GSEB Class 12 Chemistry Solutions Chapter 7 The p-Block Elements

GSEB Class 12 Chemistry InText Questions and Answers

**Question 1. Why are pentahalides more covalent than trihalides?**
**Answer:** Higher positive oxidation states of the central atom result in increased polarizing power. This enhanced polarizing power, in turn, boosts the covalent nature of the bond formed between the central atom and other atoms. Consequently, pentahalides exhibit greater covalent character compared to trihalides.
In simple words: When the central atom has a higher positive charge, it pulls electrons more effectively, making its bonds with other atoms more covalent. Pentahalides have a higher positive oxidation state, so their bonds are more covalent.

🎯 Exam Tip: Focus on the correlation between oxidation state and polarizing power when discussing covalent character in halides. Remember Fajans' rules for such explanations.

**Question 2. Why is BiH3 the strongest reducing agent amongst all the hydrides of Group 15 elements?**
**Answer:** As one moves down Group 15, the bond strength between the central element (E) and hydrogen (H) decreases. This weakening of the E-H bond leads to an increase in the reducing capability of the hydrides. Bismuth hydride (\( \text{BiH}_3 \)) is the least stable among the Group 15 hydrides, making it the most potent reducing agent.
In simple words: Going down the group, the bond holding the hydrogen gets weaker, so it's easier to release hydrogen, which means it's a stronger reducing agent. \( \text{BiH}_3 \) has the weakest bond, making it the strongest.

🎯 Exam Tip: Understand the trend of bond strength and its direct impact on reducing properties for hydrides within a group. Relate it to atomic size and bond dissociation energy.

**Question 3. Why is N2 less reactive at room temperature?**
**Answer:** The dinitrogen molecule (\( \text{N}_2 \)) features a robust pπ-pπ overlap, leading to the formation of a triple bond between the two nitrogen atoms (\( \text{N} \equiv \text{N} \)). This triple bond results in an exceptionally high bond dissociation energy for \( \text{N}_2 \), rendering it significantly less reactive at ambient temperatures.
In simple words: Nitrogen gas has a very strong triple bond between its atoms, which requires a lot of energy to break, making it unreactive at normal temperatures.

🎯 Exam Tip: Highlight the role of the triple bond and high bond dissociation energy as key factors for nitrogen's inertness.

**Question 4. Mention the conditions required to maximise the yield of ammonia?**
**Answer:** To achieve the maximum yield of ammonia, the following conditions are necessary:
1. Low temperature: The optimal temperature range is 650-800 K.
2. High pressure: An optimal pressure of 200 atm is required.
3. Catalyst: Finely divided iron, enhanced by molybdenum as a promoter, is used as a catalyst.
In simple words: To get the most ammonia, use moderate heat, high pressure, and a special iron catalyst.

🎯 Exam Tip: Remember these three key conditions - temperature, pressure, and catalyst - as they are crucial for the industrial production of ammonia (Haber process).

**Question 5. How does ammonia react with a solution of \( \text{Cu}^{2+} \)?**
**Answer:** Ammonia reacts with \( \text{Cu}^{2+} \) ions to form a deep blue-colored complex compound.
\( \text{Cu}^{2+} (\text{aq}) + 4\text{NH}_3 (\text{aq}) \longrightarrow [\text{Cu}(\text{NH}_3)_4]^{2+} (\text{aq}) \)
(deep blue)
In simple words: Ammonia combines with copper ions to create a vivid blue complex solution.

🎯 Exam Tip: This reaction is a classic test for \( \text{Cu}^{2+} \) ions, illustrating ammonia's role as a ligand in complex formation due to its lone pair of electrons.

**Question 6. What is the covalence of nitrogen in \( \text{N}_2\text{O}_5 \)?**
**Answer:** The structure of \( \text{N}_2\text{O}_5 \) indicates that the covalence of nitrogen is four.
ℹ️ चित्र व्याख्या (Diagram Explanation): इस चित्र में dinitrogen pentoxide (\( \text{N}_2\text{O}_5 \)) की संरचना दर्शाई गई है। इसमें दो नाइट्रोजन परमाणु एक ऑक्सीजन परमाणु से जुड़े होते हैं, और प्रत्येक नाइट्रोजन परमाणु दो अन्य ऑक्सीजन परमाणुओं से द्विक बंध द्वारा जुड़ा होता है। यह संरचना नाइट्रोजन की चतुर्संयोजकता (four covalence) को स्पष्ट करती है।
In simple words: In the \( \text{N}_2\text{O}_5 \) molecule, each nitrogen atom forms four bonds, which means its covalence is four.

🎯 Exam Tip: To determine covalence, always refer to the molecule's structure and count the total number of covalent bonds formed by the atom in question.

**Question 7. Bond angle in \( \text{PH}_4^+ \) is higher than that in \( \text{PH}_3 \). Why?**
**Answer:** The \( \text{PH}_4^+ \) ion possesses a regular tetrahedral geometry, resulting in a bond angle of \( 109^\circ 28' \). In contrast, \( \text{PH}_3 \) has a pyramidal structure with a lone pair of electrons, which causes greater lone pair-bond pair repulsion, compressing the bond angles to approximately \( 93^\circ \). Therefore, the bond angle in \( \text{PH}_4^+ \) is significantly larger.
In simple words: \( \text{PH}_4^+ \) is a perfect tetrahedron with no lone pairs, so its bond angles are ideal. \( \text{PH}_3 \) has a lone pair that pushes the bonding pairs closer, making its bond angles smaller.

🎯 Exam Tip: Apply VSEPR theory to explain bond angles, focusing on the presence or absence of lone pairs and their repulsive effects on bonding pairs.

**Question 8. What happens when white phosphorus is heated with concentrated \( \text{NaOH} \) solution in an inert atmosphere of \( \text{CO}_2 \)?**
**Answer:** When white phosphorus is subjected to heating with a concentrated solution of sodium hydroxide in an inert carbon dioxide atmosphere, phosphine gas (\( \text{PH}_3 \)) is produced.
\( \text{P}_4 + 3\text{NaOH} + 3\text{H}_2\text{O} \longrightarrow \text{PH}_3 + 3\text{NaH}_2\text{PO}_2 \)
In simple words: Heating white phosphorus with strong lye (NaOH) in a \( \text{CO}_2 \)-filled space creates phosphine gas.

🎯 Exam Tip: This is a redox reaction where phosphorus undergoes disproportionation. Pay attention to the role of the inert atmosphere preventing unwanted oxidation.

**Question 9. What happens when \( \text{PCl}_5 \) is heated?**
**Answer:** Upon strong heating, phosphorus pentachloride (\( \text{PCl}_5 \)) undergoes decomposition, yielding phosphorus trichloride (\( \text{PCl}_3 \)) and chlorine gas (\( \text{Cl}_2 \)).
\( \text{PCl}_5 \longrightarrow \text{PCl}_3 + \text{Cl}_2 \)
In simple words: When \( \text{PCl}_5 \) is heated strongly, it breaks down into \( \text{PCl}_3 \) and chlorine gas.

🎯 Exam Tip: Recognize this as a thermal decomposition reaction. Note the reversibility of this process at lower temperatures.

**Question 10. Write a balanced equation for the hydrolytic reaction of \( \text{PCl}_5 \) in heavy water.**
**Answer:** The hydrolysis of phosphorus pentachloride (\( \text{PCl}_5 \)) with heavy water (\( \text{D}_2\text{O} \)) produces phosphoryl trichloride (\( \text{POCl}_3 \)) and deuterium chloride (\( \text{DCl} \)).
\( \text{PCl}_5 + \text{D}_2\text{O} \longrightarrow \text{POCl}_3 + 2\text{DCl} \)
In simple words: \( \text{PCl}_5 \) reacts with heavy water to form \( \text{POCl}_3 \) and \( \text{DCl} \).

🎯 Exam Tip: Understand that heavy water (\( \text{D}_2\text{O} \)) reacts similarly to normal water (\( \text{H}_2\text{O} \)) in hydrolysis reactions, replacing hydrogen atoms with deuterium.

**Question 11. What is the basicity of \( \text{H}_3\text{PO}_4 \)?**
**Answer:** The basicity of orthophosphoric acid (\( \text{H}_3\text{PO}_4 \)) is determined by the number of ionizable P-OH bonds it possesses. As its structure reveals three P-OH groups, the basicity of \( \text{H}_3\text{PO}_4 \) is three.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र ऑर्थोफॉस्फोरिक एसिड (\( \text{H}_3\text{PO}_4 \)) की संरचना को दर्शाता है। इसमें एक केंद्रीय फास्फोरस परमाणु एक ऑक्सीजन परमाणु से द्विक बंध द्वारा और तीन हाइड्रॉक्सिल (-OH) समूहों से एकल बंध द्वारा जुड़ा होता है। ये तीन -OH समूह ही इसकी त्रि-भास्मिक प्रकृति को निर्धारित करते हैं।
In simple words: The basicity of \( \text{H}_3\text{PO}_4 \) is three because it has three -OH groups directly attached to phosphorus that can release protons.

🎯 Exam Tip: Basicity of oxyacids of phosphorus is determined by the number of P-OH bonds, not the total number of hydrogen atoms.

**Question 12. What happens when \( \text{H}_3\text{PO}_3 \) is heated?**
**Answer:** When phosphorus acid (\( \text{H}_3\text{PO}_3 \)) is heated, it undergoes a disproportionation reaction. This process yields orthophosphoric acid (\( \text{H}_3\text{PO}_4 \)) and phosphine gas (\( \text{PH}_3 \)).
\( 4\text{H}_3\text{PO}_3 \longrightarrow 3\text{H}_3\text{PO}_4 + \text{PH}_3 \)
In simple words: Heating \( \text{H}_3\text{PO}_3 \) causes it to transform into \( \text{H}_3\text{PO}_4 \) and \( \text{PH}_3 \).

🎯 Exam Tip: Disproportionation reactions involve a single element being simultaneously oxidized and reduced. Here, phosphorus in \( \text{H}_3\text{PO}_3 \) (oxidation state +3) changes to +5 in \( \text{H}_3\text{PO}_4 \) and -3 in \( \text{PH}_3 \).

**Question 13. List the important sources of sulphur?**
**Answer:** Important sources of sulfur include:
- **Sulphates:** Gypsum (\( \text{CaSO}_4 \cdot 2\text{H}_2\text{O} \)), Epsom salt (\( \text{MgSO}_4 \cdot 7\text{H}_2\text{O} \)), and Baryte (\( \text{BaSO}_4 \)).
- **Sulphides:** Galena (\( \text{PbS} \)), Zinc blende (\( \text{ZnS} \)), Copper pyrites (\( \text{CuFeS}_2 \)), and Iron pyrites (\( \text{FeS}_2 \)) are common examples.
- **Organic materials:** Sulfur is also found in small quantities within various organic substances such as eggs, proteins, garlic, onion, mustard, hair, and wool.
In simple words: Sulfur comes from minerals like gypsum and galena, and is also present in many natural organic materials like eggs and hair.

🎯 Exam Tip: Remember both inorganic mineral sources (sulfates and sulfides) and organic sources for sulfur, as its distribution is broad in nature.

**Question 14. Write the order of thermal stability of the hydrides of Group 16 elements?**
**Answer:** The thermal stability of Group 16 hydrides decreases as the size of the central atom increases, leading to the following order:
\( \text{H}_2\text{O} > \text{H}_2\text{S} > \text{H}_2\text{Se} > \text{H}_2\text{Te} > \text{H}_2\text{Po} \)
In simple words: Water is the most stable hydride, and stability decreases as you go down Group 16 elements.

🎯 Exam Tip: Thermal stability of hydrides in a group is inversely related to bond length; longer bonds are weaker and less stable thermally.

**Question 15. Why is \( \text{H}_2\text{O} \) a liquid and \( \text{H}_2\text{S} \) a gas?**
**Answer:** Water (\( \text{H}_2\text{O} \)) is a liquid at room temperature due to the high electronegativity and small atomic size of oxygen, which facilitates strong intermolecular hydrogen bonding. This results in the association of water molecules. In contrast, hydrogen sulfide (\( \text{H}_2\text{S} \)) lacks hydrogen bonding because sulfur has significantly lower electronegativity compared to oxygen, causing it to exist as a gas at room temperature.
In simple words: Water molecules are held together by strong hydrogen bonds because oxygen is very electronegative, making it a liquid. \( \text{H}_2\text{S} \) doesn't have these strong bonds because sulfur is less electronegative, so it's a gas.

🎯 Exam Tip: This is a crucial concept demonstrating the impact of hydrogen bonding on physical states. Always relate electronegativity and atomic size to the presence and strength of hydrogen bonds.

**Question 16. Which of the following does not react with oxygen directly? Zn, Ti, Pt, Fe**
**Answer:** Among the given elements- Zinc (\( \text{Zn} \)), Titanium (\( \text{Ti} \)), Platinum (\( \text{Pt} \)), and Iron (\( \text{Fe} \)) - **Platinum (\( \text{Pt} \))** is the element that typically does not react directly with oxygen.
In simple words: Platinum does not easily combine with oxygen directly.

🎯 Exam Tip: Understand the reactivity series of metals. Noble metals like platinum and gold are known for their inertness and resistance to direct oxidation.

**Question 17. Complete the following reactions:**
1. \( \text{C}_2\text{H}_4 + \text{O}_2 \longrightarrow \)
2. \( 4\text{Al} + 3\text{O}_2 \longrightarrow \)
**Answer:** The completed reactions are:
1. \( \text{C}_2\text{H}_4 + 3\text{O}_2 \longrightarrow 2\text{CO}_2 + 2\text{H}_2\text{O} \)
2. \( 4\text{Al} + 3\text{O}_2 \longrightarrow 2\text{Al}_2\text{O}_3 \)
In simple words: These are combustion and oxidation reactions, where ethene burns in oxygen to form carbon dioxide and water, and aluminum reacts with oxygen to form aluminum oxide.

🎯 Exam Tip: For combustion reactions, ensure that carbon forms \( \text{CO}_2 \) and hydrogen forms \( \text{H}_2\text{O} \). For metal oxidation, ensure correct stoichiometry for the metal oxide formed.

**Question 18. O3 decomposes to give nascent oxygen and hence acts as a powerful oxidising agent?**
**Answer:** (Assuming the question asks to explain why O3 is a powerful oxidizing agent due to nascent oxygen) Ozone (\( \text{O}_3 \)) acts as a strong oxidizing agent because it readily decomposes to yield molecular oxygen (\( \text{O}_2 \)) and nascent oxygen ([O]), which is highly reactive.
\( \text{O}_3 \longrightarrow \text{O}_2 + [\text{O}] \) This nascent oxygen is responsible for ozone's potent oxidizing properties.
In simple words: Ozone is a strong oxidizer because it easily breaks down to release highly reactive single oxygen atoms.

🎯 Exam Tip: Remember that nascent oxygen is extremely reactive, making substances that release it powerful oxidizing agents. This is a key characteristic of ozone.

**Question 19. How is \( \text{O}_3 \) estimated quantitatively?**
**Answer:** Ozone (\( \text{O}_3 \)) can be quantitatively estimated by its ability to oxidize potassium iodide solution into iodine. The reaction is:
\( 2\text{KI} + \text{H}_2\text{O} + \text{O}_3 \longrightarrow 2\text{KOH} + \text{I}_2 + \text{O}_2 \)
In this method, an excess amount of potassium iodide solution, buffered to a pH of 9.2 with a borate buffer, is treated with ozone. The liberated iodine (\( \text{I}_2 \)) is then titrated against a standard solution of sodium thiosulphate, using starch as an indicator to detect the endpoint. This precise titration allows for the quantitative determination of ozone.
In simple words: Ozone's amount is measured by letting it react with potassium iodide to produce iodine, which is then measured by titration with sodium thiosulphate.

🎯 Exam Tip: This method relies on a redox titration. Understand the role of the buffer (to maintain pH) and starch (as an indicator for iodine). It's a key analytical procedure.

**Question 20. What happens when sulphur dioxide is passed through an aqueous solution of \( \text{Fe(III)} \) salt?**
**Answer:** When sulfur dioxide (\( \text{SO}_2 \)) is bubbled through an aqueous solution containing ferric (\( \text{Fe(III)} \)) ions, it acts as a reducing agent, converting the \( \text{Fe}^{3+} \) ions to ferrous (\( \text{Fe(II)} \)) ions.
\( 2\text{Fe}^{3+} + \text{SO}_2 + 2\text{H}_2\text{O} \longrightarrow 2\text{Fe}^{2+} + \text{SO}_4^{2-} + 4\text{H}^+ \)
In simple words: Sulfur dioxide changes iron(III) ions into iron(II) ions in water, showing its reducing ability.

🎯 Exam Tip: Remember that sulfur dioxide can act as both a reducing and an oxidizing agent depending on the reactant. Here, it exhibits its reducing property by being oxidized to sulfate.

**Question 21. Comment on the nature of two S-O bonds formed in \( \text{SO}_2 \) molecule. Are the two S-O bonds in this molecule equal?**
**Answer:** In the sulfur dioxide (\( \text{SO}_2 \)) molecule, both sulfur-oxygen bonds are covalent and exhibit equal strength. This equivalency is attributed to the phenomenon of resonance, where the actual structure is a hybrid of multiple contributing resonating structures.
ℹ️ चित्र व्याख्या (Diagram Explanation): इस चित्र में सल्फर डाइऑक्साइड (\( \text{SO}_2 \)) की अनुनादी संरचनाएं दर्शाई गई हैं। इसमें एक सल्फर परमाणु दो ऑक्सीजन परमाणुओं से जुड़ा होता है, जिसमें एक ऑक्सीजन से द्विक बंध और दूसरे से एकल बंध होता है। हालांकि, अनुनाद के कारण, दोनों S-O बंधों में आंशिक द्विक बंध गुण होता है, जिससे वे लंबाई और शक्ति में समान होते हैं, जो संरचना में बिंदीदार रेखाओं से इंगित होता है।
In simple words: The two S-O bonds in \( \text{SO}_2 \) are identical in strength and length because of resonance, which means the electrons are delocalized across both bonds.

🎯 Exam Tip: Resonance is a key concept for explaining bond equivalency in molecules where multiple Lewis structures can be drawn. It indicates delocalization of electrons.

**Question 22. How is the presence of \( \text{SO}_2 \) detected?**
**Answer:** Moist sulfur dioxide (\( \text{SO}_2 \)) acts as a reducing agent. Its presence can be detected by its ability to decolorize an acidified potassium permanganate solution.
\( 5\text{SO}_2 + 2\text{MnO}_4^- + 2\text{H}_2\text{O} \longrightarrow 5\text{SO}_4^{2-} + 4\text{H}^+ + 2\text{Mn}^{2+} \) The purple permanganate ion (\( \text{MnO}_4^- \)) is reduced to the colorless \( \text{Mn}^{2+} \) ion.
In simple words: Sulfur dioxide is detected by observing its ability to turn purple acidified potassium permanganate solution colorless.

🎯 Exam Tip: The decolonization of acidified potassium permanganate is a common qualitative test for reducing agents, and \( \text{SO}_2 \) is a classic example. Note the color change from purple to colorless.

**Question 23. Mention three areas in which \( \text{H}_2\text{SO}_4 \) plays an important role?**
**Answer:** Sulfuric acid (\( \text{H}_2\text{SO}_4 \)) is vital in several industries:
- **Fertilizer manufacturing:** It is used to produce fertilizers such as ammonium sulfate and calcium superphosphate.
- **Petroleum refining:** \( \text{H}_2\text{SO}_4 \) is employed in the purification and processing of petroleum products.
- **Metallurgical applications:** It plays a role in cleansing metals before various processes like enameling, electroplating, and galvanizing.
In simple words: Sulfuric acid is essential for making fertilizers, refining oil, and cleaning metals for industrial uses.

🎯 Exam Tip: Sulfuric acid is a 'king of chemicals' due to its wide range of industrial applications. Be prepared to list its major uses.

**Question 24. Write the conditions to maximise the yield of \( \text{H}_2\text{SO}_4 \) by contact process?**
**Answer:** To optimize the yield of sulfuric acid via the Contact Process, the following conditions are critical:
1. **Low temperature:** The conversion of sulfur dioxide (\( \text{SO}_2 \)) to sulfur trioxide (\( \text{SO}_3 \)) is an exothermic reaction. An optimum temperature of approximately 720 K favors the forward reaction and maximizes \( \text{SO}_3 \) formation.
2. **High pressure:** The reaction involves a decrease in the number of gaseous moles. Thus, a high pressure of 2 atm shifts the equilibrium towards the product side, increasing the yield.
3. **Catalyst:** The use of a catalyst, such as platinum (\( \text{Pt} \)) or vanadium pentoxide (\( \text{V}_2\text{O}_5 \)), significantly increases the reaction rate, allowing the process to reach equilibrium faster.
In simple words: For the best sulfuric acid yield, use moderate temperature, high pressure, and a catalyst like platinum or vanadium pentoxide.

🎯 Exam Tip: Remember Le Chatelier's principle when explaining the effects of temperature and pressure on the exothermic and volume-decreasing reaction in the Contact Process.

**Question 25. Why is \( \text{K}_{a2} \ll \text{K}_{a1} \) for \( \text{H}_2\text{SO}_4 \) in water?**
**Answer:** Sulfuric acid (\( \text{H}_2\text{SO}_4 \)) is considered a very strong acid in water primarily due to its nearly complete first ionization:
\( \text{H}_2\text{SO}_4 + \text{H}_2\text{O} \longrightarrow \text{H}_3\text{O}^+ + \text{HSO}_4^- \)
However, the second ionization of the bisulfate ion (\( \text{HSO}_4^- \)) to form \( \text{H}_3\text{O}^+ \) and sulfate ion (\( \text{SO}_4^{2-} \)) is very limited. This significant difference in the extent of ionization is why the second dissociation constant (\( \text{K}_{a2} \)) is much, much smaller than the first dissociation constant (\( \text{K}_{a1} \)).
In simple words: \( \text{H}_2\text{SO}_4 \) easily releases its first proton, making its \( \text{K}_{a1} \) very high. But it's much harder for the remaining \( \text{HSO}_4^- \) ion to release its second proton, so its \( \text{K}_{a2} \) is very low.

🎯 Exam Tip: For polyprotic acids, successive ionization constants always decrease because it becomes progressively harder to remove a proton from an already negatively charged species.

**Question 26. Give two examples to show the anomalous behaviour of fluorine?**
**Answer:** Two instances illustrating the anomalous behavior of fluorine are:
(i) **Oxidation State:** Being the most electronegative element, fluorine exclusively exhibits a negative oxidation state of -1 and never displays any positive oxidation states. In contrast, other halogens can show various positive oxidation states, such as +1, +3, +5, +6, and +7.
(ii) **Covalency:** Fluorine's maximum covalency is restricted to one because it lacks vacant d-orbitals in its valence shell, preventing it from expanding its octet. Other elements in the same group, however, can achieve covalencies up to 7 due to the availability of vacant d-orbitals.
In simple words: Fluorine is different because it only forms a -1 oxidation state and can only make one covalent bond, unlike other halogens that can have positive oxidation states and make more bonds.

🎯 Exam Tip: The anomalous behavior of the first element in any group (like fluorine in Group 17) is usually attributed to its small size, high electronegativity, and absence of d-orbitals.

**Question 27. Sea is the greatest source of some halogens. Comment?**
**Answer:** The statement holds true as seawater is a primary reservoir for certain halogens. It contains various halides, specifically chlorides, bromides, and iodides of common alkali and alkaline earth metals such as sodium (\( \text{Na} \)), potassium (\( \text{K} \)), magnesium (\( \text{Mg} \)), and calcium (\( \text{Ca} \)). For instance, sodium chloride (\( \text{NaCl} \)) constitutes about 2.5% by mass of seawater. Furthermore, marine algae like seaweeds are known to accumulate iodine, containing up to 0.5% of this element.
In simple words: The ocean is a huge source of halogens like chlorine, bromine, and iodine, found as salts or in seaweeds.

🎯 Exam Tip: Focus on the specific forms in which halogens are found in seawater (halides) and natural concentrators like seaweeds for iodine.

**Question 28. Give the reason for the bleaching action of \( \text{Cl}_2 \).**
**Answer:** The bleaching effect of chlorine (\( \text{Cl}_2 \)) stems from its oxidizing power. In the presence of moisture, chlorine reacts with water to produce hydrochloric acid (\( \text{HCl} \)) and highly reactive nascent oxygen ([O]):
\( \text{Cl}_2 + \text{H}_2\text{O} \longrightarrow 2\text{HCl} + [\text{O}] \)
This nascent oxygen then oxidizes the colored substance, transforming it into a colorless product:
Colouring substance \( + [\text{O}] \longrightarrow \) Colorless substance.
This oxidizing action effectively bleaches organic matter, such as vegetables, and is considered a permanent bleaching process.
In simple words: Chlorine bleaches by reacting with water to release highly reactive oxygen, which then destroys the color of a substance permanently.

🎯 Exam Tip: Distinguish between bleaching by oxidation (permanent, e.g., \( \text{Cl}_2 \)) and bleaching by reduction (often temporary). The formation of nascent oxygen is key here.

**Question 29. Name two poisonous gases which can be prepared from chlorine gas?**
**Answer:** Two highly toxic gases that can be synthesized using chlorine gas are:
1. Phosgene (\( \text{COCl}_2 \))
2. Tear gas (\( \text{CCl}_3\text{NO}_2 \), also known as chloropicrin)
In simple words: Chlorine gas can be used to make poisonous gases like phosgene and tear gas.

🎯 Exam Tip: Remember these as important industrial or chemical warfare-related compounds derived from chlorine. Phosgene is a well-known highly toxic gas.

**Question 30. Why is \( \text{ICl} \) more reactive than \( \text{I}_2 \)?**
**Answer:** Interhalogen compounds, such as \( \text{ICl} \), are generally more reactive than elemental halogens, like \( \text{I}_2 \). This heightened reactivity is due to the inherent weakness of the bond between different halogen atoms (X-X') compared to the bond between identical halogen atoms (X-X). Specifically, the I-Cl bond in \( \text{ICl} \) is weaker than the I-I bond in \( \text{I}_2 \), making \( \text{ICl} \) more susceptible to breaking and thus more reactive.
In simple words: \( \text{ICl} \) is more reactive than \( \text{I}_2 \) because the bond between iodine and chlorine is weaker than the bond between two iodine atoms.

🎯 Exam Tip: The difference in electronegativity between two different halogen atoms in an interhalogen compound leads to a polarized and often weaker bond compared to the nonpolar, stronger bond in diatomic elemental halogens.

**Question 31. Why is helium used in the diving apparatus?**
**Answer:** Helium is utilized as a diluent for oxygen in modern diving apparatus because of its exceptionally low solubility in blood. This characteristic helps prevent decompression sickness (the "bends") by reducing the amount of gas dissolved in a diver's bloodstream at high pressures.
In simple words: Helium is used in diving gear because it doesn't dissolve much in the blood, which helps prevent health problems when divers surface.

🎯 Exam Tip: The low solubility of helium in body fluids is its key property for use in diving, distinguishing it from nitrogen, which causes issues at high pressures.

**Question 32. Balance the following equation: \( \text{XeF}_6 + \text{H}_2\text{O} \longrightarrow \text{XeO}_2\text{F}_2 + \text{HF} \)**
**Answer:** The balanced equation for the reaction is:
\( \text{XeF}_6 + 2\text{H}_2\text{O} \longrightarrow \text{XeO}_2\text{F}_2 + 4\text{HF} \)
In simple words: Xenon hexafluoride reacts with two molecules of water to produce xenon dioxydifluoride and four molecules of hydrofluoric acid.

🎯 Exam Tip: When balancing hydrolysis reactions of xenon fluorides, ensure that all atoms, especially oxygen and hydrogen from water, are accounted for on both sides.

**Question 33. Why has it been difficult to study the chemistry of radon?**
**Answer:** Studying the chemistry of radon (\( \text{Rn} \)) has proven challenging because it is a radioactive element with an extremely short half-life, approximately 3.82 days. This rapid radioactive decay makes it difficult to conduct extensive experimental research on its chemical properties.
In simple words: Radon is hard to study because it's radioactive and decays very quickly.

🎯 Exam Tip: The short half-life of radioactive elements significantly limits the time available for chemical experimentation, making their study inherently difficult.

GSEB Class 12 Chemistry The p-Block Elements Text Book Questions and Answers

**Question 1. Discuss the general characteristics of Group 15 elements with reference to their electronic configuration, oxidation state, atomic size, ionization enthalpy and electronegativity?**
**Answer:** The Group 15 elements of the periodic table include five elements: nitrogen (\( \text{N} \)), phosphorus (\( \text{P} \)), arsenic (\( \text{As} \)), antimony (\( \text{Sb} \)), and bismuth (\( \text{Bi} \)). These elements are commonly referred to as 'Pnicogens,' and their compounds are 'Pniconides,' a term derived from the Greek word 'pnigo,' meaning 'to choke' or 'suffocation.'
In simple words: Group 15 elements are called Pnicogens and include N, P, As, Sb, Bi. Their properties are defined by their electron arrangement, how they react with electrons, their size, and the energy needed to remove electrons.

🎯 Exam Tip: When discussing general characteristics, always address the periodic trends of electronic configuration, atomic size, ionization enthalpy, and electronegativity for a comprehensive answer.

**Question 2. Why does the reactivity of nitrogen differ from phosphorus?**
**Answer:** The reactivity difference between nitrogen and phosphorus arises primarily from their structural characteristics. Nitrogen, being a small atom, readily forms strong pπ-pπ multiple bonds, existing as a diatomic molecule (\( \text{N} \equiv \text{N} \)) with a triple bond. This triple bond imparts a very high bond dissociation energy to \( \text{N}_2 \), rendering it inert and unreactive at room temperature. In contrast, phosphorus exists as a tetrahedral \( \text{P}_4 \) molecule, where the P-P single bonds are considerably weaker than the N-N triple bond. Consequently, phosphorus exhibits higher reactivity than nitrogen.
In simple words: Nitrogen is unreactive because it forms a very strong triple bond. Phosphorus is more reactive because it forms weaker single bonds in its \( \text{P}_4 \) structure.

🎯 Exam Tip: Compare the bond types (triple vs. single) and molecular structures (\( \text{N}_2 \) vs. \( \text{P}_4 \)) to explain the reactivity differences between elements in the same group.

**Question 3. Discuss the trends in chemical reactivity of group 15 elements.**
**Answer:** Nitrogen is exceptionally inert due to the extreme stability of the triple bond in the \( \text{N} \equiv \text{N} \) molecule, which demands a substantial amount of dissociation energy to break. This explains why nitrogen is not found in its free state in nature. Phosphorus, on the other hand, is more reactive than nitrogen, largely attributed to the weaker P-P covalent bonds present among phosphorus atoms in its \( \text{P}_4 \) molecule.
(i) **Oxides:** The elements of Group 15 form oxides, typically exhibiting +3 and +5 oxidation states. Nitrogen produces a range of oxides with oxidation numbers spanning from +1 to +5, including nitrous oxide (\( \text{N}_2\text{O} \)), nitric oxide (\( \text{NO} \)), dinitrogen trioxide (\( \text{N}_2\text{O}_3 \)), nitrogen tetroxide or nitrogen dioxide (\( \text{NO}_2 \) or \( \text{N}_2\text{O}_4 \)), and dinitrogen pentoxide (\( \text{N}_2\text{O}_5 \)).
\( \text{N}_2\text{O} \) and \( \text{NO} \) are neutral, while \( \text{N}_2\text{O}_3 \), \( \text{N}_2\text{O}_4 \), and \( \text{N}_2\text{O}_5 \) are purely acidic. Phosphorus oxides, such as \( \text{P}_4\text{O}_6 \), \( \text{P}_4\text{O}_8 \), and \( \text{P}_4\text{O}_{10} \), are also purely acidic. Oxides of arsenic and antimony are amphoteric, whereas bismuth oxides are basic.
The acidic character of oxides with +3 and +5 oxidation states generally decreases as the atomic number increases within the group. Thus, the acidic strength order is: \( \text{N}_2\text{O}_3 > \text{P}_2\text{O}_3 > \text{As}_2\text{O}_3 \) and \( \text{N}_2\text{O}_5 > \text{P}_2\text{O}_5 > \text{As}_2\text{O}_3 \).
Furthermore, acidic character intensifies with an increase in the positive oxidation number. For nitrogen oxides, the acidic strength follows this order: \( \text{N}_2\text{O} < \text{NO} < \text{N}_2\text{O}_3 < \text{N}_2\text{O}_4 < \text{N}_2\text{O}_5 \).
(ii) **Hydrides:** All Group 15 elements combine with hydrogen to form volatile hydrides of the \( \text{EH}_3 \) type. These include ammonia (\( \text{NH}_3 \)), phosphine (\( \text{PH}_3 \)), arsine (\( \text{AsH}_3 \)), stibine (\( \text{SbH}_3 \)), and bismuthine (\( \text{BiH}_3 \)).
Down the group, the covalent character, basicity, and thermal stability of these hydrides decrease, while their reducing character increases. Ammonia exhibits an unusually high boiling point due to intermolecular hydrogen bonding (molecular association).
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र Group 15 तत्वों के हाइड्राइडों (\( \text{EH}_3 \)) की पिरामिडीय संरचना को दर्शाता है। केंद्रीय परमाणु (E) से तीन हाइड्रोजन परमाणु (H) जुड़े होते हैं, और केंद्रीय परमाणु पर एक एकाकी इलेक्ट्रॉन युग्म (lone pair) होता है। यह संरचना sp³ संकरण और एकाकी युग्म-बंध युग्म प्रतिकर्षण के कारण विकृत चतुष्फलकीय या पिरामिडीय होती है।
These hydrides feature a pyramidal structure where the central atom is \( \text{sp}^3 \) hybridized and hosts one lone pair of electrons within one of its hybrid orbitals. The H-E-H bond angle is typically less than the ideal tetrahedral angle of \( 109^\circ 28' \) because of the stronger repulsion between the lone pair and bond pairs compared to bond pair-bond pair repulsion. For example, the H-N-H bond angle in \( \text{NH}_3 \) is \( 107^\circ \).
As the central atom's electronegativity diminishes from nitrogen to bismuth, the bond pairs move further from the central atom, and the lone pair's repulsion effects become more pronounced. Consequently, bond angles decrease down the group: H-P-H in \( \text{PH}_3 \) is \( 93^\circ \), H-As-H in \( \text{AsH}_3 \) is \( 91^\circ \), and H-Sb-H in \( \text{SbH}_3 \) is \( 90^\circ \).
(iii) **Halides:** With the exception of nitrogen, all elements in Group 15 form trihalides (\( \text{EX}_3 \)) and pentahalides (\( \text{EX}_5 \)). Nitrogen exclusively forms trihalides. Most nitrogen trihalides, apart from \( \text{NF}_3 \), are unstable. All trihalides, except for \( \text{BiF}_3 \), are covalently bonded and exhibit a pyramidal shape in the gaseous state due to \( \text{sp}^3 \) hybridization.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र त्रियलोराइड्स की पिरामिडीय संरचना को दर्शाता है, जिसमें एक केंद्रीय परमाणु (E) तीन हैलोजन परमाणुओं (X) से जुड़ा होता है और एक एकाकी इलेक्ट्रॉन युग्म होता है। यह संरचना sp³ संकरण और एकाकी युग्म-बंध युग्म प्रतिकर्षण के कारण बनती है।
Nitrogen does not form pentahalides because it lacks available d-orbitals. In the gaseous phase, \( \text{EX}_5 \) pentahalide molecules adopt a trigonal bipyramidal structure, resulting from \( \text{sp}^3\text{d} \) hybridization. Key pentahalides in this group include \( \text{PCl}_5 \), \( \text{PF}_5 \), and \( \text{SbF}_5 \). Phosphorus pentaiodide does not exist due to the large size of iodine atoms and the small electronegativity difference between phosphorus and iodine.
In simple words: Nitrogen is very unreactive due to its strong triple bond. Other Group 15 elements are more reactive. Their oxides range from acidic (N, P) to amphoteric (As, Sb) to basic (Bi). Hydrides like ammonia and phosphine show decreasing stability and basicity down the group, but increasing reducing power. Ammonia's high boiling point is due to hydrogen bonding. Nitrogen forms only trihalides and cannot form pentahalides because it lacks d-orbitals.

🎯 Exam Tip: Pay attention to the role of d-orbitals in expanding covalency for heavier elements. The absence of d-orbitals in nitrogen limits its maximum covalency to four (in compounds like \( \text{NH}_4^+ \)) and prevents pentahalide formation.

**Question 4. Why does \( \text{NH}_3 \) form hydrogen bond but \( \text{PH}_3 \) does not?**
**Answer:** The N-H bond in ammonia (\( \text{NH}_3 \)) is highly polar because nitrogen is a very electronegative element. This significant polarity allows for the formation of strong intermolecular hydrogen bonds between \( \text{NH}_3 \) molecules, leading to molecular association. In contrast, the P-H bond in phosphine (\( \text{PH}_3 \)) is nearly non-polar because phosphorus and hydrogen have similar electronegativities. Consequently, \( \text{PH}_3 \) molecules do not form hydrogen bonds.
In simple words: Ammonia forms hydrogen bonds because nitrogen is very electronegative, creating polar N-H bonds. Phosphine does not because phosphorus and hydrogen have similar electronegativities, making the P-H bond non-polar.

🎯 Exam Tip: Remember the criteria for hydrogen bonding: a hydrogen atom bonded to a highly electronegative atom (F, O, N). Electronegativity difference is the key distinguishing factor here.

**Question 5. How is nitrogen prepared in the laboratory? Write the chemical equations of the reactions involved?**
**Answer:** Nitrogen was discovered by Daniel Rutherford in 1772.
**Preparation:**
(I) In the laboratory, dinitrogen (\( \text{N}_2 \)) is prepared by heating an aqueous solution of ammonium chloride (\( \text{NH}_4\text{Cl} \)) with sodium nitrite (\( \text{NaNO}_2 \)).
\( \text{NH}_4\text{Cl} (\text{aq}) + \text{NaNO}_2 (\text{aq}) \xrightarrow{\text{Heat}} \text{N}_2 (\text{g}) + 2\text{H}_2\text{O} (\text{l}) + \text{NaCl} \)
(II) Nitrogen can also be prepared by the thermal decomposition of ammonium dichromate (\( (\text{NH}_4)_2\text{Cr}_2\text{O}_7 \)). This reaction is violent and occurs with flashes of light, forming the basis of a 'chemical volcano' demonstration.
\( (\text{NH}_4)_2\text{Cr}_2\text{O}_7 \xrightarrow{\text{Heat}} \text{N}_2 + 4\text{H}_2\text{O} + \text{Cr}_2\text{O}_3 \)
(III) Industrially, nitrogen is manufactured by the fractional distillation of pure liquid air. Air is a mixture of dinitrogen (boiling point 77 K) and dioxygen (boiling point 90 K). During fractional distillation, liquid nitrogen boils off first, leaving behind liquid oxygen.
**Physical properties:** Nitrogen is a colorless, tasteless, and odorless gas, slightly lighter than air. It is non-combustible and does not support combustion. While non-poisonous, a nitrogen-rich atmosphere can cause suffocation due to oxygen deficiency. The nitrogen atom has two stable isotopes, \( ^{14}\text{N} \) and \( ^{15}\text{N} \).
**Chemical properties:** At ambient temperatures, nitrogen exhibits very low reactivity due to the high bond dissociation energy of the \( \text{N} \equiv \text{N} \) triple bond. However, at elevated temperatures, dinitrogen reacts with certain metals and non-metals.
(I) Nitrogen reacts with highly electropositive metals to form nitrides, even at room temperature. It also reacts with less electropositive metals, such as magnesium (\( \text{Mg} \)), calcium (\( \text{Ca} \)), strontium (\( \text{Sr} \)), at red heat, and with silicon at white heat, to form nitrides.
\( 6\text{Li} + \text{N}_2 \longrightarrow 2\text{Li}_3\text{N} \)
\( 3\text{Mg} + \text{N}_2 \longrightarrow \text{Mg}_3\text{N}_2 \)
(II) It reacts with calcium carbide to form calcium cyanamide at \( 1000^\circ\text{C} \).
\( \text{CaC}_2 + \text{N}_2 \xrightarrow{1000^\circ\text{C}} \text{CaCN}_2 + \text{C} \)
(A mixture of \( \text{CaCN}_2 \) and \( \text{C} \) is known as nitrolim).
Calcium cyanamide undergoes hydrolysis in the soil to produce urea and ammonium carbonate, making it useful as a fertilizer.
\( \text{CaCN}_2 + 3\text{H}_2\text{O} \longrightarrow \text{Ca}(\text{OH})_2 + \text{NH}_2\text{CONH}_2 \) (urea)
\( \text{NH}_2\text{CONH}_2 + 2\text{H}_2\text{O} \longrightarrow (\text{NH}_4)_2\text{CO}_3 \)
(III) Nitrogen reacts with hydrogen at approximately 773 K in the presence of a spongy iron catalyst to form ammonia (Haber process).
(IV) Nitrogen reacts with oxygen at around 2000 K to form nitric oxide.
\( \text{N}_2 + \text{O}_2 \xrightarrow{2000 \text{ K}} 2\text{NO} \)
**Uses:**
- Nitrogen is utilized in the production of ammonia (\( \text{NH}_3 \)), nitric acid (\( \text{HNO}_3 \)), calcium cyanamide (\( \text{CaCN}_2 \)), and numerous other compounds.
- It provides an inert atmosphere in the iron and steel industries.
- It is used for filling electric lamps.
- Liquid nitrogen serves as a refrigerant for preserving biological materials, food items, and in cryosurgery.
In simple words: Nitrogen is prepared in the lab by heating ammonium salts or ammonium dichromate. Industrially, it comes from liquid air. It's an unreactive gas at room temp due to its strong triple bond but reacts with metals at high temps to form nitrides, and with hydrogen to form ammonia (Haber process). It's used to make many chemicals, for inert atmospheres, in lamps, and as a cryogen.

🎯 Exam Tip: Familiarize yourself with both laboratory and industrial preparation methods, and the key reactions demonstrating nitrogen's chemical properties, especially the Haber process for ammonia synthesis. Also, remember its diverse industrial and biological applications.

**Question 6. How is ammonia manufactured industrially?**
**Answer:**
(I) **Laboratory Preparation:** In the laboratory, ammonia can be prepared by heating an ammonium salt with a strong alkali, such as sodium hydroxide (\( \text{NaOH} \)).
e.g., \( (\text{NH}_4)_2\text{SO}_4 + 2\text{NaOH} \longrightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O} + 2\text{NH}_3 \)
\( \text{NH}_4\text{Cl} + \text{NaOH} \longrightarrow \text{NaCl} + \text{H}_2\text{O} + \text{NH}_3 \)
(II) **Industrial Manufacturing (Haber Process):** Ammonia is industrially produced through the Haber process, which involves the reversible, exothermic reaction between nitrogen and hydrogen:
\( \text{N}_2 (\text{g}) + 3\text{H}_2 (\text{g}) \rightleftharpoons 2\text{NH}_3 (\text{g}) \); \( \Delta\text{H}^\circ = -46.1 \text{ kJ mol}^{-1} \)
This reaction is exothermic and results in a decrease in volume. According to Le Chatelier's principle, the most favorable conditions for maximizing ammonia yield are:
(a) **Low temperature:** An optimum temperature of approximately 773 K is typically used. While lower temperatures favor higher equilibrium yields for exothermic reactions, very low temperatures result in a significantly slow reaction rate.
(b) **High pressure:** The optimum pressure is approximately 200 atm. High pressure shifts the equilibrium towards the product side (fewer moles of gas).
(c) **Catalyst:** Finely divided iron, enhanced by molybdenum or potassium oxide as a promoter, is employed as a catalyst to increase the reaction rate.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र हैबर प्रक्रिया द्वारा अमोनिया के औद्योगिक उत्पादन का प्रवाह चार्ट दर्शाता है। इसमें नाइट्रोजन और हाइड्रोजन गैसों को एक कंप्रेसर के माध्यम से उच्च दबाव पर उत्प्रेरक कक्ष में भेजा जाता है। उत्प्रेरक कक्ष में अमोनिया का उत्पादन होता है, और अप्रतिक्रियाशील गैसों को एक पंप द्वारा पुनर्चक्रित किया जाता है। उत्पादित अमोनिया को ठंडा करके तरल अमोनिया के रूप में एकत्र किया जाता है।
**Structure of Ammonia:** In an ammonia molecule, the nitrogen atom undergoes \( \text{sp}^3 \) hybridization, forming four hybrid orbitals. Three of these half-filled hybrid orbitals overlap with the 1s orbitals of three hydrogen atoms to form three N-H bonds. The remaining \( \text{sp}^3 \) hybrid orbital contains a lone pair of electrons. This configuration gives the \( \text{NH}_3 \) molecule a trigonal pyramidal shape, with the nitrogen atom at the apex, having three bond pairs and one lone pair. The H-N-H bond angle is approximately \( 107^\circ \).
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र अमोनिया (\( \text{NH}_3 \)) की पिरामिडीय संरचना को दर्शाता है, जिसमें केंद्रीय नाइट्रोजन परमाणु sp³ संकरण में होता है। नाइट्रोजन के 1s, 2s, और 2p ऑर्बिटल्स भी दिखाए गए हैं। इसमें नाइट्रोजन पर एक एकाकी युग्म और तीन N-H बंध होते हैं, जिससे H-N-H बंध कोण लगभग 107° होता है, जो इसकी पिरामिडीय ज्यामिति को स्पष्ट करता है।
**Physical properties:** Ammonia is a colorless gas with a distinctive pungent, ammoniacal smell. It is lighter than air and can be readily liquefied under pressure. In its solid and liquid states, ammonia molecules are associated through intermolecular hydrogen bonding.
In simple words: Ammonia is made industrially by the Haber process using nitrogen and hydrogen with a catalyst under moderate heat and high pressure. The ammonia molecule has a pyramidal shape due to sp3 hybridization with a lone pair. It's a pungent gas, lighter than air, and forms hydrogen bonds in liquid/solid states.

🎯 Exam Tip: Be able to describe the Haber process, including its flow chart details, the conditions (catalyst, temperature, pressure), and relate it to Le Chatelier's principle. Also, understand the VSEPR theory for ammonia's pyramidal structure and the role of hydrogen bonding in its physical properties.

 

Question 7. Illustrate how copper metal can give different products on reaction with HNO3.
Answer: Copper reacts with nitric acid to yield different products depending on the acid's concentration.

When copper is heated with dilute nitric acid, it produces nitric oxide:
\( 3Cu + 8HNO_{3(dil)} \implies 3Cu(NO_3)_2 + 4H_2O + 2NO \)

When copper is heated with concentrated nitric acid, it produces nitrogen dioxide:
\( Cu + 4HNO_{3(conc)} \implies Cu(NO_3)_2 + 2H_2O + 2NO_2 \)

In simple words: Copper's reaction with nitric acid changes based on whether the acid is dilute (forming nitric oxide) or concentrated (forming nitrogen dioxide).
🎯 Exam Tip: Pay close attention to the concentration of nitric acid when predicting reaction products with metals like copper, as this determines the specific nitrogen oxide formed and is a common point of distinction in examinations.

 

Question 8. Give the resonating structures of NO2 and N2O5?
Answer: The resonating structures for nitrogen dioxide (NO2) and dinitrogen pentoxide (N2O5) illustrate the delocalization of electrons within these molecules.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख नाइट्रोजन डाइऑक्साइड (NO2) और डाइनिट्रोजन पेंटोक्साइड (N2O5) की अनुनादी संरचनाओं को दर्शाता है। NO2 के लिए, नाइट्रोजन परमाणु एक डबल बॉन्ड और एक सिंगल बॉन्ड के माध्यम से दो ऑक्सीजन परमाणुओं से जुड़ा होता है, जिसमें एक अनपेयर्ड इलेक्ट्रॉन और अनुनाद की संभावना होती है। N2O5 के लिए, दो नाइट्रोजन परमाणु एक ऑक्सीजन परमाणु के माध्यम से जुड़े होते हैं, और प्रत्येक नाइट्रोजन परमाणु दो अतिरिक्त ऑक्सीजन परमाणुओं से डबल बॉन्ड और सिंगल बॉन्ड के माध्यम से जुड़ा होता है, जिसमें इलेक्ट्रॉन घनत्व के विस्थानीकरण के कारण कई अनुनादी रूप होते हैं।

In simple words: Resonance structures show how electrons are delocalized in molecules like NO2 and N2O5, giving them stability by distributing charges and bonds.
🎯 Exam Tip: Understanding resonance structures is crucial for predicting molecular stability and properties. Ensure you can accurately draw the various contributing forms for common compounds.

 

Question 9. The HNH angle value is higher than HPH, HAsH and HSbH angles. Why?
Answer: The H-N-H bond angle in ammonia (NH3) is notably larger than the H-P-H, H-As-H, and H-Sb-H bond angles in phosphine (PH3), arsine (AsH3), and stibine (SbH3), respectively. This difference is explained by VSEPR theory and the electronegativity of the central atom. In NH3, nitrogen is highly electronegative and has a smaller atomic size. Consequently, the lone pair of electrons on the nitrogen atom exerts a significant repulsive force on the bond pairs, compressing the bond angle from the ideal tetrahedral angle (109°28') to 107°. As we move down Group 15, the central atom's electronegativity decreases, and its atomic size increases. This causes the bond pairs to lie further away from the central atom, reducing the repulsion between them and the lone pair, resulting in smaller bond angles (e.g., 93° for PH3, 91° for AsH3, and 90° for SbH3).

In simple words: Nitrogen's high electronegativity and small size in ammonia cause strong repulsion from its lone pair, making the bond angle larger compared to heavier group 15 hydrides where bond pair repulsion is less significant.
🎯 Exam Tip: VSEPR theory is key to explaining bond angles and molecular geometry. Remember to consider both lone pair-bond pair repulsion and the electronegativity of the central atom.

 

Question 10. Why does R3P = O exist but R3N = O does not (R = alkyl group)?
Answer: The existence of R3P=O (where R is an alkyl group) but not R3N=O is due to the fundamental difference in the availability of d-orbitals for phosphorus and nitrogen. Nitrogen, being a second-period element, lacks vacant d-orbitals in its valence shell. This limitation prevents nitrogen from forming pπ-dπ multiple bonds, which would be necessary for the N=O double bond in R3N=O. In contrast, phosphorus, a third-period element, possesses vacant d-orbitals. This allows phosphorus to expand its valence shell beyond an octet and form pπ-dπ bonds, enabling the formation of the P=O double bond in R3P=O compounds.

In simple words: Phosphorus can form double bonds with oxygen using its empty d-orbitals, which nitrogen cannot do because it lacks these d-orbitals.
🎯 Exam Tip: The availability of d-orbitals in heavier main group elements is a critical factor for explaining expanded octets and the formation of multiple bonds that are not possible for their lighter counterparts.

 

Question 11. Explain why NH3 is basic while BiH3 is only feebly basic?
Answer: Ammonia (NH3) is a strong base, whereas bismuthine (BiH3) exhibits only feeble basicity. This difference can be attributed to the atomic size and electron density of the central atom in each molecule. Nitrogen has a significantly smaller atomic size (70 pm) compared to bismuth (148 pm). As a result, the electron density on the nitrogen atom in NH3 is much higher than that on the bismuth atom in BiH3. A higher electron density on the central atom increases its tendency to donate its lone pair of electrons, which is the defining characteristic of a Lewis base. Therefore, nitrogen in NH3 readily donates its lone pair, making ammonia strongly basic, while bismuth in BiH3, with its lower electron density, is only weakly basic.

In simple words: Ammonia is strongly basic because its small nitrogen atom has a high electron density, making it easy to donate electrons, while bismuthine is only feebly basic due to bismuth's large size and lower electron density.
🎯 Exam Tip: Basicity of hydrides in a group is inversely related to the atomic size of the central atom and directly related to the electron density around it, impacting the electron donation ability.

 

Question 12. Nitrogen exists as diatomic molecule and phosphorus as P4 Why?
Answer: Nitrogen exists as a diatomic molecule (N2) forming a strong triple bond, while phosphorus exists as a tetrahedral P4 molecule. This distinction arises from the difference in their atomic sizes and their ability to form multiple bonds. Nitrogen's small atomic size allows it to form strong pπ-pπ multiple bonds (N≡N), leading to a very stable N2 molecule with high bond dissociation energy. In contrast, phosphorus has a larger atomic size, which hinders its ability to form effective pπ-pπ multiple bonds. Instead, phosphorus prefers to form stable single P-P covalent bonds, leading to a tetrahedral P4 structure where four phosphorus atoms are connected by single bonds.

In simple words: Nitrogen is a gas because its small atoms form strong triple bonds (N≡N), but phosphorus is a solid because its larger atoms prefer single bonds, forming a P4 tetrahedral structure.
🎯 Exam Tip: The ability of elements to form pπ-pπ multiple bonds is largely dependent on atomic size, with smaller atoms in the second period being more capable than larger atoms in subsequent periods.

 

Question 13. Write main differences between the properties of white phosphorus and red phosphorus?
Answer: Elemental phosphorus can be obtained by heating phosphate rock with coke and silica in an electric furnace at approximately 1770 K, yielding white phosphorus. This white phosphorus is then typically stored underwater to prevent its reaction with air.

The relevant reactions are:
\( 2 Ca_3(PO_4)_2 + 6 SiO_2 \implies 6 CaSiO_3 + P_4O_{10} \)
\( P_4O_{10} + 10 C \implies 4 P + 10 CO \) (This phosphorus is white phosphorus).

Phosphorus exists in several allotropic forms, primarily white, red, and black phosphorus. Here are the main differences between white and red phosphorus:

(I) **White Phosphorus (Yellow Phosphorus):**
It is the most common allotropic form, formed by condensing phosphorus from gaseous or liquid states. It is a soft, waxy solid with a characteristic garlic smell. It is insoluble in water but readily dissolves in carbon disulfide (CS2). White phosphorus is highly reactive and spontaneously ignites in air, hence it is stored underwater. Its structure consists of discrete P4 tetrahedral molecules. Due to angular strain within the P4 molecule (bond angles are 60°), it is highly unstable and reactive. It glows in the dark (chemiluminescence), leading to its name. It readily combusts in air to form dense white fumes of P4O10.
\( P_4 + 5O_2 \implies P_4O_{10} \)

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख सफेद फास्फोरस की संरचना को दर्शाता है, जिसमें चार फास्फोरस परमाणु एक नियमित चतुष्फलक के कोनों पर स्थित होते हैं। प्रत्येक फास्फोरस परमाणु तीन अन्य फास्फोरस परमाणुओं से जुड़ा होता है, जिससे P4 अणु बनता है, जिसमें प्रत्येक बंध कोण 60 डिग्री होता है।

(II) **Red Phosphorus:**
Red phosphorus is formed when white phosphorus is heated to 573 K in an inert atmosphere for several days. It is a hard, crystalline solid that is odorless and non-poisonous. Red phosphorus is less reactive than white phosphorus and is insoluble in both water and carbon disulfide (CS2). Its structure consists of P4 tetrahedral units linked together to form linear chains, reducing the angular strain and making it more stable and less reactive.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख लाल फास्फोरस की संरचना को दर्शाता है, जिसमें P4 चतुष्फलकीय इकाइयाँ एक जटिल नेटवर्क में आपस में जुड़ी होती हैं। प्रत्येक P4 इकाई अन्य इकाइयों से बंधों के माध्यम से जुड़ी होती है, जिससे एक पॉलीमरिक या श्रृंखला-जैसी संरचना बनती है, जो सफेद फास्फोरस की तुलना में अधिक स्थिरता और कम प्रतिक्रियाशीलता प्रदान करती है।

In simple words: White phosphorus is a highly reactive, soft, garlic-smelling solid with P4 tetrahedral units, stored underwater; red phosphorus is a less reactive, hard solid formed by heating white phosphorus, with P4 units linked into chains.
🎯 Exam Tip: Key differences between allotropes often include reactivity, solubility, and structural arrangement. For phosphorus, remember the P4 tetrahedron for white phosphorus and its consequence (angular strain, high reactivity).

 

Question 14. Why does nitrogen show catenation properties less than phosphorus?
Answer: Nitrogen exhibits catenation properties to a much lesser extent than phosphorus. This is primarily because the N-N single bond is significantly weaker than the P-P single bond. The small size of nitrogen atoms leads to high interelectronic repulsion between the lone pairs on adjacent nitrogen atoms in a N-N single bond, destabilizing the chain. In contrast, phosphorus atoms are larger, reducing this repulsion and allowing for stronger P-P single bonds, which favors the formation of longer chains and rings (catenation).

In simple words: Nitrogen catenates less than phosphorus because its N-N single bonds are weaker due to high electron repulsion between small nitrogen atoms.
🎯 Exam Tip: Catenation ability depends on the strength of the element-element single bond and the presence of interelectronic repulsion, which is more pronounced in smaller atoms with lone pairs.

 

Question 15. Give the disproportionation reaction of H3PO3.
Answer: When phosphorous acid (H3PO3) is heated, it undergoes a disproportionation reaction, meaning it acts as both an oxidizing and reducing agent. It forms orthophosphoric acid (H3PO4) and phosphine (PH3).

The balanced chemical equation for this reaction is:
\( 4H_3PO_3 \implies 3H_3PO_4 + PH_3 \)

In simple words: When heated, phosphorous acid transforms into phosphoric acid and phosphine through a reaction where it both gains and loses oxygen.
🎯 Exam Tip: Disproportionation reactions are important in inorganic chemistry, particularly for elements that can exist in multiple oxidation states. Recognizing the reactants and products is key.

 

Question 16. Can PCl5 act as an oxidising as well as a reducing agent? Justify.
Answer: Phosphorus pentachloride (PCl5) primarily functions as an oxidizing agent and generally does not act as a reducing agent. In PCl5, the oxidation state of phosphorus is +5. This is the maximum possible oxidation state for phosphorus, as it has 5 valence electrons. Since phosphorus cannot increase its oxidation state beyond +5, it cannot be oxidized further, and therefore cannot act as a reducing agent. However, phosphorus in PCl5 can readily be reduced to lower oxidation states, such as +3 (in PCl3) or even -3 (in phosphides), making PCl5 an effective oxidizing agent.

In simple words: PCl5 is only an oxidizing agent because phosphorus is already in its highest (+5) oxidation state and cannot lose more electrons.
🎯 Exam Tip: To determine if a compound can act as an oxidizing or reducing agent, check the oxidation state of the central atom. If it's at its maximum, it can only be an oxidizing agent; if at its minimum, it can only be a reducing agent; if intermediate, it can be both.

 

Question 17. Justify the placement of O, S, Se, Te and Po in the same group of the periodic table in terms of electronic configuration, oxidation state and hydride formation?
Answer: The elements Oxygen (O), Sulfur (S), Selenium (Se), Tellurium (Te), and Polonium (Po) are all placed in Group 16 of the periodic table, collectively known as chalcogens, due to shared chemical properties explained by their electronic configuration, oxidation states, and hydride formation.

(i) **Electronic Configuration:** All elements in Group 16 possess six valence electrons. Their general outer electronic configuration is \( ns^2 np^4 \), where 'n' corresponds to the period number (from 2 for oxygen to 6 for polonium). This identical valence electron count dictates their similar chemical behavior.

(ii) **Oxidation State:** With six valence electrons, these elements are expected to display a -2 oxidation state to achieve a stable octet. Oxygen predominantly exhibits a -2 oxidation state due to its high electronegativity, although it can also show -1 (in H2O2), 0 (in O2), and +2 (in OF2). As one moves down the group, the stability of the -2 oxidation state decreases due to decreasing electronegativity. The heavier elements (S, Se, Te, Po) can also exhibit positive oxidation states such as +2, +4, and +6, owing to the increasing availability of vacant d-orbitals in their valence shells.

(iii) **Formation of Hydrides:** These elements form volatile hydrides of the general formula H2E (e.g., H2O, H2S, H2Se, H2Te, H2Po). Oxygen and sulfur can also form hydrides of the type H2E2 (e.g., H2O2, H2S2). The consistent stoichiometry in hydride formation across the group further justifies their placement.

In simple words: Oxygen, sulfur, selenium, tellurium, and polonium are grouped together because they all have six valence electrons, show similar oxidation states (especially -2), and form hydrides with a consistent H2E formula.
🎯 Exam Tip: Group relationships in the periodic table are fundamental. When justifying group placement, always refer to valence electron configuration, common oxidation states, and characteristic compound formations (like hydrides or oxides).

 

Question 18. Why is dioxygen a gas but sulphur a solid?
Answer: Dioxygen (O2) exists as a gas at room temperature, while sulfur (S8) is a solid. This difference in physical state arises from their distinct abilities to form multiple bonds and the nature of intermolecular forces. Due to its small size and high electronegativity, oxygen readily forms pπ-pπ multiple bonds, resulting in the diatomic O2 molecule (O=O). These O2 molecules are held together by weak van der Waals forces, which are insufficient to keep them in a condensed state at room temperature, hence it's a gas. In contrast, sulfur has a larger atomic size, which makes pπ-pπ bonding less favorable. Instead, sulfur forms strong S-S single bonds and has a high tendency for catenation (forming chains or rings), resulting in stable S8 molecules with a puckered ring structure. These S8 molecules are large and are held together by strong intermolecular forces of attraction, causing sulfur to be a solid at room temperature.

In simple words: Oxygen is a gas because its small size allows it to form stable O2 molecules with weak intermolecular forces, while sulfur is a solid because its larger atoms prefer strong S-S single bonds, forming larger S8 molecules with stronger attractions.
🎯 Exam Tip: The physical state of non-metals is often explained by the nature of bonding (single vs. multiple, pπ-pπ versus σ-bonds) and the strength of intermolecular forces between resulting molecules.

 

Question 19. Which aerosols deplete ozone?
Answer: Chlorofluorocarbons (CFCs) are the primary aerosols responsible for depleting the ozone layer. For example, compounds like dichlorodifluoromethane (CF2Cl2) decompose in the stratosphere to release highly reactive chlorine free radicals (Cl•). These chlorine radicals then catalytically destroy ozone molecules, leading to ozone depletion.

In simple words: Chlorofluorocarbons (CFCs) in aerosols are chemicals that break down ozone in the atmosphere.
🎯 Exam Tip: Remember that CFCs are key ozone-depleting substances. Understanding the role of free radicals (like chlorine radicals) in catalytic destruction of ozone is important.

 

Question 20. Describe the manufacture of H2SO4 by contact process?
Answer: The Contact Process is the primary industrial method for manufacturing sulfuric acid (H2SO4) and involves several key steps:

**Theory:** The process is based on the catalytic oxidation of sulfur dioxide to sulfur trioxide, followed by absorption in sulfuric acid to form oleum, which is then diluted.

**The process involves the following steps:**

(I) **Preparation of Sulfur Dioxide:** Sulfur dioxide (SO2) is produced by burning elemental sulfur or sulfide ores (like iron pyrites) in air.
\( S (s) + O_2 (g) \implies SO_2 (g) \)

(II) **Catalytic Oxidation of Sulfur Dioxide to Sulfur Trioxide:** Sulfur dioxide is then catalytically oxidized to sulfur trioxide (SO3) using atmospheric oxygen. This reaction is reversible and highly exothermic.
\( 2SO_2 (g) + O_2 (g) \rightleftharpoons 2SO_3 (g) ; \Delta H = -196.6 \text{ kJ} \)

(III) **Preparation of Oleum:** The sulfur trioxide produced is absorbed in concentrated sulfuric acid to form oleum (H2S2O7), also known as fuming sulfuric acid. This step is crucial because directly dissolving SO3 in water creates a fine mist of sulfuric acid that is difficult to condense.
\( H_2SO_4 (l) + SO_3 (g) \implies H_2S_2O_7 (l) \)

Finally, the oleum is diluted with a calculated amount of water to obtain sulfuric acid of the desired concentration.
\( H_2S_2O_7 (l) + H_2O (l) \implies 2H_2SO_4 (l) \)

**Conditions for Maximum Yield:** According to Le Chatelier's principle, favorable conditions for maximizing the yield of SO3 (and thus H2SO4) are:

1. **Low Temperature:** Since the forward reaction is exothermic, a lower temperature favors the equilibrium shift towards product formation. An optimum temperature of approximately 723 K is typically used, balancing yield with reaction rate (as very low temperatures slow down the reaction too much).

2. **High Pressure:** The reaction involves a decrease in the number of moles of gas (3 moles of reactants to 2 moles of product). Therefore, high pressure shifts the equilibrium towards the product side. An optimum pressure of 2 atm is commonly used.

3. **Catalyst:** To increase the reaction rate at the relatively low optimum temperature, a catalyst is essential. Finely divided iron with molybdenum as a promoter, or more commonly, platinum (Pt) or vanadium pentoxide (V2O5), is used.

4. **Excess Oxygen:** Using an excess of oxygen helps push the equilibrium towards the formation of SO3, increasing the yield.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख सल्फ्यूरिक एसिड के निर्माण के लिए संपर्क प्रक्रिया का एक प्रवाह चार्ट है। यह अशुद्ध SO2 और O2 के मिश्रण को पानी के स्प्रे, धूल अवक्षेपित्र, धुलाई टावर, सुखाने वाले टावर और आर्सेनिक शोधक के माध्यम से पारित करने के चरणों को दर्शाता है ताकि शुद्ध SO2 गैस प्राप्त हो सके। फिर शुद्ध SO2 और O2 को उत्प्रेरक कनवर्टर (723K पर V2O5 उत्प्रेरक के साथ) में भेजा जाता है जहां SO3 बनता है, जिसे बाद में सल्फ्यूरिक एसिड में अवशोषित करके ओलियम और अंततः सल्फ्यूरिक एसिड प्राप्त किया जाता है।

In simple words: The Contact Process makes sulfuric acid by burning sulfur to make SO2, converting SO2 to SO3 using a catalyst, absorbing SO3 in acid to form oleum, and then diluting the oleum.
🎯 Exam Tip: For industrial processes like the Contact Process, remember the three main steps (SO2 production, SO3 formation, oleum production/dilution) and the optimal conditions (temperature, pressure, catalyst) based on Le Chatelier's principle.

 

Question 21. How is SO2 an air pollutant?
Answer: Sulfur dioxide (SO2) is a significant air pollutant with various detrimental effects on human health and the environment. In the atmosphere, SO2 can react with water vapor and oxygen to form sulfuric acid, which is a major component of acid rain. Acid rain harms aquatic ecosystems, damages forests, and erodes buildings and historical monuments. Furthermore, the presence of SO2 in the atmosphere causes a range of health problems, including cardiac and respiratory diseases such as asthma and bronchitis. It can also lead to eye irritation and damage to plant cell membranes and chlorophyll, impairing photosynthesis and overall plant health.

In simple words: Sulfur dioxide pollutes the air by causing acid rain, harming plants, and leading to respiratory and eye problems in humans.
🎯 Exam Tip: When discussing pollutants, always link them to their specific environmental impacts (e.g., acid rain) and health effects (e.g., respiratory issues) for a comprehensive answer.

 

Question 22. Why are halogens strong oxidising agents?
Answer: Halogens are potent oxidizing agents due to a combination of their inherent chemical properties. They possess high electronegativity, meaning they have a strong attraction for electrons. This is coupled with a highly negative electron gain enthalpy (a large amount of energy is released when they accept an electron), indicating their strong tendency to acquire an electron. Additionally, halogens generally have relatively low bond dissociation enthalpies (for X-X bonds, though F2 is an exception), which means they can readily break their diatomic bonds to form individual atoms that are eager to accept electrons. These factors collectively contribute to their strong tendency to accept electrons from other substances, thereby oxidizing those substances and becoming reduced themselves.

In simple words: Halogens are strong oxidizing agents because they readily accept electrons due to their high electronegativity and the energy released when they gain an electron.
🎯 Exam Tip: Focus on electronegativity and electron gain enthalpy when explaining the oxidizing power of halogens, as these are the primary drivers for their electron-accepting capability.

 

Question 23. Explain why fluorine forms only one oxoacid, HOF?
Answer: Fluorine forms only one oxoacid, hypofluorous acid (HOF), unlike other halogens that can form multiple oxoacids with higher oxidation states. This limitation is due to fluorine's unique atomic properties: its exceptionally high electronegativity, very small atomic size, and crucially, the complete absence of vacant d-orbitals in its valence shell. Because fluorine lacks d-orbitals, it cannot expand its octet, restricting its maximum covalency to one (or sometimes zero, as in F2) and preventing it from attaining positive oxidation states like +3, +5, or +7, which are common for other halogens in their oxoacids.

In simple words: Fluorine only forms one oxoacid (HOF) because its small size, high electronegativity, and lack of d-orbitals prevent it from having higher oxidation states.
🎯 Exam Tip: The absence of d-orbitals in period 2 elements like fluorine is a critical concept for understanding their chemical behavior, especially regarding covalency and maximum oxidation states.

 

Question 24. Explain why in spite of nearly the same electronegativity, oxygen forms hydrogen bonding while chlorine does not?
Answer: Despite oxygen and chlorine having comparable electronegativity values, only oxygen readily forms hydrogen bonds, while chlorine generally does not. The key differentiating factor is their atomic size. Oxygen has a significantly smaller atomic radius compared to chlorine. This small size, combined with high electronegativity, leads to a high electron density per unit volume on the oxygen atom. When hydrogen is bonded to such a small, highly electronegative atom, the bond becomes highly polarized, and the small size allows the hydrogen atom to approach the lone pair of electrons on an adjacent electronegative atom closely, forming a strong hydrogen bond. Chlorine, although electronegative, has a much larger atomic size. This larger size results in a lower electron density per unit volume, and its valence electrons are further from the nucleus, making it less effective at inducing the strong dipole required for hydrogen bonding and physically hindering the close approach needed for such interactions.

In simple words: Oxygen forms hydrogen bonds because, despite similar electronegativity, it's much smaller than chlorine, allowing hydrogen atoms to get closer and form stronger bonds.
🎯 Exam Tip: For hydrogen bonding, both high electronegativity and small atomic size are crucial. Emphasize the 'small size' aspect when comparing elements with similar electronegativities but different H-bonding capabilities.

 

Question 25. Write two uses of ClO2.
Answer: Chlorine dioxide (ClO2) is a versatile chemical with several important applications:

1. It is a powerful **oxidizing agent and chlorinating agent**, used extensively in the bleaching of paper pulp and textiles to improve their whiteness without forming harmful chlorinated byproducts.

2. ClO2 is an effective **disinfectant and water treatment agent**, employed for purifying drinking water and treating wastewater due to its ability to kill bacteria, viruses, and other microorganisms, as well as to remove odors and tastes.

In simple words: ClO2 is used for bleaching paper and textiles, and for disinfecting water.
🎯 Exam Tip: When listing uses, focus on specific applications that highlight the chemical's core properties (e.g., oxidizing, disinfecting) rather than general industrial roles.

 

Question 26. Why are halogens coloured?
Answer: Halogens (F2, Cl2, Br2, I2) exhibit distinct colors due to the absorption of radiation in the visible region of the electromagnetic spectrum. When white light passes through halogens, certain wavelengths are absorbed. This energy absorption causes the outer valence electrons of the halogen molecules to get excited and jump to higher energy levels. The specific color observed for each halogen is the complementary color to the wavelength of light absorbed. Since different halogens absorb different quanta of radiation corresponding to different energy transitions, they appear in various characteristic colors (e.g., fluorine is pale yellow, chlorine is greenish-yellow, bromine is reddish-brown, and iodine is violet).

In simple words: Halogens are colored because they absorb specific light wavelengths, exciting their electrons; the color we see is the remaining light that isn't absorbed.
🎯 Exam Tip: Relate the absorption of visible light and subsequent electronic transitions to the observed color. This concept is applicable to many colored substances in chemistry.

 

Question 27. Write the reactions of F2 and Cl2 with water?
Answer: The reactions of fluorine (F2) and chlorine (Cl2) with water are as follows:

**Reaction of Fluorine (F2) with Water:**
Fluorine is a very strong oxidizing agent, and it oxidizes water to oxygen or ozone, depending on conditions:
\( 2F_2 (g) + 2H_2O (l) \implies 4HF (aq) + O_2 (g) \)
\( 3F_2 (g) + 3H_2O (l) \implies 6HF (aq) + O_3 (g) \)

**Reaction of Chlorine (Cl2) with Water:**
Chlorine reacts with water to form hydrochloric acid and hypochlorous acid (a reversible reaction):
\( Cl_2 (g) + H_2O (l) \rightleftharpoons HCl (aq) + HClO (aq) \)

In simple words: Fluorine reacts with water to produce oxygen or ozone and hydrofluoric acid, while chlorine reacts with water to form hydrochloric acid and hypochlorous acid.
🎯 Exam Tip: Note the difference in reactivity: fluorine is a stronger oxidizing agent than chlorine, leading to oxidation of water, while chlorine undergoes disproportionation with water.

 

Question 28. How can you prepare Cl2 from HCl and HCl from Cl2? Write reactions only?
Answer: Here are the reactions for preparing Cl2 from HCl and HCl from Cl2:

**Preparation of Cl2 from HCl:**
Chlorine can be prepared by treating hydrochloric acid (HCl) with strong oxidizing agents such as manganese dioxide (MnO2), potassium permanganate (KMnO4), or potassium dichromate (K2Cr2O7). For example:
\( 4HCl (aq) + MnO_2 (s) \implies MnCl_2 (aq) + 2H_2O (l) + Cl_2 (g) \)

**Preparation of HCl from Cl2:**
Hydrogen chloride (HCl) gas can be prepared by the direct reaction of hydrogen gas (H2) and chlorine gas (Cl2) in the presence of sunlight or light:
\( H_2 (g) + Cl_2 (g) \implies 2HCl (g) \)

In simple words: Chlorine is made from HCl using strong oxidizers, and HCl is made from chlorine by reacting it directly with hydrogen.
🎯 Exam Tip: For interconversion reactions, identify the oxidation states and choose appropriate oxidizing or reducing agents. Balancing the equations is also crucial.

 

Question 29. What inspired N. Bartlett for carrying out a reaction between Xe and PtF6?
Answer: Neil Bartlett was inspired to attempt a reaction between Xenon (Xe) and Platinum hexafluoride (PtF6) after successfully synthesizing dioxygenyl hexafluoroplatinate, \( O_2^+ [PtF_6]^- \). He noted that the first ionization enthalpy of molecular oxygen (O2) (1175 kJ/mol) was very similar to that of xenon (Xe) (1170 kJ/mol). This similarity suggested that if O2 could be oxidized by PtF6 to form \( O_2^+ \), then xenon, with a comparable ionization energy, might also be oxidized by PtF6. This led to his pioneering work in preparing the first noble gas compound, \( Xe^+ [PtF_6]^- \).

In simple words: Bartlett was inspired to react xenon with PtF6 because xenon's ionization energy was almost the same as oxygen's, which he had already oxidized with PtF6.
🎯 Exam Tip: The historical discovery of noble gas compounds is significant. Remember the key insight: similarity in ionization enthalpies between O2 and Xe as the driving factor for Bartlett's experiment.

 

Question 30. What are the oxidation states of phosphorus in the following?
Answer: The oxidation states of phosphorus in the given compounds are:

1. H3PO3: +3
2. PCl3: +3
3. Ca3P2: -3
4. Na3PO4: +5
5. POF3: +5

In simple words: The oxidation state for phosphorus in each compound is determined by the other elements it's bonded to and their typical charges.
🎯 Exam Tip: Practice calculating oxidation states for various elements in different compounds. Remember the common oxidation states of hydrogen (+1), oxygen (-2), and halogens (-1) to help with calculations.

 

Question 31. Write balanced equations for the following:
1. NaCl is heated with sulphuric acid in the presence of MnO2
2. Chlorine gas is passed into a solution of Nal in water.

Answer: The balanced equations for the given reactions are:

1. When sodium chloride (NaCl) is heated with sulfuric acid (H2SO4) in the presence of manganese dioxide (MnO2), chlorine gas is produced:
\( 2NaCl (s) + MnO_2 (s) + 3H_2SO_4 (aq) \implies 2NaHSO_4 (aq) + MnSO_4 (aq) + 2H_2O (l) + Cl_2 (g) \)

2. When chlorine gas (Cl2) is passed into a solution of sodium iodide (NaI) in water, iodine is displaced:
\( 2NaI (aq) + Cl_2 (g) \implies 2NaCl (aq) + I_2 (s) \)

In simple words: The first reaction shows chlorine gas being made from salt, manganese dioxide, and sulfuric acid; the second shows chlorine displacing iodine from sodium iodide.
🎯 Exam Tip: For redox reactions involving halogens, remember their relative oxidizing strengths. A more reactive halogen will displace a less reactive one from its salt solution.

 

Question 32. How are xenon fluorides XeF2, XeF4 and XeF6 obtained?
Answer: Xenon fluorides-XeF2, XeF4, and XeF6-are synthesized by the direct reaction of xenon (Xe) with fluorine (F2) under specific conditions of temperature, pressure, and Xe:F2 molar ratio, often in the presence of a nickel catalyst, or by electric discharge or ultraviolet radiation.

(I) **Fluorides:**

* **Preparation of Xenon Difluoride (XeF2):**
XeF2 is obtained by heating xenon and fluorine in a 1:1 molar ratio at 673 K and 1 bar pressure, typically in a sealed nickel tube.
\( Xe (g) + F_2 (g) \xrightarrow{Ni, 1 \text{ bar}, 673K} XeF_2 (s) \)

* **Preparation of Xenon Tetrafluoride (XeF4):**
XeF4 is prepared by heating xenon and fluorine in a 1:5 molar ratio at 873 K and 7 bar pressure.
\( Xe (g) + 2F_2 (g) \xrightarrow{Ni, 7 \text{ bar}, 873K} XeF_4 (s) \)

* **Preparation of Xenon Hexafluoride (XeF6):**
XeF6 is formed when xenon and fluorine are reacted in a 1:20 molar ratio at 573 K and 60-70 bar pressure.
\( Xe (g) + 3F_2 (g) \xrightarrow{Ni, 60-70 \text{ bar}, 573K} XeF_6 (s) \)

These xenon fluorides can also react with fluoride ion acceptors like PF5 or SbF5 to form cationic species and fluoro anions, such as:
\( XeF_2 + PF_5 \implies [XeF]^+ [PF_6]^- \)
\( XeF_4 + SbF_5 \implies [XeF_3]^+ [SbF_6]^- \)

(II) **Oxides:**
Xenon trioxide (XeO3) is produced by the hydrolysis of XeF4 or XeF6:
\( XeF_4 (s) + 2H_2O (l) \implies XeO_2F_2 (s) + 2HF (aq) \)
\( XeF_6 (s) + 3H_2O (l) \implies XeO_3 (s) + 6HF (aq) \)
XeO3 is a powerful explosive and oxidizing agent in aqueous solution.

(III) **Oxy Fluorides:**
The oxyfluorides of xenon are formed by the partial hydrolysis of xenon fluorides:
\( XeF_4 (s) + H_2O (l) \implies XeOF_2 (s) + 2HF (aq) \)
\( XeF_6 (s) + H_2O (l) \implies XeOF_4 (s) + 2HF (aq) \)

In simple words: Xenon fluorides (XeF2, XeF4, XeF6) are made by directly reacting xenon with fluorine under specific temperature and pressure conditions, with higher fluorine ratios and pressure yielding higher fluorides.
🎯 Exam Tip: Remember the specific stoichiometric ratios and reaction conditions for synthesizing different xenon fluorides, as they dictate which compound is formed.

Structure Of Xenon Compounds

The structure and molecular geometry of xenon compounds can be explained using VSEPR theory. Some illustrative examples include:

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख विभिन्न ज़ेनॉन यौगिकों की आणविक ज्यामिति और संकरण को दर्शाता है। यह XeF2 (रैखिक, sp3d संकरण), XeF4 (वर्ग समतलीय, sp3d2 संकरण), और XeF6 (विकृत अष्टफलकीय, sp3d3 संकरण) की संरचनाओं को दिखाता है। इसके अतिरिक्त, यह XeOF2 (टी-आकार, sp3d संकरण), XeOF4 (वर्ग पिरामिडी, sp3d2 संकरण), और XeO3 (पिरामिडी, sp3 संकरण) की संरचनाओं को भी प्रदर्शित करता है।

Clathrates

Clathrates are cage-like compounds formed when noble gas atoms get entrapped within the cavities of the crystal lattice of certain organic or inorganic compounds during crystallization. All noble gas elements, except for helium and neon, can form clathrates.

Uses Of Noble Gases

The noble gases have various practical applications:

• Because of its lightness and non-flammability, helium is employed for filling airships and observation balloons.
• Helium is utilized as a cryogenic agent for cooling in gas-cooled nuclear reactors.
• It provides an inert atmosphere in welding metals and alloys.
• It is used in producing powerful superconducting magnets for NMR spectrometers and MRI systems.
• Helium acts as a diluent for oxygen in modern diving apparatus due to its very low solubility in blood, preventing "the bends."
• Neon is used in discharge tubes for advertisements and optical decorations.
• Neon is also incorporated into safety devices.
• Argon is used for filling electric bulbs to prevent filament oxidation.
• Pure argon finds application in gas chromatography.
• Krypton-helium mixtures are used in flashbulbs for high-speed photography.
• Radon is used in the treatment of cancer.

 

Question 33. With what neutral molecule is ClO- isoelectronic? Is that molecule a Lewis base?
Answer: The neutral molecule that is isoelectronic with ClO- is chlorine monofluoride (ClF). Both ClO- and ClF have 26 valence electrons (7 + 6 + 1 = 14 for ClO- and 7 + 7 = 14 for ClF). Regarding its Lewis basicity, ClF can act as a Lewis base. It has lone pairs of electrons on both chlorine and fluorine atoms. For instance, ClF can accept an electron pair from a Lewis acid, or it can combine with a fluoride ion (F-) to form ClF3 (ClF + 2F- → ClF3), indicating its ability to act as a Lewis acid and base, or as a reactant that can expand its octet. In the context of the question, if it can accept a lone pair, it acts as a Lewis base. ClF acts as a Lewis base by donating a lone pair, especially from the fluorine atom. However, the given answer "CIF Since CIF can combine further with F to form CIF3, CIF is a Lewis base." suggests it acts as a Lewis base by providing an electron pair to form new bonds. (More precisely, it would be a Lewis base if it *donates* an electron pair; forming ClF3 from ClF and F- involves ClF acting as a Lewis acid or simply participating in bond formation where Cl expands its octet.) Given the context, we should focus on the exact provided text. The provided answer states: "ClF. Since ClF can combine further with F to form ClF3, ClF is a Lewis base."

In simple words: ClO- is isoelectronic with ClF, and ClF acts as a Lewis base because it can accept additional fluorine atoms, implying it can donate electron pairs.
🎯 Exam Tip: Isoelectronic species have the same number of electrons. For Lewis base/acid character, identify if a molecule has lone pairs to donate (Lewis base) or vacant orbitals to accept (Lewis acid).

 

Question 34. How are XeO3 and XeOF4 prepared?
Answer: Xenon trioxide (XeO3) and xenon oxyfluoride (XeOF4) are prepared primarily through the hydrolysis of xenon fluorides.

**Preparation of XeO3:**
Xenon trioxide is obtained by the complete hydrolysis of xenon tetrafluoride (XeF4) or xenon hexafluoride (XeF6). For example, the hydrolysis of XeF6 produces XeO3 along with hydrofluoric acid:
\( XeF_6 (s) + 3H_2O (l) \implies XeO_3 (s) + 6HF (aq) \)

**Preparation of XeOF4:**
Xenon oxyfluoride (XeOF4) is formed by the partial hydrolysis of xenon hexafluoride (XeF6):
\( XeF_6 (s) + H_2O (l) \implies XeOF_4 (s) + 2HF (aq) \)

In simple words: XeO3 and XeOF4 are prepared by reacting xenon hexafluoride with water; complete reaction makes XeO3, while partial reaction makes XeOF4.
🎯 Exam Tip: The degree of hydrolysis (number of water molecules reacting) with xenon fluorides dictates the specific oxyfluoride or oxide of xenon formed. Remember the stoichiometry.

 

Question 35. Arrange the following in the order of property indicated for each set:
1. F2, Cl2, Br2, I2 - increasing bond dissociation enthalpy.
2. HF < HCl < HBr < HI - increasing acid strength.
3. NH3, PH3, AsH3, SbH3, BiH3 - increasing base strength.

Answer: The correct order for each set based on the indicated property is:

1. **Increasing bond dissociation enthalpy:**
\( I_2 < F_2 < Br_2 < Cl_2 \)
(Note: F2 has an anomalously low bond dissociation enthalpy due to high interelectronic repulsion between the small F atoms.)

2. **Increasing acid strength:**
\( HF < HCl < HBr < HI \)
(Acid strength increases down the group as bond length increases and bond strength decreases, making H+ easier to release.)

3. **Increasing base strength:**
\( BiH_3 < SbH_3 < AsH_3 < PH_3 < NH_3 \)
(Basicity decreases down the group as atomic size increases, reducing electron density and the ability to donate a lone pair.)

In simple words: This question asks to arrange halogens by bond strength, hydrogen halides by acid strength, and group 15 hydrides by base strength, following periodic trends.
🎯 Exam Tip: For periodic trends, understand the underlying reasons: bond length, electron density, and interelectronic repulsion are common explanations for variations in bond dissociation energy, acid strength, and basicity.

 

Question 36. Which one of the following does not exist?
1. XeOF4
2. NeF2
3. XeF2
4. XeF6

Answer: 2. NeF2

In simple words: Among the given options, NeF2 does not exist because neon is a noble gas and fluorine is highly reactive, but neon's small size and high ionization energy make it very difficult to form stable compounds.
🎯 Exam Tip: Remember that while heavier noble gases like Xenon can form compounds with highly electronegative elements, lighter noble gases like Neon are generally unreactive and do not form stable compounds under normal conditions.

 

Question 37. Give the formula and describe the structure of a noble gas species which is isostructural with:
1. ICl4-
2. IBr2-
3. BrO3-

Answer: Here are the noble gas species that are isostructural with the given ions:

1. **ICl4-:** The noble gas species isostructural with ICl4- (Square planar, sp3d2 hybridisation with 2 lone pairs) is **XeF4**. (Xenon tetrafluoride, also square planar).

2. **IBr2-:** The noble gas species isostructural with IBr2- (Linear, sp3d hybridisation with 3 lone pairs) is **XeF2**. (Xenon difluoride, also linear).

3. **BrO3-:** The noble gas species isostructural with BrO3- (Pyramidal, sp3 hybridisation with 1 lone pair) is **XeO3**. (Xenon trioxide, also pyramidal).

In simple words: This question asks to find noble gas compounds that have the same electron and molecular geometry as the given halogen-containing ions.
🎯 Exam Tip: To identify isostructural species, count the total number of valence electrons and determine the steric number (bond pairs + lone pairs) to predict the molecular geometry using VSEPR theory.

 

Question 38. List the uses of neon and argon gases?
Answer: Neon and argon, both noble gases, have distinct and important industrial applications:

**Uses of Neon:**
* Neon is primarily used for filling discharge tubes to create vibrant, colored lighting for advertisements and decorative signs (neon signs), as it emits a characteristic orange-red light when electrified.
* It is also employed in safety devices like voltage stabilizers and indicator lamps due to its inertness and specific electrical properties.

**Uses of Argon:**
* Argon is extensively used for filling electric light bulbs and fluorescent tubes, as its inert atmosphere prevents the oxidation of the hot filament, extending bulb life.
* It provides an inert atmosphere in high-temperature metallurgical processes, such as arc welding and cutting of metals, to prevent reaction with atmospheric gases.
* Pure argon is used as a carrier gas in gas chromatography and in spectrochemical analysis.

In simple words: Neon is used for bright signs and safety devices, while argon is used in light bulbs and welding to provide an inert atmosphere.
🎯 Exam Tip: When listing uses of noble gases, highlight the property (e.g., inertness, characteristic emission) that makes them suitable for that particular application.

GSEB Class 12 Chemistry The p-Block Elements Additional Important Questions and Answers

 

Question 1. NH3 is manufactured by passing a mixture of nitrogen and hydrogen, at a pressure of 200 atm, over a catalyst at 750 K. Under these conditions, about 15% of the gases are converted to NH3.
(a) Write an equation for the reaction and state whether it is exothermic or endothermic?
(b) Name the catalyst. Why is a catalyst used in this reaction?
(c) What happens to the gases which are not converted to ammonia?
(d) The yield of ammonia is increased if the process is carried out at
(I) even higher pressures and
(II) lower temperature. Explain why these conditions are not used in the industry.


Answer:
(a) The reaction for ammonia synthesis is:
\[N_2 + 3H_2 \rightleftharpoons 2NH_3\]
The enthalpy change for this reaction, \(\Delta H = -93.6 \, \text{kJ}\), indicating that the reaction is exothermic.
(b) The catalyst used is spongy iron, often with molybdenum or K2O as a promotor. A catalyst is crucial for increasing the reaction rate, allowing the synthesis to proceed at a practical speed.
(c) Unreacted nitrogen (N2) and hydrogen (H2) gases are recirculated back to the catalyst chamber for further conversion into ammonia. This improves the overall efficiency of the process.
(d)
(I) Higher pressures would favor the formation of ammonia because the forward reaction results in a decrease in the number of moles of gas (4 moles of reactants produce 2 moles of product). According to Le Chatelier's principle, increasing pressure shifts the equilibrium towards the side with fewer gas moles. However, extremely high pressures necessitate specialized, costly equipment and increase operational hazards, making them industrially impractical.
(II) A lower temperature would increase the equilibrium yield of ammonia, as the reaction is exothermic. Le Chatelier's principle dictates that a decrease in temperature favors an exothermic reaction. Nevertheless, significantly lowering the temperature drastically slows down the reaction rate, even with a catalyst. This would make the production process uneconomical due to prolonged reaction times.
In simple words: Ammonia production is an exothermic reaction where nitrogen and hydrogen combine. A catalyst speeds it up, and unreacted gases are recycled. While high pressure and low temperature favor ammonia formation, extreme conditions are avoided in industry due to high costs and slow reaction rates.

🎯 Exam Tip: When discussing industrial processes like the Haber-Bosch process, always mention Le Chatelier's principle and explain how it influences optimal conditions (temperature, pressure, catalyst) for maximum yield, balancing thermodynamic favorability with kinetic feasibility and economic considerations.

 

Question 2. Match the following

A	B
i. Pyramidal	a. Haber process
ii. Brown ring test	b. Ostwald process
iii. NH3	c. Nitrogen
iv. HNO3	d. Nitrous oxide
v. Rutherford	e. Nitrates
vi. Laughing gas	f. NH3


Answer:
(i) f
(ii) e
(iii) a
(iv) b
(v) c
(d) d
In simple words: This question matches chemical concepts or processes from column A to their corresponding descriptions or examples in column B, such as NH3 having a pyramidal shape or its production via the Haber process.

🎯 Exam Tip: For matching questions, carefully link each item from column A to its best fit in column B. Understanding key terms and their associated processes or properties is crucial for accuracy.

 

Question 3. Nitric acid is manufactured by the catalytic oxidation of ammonia. The reactions which take place are represented by the equations (i) and (ii) below.
(i) \(4NH_3 + 5O_2 \to 4NO + H_2O\)
(ii) \(4NO + 3O_2 + 2H_2O \to 4HNO_3\)
a. How many molecules of water would have to be formed in order to balance equation (i)?
b. Name the catalyst used in the manufacture of nitric acid.
c. Is the catalyst essential for the reaction represented by equation (i) or that represented by equation (ii), or both of these reactions?
d. Describe fully, one test for nitrate.


Answer:
a. To balance equation (i), six molecules of water (6H2O) are formed:
\(4NH_3 + 5O_2 \to 4NO + 6H_2O\).
b. The catalyst used in the manufacture of nitric acid via the Ostwald process is platinum (or platinum-rhodium gauze).
c. The catalyst is essential for the reaction represented by equation (i) (\(4NH_3 + 5O_2 \to 4NO + 6H_2O\)), as this is the catalytic oxidation step. Equation (ii) is a non-catalytic oxidation.
d. One common test for nitrate ions (\(NO_3^-\)) is the brown ring test. In this test, a small amount of the sample containing nitrate ions is mixed with freshly prepared ferrous sulfate (\(FeSO_4\)) solution. Concentrated sulfuric acid (\(H_2SO_4\)) is then carefully added down the side of the test tube, allowing it to form a layer beneath the aqueous solution. A brown ring forms at the interface between the two layers, indicating the presence of nitrate ions. This occurs because the nitrate ions are reduced to nitric oxide (NO) by \(Fe^{2+}\) ions in the presence of sulfuric acid, and the nitric oxide then reacts with excess \(Fe^{2+}\) to form a brown complex, \([Fe(H_2O)_5NO]^{2+}\).
The reactions involved are:
\[NO_3^- (aq) + 3Fe^{2+} (aq) + 4H^+ (aq) \to NO (g) + 3Fe^{3+} (aq) + 2H_2O (l)\]
\[Fe^{2+} (aq) + NO (g) + 5H_2O (l) \to [Fe(H_2O)_5NO]^{2+} (aq)\]
In simple words: For nitric acid production, 6 water molecules balance the first reaction. Platinum acts as the catalyst for the ammonia oxidation. The brown ring test detects nitrate ions by forming a distinctive brown complex at the junction of two liquid layers.

🎯 Exam Tip: Remember specific balancing coefficients for key reactions, name catalysts correctly, and for qualitative tests like the brown ring test, describe the reagents, procedure, observation, and the underlying chemical reactions, including the formula of the colored complex.

 

Question 4. For a question, to write an example of a molecule showing sp³d hybridisation, one student wrote NCl5 as the answer.
1. Do you consider it as the correct answer? Justify.
2. Give the hybridisation of 'P' in PCl3 and explain the shape of the molecule.
3. Illustrate the dehydrating property of P2O5 by using H2SO4.


Answer:
1. No, NCl5 is not a correct example for sp3d hybridization. Nitrogen, being a second-period element, lacks vacant d-orbitals in its valence shell. Therefore, it cannot expand its octet beyond four bonds (or a maximum covalency of 4), making sp3d hybridization impossible for nitrogen.
2. In PCl3, the central phosphorus atom is sp3 hybridized. Phosphorus has 5 valence electrons. In PCl3, it forms three single bonds with chlorine atoms and retains one lone pair of electrons. According to VSEPR theory, these four electron domains (three bond pairs and one lone pair) repel each other, resulting in a trigonal pyramidal molecular geometry. The bond angle in PCl3 is approximately 100°, which is slightly less than the ideal tetrahedral angle of 109.5° due to the greater lone pair-bond pair repulsion.
3. Phosphorus pentoxide (\(P_2O_5\)) is a powerful dehydrating agent. It can remove water from various compounds, including strong acids like sulfuric acid (\(H_2SO_4\)), effectively dehydrating them.
For example, when concentrated sulfuric acid is treated with \(P_2O_5\):
\[2H_2SO_4 + P_4O_{10} \to 2SO_3 + 4HPO_3\]
In this reaction, \(P_4O_{10}\) extracts water molecules from \(H_2SO_4\), leaving behind sulfur trioxide (\(SO_3\)) and metaphosphoric acid (\(HPO_3\)).
In simple words: NCl5 can't exist with sp3d hybridization because nitrogen lacks d-orbitals. PCl3 has sp3 hybridization and a trigonal pyramidal shape due to one lone pair and three bond pairs. P2O5 acts as a dehydrating agent, removing water from compounds like sulfuric acid.

🎯 Exam Tip: When explaining hybridization, always consider the availability of d-orbitals for elements. For molecular shapes, apply VSEPR theory, paying attention to lone pairs. For chemical properties, remember common reactions and illustrate with balanced equations.

 

Question 5. Does phosphorus exist in three allotropic forms?
1. Write down the names of the allotropic forms of phosphorus,
2. Draw the structure of P4 molecule.
3. The hydride of nitrogen is known as ammonia. Write the name of the hydride of phosphorus,
4. How is it prepared?


Answer:
1. Yes, phosphorus primarily exists in three main allotropic forms: white phosphorus, red phosphorus, and black phosphorus.
2. The structure of the P4 molecule (white phosphorus) is a regular tetrahedron:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह सफेद फास्फोरस की P4 टेट्राहेड्रल संरचना को दर्शाता है, जिसमें चार फास्फोरस परमाणु एक समचतुष्फलक के कोनों पर स्थित होते हैं, जो एकल सहसंयोजक बंधों से जुड़े होते हैं।
3. The hydride of nitrogen is ammonia (\(NH_3\)). The hydride of phosphorus is phosphine (\(PH_3\)).
4. Phosphine (\(PH_3\)) can be prepared by heating white phosphorus with a 40% solution of caustic soda (NaOH) in an inert atmosphere, typically CO2 or coal gas.
\[P_4 + 3NaOH + 3H_2O \to 3NaH_2PO_2 + PH_3\]
In simple words: Phosphorus exists as white, red, and black allotropes; white phosphorus is a tetrahedral P4 molecule. Ammonia is nitrogen's hydride, while phosphine is phosphorus's hydride, prepared by heating white phosphorus with sodium hydroxide.

🎯 Exam Tip: Distinguish between different allotropes by their properties and structures. For chemical preparations, remember key reactants, conditions, and balanced equations, especially for industrially important compounds like phosphine.

 

Question 6.
1. Give the structure and names of any four oxyacids of phosphorus.
2. Orthophosphoric acid is a tribasic acid while phosphorus acid is a dibasic acid. Give reason.


Answer:
1. Four oxyacids of phosphorus with their structures and names:
* **Orthophosphoric acid (\(H_3PO_4\))**
ℹ️ चित्र व्याख्या (Diagram Explanation): यह ऑर्थोफॉस्फोरिक अम्ल (\(H_3PO_4\)) की संरचना को दर्शाता है, जिसमें केंद्रीय फास्फोरस परमाणु एक ऑक्सीजन से दोहरा बंध बनाता है और तीन -OH समूहों से एकल बंध बनाता है।
* **Phosphorous acid (\(H_3PO_3\))**
ℹ️ चित्र व्याख्या (Diagram Explanation): यह फॉस्फोरस अम्ल (\(H_3PO_3\)) की संरचना को दर्शाता है, जिसमें केंद्रीय फास्फोरस परमाणु एक ऑक्सीजन से दोहरा बंध बनाता है, दो -OH समूहों से एकल बंध बनाता है, और एक हाइड्रोजन परमाणु से भी सीधा जुड़ा होता है।
* **Diphosphoric acid (Pyrophosphoric acid) (\(H_4P_2O_7\))**
ℹ️ चित्र व्याख्या (Diagram Explanation): यह डाइफॉस्फोरिक अम्ल (\(H_4P_2O_7\)) की संरचना को दर्शाता है, जिसमें दो फास्फोरस परमाणु एक ऑक्सीजन परमाणु के माध्यम से जुड़े होते हैं। प्रत्येक फास्फोरस परमाणु एक ऑक्सीजन से दोहरा बंध और दो -OH समूहों से एकल बंध बनाता है।
* **Hypophosphoric acid (\(H_4P_2O_6\))**
ℹ️ चित्र व्याख्या (Diagram Explanation): यह हाइपोफॉस्फोरिक अम्ल (\(H_4P_2O_6\)) की संरचना को दर्शाता है, जिसमें दो फास्फोरस परमाणु सीधे एक-दूसरे से जुड़े होते हैं। प्रत्येक फास्फोरस परमाणु एक ऑक्सीजन से दोहरा बंध और दो -OH समूहों से एकल बंध बनाता है।
2. The basicity of an acid is determined by the number of ionizable hydrogen atoms (protons) directly attached to oxygen atoms (as -OH groups). These are the hydrogens that can be released as \(H^+\) ions in an aqueous solution.
* **Orthophosphoric acid (\(H_3PO_4\))**: Its structure features three -OH groups attached to the phosphorus atom. All three hydrogen atoms are acidic because they are bonded to highly electronegative oxygen atoms, allowing them to dissociate. Therefore, \(H_3PO_4\) is a tribasic acid.
* **Phosphorous acid (\(H_3PO_3\))**: In contrast, the structure of \(H_3PO_3\) has only two -OH groups. The third hydrogen atom is directly bonded to the phosphorus atom (P-H bond) and is not acidic, meaning it does not ionize in water. Consequently, only the two hydrogens from the -OH groups are ionizable, making \(H_3PO_3\) a dibasic acid.
In simple words: The basicity of phosphorus oxyacids depends on the number of -OH groups. Orthophosphoric acid is tribasic because it has three ionizable -OH groups, while phosphorous acid is dibasic because it has only two ionizable -OH groups, with one hydrogen directly bonded to phosphorus, which is not acidic.

🎯 Exam Tip: For oxyacids, accurately drawing the structure is key to determining basicity. Remember that only hydrogen atoms bonded to oxygen (as -OH groups) are acidic and contribute to the acid's basicity, not those bonded directly to phosphorus.

 

Question 8. The acidic strength of hydrogen halides in aqueous solution written by three students in an examination are given below.
Appu: HF > HCl > HBr > HI
Chinnu: HI > HBr > HCl > HF
Balu: HCl > HF > HBr > HI
1. Who wrote the correct order?
2. Justify your answer?


Answer:
1. Chinnu wrote the correct order: HI > HBr > HCl > HF.
2. The acidic strength of hydrogen halides (HX) in aqueous solution increases down the group. This trend is primarily due to the decreasing bond dissociation enthalpy of the H-X bond as the size of the halogen atom increases.
* As we move from HF to HI, the size of the halogen atom (F, Cl, Br, I) increases significantly.
* This increase in size leads to a longer H-X bond length.
* A longer bond length results in weaker H-X bond dissociation energy, making it easier to break the H-X bond and release a proton (\(H^+\)) into solution.
* Therefore, HI has the weakest H-I bond and is the strongest acid, while HF has the strongest H-F bond and is the weakest acid among the hydrogen halides.
* Electronegativity also plays a role, but bond strength is the dominant factor determining acidic strength for these compounds in aqueous solution.
In simple words: Chinnu is correct; HI is the strongest acid among hydrogen halides, and HF is the weakest. This is because as you go down the halogen group, the H-X bond length increases, making it easier to break the bond and release hydrogen ions, thus increasing acidity.

🎯 Exam Tip: For acidic strength trends within a group for binary acids, remember that bond dissociation energy (influenced by atomic size and bond length) is usually the deciding factor, rather than just electronegativity alone.

 

Question 9. After discussion about the acidity of oxyacids of halogens, Ramu wrote the strength of acids in the following order.
HClO < HClO2 < HClO3 < HClO4
1. Is this order of acid strength correct or not?
2. Justify your answer.
3. Draw the structure of all these oxyacids.


Answer:
1. Yes, Ramu's order of acid strength is correct: \(HClO < HClO_2 < HClO_3 < HClO_4\).
2. The acidic strength of oxyacids of a particular halogen increases with increasing oxidation state of the central halogen atom. Alternatively, it increases with the number of oxygen atoms not bonded to hydrogen.
* In oxyacids, the hydrogen atom is bonded to an oxygen atom (-O-H). The acidity arises from the ease with which this H can be released as \(H^+\).
* The more electronegative the central atom (or the more oxygen atoms attached to it), the more it pulls electron density away from the O-H bond, weakening it and making the H more acidic.
* As the oxidation state of chlorine increases from +1 in HClO to +7 in HClO4, the number of oxygen atoms directly attached to chlorine (but not to hydrogen) increases. These additional electronegative oxygen atoms withdraw electron density from the O-H bond, making it progressively weaker and enhancing the acidity.
* According to the Lowry-Bronsted concept, a strong acid has a weak conjugate base. The stability of the conjugate bases (\(ClO^-\), \(ClO_2^-\), \(ClO_3^-\), \(ClO_4^-\)) increases with increasing delocalization of the negative charge, which is facilitated by a greater number of oxygen atoms through resonance. \(ClO_4^-\) is the most stable conjugate base (and thus the weakest base), making \(HClO_4\) the strongest acid.
3. The structures of these oxyacids are:
* **Hypochlorous acid (HClO)**
ℹ️ चित्र व्याख्या (Diagram Explanation): यह हाइपोक्लोरस अम्ल (\(HClO\)) की संरचना को दर्शाता है, जिसमें क्लोरीन परमाणु एक हाइड्रोजन और एक ऑक्सीजन परमाणु से एकल बंध बनाता है।
* **Chlorous acid (HClO2)**
ℹ️ चित्र व्याख्या (Diagram Explanation): यह क्लोरस अम्ल (\(HClO_2\)) की संरचना को दर्शाता है, जिसमें केंद्रीय क्लोरीन परमाणु एक ऑक्सीजन परमाणु से दोहरा बंध और एक -OH समूह से एकल बंध बनाता है।
* **Chloric acid (HClO3)**
ℹ️ चित्र व्याख्या (Diagram Explanation): यह क्लोरिक अम्ल (\(HClO_3\)) की संरचना को दर्शाता है, जिसमें केंद्रीय क्लोरीन परमाणु दो ऑक्सीजन परमाणुओं से दोहरा बंध और एक -OH समूह से एकल बंध बनाता है।
* **Perchloric acid (HClO4)**
ℹ️ चित्र व्याख्या (Diagram Explanation): यह परक्लोरिक अम्ल (\(HClO_4\)) की संरचना को दर्शाता है, जिसमें केंद्रीय क्लोरीन परमाणु तीन ऑक्सीजन परमाणुओं से दोहरा बंध और एक -OH समूह से एकल बंध बनाता है।
In simple words: The acid strength order is correct; acidity increases with the number of oxygen atoms attached to chlorine, as these pull electron density away from the O-H bond, making it easier for hydrogen to dissociate. This also stabilizes the resulting conjugate base.

🎯 Exam Tip: When justifying the acid strength of oxyacids, always refer to the oxidation state of the central atom and the number of terminal oxygen atoms, as these factors directly affect electron withdrawal and the stability of the conjugate base through resonance.

 

Question 10. A scientist conducted a series of experiments with xenon and fluorine. The three different conditions used are given below.
A. Ni at 400° C
B. Ni at 400° C and acetone.
C. Ni at 400° C and 200 atm.


Answer:
1. Based on the conditions, the compounds obtained are XeF2, XeF4, and XeF6, respectively.
2. The chemical equations for the formation of xenon fluorides under different conditions are:
* **For XeF2 (Condition A):** Xenon reacts with fluorine at 400°C in the presence of a nickel catalyst and low pressure (e.g., 1 bar).
\[Xe + F_2 \xrightarrow{\text{Ni, 400°C}} XeF_2\]
* **For XeF4 (Condition B):** Xenon reacts with excess fluorine at 400°C in the presence of a nickel catalyst, with acetone possibly acting as a solvent or moderator, leading to the formation of XeF4. The original text seems to have a typo "acetone" in place of higher F2 concentration or specific pressure. However, following the context, condition B usually implies a higher fluorine ratio than A. Assuming the condition meant a specific Xe:F2 ratio that yields XeF4.
\[Xe + 2F_2 \xrightarrow{\text{Ni, 400°C}} XeF_4\]
* **For XeF6 (Condition C):** Xenon reacts with a significant excess of fluorine (1:20 ratio) at 400°C and a high pressure of 200 atm (or 60-70 bar), with a nickel catalyst.
\[Xe + 3F_2 \xrightarrow{\text{Ni, 400°C, 200 atm}} XeF_6\]
3. The structures of the compounds obtained are:
* **XeF2 (linear)**: sp3d hybridization, 3 lone pairs on Xe.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह XeF2 की रैखिक संरचना को दर्शाता है, जिसमें केंद्रीय क्सीनन परमाणु पर तीन एकाकी इलेक्ट्रॉन युग्म और दो फ्लोरीन परमाणुओं से एकल बंध होते हैं। क्सीनन sp3d संकरित होता है।
* **XeF4 (square planar)**: sp3d2 hybridization, 2 lone pairs on Xe.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह XeF4 की वर्ग समतलीय संरचना को दर्शाता है, जिसमें केंद्रीय क्सीनन परमाणु पर दो एकाकी इलेक्ट्रॉन युग्म और चार फ्लोरीन परमाणुओं से एकल बंध होते हैं। क्सीनन sp3d2 संकरित होता है।
* **XeF6 (distorted octahedral)**: sp3d3 hybridization, 1 lone pair on Xe.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह XeF6 की विकृत अष्टफलकीय संरचना को दर्शाता है, जिसमें केंद्रीय क्सीनन परमाणु पर एक एकाकी इलेक्ट्रॉन युग्म और छह फ्लोरीन परमाणुओं से एकल बंध होते हैं। क्सीनन sp3d3 संकरित होता है।
In simple words: Xenon forms fluorides (XeF2, XeF4, XeF6) under specific conditions of fluorine concentration, temperature, pressure, and catalyst. Each compound has a distinct hybridization and molecular geometry: XeF2 is linear, XeF4 is square planar, and XeF6 is distorted octahedral.

🎯 Exam Tip: Remember that noble gases like xenon can form compounds with highly electronegative elements under specific conditions. Be precise with the conditions for synthesizing different fluorides and accurately describe their hybridization and molecular geometries using VSEPR theory.

 

Question 11. Account for the following:
1. Molecular nitrogen is not particularly reactive.
2. Water has a higher boiling point than H2S.
3. Nitrogen forms no pentahalide, unlike phosphorus.
4. Cl2O7 is formed but no such fluorine compound is formed.


Answer:
1. Molecular nitrogen (N2) is not particularly reactive because it exists as a diatomic molecule with a strong triple bond between the two nitrogen atoms (\(N \equiv N\)). This triple bond is highly stable and has a very high bond dissociation energy (approximately 945 kJ/mol). A large amount of energy is required to break this bond, making N2 chemically inert under ambient conditions.
2. Water (\(H_2O\)) has a significantly higher boiling point than hydrogen sulfide (\(H_2S\)) due to the presence of extensive intermolecular hydrogen bonding in water. Oxygen is much more electronegative than sulfur, and its smaller atomic size allows it to form strong hydrogen bonds with other water molecules. In contrast, sulfur's lower electronegativity and larger size prevent it from forming effective hydrogen bonds in \(H_2S\). Therefore, \(H_2S\) molecules are held together only by weaker Van der Waals forces, resulting in a much lower boiling point.
3. Nitrogen does not form pentahalides (like NCl5) because it lacks vacant d-orbitals in its valence shell. Nitrogen is a second-period element with only 2s and 2p orbitals available for bonding. It cannot expand its valence shell beyond an octet (maximum covalency of 4). Phosphorus, being a third-period element, has vacant 3d-orbitals, which it can utilize to expand its octet and achieve a +5 oxidation state, forming pentahalides like PCl5.
4. \(Cl_2O_7\) (dichlorine heptoxide) is formed, representing chlorine in its highest oxidation state of +7. However, no corresponding fluorine compound like \(F_2O_7\) exists. This is because fluorine is the most electronegative element and consistently exhibits an oxidation state of -1 in its compounds, except when forming an F-F bond. It cannot attain positive oxidation states such as +7 because it does not have vacant d-orbitals to expand its octet or participate in variable oxidation states. The small size of fluorine also contributes to its unique behavior.
In simple words: Nitrogen is unreactive due to its strong triple bond. Water boils higher than H2S because of hydrogen bonding. Nitrogen doesn't form pentahalides as it lacks d-orbitals, unlike phosphorus. Chlorine forms Cl2O7 with a +7 oxidation state, but fluorine cannot form a similar compound because it only shows a -1 oxidation state.

🎯 Exam Tip: When explaining differences in reactivity, boiling points, or bonding behavior, always consider key factors like bond strength, electronegativity, atomic size, and the availability of d-orbitals. These fundamental chemical principles govern the observed properties.

 

Question 12.
1. Draw the structure of H2SO4
2. Write a chemical reaction which shows the dehydrating nature of H2SO4.
3. Why is H2SO4 called the king of chemicals?


Answer:
1. The structure of \(H_2SO_4\) (sulfuric acid) is:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह सल्फ्यूरिक अम्ल (\(H_2SO_4\)) की संरचना को दर्शाता है, जिसमें केंद्रीय सल्फर परमाणु दो ऑक्सीजन परमाणुओं से दोहरा बंध और दो -OH समूहों से एकल बंध बनाता है।
2. Concentrated sulfuric acid is a potent dehydrating agent. It can remove water from organic compounds, such as glucose. When glucose (\(C_6H_{12}O_6\)) is treated with concentrated \(H_2SO_4\), it gets charred, turning black as carbon is left behind due to the removal of water.
\[C_6H_{12}O_6 \xrightarrow{\text{Conc. H2SO4}} 6C + 6H_2O\]
3. Sulfuric acid (\(H_2SO_4\)) is often referred to as the "king of chemicals" because of its immense importance and extensive applications across various industries. It is used in:
* **Manufacturing fertilizers**: like ammonium sulfate and superphosphate.
* **Petroleum refining**: for purification processes.
* **Metallurgy**: for cleaning metals (pickling) before electroplating, enameling, or galvanizing.
* **Storage batteries**: as an electrolyte.
* **Chemical synthesis**: in the production of other acids, explosives, dyes, detergents, and many other chemicals.
Its versatility and crucial role in industrial chemistry earn it this distinguished title.
In simple words: Sulfuric acid's structure has a central sulfur double-bonded to two oxygens and single-bonded to two -OH groups. It's a strong dehydrating agent, turning glucose into carbon. It's called the "king of chemicals" because it's vital in countless industries, from fertilizers to batteries.

🎯 Exam Tip: For acids like H2SO4, know both its structural representation and its key chemical properties (e.g., dehydrating action with an example). Also, be ready to list its major industrial applications to justify its significance.

 

Question 13.
a. What are interhalogen compounds?
b. Give the method of preparation and structure of


Answer:
a. Interhalogen compounds are binary compounds formed between two different halogens. They can be represented by the general formulas AX, AX3, AX5, and AX7, where A is the larger and less electronegative halogen, and X is the smaller and more electronegative halogen. These compounds are generally more reactive than pure halogens (except \(F_2\)) because the A-X bond is weaker than the X-X bond in diatomic halogens, due to the difference in electronegativity between the combining halogen atoms. Examples include \(ClF_3\), \(IF_7\), and \(BrF_5\).
b. Method of preparation and structure for specific interhalogen compounds:
* **Preparation of \(ClF_3\):** Chlorine reacts with fluorine in a controlled ratio.
\[Cl_2 + 3F_2 \to 2ClF_3\]
* **Preparation of \(IF_7\):** Iodine pentafluoride reacts with excess fluorine.
\[F_2 + IF_5 \to IF_7\]
* **Preparation of \(ClF_5\):** Chlorine trifluoride reacts with fluorine.
\[ClF_3 + F_2 \to ClF_5\]
* **Structure of \(ClF_3\):** It has a T-shaped geometry. The central chlorine atom is sp3d hybridized, with two lone pairs and three bond pairs (three Cl-F bonds). The lone pairs occupy equatorial positions to minimize repulsion.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह ClF3 की T-आकार की संरचना को दर्शाता है, जिसमें केंद्रीय क्लोरीन परमाणु पर दो एकाकी इलेक्ट्रॉन युग्म और तीन फ्लोरीन परमाणुओं से एकल बंध होते हैं। यह sp3d संकरण को प्रदर्शित करता है।
* **Structure of \(IF_7\):** It has a pentagonal bipyramidal geometry. The central iodine atom is sp3d3 hybridized, with seven I-F bond pairs and no lone pairs.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह IF7 की पंचकोणीय द्विपिरामिडी संरचना को दर्शाता है, जिसमें केंद्रीय आयोडीन परमाणु सात फ्लोरीन परमाणुओं से एकल बंध बनाता है और कोई एकाकी इलेक्ट्रॉन युग्म नहीं होता है। यह sp3d3 संकरण को प्रदर्शित करता है।
* **Structure of \(ClF_5\):** It has a square pyramidal geometry. The central chlorine atom is sp3d2 hybridized, with one lone pair and five bond pairs (five Cl-F bonds).
ℹ️ चित्र व्याख्या (Diagram Explanation): यह ClF5 की वर्ग पिरामिडी संरचना को दर्शाता है, जिसमें केंद्रीय क्लोरीन परमाणु पर एक एकाकी इलेक्ट्रॉन युग्म और पांच फ्लोरीन परमाणुओं से एकल बंध होते हैं। यह sp3d2 संकरण को प्रदर्शित करता है।
In simple words: Interhalogen compounds are formed between two different halogens and are generally more reactive than individual halogens. Their preparation involves specific halogen combinations, and their structures (like T-shaped for ClF3, pentagonal bipyramidal for IF7, and square pyramidal for ClF5) are determined by VSEPR theory and hybridization.

🎯 Exam Tip: For interhalogen compounds, understand their general definition and reactivity relative to halogens. For specific examples, recall the preparation methods and use VSEPR theory to deduce their hybridization and molecular geometries accurately, including the positions of lone pairs.