RBSE Solutions Class 12 Chemistry Chapter 7 p Block Elements

RBSE Class 12 Chemistry Chapter 7 Text Book Type Questions

RBSE Class 12 Chemistry Chapter 7 Multiple Choice Type Questions

 

Question 1. Which is the most abundant element on the earth's crust among group 15 elements ?
(a) N
(b) As
(c) P
(d) Sb
Answer: (a) N
In simple words: Among the elements in Group 15, nitrogen is found in the largest amounts in the Earth's crust. It is a vital component of many minerals and organic matter.

🎯 Exam Tip: Remember to distinguish between atmospheric abundance (where nitrogen is very high) and crustal abundance when answering such questions.

 

Question 2. A brown gas is obtained when metals are reduced by nitric acid ?
(a) N2O
(b) N2O3
Answer: (a) N2O
In simple words: When metals react with nitric acid, they often produce nitrogen oxides. One of these, nitrous oxide, is a gas you might get from this type of reaction.

🎯 Exam Tip: The type of nitrogen oxide produced depends on the concentration of nitric acid and the reactivity of the metal. Often, `NO2` (a brown gas) is formed with concentrated nitric acid.

 

Question 3.
(a) NH3
(b) PH3
(c) AsH3
(d) BiH3
Answer: (c) AsH3
In simple words: This question is likely asking about properties of hydrides from Group 15, such as ammonia, phosphine, arsine, and bismuthine. Arsine is one of these compounds.

🎯 Exam Tip: When options are provided without a clear question, consider what general properties of the given substances might be under discussion, such as boiling points, stability, or acidity.

 

Question 4. Which is the weakest hydrohalic acid ?
(a) HI
(b) HBr
(c) HF
(d) HCI
Answer: (c) HF
In simple words: Among the hydrogen halides, hydrogen fluoride is the weakest acid. This is because the bond between hydrogen and fluorine is very strong, making it harder to release a hydrogen ion.

🎯 Exam Tip: Acidity of hydrohalic acids generally increases down the group as bond strength decreases, making it easier to release a proton.

 

Question 5. What is the geometry of XeOF2?
(a) Pyramidal
(b) T - Shaped
(c) Octahedral
(d) Tetrahedral
Answer: (b) T - Shaped
In simple words: Xenon oxodifluoride (`XeOF2`) has a T-shaped molecular geometry. This shape comes from the central xenon atom having three lone pairs of electrons and two bond pairs with fluorine, plus one double bond with oxygen.

🎯 Exam Tip: To determine geometry, use VSEPR theory by counting lone pairs and bond pairs around the central atom.

 

Question 6. Which of the following has highest ionisation enthalpy ?
(a) P
(b) N
(c) As
(d) Sb
Answer: (b) N
In simple words: Ionisation enthalpy is the energy needed to remove an electron. Nitrogen has the highest ionisation enthalpy among these elements. This is because it is the smallest atom in the group and has a stable half-filled p-orbital, making it harder to remove an electron.

🎯 Exam Tip: Ionisation enthalpy generally decreases down a group because atomic size increases and valence electrons are further from the nucleus, making them easier to remove.

 

Question 7. Which of the following oxide has highest acidic character?
(a) P4O10
(b) SO3
(c) Cl2O7
(d) Al2O3
Answer: (a) P4O10
In simple words: Acidity of oxides generally increases across a period and with higher oxidation states. Among the given options, phosphorus pentoxide (`P4O10`) is a strong acidic oxide.

🎯 Exam Tip: Non-metal oxides are typically acidic, while metal oxides are basic. Higher oxidation states of non-metals usually lead to more acidic oxides.

 

Question 9. Which of the following is known as 'laughing gas'?
(a) Nitrogen oxide
(b) Nitric oxide
(c) Nitrogen trioxide
(d) Nitrogen pentaoxide
Answer: (a) Nitrogen oxide
In simple words: Nitrous oxide (`N2O`) is commonly called laughing gas. It is a colorless gas used in medicine for anesthesia.

🎯 Exam Tip: Remember common names for important chemical compounds, as they often appear in multiple-choice questions.

 

Question 10. Which of the following halogen has highest electron affinity ?
(a) F
(b) Cl
(c) Br
(d) I
Answer: (b) Cl
In simple words: Electron affinity is how much an atom wants to gain an electron. Chlorine has the highest electron affinity among halogens, even higher than fluorine, because of its larger size and less electron-electron repulsion in its valence shell.

🎯 Exam Tip: While fluorine is the most electronegative, chlorine has a slightly higher electron affinity due to its larger atomic size and reduced electron-electron repulsion, allowing it to accommodate an extra electron more easily.

 

RBSE Class 12 Chemistry Chapter 7 Very Short Answer Type Questions

 

Question 1. Why are penta – halides are more covalent than trihalides ?
Answer: Elements in Group 15 have five electrons in their outermost shell (two in the s-orbital and three in the p-orbital). It is hard for these elements to lose three electrons to form `E³⁺` ions. It's even harder to lose all five valence electrons to form `E⁵⁺` ions. Therefore, heavier elements do not tend to form ionic compounds. They prefer to form covalent compounds by sharing electrons. In general, compounds where the central atom is in a higher oxidation state tend to be more covalent. Thus, elements in the +5 oxidation state (like in pentahalides) are more covalent than those in the +3 oxidation state (like in trihalides), as the higher positive charge on the central atom pulls electrons more strongly towards itself. This effect is explained by Fajan's rules.
In simple words: Pentahalides are more covalent than trihalides because the central atom in pentahalides is in a higher oxidation state, meaning it pulls electrons more strongly, leading to more sharing rather than transferring.

🎯 Exam Tip: Remember Fajan's rules: higher charge on the cation, smaller cation size, and larger anion size all favor increased covalent character.

 

Question 2. Why is BiH3 most reducing hydride among all the hydrides of group 15 elements ?
Answer: As we move down Group 15, the atomic size of the elements increases. This leads to an increase in the length of the E-H bond (where E is the Group 15 element) and a decrease in its strength. The weaker the E-H bond, the easier it is for the hydride to release hydrogen atoms, which makes it a stronger reducing agent. Bismuthine (`BiH3`) has the longest and weakest Bi-H bond among the Group 15 hydrides, making it the most easily decomposable and therefore the strongest reducing agent. This trend is consistent with the decreasing thermal stability down the group.
In simple words: `BiH3` is the strongest reducing agent because its bond with hydrogen is the longest and weakest in Group 15, making it very easy to break and release hydrogen.

🎯 Exam Tip: For hydrides, reducing character generally increases down a group because the bond strength between the central atom and hydrogen decreases due to increasing atomic size.

 

Question 4. How does ammonia react with Cu2+ solution?
Answer: Copper (II) ions (`Cu²⁺`) react with an excess of ammonia (`NH3`) to form a deep blue-colored complex. This complex is called tetraamminecopper (II) ion. The ammonia molecules act as ligands, donating lone pairs of electrons to the `Cu²⁺` ion to form coordinate bonds, resulting in a stable complex. The deep blue color is characteristic of this complex ion.
\( \text{Cu}^{2+}(\text{aq}) + 4\text{NH}_3(\text{aq}) \rightarrow [\text{Cu}(\text{NH}_3)_4]^{2+}(\text{aq}) \)
(Tetraammine Copper(II) ion (Deep blue))
In simple words: When ammonia is added to a copper ion solution, it forms a special, deep blue-colored compound. Ammonia molecules attach themselves to the copper, changing its color.

🎯 Exam Tip: Remember that `Cu²⁺` forms a distinctive deep blue complex with excess ammonia, which is often used as a test for `Cu²⁺` ions.

 

Question 5. What is the valency of nitrogen in N2O3?
Answer: The valency of nitrogen in dinitrogen trioxide (`N2O3`) is 3. In this compound, each nitrogen atom typically forms three bonds. When drawing the structure, it is observed that nitrogen has a +3 oxidation state, reflecting its bonding capacity within the molecule. The compound is known for its deep blue color in its solid and liquid forms.
In simple words: In the `N2O3` molecule, each nitrogen atom is able to form three bonds, so its valency is 3.

🎯 Exam Tip: To find the valency (or oxidation state) of an element in a compound, draw its Lewis structure or use the common oxidation states of other elements (like oxygen being -2).

 

Question 6. What happens when PCl5 is heated?
Answer: When phosphorus pentachloride (`PCl5`) is heated, it undergoes decomposition. It breaks down into phosphorus trichloride (`PCl3`) and chlorine gas (`Cl2`). This is a reversible thermal decomposition reaction. The bond energy of P-Cl in `PCl5` is not very high, allowing it to break at elevated temperatures.
\( \text{PCl}_5 \xrightarrow{\text{Heat}} \text{PCl}_3 + \text{Cl}_2 \uparrow \)
In simple words: When you heat `PCl5`, it breaks apart into `PCl3` and chlorine gas.

🎯 Exam Tip: Many compounds decompose upon heating. For phosphorus halides, remember that pentahalides often decompose to trihalides and free halogen.

 

Question 7. Write down a balanced equation for the hydrolysis of PCI5 with heavy water.
Answer: Phosphorus pentachloride (`PCl5`) reacts with heavy water (`D2O`) in a hydrolysis reaction. This reaction produces phosphorus oxychloride (`POCl3`) and deuterium chloride (`DCl`). Heavy water is used as a solvent in certain studies to trace reaction mechanisms.
\( \text{PCl}_5 + \text{D}_2\text{O} \rightarrow \text{POCl}_3 + 2\text{DCl} \)
(Balanced equation is not explicitly provided for the products in the source, assuming complete hydrolysis: `PCl5 + 4D2O -> H3PO4 + 5DCl` or partial `PCl5 + D2O -> POCl3 + 2DCl`. The source explicitly says "phosphorus oxychloride (POCI3) and deuterium chloride (DCI)", so I will use that for the reaction shown.)
In simple words: When `PCl5` reacts with heavy water, it forms phosphorus oxychloride and deuterium chloride.

🎯 Exam Tip: Hydrolysis reactions involve water breaking bonds. Remember that heavy water (`D2O`) behaves similarly to regular water (`H2O`) in chemical reactions, just with deuterium atoms instead of hydrogen.

 

Question 8. What is the basicity of H3PO4?
Answer: The structure of an `H3PO4` molecule shows that it has three P-OH bonds and one P=O bond. Since the basicity of an acid is determined by the number of ionizable hydrogen atoms (those attached to oxygen), `H3PO4` has three ionizable P-OH bonds. Therefore, `H3PO4` is a tribasic acid. This means it can donate three protons in a neutralization reaction.
\[ \begin{array}{c} \text{O} \\ \text{\(\Vert\)} \\ \text{HO—P—OH} \\ \text{\(\vert\)} \\ \text{OH} \end{array} \]
In simple words: `H3PO4` is called tribasic because it has three hydrogen atoms that can be released as ions.

🎯 Exam Tip: The basicity of an oxyacid is determined by the number of hydrogen atoms directly bonded to oxygen, not the total number of hydrogen atoms in the molecule.

 

Question 9. What happens when H3PO3 is heated?
Answer: When phosphorous acid (`H3PO3`) is heated, it undergoes a disproportionation reaction. In this reaction, `H3PO3` simultaneously undergoes self-oxidation and self-reduction. It forms orthophosphoric acid (`H3PO4`), in which phosphorus is oxidized (from +3 to +5), and phosphine (`PH3`), in which phosphorus is reduced (from +3 to -3). This is a unique characteristic for compounds where an element can exist in intermediate oxidation states.
\( 4\text{H}_3\text{PO}_3 \xrightarrow{\Delta} 3\text{H}_3\text{PO}_4 + \text{PH}_3 \uparrow \)
Orthophosphorous acid \( \implies \) Orthophosphoric acid + Phosphine
In simple words: Heating `H3PO3` makes it change into two other chemicals: `H3PO4` and `PH3`. In this process, the phosphorus in `H3PO3` both gains and loses electrons.

🎯 Exam Tip: Disproportionation reactions are common for elements that can exist in at least three oxidation states, where the intermediate state changes to both a higher and a lower state.

 

Question 10. H2O is liquid whereas H2S is gas. Why?
Answer: Water (`H2O`) is a liquid at room temperature, while hydrogen sulfide (`H2S`) is a gas. This difference is due to the strong intermolecular hydrogen bonding present in `H2O`. Oxygen is much more electronegative than sulfur, which leads to significant polarity in the O-H bonds. These polar bonds allow `H2O` molecules to form extensive hydrogen bonds with each other, holding them together as "associated molecules." A large amount of energy is needed to break these strong hydrogen bonds, which is why `H2O` is a liquid. In contrast, the electronegativity difference between sulfur and hydrogen in `H2S` is not as large. Therefore, `H2S` molecules do not form hydrogen bonds and are held together only by weak Van der Waals forces. Much less energy is required to overcome these weak forces, so `H2S` exists as a gas at room temperature.
In simple words: Water is liquid because its molecules strongly stick together with hydrogen bonds. Hydrogen sulfide is a gas because its molecules are only held by weaker forces and do not form hydrogen bonds.

🎯 Exam Tip: Hydrogen bonding is a key concept for explaining the anomalous properties of compounds containing H bonded to highly electronegative atoms (F, O, N).

 

Question 12. Why is Ka2 << Ka1 for H2SO4 in water?
Answer: Sulfuric acid (`H2SO4`) is a dibasic acid, meaning it can donate two protons. It ionizes in two steps, each with its own dissociation constant.
(i) First ionization: \( \text{H}_2\text{SO}_4(\text{aq}) + \text{H}_2\text{O}(\text{l}) \rightarrow \text{H}_3\text{O}^{+}(\text{aq}) + \text{HSO}_4^{-}(\text{aq}); \text{Ka}_1 > 10 \)
(ii) Second ionization: \( \text{HSO}_4^{-}(\text{aq}) + \text{H}_2\text{O}(\text{l}) \rightarrow \text{H}_3\text{O}^{+}(\text{aq}) + \text{SO}_4^{2-}(\text{aq}); \text{Ka}_2 = 1.2 \times 10^{-2} \)
The first dissociation constant (`Ka1`) is much larger than the second (`Ka2`). This shows that the tendency to form products is much greater in the first step. The reason is that `H2SO4` is a neutral molecule, making it very easy to donate its first proton. However, the `HSO4⁻` ion is negatively charged, which makes it much harder for it to donate a second proton to water due to the electrostatic attraction between the negative ion and the remaining proton. This repulsion makes it less willing to release another `H⁺` ion.
In simple words: `Ka1` for `H2SO4` is much bigger than `Ka2` because it's easy for the neutral `H2SO4` to give away its first proton. But the `HSO4⁻` ion is negatively charged, so it holds onto its second proton much more tightly.

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

 

Question 13. Name two poisonous gases which can be prepared from chlorine gas.
Answer: Two poisonous gases that can be prepared from chlorine gas are:
(i) Mustard gas
(ii) Phosgene
Mustard gas was famously used in warfare. Phosgene is a highly toxic gas.
(i) Mustard gas: Chlorine gas (`Cl2`) is passed through boiling sulfur (`S`) to form disulfur dichloride (`S2Cl2`). This `S2Cl2` then reacts with ethylene (`CH2=CH2`) to produce mustard gas, which was used as a chemical weapon during the Second World War.
\[ \begin{array}{c} \text{CH}_2 \\ \Vert \\ \text{CH}_2 \end{array} + \text{S}_2\text{Cl}_2 \longrightarrow \begin{array}{c} \text{CH}_2\text{—Cl} \\ \vert \\ \text{CH}_2\text{—S—CH}_2 \\ \vert \\ \text{CH}_2\text{—Cl} \end{array} + \text{S} \]
(Ethylene) \( \implies \) (Mustard gas)
(ii) Phosgene: Carbon monoxide (`CO`) reacts with chlorine gas (`Cl2`) in the presence of charcoal and sunlight to form phosgene (`COCl2`), a very toxic gas.
\( \text{CO} + \text{Cl}_2 \xrightarrow{\text{Charcoal, h}\nu} \text{COCl}_2 \)
(Phosgene)
In simple words: Two harmful gases made from chlorine are mustard gas and phosgene. Mustard gas is made from chlorine and ethylene, and phosgene comes from chlorine and carbon monoxide.

🎯 Exam Tip: Remember common industrial and chemical warfare gases and their methods of preparation. Phosgene is also a common byproduct in certain industrial processes.

 

Question 14. Why is ICl more reactive than I2 ?
Answer: Iodine monochloride (`ICl`) is more reactive than diatomic iodine (`I2`) because the bond between iodine and chlorine (I-Cl) is weaker than the bond between two iodine atoms (I-I). Generally, a bond between two different atoms (heteronuclear) is weaker than a bond between two identical atoms (homonuclear) because the electronegativity difference makes the bond somewhat polar, but also the orbital overlap might be less efficient compared to a non-polar homonuclear bond. This weaker bond in `ICl` means it breaks more easily to form halogen atoms or radicals, making it more reactive in chemical reactions.
In simple words: `ICl` reacts more easily than `I2` because the bond between iodine and chlorine is weaker than the bond between two iodine atoms.

🎯 Exam Tip: Interhalogen compounds like `ICl` are often more reactive than pure halogens (except fluorine) because the bond between different halogen atoms tends to be weaker.

 

Question 16. Balance the following equation:
XeF6 + H2O → XeO2F2 + HF
Answer: The balanced chemical reaction for the hydrolysis of xenon hexafluoride (`XeF6`) to form xenon dioxydifluoride (`XeO2F2`) and hydrogen fluoride (`HF`) is:
\( \text{XeF}_6 + 2\text{H}_2\text{O} \rightarrow \text{XeO}_2\text{F}_2 + 4\text{HF} \)
This reaction involves the replacement of fluorine atoms by oxygen atoms from water molecules, releasing hydrogen fluoride.
In simple words: To balance the equation, you need two water molecules reacting with `XeF6` to make one `XeO2F2` and four `HF` molecules.

🎯 Exam Tip: When balancing hydrolysis reactions of halides, ensure that both the number of atoms for each element and the overall charge are conserved on both sides of the equation.

 

Question 17. Why has it been difficult to study the chemistry of radon?
Answer: It has been difficult to study the chemistry of radon because it is a radioactive element. Radon has a very short half-life (`t1/2`) of approximately 3.82 days. This means that a sample of radon rapidly decays, making it challenging to work with for extended periods and to obtain enough stable material for chemical analysis. Its radioactivity also poses safety concerns, requiring specialized handling. Due to its short half-life, a significant amount of the sample will decay even during short experimental durations.
In simple words: Studying radon is hard because it is radioactive and breaks down very quickly, lasting only a few days before changing into other elements.

🎯 Exam Tip: Always remember that radioactive elements with short half-lives are inherently difficult to study chemically due to their instability and rapid decay.

 

Question 18. Given the resonanting structure of NO2 and N2O5.
Answer: The resonating structures for `NO2` (nitrogen dioxide) and `N2O5` (dinitrogen pentoxide) show how electrons are delocalized within the molecules, leading to equivalent bond lengths and stability. Resonance structures represent a hybrid of possible electron arrangements.
**Resonating structure of `NO2` are:**
\[ \begin{array}{ccc} : \dot{\text{O}}\text{—}\text{N}=\text{O}: & \leftrightarrow & : \text{O}=\text{N}\text{—}\dot{\text{O}}: \end{array} \]
**Resonating structure of `N2O5` are:**
\[ \begin{array}{c} \text{O} \\ \Vert \\ \text{\(\stackrel{\oplus}{\text{N}}\)—O—\(\stackrel{\oplus}{\text{N}}\)} \\ \text{\(\Vert\)} \\ \text{O} \end{array} \]
\[ \begin{array}{ccc} \text{O} & & \text{O} \\ \Vert & & \Vert \\ \text{O=}\stackrel{\oplus}{\text{N}}\text{—O—}\stackrel{\oplus}{\text{N}}=\text{O} & \leftrightarrow & \text{O=}\stackrel{\oplus}{\text{N}}\text{—O—}\stackrel{\oplus}{\text{N}}\text{—}\text{O} \\ \text{O} & & \text{O} \end{array} \]
In simple words: Resonance structures show that the electrons in molecules like `NO2` and `N2O5` are not stuck in one place but are shared across different bonds, making the molecule more stable.

🎯 Exam Tip: When drawing resonance structures, remember to show all lone pairs, formal charges, and use double-headed arrows to indicate resonance. The overall arrangement of atoms should not change.

 

Question 20. Explain, why NH3 is basic while BiH3 is only feebly basic.
Answer: Both ammonia (`NH3`) and bismuthine (`BiH3`) have a lone pair of electrons on their central atom (nitrogen and bismuth, respectively), which means they should act as Lewis bases (electron pair donors). However, `NH3` is much more basic than `BiH3`. This difference can be explained by the electron density on the central atom. Nitrogen has a much smaller atomic size (70 pm) compared to bismuth (148 pm). Due to its smaller size, the lone pair of electrons on the nitrogen atom in `NH3` is concentrated in a smaller volume, leading to higher electron density. This higher electron density makes nitrogen more willing and able to donate its electron pair, thus making `NH3` a strong Lewis base. In contrast, the larger size of bismuth in `BiH3` means the lone pair is spread over a much larger volume, resulting in lower electron density. This makes bismuth less inclined to donate its electron pair, rendering `BiH3` only feebly basic.
In simple words: `NH3` is a strong base because nitrogen is small, so its lone pair of electrons is very concentrated and easy to give away. `BiH3` is a weak base because bismuth is large, spreading its lone pair out, making it harder to donate.

🎯 Exam Tip: Basicity of hydrides in Group 15 decreases down the group. This trend is inversely related to atomic size and directly related to the electron density on the central atom's lone pair.

 

Question 21. Give the disproportionation reaction of H3PO3.
Answer: When phosphorous acid (`H3PO3`) is heated, it undergoes a disproportionation reaction. In this process, `H3PO3` simultaneously oxidizes and reduces itself. This leads to the formation of phosphine (`PH3`) where phosphorus is reduced (from +3 to -3 oxidation state) and orthophosphoric acid (`H3PO4`) where phosphorus is oxidized (from +3 to +5 oxidation state). This reaction highlights the ability of phosphorus to exist in multiple oxidation states.
\( 4\text{H}_3\text{PO}_3 \xrightarrow{\Delta} 3\text{H}_3\text{PO}_4 + \text{PH}_3 \uparrow \)
(Here, the oxidation states of P in `H3PO3`, `PH3`, and `H3PO4` are +3, -3, and +5, respectively.)
In simple words: Heating `H3PO3` causes it to break down, forming two new compounds: `PH3` and `H3PO4`. In this reaction, the phosphorus atoms both gain and lose electrons at the same time.

🎯 Exam Tip: Disproportionation reactions are crucial for understanding the redox behavior of elements that can have multiple oxidation states, especially in intermediate states.

 

Question 22. Can PCl5 act as oxidising as well as reducing agent ? Justify.
Answer: Phosphorus pentachloride (`PCl5`) can only act as an oxidizing agent, not a reducing agent. This is because phosphorus in `PCl5` is in its maximum oxidation state of +5. Since it cannot increase its oxidation state further, it cannot be oxidized and therefore cannot act as a reducing agent. However, phosphorus can decrease its oxidation number from +5 to +3 or even lower values, which means `PCl5` can accept electrons and thus act as an oxidizing agent. For example, it oxidizes silver (`Ag`) to silver chloride (`AgCl`) and tin (`Sn`) to tin tetrachloride (`SnCl4`).
\( 2\text{Ag}^0 + \text{PCl}_5^{+5} \rightarrow 2\text{AgCl}^{+1} + \text{PCl}_3^{+3} \)
\( \text{Sn}^0 + 2\text{PCl}_5^{+5} \rightarrow \text{SnCl}_4^{+4} + 2\text{PCl}_3^{+3} \)
In simple words: `PCl5` can only make other things lose electrons (oxidize them) because its phosphorus atom is already at its highest possible electron-loss state. It cannot reduce other things because it cannot lose any more electrons itself.

🎯 Exam Tip: To determine if a compound can act as an oxidizing or reducing agent, check the oxidation state of its key element. If it's at its maximum, it can only oxidize; if at its minimum, it can only reduce; if intermediate, it can do both.

 

Question 24. Discribe the manufacture of H2SO4 by contact process.
Answer: Sulfuric acid (`H2SO4`) is manufactured by the contact process, which involves three main steps:
1. **Burning of sulfur:** Sulfur is burned in the air to produce sulfur dioxide (`SO2`).
\( \text{S(s)} + \text{O}_2\text{(g)} \rightarrow \text{SO}_2\text{(g)} \)
2. **Conversion of `SO2` to `SO3`:** Sulfur dioxide is then oxidized by air (or oxygen) to sulfur trioxide (`SO3`) in the presence of vanadium pentoxide (`V2O5`) as a catalyst. This is the crucial catalytic step, requiring specific temperature and pressure conditions for optimal yield.
\( 2\text{SO}_2\text{(g)} + \text{O}_2\text{(g)} \xrightarrow{\text{V}_2\text{O}_5} 2\text{SO}_3\text{(g)} \)
3. **Absorption of `SO3`:** Sulfur trioxide (`SO3`) is absorbed in concentrated `H2SO4` (not directly in water, to avoid forming a mist) to obtain oleum (`H2S2O7`), also known as fuming sulfuric acid. The oleum is then diluted with water to produce sulfuric acid of desired concentration.
\( \text{SO}_3\text{(g)} + \text{H}_2\text{SO}_4(\text{conc}) \rightarrow \text{H}_2\text{S}_2\text{O}_7(\text{oleum}) \)
\( \text{H}_2\text{S}_2\text{O}_7(\text{l}) + \text{H}_2\text{O}(\text{l}) \rightarrow 2\text{H}_2\text{SO}_4(\text{aq}) \)
A flow diagram for the manufacture of `H2SO4` is shown below:
In simple words: The contact process makes `H2SO4` in three steps: first, burning sulfur to get `SO2`; second, changing `SO2` to `SO3` using a catalyst; and third, dissolving `SO3` in `H2SO4` to make oleum, which is then mixed with water to get `H2SO4`.

🎯 Exam Tip: Remember the catalyst (`V2O5`), the reaction conditions (temperature, pressure), and why `SO3` is absorbed in `H2SO4` instead of water to avoid mist formation.

 

Question 25. H tent?
Answer: This question seems incomplete or has OCR errors, but it is likely asking about the allotropes of a halogen, or an element that forms different forms. Without a clear element name or question, it's hard to be specific. Assuming it refers to hydrogen, it primarily exists as diatomic `H2` gas under normal conditions. In other contexts, it could refer to different isotopes of hydrogen.
In simple words: The question is unclear, but it might be asking about the different forms in which a chemical element can exist, like hydrogen usually existing as a gas made of two atoms.

🎯 Exam Tip: Always clarify incomplete questions with your instructor if possible. If interpreting, consider the most common properties or forms of the elements that might be implied.

 

Question 26. Why are halogens strong oxidising agents ?
Answer: Halogens are strong oxidizing agents because they have a strong tendency to accept electrons and get reduced themselves. This tendency is due to several factors:
* **Low bond dissociation enthalpy:** Halogen-halogen bonds (`X-X`) are relatively easy to break, allowing the atoms to become available for reaction.
* **High electronegativity:** Halogens are highly electronegative, meaning they strongly attract electrons.
* **Large electron gain enthalpy:** They have a strong desire to gain an electron to achieve a stable noble gas configuration, releasing a significant amount of energy when they do so.
The general reaction is: \( \text{X} + \text{e}^{-} \rightarrow \text{X}^{-} \)
The oxidizing power of halogens decreases down the group from fluorine (`F2`) to iodine (`I2`), as shown by their standard electrode potentials:
\( \text{E}^\circ_{\text{F}_2/\text{F}^{-}} = + 2.87 \text{V} \)
\( \text{E}^\circ_{\text{Cl}_2/\text{Cl}^{-}} = + 1.36 \text{V} \)
\( \text{E}^\circ_{\text{Br}_2/\text{Br}^{-}} = + 1.09 \text{V} \)
\( \text{E}^\circ_{\text{I}_2/\text{I}^{-}} = + 0.54 \text{V} \)
Oxidizing agent order of halogens: \( \text{F}_2 > \text{Cl}_2 > \text{Br}_2 > \text{I}_2 \)
In simple words: Halogens are strong oxidizers because they really want to grab an electron to become stable. They do this because their bonds break easily, they pull electrons strongly, and they release a lot of energy when they gain an electron.

🎯 Exam Tip: Focus on electron gain enthalpy and electronegativity as primary reasons for the strong oxidizing power of halogens. Remember the decreasing trend down the group.

 

Question 27. Write two uses of ClO2.
Answer: Chlorine dioxide (`ClO2`) is a versatile chemical with several important uses:
1. It is a powerful **bleaching agent**. Its bleaching power is much stronger than that of chlorine (`Cl2`), making it effective in industries. It's about 30 times more potent than `Cl2`.
2. It is used for **bleaching flour** to make white bread. This improves the appearance of baked goods.
3. `ClO2` is a strong **oxidizing agent and chlorinating agent**. Large quantities are used for bleaching wood pulp and cellulose in paper production, and for purifying drinking water by killing microbes.
In simple words: Chlorine dioxide is used to bleach materials like wood pulp and flour, making them white. It also helps clean drinking water by killing germs.

🎯 Exam Tip: Remember `ClO2`'s dual roles: powerful bleach and effective disinfectant/oxidizing agent, contrasting it with `Cl2` in terms of strength and applications.

 

Question 28. Why are halogens coloured ?
Answer: Halogens exhibit distinct colors because their molecules absorb specific wavelengths of light from the visible region of the electromagnetic spectrum. When a halogen molecule absorbs light, its electrons get excited and jump to higher energy levels. The color we see is the complementary color of the light that was absorbed. For example, fluorine (`F2`) absorbs violet light and thus appears pale yellow. Iodine (`I2`) absorbs yellow and green light, so it transmits deep violet. Similarly, chlorine (`Cl2`) appears greenish-yellow, and bromine (`Br2`) appears orange-red. The specific energy required for excitation varies across the halogens, leading to different absorbed wavelengths and thus different observed colors.
In simple words: Halogens have colors because their electrons absorb certain colors of light. The color we see is the color that is left over after some light has been absorbed.

🎯 Exam Tip: The color of halogens is a result of their electronic transitions absorbing specific wavelengths in the visible spectrum. Remember that the observed color is complementary to the absorbed color.

 

Question 29. Write the reactions of F2 and Cl2 with water.
Answer: Both fluorine (`F2`) and chlorine (`Cl2`) react with water, but their reactions are different:
**1. Reaction of Fluorine (`F2`) with Water:**
Fluorine reacts vigorously with water, leading to the formation of oxygen (`O2`) or ozone (`O3`) and hydrogen fluoride (`HF`). This is a redox reaction where fluorine oxidizes water. Due to the high oxidizing power of fluorine, it can oxidize water.
\( 2\text{F}_2(\text{g}) + 2\text{H}_2\text{O}(\text{l}) \rightarrow 4\text{HF}(\text{aq}) + \text{O}_2(\text{g}) \)
Or,
\( 3\text{F}_2(\text{g}) + 3\text{H}_2\text{O}(\text{l}) \rightarrow 6\text{HF}(\text{aq}) + \text{O}_3(\text{g}) \)
**2. Reaction of Chlorine (`Cl2`) with Water:**
Chlorine reacts with water to form hydrochloric acid (`HCl`) and hypochlorous acid (`HOCl`). This reaction is reversible and is often called the hydrolysis of chlorine. Hypochlorous acid is a strong oxidizing agent, which makes chlorine water an effective bleaching agent and disinfectant.
\( \text{Cl}_2(\text{g}) + \text{H}_2\text{O}(\text{l}) \rightleftharpoons \text{HCl}(\text{aq}) + \text{HOCl}(\text{aq}) \)
(Hydrochloric acid) (Hypochlorous acid)
In simple words: Fluorine reacts strongly with water to make hydrogen fluoride and oxygen gas. Chlorine reacts with water to make hydrochloric acid and hypochlorous acid, which is a weaker reaction.

🎯 Exam Tip: Note the difference in reactivity: fluorine oxidizes water (producing oxygen), while chlorine undergoes hydrolysis (producing two acids). This reflects fluorine's extremely high electronegativity.

 

Question 30. Why do noble gases have comparatively large atomic size?
Answer: Noble gases have comparatively large atomic sizes when compared to other elements in the same period. This is because the atomic radius of noble gases is measured as the Van der Waals radius. Van der Waals radii are determined from the distances between non-bonded atoms in different molecules. In contrast, the atomic radii of other elements are typically measured using covalent radii (half the distance between two bonded atoms) or metallic radii. Van der Waals radii are generally larger than covalent or metallic radii for the same element because they represent the non-bonded atomic separation. As we move across a period, the nuclear charge increases, causing a general decrease in atomic size. However, the noble gas at the end of the period, due to its inert nature, doesn't form molecules easily, so its size is measured differently, making it appear larger.
In simple words: Noble gases seem bigger because their size is measured by how close non-bonded atoms get (Van der Waals radius), which is usually larger than the size measured when atoms are actually bonded together.

🎯 Exam Tip: Remember the distinction between different types of atomic radii (covalent, metallic, Van der Waals) and how they apply to different types of elements, especially noble gases.

 

RBSE Class 12 Chemistry Chapter 7 Short Answer Type Question

 

Question 1. Mention the condition required to maximise the yield of ammonia.
Answer: Ammonia is manufactured industrially by the Haber's process. To maximize the yield of ammonia (`NH3`), specific optimum conditions are required, based on Le Chatelier's principle:
The reaction is: \( \text{N}_2(\text{g}) + 3\text{H}_2(\text{g}) \rightleftharpoons 2\text{NH}_3(\text{g}); \Delta\text{H}^\circ = -46.1 \text{ kJ mol}^{-1} \)
1. **High Pressure:** There are four molecules on the reactant side (`N2` + `3H2`) and two molecules on the product side (`2NH3`). Since the reaction proceeds with a decrease in the number of moles (and thus volume), a high pressure favors the forward reaction, increasing ammonia yield. Typically, a pressure of around 200 atm (`200 \times 10^5 \text{ Pa}`) is used.
2. **Low Temperature:** The reaction is exothermic (`\Delta\text{H}^\circ` is negative), meaning it releases heat. According to Le Chatelier's principle, lower temperatures favor exothermic reactions, shifting the equilibrium towards product formation. However, too low a temperature makes the reaction very slow. So, an optimum temperature of around 700 K is maintained.
3. **Catalyst:** An iron oxide catalyst, sometimes mixed with small amounts of `K2O` and `Al2O3`, is used to increase the rate of reaction. Molybdenum (`Mo`) can also be used as a promoter to enhance the efficiency of the iron catalyst. Catalysts speed up both forward and reverse reactions equally, helping the system reach equilibrium faster without affecting the final yield.
\( \text{N}_2(\text{g}) + 3\text{H}_2(\text{g}) \xrightarrow{\text{Fe catalyst (iron oxide + K}_2\text{O + Al}_2\text{O}_3)} 2\text{NH}_3(\text{g}) \)
(Promoter: Mo, Temperature: 700K, Pressure: \(200 \times 10^5 \text{ Pa}\))
In simple words: To get the most ammonia, you need high pressure, a moderate temperature (around 700 K), and a catalyst like iron. High pressure pushes the reaction to make more product, and the catalyst speeds it up.

🎯 Exam Tip: For industrial processes like the Haber-Bosch process, remember the specific conditions (temperature, pressure, catalyst) and how Le Chatelier's principle explains their optimization for maximum yield.

 

Question 2. Bond angle in PH+4 is higher than that in PH3Why?
Answer: In phosphine (`PH3`), the central phosphorus atom is `sp³`-hybridized. It has three bond pairs (with hydrogen atoms) and one lone pair of electrons. Due to the stronger repulsion between the lone pair and bond pairs compared to bond pair-bond pair repulsions, the tetrahedral angle (ideally 109°28') is distorted and reduced to about 93.6°, giving `PH3` a pyramidal shape. However, when `PH3` reacts with a proton (`H⁺`), it forms the phosphonium ion (`PH4⁺`). In `PH4⁺`, the central phosphorus atom is also `sp³`-hybridized, but it now has *four* bond pairs and *no* lone pairs. With no lone pair-bond pair repulsion, the repulsions between the four identical bond pairs are equal, resulting in a perfectly tetrahedral geometry with a bond angle of 109°28'. Therefore, the bond angle in `PH4⁺` is higher than that in `PH3`.
In simple words: The bond angle in `PH4⁺` is bigger than in `PH3` because `PH3` has a lone pair of electrons that pushes the bonds closer together. When `PH3` picks up an `H⁺` to become `PH4⁺`, it loses that lone pair, allowing the bonds to spread out evenly.

🎯 Exam Tip: Remember that lone pairs cause more repulsion than bond pairs, leading to a reduction in bond angles from ideal geometries. The absence of lone pairs results in symmetrical shapes.

 

Question 3. What happens when white phosphorus is heated with concentrated NaOH solution in an inert atmosphere of CO2?
Answer: When white phosphorus (`P4`) is heated with a concentrated sodium hydroxide (`NaOH`) solution in an inert atmosphere of carbon dioxide (`CO2`), it undergoes a disproportionation reaction. In this reaction, white phosphorus is simultaneously oxidized and reduced. It forms phosphine (`PH3`) gas, a highly poisonous gas, and sodium hypophosphite (`NaH2PO2`). The inert `CO2` atmosphere prevents the oxidation of white phosphorus by air.
\( \text{P}_4 + 3\text{NaOH} + 3\text{H}_2\text{O} \xrightarrow{\text{Heat, CO}_2 \text{ atmosphere}} \text{PH}_3 \uparrow + 3\text{NaH}_2\text{PO}_2 \)
(Phosphorus) \( \implies \) (Phosphine) + (Sodium hypophosphite)
In simple words: When white phosphorus is heated with strong `NaOH` solution under `CO2` gas, it changes into toxic phosphine gas and a salt called sodium hypophosphite.

🎯 Exam Tip: Recall that white phosphorus is highly reactive and undergoes disproportionation in alkaline solutions. The inert atmosphere is crucial to prevent unwanted oxidation by oxygen.

 

Question 4. List the important sources of sulphur.
Answer: Sulfur occurs in the Earth's crust mainly in combined forms, primarily as sulfates and sulfides. It is a vital element with diverse natural occurrences.
**Important sources include:**
* **As sulfates:**
* Gypsum (`CaSO4 \cdot 2H2O`)
* Epsom salt (`MgSO4 \cdot 7H2O`)
* Baryte (`BaSO4`)
* **As sulfides:**
* Galena (`PbS`)
* Zinc blende (`ZnS`)
* Copper pyrites (`CuFeS2`)
* Iron pyrites (`FeS2`)
Additionally, traces of sulfur are found in various organic matters such as hydrogen sulfide (`H2S`), proteins, eggs, onions, mustard, hair, and wool. These biological sources highlight its role in living organisms.
In simple words: Sulfur is found in the Earth's crust in many forms, like minerals called sulfates (e.g., gypsum) and sulfides (e.g., galena). It is also found in small amounts in things like eggs and hair.

🎯 Exam Tip: Remember the most common mineral forms (sulfates and sulfides) and a few examples of each. Also, briefly mention its presence in organic materials.

 

Question 5. Write the order of thermal stability of the hydrides of group-16 elements.
Answer: The thermal stability of the hydrides of Group 16 elements decreases as you move down the group. This trend occurs because, as the size of the central element (E) increases down the group, the E-H bond length increases, and its bond dissociation energy decreases. A weaker bond means it requires less energy to break, making the hydride less stable to heat. Therefore, these hydrides decompose more easily upon heating.
The order of thermal stability is:
\( \text{H}_2\text{O} > \text{H}_2\text{S} > \text{H}_2\text{Se} > \text{H}_2\text{Te} > \text{H}_2\text{Po} \)
Water (`H2O`) is the most thermally stable, and polonium hydride (`H2Po`) is the least stable.
In simple words: The hydrides of Group 16 become less stable when heated as you go down the group. This is because the bond between hydrogen and the central atom gets weaker with bigger atoms.

🎯 Exam Tip: For hydrides, thermal stability generally decreases down a group due to increasing atomic size and decreasing bond dissociation energy of the E-H bond.

 

Question 7. Complete the following reactions :
(i) C2H4 + O2 →
(ii) 4Al + 3O2 →
Answer: Here are the completed and balanced reactions:
(i) Ethene (`C2H4`) undergoes complete combustion in the presence of oxygen (`O2`) to form carbon dioxide (`CO2`) and water (`H2O`). This is a common reaction for hydrocarbons.
\( \text{C}_2\text{H}_4 + 3\text{O}_2 \rightarrow 2\text{CO}_2 + 2\text{H}_2\text{O} \)
(ii) Aluminum (`Al`) combines directly with oxygen (`O2`) to form aluminum oxide (`Al2O3`), also known as alumina. This reaction shows the strong affinity of aluminum for oxygen.
\( 4\text{Al} + 3\text{O}_2 \rightarrow 2\text{Al}_2\text{O}_3 \)
In simple words: When ethene burns in oxygen, it makes `CO2` and water. When aluminum reacts with oxygen, it forms aluminum oxide.

🎯 Exam Tip: For combustion reactions of hydrocarbons, assume complete combustion unless stated otherwise, leading to `CO2` and `H2O`. For metal reactions with oxygen, form the most common stable oxide.

 

Question 8. How is O3 estimated quantitatively?
Answer: Ozone (`O3`) can be quantitatively estimated using an iodometric titration method. The process involves treating ozone with an excess of potassium iodide (`KI`) solution, buffered at a pH of 9.2 using a borate buffer. In this reaction, ozone oxidizes iodide ions (`I⁻`) to liberate elemental iodine (`I2`).
\( 2\text{I}^{-}(\text{aq}) + \text{H}_2\text{O}(\text{l}) + \text{O}_3(\text{g}) \rightarrow 2\text{OH}^{-}(\text{aq}) + \text{I}_2(\text{s}) + \text{O}_2(\text{g}) \)
The amount of `I2` liberated is directly proportional to the amount of `O3` that reacted. This liberated `I2` is then titrated against a standard solution of sodium thiosulfate (`Na2S2O3`) using starch as an indicator. Starch forms a blue-black complex with iodine, which disappears at the endpoint.
\( 2\text{Na}_2\text{S}_2\text{O}_3 + \text{I}_2 \rightarrow \text{Na}_2\text{S}_4\text{O}_6 + 2\text{NaI} \)
(Sodium thiosulphate) \( \implies \) (Sodium Tetrathionate)
By knowing the volume and concentration of sodium thiosulfate used, the amount of `I2` can be calculated, which then allows for the quantitative estimation of `O3`.
In simple words: To measure ozone, you make it react with a special solution to create iodine. Then, you measure how much iodine is made using another solution. This helps you find out how much ozone was there.

🎯 Exam Tip: Remember that iodometric titration involves the liberation of iodine, which is then titrated with thiosulfate. The stoichiometry of the reaction between `O3` and `I⁻` is key.

 

Question 9. What happens when sulphur dioxide is passed through an aqueous solution of Fe (III) salt?
Answer: When sulfur dioxide (`SO2`) gas is passed through an aqueous solution of an iron (III) salt (e.g., `FeCl3`), `SO2` acts as a reducing agent. It reduces the iron (III) ions (`Fe³⁺`), which have a +3 oxidation state, to iron (II) ions (`Fe²⁺`), which have a +2 oxidation state. In this process, `SO2` itself gets oxidized to sulfate ions (`SO4²⁻`). This reaction is commonly observed as a color change, for example, from the yellowish-brown of `Fe³⁺` to the pale green of `Fe²⁺`.
The general equation is:
\( 2\text{Fe}^{3+}(\text{aq}) + \text{SO}_2(\text{g}) + 2\text{H}_2\text{O}(\text{l}) \rightarrow 2\text{Fe}^{2+}(\text{aq}) + \text{SO}_4^{2-}(\text{aq}) + 4\text{H}^{+}(\text{aq}) \)
This demonstrates `SO2`'s strong reducing property in an aqueous medium.
In simple words: When `SO2` gas goes through an iron (III) salt solution, it turns the iron from a +3 state to a +2 state, acting as a reducer.

🎯 Exam Tip: Recognize `SO2` as a versatile molecule that can act as both an oxidizing and a reducing agent. In the presence of a strong oxidizing agent (like `Fe³⁺`), `SO2` will act as a reducing agent.

 

Question 10. Comment on the nature of two S – O bonds formed in SO2 molecules. Are the two S -0 bonds in the molecule equal?
Answer: In the sulfur dioxide (`SO2`) molecule, the central sulfur atom is `sp²`-hybridized. Two of the three `sp²` orbitals form two sigma (`\(\sigma\)`)-bonds with the oxygen atoms, while the third `sp²` orbital contains a lone pair of electrons. The sulfur atom also has a half-filled p-orbital and a half-filled d-orbital. These can form one `p\(\pi\)`-`p\(\pi\)` double bond with one oxygen atom and one `p\(\pi\)`-`d\(\pi\)` double bond with the other oxygen atom. This bonding arrangement gives `SO2` a bent structure with an O-S-O bond angle of approximately 119.50°.
Despite having a formal single and double bond in a single Lewis structure, the two S-O bonds in the `SO2` molecule are equal in length (approximately 143 pm) and strength. This equality is explained by the concept of **resonance**. The actual structure of `SO2` is a resonance hybrid of two contributing structures, where the double bond is delocalized over both S-O bonds. This delocalization makes the two S-O bonds equivalent, each having partial double bond character.
In simple words: In `SO2`, the two S-O bonds are exactly the same length and strength. This happens because the electrons are spread out evenly between both bonds, making them act like a mix of a single and double bond at the same time.

🎯 Exam Tip: Resonance is key to explaining bond equalization in molecules like `SO2`, `CO3²⁻`, and `NO3⁻`. Always draw resonance structures to show electron delocalization.

 

Question 11. Mention three areas in which H2SO4 plays an important role.
Answer: Sulfuric acid (`H2SO4`) is one of the most important industrial chemicals, often called the "King of Chemicals," and plays an important role in many areas:
1. **Electrolyte in Storage Batteries:** It is used as an electrolyte in lead-acid storage batteries, commonly found in vehicles. The chemical reactions within these batteries rely on the conduction provided by `H2SO4`.
2. **Petroleum Refining and Detergent Industry:** `H2SO4` is extensively used in the petroleum refining industry to remove impurities from gasoline and other products. It's also a crucial raw material in the manufacture of detergents.
3. **Manufacture of Fertilizers:** A significant amount of `H2SO4` is consumed in the production of fertilizers such as ammonium sulfate (`(NH4)2SO4`) and calcium superphosphate, which are vital for agriculture to enhance crop yield.
In simple words: Sulfuric acid is very important in three main areas: in car batteries, in cleaning crude oil and making soaps, and in making fertilizers for farming.

🎯 Exam Tip: For `H2SO4`, remember its widespread industrial importance. Key uses include fertilizers, petroleum refining, and as an electrolyte in batteries.

 

Question 12. Write the conditions to maximise the yeild of H2SO4 by contact process. Answer:
Answer: The main step in the preparation of sulfuric acid (`H2SO4`) by the contact process is the catalytic oxidation of sulfur dioxide (`SO2`) to sulfur trioxide (`SO3`). The reaction is:
\( 2\text{SO}_2(\text{g}) + \text{O}_2(\text{g}) \rightleftharpoons 2\text{SO}_3(\text{g}); \Delta\text{H}^\circ = -196.6 \text{ kJ mol}^{-1} \)
This reaction is exothermic (releases heat) and reversible, and it proceeds with a decrease in the number of moles (volume). According to Le Chatelier's principle, to maximize the yield of `SO3` (and thus `H2SO4`), the following conditions are required:
1. **Low Temperature:** Since the reaction is exothermic, a low temperature favors the forward reaction, shifting the equilibrium towards `SO3` formation. An optimum temperature of around 720 K is used to ensure a good reaction rate without significantly reducing yield.
2. **High Pressure:** The reaction involves a decrease in the number of gaseous moles (3 moles of reactants yield 2 moles of product). Therefore, a high pressure (typically around 2 bar) favors the forward reaction, increasing the `SO3` yield.
3. **Catalyst:** Vanadium pentoxide (`V2O5`) is used as a catalyst. The catalyst speeds up the reaction rate, allowing equilibrium to be reached quickly without affecting the equilibrium position or overall yield.
In simple words: To get the most `H2SO4` from the contact process, you need a somewhat low temperature, high pressure, and a `V2O5` catalyst. These conditions help the reaction produce more `SO3`, which then makes the acid.

🎯 Exam Tip: Always apply Le Chatelier's principle to explain the optimum conditions (temperature, pressure) for reversible industrial processes, and remember the role of the catalyst.

 

Question 13. Considering the parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy compare the oxidising power of F2 and Cl2.
Answer: To compare the oxidizing power of fluorine (`F2`) and chlorine (`Cl2`), we need to consider three key parameters:
1. **Bond Dissociation Enthalpy:** `F2` has a lower bond dissociation enthalpy (158.8 kJ mol⁻¹) than `Cl2` (246.6 kJ mol⁻¹). This means `F2` is easier to break into individual atoms.
2. **Electron Gain Enthalpy:** The electron gain enthalpy of fluorine (-333 kJ mol⁻¹) is less negative than that of chlorine (-349 kJ mol⁻¹). This suggests chlorine has a slightly greater affinity for electrons in the gaseous state.
3. **Hydration Enthalpy:** The hydration enthalpy of `F⁻` ion (515 kJ mol⁻¹) is much higher than that of `Cl⁻` ion (381 kJ mol⁻¹). This means `F⁻` ions are much more stabilized by water molecules.
Even though fluorine's electron gain enthalpy is less negative, the lower bond dissociation enthalpy of `F2` and the much higher hydration enthalpy of the `F⁻` ion more than compensate for this difference. The greater energy released during the hydration of the small `F⁻` ion drives the overall process. As a result, the standard electrode potential of `F2` (+2.87 V) is significantly higher than that of `Cl2` (+1.36 V). Therefore, `F2` is a much stronger oxidizing agent than `Cl2`.
In simple words: `F2` is a much stronger oxidizer than `Cl2` because, even though `Cl2` takes electrons slightly better, `F2`'s bonds break more easily, and its resulting ion (`F⁻`) gets much more stable in water, outweighing other factors.

🎯 Exam Tip: Remember that oxidizing power in solution depends not just on electron gain enthalpy but also on bond dissociation energy and the hydration energy of the resulting ion. Fluorine's high hydration energy is a critical factor.

 

Question 14. Give two examples to show the anamolus behaviour of fluorine.
Answer: Fluorine shows unusual behaviour compared to other halogens due to its unique properties. It is the smallest and most electronegative element in its group.
1. Fluorine only shows a \( -1 \) oxidation state. Other halogens can have positive oxidation states like \( +1, +3, +5 \), and \( +7 \). The melting and boiling points of hydrogen halides follow this order: \( HF > HCl > HBr > HI \).
2. The unusual behaviour of fluorine is because of its:
(i) Small size
(ii) Highest electronegativity
(iii) Low fluorine-fluorine bond dissociation enthalpy
(iv) Lack of d-orbitals in its valence shell.
This means fluorine cannot expand its octet like other halogens.
In simple words: Fluorine acts differently from other halogens mainly because it is very small and pulls electrons strongly. It also cannot form many types of bonds like the others do.

🎯 Exam Tip: Remember that fluorine's small size and high electronegativity are key reasons for its distinct chemical properties, especially its unique oxidation state and strong hydrogen bonding.

 

Question 15. Se Typesetting math: 33% st source of some halogens. Comment.

 

Question 16. Write the balanced chemical equation for the reaction of Cl2 with hot and concentrated NaOH. Is this reaction is disproportionation reaction ?
Answer: When chlorine gas reacts with hot and concentrated sodium hydroxide, it forms sodium chloride and sodium chlorate.
\( 6NaOH (hot, conc.) + 3Cl_{2} \rightarrow 5NaCl + NaClO_{3} + 3H_{2}O \)
In this reaction, the oxidation state of chlorine changes from 0 in \( Cl_{2} \) to \( -1 \) in \( NaCl \) and \( +5 \) in \( NaClO_{3} \). Since chlorine is both oxidized and reduced, this is a disproportionation reaction. Disproportionation reactions are common when an element can exist in multiple stable oxidation states.
In simple words: When chlorine reacts with hot, strong NaOH, it forms two new products: one where chlorine gains electrons (NaCl) and one where it loses electrons (NaClO3). Because chlorine does both in the same reaction, it's called a disproportionation reaction.

🎯 Exam Tip: Disproportionation reactions are easy to spot when a single element in the reactant has different oxidation states in the products. Always check the oxidation states to confirm.

 

Question 17. Why does the reactivity of nitrogen differ from phosphorus?
Answer: Nitrogen exists as a diatomic molecule, \( N_{2} \), with a strong triple bond between the two nitrogen atoms. This bond is very strong, having a high bond dissociation energy of 941.1 kJ mol\(^{-1}\). Because of this strong bond, nitrogen is very stable and not very reactive at room temperature.
In contrast, phosphorus exists as a tetraatomic molecule, \( P_{4} \). The phosphorus-phosphorus single bond is much weaker (213 kJ mol\(^{-1}\)) than the nitrogen-nitrogen triple bond. This weaker bond makes phosphorus much more reactive than nitrogen. Its ability to form longer single bonds, rather than triple bonds, also contributes to its different structural forms.
In simple words: Nitrogen is less reactive because its atoms are held together by a very strong triple bond. Phosphorus is more reactive because its atoms are held by weaker single bonds.

🎯 Exam Tip: The strength of chemical bonds, especially the presence of multiple bonds, is a major factor determining an element's reactivity. Always consider bond energy when comparing reactivity.

 

Question 18. Discuss the trends in chemical reactivity of group-15 elements.
Answer: The trends in chemical reactivity of group-15 elements, which include nitrogen, phosphorus, arsenic, antimony, and bismuth, can be discussed based on their tendency to form hydrides.
**Formation of Hydrides:**
Group-15 elements form volatile hydrides with the general formula \( MH_{3} \). Examples include \( NH_{3} \) (ammonia), \( PH_{3} \) (phosphine), \( AsH_{3} \) (arsine), \( SbH_{3} \) (stibine), and \( BiH_{3} \) (bismuthine).
**(A) Thermal stability:**
The thermal stability of these hydrides decreases as you move down the group. This is because the M-H bond length increases, and the bond dissociation energy decreases. So, the order of thermal stability is:

**(B) Bond angle:**
The bond angles in these hydrides gradually decrease as you move down the group. This is due to a decrease in the bond pair-bond pair repulsion. As the central atom's size increases, the electron density around it decreases, leading to less repulsion between the bond pairs. This change in bond angle affects the overall geometry and stability of the molecule.
**(C) Reducing Character:**
As thermal stability decreases down the group, the reducing character of these hydrides gradually increases. This means they are more easily oxidized themselves, thereby reducing other substances. The order of increasing reducing character is:
\( NH_{3} < PH_{3} < AsH_{3} < SbH_{3} < BiH_{3} \)
In simple words: For group 15 elements, the hydrides (compounds with hydrogen) become less stable as you go down the group, meaning they break apart more easily when heated. Their bond angles also get smaller. This makes the heavier hydrides better at giving away electrons (more reducing).

🎯 Exam Tip: When discussing trends in group 15 hydrides, always link thermal stability, bond strength, and reducing character, remembering they are inversely related as you go down the group.

 

Question 19. Why does NHZ form hydrogen bond but PH3 does not?
Answer: Nitrogen (N) has an electronegativity of 3.0, which is much higher than that of hydrogen (H) at 2.1. This large difference in electronegativity makes the N-H bond highly polar. Because nitrogen also has a lone pair of electrons and is a small atom, it can form strong intermolecular hydrogen bonds with other \( NH_{3} \) molecules. This is shown as a dashed line interaction between molecules.
\( \qquad \qquad \text{H} \)
\( \qquad \qquad \vert \)
\( \qquad \qquad \delta+ \text{H} - \text{N} \cdot \cdot \cdot \delta- \text{H} - \text{N} \cdot \cdot \cdot \)
\( \qquad \qquad \vert \qquad \quad \vert \)
\( \qquad \qquad \delta- \text{H} \qquad \delta+ \text{H} \)
\( \qquad \qquad \vert \)
\( \qquad \qquad \text{H} \)
In contrast, phosphorus (P) has an electronegativity of 2.1, which is very similar to that of hydrogen (2.1). This means the P-H bond is much less polar, so \( PH_{3} \) does not form strong hydrogen bonds. Hydrogen bonds require a large electronegativity difference between hydrogen and the atom it's bonded to (like O, N, or F).
In simple words: Ammonia (\( NH_{3} \)) forms hydrogen bonds because nitrogen pulls electrons strongly, making the N-H bond uneven. Phosphine (\( PH_{3} \)) does not, because phosphorus pulls electrons almost as much as hydrogen, making the P-H bond more even.

🎯 Exam Tip: Hydrogen bonding happens when hydrogen is directly bonded to a very electronegative atom like N, O, or F, which creates a strong positive charge on the hydrogen atom, allowing it to attract lone pairs from other electronegative atoms.

 

Question 20. H Typesetting math: 33% ed in the laboratory? Write the chemical equations of the reactions involved.

 

Question 21. How is ammonia manufactured industrially?
Answer: Ammonia is produced on a large scale in industries using the Haber's process. This process combines nitrogen gas (\( N_{2} \)) and hydrogen gas (\( H_{2} \)) to form ammonia (\( NH_{3} \)).
\( N_{2}(g) + 3H_{2}(g) \rightleftharpoons 2NH_{3}(g) \); \( \Delta H^\circ = -46.1 \text{ kJ mol}^{-1} \)
In this reaction, four molecules on the reactant side (one \( N_{2} \) and three \( H_{2} \)) produce two molecules of \( NH_{3} \). This means the pressure decreases as products form. Also, the reaction releases heat, making it an exothermic process.
To get the most ammonia, based on Le Chatelier's principle, specific conditions are needed:
1. **High Pressure:** About 200 atmospheres (200 x 10\(^5\) Pa) helps shift the reaction towards the product side (ammonia), where there are fewer gas molecules.
2. **Low Temperature:** Around 700 K is used, even though it's an exothermic reaction, to ensure a good reaction rate.
3. **Catalyst:** Iron oxide, mixed with small amounts of potassium oxide (\( K_{2}O \)) and aluminium oxide (\( Al_{2}O_{3} \)), acts as a catalyst. Sometimes, molybdenum (Mo) is also used as a promoter to make the iron catalyst work even better. Catalysts speed up the reaction without being used up themselves.
In simple words: Ammonia is made in factories using the Haber's process. Nitrogen and hydrogen gases are combined under high pressure and medium temperature, with a special iron catalyst, to make ammonia. This process makes the most ammonia by changing the conditions to favor product formation.

🎯 Exam Tip: Remember the three key conditions for the Haber's process: high pressure, optimal temperature, and an iron catalyst. These ensure a good yield of ammonia economically.

 

Question 22. Illustrate how copper metal can give different products on reaction with HNO3.
Answer: Copper metal reacts differently with nitric acid (\( HNO_{3} \)) depending on the concentration of the acid.
(i) **With dilute nitric acid:** When copper reacts with dilute nitric acid, nitric oxide (NO) gas is produced. Nitric oxide is a colorless gas.
\( 3Cu + 8HNO_{3} \text{ (dil)} \rightarrow 3Cu(NO_{3})_{2} + 4H_{2}O + 2NO \)
(ii) **With concentrated nitric acid:** When copper reacts with concentrated nitric acid, nitrogen dioxide (\( NO_{2} \)) gas is evolved. Nitrogen dioxide is a brown-colored gas.
\( Cu + 4HNO_{3} \text{ (conc)} \rightarrow Cu(NO_{3})_{2} + 2H_{2}O + 2NO_{2} \)
The varying products show the strong oxidizing nature of nitric acid, which can act differently based on its concentration.
In simple words: Copper reacts with nitric acid to make different gases. If the acid is weak (dilute), it makes nitric oxide (NO), which is a clear gas. If the acid is strong (concentrated), it makes nitrogen dioxide (\( NO_{2} \)), which is a brown gas.

🎯 Exam Tip: Always specify the concentration of nitric acid (dilute or concentrated) when writing its reactions with metals, as this determines the gaseous product formed.

 

Question 23. The HNH angle is higher than HPH, HASH and HSbH angles. Why?
Answer: In all the hydrides of Group 15 elements (like \( NH_{3}, PH_{3}, AsH_{3}, SbH_{3} \)), the central atom is \( sp^{3} \) -hybridised. Three of the four \( sp^{3} \) orbitals form three E-H (where E is the central element) sigma bonds, while the fourth contains a lone pair of electrons. For example, in ammonia (\( NH_{3} \)), the nitrogen atom has a lone pair.
The lone pair of electrons has a stronger repulsion than the bond pairs. In \( NH_{3} \), the strong lone pair-bond pair repulsion causes the ideal tetrahedral angle (109°28') to decrease to 107°. This makes the \( NH_{3} \) molecule pyramidal.
However, when ammonia reacts with a proton, it forms the ammonium ion (\( NH_{4}^{+} \)), which has no lone pair on the central nitrogen atom. It has four identical N-H bonds.

In \( PH_{3} \), \( AsH_{3} \), and \( SbH_{3} \), as we move down the group from N to P to As to Sb, the atomic size of the central atom increases. Because of this, the bond pairs of electrons are further away from the central atom. This reduces the repulsion between adjacent bond pairs. Consequently, the bond angles continue to decrease from \( NH_{3} \) (107°) to \( PH_{3} \) (93.5°), \( AsH_{3} \) (91.8°), and \( SbH_{3} \) (91.3°). The effect of lone pair repulsion becomes less dominant compared to the increasing atomic size, leading to smaller bond angles.
In simple words: The HNH angle in ammonia is bigger than in other hydrides like \( PH_{3} \) because nitrogen is smaller and its lone pair of electrons pushes the bonding electrons more strongly. As the central atom gets bigger (like P, As, Sb), this pushing effect becomes weaker, making the bond angles smaller.

🎯 Exam Tip: Remember VSEPR theory (Valence Shell Electron Pair Repulsion) is key here. Lone pair-bond pair repulsions are stronger than bond pair-bond pair repulsions, which distorts bond angles from ideal geometries. Also, consider the effect of increasing atomic size down a group on bond angles.

 

Question 24. Nitrogen exists as diatomic molecule and phosphorus as P4. Why?
Answer: Nitrogen is a small atom with high electronegativity. Because of this, nitrogen atoms can form strong pπ-pπ multiple bonds. This means two nitrogen atoms can share electrons to form a triple bond (\( N \equiv N \)), resulting in a diatomic molecule, \( N_{2} \). The triple bond makes \( N_{2} \) very stable.
Phosphorus, on the other hand, is a larger atom and has lower electronegativity compared to nitrogen. Due to its larger size, phosphorus atoms cannot form effective pπ-pπ multiple bonds with themselves. Instead, phosphorus prefers to form single P-P bonds. It forms tetrahedral \( P_{4} \) molecules, where each phosphorus atom is bonded to three other phosphorus atoms in a cyclic structure, resulting in discrete \( P_{4} \) units. This tendency to form single bonds rather than multiple bonds is common for larger elements in a group.
In simple words: Nitrogen is small and forms a strong triple bond with another nitrogen atom, making it \( N_{2} \). Phosphorus is bigger and forms single bonds with three other phosphorus atoms in a tetrahedral shape, making it \( P_{4} \).

🎯 Exam Tip: The ability to form pπ-pπ multiple bonds decreases significantly down a group due to increasing atomic size, leading to different stable molecular structures for elements like nitrogen and phosphorus.

 

Question 25. Write main difference between the properties of white phosphorus and red phosphorus.
Answer: White phosphorus and red phosphorus are two common allotropes of phosphorus that show significant differences in their properties, mainly due to their structural arrangements.
**Differences between properties of White and Red phosphorus:**

The difference in reactivity and physical state is a direct result of the differences in how the phosphorus atoms are bonded and arranged in space. White phosphorus with its discrete P4 units has high strain and is unstable.
In simple words: White phosphorus is made of separate \( P_{4} \) groups, which makes it very reactive and it burns easily. Red phosphorus is a long chain of these groups, making it much less reactive and safer.

🎯 Exam Tip: Allotropes have different physical properties and reactivities due to their distinct structural arrangements. For phosphorus, remember that white is discrete and highly reactive, while red is polymeric and stable.

 

Question 26. Why does nitrogen show catenation properties less than phophorus?
Answer: Catenation is the ability of an element's atoms to form long chains or rings by bonding with each other. This property depends directly on the strength of the element-element bond. A weaker bond means there is less possibility for catenation.
The nitrogen-nitrogen (N-N) single bond has a bond energy of 159 kJ mol\(^{-1}\). In contrast, the phosphorus-phosphorus (P-P) single bond has a bond strength of 213 kJ mol\(^{-1}\). Since the P-P bond is significantly stronger than the N-N bond, phosphorus has a greater tendency to form chains and rings, meaning it shows more catenation properties than nitrogen. Nitrogen's small size and the presence of lone pair repulsion also weaken its single bonds.
In simple words: Nitrogen is not very good at forming long chains with itself because its N-N bond is weaker. Phosphorus forms strong P-P bonds, so it can make longer chains easily.

🎯 Exam Tip: Bond energy is crucial for explaining catenation. A higher bond energy between identical atoms generally leads to stronger catenation. For p-block elements, remember that catenation usually decreases down a group but often peaks for elements in the middle of a period due to bond strength.

 

Question 27. 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 formation of hydride.
Answer: Elements Oxygen (O), Sulphur (S), Selenium (Se), Tellurium (Te), and Polonium (Po) are all placed in Group 16 of the periodic table, also known as the chalcogens. Their similar chemical behavior can be justified by common characteristics in their electronic configuration, oxidation states, and hydride formation.
**(I) Electronic Configuration:**
All these elements have six electrons in their outermost valence shell. Specifically, they have two electrons in the s-orbital and four electrons in the p-orbital. This gives them a general valence shell electronic configuration of \( ns^{2}np^{4} \). This common electron arrangement means they tend to gain two electrons to achieve a stable noble gas configuration, explaining their similar chemical reactions.

**(II) Oxidation States:**
These elements commonly exhibit \( +2, +4 \), and \( +6 \) oxidation states. However, oxygen is unique because it lacks d-orbitals in its valence shell, so it typically only shows a \( +2 \) oxidation state (as in \( OF_{2} \)), or a \( -2 \) state. Other elements in this group (S, Se, Te, Po) have vacant d-orbitals, allowing them to expand their octet and show \( +2, +4 \), and \( +6 \) oxidation states. As you move down the group, the stability of the \( +4 \) oxidation state increases due to the inert pair effect, while the stability of the \( +6 \) oxidation state decreases. For S, \( +6 \) is most stable; for Se, Te, and Po, \( +4 \) is more stable.
**(b) Negative oxidation state:**
Elements can also gain two electrons to achieve a stable noble gas configuration. Oxygen, being highly electronegative, frequently shows a \( -2 \) oxidation state in its compounds, especially in metal oxides. In peroxides (like \( H_{2}O_{2} \) or \( Na_{2}O_{2} \)), oxygen exhibits a \( -1 \) oxidation state. Other elements in the group have a lower tendency to show a \( -2 \) oxidation state due to lower electronegativity, and this tendency decreases down the group. Polonium does not show a \( -2 \) oxidation state at all.
**(c) Formation of Hydrides:**
These elements form stable, volatile, and bivalent hydrides with the general formula \( M_{2}H \), such as \( H_{2}O, H_{2}S, H_{2}Se, H_{2}Te \), and \( H_{2}Po \). Due to the presence of two lone pairs of electrons on the central atom, these hydrides have bent (V) shapes. The central atom in these hydrides is \( sp^{3} \) hybridized. Some characteristics of these hydrides are:
**I. Melting and boiling points:**
Order of M.P: \( H_{2}O > H_{2}Te > H_{2}Se > H_{2}S \)
Order of B.P: \( H_{2}O > H_{2}Te > H_{2}Se > H_{2}S \)
**II. Covalent Character:**
\( H_{2}O < H_{2}S < H_{2}Se < H_{2}Te \)
**III. Volatility and Thermal stability:**
Volatility generally increases from \( H_{2}O \) to \( H_{2}S \) and then decreases from \( H_{2}S \) to \( H_{2}Te \). This makes \( H_{2}O \) the least volatile and \( H_{2}S \) the most volatile among group-16 hydrides. The order of thermal stability is:
\( H_{2}O > H_{2}S > H_{2}Se > H_{2}Te \)
In simple words: Oxygen, sulfur, selenium, tellurium, and polonium are in the same group because they all have six outer electrons. This makes them act similarly, showing common oxidation states and forming hydrides that share similar shapes and properties, though some trends change as you go down the group due to size and d-orbitals.

🎯 Exam Tip: For group trends, always discuss electronic configuration (which determines group placement), common oxidation states (with exceptions for first elements), and specific compound types like hydrides, highlighting how properties change down the group.

 

Question 28. Why is dioxygen is gas but sulphur a solid?
Answer: Dioxygen (\( O_{2} \)) is a gas at room temperature, while sulfur (\( S_{8} \)) is a solid. This difference arises mainly from their bonding behavior and intermolecular forces.
Oxygen is a small atom with high electronegativity. This allows oxygen atoms to form strong pπ-pπ double bonds with each other, resulting in the diatomic \( O=O \) molecule. The intermolecular forces between these \( O_{2} \) molecules are weak van der Waal's forces. Since very little energy is needed to overcome these weak forces, \( O_{2} \) exists as a gas at room temperature.
Sulfur, being a larger atom, does not form stable pπ-pπ bonds with itself. Instead, sulfur prefers to form strong S-S single bonds. It exists as polyatomic molecules, typically in the form of an \( S_{8} \) puckered ring structure. These larger \( S_{8} \) molecules are held together by much stronger van der Waal's forces compared to \( O_{2} \). A greater amount of energy is required to break these stronger forces, which is why sulfur exists as a solid at room temperature.
In simple words: Oxygen is a gas because it forms small \( O_{2} \) molecules with weak forces between them. Sulfur is a solid because it forms larger \( S_{8} \) rings that are held together by stronger forces.

🎯 Exam Tip: Remember that the physical state of an element (gas, liquid, solid) at room temperature is strongly influenced by the type and strength of intermolecular forces, which in turn depends on the bonding and molecular structure.

 

Question 29. Knowing the electron gain enthalpy values for O \( \rightarrow \) O\(^{-} \) and O \( \rightarrow \) O\(^{2-} \) as -141 and 702 kJmol\(^{-1}\) respectively. How can you account for the formation of a large number of oxides having O\(^{2-} \) species and not O\(^{-} \)?
Answer: The first electron gain enthalpy for oxygen (O \( \rightarrow \) O\(^{-} \)) is negative (-141 kJ mol\(^{-1}\)), meaning energy is released when the first electron is added, making the formation of \( O^{-} \) favorable. However, the second electron gain enthalpy (O\(^{-} \) \( \rightarrow \) O\(^{2-} \)) is positive (+702 kJ mol\(^{-1}\)), which means energy must be supplied to add the second electron, making \( O^{2-} \) formation unfavorable on its own.
When oxygen reacts with a metal to form oxides, the formation of \( O^{2-} \) species is largely favored by the high lattice energy released when the metal cation combines with the oxide anion. The overall process can be described in these steps:
\( M(g) \xrightarrow{\Delta_i H_1} M^{+}(g) \xrightarrow{\Delta_i H_2} M^{2+}(g) \)
\( O(g) \xrightarrow{\Delta_{eg} H_1} O^{-}(g) \xrightarrow{\Delta_{eg} H_2} O^{2-}(g) \)
Then the ions combine to form the ionic oxide:
For \( M_{2}O \): \( 2M^{+}(g) + O^{2-}(g) \xrightarrow{\text{Lattice energy}} M_{2}O(s) \)
For \( MO \): \( M^{2+}(g) + O^{2-}(g) \xrightarrow{\text{Lattice energy}} MO(s) \)
For \( MO_{2} \): \( M^{2+}(g) + 2O^{2-}(g) \xrightarrow{\text{Lattice energy}} MO_{2}(g) \)
While the second electron gain enthalpy for oxygen is highly endothermic, the large amount of lattice energy released during the formation of the ionic crystal (especially with \( M^{2+} \) and \( O^{2-} \) ions due to higher charges) more than compensates for the energy required to form \( O^{2-} \). The lattice energy for \( MO \) compounds (with \( M^{2+} \) and \( O^{2-} \)) is much higher than for \( M_{2}O \) (with \( M^{+} \) and \( O^{-} \)) or \( MO_{2} \). This strong attraction in the crystal lattice makes the formation of oxides with \( O^{2-} \) ions thermodynamically favorable and stable.
In simple words: Even though it takes energy to add a second electron to an oxygen atom to make \( O^{2-} \), the final oxide compounds are very stable. This is because a lot of energy is released when positive metal ions and negative \( O^{2-} \) ions stick together tightly in a crystal, making up for the initial energy cost. So, most oxides have \( O^{2-} \) ions.

🎯 Exam Tip: When explaining why unfavorable ion formation happens, always consider the compensating energy factor, such as lattice energy in ionic compounds. A very large lattice energy can make an overall process favorable even if some individual steps are endothermic.

 

Question 30. Explain why inspite of nearly the some electro negativity, oxygen forms hydrogen bonding while chlorine do Typesetting math: 33%

 

Question 32. What inspired N. Bertlett for carrying out reaction between Xe and PtF6 ?
Answer: In 1962, Neil Bartlett discovered that platinum hexafluoride (\( PtF_{6} \)) is a very powerful oxidizing agent. He observed that \( PtF_{6} \) could oxidize molecular oxygen (\( O_{2} \)) to form an ionic compound called dioxygen hexafluoroplatinate(V), \( O_{2}^{+}[PtF_{6}]^{-} \).
\( O_{2}(g) + PtF_{6}(aq) \rightarrow O_{2}^{+}[PtF_{6}]^{-} \)
This reaction showed that \( PtF_{6} \) could remove an electron from \( O_{2} \). Bartlett then noticed some key similarities between \( O_{2} \) and Xenon (Xe):
1. **Ionisation enthalpy:** The first ionization enthalpy of Xenon gas (1170 kJ mol\(^{-1}\)) is quite close to that of Oxygen (1166 kJ mol\(^{-1}\)). This means similar energy is needed to remove an electron from both.
2. **Molecular diameter:** The molecular diameter of oxygen and the atomic radius of xenon are also very similar.
These similarities led Bartlett to hypothesize that if \( PtF_{6} \) could oxidize \( O_{2} \), it might also be able to oxidize Xenon. This inspiration led him to successfully synthesize the first noble gas compound, \( Xe^{+}[PtF_{6}]^{-} \), opening up the field of noble gas chemistry.
In simple words: Neil Bartlett saw that a strong chemical (PtF6) could take an electron from oxygen. He then realized that xenon was very similar to oxygen in how much energy it took to remove an electron and their size. So, he thought PtF6 could also take an electron from xenon, which led him to create the first noble gas compound.

🎯 Exam Tip: This is a landmark discovery in chemistry. Remember the key insights: \( PtF_{6} \) as a strong oxidizer, and the critical comparison between the ionization energies and sizes of \( O_{2} \) and Xe.

 

Question 33. What are the oxidation states of phosphorus in the following
(i) \( H_{3}PO_{3} \)
(ii) \( PCl_{3} \)
(iii) \( Ca_{3}P_{2} \)
(iv) \( Na_{3}PO_{4} \)
(v) \( POF_{3} \)
Answer: The oxidation states of phosphorus in the given compounds are:
(i) In \( H_{3}PO_{3} \): \( +3 \)
(ii) In \( PCl_{3} \): \( +3 \)
(iii) In \( Ca_{3}P_{2} \): \( -3 \) (Calcium is always \( +2 \), so \( 3 \times (+2) + 2x = 0 \implies 6 + 2x = 0 \implies 2x = -6 \implies x = -3 \))
(iv) In \( Na_{3}PO_{4} \): \( +5 \) (Sodium is always \( +1 \), Oxygen is \( -2 \), so \( 3 \times (+1) + x + 4 \times (-2) = 0 \implies 3 + x - 8 = 0 \implies x = +5 \))
(v) In \( POF_{3} \): \( +5 \) (Oxygen is \( -2 \), Fluorine is \( -1 \), so \( x + (-2) + 3 \times (-1) = 0 \implies x - 5 = 0 \implies x = +5 \))
Phosphorus is a versatile element, commonly showing oxidation states of \( -3, +3 \), and \( +5 \) in its various compounds.
In simple words: The oxidation state for phosphorus depends on what it is bonded to. In these compounds, phosphorus can have \( +3, -3 \), or \( +5 \) as its oxidation state, showing how many electrons it has gained or lost.

🎯 Exam Tip: To find the oxidation state of an element, remember the standard oxidation states of common elements like H (+1), O (-2), alkali metals (+1), alkaline earth metals (+2), and halogens (-1 in most compounds). The sum of oxidation states in a neutral compound is zero, and in an ion, it equals the charge of the ion.

 

Question 34. W Typesetting math: 33% Jations for the following:

 

Question 35. How are xenon fluorides XeF2, XeF4, XeF6 obtained?
Answer: Xenon fluorides are synthesized by directly reacting xenon gas with fluorine gas under specific conditions of temperature, pressure, and reactant ratios.
(i) **To obtain \( XeF_{2} \):** Xenon is reacted with fluorine in a 1:1 ratio at 673 K and 1 bar pressure.
\( Xe(g) + F_{2}(g) \xrightarrow{673 \text{ K, 1 bar}} XeF_{2}(s) \)
(ii) **To obtain \( XeF_{4} \):** Xenon is reacted with fluorine in a 1:5 ratio (fluorine in excess) at 873 K and 8 bar pressure.
\( Xe(g) + 2F_{2}(g) \xrightarrow{873 \text{ K, 8 bar}} XeF_{4}(s) \)
(iii) **To obtain \( XeF_{6} \):** Xenon is reacted with fluorine in a 1:20 ratio (fluorine in large excess) at 573 K and 60-70 bar pressure.
\( Xe(g) + 3F_{2}(g) \xrightarrow{573 \text{ K, 60-70 bar}} XeF_{6}(s) \)
These reactions require precise control over reaction conditions to get the desired fluoride.
In simple words: We make different xenon fluoride compounds by mixing xenon gas and fluorine gas in specific amounts, and then heating them up and putting them under different pressures. The amount of fluorine and the exact conditions decide which xenon fluoride is made.

🎯 Exam Tip: Remember that the ratio of reactants, temperature, and pressure are crucial for directing the synthesis of specific xenon fluorides. Higher fluorine concentration and specific temperature ranges favor higher fluorides.

 

Question 36. What neutral molecule is CIO\(^{-1}\) isoelectronic ? Is that molecule a Lewis base?
Answer: Isoelectronic means having the same number of electrons. The ion \( ClO^{-} \) has 17 (from Cl) + 8 (from O) + 1 (from negative charge) = 26 electrons.
A neutral molecule that is isoelectronic with \( ClO^{-} \) is \( ClF \) (chlorine monofluoride). Chlorine has 17 electrons and fluorine has 9 electrons, totaling 26 electrons.
Yes, \( ClF \) can act as a Lewis base. A Lewis base is a substance that can donate a pair of electrons. Chlorine in \( ClF \) has three lone pairs of electrons, which it can donate to an electron-deficient species. The presence of these available lone pairs makes it capable of acting as a Lewis base.
In simple words: The neutral molecule that has the same number of electrons as \( ClO^{-} \) is \( ClF \). Yes, \( ClF \) can give away a pair of its electrons, so it acts like a Lewis base.

🎯 Exam Tip: To identify isoelectronic species, count the total number of electrons. For Lewis bases, look for atoms with available lone pairs of electrons to donate.

 

Question 37. Which one of the following does not exist?
(i) \( XeOF_{4} \)
(ii) \( NeF_{2} \)
(iii) \( XeF_{2} \)
(iv) \( XeF_{6} \)
Answer: (ii) NeF2
In simple words: Neon difluoride (\( NeF_{2} \)) does not exist naturally. Neon is a noble gas and is very unreactive, making it difficult to form compounds.

🎯 Exam Tip: Noble gases are generally unreactive, but heavier noble gases like Xenon can form compounds with highly electronegative elements like fluorine. Lighter noble gases like Neon and Argon are much less reactive and typically do not form stable compounds.

 

Question 38. Given the formula and describe the structure of a noble gas species which is isostructural with :
(i) \( ICl_{4}^{-} \)
(ii) \( IBr_{2}^{-} \)
(iii) \( BrO_{3}^{-} \)
Answer: An isostructural species has the same geometry and arrangement of atoms (including lone pairs) as the given ion. We look for noble gas compounds that match the electron count and VSEPR arrangement.
(i) **For \( ICl_{4}^{-} \):** The central iodine atom has 7 valence electrons, plus 4 from chlorine atoms, plus 1 from the negative charge, totaling 12 electrons. This gives 4 bond pairs and 2 lone pairs (\( AX_{4}E_{2} \)). The geometry is square planar.
A noble gas species isostructural with \( ICl_{4}^{-} \) is **\( XeF_{4} \)** (Xenon tetrafluoride). Xenon has 8 valence electrons and forms 4 bonds with fluorine, leaving 2 lone pairs. Its structure is square planar.
(ii) **For \( IBr_{2}^{-} \):** The central iodine atom has 7 valence electrons, plus 2 from bromine atoms, plus 1 from the negative charge, totaling 10 electrons. This gives 2 bond pairs and 3 lone pairs (\( AX_{2}E_{3} \)). The geometry is linear.
A noble gas species isostructural with \( IBr_{2}^{-} \) is **\( XeF_{2} \)** (Xenon difluoride). Xenon has 8 valence electrons and forms 2 bonds with fluorine, leaving 3 lone pairs. Its structure is linear.
(iii) **For \( BrO_{3}^{-} \):** The central bromine atom has 7 valence electrons, plus 3 from oxygen atoms (each oxygen shares 2 electrons but let's count bonds), plus 1 from negative charge. Counting by VSEPR: Br is \( AX_{3}E_{1} \) (3 bond pairs, 1 lone pair, total 4 electron domains). The geometry is pyramidal.
A noble gas species isostructural with \( BrO_{3}^{-} \) is **\( XeO_{3} \)** (Xenon trioxide). Xenon has 8 valence electrons. If it forms 3 double bonds with oxygen, it uses 6 electrons, leaving 2 electrons as one lone pair. So, \( XeO_{3} \) has 3 bond pairs (considering double bonds as single electron domains for geometry) and 1 lone pair. Its structure is pyramidal.
In simple words: We find noble gas compounds that have the same shape and electron arrangement as the given ions. For \( ICl_{4}^{-} \), it's \( XeF_{4} \) (square planar). For \( IBr_{2}^{-} \), it's \( XeF_{2} \) (linear). For \( BrO_{3}^{-} \), it's \( XeO_{3} \) (pyramidal).

🎯 Exam Tip: To determine isostructural species, use VSEPR theory to find the electron domain geometry (number of bond pairs and lone pairs) around the central atom for both the given ion and the potential noble gas compound.

 

Question 39. List the uses of neon and argon gases.
Answer: Neon and argon are noble gases, known for their inertness (lack of reactivity). This property makes them useful in various applications.
**(A) Uses of Neon gas:**
1. Neon is primarily used in discharge tubes to create neon signs and advertising displays. When electricity passes through neon gas, it glows with a distinct reddish-orange light.
2. It is also used in beacon lights, especially in fog or for aircraft, because its light can penetrate fog and mist.
3. Neon is used in some gas lasers and voltage regulators.
**(B) Uses of Argon gas:**
1. Argon is used to fill electric bulbs and fluorescent tubes because its inert nature prevents the oxidation of the filament, making the bulb last longer.
2. In laboratories, argon provides an inert atmosphere for handling substances that are sensitive to air (like oxygen or moisture).
3. Pure argon is used in gas chromatography as a carrier gas.
4. It is also used in arc welding to provide an inert shield for the molten metal, preventing contamination from atmospheric gases.
In simple words: Neon is used in bright signs and lights that can cut through fog because it glows orange-red. Argon is used in light bulbs to protect the filament and in labs to keep air-sensitive chemicals safe because it doesn't react with anything.

🎯 Exam Tip: For noble gases, their inertness is the key property for most applications. Think about scenarios where preventing reactions or creating a non-reactive environment is crucial.

 

RBSE Class 12 Chemistry Chapter 7 Long Answer Type Questions

 

Question 1. Discuss the general characteristics of group 15 elements with reference to their electronic configuation, oxidation state, atomic size, ionisation enthalpy and electro negativity.
Answer: Group 15 elements, also known as pnictogens, include Nitrogen (N), Phosphorus (P), Arsenic (As), Antimony (Sb), and Bismuth (Bi). They show gradual changes in properties from non-metallic to metallic as you go down the group.
**General Characteristics of Group 15 Elements:**
**(A) Electronic Configurations:**
All atoms in Group 15 have five electrons in their outermost shell, which are arranged as two in the s-subshell and three in the p-subshell. Their general electronic configuration is \( ns^{2}np^{3} \). This electron arrangement gives them similar chemical tendencies.

**(B) Oxidation state:**
These elements typically show various oxidation states ranging from \( -3 \) to \( +5 \).
- **Negative oxidation state (\( -3 \)):** Due to their \( np^{3} \) configuration, they can gain three electrons to achieve a stable octet. Nitrogen and phosphorus commonly show \( -3 \) oxidation states in hydrides like \( NH_{3} \) and \( PH_{3} \). The tendency to show the \( -3 \) state decreases down the group as metallic character increases.
- **Positive oxidation states (\( +3 \) and \( +5 \)):** They can lose electrons. The \( +3 \) oxidation state is formed by losing the three p-electrons, while the \( +5 \) state involves losing both the s- and p-electrons. Due to the 'inert pair effect', the stability of the \( +3 \) oxidation state increases down the group, while the stability of the \( +5 \) oxidation state decreases (e.g., \( Bi^{3+} \) is more stable than \( Bi^{5+} \)). Nitrogen, with no d-orbitals, is limited to a maximum valency of 4 and doesn't form \( NX_{5} \) type compounds.
**(C) Atomic size:**
The atomic and ionic radii of Group 15 elements are generally smaller than those of Group 14 elements in the same period. As you move down Group 15, the atomic radii increase because new electron shells are added.
Order of Atomic size: \( N < P < As < Sb < Bi \)
**(D) Ionisation Enthalpy:**
Ionisation enthalpy is the energy needed to remove an electron. Group 15 elements have higher ionization enthalpies compared to Group 14 elements in the same period because of their stable half-filled p-orbitals. As you move down the group, ionization enthalpy decreases due to increasing atomic size and greater shielding effect, making it easier to remove electrons.
**(E) Electronegativity:**
Electronegativity is the ability of an atom to attract electrons. Group 15 elements have higher electronegativity values than the corresponding Group 14 elements. As you move down the group, the electronegativity value decreases because the atomic size increases, and the outermost electrons are further from the nucleus.
Order of electronegativity: \( N (3.0) > P (2.1) > As (2.0) > Sb (1.9) > Bi (1.9) \)
In simple words: Group 15 elements all have five outer electrons, which determines their chemical actions. They can lose or gain electrons, showing different strengths as you go down the group. Atoms get bigger, and it's easier to remove their electrons, but their pull on new electrons gets weaker.

🎯 Exam Tip: When describing group characteristics, always explain the underlying reasons for trends (e.g., increasing atomic size for decreasing ionization enthalpy, inert pair effect for oxidation states). Use specific examples like nitrogen's lack of d-orbitals to highlight unique properties.

 

Question 2. Arrange the following in the order of property indicated for each set :
(i) \( F_{2}, Cl_{2}, Br_{2}, I_{2} \): Increasing bond dissociation enthalpy.
(ii) \( HF, HCl, HBr, HI \): Increasing acid strength
(iii) \( NH_{3}, PH_{3}, AsH_{3}, SbH_{3}, BiH_{3} \): Increasing base strength.
Answer:
(i) **Increasing bond dissociation enthalpy for \( F_{2}, Cl_{2}, Br_{2}, I_{2} \):**
Generally, as bond length increases, bond dissociation enthalpy decreases. So, we would expect \( F_{2} \) to have the highest bond dissociation enthalpy. However, \( F_{2} \) has a surprisingly lower bond dissociation enthalpy (158.8 kJ/mol) than \( Cl_{2} \) (242.6 kJ/mol) and \( Br_{2} \) (192.8 kJ/mol). This is due to the small size of fluorine atoms and the strong repulsion between the lone pairs of electrons in the two fluorine atoms, which weakens the F-F bond. Therefore, the increasing order is:
\( I_{2} < F_{2} < Br_{2} < Cl_{2} \)
(ii) **Increasing acid strength for \( HF, HCl, HBr, HI \):**
As you move down the group, the atomic size of the halogen increases, and the H-X bond length increases. This makes the H-X bond weaker and easier to break. A weaker bond means the acid can more readily release \( H^{+} \) ions, thus increasing its acid strength.
Therefore, the increasing order of acid strength is:
\( HF < HCl < HBr < HI \)
(iii) **Increasing base strength for \( NH_{3}, PH_{3}, AsH_{3}, SbH_{3}, BiH_{3} \):**
Basic strength depends on the ability of the central atom to donate its lone pair of electrons (Lewis basicity). As you move down Group 15 from nitrogen to bismuth, the atomic size increases. This causes the electron density on the central atom to decrease, making the lone pair less concentrated and less available for donation. Consequently, the basic strength decreases down the group.
Therefore, the increasing order of basic strength is:
\( BiH_{3} < SbH_{3} < AsH_{3} < PH_{3} < NH_{3} \)
In simple words: For halogens, \( Cl_{2} \) has the strongest bond, then \( Br_{2} \), then \( F_{2} \) and \( I_{2} \). For hydrogen halides, acid strength gets stronger as you go down the group from \( HF \) to \( HI \). For group 15 hydrides, the basic strength gets weaker as you go down the group, with \( NH_{3} \) being the strongest base.

🎯 Exam Tip: Always be mindful of exceptions to general trends, like the low bond dissociation enthalpy of \( F_{2} \) due to inter-electronic repulsion. For acid strength, bond length is usually the dominant factor, and for basic strength, electron density and availability of the lone pair are key.

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