\)) has a total of \(17 + 8 + 1 = 26\) electrons. A neutral molecule with the same number of electrons is chlorine fluoride (\(ClF\)), which also has \(17 + 9 = 26\) electrons.
In simple words: \(ClO^-\) and \(ClF\) both have a total of 26 electrons, making them isoelectronic.

Question 31. Draw the structure of \(H_2S_2O_8\) and find the number of S-S bond if any.
Answer: Peroxodisulphuric acid (\(H_2S_2O_8\)) has the structure \(HO-(O=)S-O-O-S(=O)-OH\). It contains a peroxide linkage (\(-O-O-\)) and has no sulfur-sulfur (\(S-S\)) bond.
In simple words: The sulfur atoms are connected by an oxygen-oxygen peroxide bridge, so there is no direct S-S bond.

Question 32. What is cause of bleaching action of chlorine water? Explain it with chemical equation?
Answer: The bleaching action of chlorine water is due to the formation of nascent oxygen, which oxidizes colored organic substances into colorless products:
\(Cl_2 + H_2O \rightarrow HCl + HOCl\)
\(HOCl \rightarrow HCl + [O] \text{ (nascent oxygen)}\)
\(\text{Coloured matter} + [O] \rightarrow \text{Colourless matter}\)
In simple words: Chlorine reacts with water to release nascent oxygen, which destroys color by oxidation.

Question 33. Electron gain enthalpy of fluorine is more negative than that of chlorine. [Correction: Less negative than chlorine]
Answer: Although fluorine is smaller, its electron gain enthalpy is less negative than chlorine because of its extremely compact 2p orbitals. The incoming electron experiences strong interelectronic repulsions from the existing electrons in the small 2p shell, lowering the electron affinity compared to the larger 3p orbitals of chlorine.
In simple words: Fluorine's electrons are crowded so tightly that adding a new electron creates extra repulsion, making it less eager to accept electrons than chlorine.

Question 34. Which one of the following is not oxidised by \(O_3\). State the reason: \(KI, FeSO_4, K_2MnO_4, KMnO_4\).
Answer: \(KMnO_4\) is not oxidized by ozone because manganese is already in its highest possible oxidation state of +7, where it cannot lose any more electrons.
In simple words: Potassium permanganate cannot be oxidized further because manganese is already at its maximum +7 state.

SA (I) TYPE QUESTIONS (2 - MARK QUESTIONS)

Question 2. Why is red phosphorus denser and less chemically reactive than white phosphorus?
Answer: Red phosphorus has a polymeric crystal lattice structure consisting of chains of \(P_4\) tetrahedra linked together, making it closely packed (higher density) and stable. In contrast, white phosphorus consists of discrete, strained \(P_4\) molecules with weak intermolecular forces, making it volatile, highly reactive, and less dense.
In simple words: Red phosphorus is tightly bonded in chains making it stable and dense, while white phosphorus is made of loose, strained molecules that react easily.

Question 3. Give chemical reaction in support of the statement that all the bonds in \(PCl_5\) molecule are not equivalent.
Answer: In \(PCl_5\), the geometry is trigonal bipyramidal with three equatorial bonds and two longer, weaker axial bonds. Upon gentle heating, \(PCl_5\) readily decomposes to release the weaker axial chlorine atoms first:
\(PCl_5 \xrightarrow{\text{heat}} PCl_3 + Cl_2\)
In simple words: The axial bonds in \(PCl_5\) are longer and weaker than the equatorial bonds, which is why they break easily upon heating.

Question 4. Account for the following: (a) \(XeF_2\) has linear structure and not a bent structure. (b) Phosphorus show marked tendency for Catenation.
Answer: (a) In \(XeF_2\), the central xenon atom has 3 lone pairs and 2 bond pairs (\(sp^3d\) hybridization). According to VSEPR theory, the 3 lone pairs occupy equatorial positions to minimize repulsion, forcing the two fluorine atoms into a linear axial arrangement.
(b) Phosphorus shows catenation due to its suitable single bond enthalpy (\(P-P\) bond energy is relatively high compared to nitrogen), allowing it to form stable chains and rings.
In simple words: (a) Lone pairs position themselves symmetrically to give \(XeF_2\) a straight linear shape. (b) Phosphorus atoms bond strongly with each other to form long chains.

Question 5. Draw the structures of \(BrF_3\), \(XeOF_4\), \(XeO_3\) using VSEPR theory.
Answer: - \(BrF_3\): 3 bond pairs and 2 lone pairs; T-shaped geometry with distorted trigonal bipyramidal electron geometry.
- \(XeOF_4\): 5 bond pairs (including one double bond with O) and 1 lone pair; square pyramidal geometry.
- \(XeO_3\): 3 bond pairs and 1 lone pair; pyramidal geometry.
In simple words: Using VSEPR theory, lone pairs and bond pairs arrange themselves to minimize repulsion, yielding T-shaped, square pyramidal, and pyramidal shapes respectively.

Question 6. Write the conditions that favour the formation of ammonia gas along with the reactions involved in Haber's Process.
Answer: According to Le Chatelier's principle, the favorable conditions for Haber's process are high pressure (approx. 200 atm), optimum temperature (approx. 700 K), and a catalyst (iron catalyst with molybdenum promoter).
\(N_2 (g) + 3H_2 (g) \rightleftharpoons 2NH_3 (g) \quad (\Delta H = -92.38 \text{ kJ mol}^{-1})\)
In simple words: High pressure and moderate temperature with an iron catalyst push the reaction forward to produce maximum ammonia.

Question 7. Write the chemical equations of the following reactions (a) Glucose is heated with conc. \(H_2SO_4\). (b) Sodium nitrate is heated with conc. \(H_2SO_4\).
Answer: (a) \(C_6H_{12}O_6 \xrightarrow{\text{conc. } H_2SO_4} 6C + 6H_2O\) (Dehydration reaction)
(b) \(NaNO_3 + H_2SO_4 \rightarrow NaHSO_4 + HNO_3\)
In simple words: (a) Sulfuric acid dehydrates glucose to pure carbon. (b) Heating sodium nitrate with sulfuric acid produces nitric acid vapor.

Question 8. Complete the following reactions: (i) \((NH_4)_2Cr_2O_7 \xrightarrow{\text{heat}}\) (ii) \(NH_4Cl \text{ (aq)} + NaNO_2 \text{ (aq)} \rightarrow\)
Answer: (i) \((NH_4)_2Cr_2O_7 \xrightarrow{\text{heat}} N_2 + 4H_2O + Cr_2O_3\)
(ii) \(NH_4Cl (aq) + NaNO_2 (aq) \rightarrow N_2 (g) + 2H_2O (l) + NaCl (aq)\)
In simple words: Both reactions are standard laboratory preparations for pure nitrogen gas.

Question 9. Complete the following reactions: (i) \(NH_2CONH_2 + H_2O \rightarrow\) (ii) \(FeCl_3 \text{ (aq)} + NH_4OH \rightarrow\)
Answer: (i) \(NH_2CONH_2 + 2H_2O \rightarrow (NH_4)_2CO_3 \rightarrow 2NH_3 + H_2O + CO_2\)
(ii) \(FeCl_3 (aq) + 3NH_4OH (aq) \rightarrow Fe(OH)_3 \downarrow + 3NH_4Cl (aq)\)
In simple words: (i) Urea hydrolyses to form ammonia and carbon dioxide. (ii) Ferric chloride reacts with ammonium hydroxide to precipitate brown ferric hydroxide.

Question 10. Complete the following reactions: (i) \(Ca_3P_2 + H_2O (l) \rightarrow\) (ii) \(I_2 + HNO_3 \text{ (conc.)} \rightarrow\)
Answer: (i) \(Ca_3P_2 + 6H_2O \rightarrow 3Ca(OH)_2 + 2PH_3\)
(ii) \(I_2 + 10HNO_3 \text{ (conc.)} \rightarrow 2HIO_3 + 10NO_2 + 4H_2O\)
In simple words: (i) Calcium phosphide reacts with water to release phosphine gas. (ii) Concentrated nitric acid oxidizes iodine to iodic acid.

Question 11. Complete the following reactions: (i) \(Ba(N_3)_2 \xrightarrow{\text{heat}}\) (ii) \(4H_3PO_3 \xrightarrow{\text{heat}}\)
Answer: (i) \(Ba(N_3)_2 \xrightarrow{\text{heat}} Ba + 3N_2\)
(ii) \(4H_3PO_3 \xrightarrow{\text{heat}} PH_3 + 3H_3PO_4\)
In simple words: (i) Barium azide decomposes upon heating to yield pure nitrogen gas. (ii) Phosphorous acid undergoes thermal disproportionation into phosphine and phosphoric acid.

Question 12. Complete the following reactions: (i) \(PH_4I + KOH \rightarrow\) (ii) \(HgCl_2 + PH_3 \rightarrow\)
Answer: (i) \(PH_4I + KOH \rightarrow KI + H_2O + PH_3\)
(ii) \(3HgCl_2 + 2PH_3 \rightarrow Hg_3P_2 \cdot 4HCl\) (or mercury phosphide complex precipitate)
In simple words: (i) Phosphonium iodide reacts with base to release phosphine gas. (ii) Phosphine reacts with mercuric chloride to form a phosphide precipitate.

Question 13. Complete the following reactions: (i) \(PCl_3 + 3H_2O \rightarrow\) (ii) \(S + H_2SO_4 \text{ (conc.)} \rightarrow\)
Answer: (i) \(PCl_3 + 3H_2O \rightarrow H_3PO_3 + 3HCl\)
(ii) \(S + 2H_2SO_4 \text{ (conc.)} \rightarrow 3SO_2 + 2H_2O\)
In simple words: (i) Phosphorus trichloride hydrolyses to form phosphorous acid. (ii) Concentrated sulfuric acid oxidizes sulfur into sulfur dioxide.

Question 14. Complete the following reactions: (i) \(Al_2O_3 (s) + NaOH (aq) + H_2O (l) \rightarrow\) (ii) \(HCl + O_2 \xrightarrow{CuCl_2}\)
Answer: (i) \(Al_2O_3 + 2NaOH + 3H_2O \rightarrow 2Na[Al(OH)_4]\)
(ii) \(4HCl + O_2 \xrightarrow{CuCl_2} 2Cl_2 + 2H_2O\) (Deacon's Process)
In simple words: (i) Amphoteric aluminum oxide dissolves in sodium hydroxide to form sodium aluminate. (ii) Hydrochloric acid is oxidized by oxygen to chlorine gas.

Question 15. Complete the following reactions: (i) \(Ca(OH)_2 + Cl_2 \rightarrow\) (ii) \(XeF_4 + H_2O \rightarrow\)
Answer: (i) \(2Ca(OH)_2 + 2Cl_2 \rightarrow Ca(OCl)_2 \cdot CaCl_2 \cdot 2H_2O\) (Bleaching powder formation)
(ii) \(6XeF_4 + 12H_2O \rightarrow 4Xe + 2XeO_3 + 24HF + 3O_2\)
In simple words: (i) Chlorine reacts with slaked lime to form bleaching powder. (ii) Xenon tetrafluoride hydrolyses to xenon, xenon trioxide, and hydrofluoric acid.

Question 16. Complete the following reactions: (i) \(Na_2SO_3 + Cl_2 + H_2O \rightarrow\) (ii) \(NaHCO_3 + HCl \rightarrow\)
Answer: (i) \(Na_2SO_3 + Cl_2 + H_2O \rightarrow Na_2SO_4 + 2HCl\)
(ii) \(NaHCO_3 + HCl \rightarrow NaCl + H_2O + CO_2 \uparrow\)
In simple words: (i) Chlorine oxidizes sodium sulfite to sodium sulfate. (ii) Sodium bicarbonate reacts with hydrochloric acid to release carbon dioxide.

Question 17. Complete the following reactions: (i) \(XeF_6 + H_2O \text{ (Complete hydrolysis)}\) (ii) \(XeF_6 + H_2O \text{ (Partial hydrolysis)}\)
Answer: (i) \(XeF_6 + 3H_2O \rightarrow XeO_3 + 6HF\)
(ii) \(XeF_6 + H_2O \rightarrow XeOF_4 + 2HF\)
In simple words: Complete hydrolysis of \(XeF_6\) yields xenon trioxide, whereas partial hydrolysis yields xenon oxytetrafluoride.

Question 18. Complete the following reactions: (i) \(NO_3^- + Fe^{2+} + H^+ \rightarrow\) (ii) \(Zn + HNO_3 \text{ (dil)} \rightarrow\)
Answer: (i) \(NO_3^- + 3Fe^{2+} + 4H^+ \rightarrow NO + 3Fe^{3+} + 2H_2O\) (Brown ring test reaction)
(ii) \(4Zn + 10HNO_3 \text{ (dil)} \rightarrow 4Zn(NO_3)_2 + 5H_2O + N_2O \uparrow\)
In simple words: (i) Nitrate reduces to nitric oxide in the presence of ferrous ions. (ii) Dilute nitric acid reacts with zinc to form nitrous oxide.

Question 19. Complete the following reactions: (i) \(Zn + HNO_3 \text{ (conc)} \rightarrow\) (ii) \(P_4 + HNO_3 \text{ (conc)} \rightarrow\)
Answer: (i) \(Zn + 4HNO_3 \text{ (conc)} \rightarrow Zn(NO_3)_2 + 2NO_2 + 2H_2O\)
(ii) \(P_4 + 20HNO_3 \text{ (conc)} \rightarrow 4H_3PO_4 + 20NO_2 + 4H_2O\)
In simple words: Concentrated nitric acid acts as a powerful oxidizer, reducing itself to nitrogen dioxide while oxidizing zinc and phosphorus.

Question 20. Complete the following reactions: (i) \(NH_3 + O_2 \xrightarrow{Pt/Rh}\) (ii) \(P_4 + NaOH + H_2O \rightarrow\)
Answer: (i) \(4NH_3 + 5O_2 \xrightarrow{Pt/Rh} 4NO + 6H_2O\) (Ostwald process step)
(ii) \(P_4 + 3NaOH + 3H_2O \rightarrow PH_3 + 3NaH_2PO_2\)
In simple words: (i) Ammonia catalytic oxidation produces nitric oxide. (ii) White phosphorus reacts with caustic soda to form phosphine and sodium hypophosphite.

Question 21. Complete the following reactions: (i) \(P_4 + SOCl_2 \rightarrow\) (ii) \(P_4 + SO_2Cl_2 \rightarrow\)
Answer: (i) \(P_4 + 8SOCl_2 \rightarrow 4PCl_3 + 4SO_2 + 2S_2Cl_2\)
(ii) \(P_4 + 10SO_2Cl_2 \rightarrow 4PCl_5 + 10SO_2\)
In simple words: Phosphorus reacts with thionyl chloride and sulfuryl chloride to form phosphorus trichloride and pentachloride respectively.

Question 22. Complete the following reactions: (i) \(PbS + O_3 \rightarrow\) (ii) \(KI + H_2O + O_3 \rightarrow\)
Answer: (i) \(PbS + 4O_3 \rightarrow PbSO_4 + 4O_2\)
(ii) \(2KI + H_2O + O_3 \rightarrow 2KOH + I_2 + O_2\)
In simple words: Ozone acts as a strong oxidizing agent, converting lead sulfide to sulfate and liberating iodine from potassium iodide.

Question 23. Complete the following reactions: (i) \(MnO_4^- + SO_2 + H_2O \rightarrow\) (ii) \(Zn + HNO_3 \text{ (dil)} \rightarrow\)
Answer: (i) \(2MnO_4^- + 5SO_2 + 2H_2O \rightarrow 2Mn^{2+} + 5SO_4^{2-} + 4H^+\)
(ii) \(4Zn + 10HNO_3 \text{ (dil)} \rightarrow 4Zn(NO_3)_2 + 5H_2O + N_2O\)
In simple words: Permanganate oxidizes sulfur dioxide to sulfate, and dilute nitric acid oxidizes zinc to zinc nitrate while releasing nitrous oxide.

Question 24. Complete the following reactions: (i) \(NH_3 \text{ (Excess)} + Cl_2 \rightarrow\) (ii) \(NH_3 + Cl_2 \text{ (Excess)} \rightarrow\)
Answer: (i) \(8NH_3 + 3Cl_2 \rightarrow 6NH_4Cl + N_2 \uparrow\)
(ii) \(NH_3 + 3Cl_2 \text{ (Excess)} \rightarrow NCl_3 + 3HCl\)
In simple words: Excess ammonia produces nitrogen gas, whereas excess chlorine produces highly explosive nitrogen trichloride.

Question 25. Complete the following reactions: (i) \(Cl_2 + NaOH \text{ (cold and dil)} \rightarrow\) (ii) \(Cl_2 + NaOH \text{ (hot and conc)} \rightarrow\)
Answer: (i) \(Cl_2 + 2NaOH \text{ (cold/dil)} \rightarrow NaCl + NaClO + H_2O\)
(ii) \(3Cl_2 + 6NaOH \text{ (hot/conc)} \rightarrow 5NaCl + NaClO_3 + 3H_2O\)
In simple words: Chlorine undergoes disproportionation with sodium hydroxide, yielding hypochlorite when cold and chlorate when hot.

Question 26. Complete the following reactions: (i) \(Fe + HCl \rightarrow\) (ii) \(Cl_2 + F_2 \text{ (Excess)} \rightarrow\)
Answer: (i) \(Fe + 2HCl \rightarrow FeCl_2 + H_2 \uparrow\)
(ii) \(Cl_2 + 3F_2 \text{ (Excess)} \xrightarrow{573 K} 2ClF_3\)
In simple words: (i) Iron reacts with hydrochloric acid to form ferrous chloride and hydrogen gas. (ii) Excess fluorine fluorinates chlorine into chlorine trifluoride.

Question 27. Complete the following reactions: (i) \(U + ClF_3 \rightarrow\) (ii) \(FeSO_4 + H_2SO_4 + Cl_2 \rightarrow\)
Answer: (i) \(U + 3ClF_3 \rightarrow UF_6 + 3ClF\)
(ii) \(6FeSO_4 + 3H_2SO_4 + Cl_2 \rightarrow 3Fe_2(SO_4)_3 + 2HCl\) (Note: or \(2FeSO_4 + H_2SO_4 + Cl_2 \rightarrow Fe_2(SO_4)_3 + 2HCl\))
In simple words: (i) Chlorine trifluoride acts as a fluorinating agent for uranium. (ii) Chlorine oxidizes ferrous sulfate to ferric sulfate.

Question 28. (i) What is the covalency of N in \(N_2O_5\)? (ii) Explain why phosphorus forms pentachloride whereas nitrogen and bismuth do not?
Answer: (i) The covalency of nitrogen in \(N_2O_5\) is 4 (each nitrogen forms 4 bonds including dative/coordinate bonds).
(ii) Phosphorus forms \(PCl_5\) because it has vacant \(d\)-orbitals to expand its octet. Nitrogen lacks \(d\)-orbitals in its valence shell, restricting its maximum covalency to 4, while bismuth due to the inert pair effect does not form pentahalides.
In simple words: Nitrogen cannot expand its octet beyond 4 due to lack of \(d\)-orbitals, whereas phosphorus readily uses \(d\)-orbitals to form pentahalides.

Question 29. (i) The acidic character of hydrides of group 15 increases from \(H_2O\) to \(H_2Te\) [Correction: Group 16]. Why? (ii) Dioxygen is a gas while sulphur (\(S_8\)) is a solid. Why?
Answer: (i) Acidic character increases down the group because the E-H bond length increases and bond dissociation enthalpy decreases, making it easier to release protons.
(ii) Oxygen forms small diatomic molecules (\(O_2\)) held by weak van der Waals forces, making it a gas. Sulfur forms large puckered crown-shaped \(S_8\) rings with greater molecular mass and stronger intermolecular forces, making it a solid.
In simple words: (i) Weaker E-H bonds down the group release protons more easily. (ii) Small oxygen molecules fly freely as a gas, while large sulfur rings pack together as a solid.

Question 30. (i) Interhalogen compounds are more reactive than halogens except \(F_2\). Why? (ii) Give one important use of \(ClF_3\).
Answer: (i) Interhalogen bonds (\(X-X'\)) are weaker than halogen-halogen bonds (\(X-X\)) because of difference in electronegativities, making them more reactive (except fluorine which has an exceptionally weak bond).
(ii) \(ClF_3\) is used as a fluorinating agent and as a propellant in rockets.
In simple words: (i) Bonds between different halogen atoms are polarized and weaker, so they break and react faster. (ii) \(ClF_3\) is used to fluorinate uranium during nuclear fuel processing.

Question 31. (i) Write the composition of bleaching powder. (ii) What happens when NaCl is heated with conc. \(H_2SO_4\) in the presence of \(MnO_2\). Write the chemical equation.
Answer: (i) Bleaching powder is a complex mixture of calcium hypochlorite, calcium chloride, and calcium hydroxide hydrates: \(Ca(OCl)_2 \cdot CaCl_2 \cdot Ca(OH)_2 \cdot 2H_2O\)
(ii) Chlorine gas is liberated:
\(4NaCl + MnO_2 + 4H_2SO_4 \rightarrow MnCl_2 + 4NaHSO_4 + 2H_2O + Cl_2 \uparrow\)
In simple words: (i) Bleaching powder is a mixed salt of calcium. (ii) NaCl heated with manganese dioxide and sulfuric acid produces chlorine gas.

Question 32. Arrange the following in the decreasing order of their basicity. Assign the reason: \(PH_3, NH_3, SbH_3, AsH_3, BiH_3\).
Answer: Decreasing order of basicity: \(NH_3 > PH_3 > AsH_3 > SbH_3 > BiH_3\)
Reason: As size increases down the group, the lone pair of electrons is spread over a larger volume, reducing electron density and making the donation of the lone pair increasingly difficult.
In simple words: Smaller nitrogen concentrates its lone pair better than heavier atoms, making ammonia the strongest base in the series.

Question 33. A colourless and a pungent smelling gas which easily liquifies to a colourless liquid and freezes to a white crystalline solid, gives dense white fumes with ammonia. Identify the gas and write the chemical equation for its laboratory preparation.
Answer: The gas is hydrogen chloride (\(HCl\)).
Laboratory preparation:
\(NaCl + H_2SO_4 \text{ (conc.)} \xrightarrow{\le 200^\circ\text{C}} NaHSO_4 + HCl \uparrow\)
In simple words: The gas is HCl, which forms dense white ammonium chloride fumes with ammonia.

Question 34. Complete following disproportionation reactions: (a) \(P_4 + NaOH + H_2O \rightarrow\) (b) \(HNO_2 \xrightarrow{H^+}\)
Answer: (a) \(P_4 + 3NaOH + 3H_2O \rightarrow PH_3 + 3NaH_2PO_2\)
(b) \(3HNO_2 \rightarrow HNO_3 + 2NO + H_2O\)
In simple words: Both phosphorus and nitrous acid undergo disproportionation when treated accordingly.

Question 35. Arrange the following trichlorides in decreasing order of bond angle \(NCl_3, PCl_3, AsCl_3, SbCl_3\).
Answer: Decreasing order of bond angle: \(NCl_3 > PCl_3 > AsCl_3 > SbCl_3\)
Reason: As the electronegativity of the central atom decreases down the group, the bond pair electrons shift further away from the central atom, reducing bond-bond repulsion and decreasing the bond angle.
In simple words: Smaller central atoms crowd bond pairs closer together, resulting in wider bond angles.

Question 36. Suggest reason why only known binary compounds of noble gases are fluorides and oxides of Krypton, Xenon.
Answer: Fluorine and oxygen are the two most electronegative and powerful oxidizing elements in the periodic table. Krypton and Xenon have relatively low ionization enthalpies, allowing F and O to oxidize them and stabilize their binary compounds.
In simple words: Only powerful oxidizers like fluorine and oxygen are electronegative enough to strip electrons from heavy noble gases.

Question 37. Which fluorinating agent are oftenly used instead of \(F_2\)? Write two chemical equations showing their use as fluorinating agents.
Answer: Halogen fluorides like \(BrF_5\) and \(IF_7\) are often used as safer fluorinating agents instead of elemental fluorine.
Equations:
\(BrF_5 + 3H_2O \rightarrow HBrO_3 + 5HF\)
\(2IF_7 + SiO_2 \rightarrow 2IOF_5 + SiF_4\)
In simple words: Compounds like \(BrF_5\) and \(IF_7\) supply fluorine atoms safely without handling dangerous fluorine gas.

Question 38. (a) Hydrolysis of \(XeF_6\) is not regarded as a redox reaction. Why? (b) Write a chemical equation to represent the oxidising nature of \(XeF_4\).
Answer: (a) During the hydrolysis of \(XeF_6\), the oxidation state of xenon remains +6 and fluorine remains -1 throughout, with no change in oxidation numbers.
(b) \(XeF_4 + 2H_2 \rightarrow Xe + 4HF\)
In simple words: (a) No electrons are transferred during \(XeF_6\) hydrolysis. (b) \(XeF_4\) oxidizes hydrogen to HF while getting reduced to xenon metal.

Question 39. Write Chemical equation: (a) \(XeF_2\) is hydrolysed (b) \(PtF_6\) and Xenon are mixed together.
Answer: (a) \(2XeF_2 + 2H_2O \rightarrow 2Xe + 4HF + O_2\)
(b) \(Xe + PtF_6 \rightarrow Xe^+[PtF_6]^-\) (Or \(XePtF_6\))
In simple words: (a) Hydrolysis of xenon difluoride yields xenon, HF, and oxygen gas. (b) Platinum hexafluoride oxidizes xenon to form a red crystalline ionic complex.

SA (II) TYPE QUESTIONS (3 - MARK QUESTIONS)

Question 1. (i) How is \(HNO_3\) prepared commercially? (ii) Write chemical equations of the reactions involved. (iii) What concentration by mass of \(HNO_3\) is obtained?
Answer: (i) Commercially, nitric acid is prepared by Ostwald's process.
(ii) Reactions:
1. \(4NH_3 + 5O_2 \xrightarrow{Pt/Rh, 500K, 9\text{ bar}} 4NO + 6H_2O\)
2. \(2NO + O_2 \rightleftharpoons 2NO_2\)
3. \(3NO_2 + H_2O \rightarrow 2HNO_3 + NO\)
(iii) The resulting nitric acid is approximately 68% concentrated by mass, which can be concentrated up to 98% by dehydration with concentrated sulfuric acid.
In simple words: Nitric acid is manufactured by catalytic oxidation of ammonia followed by absorption of nitrogen dioxide in water.

Question 2. (i) How does \(O_3\) react with lead sulphide? Write chemical equation. (ii) What happens when \(SO_2\) is passed in acidified \(KMnO_4\) solution? (iii) \(SO_2\) behaves with lime water similar to \(CO_2\).
Answer: (i) Ozone oxidizes black lead sulfide to white lead sulfate:
\(PbS + 4O_3 \rightarrow PbSO_4 + 4O_2\)
(ii) Sulfur dioxide decolorizes acidified potassium permanganate by reducing purple \(MnO_4^-\) to colorless \(Mn^{2+}\):
\(2MnO_4^- + 5SO_2 + 2H_2O \rightarrow 2Mn^{2+} + 5SO_4^{2-} + 4H^+\)
(iii) \(SO_2\) turns lime water milky due to the formation of insoluble calcium sulfite, just like \(CO_2\) forms calcium carbonate:
\(Ca(OH)_2 + SO_2 \rightarrow CaSO_3 \downarrow + H_2O\)
In simple words: Ozone acts as a powerful oxidizer, sulfur dioxide decolourizes purple permanganate, and both acidic gases turn lime water milky.

Question 3. Assign reason for the following: (i) Sulphur in vapour state exhibits paramagnetism. (ii) \(F_2\) is strongest oxidising agent among halogens. (iii) In spite of having same electronegativity, oxygen forms hydrogen bond while chlorine does not.
Answer: (i) In the vapor state, sulfur primarily exists as \(S_2\) molecules, which contain two unpaired electrons in anti-bonding \(\pi^*\) orbitals, similar to oxygen (\(O_2\)), making it paramagnetic.
(ii) Fluorine has a low bond dissociation enthalpy and high negative electron gain enthalpy, making it the strongest oxidizing agent.
(iii) Oxygen has a much smaller atomic radius than chlorine, allowing higher charge density and localized concentration of electrons necessary for strong hydrogen bonding.
In simple words: (i) \(S_2\) vapor has unpaired electrons. (ii) Fluorine easily accepts electrons. (iii) Oxygen's smaller size gives it high charge density for H-bonding.

Question 4. Give appropriate reason for each of the following: (i) Metal fluorides are more ionic than metal chlorides. (ii) Perchloric acid is stronger than sulphuric acid. (iii) Addition of chlorine to KI solution gives it a brown colour but excess of \(Cl_2\) makes it colourless.
Answer: (i) According to Fajan's rules, smaller fluoride ions are less polarized by cations than larger chloride ions, imparting more ionic character to fluorides.
(ii) Perchlorate ion (\(ClO_4^-\)) is more resonance stabilized than sulfate ion (\(SO_4^{2-}\)), making perchloric acid release protons more readily.
(iii) Chlorine liberates brown iodine from KI, but excess chlorine oxidizes iodine further into colorless iodic acid (\(HIO_3\)):
\(2KI + Cl_2 \rightarrow 2KCl + I_2\) (Brown)
\(5Cl_2 + I_2 + 6H_2O \rightarrow 2HIO_3 + 10HCl\) (Colourless)
In simple words: (i) Fluoride ions resist polarization. (ii) Perchlorate resonance stabilization enhances acid strength. (iii) Excess chlorine over-oxidizes iodine into colorless iodic acid.

Question 5. Explain why: (i) No chemical compound of helium is known. (ii) Bond dissociation energy of fluorine is less than that of chlorine. (iii) Two S-O bonds in \(SO_2\) are identical.
Answer: (i) Helium has a small, tightly held duplet electronic configuration (\(1s^2\)), extremely high ionization enthalpy, and zero electron gain enthalpy, making it chemically inert.
(ii) High interelectronic repulsions between lone pairs in the extremely small fluorine atoms weaken the F-F bond.
(iii) The two S-O bonds in \(SO_2\) are identical due to resonance, which averages out single and double bond characters.
In simple words: (i) Helium's stable duplet makes it completely unreactive. (ii) Lone pair crowding weakens fluorine's bond. (iii) Resonance makes both sulfur-oxygen bonds equal in length.

Question 6. Out of the following hydrides of group 16 elements, which will have: (i) lowest boiling point (ii) highest bond angle (iii) highest electropositive hydrogen.
Answer: Hydrides: \(H_2S, H_2O, H_2Te\).
(i) Lowest boiling point: \(H_2S\) (due to absence of hydrogen bonding and lower molar mass than \(H_2Te\)).
(ii) Highest bond angle: \(H_2O\) (\(104.5^\circ\), due to high electronegativity of oxygen pushing bond pairs closer).
(iii) Highest electropositive hydrogen: \(H_2Te\) (due to lowest electronegativity of tellurium, making the Te-H bond most polar and acidic).
In simple words: (i) \(H_2S\) boils at the lowest temperature. (ii) Water has the widest bond angle. (iii) Tellurium hydride has the most acidic/electropositive hydrogen.

Question 7. (i) How is \(XeO_3\) prepared from \(XeF_6\)? Write the chemical equation for the reaction. (ii) Draw the structure of \(XeF_4\).
Answer: (i) Complete hydrolysis of xenon hexafluoride yields xenon trioxide:
\(XeF_6 + 3H_2O \rightarrow XeO_3 + 6HF\)
(ii) Structure of \(XeF_4\): Square planar geometry with 4 bond pairs and 2 lone pairs arranged in an octahedral electron geometry.
In simple words: (i) Treating \(XeF_6\) with water produces \(XeO_3\). (ii) \(XeF_4\) has a flat square planar molecular shape.

Question 8. (i) Thermal stability of hydrides of group 16 elements decreases down the group. Why? (ii) Compare the oxidising powers of \(F_2\) and \(Cl_2\) on the basis of bond dissociation enthalpy, electron gain enthalpy of halogens and hydration enthalpy of halide ions. (iii) Write the chemical equation for the reaction of copper metal with conc. \(HNO_3\).
Answer: (i) As size increases down group 16, M-H bond length increases and bond dissociation energy decreases, reducing thermal stability.
(ii) Fluorine is a stronger oxidizing agent than chlorine because its exceptionally low bond dissociation enthalpy and high hydration enthalpy outweigh its less negative electron gain enthalpy.
(iii) \(Cu + 4HNO_3 \text{ (conc.)} \rightarrow Cu(NO_3)_2 + 2NO_2 + 2H_2O\)
In simple words: (i) Heavier hydrides break apart easier upon heating. (ii) Fluorine wins as an oxidizer due to easy bond cleavage and high hydration energy. (iii) Copper dissolves in concentrated nitric acid releasing brown nitrogen dioxide.

Question 9. An unknown salt X reacts with hot conc. \(H_2SO_4\) to produce a brown coloured gas which intensifies on addition on copper turnings. On adding dilute ferrous sulphate solution to an aqueous solution of X and then carefully adding conc. \(H_2SO_4\) along the sides of the test tube, a brown complex Y is formed at the interface between the solution and \(H_2SO_4\). Identify X and Y and write the chemical equation involved in the reaction.
Answer: - Salt X is Nitrate salt (\(NaNO_3\) or \(KNO_3\)), which releases brown \(NO_2\) gas with conc. \(H_2SO_4\), intensifying upon copper addition due to NO production.
- Complex Y is the brown ring complex: \([Fe(H_2O)_5(NO)]SO_4\).
Equation for brown ring test:
\(NO_3^- + 3Fe^{2+} + 4H^+ \rightarrow NO + 3Fe^{3+} + 2H_2O\)
\([Fe(H_2O)_6]^{2+} + NO \rightarrow [Fe(H_2O)_5(NO)]^2+ (Brown \text{ ring}) + H_2O\)
In simple words: X is a nitrate salt producing brown nitrogen dioxide, and Y is the classic nitroprusside brown ring complex.

Question 10. Assign reason to the following: (i) Noble gases have large positive values of electron gain enthalpy. (ii) Helium is used by scuba divers.
Answer: (i) Noble gases have completely filled stable valence shell configurations (\(ns^2 np^6\)). Adding an extra electron requires putting it into a higher energy shell, which is highly unfavorable.
(ii) Helium has very low solubility in blood under high pressure, preventing "bends" (decompression sickness) when deep-sea divers surface.
In simple words: (i) Noble gases already have full octets so they resist taking extra electrons. (ii) Helium's low blood solubility makes it safe for scuba breathing gas mixtures.

Question 11. Arrange the following in the order of the property indicated for each set: (a) \(F_2, Cl_2, Br_2, I_2\) (Increasing bond dissociation energy). (b) \(HF, HCl, HBr, HI\) (decreasing acid strength). (c) \(NH_3, PH_3, AsH_3, SbH_3, BiH_3\) (decreasing base strength).
Answer: (a) \(F_2 < I_2 < Br_2 < Cl_2\) (Note: Fluorine is exceptionally low due to lone-pair repulsion)
(b) \(HF < HCl < HBr < HI\) (Wait, decreasing acid strength order is \(HI > HBr > HCl > HF\))
(c) \(NH_3 > PH_3 > AsH_3 > SbH_3 > BiH_3$
In simple words: Arranged according to fundamental periodic trends in bond energy, acidity, and basicity.

Question 12. A translucent white waxy solid (A) on heating in an inert atmosphere is converted to its allotropic form (B). Allotrope (A) on reaction with very dilute aqueous NaOH liberates a highly poisonous gas (C) having a rotten fish smell, with excess of chlorine forms D which hydrolyses to form compound (E). Identify the compounds (A) to (E).
Answer: - A: White phosphorus (\(P_4\))
- B: Red phosphorus
- C: Phosphine (\(PH_3\))
- D: Phosphorus pentachloride (\(PCl_5\)) or trichloride (\(PCl_3\))
- E: Phosphoric acid (\(H_3PO_4\))
In simple words: A is white phosphorus, converting to red phosphorus (B), releasing phosphine gas (C), and reacting with chlorine to form phosphorus chlorides (D) which hydrolyze to phosphoric acid (E).

Question 13. Write balanced equation for the following reactions: (a) Zn is treated with dilute \(HNO_3\). (b) NaCl is heated with \(H_2SO_4\) in the presence of \(MnO_2\). (c) Iodine is treated with conc. \(HNO_3\).
Answer: (a) \(4Zn + 10HNO_3 \text{ (dil)} \rightarrow 4Zn(NO_3)_2 + 5H_2O + N_2O\)
(b) \(4NaCl + MnO_2 + 4H_2SO_4 \rightarrow MnCl_2 + 4NaHSO_4 + 2H_2O + Cl_2\)
(c) \(I_2 + 10HNO_3 \text{ (conc)} \rightarrow 2HIO_3 + 10NO_2 + 4H_2O\)
In simple words: Balanced chemical equations representing reduction-oxidation behavior with nitric acid and manganese dioxide.

Question 14. \(X_2\) is a greenish yellow gas with pungent offensive smell used in purification of water. It partially dissolves in \(H_2O\) to give a solution which turns blue litmus red. When \(X_2\) is passed through NaBr Solution, \(Br_2\) is obtained. (a) Identify \(X_2\), name the group to which it belongs. (b) What are the products obtained when \(X_2\) reacts with \(H_2O\)? Write chemical equation. (c) What happens when \(X_2\) reacts with hot and conc. NaOH? Give equation.
Answer: (a) \(X_2\) is Chlorine gas (\(Cl_2\)), belonging to Group 17 (Halogens).
(b) Products are hydrochloric acid and hypochlorous acid:
\(Cl_2 + H_2O \rightarrow HCl + HOCl\)
(c) Reacts to form sodium chloride, sodium chlorate, and water:
\(3Cl_2 + 6NaOH \text{ (hot/conc)} \rightarrow 5NaCl + NaClO_3 + 3H_2O\)
In simple words: Chlorine gas bleaches, water-purifies, and undergoes disproportionation with hot sodium hydroxide.

Question 16. Assign the appropriate reason for the following: (a) Nitrogen exists as diatomic molecule and phorphorous as \(P_4\), Why? (b) Why does \(R_3P = O\) exist but \(R_3N = O\) does not? (\(R =\) an alkyl group). (c) Explain why fluorine forms only one oxoacid, HOF.
Answer: (a) Small nitrogen forms stable \(p\pi - p\pi\) multiple bonds (\(N \equiv N\)), whereas phosphorus forms single covalent bonds and tetraatomic structures (\(P_4\)).
(b) Nitrogen lacks vacant \(d\)-orbitals and cannot expand its covalence beyond 4, so \(R_3N=O\) (covalence 5) cannot exist. Phosphorus has vacant \(d\)-orbitals and easily forms \(R_3P=O\).
(c) Fluorine is extremely electronegative and lacks \(d\)-orbitals, preventing higher oxidation states (+3, +5, +7) required for other oxoacids.
In simple words: (a) Nitrogen forms triple bonds while phosphorus forms single bonds. (b) Nitrogen cannot expand its octet beyond 4. (c) Fluorine's high electronegativity and lack of \(d\)-orbitals restrict it to a single oxoacid.

LONG ANSWER TYPE QUESTIONS (5 - MARK QUESTIONS)

Question 1. How is \(PH_3\) prepared in the laboratory? How is it purified? How does the solution of \(PH_3\) in water react on irradiation with light and on absorption in \(CuSO_4\)? How can you prove that \(PH_3\) is basic in nature? Write the chemical equations for all the reactions involved.
Answer: - Laboratory Preparation: By heating white phosphorus with concentrated aqueous sodium hydroxide in an inert atmosphere of \(\text{CO}_2\):
\(P_4 + 3NaOH + 3H_2O \rightarrow PH_3 + 3NaH_2PO_2\)
- Purification: Phosphine is contaminated with \(P_2H_4\) vapor, which is removed by passing it through hydriodic acid (\(HI\)) to form phosphonium iodide (\(PH_4I\)), which reacts with KOH to regenerate pure \(PH_3\).
- Reaction with light: Aqueous phosphine solution decomposes into red phosphorus and hydrogen upon irradiation with light:
\(PH_3 \xrightarrow{h\nu} P_{red} + 3H_2\)
- Reaction with \(CuSO_4\): Precipitates copper phosphide:
\(3CuSO_4 + 2PH_3 \rightarrow Cu_3P_2 \downarrow + 3H_2SO_4\)
- Basic nature test: Phosphine reacts with hydrogen halides like HI to form phosphonium iodide (\(PH_4I\)), proving its weak Lewis basic character:
\(PH_3 + HI \rightarrow PH_4I\)
In simple words: Phosphine is prepared using white phosphorus and alkali, purified via phosphonium salts, and behaves as a weak base by combining with acids.

Question 2. Assign a possible reason for the following: (a) Stability of +5 oxidation state decreases and that of +3 oxidation state increases down the group 15 elements. (b) \(H_2O\) is less acidic than \(H_2S\). (c) \(SF_6\) is inert while \(SF_4\) is highly reactive towards hydrolysis. (d) \(H_3PO_2\) and \(H_3PO_3\) act as good reducing agents while \(H_3PO_4\) does not. (e) Noble gases have comparatively large size in their respective periods.
Answer: (a) Due to the inert pair effect, \(ns^2\) electrons are reluctant to participate in bonding in heavier group 15 elements.
(b) The S-H bond is longer and weaker than the O-H bond, making \(H_2S\) dissociate protons more readily than \(H_2O\).
(c) Sulfur in \(SF_6\) is sterically crowded by six fluorine atoms and coordinatively saturated, whereas \(SF_4\) is less crowded and vulnerable to water attack.
(d) \(H_3PO_2\) and \(H_3PO_3\) contain P-H bonds which act as reducing centers. \(H_3PO_4\) lacks P-H bonds.
(e) Noble gases are measured by their van der Waals radii rather than covalent radii, which are inherently larger than the bonding radii of other elements in the same period.
In simple words: Answers rely on the inert pair effect, bond strength, steric shielding, presence of P-H reducing bonds, and van der Waals atomic sizing.

Question 3. (a) How is \(XeF_6\) prepared from the \(XeF_4\)? Write the chemical equation for the reaction. (b) Deduce the structure of \(XeF_6\) using VSEPR theory. (c) How does \(XeF_2\) reacts with \(PF_5\)? (d) Give one use each of helium and neon. (e) Write the chemical equation for the hydrolysis of \(XeF_4\).
Answer: (a) \(XeF_4 + O_2F_2 \rightarrow XeF_6 + O_2\) (at 143 K)
(b) \(XeF_6\) has 6 bond pairs and 1 lone pair (total 7 electron pairs, \(sp^3d^3\) hybridization), resulting in a distorted octahedral geometry.
(c) \(XeF_2 + PF_5 \rightarrow [XeF]^+[PF_6]^-\)
(d) Helium is used in filling meteorological balloons and diving gas cylinders; neon is used in discharge glow indicator tubes and advertising neon signs.
(e) \(6XeF_4 + 12H_2O \rightarrow 4Xe + 2XeO_3 + 24HF + 3O_2\)
In simple words: Details reactions, distorted octahedral structure of \(XeF_6\), adduct formation with \(PF_5\), commercial uses, and hydrolysis.

Question 4. (a) Why does nitrogen show anomalous behaviour? Discuss the trend of chemical reactivity of group 15 elements with: (a) oxygen (b) halogens (c) metals. (b) \(H_3PO_3\) is a dibasic acid. Why?
Answer: (a) Nitrogen shows anomalous behavior due to its small size, high electronegativity, high ionization enthalpy, and absence of \(d\)-orbitals.
Reactivity trends in Group 15:
- With Oxygen: Form oxides in +3 and +5 oxidation states (\(E_2O_3\) and \(E_2O_5\)). Acidity of oxides decreases down the group.
- With Halogens: Form trihalides (\(EX_3\)) and pentahalides (\(EX_5\)).
- With Metals: React with metals to form binary compounds in -3 oxidation state (e.g., \(Ca_3N_2\), \(Ca_3P_2\)).
(b) \(H_3PO_3\) is dibasic because it contains two ionizable -OH groups attached directly to phosphorus, while the third hydrogen is bonded directly to phosphorus (P-H bond) and is non-ionizable.

In simple words: Nitrogen is anomalous due to small size and no \(d\)-orbitals. Group 15 forms oxides, halides, and metal phosphides/nitrides, and \(H_3PO_3\) has only two replaceable protons.

Question 5. (a) Arrange the following in the order of their increasing acid strength: \(Cl_2O_7, SO_2, P_4O_10\). (b) How is \(N_2O\) gas prepared? And draw its structure. (c) Give one chemical reaction to show \(O_3\) is an oxidising agent.
Answer: (a) Increasing order of acid strength of oxides: \(P_4O_10 < SO_2 < Cl_2O_7\) (Acid strength of oxides increases with an increase in electronegativity and oxidation state of the central atom).
(b) Preparation of \(N_2O\) (Nitrous oxide):
\(NH_4NO_3 \xrightarrow{\text{heat}} N_2O + 2H_2O\)
Structure: Linear resonance hybrid structure: \(\ddot{N} = N = \ddot{O} \leftrightarrow \text{ } ^-:\!\ddot{N}-N \equiv O^+\)
(c) Ozone oxidizes potassium iodide to iodine:
\(2KI + H_2O + O_3 \rightarrow 2KOH + I_2 + O_2\)
In simple words: (a) Oxides get more acidic as the central atom's oxidation state rises. (b) Heating ammonium nitrate yields \(N_2O\). (c) Ozone oxidizes iodide ions to free iodine.

Question 6. Identify A, B, C, D and E in the following sequence of reactions... [Sequence involves \(P_4 \xrightarrow{Cl_2} B \xrightarrow{H_2O} C \text{ etc.}\)]
Answer: - A: White phosphorus (\(P_4\))
- B: Phosphorus pentachloride (\(PCl_5\)) or trichloride (\(PCl_3\))
- C: Phosphoric acid (\(H_3PO_4\)) or phosphorous acid (\(H_3PO_3\))
- D: Yellowish white solid / derivatives
- E: Hydrolysis product
In simple words: Tracing the chemical reactions starting from white phosphorus through chlorination and hydrolysis yields corresponding phosphorus halides and acids.

Question 7. A white waxy, translucent solid, M, insoluble in water but soluble in \(CS_2\), glows in dark. M dissolves in NaOH in an inert atmosphere giving a poisonous gas (N). Also M catches fire to give dense white fumes of Q: (a) Identify M, N and Q and write the chemical equations of the reactions involved. (b) M exists in the form of discrete tetrahedral molecules. Draw its structure. (c) M on heating at 573 K is changed into other less reactive form, Q, which is non-poisonous, insoluble in water as well as in \(CS_2\) and does not glow in dark, Identify Q and draw its structure.
Answer: (a) - M: White phosphorus (\(P_4\))
- N: Phosphine (\(PH_3\))
- Q: Phosphorus pentoxide (\(P_4O_{10}\))
Equations:
\(P_4 + 3NaOH + 3H_2O \rightarrow PH_3 + 3NaH_2PO_2\)
\(P_4 + 5O_2 \rightarrow P_4O_{10}\)
(b) White phosphorus consists of discrete \(P_4\) tetrahedra with bond angles of \(60^\circ\).
(c) Q is red phosphorus, which consists of polymeric chains of linked \(P_4\) tetrahedra.
In simple words: M is white phosphorus, N is phosphine, and heating transforms white phosphorus into polymeric red phosphorus.

Question 8. Write the structure of A, B, C, D and E in the following sequence of reactions... [Ostwald process sequence: \(NH_3 + O_2 \xrightarrow{Pt/Rh} A \rightarrow B \rightarrow C + A\)]
Answer: - A: Nitric oxide (\(NO\))
- B: Nitrogen dioxide (\(NO_2\))
- C: Nitric acid (\(HNO_3\)) - obtained via the Ostwald process.
- D: Iodine liberated or related oxidation product.
In simple words: Identifies the intermediates of the commercial Ostwald process for manufacturing nitric acid from ammonia.

Question 9. Give reason for each of the following: (a) \(NH_3\) is more basic than \(PH_3\). (b) Ammonia is a good complexing agent. (c) Bleaching by \(SO_2\) is temporary. (d) \(PCl_5\) is ionic in solid state. (e) Sulphur in vapour state exhibits paramagnetism.
Answer: (a) Nitrogen is smaller and has higher electron density, making its lone pair more available than in larger phosphorus.
(b) Ammonia has a lone pair on nitrogen that it donates to transition metal ions to form stable coordination complexes.
(c) Bleaching by sulfur dioxide involves reduction, which is air-oxidized back to a colored state over time.
(d) In solid state, \(PCl_5\) exists as an ionic crystal lattice: \([PCl_4]^+[PCl_6]^-\).
(e) Sulfur vapor contains \(S_2\) molecules with two unpaired electrons in antibonding orbitals.
In simple words: Covers basicity, complex formation, temporary reduction bleaching, ionic solid structure, and sulfur vapor paramagnetism.

Question 10. Knowing the electrons gain enthalpy value for \(O \rightarrow O^-\) and \(O^- \rightarrow O^{2-}\) as -141 and \(+720 \text{ kJ mol}^{-1}\) respectively, how can you account for the formation of large number of oxides having \(O^{2-}\) species and not \(O^-\)?
Answer: Although adding the second electron to \(O^-\) is endothermic (\(+720 \text{ kJ mol}^{-1}\)) due to electrostatic repulsion, the high lattice enthalpy released during the formation of ionic metal oxides with \(O^{2-}\) ions more than compensates for this energy penalty, stabilizing the divalent oxide lattice.
In simple words: Even though adding a second electron takes energy, the huge amount of lattice energy released when crystals form makes \(O^{2-}\) oxides stable.