Samacheer Kalvi Class 12 Chemistry Solutions Chapter 3 pBlock Elements II

Samacheer Kalvi 12th Chemistry Solutions Chapter 3 p-Block Elements – II

12th Chemistry Guide p-Block Elements – II Text Book Questions and Answers

Part - I - Text Book Evaluation

I. Choose the correct answer

 

Question 1. In which of the following, \( \text{NH}_3 \) is not used?
(a) Nessler's reagent
(b) Reagent for the analysis of IV group basic radical
(c) Reagent for the analysis of III group basic radical
(d) Tollen's reagent
Answer: (d) Tollen's reagent
In simple words: Ammonia is used in many chemical tests and reactions, but it is not used in Tollen's reagent. Tollen's reagent is used to test for aldehydes.

🎯 Exam Tip: Remember the specific applications of common chemical reagents. Knowing what a reagent is *not* used for can also help differentiate its functions.

 

Question 2. Which is true regarding nitrogen?
(a) least electronegative element
(b) has low ionisation enthalpy than oxygen
(c) d – orbitals available
(d) ability to form \( \text{p}\pi \)- \( \text{p}\pi \) bonds with itself
Answer: (d) ability to form \( \text{p}\pi \)- \( \text{p}\pi \) bonds with itself
In simple words: Nitrogen atoms can easily create double or triple bonds with other nitrogen atoms. This unique ability is due to its small size and strong tendency to share p-orbital electrons.

🎯 Exam Tip: Recall the key properties of nitrogen, such as its ability to form multiple bonds, which distinguishes it from heavier group 15 elements.

 

Question 3. An element belongs to group 15 and 3rd period of the periodic table, its electronic configuration would be
(a) \( \text{1s}^2 \text{2s}^2 \text{2p}^4 \)
(b) \( \text{1s}^2 \text{2s}^2 \text{2p}^3 \)
(c) \( \text{1s}^2 \text{2s}^2 \text{2p}^6 \text{3s}^2 \text{3p}^2 \)
(d) \( \text{1s}^2 \text{2s}^2 \text{2p}^6 \text{3s}^2 \text{3p}^3 \)
Answer: (d) \( \text{1s}^2 \text{2s}^2 \text{2p}^6 \text{3s}^2 \text{3p}^3 \)
In simple words: An element in Group 15 needs 5 valence electrons, and being in the 3rd period means its outermost electrons are in the 3s and 3p orbitals. This configuration shows a full inner shell and 5 valence electrons, matching the element phosphorus.

🎯 Exam Tip: To determine electronic configuration from group and period, identify the principal quantum number (period) and the number of valence electrons (group). For p-block elements, Group number = 10 + valence electrons.

 

Question 4. Solid (A) reacts with strong aqueous NaOH liberating a foul smelling gas(B) which spontaneously burn in air giving smoky rings. A and B are respectively
(a) \( \text{P}_4 \)(red) and \( \text{PH}_3 \)
(b) \( \text{P}_4 \) (white) and \( \text{PH}_3 \)
(c) \( \text{S}_8 \) and \( \text{H}_2 \text{S} \)
(d) \( \text{P}_4 \)(white) and \( \text{H}_2 \text{S} \)
Answer: (b) \( \text{P}_4 \) (white) and \( \text{PH}_3 \)
In simple words: When white phosphorus reacts with strong sodium hydroxide, it releases phosphine gas. This gas has a very bad smell and catches fire on its own in the air, creating smoke.

🎯 Exam Tip: This question tests knowledge of characteristic reactions and properties of phosphorus allotropes, particularly the reaction of white phosphorus with strong bases and the properties of phosphine gas.

 

Question 5. On hydrolysis, \( \text{PCl}_3 \) gives
(a) \( \text{H}_3 \text{PO}_3 \)
(b) \( \text{PH}_3 \)
(c) \( \text{H}_3 \text{PO}_4 \)
(d) \( \text{POCl}_3 \)
Answer: (a) \( \text{H}_3 \text{PO}_3 \)
In simple words: When phosphorus trichloride reacts with water, it breaks down and forms phosphorous acid. Water molecules effectively replace the chlorine atoms.

🎯 Exam Tip: Know the hydrolysis products of common non-metal chlorides, as they often form acids corresponding to the non-metal's oxidation state.

 

Question 6. \( \text{P}_4 \text{O}_6 \) reacts with cold water to give
(a) \( \text{H}_3 \text{PO}_3 \)
(b) \( \text{H}_4 \text{P}_2 \text{O}_7 \)
(c) \( \text{HPO}_3 \)
(d) \( \text{H}_3 \text{PO}_4 \)
Answer: (a) \( \text{H}_3 \text{PO}_3 \)
In simple words: When phosphorus trioxide reacts with cold water, it forms phosphorous acid. This is a common way to produce this specific acid.

🎯 Exam Tip: Distinguish between the hydrolysis products of phosphorus trioxide (\( \text{P}_4 \text{O}_6 \)) and phosphorus pentoxide (\( \text{P}_4 \text{O}_{10} \)), which form phosphorous acid and phosphoric acid, respectively.

 

Question 7. The basicity of pyrophosphorous acid \( (\text{H}_4 \text{P}_2 \text{O}_3) \) is
(a) 1
(b) 2
(c) 3
(d) 5
Answer: (b) 2
In simple words: Pyrophosphorous acid has two hydrogen atoms that can be replaced by a base, making its basicity 2. This means it can donate two protons.

🎯 Exam Tip: The basicity of an oxyacid of phosphorus is determined by the number of \( \text{P}-\text{OH} \) bonds, not the total number of hydrogen atoms. Visualize the structure to confirm.

 

Question 8. The molarity of given orthophosphoric acid solution is 2M. Its normality is
(a) 6N
(b) 4N
(c) 3N
(d) none of the options
Answer: (a) 6N
Normality = M x basicity \( = 2 \times 3 = 6 \)
In simple words: Orthophosphoric acid can donate three hydrogen ions (it is tribasic), so its normality is three times its molarity. If the molarity is 2M, the normality is 6N.

🎯 Exam Tip: Remember the relationship: Normality = Molarity x Basicity (for acids) or Acidity (for bases). For orthophosphoric acid \( (\text{H}_3 \text{PO}_4) \), the basicity is 3.

 

Question 9. Assertion : bond dissociation energy of fluorine is greater than chlorine gas Reason: chlorine has more electronic repulsion than flourine
(a) Both assertion and reason are true and reason is the correct explanation of assertion.
(b) Both assertion and reason are true but reason is not the correct explanation of assertion
(c) Assertion is true but reason is false
(d) Both assertion and reason are false
Answer: (d) Both assertion and reason are false The converse is true
In simple words: The statement about fluorine's bond energy being greater is incorrect. Also, the reason given about chlorine's electronic repulsion is not right. Actually, fluorine has unusually low bond dissociation energy due to high electron-electron repulsion in its small atomic size, and chlorine has a higher bond dissociation energy.

🎯 Exam Tip: Pay attention to the exceptions and trends in bond dissociation energies within a group, especially for the first element like fluorine, which often behaves anomalously due to its small size and high electron density.

 

Question 10. Among the following, which is the strongest oxidizing agent?
(a) \( \text{Cl}_2 \)
(b) \( \text{F}_2 \)
(c) \( \text{Br}_2 \)
(d) \( \text{I}_2 \)
Answer: (b) \( \text{F}_2 \)
In simple words: Fluorine is the most powerful oxidizing agent among all halogens. It readily gains electrons from other substances because it is the most electronegative element.

🎯 Exam Tip: Oxidizing power of halogens decreases down the group. Fluorine is the strongest oxidizing agent due to its high electronegativity and small size, which makes it very eager to gain an electron.

 

Question 11. The correct order of the thermal stability of hydrogen halide is (PTA – 4)
(a) \( \text{HI} > \text{HBr} > \text{HCl} > \text{HF} \)
(b) \( \text{HF} > \text{HCl} > \text{HBr} > \text{HI} \)
(c) \( \text{HCl} > \text{HF} > \text{HBr} > \text{HI} \)
(d) \( \text{HI} > \text{HCl} > \text{HF} > \text{HBr} \)
Answer: (b) \( \text{HF} > \text{HCl} > \text{HBr} > \text{HI} \)
In simple words: Hydrogen fluoride is the most stable of these acids when heated, and hydrogen iodide is the least stable. This is because the bond strength decreases as you go down the halogen group.

🎯 Exam Tip: Thermal stability of hydrogen halides decreases as the bond length increases and bond strength decreases down the group. \( \text{HF} \) has the strongest bond, while \( \text{HI} \) has the weakest.

 

Question 12. Which one of the following compounds is not formed?
(a) \( \text{XeOF}_4 \)
(b) \( \text{XeO}_3 \)
(c) \( \text{XeF}_2 \)
(d) \( \text{NeF}_2 \)
Answer: (d) \( \text{NeF}_2 \)
In simple words: Xenon can form compounds with oxygen and fluorine, but neon, being a much smaller noble gas, is generally unreactive and does not form stable compounds like \( \text{NeF}_2 \). Neon's small size makes it very difficult for it to lose or share electrons.

🎯 Exam Tip: Remember that reactivity of noble gases increases down the group. Xenon is known to form compounds, especially with highly electronegative elements like fluorine and oxygen, while lighter noble gases like neon are much less reactive.

 

Question 13. Most easily liquefiable gas is
(a) Ar
(b) Ne
(c) He
(d) Kr
Answer: (d) Kr
In simple words: Krypton is the easiest to turn into a liquid among these noble gases. Larger atoms have stronger intermolecular forces, making them liquefy more easily.

🎯 Exam Tip: The ease of liquefaction for noble gases increases with increasing atomic mass because the London dispersion forces become stronger, leading to higher boiling points.

 

Question 14. \( \text{XeF}_6 \) on complete hydrolysis produces
(a) \( \text{XeOF}_4 \)
(b) \( \text{XeO}_2 \text{F}_2 \)
(c) \( \text{XeO}_3 \)
(d) \( \text{XeO}_2 \)
Answer: (c) \( \text{XeO}_3 \)
In simple words: When xenon hexafluoride reacts completely with water, all its fluorine atoms are replaced by oxygen, leading to the formation of xenon trioxide. This reaction shows how highly fluorinated xenon compounds can be converted into oxides.

🎯 Exam Tip: Understand the hydrolysis reactions of xenon fluorides; partial hydrolysis yields oxyfluorides, while complete hydrolysis yields xenon trioxide (\( \text{XeO}_3 \)).

 

Question 15. Which of the following is strongest acid among all?
(a) HI
(b) HF
(c) HBr
(d) HCl
Answer: (a) HI
In simple words: Among these hydrogen halides, hydroiodic acid (HI) is the strongest acid. This is because the bond between hydrogen and iodine is the longest and weakest, making it easiest for the hydrogen ion to be released.

🎯 Exam Tip: Acid strength of hydrogen halides in aqueous solution increases down the group (\( \text{HF} < \text{HCl} < \text{HBr} < \text{HI} \)), mainly due to decreasing bond dissociation enthalpy (H-X bond strength) as the size of the halogen increases.

 

Question 16. Which one of the following orders is correct for the bond dissociation enthalpy of halogen molecules? (NEET)
(a) \( \text{Br}_2 > \text{I}_2 > \text{F}_2 > \text{Cl}_2 \)
(b) \( \text{F}_2 > \text{Cl}_2 > \text{Br}_2 > \text{I}_2 \)
(c) \( \text{I}_2 > \text{Br}_2 > \text{Cl}_2 > \text{F}_2 \)
(d) \( \text{Cl}_2 > \text{Br}_2 > \text{F}_2 > \text{I}_2 \)
Answer: (d) \( \text{Cl}_2 > \text{Br}_2 > \text{F}_2 > \text{I}_2 \)
In simple words: Chlorine gas has the strongest bond, followed by bromine. Fluorine's bond is surprisingly weaker than both chlorine and bromine, and iodine has the weakest bond among them. This difference in fluorine is due to electron repulsion.

🎯 Exam Tip: Remember the anomalous trend for \( \text{F}_2 \) bond dissociation enthalpy. Due to the small size and high electron-electron repulsion of fluorine atoms, the \( \text{F}-\text{F} \) bond is weaker than expected and weaker than the \( \text{Cl}-\text{Cl} \) and \( \text{Br}-\text{Br} \) bonds.

 

Question 17. Among the following the correct order of acidity is (NEET)
(a) \( \text{HClO}_2 < \text{HClO} < \text{HClO}_3 < \text{HClO}_4 \)
(b) \( \text{HClO}_4 < \text{HClO}_2 < \text{HClO} < \text{HClO}_3 \)
(c) \( \text{HClO}_3 < \text{HClO}_4 < \text{HClO}_2 < \text{HClO} \)
(d) \( \text{HClO} < \text{HClO}_2 < \text{HClO}_3 < \text{HClO}_4 \)
Answer: (d) \( \text{HClO} < \text{HClO}_2 < \text{HClO}_3 < \text{HClO}_4 \)
In simple words: The strength of oxyacids increases as the number of oxygen atoms not bonded to hydrogen increases. Perchloric acid (\( \text{HClO}_4 \)) is the strongest because it has the most non-hydrogen-bonded oxygen atoms.

🎯 Exam Tip: For oxyacids of the same central atom, acidity increases with the increasing number of oxygen atoms not bonded to hydrogen (or with increasing oxidation state of the central atom).

 

Question 18. When copper is heated with cone HNOs it produces
(a) \( \text{Cu} (\text{NO}_3)_2 \), NO and \( \text{NO}_2 \)
(b) \( \text{Cu} (\text{NO}_3)_2 \) and \( \text{N}_2 \text{O} \)
(c) \( \text{Cu} (\text{NO}_3)_2 \) and \( \text{NO}_2 \)
(d) \( \text{Cu} (\text{NO}_3)_2 \) and NO
Answer: (c) \( \text{Cu} (\text{NO}_3)_2 \) and \( \text{NO}_2 \)
In simple words: When copper metal reacts with concentrated nitric acid and is heated, it forms copper nitrate and nitrogen dioxide gas. Nitrogen dioxide is a brown gas.

🎯 Exam Tip: Remember that the reaction products of metals with nitric acid depend on the concentration of the acid. Concentrated nitric acid typically yields \( \text{NO}_2 \), while dilute nitric acid yields NO or \( \text{N}_2 \text{O} \).

 

Question 1. What is the inert pair effect?
Answer: In p-block elements, as we go down a group, the two electrons in the outermost s-orbital sometimes become "lazy" and do not take part in bonding. This means only the p-orbital electrons are available for chemical bonding. This phenomenon is called the inert pair effect, and it leads to lower oxidation states becoming more stable for heavier elements in the group.
In simple words: As you go down a group in the p-block, the two outermost s-electrons become less likely to participate in chemical reactions. This 'laziness' is called the inert pair effect.

🎯 Exam Tip: Define the inert pair effect clearly by mentioning the s-electrons, their non-participation in bonding, and its prevalence in heavier p-block elements, leading to stability of lower oxidation states.

 

Question 2. Chalcogens belong to p-block. Give reason.
Answer:

• Chalcogens are elements in Group 16 of the periodic table, such as Oxygen, Sulfur, Selenium, Tellurium, and Polonium.
• Their outer electronic configuration is \( \text{ns}^2 \text{np}^4 \). This means their last electron enters a p-orbital.
• Because the differentiating electron (the last electron added) occupies a p-orbital, these elements are classified as p-block elements. They exhibit a range of properties due to this p-orbital electron configuration.
• Therefore, they belong to p-block elements.

In simple words: Chalcogens are in the p-block because their outermost electron fills up a p-orbital. This electron arrangement is how scientists group elements in the periodic table.

🎯 Exam Tip: To explain why an element belongs to a specific block, always refer to its general outer electronic configuration and specify which orbital the last electron enters.

 

Question 3. Explain why fluorine always exhibits an oxidation state of -1?
Answer: Fluorine is the most electronegative element among all elements, including other halogens. This extreme electronegativity means it strongly attracts electrons. Also, fluorine does not have any d-orbitals available to expand its octet. This lack of d-orbitals prevents it from exhibiting positive oxidation states. Therefore, fluorine always shows an oxidation state of -1, while other halogens can show positive oxidation states like +1, +3, +5, and +7. This makes fluorine unique in its group regarding oxidation states.
In simple words: Fluorine always has an oxidation state of -1 because it is the most electron-hungry element and cannot share its electrons in a way that would give it a positive charge. It always pulls electrons towards itself.

🎯 Exam Tip: Highlight fluorine's exceptional electronegativity and the absence of d-orbitals as the primary reasons for its unique -1 oxidation state. Contrast this with other halogens that can expand their octet using d-orbitals.

 

Question 4. Give the oxidation state of halogen in the following molecules:
(a) \( \text{OF}_2 \)
(b) \( \text{O}_2 \text{F}_2 \)
(c) \( \text{Cl}_2 \text{O}_3 \)
(d) \( \text{I}_2 \text{O}_4 \)
Answer:

In simple words: The oxidation state shows how many electrons an atom has gained or lost. In compounds with fluorine, fluorine always takes electrons, so its oxidation state is -1. For other halogens with oxygen, oxygen usually takes electrons, so the halogen will have a positive oxidation state.

🎯 Exam Tip: When calculating oxidation states, remember that fluorine always has an oxidation state of -1 in its compounds because it is the most electronegative element. Oxygen typically has -2, but in peroxides or compounds with fluorine, it can be different.

 

Question 5. What are interhalogen compounds? Give examples
Answer: Interhalogen compounds are special compounds formed when two different halogen elements react with each other. These compounds typically have a general formula of \( \text{XX}'_{\text{n}} \), where X is the larger, less electronegative halogen, and X' is the smaller, more electronegative halogen. For example, bromine fluoride (\( \text{BrF}_3 \)) and iodine monochloride (\( \text{ICl} \)) are interhalogen compounds. Fluorine reacts readily with other halogens, even oxygen, forming compounds like difluorine oxide (\( \text{F}_2 \text{O} \)) and difluorine dioxide (\( \text{F}_2 \text{O}_2 \)).
In simple words: Interhalogen compounds are chemical compounds made from two different halogens mixed together, like chlorine and fluorine. They are formed when halogens react with other halogens.

🎯 Exam Tip: Define interhalogen compounds clearly, stating they are formed between two different halogens, and provide examples illustrating the \( \text{XX}'_{\text{n}} \) formula where X is the larger halogen.

 

Question 6. Why fluorine is more reactive than other halogens? (PTA – 1, 3)
Answer: Fluorine is the most reactive element among all halogens. Its high reactivity is primarily due to the very low bond dissociation energy of the \( \text{F}-\text{F} \) bond. This weak bond allows fluorine molecules to easily break apart and react with other substances. Additionally, fluorine has a small atomic size and the highest electronegativity, which makes it strongly attract electrons and readily participate in chemical reactions. This combination makes it an extremely powerful oxidizing agent.
In simple words: Fluorine reacts more easily than other halogens because the bond holding its two atoms together is very weak. Also, it is very small and pulls electrons strongly, making it quick to react.

🎯 Exam Tip: Emphasize the unexpectedly low \( \text{F}-\text{F} \) bond dissociation energy, along with its high electronegativity and small size, as the main reasons for fluorine's extreme reactivity.

 

Question 7. Write the uses of helium. (PTA – 2)
Answer:
1. Helium mixed with oxygen is used by deep-sea divers instead of an air-oxygen mixture. This prevents a painful and dangerous condition called 'bends' caused by nitrogen dissolving in the blood under high pressure.
2. Helium is used to provide an inert atmosphere for processes like electric arc welding of metals. This protects the metals from reacting with air.
3. Helium has the lowest boiling point of all elements, so it is used in cryogenics, which is the science of very low temperatures.
4. It is much less dense than air, which makes it ideal for filling air balloons and airships.
In simple words: Helium helps divers avoid a painful condition, provides a safe environment for welding, is used for very cold experiments, and fills balloons because it is lighter than air.

🎯 Exam Tip: For uses of elements, try to link each use to a specific unique property (e.g., low density for balloons, inertness for welding, low boiling point for cryogenics).

 

Question 8. What is the hybridisation of iodine in \( \text{IF}_7 \)? Give its structure. (PTA – 5)
Answer: In \( \text{IF}_7 \), the hybridization of iodine is \( \text{sp}^3 \text{d}^3 \). The molecule has a pentagonal bipyramidal structure. The iodine atom is at the center, surrounded by seven fluorine atoms, five in a plane forming a pentagon and two at the axial positions.

In simple words: The central iodine atom in \( \text{IF}_7 \) uses a mix of s, p, and d orbitals for bonding, giving it \( \text{sp}^3 \text{d}^3 \) hybridization. Its shape is like two pyramids joined at their bases, with a five-sided base.

🎯 Exam Tip: To determine hybridization and geometry, first count the number of bond pairs and lone pairs around the central atom. For \( \text{IF}_7 \), iodine forms 7 bonds and has no lone pairs, leading to \( \text{sp}^3 \text{d}^3 \) hybridization and a pentagonal bipyramidal geometry.

 

Question 9. Give the balanced equation for the reaction between chlorine with cold NaOH and hot NaOH
Answer:
Chlorine reacts with cold NaOH to give sodium hypochlorite:
\( \text{Cl}_2 + 2\text{NaOH} \rightarrow \text{NaOCl} + \text{NaCl} + \text{H}_2 \text{O} \)
Chlorine reacts with hot NaOH to give sodium chlorate:
\( 3\text{Cl}_2 + 6\text{NaOH} \rightarrow \text{NaClO}_3 + 5\text{NaCl} + 3\text{H}_2 \text{O} \)
In simple words: Chlorine gas reacts differently with cold and hot sodium hydroxide. With cold solution, it forms sodium hypochlorite, but with hot solution, it creates sodium chlorate. These are different products because of the temperature difference.

🎯 Exam Tip: It is important to remember that the reaction of halogens like chlorine with alkali solutions is temperature-dependent, leading to different oxidation products (hypohalite with cold, halate with hot).

 

Question 10. How will you prepare chlorine in the laboratory? (PTA – 2)
Answer:
1. Chlorine can be prepared by reacting concentrated sulfuric acid with chlorides in the presence of manganese dioxide. This method uses common chemicals to generate the gas.
\( 4\text{NaCl} + \text{MnO}_2 + 4\text{H}_2 \text{SO}_4 \rightarrow \text{Cl}_2 + \text{MnCl}_2 + 4\text{NaHSO}_4 + 2\text{H}_2 \text{O} \)
2. It can also be prepared by oxidizing hydrochloric acid using various oxidizing agents such as manganese dioxide (\( \text{MnO}_2 \)), lead dioxide (\( \text{PbO}_2 \)), potassium permanganate (\( \text{KMnO}_4 \)), or dichromate (\( \text{K}_2 \text{Cr}_2 \text{O}_7 \)). These strong oxidizers help to release chlorine from the acid.
\( \text{PbO}_2 + 4\text{HCl} \rightarrow \text{PbCl}_2 + 2\text{H}_2 \text{O} + \text{Cl}_2 \)
\( \text{MnO}_2 + 4\text{HCl} \rightarrow \text{MnCl}_2 + 2\text{H}_2 \text{O} + \text{Cl}_2 \)
\( 2\text{KMnO}_4 + 16\text{HCl} \rightarrow 2\text{KCl} + 2\text{MnCl} + 8\text{H}_2 \text{O} + 5\text{Cl}_2 \)
\( \text{K}_2 \text{Cr}_2 \text{O}_7 + 14\text{HCl} \rightarrow 2\text{KCl} + 2\text{CrCl}_3 + 7\text{H}_2 \text{O} + 3\text{Cl}_2 \)
3. When bleaching powder is treated with mineral acids, chlorine gas is liberated. This is a practical way to get chlorine from a common household chemical.
\( \text{CaOCl}_2 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2 \text{O} + \text{Cl}_2 \)
\( \text{CaOCl}_2 + \text{H}_2 \text{SO}_4 \rightarrow \text{CaSO}_4 + \text{H}_2 \text{O} + \text{Cl}_2 \)
In simple words: Chlorine can be made in the lab by mixing certain chemicals with acids, especially hydrochloric acid, using strong agents that take electrons away. Bleaching powder can also release chlorine when it reacts with acids.

🎯 Exam Tip: Be prepared to list at least two common laboratory methods for chlorine preparation, including their balanced chemical equations. Emphasize the role of oxidizing agents in these reactions.

 

Question 11. Give the uses of sulphuric acid.
Answer:
Sulphuric acid is a very important chemical with many industrial uses:

• It is used in making fertilizers, such as ammonium sulfate and superphosphates, which help plants grow better.
• It is essential for manufacturing other important chemicals like hydrochloric acid and nitric acid.
• Sulphuric acid acts as a drying agent because it absorbs water very well.
• It is used in preparing pigments (colors for paints), explosives, and many other products.

In simple words: Sulphuric acid is used to make farm fertilizers, other acids, and explosives. It also helps to dry things by absorbing water.

🎯 Exam Tip: Focus on the major industrial applications of sulfuric acid, often summarized by the acronym "FAME" (Fertilizers, Acids, Metals, Explosives). Also, include its property as a dehydrating agent.

 

Question 12. Give a reason to support that sulphuric acid is a dehydrating agent. (PTA – 1)
Answer: Sulphuric acid is a powerful dehydrating agent because it has a very strong attraction for water molecules. When it comes into contact with substances containing water, it readily removes the water. For example, when dissolved in water, it forms hydrates like \( \text{H}_2 \text{SO}_4 \cdot \text{H}_2 \text{O} \) and \( \text{H}_2 \text{SO}_4 \cdot 2\text{H}_2 \text{O} \), and this reaction releases a lot of heat. It can even remove water from organic compounds, as seen in the reactions below:
\( \text{C}_{12} \text{H}_{22} \text{O}_{11} + \text{H}_2 \text{SO}_4 \rightarrow 12\text{C} + \text{H}_2 \text{SO}_4 \cdot 11\text{H}_2 \text{O} \)
\( \text{HCOOH} + \text{H}_2 \text{SO}_4 \rightarrow \text{CO} + \text{H}_2 \text{SO}_4 \cdot \text{H}_2 \text{O} \)
In simple words: Sulphuric acid loves water very much and pulls it out of other things. It can even take water out of sugar and formic acid, leaving behind carbon or carbon monoxide.

🎯 Exam Tip: Explain the dehydrating property by mentioning sulfuric acid's strong affinity for water and illustrate with an example like charring of sugar, showing the removal of water elements from organic compounds.

 

Question 13. Write the reason for the anomalous behaviour of Nitrogen.
Answer:
1. Nitrogen behaves differently from other elements in its group mainly due to its small size, very high electronegativity, high ionization enthalpy, and the complete absence of d-orbitals. These factors make it quite unique.
2. Nitrogen has a unique ability to form \( \text{p}\pi – \text{p}\pi \) multiple bonds (like in \( \text{N}_2 \)) with other nitrogen atoms. However, the heavier members of Group 15 cannot form these types of bonds effectively because their atomic orbitals are too large and spread out, preventing good overlapping.
3. Nitrogen exists as a diatomic molecule (\( \text{N}_2 \)) with a strong triple bond between its two atoms. In contrast, other elements in the group form single bonds in their elemental state (e.g., \( \text{P}_4 \)).
4. Nitrogen cannot form \( \text{d}\pi – \text{p}\pi \) bonds because it lacks d-orbitals. Other elements in the group, however, can use their available d-orbitals to form such bonds. This is crucial for their bonding behavior.
In simple words: Nitrogen is special because it is tiny, pulls electrons very strongly, and doesn't have d-orbitals. This allows it to form strong double and triple bonds with itself, unlike the larger elements in its group.

🎯 Exam Tip: When discussing anomalous behavior, focus on small size, high electronegativity, absence of d-orbitals, and the ability to form \( \text{p}\pi – \text{p}\pi \) multiple bonds as key differentiating points for the first element in a group.

 

Question 14. Write the molecular formula and structural formula for the following molecules
(a) Nitric acid
(b) Dinitrogen pentoxide
(c) Phosphoric acid
(d) Phosphine
Answer:

In simple words: Each chemical compound has a unique recipe (molecular formula) and a specific way its atoms are connected (structural formula). For instance, nitric acid has one hydrogen, one nitrogen, and three oxygens, arranged in a flat shape.

🎯 Exam Tip: Be able to draw or describe the molecular and structural formulas for common inorganic compounds. Pay attention to bond types (single, double, coordinate) and geometry when representing structures.

 

Question 15. Give the uses of argon.
Answer: Argon is a noble gas, meaning it is very unreactive. This property makes it useful in situations where an inert atmosphere is needed. For example, argon is used to fill incandescent light bulbs because it prevents the hot filament from reacting with oxygen and burning out quickly, thereby making the bulbs last longer. Its inertness protects the sensitive materials inside.
In simple words: Argon is used in light bulbs to protect the hot wire from breaking. It does this because it does not react with anything, keeping the wire safe.

🎯 Exam Tip: The primary use of argon (and other noble gases) stems from its inert nature. Focus on applications where an unreactive atmosphere is required.

 

Question 16. Write the valence shell electronic configuration of group -15 elements.
Answer: The valence shell electronic configuration of Group 15 elements is \( \text{ns}^2 \text{np}^3 \). This means they all have two electrons in their outermost s-orbital and three electrons in their outermost p-orbital, totaling five valence electrons. This configuration is key to understanding their chemical properties and reactivity across the group.

In simple words: All elements in Group 15 have two electrons in their outer 's' orbital and three electrons in their outer 'p' orbital. This general electron arrangement helps explain how they behave in chemical reactions.

🎯 Exam Tip: Always state the general valence shell electronic configuration for a given group, and if possible, provide specific examples for elements within that group.

 

Question 17. Give two equations to illustrate the chemical behaviour of phosphine.
Answer:
Phosphine (\( \text{PH}_3 \)) shows several important chemical behaviors:
Basic Nature:
Phosphine is a weakly basic compound and can react with acids to form phosphonium salts. For example, it reacts with hydroiodic acid to form phosphonium iodide. The lone pair on phosphorus enables this basic behavior.
\( \text{PH}_3 + \text{HI} \rightarrow \text{PH}_4 \text{I} \)
\( \text{PH}_4 \text{I} + \text{H}_2 \text{O} \underrightarrow{\Delta} \text{PH}_3 + \text{H}_3 \text{O}^+ + \text{I}^- \)
It can also react with halogens to give phosphorus pentahalides:
\( \text{PH}_3 + 4\text{Cl}_2 \rightarrow \text{PCl}_5 + 3\text{HCl} \)
Combustion:
When phosphine is heated in the presence of air or oxygen, it readily burns to produce metaphosphoric acid. This reaction demonstrates its flammability.
\( 4\text{PH}_3 + 8\text{O}_2 \xrightarrow{\Delta} \text{P}_4 \text{O}_{10} + 6\text{H}_2 \text{O} \)
\( \text{P}_4 \text{O}_{10} + 6\text{H}_2 \text{O} \rightarrow 4\text{H}_3 \text{PO}_4 \)
In simple words: Phosphine can act like a weak base, combining with acids to form salts. It also burns when heated with air or oxygen, creating metaphosphoric acid.

🎯 Exam Tip: For illustrating chemical behavior, choose reactions that clearly demonstrate a specific property (e.g., basicity, combustion). Ensure equations are balanced and conditions (like heat) are indicated.

 

Question 18. Give a reaction between nitric acid and a basic oxide.
Answer: Nitric acid is an acid, and it reacts with basic oxides (like iron(II) oxide) to form a salt and water. This is a typical acid-base neutralization reaction where the basic oxide neutralizes the acid. The reaction often involves redox processes when the metal can exist in multiple oxidation states.
\( 3\text{FeO} + 10\text{HNO}_3 \rightarrow 3\text{Fe} (\text{NO}_3)_3 + \text{NO} + 5\text{H}_2 \text{O} \)
In simple words: Nitric acid reacts with a basic oxide, like iron(II) oxide, to make a salt and water. This is a common acid-base reaction.

🎯 Exam Tip: Remember that basic oxides react with acids to form salt and water. When the metal in the basic oxide can be oxidized, nitric acid (being an oxidizing agent) may also cause a redox reaction, producing nitrogen oxides.

 

Question 19. What happens when \( \text{PCl}_5 \) is heated?
Answer: When phosphorus pentachloride (\( \text{PCl}_5 \)) is heated, it undergoes thermal decomposition. This means it breaks down into simpler compounds. Specifically, it decomposes into phosphorus trichloride (\( \text{PCl}_3 \)) and chlorine gas (\( \text{Cl}_2 \)). This is a reversible reaction, meaning \( \text{PCl}_3 \) and \( \text{Cl}_2 \) can recombine to form \( \text{PCl}_5 \) under different conditions.
\( \text{PCl}_5 \text{(g)} \rightleftharpoons \text{PCl}_3 \text{(g)} + \text{Cl}_2 \text{(g)} \)
In simple words: When you heat phosphorus pentachloride, it breaks apart into phosphorus trichloride and chlorine gas. It's like taking a complex molecule and splitting it into two simpler ones.

🎯 Exam Tip: Remember the thermal decomposition of \( \text{PCl}_5 \) as a classic example of a reversible gas-phase reaction. Be able to write the balanced equation.

 

Question 20. Suggest a reason why HF is a weak acid, whereas binary acids of all other halogens are strong acids.
Answer:
Hydrogen fluoride (\( \text{HF} \)) is considered a weak acid, unlike \( \text{HCl} \), \( \text{HBr} \), and \( \text{HI} \), which are strong acids. The primary reasons for this difference are:

• \( \text{HF} \) is only slightly ionized in water. This means it does not fully break apart into \( \text{H}^+ \) and \( \text{F}^- \) ions.
• Other halogen acids like \( \text{HCl} \), \( \text{HBr} \), and \( \text{HI} \) are almost completely ionized in water, releasing nearly all their \( \text{H}^+ \) ions.
• Among halogen acids, the electronegativity difference is highest (1.9) in \( \text{HF} \) acid. This makes the \( \text{H}-\text{F} \) bond very polar and strong.
• Because the bond between hydrogen and fluorine is very strong, it is difficult for water molecules to break it and release the \( \text{H}^+ \) ion, making \( \text{HF} \) a weaker acid. The strong hydrogen bonding in \( \text{HF} \) also plays a significant role in its properties.

In simple words: Hydrogen fluoride is a weak acid because the bond between hydrogen and fluorine is very strong and hard for water to break. Other halogen acids have weaker bonds, so they release their hydrogen easily and are strong acids.

🎯 Exam Tip: Key factors for the weakness of \( \text{HF} \) are the high electronegativity of fluorine leading to a very strong \( \text{H}-\text{F} \) bond and extensive hydrogen bonding, which hinders the release of \( \text{H}^+ \) ions.

 

Question 21. Deduce the oxidation number of oxygen in hypofluorous acid – HOF.
Answer: To find the oxidation number of oxygen in hypofluorous acid (\( \text{HOF} \)), we use the known oxidation states of hydrogen and fluorine. Hydrogen typically has an oxidation number of +1, and fluorine, being the most electronegative element, always has an oxidation number of -1 in its compounds. Let x be the oxidation number of oxygen.
Oxidation number of \( \text{F} = -1 \)
Oxidation number of \( \text{H} = +1 \)
Oxidation number of \( \text{O} \) in \( \text{HOF} = \text{x} \)
For a neutral compound, the sum of oxidation numbers is 0:
\( +1 + \text{x} + (-1) = 0 \)
\( \text{x} = 0 \)
Therefore, the oxidation number of \( \text{O} \) in \( \text{HOF} = 0 \). Oxygen here is not oxidized or reduced relative to its elemental state, making this compound quite unique.
In simple words: In the chemical \( \text{HOF} \), hydrogen has a charge of +1 and fluorine has -1. When you add these up, oxygen must have a charge of 0 to make the whole molecule neutral.

🎯 Exam Tip: Remember that fluorine always takes precedence for the -1 oxidation state. In compounds with fluorine, oxygen may have a positive oxidation state (as in \( \text{OF}_2 \)) or even 0, as calculated for \( \text{HOF} \), which is unusual for oxygen.

 

Question 22. What type of hybridisation occur in (PTA - 5)
a) \( BrF_5 \)
b) \( BrF_3 \)
Answer: This question asks about the hybridization type. The answer should specify the hybridization for each compound listed. For \( BrF_5 \), the central atom Bromine undergoes \( sp^3d^2 \) hybridization, leading to a square pyramidal shape. For \( BrF_3 \), Bromine undergoes \( sp^3d \) hybridization, resulting in a T-shaped molecule. Hybridization helps predict the geometric arrangement of atoms in a molecule.
In simple words: This question asks what kind of mixing of atomic orbitals happens in \( BrF_5 \) and \( BrF_3 \).

🎯 Exam Tip: To determine hybridization, count the number of sigma bonds and lone pairs around the central atom. This sum corresponds to the number of hybrid orbitals.

 

Question 23. Complete the following reactions
1. \( NaCl + MnO_2 + H_2SO_4 \rightarrow \)
2. \( NaNO_2 + HCl \rightarrow \)
3. \( P_4 + NaOH + H_2O \rightarrow \)
4. \( AgNO_3 + PH_3 \rightarrow \)
5. \( Mg + HNO_3 \rightarrow \)
6. \( KClO_3 \stackrel{\triangle}{\rightarrow} \)
7. \( Cu + H_2SO_4 \stackrel{\text{Hot conc.}}{\rightarrow} \)
8. \( Sb + Cl_2 \rightarrow \)
9. \( HBr + H_2SO_4 \rightarrow \)
10. \( XeF_6 + H_2O \rightarrow \)
11. \( XeO_6^{4-} + Mn^{2+} + H^+ \rightarrow \)
12. \( XeOF_4 + SiO_2 \rightarrow \)
13. \( Xe + F_2 \stackrel{\text{Ni/200 atm}}{\text{400°C}} \rightarrow \)
Answer:
1. \( 4NaCl + MnO_2 + 4H_2SO_4 \rightarrow Cl_2 + MnCl_2 + 4NaHSO_4 + 2H_2O \)
2. \( NaNO_2 + HCl \rightarrow NaCl + HNO_2 \)
3. \( P_4 + 3NaOH + 3H_2O \rightarrow 3NaH_2PO_2 + PH_3 \uparrow \) (sodium hypophosphite and phosphine)
4. \( 6AgNO_3 + PH_3 + 3H_2O \rightarrow 6Ag + 6HNO_3 + H_3PO_3 \)
5. \( 4Mg + 10HNO_3 \rightarrow 4Mg(NO_3)_2 + NH_4NO_3 + 3H_2O \)
5. \( 4Mg + 10HNO_3 \rightarrow 4Mg(NO_3)_2 + N_2O + 5H_2O \)
6. \( 2KClO_3 \stackrel{\triangle}{\rightarrow} 2KCl + 3O_2 \uparrow \)
7. \( Cu + 2H_2SO_4 \stackrel{\text{hot, conc}}{\rightarrow} CuSO_4 + 2H_2O + 2SO_2 \uparrow \)
8. \( 2Sb + 3Cl_2 \rightarrow 2SbCl_3 \)
9. \( 2HBr + H_2SO_4 \rightarrow 2H_2O + Br_2 + SO_2 \)
10. \( XeF_6 + 3H_2O \rightarrow XeO_3 + 6HF \)
11. \( 5XeO_6^{4-} + 2Mn^{2+} + 14H^+ \rightarrow 2MnO_4^- + 5XeO_3 + 7H_2O \)
12. \( 2XeOF_4 + SiO_2 \rightarrow 2XeO_2F_2 + SiF_4 \)
13. \( Xe + 3F_2 \stackrel{\text{Ni/200 atm}}{\text{400°C}} XeF_6 \)
In simple words: These are chemical equations that show how different substances react together under certain conditions. The products formed are shown on the right side of the arrow, and the conditions for the reaction are sometimes written above the arrow.

🎯 Exam Tip: When completing reactions, always balance the equation and include states of matter and reaction conditions if specified to ensure full marks.

 

III. Evaluate Yourself

 

Question 1. Write the products formed in the reaction of nitric acid (both dilute and concentrated) with zinc.
Answer: When nitric acid reacts with zinc, the products depend on whether the acid is dilute or concentrated. For dilute nitric acid, ammonium nitrate and water are formed, while concentrated nitric acid produces zinc nitrate, nitrogen dioxide, and water. These reactions highlight the varying oxidizing power of nitric acid.
\( 4Zn + 10HNO_3 \text{ (Dilute)} \rightarrow 4Zn(NO_3)_2 + NH_4NO_3 + 3H_2O \)
\( 4Zn + 10HNO_3 \text{ (Very dilute)} \rightarrow 4Zn(NO_3)_2 + N_2O + 5H_2O \)
\( Zn + 4HNO_3 \text{ (Conc)} \rightarrow Zn(NO_3)_2 + 2NO_2 + 2H_2O \)
In simple words: When zinc metal mixes with nitric acid, the outcome changes based on how strong the acid is. With weaker acid, it makes ammonium nitrate and water. With stronger acid, it makes zinc nitrate, nitrogen dioxide gas, and water.

🎯 Exam Tip: Remember that nitric acid is a strong oxidizing agent, and its reaction products with metals vary significantly with its concentration, often producing different nitrogen oxides.

 

Part II – Additional Questions

 

I. Match the Following

 

Question 1.

Answer:
In simple words: This table shows the "oxidation state" of nitrogen in different chemical compounds. The oxidation state tells us how many electrons nitrogen seems to have gained or lost when it joins with other atoms.

🎯 Exam Tip: To find the oxidation state, remember that the sum of oxidation states in a neutral compound is zero, and in an ion, it equals the ion's charge. Assign known values for common elements like hydrogen (+1) and oxygen (-2).

 

Question 2.

Answer:
In simple words: This table matches the names of different sulfur-based acids with their correct chemical formulas. Each acid has a unique combination of hydrogen, sulfur, and oxygen atoms.

🎯 Exam Tip: Pay close attention to prefixes like "thio-", "pyro-", and "peroxo-" as they indicate specific structural features and different numbers of sulfur and oxygen atoms in the acid's formula.

 

Question 3.

Answer:
In simple words: This table links different "interhalogen compounds" with their "hybridization" type. Hybridization tells us how the atomic orbitals of the central atom mix to form new orbitals for bonding. Each compound has a specific type of hybridization that explains its shape.

🎯 Exam Tip: Remember to calculate the steric number (sum of lone pairs and sigma bonds) for the central atom to correctly identify the hybridization type for each interhalogen compound.

 

Question 4.

Answer:
In simple words: This table shows the shapes of different compounds made with Xenon. Each Xenon compound has a unique structure, like T-shaped or pyramidal, which describes how its atoms are arranged in space.

🎯 Exam Tip: Practice drawing the Lewis structures and using VSEPR theory to accurately predict the molecular geometry (shape) of xenon compounds, paying attention to lone pairs of electrons.

 

II. Assertion and Reason

 

Question 1. Assertion (A) : Aqueous solution of potash Alum is acidic
Reason (R) : Aluminium sulphate undergo hydrolysis. (PTA - 2)
Answer: The correct statement is that both the Assertion (A) and the Reason (R) are true, and Reason (R) correctly explains Assertion (A). Potash alum's aqueous solution is acidic because aluminium sulphate, a component of alum, undergoes hydrolysis in water. Hydrolysis means it reacts with water, and this reaction releases hydrogen ions, which makes the solution acidic. This is why many metal salts can affect the pH of water.
In simple words: Potash alum solution is sour (acidic) because a part of it, aluminum sulfate, mixes with water in a way that creates acid.

🎯 Exam Tip: In assertion-reason questions, first determine if A and R are individually true, then check if R logically explains A. Hydrolysis of metal salts often leads to acidic or basic solutions.

 

Question 2. Assertion (A) : Elements belonging to group 16 are called chalcogens
Reason (R) : Group 16 elements are saltforming elements
Answer: The Assertion (A) is true: elements in Group 16 are indeed called chalcogens. However, the Reason (R) is false. Group 16 elements are actually known as "ore-forming elements" because many ores are metal oxides or sulfides. The term "salt-forming elements" typically refers to halogens (Group 17).
In simple words: The first part is true: Group 16 elements are called chalcogens. But the reason given is wrong. These elements actually form ores, not salts, with metals.

🎯 Exam Tip: Distinguish between group names and their common properties. Chalcogens are "ore-forming," while halogens are "salt-forming."

 

Question 3. Assertion (A) : Among halogen acids, HF has low melting and boiling points
Reason (R) : In HF hydrogen bond is present.
Answer: The Assertion (A) states that HF has low melting and boiling points among halogen acids, which is incorrect. In fact, HF has unusually *high* melting and boiling points compared to other hydrogen halides (HCl, HBr, HI). The Reason (R) is correct: hydrogen bonding *is* present in HF. The strong hydrogen bonds between HF molecules require more energy to break, leading to its higher boiling and melting points. Therefore, A is wrong, and R is correct, but A is incorrect in its claim of low melting/boiling points for HF.
In simple words: The first statement is wrong because HF actually has high melting and boiling points. The reason given, that HF has hydrogen bonds, is true, and these strong bonds cause its high melting and boiling points.

🎯 Exam Tip: Remember that hydrogen bonding significantly increases boiling and melting points due to the extra energy needed to overcome these strong intermolecular forces.

 

Question 4. Assertion (A) : A small piece of Zinc dissolved in dilute nitric acid but hydrogen gas is not evolved. (PTA - 3)
Reason (R) : \( HNO_3 \) is an oxidising agent and this oxidizes hydrogen.
Answer: The Assertion (A) is correct: when zinc reacts with dilute nitric acid, hydrogen gas is generally not produced. The Reason (R) is also correct: nitric acid is indeed a strong oxidizing agent. However, the reason does not fully explain *why* hydrogen is not evolved. Instead of simply oxidizing hydrogen that *would* be formed, nitric acid itself gets reduced to various nitrogen oxides (like \( NO \), \( N_2O \), or \( NH_4NO_3 \)) because it is a powerful oxidant. This is a common characteristic for metals reacting with oxidizing acids. So, while A is true and R is true, R is not the *correct* explanation for A in this specific context, because it implies hydrogen is formed and then oxidized, rather than not being formed at all in the first place due to \( HNO_3 \)'s direct action.
In simple words: Zinc dissolves in weak nitric acid without making hydrogen gas. This is true. The reason is that nitric acid is a strong helper in chemical reactions (an oxidizing agent) and it changes any hydrogen that might form.

🎯 Exam Tip: For assertion-reason questions involving oxidizing acids, remember that highly oxidizing acids like nitric acid typically produce nitrogen oxides (or other reduced forms of nitrogen) rather than hydrogen gas when reacting with most metals.

 

III. Pick Out the Correct Statement

 

Question 1.
i) Oxygen is diamagnetic
ii) Oxygen forms hydrogen bonds
iii) Oxygen exists in two allotropic forms
iv) Oxygen exists as a triatomic gas
a) (i) &(ii)
b) (ii) &(iii)
c) (iii) &(iv)
d) (i) &(iv)
Answer: (b) (ii) &(iii)
The correct statements are that oxygen forms hydrogen bonds and exists in two allotropic forms. Oxygen is actually paramagnetic (not diamagnetic) and exists as a diatomic gas, not a triatomic gas, although ozone (triatomic) is an allotrope. The two common allotropic forms are diatomic oxygen (\( O_2 \)) and triatomic ozone (\( O_3 \)). Oxygen does not form hydrogen bonds itself; however, water (\( H_2O \)), which contains oxygen, does form hydrogen bonds. This statement is subtly tricky and refers to the ability of oxygen *atoms* to participate in hydrogen bonding when part of molecules like water or alcohols.
In simple words: The correct option says oxygen can make hydrogen bonds and comes in two different forms. Oxygen itself is usually found as two atoms joined together, and it has magnetic properties.

🎯 Exam Tip: Be precise with chemical properties. Oxygen gas (\( O_2 \)) is paramagnetic and diatomic. Ozone (\( O_3 \)) is an allotrope. Oxygen *atoms* in compounds can participate in hydrogen bonding, but the \( O_2 \) molecule itself does not form hydrogen bonds.

 

Question 2.
i) Sulphur exists in crystalline as well as an amorphous form
ii) Rhombic sulphur has a characteristic yellow colour and composed of \( S_8 \)
iii) When heated slowly above 96° C monoclinic sulphur is converted into Rhombic sulphur
iv) At around 140°C the monoclinic sulphur melts to form mobile pale yellow liquid called X sulphur
Answer: (a) (i) &(ii)
The correct statements are (i) Sulfur exists in both crystalline and amorphous forms, and (ii) Rhombic sulfur has a characteristic yellow color and is composed of \( S_8 \) rings. Statement (iii) is incorrect because heating rhombic sulfur *above* 96°C converts it to monoclinic sulfur, not the other way around. Statement (iv) is also incorrect, as rhombic sulfur is converted to monoclinic sulfur, and then monoclinic sulfur melts around 119°C, not 140°C, to form mobile pale yellow liquid, \( S_{\lambda} \) or lambda sulfur. So the correct statements are (i) and (ii).
In simple words: Sulfur can be found in two main types: crystals or a shapeless form. The common yellow crystal type of sulfur is made of \( S_8 \) rings.

🎯 Exam Tip: Remember the allotropes of sulfur: rhombic (alpha) is stable at room temperature, converting to monoclinic (beta) above 95.6°C. Both consist of \( S_8 \) rings.

 

Question 3.
i) \( H_2SO_4 \) is a dibasic acid
ii) \( H_3PO_3 \) is a tribasic acid
iii) \( H_3PO_4 \) is a dibasic acid
iv) \( H_3PO_2 \) is a monobasic acid
a) (i) &(ii)
b) (ii) &(iii)
c) (iii) &(iv)
d) (i) &(iv)
Answer: (d) (i) &(iv)
The correct statements are (i) \( H_2SO_4 \) is a dibasic acid, meaning it can donate two protons, and (iv) \( H_3PO_2 \) is a monobasic acid, meaning it can donate one proton. Statement (ii) is incorrect because \( H_3PO_3 \) (phosphorous acid) is a dibasic acid, not tribasic, as it has two P-OH bonds. Statement (iii) is incorrect because \( H_3PO_4 \) (orthophosphoric acid) is a tribasic acid, not dibasic, as it has three P-OH bonds. The basicity depends on the number of dissociable hydrogen atoms directly attached to oxygen.
In simple words: The correct options are that \( H_2SO_4 \) can give away two hydrogen ions, and \( H_3PO_2 \) can give away one hydrogen ion. These numbers tell us how "strong" the acid can act in a reaction.

🎯 Exam Tip: Basicity of oxyacids of phosphorus (like \( H_3PO_3 \), \( H_3PO_4 \), \( H_3PO_2 \)) is determined by the number of hydrogen atoms directly bonded to oxygen (P-OH bonds), not the total number of hydrogen atoms.

 

Question 4.
i) Krypton is used in cryogenics.
ii) Neon is used in high-speed electronic flashbulbs used by photographers.
iii) Helium is used to provide an inert atmosphere in electric arc welding of metals.
iv) Radon is used as a source of gamma rays.
a) (i) &(ii)
b) (ii) &(iii)
c) (iii) &(iv)
d) (i) &(iv)
Answer: (c) (iii) &(iv)
The correct statements are (iii) Helium is used to provide an inert atmosphere in electric arc welding of metals, and (iv) Radon is used as a source of gamma rays, especially in medical treatments like radiotherapy. Statement (i) is incorrect; liquid helium is used in cryogenics, not krypton. Statement (ii) is also incorrect; Xenon is used in high-speed electronic flashbulbs, not neon. Neon is commonly used in advertising signs for its distinctive glow.
In simple words: The true facts are that helium helps keep things from reacting during welding, and radon is used to make gamma rays. Krypton is not used in cryogenics, and neon is not used in high-speed flashbulbs.

🎯 Exam Tip: Know the specific uses of each noble gas. Helium is for cryogenics/inert atmosphere, neon for signs, argon for inert atmosphere, krypton for some special lamps, xenon for flashbulbs, and radon for radiotherapy.

 

IV. Pick Out the Incorrect Statement

 

Question 1.
i) In inter halogen compounds the central atom will be the smaller halogen
ii) Interhalogen compounds can be formed only between two halogen atoms.
iii) Flourine can act as a central atom.
iv) Interhalogens are strong oxidising agents
a) (i)
b) (i) &(iii)
c) (ii) &(iii)
d) (i) &(iv)
Answer: (b) (i) &(iii)
The incorrect statements are (i) and (iii). Statement (i) is incorrect because in interhalogen compounds, the central atom is typically the *larger* halogen, as it can accommodate more smaller halogen atoms around it. Statement (iii) is incorrect because fluorine, being the most electronegative and smallest halogen, *cannot* act as a central atom in interhalogen compounds; it always acts as the terminal atom. Statements (ii) and (iv) are correct; interhalogen compounds are formed between two different halogens (e.g., \( ClF_3 \), \( BrF_5 \)) and they are generally strong oxidizing agents due to the difference in electronegativity.
In simple words: The wrong statements are: the central atom in interhalogen compounds is always the smaller one, and fluorine can be a central atom. Actually, the central atom is usually the bigger halogen, and fluorine is always on the outside because it's so small and reactive.

🎯 Exam Tip: Remember that in interhalogen compounds, the larger, less electronegative halogen acts as the central atom, while the smaller, more electronegative halogen(s) act as terminal atoms.

 

Question 2.
i) Nitrogen reacts with group 2 metals to form ionic nitrides.
ii) Ammonia is less soluble in water.
iii) Liquid nitrogen is used in biological preservation.
iv) In the conversion of metal oxides to metal ammonia acts as an oxidising agent.
a) (i) &(ii)
b) (ii) &(iii)
c) (i) &(iii)
d) (ii) &(iv)
Answer: (d) (ii) &(iv)
The incorrect statements are (ii) and (iv). Statement (ii) is incorrect because ammonia is *extremely soluble* in water due to its ability to form hydrogen bonds with water molecules. Statement (iv) is incorrect because in the conversion of metal oxides to metals, ammonia acts as a *reducing agent* (it removes oxygen), not an oxidizing agent. Statements (i) and (iii) are correct: nitrogen reacts with Group 2 metals (like Mg, Ca) to form ionic nitrides (e.g., \( Mg_3N_2 \)), and liquid nitrogen is widely used for biological preservation (cryopreservation) due to its very low temperature.
In simple words: The wrong statements are that ammonia does not dissolve well in water and that ammonia helps metals become metal oxides by taking oxygen. Actually, ammonia dissolves very well in water, and it helps remove oxygen from metal oxides.

🎯 Exam Tip: Ammonia's high solubility in water and its reducing properties are key characteristics. Remember that hydrogen bonding makes ammonia highly soluble, and it reduces metal oxides by donating hydrogen to form water or nitrogen gas.

 

Question 3.
i) When reacted with metals, nitric acid liberates hydrogen
ii) Chromium when reacted with nitric acid becomes passive due to the formation of nitrate on its surface.
iii) In most of the reactions nitric acid acts as an oxidising agent
iv) Fuming nitric acid contains oxides of nitrogen
a) (i) &(ii)
b) (ii) &(iii)
c) (ii) &(iv)
d) (i) &(iv)
Answer: (a) (i) &(ii)
The incorrect statements are (i) and (ii). Statement (i) is incorrect because nitric acid, being a strong oxidizing agent, rarely liberates hydrogen when reacting with metals. Instead, it typically produces various nitrogen oxides. Statement (ii) is incorrect; chromium does become passive with nitric acid, but this is due to the formation of a protective *oxide* layer on its surface, not a nitrate layer. Statements (iii) and (iv) are correct: nitric acid primarily acts as an oxidizing agent in most reactions, and fuming nitric acid is known to contain dissolved oxides of nitrogen, which give it a yellowish-brown color.
In simple words: The false statements are: nitric acid makes hydrogen gas when it reacts with metals, and chromium gets a protective nitrate layer from nitric acid. In reality, nitric acid usually makes other nitrogen gases, and chromium gets a protective oxide layer.

🎯 Exam Tip: Nitric acid's strong oxidizing nature prevents hydrogen liberation with most metals. The passivation of metals like Cr, Al, Fe by nitric acid is due to the formation of a thin, non-reactive *oxide* film on their surface.

 

Question 4.
i) The rate of decomposition of ozone increases sharply in alkaline solution.
ii) In acidic solution ozone exceeds the oxidising power of fluorine and atomic oxygen
iii) Considerable amount of ozone is formed in the upper atmosphere by the action of UV light
iv) The shape of the Ozone molecule is linear
a) (i) &(ii)
b) (ii) &(iii)
c) (iii) &(iv)
d) (i) &(iv)
Answer: (d) (i) &(iv)
The incorrect statements are (i) and (iv). Statement (i) is incorrect because the rate of decomposition of ozone *drops* sharply in alkaline solutions, it does not increase. Statement (iv) is incorrect; the shape of the ozone molecule (\( O_3 \)) is *bent*, not linear, due to the presence of lone pairs on the central oxygen atom. Statements (ii) and (iii) are correct: ozone is a very strong oxidizing agent, and in acidic solutions, its oxidizing power can even exceed that of fluorine and atomic oxygen. Also, a considerable amount of ozone is indeed formed in the upper atmosphere when oxygen molecules absorb ultraviolet (UV) light.
In simple words: The false statements are that ozone breaks down quickly in basic water and that its shape is straight. Actually, ozone breaks down slower in basic water and has a bent shape.

🎯 Exam Tip: Remember that ozone is a bent molecule (not linear) and its decomposition rate is influenced by pH. Its powerful oxidizing nature is also a key property.

 

V. Pick Out the Odd Man Out

 

Question 1. w.r.t oxidation number pick the odd man out.
a) \( HPO_3 \)
b) \( H_3PO_3 \)
c) \( H_3PO_4 \)
d) \( H_4P_2O_7 \)
Answer: (b) \( H_3PO_3 \)
In \( HPO_3 \) (metaphosphoric acid), the oxidation number of phosphorus is +5. In \( H_3PO_4 \) (orthophosphoric acid), it is +5. In \( H_4P_2O_7 \) (pyrophosphoric acid), it is also +5. However, in \( H_3PO_3 \) (phosphorous acid), the oxidation number of phosphorus is +3. Therefore, \( H_3PO_3 \) is the odd one out because its phosphorus has a different oxidation state compared to the others. This difference arises from its structure with one P-H bond.
In simple words: When we look at how many electrons phosphorus seems to lose or gain in these chemicals, \( H_3PO_3 \) is different. In most of them, phosphorus has a +5 charge, but in \( H_3PO_3 \), it only has a +3 charge.

🎯 Exam Tip: To find the oxidation number of phosphorus in oxyacids, remember to consider that oxygen usually has an oxidation number of -2 and hydrogen is +1. For phosphorus acids, specifically note if there are P-H bonds, as those hydrogens do not contribute to the basicity or the calculation in the same way as P-OH hydrogens.

 

Question 2. w.r.t reaction with sulphuric acid pick the odd man out
a) Gold
b) Silver
c) Platinum
d) Copper
Answer: (d) Copper
Copper is the odd one out because it reacts with sulphuric acid, especially concentrated sulphuric acid, to produce copper sulphate and sulphur dioxide. Gold, silver, and platinum are all noble metals and generally do not react with sulphuric acid under normal conditions. This difference in reactivity is due to their position in the electrochemical series. Noble metals have very low reactivity.
In simple words: Copper is the different one here because it mixes with sulfuric acid. Gold, silver, and platinum metals do not react with sulfuric acid.

🎯 Exam Tip: Remember that noble metals like gold, silver, and platinum are largely unreactive with common acids, including sulphuric acid, which is a key distinction from more reactive metals like copper.

 

Question 3. w.r.t reactivity pick the odd man out
a) \( F_2 \)
b) \( Cl_2 \)
c) \( Br_2 \)
d) \( I_2 \)
Answer: (a) \( F_2 \)
Fluorine (\( F_2 \)) is the odd one out because it is the most reactive of all halogens. Its high reactivity is due to the very low \( F-F \) bond dissociation energy and its high electronegativity. Chlorine (\( Cl_2 \)), bromine (\( Br_2 \)), and iodine (\( I_2 \)) are progressively less reactive as you go down the group. This trend explains why fluorine always exists in a -1 oxidation state in compounds.
In simple words: Fluorine is the standout here because it is the most eager to react among all the listed elements. It has a weaker bond within itself and likes to grab electrons a lot.

🎯 Exam Tip: Remember the trend of reactivity for halogens: Fluorine is the most reactive, and reactivity decreases down the group from \( F_2 \) to \( I_2 \).

 

Question 4. w.r.t the ability to form oxoacids pick the odd man out
a) fluorine
b) chlorine
c) bromine
d) iodine
Answer: (a) Fluorine
Fluorine is the odd one out because it forms only one oxoacid, hypofluorous acid (HOF). All other halogens (chlorine, bromine, and iodine) can form multiple oxoacids, such as hypohalous, halous, halic, and perhalic acids, because they have available d-orbitals for expanded valence shells. Fluorine lacks these d-orbitals, limiting its ability to form various oxoacids. This unique property sets it apart from other halogens.
In simple words: Fluorine is the different one because it can only make one type of acid with oxygen (an oxoacid). The other elements listed can make many different types of acids with oxygen.

🎯 Exam Tip: Remember that fluorine forms only one oxoacid (HOF) due to the absence of d-orbitals in its valence shell, preventing it from showing higher oxidation states.

 

VI. Choose the Best Answer.

 

Question 1. The principal gas present in atmosphere is
a) \( O_2 \)
b) \( N_2 \)
c) \( H_2 \)
d) \( CO_2 \)
Answer: (b) \( N_2 \)
The principal gas present in the atmosphere is nitrogen (\( N_2 \)). It makes up approximately 78% of the Earth's atmosphere. Oxygen (\( O_2 \)) is the second most abundant at about 21%, while carbon dioxide (\( CO_2 \)) and hydrogen (\( H_2 \)) are present in much smaller amounts. Nitrogen's abundance is critical for various biological and chemical processes.
In simple words: The main gas in the air we breathe is nitrogen. It makes up most of our atmosphere.

🎯 Exam Tip: Remember the approximate percentages of major gases in Earth's atmosphere: Nitrogen (78%), Oxygen (21%), Argon (0.9%), Carbon dioxide (0.04%).

 

Question 2. The basicity of hypophosphorous acid is (PTA - 2)
a) 1
b) 2
c) 3
d) 4
Answer: (a) 1
The basicity of hypophosphorous acid (\( H_3PO_2 \)) is 1. This means it can donate only one proton (\( H^+ \)) in a reaction. Although it has three hydrogen atoms, only one is directly bonded to an oxygen atom (P-OH), making it dissociable. The other two hydrogens are directly bonded to phosphorus (P-H) and are not acidic. Understanding its structure helps determine its basicity.
In simple words: Hypophosphorous acid can give away only one hydrogen ion in a reaction, so its basicity is 1.

🎯 Exam Tip: For oxyacids of phosphorus, basicity is determined by the number of P-OH bonds, not the total number of hydrogen atoms. P-H bonds are generally non-acidic.

 

Question 3. Chile salt petre is
a) \( NaNO_2 \)
b) \( NaNO_3 \)
c) \( KNO_2 \)
d) \( KNO_3 \)
Answer: (b) \( NaNO_3 \)
Chile saltpetre is the common name for sodium nitrate (\( NaNO_3 \)). It is a naturally occurring mineral found in large deposits in Chile, hence the name. It is primarily used as a fertilizer and in the production of gunpowder and other chemicals. Nitrites and other nitrates are different compounds with distinct uses.
In simple words: Chile saltpetre is just another name for sodium nitrate, which is a chemical with the formula \( NaNO_3 \).

🎯 Exam Tip: Distinguish between common names and chemical formulas for important compounds like Chile saltpetre (\( NaNO_3 \)) and Indian saltpetre (\( KNO_3 \)).

 

Question 4. Indian salt petre is
a) \( NaNO_2 \)
b) \( NaNO_3 \)
c) \( KNO_2 \)
d) \( KNO_3 \)
Answer: (d) \( KNO_3 \)
Indian saltpetre is the common name for potassium nitrate (\( KNO_3 \)). It is historically significant and has been used in various applications, including as a component of gunpowder and as a fertilizer. It is distinct from sodium nitrate (Chile saltpetre) due to the presence of potassium instead of sodium. Its chemical properties make it valuable in many industries.
In simple words: Indian saltpetre is the common name for the chemical potassium nitrate, which has the formula \( KNO_3 \).

🎯 Exam Tip: Make sure to differentiate between "Chile saltpetre" (\( NaNO_3 \)) and "Indian saltpetre" (\( KNO_3 \)) as they are common distractors in questions.

 

Question 5. Inert character of nitrogen is due to its
a) high ionisation energy
b) low electro negativity
c) high bonding energy
d) low bonding energy
Answer: (c) high bonding energy
The inert character of nitrogen gas (\( N_2 \)) is primarily due to its high bonding energy. Nitrogen molecules have a very strong triple bond between the two nitrogen atoms (\( N \equiv N \)). This triple bond requires a large amount of energy to break, making nitrogen relatively unreactive at room temperature. While nitrogen has high ionization energy and moderate electronegativity, the strength of its triple bond is the main reason for its inertness. This strong bond contributes to its stability.
In simple words: Nitrogen gas doesn't react easily because the two nitrogen atoms are held together very tightly by a strong triple bond, which needs a lot of energy to break.

🎯 Exam Tip: Remember that the inertness of diatomic nitrogen (\( N_2 \)) is directly linked to the high bond dissociation energy of its triple bond, which makes it stable and less reactive.

 

Question 9. Urea on hydrolysis gives
(a) \( \text{NO}_2 \)
(b) \( \text{HNO}_3 \)
(c) \( \text{NH}_3 \)
(d) \( \text{N}_2\text{O} \)
Answer: (c) \( \text{NH}_3 \)
In simple words: When urea reacts with water, it breaks down and produces ammonia gas. This reaction is common in biological processes.

🎯 Exam Tip: Remember that hydrolysis means a compound reacts with water to break down. For urea, this process forms ammonia, a key compound in the nitrogen cycle.

 

Question 10. The catalyst used in Haber's process is
(a) Ni
(b) Fe
(c) Co
(d) Pt
Answer: (b) Fe
In simple words: In the Haber process, iron is used as a special helper to speed up the reaction that makes ammonia. It helps nitrogen and hydrogen combine faster.

🎯 Exam Tip: The Haber process uses finely divided iron as a catalyst, often with molybdenum or aluminum oxide as promoters, to increase the yield of ammonia.

 

Question 11. The smell of ammonia is
(a) rotten egg
(b) rotten fish
(c) pungent
(d) garlic
Answer: (c) pungent
In simple words: Ammonia has a very strong, sharp, and irritating smell. Many cleaning products use ammonia, and its scent is easily recognizable.

🎯 Exam Tip: Describing smells in chemistry often involves using sensory words. "Pungent" is a common term for strong, sharp, and unpleasant odors, fitting for ammonia.

 

Question 12. Like water, ammonia is a fairly good ionising solvent, because its dielectric constant is
(a) considerably low
(b) considerably high
(c) equal to zero
Answer: (b) considerably high
In simple words: Ammonia is good at separating charged particles, just like water, because it has a high dielectric constant. This means it can easily dissolve many ionic compounds.

🎯 Exam Tip: A high dielectric constant indicates a solvent's ability to reduce the electrostatic forces between dissolved ions, thereby facilitating their dissolution and ionization.

 

Question 13. The process used for the manufacture of nitric acid is known as
(a) Haber's process
(b) Deacon's process
(c) contact process
(d) Ostwald's process
Answer: (d) Ostwald's process
In simple words: Nitric acid is made using the Ostwald process. This method involves several steps to turn ammonia into nitric acid.

🎯 Exam Tip: The Ostwald process starts with the catalytic oxidation of ammonia to nitric oxide, which is then further oxidized to nitrogen dioxide and finally absorbed in water to form nitric acid.

 

Question 14. With excess of chlorine, ammonia reacts to give an explosive substance
(a) \( \text{N}_2 \)
(b) \( \text{NH}_4\text{NO}_3 \)
(c) \( \text{NH}_4\text{Cl} \)
(d) \( \text{NCl}_3 \)
Answer: (d) \( \text{NCl}_3 \)
In simple words: When there's a lot of chlorine, it reacts with ammonia to create nitrogen trichloride, which is a very unstable and explosive compound. This reaction needs to be handled with extreme care.

🎯 Exam Tip: Remember that the ratio of reactants is critical in chemical reactions. Excess chlorine reacting with ammonia leads to the formation of highly explosive nitrogen trichloride, \( \text{NCl}_3 \).

 

Question 15. The deep blue colour compound formed when excess of ammonia is added to aqueous solution of copper sulphate is
(a) \( \text{Cu(NO}_3\text{)}_2 \)
(b) \( \text{Cu(NH}_3\text{)}_2^{2+} \)
(c) \( \text{Cu(NH}_3\text{)}_4^{2+} \)
(d) \( \text{Cu(NH}_3\text{)}_2^{+} \)
Answer: (c) \( \text{Cu(NH}_3\text{)}_4^{2+} \)
In simple words: When you add a lot of ammonia to copper sulphate solution, it makes a special blue compound called tetraamminecopper(II) ion. This compound has a very striking deep blue color.

🎯 Exam Tip: The formation of the deep blue \( \text{[Cu(NH}_3\text{)}_4]^{2+} \) complex ion is a classic qualitative test for copper(II) ions in the presence of excess ammonia.

 

Question 16. The shape of ammonia molecule is
(a) linear
(b) pyramidal
(c) square planar
(d) octahedral
Answer: (b) pyramidal
In simple words: The ammonia molecule has a pyramid-like shape. This is because the central nitrogen atom has three bonds to hydrogen atoms and one lone pair of electrons, pushing the hydrogen atoms into a triangular base.

🎯 Exam Tip: The pyramidal shape of ammonia (and other molecules with three bond pairs and one lone pair) arises from the \( \text{sp}^3 \) hybridization of the central atom and the repulsion caused by the lone pair of electrons.

 

Question 17. The bond angle in ammonia is
(a) \( 104^{\circ} \)
(b) \( 104^{\circ}28' \)
(c) \( 107^{\circ} \)
(d) \( 180^{\circ} \)
Answer: (c) \( 107^{\circ} \)
In simple words: The hydrogen atoms in an ammonia molecule are bent at an angle of about \( 107^{\circ} \). This is slightly less than the ideal tetrahedral angle because of the lone pair of electrons pushing the bonds closer.

🎯 Exam Tip: Lone pair-bond pair repulsion is stronger than bond pair-bond pair repulsion, causing the bond angles in molecules like ammonia to be slightly smaller than the ideal tetrahedral angle of \( 109.5^{\circ} \).

 

Question 18. The colour of Pure nitric acid is
(a) colourless
(b) brown
(c) pale green
(d) green
Answer: (a) colourless
In simple words: Pure nitric acid, when freshly made, has no color. It looks clear like water. However, it can change color over time.

🎯 Exam Tip: While pure nitric acid is colourless, it often appears yellowish or brownish due to the decomposition of \( \text{HNO}_3 \) into nitrogen dioxide \( (\text{NO}_2) \) when exposed to light, with the \( \text{NO}_2 \) dissolving in the acid.

 

Question 19. Fuming nitric acid contains oxides of
(a) sulphur
(b) hydrogen
(c) nitrogen
(d) carbon
Answer: (c) nitrogen
In simple words: Fuming nitric acid contains extra nitrogen oxides dissolved in it. These oxides are what give it its fuming property and a reddish-brown color.

🎯 Exam Tip: Fuming nitric acid is a highly concentrated form of nitric acid that contains dissolved nitrogen dioxide \( (\text{NO}_2) \), which makes it release reddish-brown fumes when exposed to air.

 

Question 20. Nitric acid can act as
(a) an acid
(b) an oxidising agent
(c) nitrating agent
(d) All of the options
Answer: (d) All of the options
In simple words: Nitric acid can do many things: it acts as a strong acid, it helps other substances lose electrons (oxidizing agent), and it can add a nitro group to organic compounds (nitrating agent). It's a very versatile chemical.

🎯 Exam Tip: Nitric acid is known for its strong acidic nature, powerful oxidizing properties (especially when concentrated), and its ability to act as a nitrating agent in organic reactions.

 

Question 21. The formula of hyponitrous acid is (MARCH 2020)
(a) \( \text{H}_2\text{N}_2\text{O}_2 \)
(b) \( \text{H}_4\text{N}_2\text{O}_4 \)
(c) HOONO
(d) \( \text{HNO}_2 \)
Answer: (a) \( \text{H}_2\text{N}_2\text{O}_2 \)
In simple words: Hyponitrous acid has the chemical formula \( \text{H}_2\text{N}_2\text{O}_2 \). This is a different acid of nitrogen compared to more common ones like nitric acid.

🎯 Exam Tip: Be careful with the different oxoacids of nitrogen, as they have similar-sounding names but distinct formulas and oxidation states. Hyponitrous acid is an example with a lower oxidation state for nitrogen.

 

Question 22. The oxidising power of oxo acids follows the order
(a) \( \text{HOX} > \text{HXO}_2 > \text{HXO}_3 > \text{HXO}_4 \)
(b) \( \text{HXO}_4 > \text{HXO}_3 > \text{HXO}_2 > \text{HOX} \)
(c) \( \text{HXO}_3 > \text{HXO}_4 > \text{HXO}_2 > \text{HOX} \)
(d) \( \text{HOX} > \text{HXO}_4 > \text{HXO}_3 > \text{HXO}_2 \)
Answer: (a) \( \text{HOX} > \text{HXO}_2 > \text{HXO}_3 > \text{HXO}_4 \)
In simple words: The ability of oxoacids to oxidize other substances becomes weaker as more oxygen atoms are added. So, an acid with fewer oxygen atoms is a stronger oxidizer than one with many oxygen atoms.

🎯 Exam Tip: For oxoacids of halogens (like chlorine, bromine, iodine), the oxidizing power decreases as the number of oxygen atoms increases due to the increasing stability of the central halogen in higher oxidation states.

 

Question 23. White phosphorous is kept under
(a) kerosene
(b) water
(c) alcohol
(d) ether
Answer: (b) water
In simple words: White phosphorus is kept submerged in water. This is because it reacts very easily with air and catches fire quickly, but it does not react with water.

🎯 Exam Tip: White phosphorus is highly reactive with atmospheric oxygen and is pyrophoric (spontaneously ignites in air). Storing it under water prevents contact with air, as it is insoluble in water and does not react with it.

 

Question 24. White phosphorous becomes yellow phosphorous due to
(a) hydrolysis
(b) reduction
(c) oxidation
(d) displacement
Answer: (c) oxidation
In simple words: White phosphorus slowly turns yellow when it reacts with air. This change happens because of a slow oxidation process.

🎯 Exam Tip: The conversion of white phosphorus to yellow phosphorus is a slow oxidation process, which forms a thin layer of oxides on its surface, giving it a yellowish tint.

 

Question 25. In the conversion of yellow phosphorous into phosphine, phosphorous acts as
(a) oxidising agent
(b) reducing agent
(c) catalyst
(d) hydrolysing agent
Answer: (b) reducing agent
In simple words: When yellow phosphorus is turned into phosphine, the phosphorus itself helps another substance gain electrons, meaning it acts as a reducing agent.

🎯 Exam Tip: In this reaction, phosphorus undergoes disproportionation, where it is both oxidized and reduced. However, its primary role in forming phosphine from an initial higher oxidation state often involves it acting as a reducing agent.

 

Question 26. In the conversion of phosphorous into orthophosphoric acid, the catalyst used is
(a) \( \text{Cl}_2 \)
(b) \( \text{Br}_2 \)
(c) \( \text{I}_2 \)
(d) \( \text{F}_2 \)
Answer: (c) \( \text{I}_2 \)
In simple words: When phosphorus is changed into orthophosphoric acid, iodine is used as a helper chemical to make the reaction happen. It speeds up the process.

🎯 Exam Tip: The oxidation of phosphorus to orthophosphoric acid typically uses concentrated nitric acid as an oxidizing agent, with a small amount of iodine often added as a catalyst to accelerate the reaction.

 

Question 27. Which is used in match boxes?
(a) White phosphorous
(b) Red phosphorous
(c) Black phosphorous
(d) Scarlet phosphorous
Answer: (b) Red phosphorous
In simple words: Red phosphorus is the type of phosphorus found on the side of matchboxes. It is safer to use than white phosphorus because it is not as reactive.

🎯 Exam Tip: Red phosphorus is used in matchboxes because it is less reactive and less toxic than white phosphorus, making it safe for everyday use. It ignites upon friction with an oxidizing agent like potassium chlorate.

 

Question 28. The acid having O-O bond in its structure (PTA - 6)
(a) \( \text{H}_2\text{SO}_3 \)
(b) \( \text{H}_2\text{S}_2\text{O}_6 \)
(c) \( \text{H}_2\text{S}_2\text{O}_8 \)
(d) \( \text{H}_2\text{S}_4\text{O}_6 \)
Answer: (c) \( \text{H}_2\text{S}_2\text{O}_8 \)
In simple words: Persulphuric acid, which is \( \text{H}_2\text{S}_2\text{O}_8 \), has a special bond between two oxygen atoms directly. This is called a peroxide linkage.

🎯 Exam Tip: Oxoacids containing a peroxide (O-O) bond are known as peroxoacids. \( \text{H}_2\text{S}_2\text{O}_8 \), also called Marshall's acid, is a common example with a direct oxygen-oxygen linkage.

 

Question 29. The smell of phosphine is
(a) rotten egg
(b) rotten fish
(c) pungent
(d) garlic
Answer: (b) rotten fish
In simple words: Phosphine gas has a very unpleasant smell that is often compared to the smell of decaying fish. This strong, distinctive odor helps in detecting its presence.

🎯 Exam Tip: Phosphine \( (\text{PH}_3) \) is a highly toxic gas known for its characteristic "rotten fish" odor. This smell is a crucial identifier in laboratory settings.

 

Question 30. The compounds used in Holme's signal are
(a) \( \text{CaC}_2 \) & \( \text{Ca}_3\text{P}_2 \)
(b) AlP & \( \text{Ca}_3\text{P}_2 \)
(c) \( \text{CaC}_2 \) & \( \text{P}_4 \)
(d) AlP & \( \text{P}_4 \)
Answer: (a) \( \text{CaC}_2 \) & \( \text{Ca}_3\text{P}_2 \)
In simple words: Holme's signal uses a mix of calcium carbide and calcium phosphide. When these chemicals react with water, they produce gases that catch fire, creating a bright signal.

🎯 Exam Tip: Holme's signal is a pyrotechnic device used by ships as a distress signal. It works by releasing acetylene \( (\text{C}_2\text{H}_2) \) from \( \text{CaC}_2 \) and phosphine \( (\text{PH}_3) \) from \( \text{Ca}_3\text{P}_2 \) upon contact with water, with phosphine igniting spontaneously and igniting acetylene, producing a bright flare.

 

Question 31. The gases liberated in Holme's signal are
(a) \( \text{C}_2\text{H}_2 \) & \( \text{CH}_4 \)
(b) \( \text{C}_2\text{H}_2 \) & \( \text{PH}_3 \)
(c) \( \text{C}_2\text{H}_4 \) & \( \text{PH}_3 \)
(d) \( \text{CH}_4 \) & \( \text{PH}_3 \)
Answer: (b) \( \text{C}_2\text{H}_2 \) & \( \text{PH}_3 \)
In simple words: Holme's signal releases two main gases: acetylene and phosphine. These gases are responsible for the fire and smoke seen as a signal.

🎯 Exam Tip: Calcium carbide \( (\text{CaC}_2) \) reacts with water to produce acetylene \( (\text{C}_2\text{H}_2) \), and calcium phosphide \( (\text{Ca}_3\text{P}_2) \) reacts with water to produce phosphine \( (\text{PH}_3) \). Phosphine's spontaneous ignition then lights the acetylene.

 

Question 32. The formula of pyrophosphoric acid is
(a) \( \text{H}_4\text{P}_2\text{O}_6 \)
(b) \( \text{H}_4\text{P}_2\text{O}_7 \)
(c) \( \text{H}_3\text{PO}_2 \)
(d) \( \text{H}_3\text{PO}_3 \)
Answer: (b) \( \text{H}_4\text{P}_2\text{O}_7 \)
In simple words: Pyrophosphoric acid has the chemical formula \( \text{H}_4\text{P}_2\text{O}_7 \). It's formed when two molecules of orthophosphoric acid lose one molecule of water.

🎯 Exam Tip: Pyrophosphoric acid is an example of a polyphosphoric acid, formed by the condensation of two phosphoric acid molecules. Its structure involves a P-O-P linkage.

 

Question 33. Thermodynamically stable allotrophic form of sulphur is
(a) Rhombic sulphur
(b) Monoclinic sulphur
(c) Plastic sulphur
(d) Colloidal sulphur
Answer: (a) Rhombic sulphur
In simple words: Rhombic sulphur is the most stable form of sulphur at normal temperatures. It is the type of sulphur that you usually find.

🎯 Exam Tip: Rhombic sulphur (alpha-sulphur) is stable below \( 95.6^{\circ}\text{C} \) and is the most common and stable allotrope. Monoclinic sulphur (beta-sulphur) is stable above this temperature.

 

Question 34. The gas found in volcanic eruptions is
(a) \( \text{NO}_2 \)
(b) NO
(c) \( \text{SO}_2 \)
(d) \( \text{SO}_3 \)
Answer: (c) \( \text{SO}_2 \)
In simple words: Sulfur dioxide is a common gas that comes out of volcanoes. It smells sharp and can affect the air.

🎯 Exam Tip: Volcanic gases primarily consist of water vapor, carbon dioxide, and sulfur dioxide. \( \text{SO}_2 \) is a major component and a significant contributor to volcanic smog and acid rain.

 

Question 35. The hybridisation of sulphur in \( \text{SO}_2 \) is
(a) sp
(b) \( \text{sp}^2 \)
(c) \( \text{sp}^3 \)
(d) \( \text{dsp}^2 \)
Answer: (b) \( \text{sp}^2 \)
In simple words: In sulfur dioxide, the central sulfur atom uses \( \text{sp}^2 \) hybrid orbitals to form its bonds. This hybridization helps give the molecule its bent shape.

🎯 Exam Tip: To determine hybridization, count the number of sigma bonds and lone pairs around the central atom. For \( \text{SO}_2 \), sulfur has two sigma bonds and one lone pair, totaling three electron domains, leading to \( \text{sp}^2 \) hybridization and a bent molecular geometry.

 

Question 36. The gas liberated when dilute sulphuric acid reacts with metals is
(a) \( \text{SO}_2 \)
(b) \( \text{SO}_3 \)
(c) \( \text{H}_2 \)
(d) \( \text{O}_2 \)
Answer: (c) \( \text{H}_2 \)
In simple words: When a weak sulfuric acid reacts with metals, it usually produces hydrogen gas. This happens as the metal replaces the hydrogen in the acid.

🎯 Exam Tip: Dilute sulphuric acid reacts with most reactive metals (those above hydrogen in the reactivity series) to produce hydrogen gas and a metal sulfate. Stronger oxidizing agents like concentrated sulphuric acid yield \( \text{SO}_2 \) instead of \( \text{H}_2 \).

 

Question 37. The gas liberated when cone, sulphuric acid reacts with metals is
(a) \( \text{SO}_2 \)
(b) \( \text{SO}_3 \)
(c) \( \text{H}_2 \)
(d) \( \text{O}_2 \)
Answer: (a) \( \text{SO}_2 \)
In simple words: When strong sulfuric acid reacts with metals, it releases sulfur dioxide gas. This shows that concentrated sulfuric acid is a strong oxidizer.

🎯 Exam Tip: Concentrated sulphuric acid is a powerful oxidizing agent. When it reacts with metals, it gets reduced, typically producing sulfur dioxide \( (\text{SO}_2) \), while the metal is oxidized.

 

Question 38. When sulphuric acid reacts with barium chloride solution, the white precipitate formed is
(a) \( \text{PbSO}_4 \)
(b) \( \text{BaSO}_4 \)
(c) \( \text{(CH}_3\text{COO)}_2\text{SO}_4 \)
Answer: (b) \( \text{BaSO}_4 \)
In simple words: When sulfuric acid mixes with barium chloride, it forms a white solid that doesn't dissolve. This solid is barium sulfate. This reaction is often used to test for sulfate ions.

🎯 Exam Tip: The formation of a white precipitate of barium sulfate \( (\text{BaSO}_4) \) upon adding barium chloride solution to an unknown solution is a characteristic test for the presence of sulfate ions \( (\text{SO}_4^{2-}) \).

 

Question 39. The halogen which exists as a liquid is
(a) flourine
(b) chlorine
(c) bromine
(d) iodine
Answer: (c) bromine
In simple words: Bromine is the only halogen element that is a liquid at room temperature. It has a reddish-brown color.

🎯 Exam Tip: Recall the physical states of halogens at room temperature: fluorine \( (\text{F}_2) \) and chlorine \( (\text{Cl}_2) \) are gases, bromine \( (\text{Br}_2) \) is a liquid, and iodine \( (\text{I}_2) \) and astatine \( (\text{At}_2) \) are solids.

 

Question 40. The halogen which exists as a solid is
(a) flourine
(b) chlorine
(c) bromine
(d) iodine
Answer: (d) iodine
In simple words: Iodine is a halogen that exists as a solid at normal temperatures. It forms dark purple crystals.

🎯 Exam Tip: As you go down Group 17 (halogens), the melting and boiling points increase, leading to fluorine and chlorine being gases, bromine a liquid, and iodine a solid at room temperature.

 

Question 41. Chlorine is manufactured by
(a) Haber's process
(b) Deacon's process
(c) Contact process
(d) Ostwald's process
Answer: (b) Deacon's process
In simple words: The Deacon's process is a way to make chlorine gas. It uses a special chemical reaction to produce chlorine from hydrogen chloride and oxygen.

🎯 Exam Tip: The Deacon's process involves the catalytic oxidation of hydrogen chloride (HCl) gas with atmospheric oxygen, usually using \( \text{CuCl}_2 \) as a catalyst, to produce chlorine gas.

 

Question 43. Aqua regia is a mixture of cone. HCl and cone. \( \text{HNO}_3 \) in the ratio
(a) 1:3
(b) 3:1
(c) 2:3
(d) 3:2
Answer: (b) 3:1
In simple words: Aqua regia is a strong mixture made by combining three parts of concentrated hydrochloric acid with one part of concentrated nitric acid. This mix is known for dissolving noble metals like gold.

🎯 Exam Tip: Aqua regia is highly corrosive and is notable for its ability to dissolve gold and platinum, which are unreactive to most individual acids. The 3:1 ratio of HCl to \( \text{HNO}_3 \) is crucial for its unique properties.

 

Question 44. The halogen acid which forms hydrogen bond is
(a) HF
(b) HCl
(c) HBr
(d) HI
Answer: (a) HF
In simple words: Hydrogen fluoride \( (\text{HF}) \) is the only halogen acid that forms strong hydrogen bonds. This happens because fluorine is very electronegative, pulling electrons away from hydrogen.

🎯 Exam Tip: Hydrogen bonding occurs when hydrogen is directly bonded to a highly electronegative atom like fluorine (F), oxygen (O), or nitrogen (N). Among the hydrogen halides, only HF exhibits significant hydrogen bonding.

 

Question 45. Among halogen acids, the strongest bond is present in
(a) HF
(b) HCl
(c) HBr
(d) HI
Answer: (a) HF
In simple words: Out of all the halogen acids, the bond in hydrogen fluoride \( (\text{HF}) \) is the strongest. This is because fluorine is a small atom and forms a very tight bond with hydrogen.

🎯 Exam Tip: Bond strength generally decreases down a group due to increasing atomic size and bond length. Thus, the H-F bond in HF is the strongest among hydrogen halides, despite HF being the weakest acid among them in aqueous solution.

 

Question 46. Among halogen acids, the weakest bond is present in
(a) HF
(b) HCl
(c) HBr
(d) HI
Answer: (d) HI
In simple words: The bond in hydrogen iodide \( (\text{HI}) \) is the weakest among the halogen acids. This is because iodine is a much larger atom, making the bond with hydrogen longer and easier to break.

🎯 Exam Tip: Bond strength in hydrogen halides decreases as the size of the halogen atom increases (F to I). A weaker bond means it's easier to break, which is why HI is the strongest acid among them.

 

Question 47. Among halogen acids, the strongest acid is
(a) HF
(b) HCl
(c) HBr
(d) HI
Answer: (d) HI
In simple words: Hydrogen iodide \( (\text{HI}) \) is the strongest acid among the halogen acids. Its large size means its bond with hydrogen is easily broken, releasing hydrogen ions.

🎯 Exam Tip: Acidity of hydrogen halides increases down the group \( (\text{HF} < \text{HCl} < \text{HBr} < \text{HI}) \) because the bond strength decreases, making it easier for the H-X bond to break and release a proton.

 

Question 48. Among halogen acids, the weakest acid is
(a) HF
(b) HCl
(c) HBr
(d) HI
Answer: (a) HF
In simple words: Hydrogen fluoride \( (\text{HF}) \) is the weakest acid among the halogen acids. This is mainly because it forms very strong hydrogen bonds, which makes it harder for the acid to release its hydrogen ion.

🎯 Exam Tip: The anomalous weakness of HF as an acid, despite fluorine being the most electronegative halogen, is primarily due to its strong hydrogen bonding and the high bond dissociation energy of the H-F bond.

 

Question 49. The correct order of acid strength is
(a) \( \text{HF} > \text{HCl} > \text{HBr} > \text{HI} \)
(b) \( \text{HF} < \text{HCl} < \text{HBr} < \text{HI} \)
(c) \( \text{HF} > \text{HCl} < \text{HBr} > \text{HI} \)
(d) \( \text{HF} < \text{HCl} > \text{HBr} < \text{HI} \)
Answer: (b) \( \text{HF} < \text{HCl} < \text{HBr} < \text{HI} \)
In simple words: The strength of halogen acids increases as you go down the group from fluorine to iodine. This means \( \text{HI} \) is the strongest, and \( \text{HF} \) is the weakest.

🎯 Exam Tip: For binary acids of elements in the same group (like HX), acidity increases with increasing atomic size of the non-metal, as bond strength decreases, making proton release easier. This is why HI is the strongest and HF the weakest.

 

Question 50. Which is more reactive towards hydrogen?
(a) flourine
(b) chlorine
(c) bromine
(d) iodine
Answer: (a) flourine
In simple words: Fluorine is the most eager to react with hydrogen among all the halogens. This is because fluorine is very reactive and forms strong bonds.

🎯 Exam Tip: Reactivity of halogens generally decreases down the group. Fluorine is the most reactive halogen due to its small size and high electronegativity, readily reacting with hydrogen even in the dark.

 

Question 51. The number of bond pair and lone pair of electrons present in the interhalogen compound \( \text{BrF}_3 \) is
(a) 1 & 3
(b) 3 & 2
(c) 5 & 1
(d) 7 & 0
Answer: (b) 3 & 2
In simple words: In the compound \( \text{BrF}_3 \), the central bromine atom forms three bonds with fluorine atoms and has two pairs of electrons that are not involved in bonding. These unshared pairs affect its shape.

🎯 Exam Tip: For \( \text{BrF}_3 \), the central bromine atom has 7 valence electrons. Three electrons form single bonds with three fluorine atoms, leaving \( 7 - 3 = 4 \) electrons as two lone pairs. So, there are 3 bond pairs and 2 lone pairs, resulting in a T-shaped molecular geometry.

 

Question 52. The oxidation number of oxygen in \( \text{F}_2\text{O} \) is
(a) -2
(b) -1
(c) +2
(d) +1
Answer: (c) +2
In simple words: In the compound \( \text{F}_2\text{O} \), oxygen has an oxidation number of +2. This is unusual because oxygen is usually -2, but fluorine is more electronegative than oxygen.

🎯 Exam Tip: Fluorine is the most electronegative element, so in compounds with fluorine, fluorine always takes an oxidation state of -1. In \( \text{F}_2\text{O} \), since there are two fluorine atoms, the oxygen atom must have a +2 oxidation state to balance the charges.

 

Question 53. The oxidation number of chlorine in \( \text{Cl}_2\text{O}_7 \) is
(a) +1
(b) +5
(c) +6
(d) +7
Answer: (d) +7
In simple words: In the compound \( \text{Cl}_2\text{O}_7 \), each chlorine atom has an oxidation number of +7. This is the highest oxidation state chlorine can achieve, given it is bonded to many oxygen atoms.

🎯 Exam Tip: To calculate the oxidation number, remember that oxygen typically has an oxidation state of -2. In \( \text{Cl}_2\text{O}_7 \), let the oxidation state of chlorine be \( x \). Then \( 2x + 7(-2) = 0 \Rightarrow 2x - 14 = 0 \Rightarrow 2x = 14 \Rightarrow x = +7 \).

 

Question 54. The strongest oxidising agent among the following is
(a) chlorous acid
(b) chloric acid
(c) hypochlorous acid
(d) perchloric acid
Answer: (c) hypochlorous acid
In simple words: Hypochlorous acid is the strongest oxidizer among these chlorine acids. It has the fewest oxygen atoms, making it more willing to gain electrons.

🎯 Exam Tip: For oxoacids of a given element, the oxidizing power generally decreases as the oxidation state of the central atom increases (i.e., as the number of oxygen atoms increases). Therefore, hypochlorous acid \( (\text{HClO}) \), with chlorine in the +1 oxidation state, is the strongest oxidizing agent among the options.

 

Question 55. The first ionisation energy of noble gases is in the order
(a) \( \text{He} < \text{Ne} < \text{Ar} < \text{Kr} \)
(b) \( \text{He} > \text{Ne} > \text{Ar} > \text{Kr} \)
(c) \( \text{He} < \text{Ne} > \text{Ar} < \text{Kr} \)
(d) \( \text{He} > \text{Ne} < \text{Ar} > \text{Kr} \)
Answer: (b) \( \text{He} > \text{Ne} > \text{Ar} > \text{Kr} \)
In simple words: The energy needed to remove an electron from a noble gas gets smaller as you go down the group. This means helium needs the most energy, and krypton needs less.

🎯 Exam Tip: Ionization energy generally decreases down a group due to increasing atomic size and greater shielding effect, which makes it easier to remove valence electrons. Thus, helium has the highest ionization energy among noble gases.

 

Question 56. Among noble gases, chemical reactivity is shown by
(a) He & Ne
(b) Ar & Kr
(c) Kr & Xe
(d) Xe & Rn
Answer: (c) Kr & Xe
In simple words: Krypton and Xenon are the noble gases that can sometimes react with other chemicals. They are less stable than the lighter noble gases and can form compounds.

🎯 Exam Tip: While noble gases are generally unreactive, the larger noble gases like Krypton and Xenon have lower ionization energies and can form compounds, especially with highly electronegative elements like fluorine and oxygen.

 

Question 57. Which among the following is used in cryogenics?
(a) Ar
(b) Ne
(c) He
(d) Kr
Answer: (c) He
In simple words: Helium is used in cryogenics, which is the science of very low temperatures. It is very good at making things extremely cold because it has the lowest boiling point of any element.

🎯 Exam Tip: Helium has the lowest boiling point of any element and remains a liquid at extremely low temperatures, making it ideal for cryogenics, such as cooling superconducting magnets in MRI machines.

 

Question 58. Which is used for filling air balloons?
(a) He
(b) Ne
(c) Ar
(d) Kr
Answer: (a) He
In simple words: Helium is used to fill air balloons because it is very light and does not burn. This makes balloons float and keeps them safe.

🎯 Exam Tip: Helium is preferred over hydrogen for filling balloons and airships because, although hydrogen is lighter, helium is non-flammable and therefore much safer.

 

Question 59. Which is used in advertisement sign boards?
(a) He
(b) Ne
(c) Ar
(d) Kr
Answer: (b) Ne
In simple words: Neon gas is used in advertising signs because it glows with a bright red-orange light when electricity passes through it. This makes signs very noticeable.

🎯 Exam Tip: Neon lights produce a characteristic reddish-orange glow. Different colors in "neon" signs are achieved by using other noble gases or by coating the inside of the glass tubes with phosphors.

 

Question 60. Lamps used in airports as approaching lights is filled with
(a) He
(b) Ne
(c) Ar
(d) Kr
Answer: (c) Ar
In simple words: Lamps for airport approach lights are often filled with argon gas. Argon helps protect the bulb's filament and makes the light last longer and glow brightly.

🎯 Exam Tip: Argon is commonly used as an inert filling gas in incandescent light bulbs, including high-intensity airport lights, to prevent the tungsten filament from oxidizing and to slow down its evaporation, thereby extending the bulb's lifespan and improving efficiency.

 

Question 58. Which is used for filling air balloons?
(a) He
(b) Ne
(c) Ar
(d) Kr
Answer: (a) He
In simple words: Helium is a very light gas and does not burn, making it perfect for filling balloons safely.

🎯 Exam Tip: Remember that helium is much safer than hydrogen for balloons because it is non-flammable.

 

Question 59. Which is used in advertisement sign boards?
(a) He
(b) Ne
(c) Ar
(d) Kr
Answer: (b) Ne
In simple words: Neon gas glows brightly when electricity passes through it, which is why it's used to make colourful advertisement signs.

🎯 Exam Tip: Different noble gases glow with distinct colours when electrified, making them suitable for various lighting applications.

 

Question 60. Lamps used in airports as approaching lights is filled with
(a) He
(b) Ne
(c) Ar
(d) Kr
Answer: (b) Ne
In simple words: Neon gas produces a bright, reddish-orange light when electricity passes through it. This makes it ideal for airport approach lights, as the light can easily penetrate fog.

🎯 Exam Tip: The ability of neon lights to cut through atmospheric conditions like fog is a key reason for their use in aviation safety.

 

VII. Two Mark Questions

 

Question 1. How is pure nitrogen gas prepared?
Answer: Pure nitrogen gas can be made by heating sodium azide to about 575 Kelvin. This process breaks down the sodium azide into sodium metal and nitrogen gas. This method produces very high-purity nitrogen.
\( \ce{2NaN3 \underrightarrow{573K} 2Na + 3N2} \)
In simple words: To get pure nitrogen, you heat a chemical called sodium azide, and it splits into sodium metal and nitrogen gas.

🎯 Exam Tip: Remember the specific reactant (sodium azide) and the approximate temperature for this thermal decomposition reaction.

 

Question 2. Nitrogen does not form any penta halides like phosphorus, why?
Answer: Nitrogen cannot create compounds with five halogen atoms (pentahalides) in the same way phosphorus does. This is because nitrogen does not have any empty 'd-orbitals' available in its valence shell. Without these d-orbitals, nitrogen cannot perform 'sp3d hybridization,' which is a necessary step for forming such compounds and accommodating more than four bonds.
In simple words: Nitrogen cannot make five bonds with halogens because it does not have enough empty electron shells (d-orbitals) like phosphorus does.

🎯 Exam Tip: Focus on the absence of d-orbitals and the inability to undergo sp3d hybridization as the key reasons.

 

Question 3. What is Haber's process?
Answer: Haber's process is a method to create ammonia (\( \ce{NH3} \)). It involves combining nitrogen gas (\( \ce{N2} \)) and hydrogen gas (\( \ce{H2} \)) under very high pressure and at an ideal temperature, with an iron catalyst to speed up the reaction. This reaction is important for making fertilizers and other chemicals, supporting global food production.
\( \ce{N2 + 3H2 \rightleftharpoons 2NH3} \)
\( \Delta H_f = -46.2 \text{ Kjmol}^{-1} \)
In simple words: Haber's process makes ammonia by mixing nitrogen and hydrogen gases under high pressure and heat, using an iron catalyst.

🎯 Exam Tip: Be sure to mention high pressure, optimum temperature, and the iron catalyst as critical conditions for the Haber process.

 

Question 4. Write the uses of nitrogen
Answer: Nitrogen is used in many ways. It is very important for making ammonia, nitric acid, and calcium cyanamide. It is also used to create very low temperatures, which is useful in cryosurgery (surgery using cold) and for keeping biological samples safe (biological preservation). Nitrogen gas helps keep things fresh by replacing oxygen.
In simple words: Nitrogen helps make ammonia and nitric acid, creates very cold temperatures for surgery, and preserves things.

🎯 Exam Tip: Categorize the uses into industrial production and cryogenic applications to ensure a comprehensive answer.

 

Question 5. How is ammonia prepared in the laboratory?
Answer: In a laboratory, ammonia is made by warming an ammonium salt with a chemical base. For instance, if you heat ammonium chloride (\( \ce{NH4Cl} \)) with calcium oxide (\( \ce{CaO} \)), you get calcium chloride (\( \ce{CaCl2} \)), ammonia gas (\( \ce{NH3} \)), and water (\( \ce{H2O} \)). This method provides a simple way to produce ammonia for experiments.
\( \ce{2NH4Cl + CaO \rightarrow CaCl2 + 2NH3 + H2O} \)
In simple words: You can make ammonia in the lab by heating an ammonium salt with a base, which gives off ammonia gas.

🎯 Exam Tip: Remember that ammonium compounds react with strong bases to release ammonia gas.

 

Question 6. Write about the reducing property of ammonia.
Answer: Ammonia can act as a reducing agent. This means it can take oxygen away from other substances, causing them to be reduced. When ammonia gas is passed over hot metal oxides, it changes the metal oxides back into pure metals. For example, ammonia can reduce lead oxide (\( \ce{3PbO} \)) to lead metal (\( \ce{3Pb} \)).
\( \ce{3PbO + 2NH3 \rightarrow 3Pb + N2 + 3H2O} \)
In simple words: Ammonia can remove oxygen from hot metal oxides, turning them back into pure metals.

🎯 Exam Tip: Highlight that ammonia's reducing property is typically observed with heated metal oxides, where it itself gets oxidized to nitrogen gas.

 

Question 7. What happens when ammonia reacts with excess of chlorine?
Answer: When ammonia reacts with an excess of chlorine, it produces an explosive substance called nitrogen trichloride (\( \ce{NCl3} \)). This reaction shows the strong oxidizing nature of chlorine.
\( \ce{NH3 + 3Cl2 \rightarrow NCl3 + 3HCl} \)
In simple words: If you mix ammonia with too much chlorine, it makes a dangerous, explosive chemical called nitrogen trichloride.

🎯 Exam Tip: Distinguish this from the reaction where ammonia is in excess, which produces nitrogen gas instead of nitrogen trichloride.

 

Question 8. On standing nitric acid becomes yellow in colour why?
Answer: When left to stand, pure nitric acid turns yellow. This happens because nitric acid is not very stable and breaks down when exposed to sunlight or heat. It breaks into nitrogen dioxide (\( \ce{NO2} \)), water, and oxygen. The nitrogen dioxide is a reddish-brown gas, and its presence causes the nitric acid to appear yellow. This is why it should be stored in dark bottles.
\( \ce{4HNO3 \rightarrow 4NO2 + 2H2O + O2} \)
In simple words: Nitric acid turns yellow over time because it breaks down in light to make brown nitrogen dioxide gas, which colours the acid.

🎯 Exam Tip: Emphasize the decomposition of nitric acid by light/heat and the formation of nitrogen dioxide as the reason for the yellow colour.

 

Question 9. Prove that nitric acid is an oxidising agent.
Answer: Nitric acid is a strong oxidising agent. This means it can cause other substances to lose electrons (get oxidised). For example, it oxidises non-metals like carbon and sulfur. When carbon reacts with concentrated nitric acid, it produces carbon dioxide, nitrogen dioxide, and water. Similarly, sulfur reacts to form sulfuric acid and nitrogen monoxide. These reactions clearly demonstrate how nitric acid acts to oxidise other elements.
\( \ce{C + 4HNO3 \rightarrow CO2 + 4NO2 + 2H2O} \)
\( \ce{S + 2HNO3 \rightarrow H2SO4 + 2NO} \)
In simple words: Nitric acid is an oxidising agent because it makes carbon and sulfur lose electrons, turning them into carbon dioxide and sulfuric acid.

🎯 Exam Tip: Provide at least two clear examples with balanced equations showing nitric acid oxidizing other substances.

 

Question 10. Prove that nitric acid is a nitrating agent.
Answer: Nitric acid acts as a nitrating agent by adding a nitro group (\( \ce{NO2^+} \)) to organic compounds, replacing a hydrogen atom. This process is called nitration. A classic example is the nitration of benzene, where a hydrogen atom on the benzene ring is replaced by a nitro group to form nitrobenzene. Sulfuric acid is often used as a catalyst in this reaction to generate the nitronium ion (\( \ce{NO2^+} \)).
\[ \ce{C6H6 + HNO3 \xrightarrow{H2SO4} C6H5NO2 + H2O} \]
In simple words: Nitric acid is a nitrating agent because it adds a nitro group to other chemicals, like turning benzene into nitrobenzene.

🎯 Exam Tip: For nitration, remember the role of sulfuric acid as a catalyst and the formation of the nitronium ion (\( \ce{NO2^+} \)).

 

Question 11. Write the uses of nitric acid is used
Answer: Nitric acid has many important uses. It acts as an oxidising agent in various chemical reactions. It is also a key component in preparing aqua regia, a powerful acid mixture that can dissolve noble metals like gold. Furthermore, salts made from nitric acid, such as silver nitrate (\( \ce{AgNO3} \)) and sodium nitrate (\( \ce{NaNO3} \)), are used in photography and in making gunpowder for weapons.
In simple words: Nitric acid is used for oxidising reactions, making aqua regia, and its salts are used in photography and gunpowder.

🎯 Exam Tip: Mention both its chemical properties (oxidizing agent) and its applications (aqua regia, salts for specific industries).

 

Question 12. How is nitrous oxide prepared?
Answer: Nitrous oxide is created by heating ammonium nitrate carefully. When ammonium nitrate (\( \ce{NH4NO3} \)) is heated, it breaks down into nitrous oxide gas (\( \ce{N2O} \)) and water (\( \ce{H2O} \)). This is a common laboratory method to produce this gas.
\( \ce{NH4NO3 \rightarrow N2O + 2H2O} \)
In simple words: You prepare nitrous oxide by gently heating ammonium nitrate, which breaks it down into nitrous oxide and water.

🎯 Exam Tip: Note the careful heating condition as ammonium nitrate can be explosive if heated too rapidly.

 

Question 13. How is nitrous acid prepared?
Answer: Nitrous acid can be made by reacting a nitrite salt with an acid. For example, if you treat barium nitrite (\( \ce{Ba(NO2)2} \)) with sulfuric acid (\( \ce{H2SO4} \)), you will get nitrous acid (\( \ce{2HNO2} \)) and barium sulfate (\( \ce{BaSO4} \)). This method allows for the preparation of nitrous acid in a controlled manner.
\( \ce{Ba(NO2)2 + H2SO4 \rightarrow 2HNO2 + BaSO4} \)
In simple words: To make nitrous acid, you react a nitrite salt with another acid.

🎯 Exam Tip: Remember that nitrous acid is generally unstable and is often prepared just before use.

 

Question 14. What is phosphorescence?
Answer: Phosphorescence is a special type of glow. It happens when white phosphorus slowly reacts with oxygen in the air without burning, a process called slow oxidation. This slow reaction makes it give off a greenish-yellow light that can be seen in the dark. The main substance formed during this slow oxidation is phosphorus trioxide (\( \ce{P2O3} \)).
In simple words: Phosphorescence is the greenish-yellow light that white phosphorus makes when it slowly reacts with air in the dark.

🎯 Exam Tip: Differentiate phosphorescence from fluorescence by its longer-lasting glow after the light source is removed.

 

Question 15. How is phosphine prepared?
Answer: Phosphine (\( \ce{PH3} \)) is prepared by the action of sodium hydroxide with white phosphorus in an inert atmosphere, such as carbon dioxide. This prevents the phosphine from spontaneously igniting in the air.
\( \ce{P4 + 3NaOH + 3H2O \rightarrow PH3 + 3NaH2PO2} \)
In simple words: Phosphine is made by reacting white phosphorus with sodium hydroxide in an environment without air.

🎯 Exam Tip: Emphasize the inert atmosphere (e.g., carbon dioxide) as crucial to prevent the highly flammable phosphine from catching fire.

 

Question 16. How is orthophosphoric acid prepared in ' the laboratory?
Answer: Orthophosphoric acid (\( \ce{H3PO4} \)) can be prepared in the laboratory. This is done by reacting phosphorus with concentrated nitric acid. A small amount of iodine acts as a catalyst to help speed up this oxidation reaction, turning phosphorus into orthophosphoric acid.
In simple words: You can prepare orthophosphoric acid by reacting phosphorus with strong nitric acid, using iodine to help the reaction.

🎯 Exam Tip: Note that concentrated nitric acid is a strong oxidizing agent, essential for oxidizing phosphorus to its highest oxidation state in orthophosphoric acid.

 

Question 17. Write the uses of phosphorous
Answer: Phosphorus is used to make special types of metal mixtures called alloys. One example is phosphor bronze, an alloy that is stronger and more resistant to wear than ordinary bronze, making it suitable for gears and springs. This makes phosphorus valuable in manufacturing.
In simple words: Phosphorus is used to create strong alloys, like phosphor bronze.

🎯 Exam Tip: Focus on its application in metallurgy, specifically for improving the properties of other metals.

 

Question 18. What happens when phosphine is heated in the absence of air?
Answer: When phosphine is heated without any air present, it breaks down into its basic elements. At a temperature of 317 Kelvin, phosphine (\( \ce{4PH3} \)) will decompose to form phosphorus (\( \ce{P4} \)) and hydrogen gas (\( \ce{6H2} \)). This reaction shows how unstable phosphine is under heat.
\( \ce{4PH3 \underrightarrow{317K} P4 + 6H2} \)
In simple words: If you heat phosphine without air, it breaks down into phosphorus and hydrogen gas.

🎯 Exam Tip: Remember that thermal decomposition is a common way to break down unstable compounds into their simpler components.

 

Question 19. Write about the reducing property of phosphine?
Answer: Phosphine has a reducing property, meaning it can cause other substances to gain electrons. For example, phosphine can reduce silver nitrate. In this reaction, phosphine changes silver nitrate (\( \ce{6AgNO3} \)) into pure silver (\( \ce{6Ag} \)), while also forming nitric acid (\( \ce{6HNO3} \)) and phosphorous acid (\( \ce{H3PO3} \)).
\( \ce{PH3 + 6AgNO3 + 3H2O \rightarrow 6Ag + 6HNO3 + H3PO3} \)
In simple words: Phosphine can reduce silver nitrate into silver, showing its ability to give electrons to other compounds.

🎯 Exam Tip: Recognize that a reducing agent like phosphine itself gets oxidized during the reaction.

 

Question 20. How is phosphorous trichloride prepared?
Answer: Phosphorous trichloride (\( \ce{PCl3} \)) can be made in two main ways. One way is to slowly pass chlorine gas over white phosphorus, which then forms phosphorus trichloride. Another method involves reacting white phosphorus (\( \ce{P4} \)) with thionyl chloride (\( \ce{8SOCl2} \)) to produce phosphorus trichloride (\( \ce{4PCl3} \)), along with sulfur dioxide (\( \ce{4SO2} \)) and disulfur dichloride (\( \ce{2S2Cl2} \)).
\( \ce{P4 + 8SOCl2 \rightarrow 4PCl3 + 4SO2 + 2S2Cl2} \)
In simple words: Phosphorus trichloride is made either by passing chlorine gas over white phosphorus or by reacting white phosphorus with thionyl chloride.

🎯 Exam Tip: Remember both methods, as they represent direct halogenation and a reaction with a chlorinating agent, respectively.

 

Question 21. Ozone (\( \ce{O3} \)) acts as a powerful oxidizing agent why? (PTA - 5)
Answer: Ozone (\( \ce{O3} \)) is a strong oxidising agent because it is not very stable. Under normal conditions or when heated, ozone easily breaks down into a molecule of oxygen (\( \ce{O2} \)) and a single, highly reactive oxygen atom (called nascent oxygen). This nascent oxygen is a free radical and quickly reacts with other substances, causing them to oxidise. Its instability drives its powerful oxidizing capabilities.
\( \ce{O3 \underrightarrow{\triangle} O2 + [O]} \)
In simple words: Ozone is a strong oxidising agent because it is unstable and easily breaks down to release a very reactive oxygen atom.

🎯 Exam Tip: The key to ozone's oxidizing power lies in its decomposition into oxygen and highly reactive nascent oxygen.

 

Question 22. Write the uses of oxygen
Answer: Oxygen is vital for life on Earth, as it's essential for all living organisms to survive through respiration. Beyond that, oxygen has important industrial uses. It is used in oxyacetylene welding to create very hot flames for cutting and joining metals. Liquid oxygen also serves as a powerful fuel in rockets, helping them launch into space.
In simple words: Oxygen is needed for all living things to breathe, used in welding metals, and as rocket fuel.

🎯 Exam Tip: Think of both biological and industrial applications when listing the uses of oxygen.

 

Question 23. How is sulphur dioxide prepared in the laboratory?
Answer: Sulfur dioxide (\( \ce{SO2} \)) can be prepared in a laboratory by reacting either a metal or a metal sulfite with sulfuric acid. For instance, when copper (\( \ce{Cu} \)) reacts with concentrated sulfuric acid (\( \ce{H2SO4} \)), it produces copper sulfate, sulfur dioxide gas, and water. Alternatively, a sulfite salt, like sodium sulfite (\( \ce{Na2SO3} \)), will react with an acid to release sulfur dioxide gas.
\( \ce{Cu + 2H2SO4 \rightarrow CuSO4 + SO2 + 2H2O} \)
\( \ce{SO3^2- + 2H+ \rightarrow SO2 + H2O} \)
In simple words: You can make sulfur dioxide in the lab by reacting a metal with sulfuric acid, or by reacting a sulfite salt with an acid.

🎯 Exam Tip: Distinguish between the reaction with a metal (like copper) and the reaction with a sulfite salt, as both are common laboratory methods.

 

Question 24. Illustrate the oxidising property of SO2.
Answer: Sulfur dioxide can act as an oxidising agent, meaning it causes other substances to lose electrons. For example, it oxidises hydrogen sulfide (\( \ce{H2S} \)) to form sulfur (\( \ce{S} \)) and water. It also oxidises magnesium metal (\( \ce{Mg} \)) to produce magnesium oxide (\( \ce{MgO} \)) and elemental sulfur. These reactions show sulfur dioxide taking electrons from other compounds.
\( \ce{2H2S + SO2 \rightarrow 3S + 2H2O} \)
\( \ce{2Mg + SO2 \rightarrow 2MgO + S} \)
In simple words: Sulfur dioxide oxidises things like hydrogen sulfide to sulfur and magnesium to magnesium oxide.

🎯 Exam Tip: Remember that sulfur dioxide can both oxidise and reduce, depending on the other reactant. Here, focus on its role as an oxidising agent.

 

Question 25. Write about contact process.
Answer: The Contact Process is the main industrial method for making sulfuric acid. In this process, sulfur dioxide gas (\( \ce{SO2} \)) is reacted with oxygen gas (\( \ce{O2} \)) to form sulfur trioxide (\( \ce{SO3} \)). This reaction occurs at about 450°C and uses vanadium pentoxide (\( \ce{V2O5} \)) as a catalyst to speed it up. Sulfur trioxide is then further processed to produce sulfuric acid, which is a key industrial chemical.
\( \ce{2SO2(g) + O2(g) \underrightarrow{V2O5 \\ 450°C} 2SO3(g)} \)
In simple words: The contact process is how sulfuric acid is made; sulfur dioxide reacts with oxygen using a catalyst to form sulfur trioxide, then more steps make the acid.

🎯 Exam Tip: Key aspects to remember are the reactants, catalyst (V2O5), and temperature range (450°C).

 

Question 26. Write the uses of sulphurdioxide.
Answer: Sulfur dioxide has several important uses. It is commonly used as a bleaching agent for materials like hair, silk, and wool. Additionally, in agriculture, sulfur dioxide helps disinfect crops and plants, protecting them from pests and diseases. It also plays a role in food preservation by inhibiting the growth of microbes.
In simple words: Sulfur dioxide is used to bleach things like hair, clean crops, and keep plants healthy.

🎯 Exam Tip: Focus on its applications in bleaching and as a disinfectant/preservative.

 

Question 27. Write about the structure of sulphr dioxide.
Answer: Sulfur dioxide (\( \ce{SO2} \)) has a bent shape, not a straight one. The central sulfur atom undergoes \( \ce{sp^2} \) hybridization. It forms one double bond and one single bond with the two oxygen atoms, with the electrons delocalised across both S-O bonds due to resonance. There is also a lone pair of electrons on the sulfur atom, which causes the molecule to have its bent shape and contributes to its reactivity.

In simple words: Sulfur dioxide has a bent shape because the central sulfur atom has a lone pair of electrons and forms bonds with two oxygen atoms.

🎯 Exam Tip: When describing molecular structures, always mention the hybridization of the central atom and explain how lone pairs influence the final geometry.

 

Question 28. Illustrate the dehydrating property of sulphuric acid.
Answer: Sulfuric acid is a powerful dehydrating agent, meaning it can remove water from other substances. For example, when concentrated sulfuric acid is added to sugar (sucrose, \( \ce{C12H22O11} \)), it removes all the water molecules, leaving behind black carbon. It also removes water from formic acid (\( \ce{HCOOH} \)) to produce carbon monoxide (\( \ce{CO} \)) and from oxalic acid (\( \ce{(COOH)2} \)) to yield carbon monoxide, carbon dioxide (\( \ce{CO2} \)), and water. These reactions highlight sulfuric acid's strong ability to pull out water.
\( \ce{C12H22O11 + H2SO4 \rightarrow 12C + H2SO4.11H2O} \)
\( \ce{HCOOH + H2SO4 \rightarrow CO + H2SO4.H2O} \)
\( \ce{(COOH)2 + H2SO4 \rightarrow CO + CO2 + H2SO4.H2O} \)
In simple words: Sulfuric acid removes water from other chemicals, like turning sugar into carbon or changing formic acid into carbon monoxide.

🎯 Exam Tip: Provide diverse examples, such as dehydration of carbohydrates and organic acids, to fully illustrate this property.

 

Question 29. Sulphuric acid is a dibasic acid. Illustrate with two equations.
Answer: Sulfuric acid (\( \ce{H2SO4} \)) is called a dibasic acid because it can donate two hydrogen ions (\( \ce{H+} \)) in a reaction. This is shown by its ability to form two different types of salts when reacting with a base like sodium hydroxide. First, it can react with one molecule of sodium hydroxide to form sodium bisulfate and water. Second, it can react with two molecules of sodium hydroxide to form sodium sulfate and water. This stepwise proton donation makes it dibasic.
\( \ce{H2SO4 + NaOH \rightarrow NaHSO4 + H2O} \)
\( \ce{H2SO4 + 2NaOH \rightarrow Na2SO4 + 2H2O} \)
In simple words: Sulfuric acid is called dibasic because it can give away two hydrogen ions, making two types of salts with a base.

🎯 Exam Tip: Clearly show the two stages of neutralization by reacting sulfuric acid with one and then two moles of a base.

 

Question 30. How is chlorine is manufactured by Deacon's process?
Answer: The Deacon's process is a method for making chlorine gas. In this process, a mixture of air and hydrochloric acid vapor is passed through a special chamber. This chamber contains pumice stones that are soaked in cuprous chloride (\( \ce{Cu2Cl2} \)), which acts as a catalyst. The reaction takes place at around 723 Kelvin, producing chlorine gas and water. The chlorine gas produced is often dilute and used in making bleaching powder.
\( \ce{4HCl + O2 \underrightarrow{Cu2Cl2 \\ 400°C} 2H2O + 2Cl2} \)
In simple words: Chlorine gas is made in the Deacon's process by reacting hydrochloric acid with air over a special catalyst at high temperatures.

🎯 Exam Tip: Remember the reactants (HCl and O2), the catalyst (cuprous chloride), and the temperature as crucial for the Deacon's process.

 

Question 31. Write about the bleaching action of chlorine.
Answer: Chlorine acts as a strong bleaching agent because it releases very reactive "nascent oxygen." When chlorine reacts with water, it forms hydrochloric acid and hypochlorous acid (\( \ce{HOCl} \)). Hypochlorous acid is unstable and breaks down to produce hydrochloric acid and nascent oxygen (\( \ce{[O]} \)). This powerful nascent oxygen then reacts with coloured substances, turning them into colourless products. Because the colour is destroyed by oxidation, chlorine's bleaching effect is permanent.
\( \ce{H2O + Cl2 \rightarrow HCl + HOCl} \)
\( \ce{HOCl \rightarrow HCl + [O]} \)
\( \text{Colouring matter} + \ce{[O] \rightarrow \text{Colourless oxidation product}} \)
In simple words: Chlorine bleaches by releasing a very reactive type of oxygen that destroys colours, making the bleaching permanent.

🎯 Exam Tip: Highlight the formation of nascent oxygen from hypochlorous acid as the active bleaching species, leading to irreversible colour removal.

 

Question 32. How is bleaching powder prepared? (MARCH 2020)
Answer: Bleaching powder (\( \ce{CaOCl2} \)) is made by passing chlorine gas through dry slaked lime, which is calcium hydroxide (\( \ce{Ca(OH)2} \)). This reaction produces calcium oxychloride, which is bleaching powder, and water. This is a common industrial method to produce this important disinfectant and bleaching agent.
\( \ce{Ca(OH)2 + Cl2 \rightarrow CaOCl2 + H2O} \)
In simple words: You make bleaching powder by letting chlorine gas react with dry slaked lime.

🎯 Exam Tip: Remember the specific reactants: dry slaked lime and chlorine gas for bleaching powder preparation.

 

Question 33. Write the uses of chlorine
Answer: Chlorine has several important uses. It is widely used to purify drinking water, killing harmful bacteria and making it safe to consume. It is also a key agent in the bleaching of cotton textiles, paper, and rayon, making them whiter. Additionally, chlorine plays a role in the process of extracting valuable metals like gold and platinum.
In simple words: Chlorine is used to clean drinking water, bleach fabrics and paper, and help extract precious metals.

🎯 Exam Tip: Focus on chlorine's roles as a disinfectant, bleaching agent, and in metallurgical processes.

 

Question 34. How is hydrochloric acid prepared in the laboratory?
Answer: In the laboratory, hydrochloric acid is prepared by reacting sodium chloride (\( \ce{NaCl} \)) with concentrated sulfuric acid (\( \ce{H2SO4} \)). This reaction first produces sodium bisulfate and hydrogen chloride gas at a lower temperature. If more sodium chloride is added and heated strongly, sodium sulfate is formed along with more hydrogen chloride. The hydrogen chloride gas is then dried by passing it through concentrated sulfuric acid to obtain pure, dry hydrochloric acid.
\( \ce{NaCl + H2SO4 \rightarrow NaHSO4 + HCl} \)
\( \ce{NaHSO4 + NaCl \rightarrow Na2SO4 + HCl} \)
In simple words: Hydrochloric acid is made in the lab by reacting table salt with concentrated sulfuric acid, which releases hydrogen chloride gas.

🎯 Exam Tip: Remember this is a displacement reaction where the less volatile sulfuric acid displaces the more volatile HCl.

 

Question 35. How is xenon trioxide prepared?
Answer: Xenon trioxide (\( \ce{XeO3} \)) can be prepared through the complete hydrolysis of xenon hexafluoride (\( \ce{XeF6} \)). This means reacting xenon hexafluoride with water. The reaction is: \( \ce{XeF6 + 3H2O \rightarrow XeO3 + 6HF} \). It can also be formed in a multi-step process by reacting xenon hexafluoride with silicon dioxide, then further reacting the products with silicon dioxide.
In simple words: Xenon trioxide is prepared by letting xenon hexafluoride react completely with water.

🎯 Exam Tip: Hydrolysis of xenon fluorides is a common method to obtain xenon oxides; remember the stoichiometric ratio for complete hydrolysis.

 

Question 36. How is sodium per xenate obtained?
Answer: Sodium perxenate (\( \ce{Na4XeO6} \)) is obtained by reacting xenon hexafluoride (\( \ce{XeF6} \)) with a concentrated solution of sodium hydroxide (\( \ce{NaOH} \)). Specifically, when xenon hexafluoride reacts with 2.5 M sodium hydroxide, it produces sodium perxenate, along with xenon gas, oxygen gas, sodium fluoride, and water. This is an important reaction for synthesizing hypervalent xenon compounds.
\( \ce{2XeF6 + 16NaOH \rightarrow Na4XeO6 + Xe + O2 + 12NaF + 8H2O} \)
In simple words: You get sodium perxenate by mixing xenon hexafluoride with a strong solution of sodium hydroxide.

🎯 Exam Tip: Note that perxenates are powerful oxidizing agents due to xenon's high oxidation state.

 

Question 37. Show that sodium per xenate is a strong oxidising agent
Answer: Sodium perxenate is a very strong oxidising agent. This is shown by its ability to oxidise manganese(II) ions (\( \ce{Mn^2+} \)) into permanganate ions (\( \ce{MnO4^-} \)). This powerful oxidation happens even without a catalyst, highlighting the strong oxidising nature of sodium perxenate and its high oxidation state of xenon (+8).
\( \ce{5XeO6^4- + 2Mn^2+ + 14H+ \rightarrow 2MnO4^- + 5XeO3 + 7H2O} \)
In simple words: Sodium perxenate is a strong oxidising agent because it can turn manganese(II) into permanganate, even without help from a catalyst.

🎯 Exam Tip: The conversion of \( \ce{Mn^2+} \) to \( \ce{MnO4^-} \) is a classic test for strong oxidizing agents in acidic medium.

 

Question 38. Give reason: \( \ce{ICl} \) is more reactive than \( \ce{I2} \) (PTA - 3)
Answer: Iodine monochloride (\( \ce{ICl} \)) is more reactive than iodine (\( \ce{I2} \)). This is because interhalogen compounds like \( \ce{ICl} \) generally have weaker bonds between the different halogen atoms compared to the bonds in a diatomic halogen molecule like \( \ce{I2} \). A weaker bond means it's easier to break, making \( \ce{ICl} \) less stable and thus more reactive than \( \ce{I2} \). The polarity of the \( \ce{I-Cl} \) bond also contributes to its reactivity.
In simple words: \( \ce{ICl} \) is more reactive than \( \ce{I2} \) because the bond between iodine and chlorine is weaker and easier to break than the bond between two iodine atoms.

🎯 Exam Tip: Remember that the difference in electronegativity between two different halogens in an interhalogen compound leads to a more polar and often weaker bond compared to a non-polar bond between identical halogen atoms.

 

VIII. Three Mark Questions

 

Question 1. Write about the structure of ammonia.
Answer: Ammonia (\( \ce{NH3} \)) has a pyramidal molecular shape. The nitrogen atom in ammonia undergoes \( \ce{sp^3} \) hybridization. Although this hybridization typically leads to a tetrahedral electron geometry, one of the positions is occupied by a lone pair of electrons. This lone pair pushes the three N-H bonds closer together, resulting in a pyramidal molecular shape with a bond angle of about 107°. Each N-H bond has a length of approximately 1.016 Å. This unique shape gives ammonia its characteristic properties.

In simple words: Ammonia has a pyramid shape. The central nitrogen atom has one pair of electrons that don't bond, pushing the other three hydrogen atoms down, making the bond angle around 107 degrees.

🎯 Exam Tip: For molecular structures, always draw a clear diagram and label the hybridization, bond angles, and any lone pairs present.

 

Question 2. How does red phosphorous react with oxygen?
Answer: When red phosphorus is heated in the presence of oxygen, it reacts to form oxides of phosphorus. Depending on the amount of oxygen available, it can produce phosphorus trioxide (\( \ce{P4O6} \)) if oxygen is limited, or phosphorus pentoxide (\( \ce{P4O10} \)) if there is enough oxygen. These are both white, solid compounds. The burning of red phosphorus is a vigorous exothermic reaction.
\( \ce{P4 + 3O2 \rightarrow P4O6} \) (limited oxygen)
\( \ce{P4 + 5O2 \rightarrow P4O10} \) (excess oxygen)
In simple words: When red phosphorus is heated with oxygen, it burns to form either phosphorus trioxide or phosphorus pentoxide, depending on how much oxygen is present.

🎯 Exam Tip: Note the difference in products based on the availability of oxygen: trioxide with limited oxygen and pentoxide with excess oxygen.

 

Question 3. How is pure phosphine prepared?
Answer: Pure phosphine (\( \ce{PH3} \)) can be prepared in two main ways. One method involves heating phosphorous acid (\( \ce{4H3PO3} \)), which decomposes to yield orthophosphoric acid (\( \ce{3H3PO4} \)) and phosphine gas. Another method is to react phosphonium iodide (\( \ce{PH4I} \)) with a caustic soda (sodium hydroxide) solution. This reaction produces pure phosphine gas along with sodium iodide and water. Phosphine is a toxic gas, so its preparation must be done carefully.
\( \ce{4H3PO3 \underrightarrow{\triangle} 3H3PO4 + PH3 \uparrow} \)
\( \ce{PH4I + NaOH \rightarrow PH3 \uparrow + NaI + H2O} \)
In simple words: Pure phosphine can be made by heating phosphorous acid or by reacting phosphonium iodide with sodium hydroxide solution.

🎯 Exam Tip: Remember both decomposition and hydrolysis reactions, and the specific reactants for each method.

 

Question 4. What happens when phosphine is heated with air?
Answer: When phosphine is heated in the presence of air or oxygen, it burns (combusts). This combustion reaction first produces phosphorus pentoxide (\( \ce{P4O10} \)) and water. The phosphorus pentoxide then reacts further with any available water to form metaphosphoric acid (\( \ce{HPO3} \)). This process demonstrates phosphine's flammability and its reaction with oxygen, producing an acidic product.
\( \ce{4PH3 + 8O2 \underrightarrow{\triangle} P4O10 + 6H2O} \)
\( \ce{P4O10 + 6H2O \rightarrow 4HPO3 + 4H2O} \)
In simple words: When phosphine is heated in air, it burns to make phosphorus pentoxide and water, which then combine to form metaphosphoric acid.

🎯 Exam Tip: Note that phosphine is highly flammable, igniting easily in air to form oxides of phosphorus.

 

Question 5. Write about Holmes signal
Answer: Holmes' signal is an emergency signal used at sea. It works by placing a mixture of calcium carbide and calcium phosphide in a container. When a ship is in trouble, this container is pierced and thrown into the sea. The mixture then reacts with seawater to release two gases: acetylene (\( \ce{C2H2} \)) and phosphine (\( \ce{PH3} \)). The phosphine gas spontaneously catches fire in the air, which in turn ignites the acetylene. These burning gases produce a bright flame and a lot of smoke, creating a visible signal that helps guide rescue ships.
In simple words: Holmes' signal uses a chemical mixture that, when thrown into the sea, releases gases that catch fire and create smoke to signal for help.

🎯 Exam Tip: The spontaneous ignition of phosphine and the burning of acetylene are key to the effectiveness of Holmes' signal.

 

Question 6. Write about the structure of phosphine
Answer: Phosphine (\( \ce{PH3} \)) has a pyramidal molecular shape. The central phosphorus atom undergoes \( \ce{sp^3} \) hybridization. In this arrangement, three orbitals are used to form bonds with hydrogen atoms, containing bond pair electrons. The fourth orbital holds a lone pair of electrons. This lone pair exerts a stronger repulsion than bonding pairs, which pushes the P-H bonds closer together, giving the molecule its characteristic pyramidal geometry with a bond angle of about 94°.

In simple words: Phosphine has a pyramid shape because the central phosphorus atom has a non-bonding electron pair that pushes the other three hydrogen atoms away, resulting in a bond angle of about 94 degrees.

🎯 Exam Tip: The lone pair on the phosphorus atom is key to understanding the pyramidal geometry and the deviation from an ideal tetrahedral bond angle.

 

Question 7. How is oxygen prepared in the laboratory?
Answer: Oxygen can be prepared in the laboratory using a couple of methods. One common way is to decompose hydrogen peroxide (\( \ce{H2O2} \)) in the presence of manganese dioxide (\( \ce{MnO2} \)) as a catalyst. This reaction quickly breaks down hydrogen peroxide into water and oxygen gas. Another method involves the oxidation of potassium permanganate (\( \ce{KMnO4} \)) in an acidic medium, typically with hydrogen peroxide, which also releases oxygen.
\( \ce{2H2O2 \underrightarrow{MnO2} 2H2O + O2} \)
\( \ce{2KMnO4 + 5H2O2 + 3H2SO4 \rightarrow K2SO4 + 2MnSO4 + 8H2O + 5O2} \)
In simple words: Oxygen can be made in the lab by breaking down hydrogen peroxide with manganese dioxide, or by reacting potassium permanganate with hydrogen peroxide in an acid.

🎯 Exam Tip: Remember that \( \ce{MnO2} \) acts as a catalyst in the decomposition of \( \ce{H2O2} \), speeding up the oxygen production without being consumed.

 

Question 8. Write about ozone
Answer: Ozone (\( \ce{O3} \)) is a special form of oxygen, known as an allotrope, and it exists as a triatomic gas (meaning it has three oxygen atoms). While very little ozone is found at sea level, it is naturally formed in the Earth's upper atmosphere when oxygen molecules absorb ultraviolet (UV) light. In a laboratory, ozone can be prepared by passing a silent electrical discharge through oxygen gas. This process converts about 10% of the oxygen into ozone, creating a mixture called ozonised oxygen. Ozone has a characteristic pungent smell.
In simple words: Ozone is a three-atom form of oxygen, found in the upper atmosphere from UV light, and can be made in the lab using electricity on oxygen.

🎯 Exam Tip: Emphasize ozone's allotropic nature, its formation in the atmosphere via UV, and its laboratory preparation using silent electric discharge.

 

Question 9. Write about the reducing property of sulphur dioxide
Answer: Sulphur dioxide can act as a reducing agent in several reactions, meaning it can change other substances by donating electrons. For example, it reacts with chlorine in water to form sulphuric acid and hydrochloric acid:
\( SO_2 + 2H_2O + Cl_2 \rightarrow H_2SO_4 + 2HCl \)
It also reduces potassium permanganate (which is purple) into colourless manganese sulphate:
\( 2KMnO_4 + 5SO_2 + 2H_2O \rightarrow K_2SO_4 + 2MnSO_4 + 2H_2SO_4 \)
Additionally, it reduces potassium dichromate (which is orange) to green chromic sulphate:
\( K_2Cr_2O_7 + 3SO_2 + H_2SO_4 \rightarrow K_2SO_4 + Cr_2(SO_4)_3 + H_2O \)
This versatility makes sulphur dioxide a useful reducing agent in various chemical processes.
In simple words: Sulphur dioxide is a reducing agent, meaning it makes other chemicals gain electrons. It can turn chlorine into hydrochloric acid and change coloured solutions like permanganate and dichromate.

🎯 Exam Tip: Remember that reducing agents get oxidized themselves. In these reactions, sulphur dioxide is oxidized from \(+4\) to \(+6\) oxidation state.

 

Question 10. Write about the bleaching action of sulphur dioxide.
Answer: Sulphur dioxide has a bleaching effect, especially when water is present. It can remove the color from materials such as wool, silk, and sponges. This happens because sulphur dioxide combines with water to form sulphurous acid, which then acts as a reducing agent, converting the colored substance into a colorless one. The bleaching action of sulphur dioxide is temporary, as the color can return if the bleached material is exposed to air, where it can be re-oxidized. This makes it different from chlorine, which provides permanent bleaching.
In simple words: Sulphur dioxide can make colored materials like wool and silk turn white, but only if water is also present. This bleaching is not permanent, and the color can come back if the material is exposed to air.

🎯 Exam Tip: Distinguish between the temporary bleaching of sulphur dioxide (due to reduction) and the permanent bleaching of chlorine (due to oxidation).

IX. Five Mark Questions

 

Question 1. How is nitric acid manufactured using Ostwald's process?
Answer: Nitric acid is made industrially using the Ostwald's process, which involves three main stages:
1. **Catalytic Oxidation of Ammonia:** Ammonia gas, typically produced by the Haber's process, is mixed with about 10 times its volume of air. This mixture is preheated and then passed over a platinum-rhodium gauze catalyst at a very high temperature of around 1275 Kelvin. In this catalytic oxidation step, ammonia quickly reacts with oxygen to form nitric oxide (\(NO\)) and water. This reaction is highly exothermic, meaning it releases a lot of heat.
\( 4NH_3 + 5O_2 \rightarrow 4NO + 6H_2O + 120 \text{ KJ} \)
2. **Oxidation of Nitric Oxide:** The nitric oxide produced is then further oxidized by the remaining oxygen from the air to form nitrogen dioxide (\(NO_2\)). This reaction occurs at a lower temperature.
\( 2NO + O_2 \rightarrow 2NO_2 \)
3. **Absorption of Nitrogen Dioxide:** Finally, the nitrogen dioxide is passed through a series of absorption towers where it reacts with water to produce nitric acid. Any unreacted nitric oxide can be re-oxidized and recycled. This process is crucial for making fertilizers, explosives, and other industrial chemicals.
\( 3NO_2 + H_2O \rightarrow 2HNO_3 + NO \)
In simple words: Nitric acid is made using the Ostwald's process. First, ammonia gas is mixed with air and heated over a special metal gauze to make nitric oxide. This nitric oxide then turns into nitrogen dioxide, which is finally dissolved in water to make nitric acid.

🎯 Exam Tip: Remember the three main steps: oxidation of ammonia, oxidation of nitric oxide, and absorption of nitrogen dioxide. The platinum-rhodium catalyst and high temperature (1275 K) are key details for full marks.

 

Question 2. Explain the action of nitric acid on metals with one example.
Answer: Nitric acid reacts with metals, and the specific reaction products largely depend on the acid's concentration. Generally, a primary reaction occurs where the metal forms a nitrate and releases nascent hydrogen atoms. These highly reactive nascent hydrogen atoms then participate in secondary reactions, reducing the nitric acid itself to various nitrogen compounds.
**Example with Copper:**
**1. With Dilute Nitric Acid:** When dilute nitric acid reacts with copper, the primary reaction releases nascent hydrogen:
\( 3Cu + 6HNO_3 \rightarrow 3Cu(NO_3)_2 + 6(H) \)
The nascent hydrogen then reduces the nitric acid to form nitrous acid, which further decomposes:
\( 6(H) + 3HNO_3 \rightarrow 3HNO_2 + 3H_2O \)
\( 3HNO_2 \rightarrow HNO_3 + 2NO + H_2O \)
The overall reaction is:
\( 3Cu + 8HNO_3 (\text{dilute}) \rightarrow 3Cu(NO_3)_2 + 2NO + 4H_2O \)
Here, copper nitrate, nitric oxide, and water are formed.
**2. With Concentrated Nitric Acid:** When concentrated nitric acid reacts with copper, the nascent hydrogen produced similarly reduces the nitric acid. However, the reduction product is different due to the higher concentration:
\( Cu + 4HNO_3 (\text{conc}) \rightarrow Cu(NO_3)_2 + 2NO_2 + 2H_2O \)
In this case, copper nitrate, nitrogen dioxide, and water are formed. Nitric acid acts as a strong oxidizing agent, but its specific behavior varies significantly with concentration.
In simple words: Nitric acid reacts with metals, but what it makes depends on how strong the acid is. First, it forms metal nitrate and fresh hydrogen. This hydrogen then changes the nitric acid into other things. For example, with copper, if the acid is weak, it makes nitric oxide. If the acid is strong, it makes nitrogen dioxide.

🎯 Exam Tip: Always specify the concentration of nitric acid (dilute or concentrated) when discussing its reaction with metals, as this dictates the nitrogen-containing product formed.

 

Question 3. Write the preparation of nitrogen oxides.
Answer: Nitrogen forms several oxides, each with unique properties and methods of preparation:
1. **Dinitrogen Oxide (Nitrous Oxide, \(N_2O\)):** This oxide is prepared by carefully heating ammonium nitrate (\(NH_4NO_3\)).
\( NH_4NO_3 \rightarrow N_2O + 2H_2O \)
2. **Nitric Oxide (\(NO\)):** Nitric oxide is primarily formed during the catalytic oxidation of ammonia in the Ostwald's process, as seen in nitric acid manufacturing. It can also be formed by the reaction of dilute nitric acid with copper.
3. **Dinitrogen Trioxide (\(N_2O_3\)):** This oxide is formed by cooling an equimolar mixture of nitric oxide (\(NO\)) and nitrogen dioxide (\(NO_2\)) to a low temperature.
\( NO + NO_2 \leftrightarrow N_2O_3 \)
4. **Nitrogen Dioxide (\(NO_2\)):** Nitrogen dioxide is formed by heating lead nitrate or by the catalytic oxidation of nitric oxide with oxygen. It is a reddish-brown gas.
\( 2Pb(NO_3)_2 \rightarrow 4NO_2 + 2PbO + O_2 \)
5. **Dinitrogen Tetroxide (\(N_2O_4\)):** This oxide exists in equilibrium with nitrogen dioxide. It is formed by cooling nitrogen dioxide.
\( 2NO_2 \leftrightarrow N_2O_4 \)
6. **Dinitrogen Pentoxide (\(N_2O_5\)):** Dinitrogen pentoxide is prepared by dehydrating concentrated nitric acid with phosphorus pentoxide (\(P_2O_5\)).
\( 2HNO_3 + P_2O_5 \rightarrow N_2O_5 + 2HPO_3 \)
Each method uses specific reactants and conditions to isolate the desired nitrogen oxide, which plays various roles in chemistry and industry.
In simple words: Nitrogen oxides are made in different ways. Heating ammonium nitrate makes nitrous oxide. Nitric oxide is made from ammonia. Mixing nitric oxide and nitrogen dioxide makes dinitrogen trioxide. Heating lead nitrate makes nitrogen dioxide. Cooling nitrogen dioxide makes dinitrogen tetroxide. Finally, removing water from nitric acid using phosphorus pentoxide makes dinitrogen pentoxide.

🎯 Exam Tip: Pay attention to the specific reactants and conditions (e.g., heating, cooling, catalysts) for each nitrogen oxide, as these are crucial for correct preparation.

 

Question 4. Write the preparation of oxoacids of nitrogen
Answer: Oxoacids of nitrogen are compounds containing nitrogen, oxygen, and hydrogen, and each can be prepared through specific chemical reactions:
(i) **Hyponitrous Acid (\(H_2N_2O_2\)):** This acid is prepared by reacting silver hyponitrite (\(Ag_2N_2O_2\)) with hydrochloric acid.
\( Ag_2N_2O_2 + 2HCl \rightarrow 2AgCl + H_2N_2O_2 \)
(ii) **Nitrous Acid (\(HNO_2\)):** Nitrous acid can be formed by treating barium nitrite (\(Ba(NO_2)_2\)) with sulphuric acid.
\( Ba(NO_2)_2 + H_2SO_4 \rightarrow 2HNO_2 + BaSO_4 \)
(iii) **Nitric Acid (\(HNO_3\)):** Nitric acid is commonly prepared in the laboratory by reacting potassium nitrate (\(KNO_3\)) with concentrated sulphuric acid. Industrially, it is made by the Ostwald's process.
\( KNO_3 + H_2SO_4 \rightarrow KHSO_4 + HNO_3 \)
(iv) **Peroxynitrous Acid (\(HNO_3\)):** This acid, containing a peroxide linkage, can be formed from hydrogen peroxide and nitrous acid. Its structure is \(ON(OOH)\).
\( H_2O_2 + HNO_2 \rightarrow ON(OOH) + H_2O \)
(v) **Pernitric Acid (\(HNO_4\)):** This acid, also with a peroxide linkage, can be obtained from hydrogen peroxide and dinitrogen pentoxide. Its structure is \(NO_2OOH\).
\( H_2O_2 + N_2O_5 \rightarrow NO_2OOH + HNO_3 \)
These various methods allow for the synthesis of different nitrogen oxoacids, each with distinct properties and uses.
In simple words: Different nitrogen oxoacids are made in specific ways. Hyponitrous acid comes from silver hyponitrite and hydrochloric acid. Nitrous acid is made from barium nitrite and sulphuric acid. Nitric acid is made from potassium nitrate and sulphuric acid. Other methods use hydrogen peroxide to make peroxynitrous acid and pernitric acid.

🎯 Exam Tip: Note the valency and bonding in each oxoacid; for example, the `N-N` bond in hyponitrous acid and the `O-O` peroxide bond in peroxoacids are key structural features.

 

Question 5. Explain the structure of oxides of phosphorus
Answer: The oxides of phosphorus, such as phosphorus trioxide (\(P_4O_6\)) and phosphorus pentoxide (\(P_4O_{10}\)), have complex molecular structures that are often described as cage-like.
**1. Phosphorus Trioxide (\(P_4O_6\)):**
* In \(P_4O_6\), four phosphorus atoms are situated at the corners of a tetrahedron, and six oxygen atoms are positioned along the edges, bridging pairs of phosphorus atoms. This forms a P-O-P framework.
* The P-O bond distance in \(P_4O_6\) is approximately 165.6 pm, which is shorter than a typical single P-O bond (around 184 pm). This shorter distance suggests that there is some double bond character due to \(p\pi-d\pi\) bonding. This electronic interaction contributes to the stability and geometry of the molecule.
* The molecule thus exhibits considerable double bond character in its structure.

**2. Phosphorus Pentoxide (\(P_4O_{10}\)):**
* The structure of \(P_4O_{10}\) is derived from the \(P_4O_6\) framework, with four additional terminal oxygen atoms double-bonded to each phosphorus atom. So, each phosphorus atom in \(P_4O_{10}\) is single-bonded to three bridging oxygen atoms and double-bonded to one terminal oxygen atom.
* The P-O bond length in \(P_4O_{10}\) is approximately 143 pm, which is even shorter than in \(P_4O_6\). This indicates a stronger double bond character. This phenomenon is often attributed to the lateral overlap of filled \(p\)-orbitals of the oxygen atom with empty \(d\)-orbitals on the phosphorus atom.
* \(P_4O_{10}\) is widely known for its strong dehydrating properties, often used to remove water from compounds.

In simple words: Phosphorus oxides like \(P_4O_6\) and \(P_4O_{10}\) have cage-like structures. In \(P_4O_6\), four phosphorus atoms are at the corners of a tetrahedron, and six oxygen atoms are on the edges. In \(P_4O_{10}\), it's similar but with four extra oxygen atoms double-bonded to each phosphorus atom. These extra bonds make the P-O bonds shorter than usual.

🎯 Exam Tip: When drawing these structures, ensure that each phosphorus atom in \(P_4O_6\) is bonded to three oxygens, and in \(P_4O_{10}\), it's bonded to three bridging oxygens and one terminal double-bonded oxygen. Pay attention to the P-O bond lengths as well.

 

Question 6. Write the structure of and basicity oxoacids of phosphorous.
Answer: The oxoacids of phosphorus are compounds that contain phosphorus, oxygen, and hydrogen, each possessing a unique molecular structure and specific basicity (the number of replaceable hydrogen atoms or H+ ions they can donate). The basicity depends on the number of P-OH bonds present in the structure.

The basicity of these acids is determined by the number of hydrogen atoms directly attached to an oxygen atom (P-OH bonds), which are typically ionizable. Hydrogen atoms directly bonded to phosphorus (P-H bonds) are generally not ionizable.
In simple words: Phosphorus forms many acids with oxygen and hydrogen. Each acid has a different shape and a specific basicity, which is the number of acidic hydrogen atoms. The basicity depends on how many hydroxyl (OH) groups are connected to the phosphorus.

🎯 Exam Tip: To determine basicity, always count the number of P-OH bonds, not the total number of hydrogen atoms, as P-H bonds are usually non-acidic.

 

Question 7. Write the preparation of oxoacids of phosphorous
Answer: Oxoacids of phosphorus can be prepared through various chemical reactions, each method tailored to produce a specific acid:
(i) **Hypophosphorous acid (\(H_3PO_2\)):** This acid is formed when white phosphorus (\(P_4\)) reacts with water under specific conditions, often in the presence of an alkali.
\( P_4 + 6H_2O \rightarrow 3H_3PO_2 + PH_3 \)
(ii) **Orthophosphorous acid (\(H_3PO_3\)):** This acid can be obtained by the hydrolysis of phosphorus trioxide (\(P_4O_6\)), where water reacts with the oxide.
\( P_4O_6 + 6H_2O \rightarrow 4H_3PO_3 \)
(iii) **Hypophosphoric acid (\(H_4P_2O_6\)):** This acid is prepared by the slow oxidation of white phosphorus with oxygen and water, for example, by passing air slowly over white phosphorus partially immersed in water.
\( 2P + 2O_2 + 2H_2O \rightarrow H_4P_2O_6 \)
(iv) **Orthophosphoric acid (\(H_3PO_4\)):** This is one of the most common phosphorus oxoacids and is primarily produced by the hydrolysis of phosphorus pentoxide (\(P_4O_{10}\)).
\( P_4O_{10} + 6H_2O \rightarrow 4H_3PO_4 \)
(v) **Pyrophosphoric acid (\(H_4P_2O_7\)):** This acid can be synthesized by heating orthophosphoric acid, which results in the elimination of a water molecule between two molecules of orthophosphoric acid.
\( 2H_3PO_4 \rightarrow H_4P_2O_7 + H_2O \)
These methods highlight the different pathways to create these important phosphorus compounds, each with distinct chemical properties.
In simple words: Phosphorus oxoacids are made in different ways. Hypophosphorous acid is made from phosphorus and water. Orthophosphorous acid comes from phosphorus trioxide and water. Hypophosphoric acid is made by slowly reacting phosphorus with air and water. Orthophosphoric acid is made from phosphorus pentoxide and water. Pyrophosphoric acid is made by heating orthophosphoric acid.

🎯 Exam Tip: Pay close attention to the specific phosphorus oxide reactant (e.g., \(P_4\), \(P_4O_6\), \(P_4O_{10}\)) and the role of water in each preparation method.

 

Question 8. Write the structure of oxo acids of sulphur.
Answer: Sulphur forms several oxoacids, each possessing a unique molecular structure and chemical properties. These structures often involve sulphur atoms bonded to oxygen and hydrogen, sometimes with direct sulphur-sulphur bonds or oxygen bridges.

The arrangement of sulphur, oxygen, and hydrogen atoms in these structures dictates their reactivity, acidity, and industrial applications. Understanding these structures is key to predicting their chemical behavior.
In simple words: Sulphur makes many different acids with oxygen and hydrogen, and each has its own shape. For example, sulphuric acid has sulphur with two double-bonded oxygens and two single-bonded hydroxyls. Thiosulphuric acid is like sulphuric acid, but one oxygen is replaced by sulphur. Dithionous acid has two sulphur atoms directly linked. The way these atoms are connected makes each acid special.

🎯 Exam Tip: Pay attention to the oxidation states of sulphur in each acid, as well as the presence of S-S bonds or S-O-S bridges, which are distinctive features.

 

Question 9. List any five compounds of Xenon and mention the type of hybridization and structure of the compounds

 

Question 9. List any five compounds of Xenon and mention the type of hybridization and structure of the compounds (PTA - 6)
Answer: Xenon forms various compounds with different hybridizations and structures. Here are some examples:

In simple words: Xenon, a noble gas, can form compounds by sharing electrons. The way its electrons are arranged (hybridization) affects the final shape of the molecule. For example, XeF4 has a flat, square shape.

🎯 Exam Tip: When listing compounds, always specify the correct hybridization and corresponding molecular geometry as it's crucial for understanding their chemical behavior.