Maharashtra Board Class 11 Chemistry Chapter 8 Elements of Group 1 and 2 Solutions

Elements Of Group 1 And 2 Class 11 Exercise Question Answers Solutions Maharashtra Board

Class 11 Chemistry Chapter 8 Exercise Solutions Maharashtra Board

Chemistry Class 11 Chapter 8 Exercise Solutions

1. Explain The Following

 

Question A. Hydrogen shows similarity with alkali metals as well as halogens.
Answer:

• Electronic configuration of hydrogen is 1s¹ which is similar to the outer electronic configuration of alkali metals of group 1 i.e., ns¹.
• However, 1s¹ also resembles the outer electronic configuration of group 17 elements i.e., ns² np⁵.
• By adding one electron to H, it will attain electronic configuration of the inert gas He which is 1s² and by adding one electron to ns² np⁵ we get ns² np⁶ which is the outer electronic configuration of the remaining inert gases.
• Therefore, some chemical properties of hydrogen are similar to those of alkali metals while some resemble halogens.

Hence, hydrogen shows similarity with alkali metals as well as halogens.
In simple words: Hydrogen has properties similar to both alkali metals (like having one valence electron) and halogens (like needing one electron to complete its shell). This unique behavior explains why it shares characteristics with both groups.

🎯 Exam Tip: When explaining hydrogen's dual similarity, always mention its electronic configuration and the tendency to gain or lose electrons, linking these to the characteristics of both alkali metals and halogens.

 

Question B. Standard reduction potential of alkali metals have high negative values.
Answer:

• The general outer electronic configuration of alkali metals is ns¹.
• They readily lose one valence shell electron to achieve stable noble gas configuration and hence, they are highly electropositive and are good reducing agents.

Hence, standard reduction potentials of alkali metals have high negative values.
In simple words: Alkali metals easily lose an electron to become stable, making them highly reactive and strong reducing agents. This strong tendency to lose electrons results in very negative standard reduction potentials.

🎯 Exam Tip: Focus on the ease of electron loss and the resulting electropositivity and reducing nature as key factors for high negative standard reduction potentials in alkali metals.

 

Question C. Alkaline earth metals have low values of electronegativity; which decrease down the group.
Answer:

• Electronegativity represents attractive force exerted by the nucleus on shared electrons.
• The general outer electronic configuration of alkaline earth metals is ns². They readily lose their two valence shell electrons to achieve stable noble gas configuration. They are electropositive and hence, they have low values of electronegativity.

In simple words: Alkaline earth metals have low electronegativity because they easily lose their two valence electrons to achieve a stable electron configuration. As you go down the group, atomic size increases, and the outermost electrons are further from the nucleus, making it even easier to lose them, thus decreasing electronegativity.

🎯 Exam Tip: To score well, relate electronegativity directly to the tendency to lose electrons and explain how atomic size influences this trend down the group.

 

Question D. Sodium dissolves in liquid ammonia to form a solution which shows electrical conductivity.
Answer:
(i) Sodium dissolves in liquid ammonia giving deep blue coloured solutions which is electrically conducting in nature. \[\text{Na} + (\text{x} + \text{y}) \text{NH}_3 \to [\text{Na}(\text{NH}_3)_\text{x}]^+ + [\text{e}(\text{NH}_3)_\text{y}]^-\]
(ii) Due to formation of ions, the solution shows electrical conductivity.
In simple words: When sodium dissolves in liquid ammonia, it forms solvated sodium ions and solvated electrons. These free-moving charged particles (ions and electrons) allow the solution to conduct electricity.

🎯 Exam Tip: Remember the two key products of alkali metal dissolution in liquid ammonia: solvated metal ions and solvated electrons. Their presence is crucial for explaining the electrical conductivity.

 

Question E. BeCl2 is covalent while MgCl2 is ionic.
Answer:

• Be²⁺ ion has very small ionic size and therefore, it has very high charge density.
• Due to this, it has high tendency to distort the electron cloud around the negatively charged chloride ion (Cl⁻) which is larger in size.
• This results in partial covalent character of the bond in BeCl₂.
• Mg²⁺ ion has very less tendency to distort the electron cloud of Cl⁻ due to the bigger size of Mg²⁺ as compared to Be²⁺.

Hence, BeCl₂ is covalent while MgCl₂ is ionic.
In simple words: Beryllium's small size and high charge density (high polarizing power) significantly distort the electron cloud of the larger chloride ion, leading to covalent character in BeCl₂. Magnesium, being larger, has less polarizing power, so MgCl₂ remains ionic.

🎯 Exam Tip: This question relates to Fajan's rules. Emphasize the small size and high charge density (polarizing power) of Be²⁺ and the larger size of the anion (Cl⁻) to explain covalent character.

 

Question F. Lithium floats an water while sodium floats and catches fire when put in water.
Answer:

• When lithium and sodium react with water, hydrogen gas is released. Due to these hydrogen gas bubbles, lithium and sodium floats on water. eg. \(2\text{Na} + 2\text{H}_2\text{O} \to 2\text{Na}^+ + 2\text{OH}^- + \text{H}_2\uparrow\)
• The reactivity of group 1 metals increases with increasing atomic radius and lowering of ionization enthalpy down the group.
• Thus, sodium having lower ionization enthalpy, is more reactive than lithium.
• Hence, lithium reacts slowly while sodium reacts vigorously with water.
• Since the reaction of sodium with water is highly exothermic, it catches fire when put in water.

In simple words: Both lithium and sodium float because the hydrogen gas produced in their reaction with water forms bubbles, providing buoyancy. Sodium reacts much more vigorously than lithium due to its lower ionization enthalpy, and this highly exothermic reaction generates enough heat to ignite the hydrogen gas, causing it to catch fire.

🎯 Exam Tip: Key points for this answer are the role of hydrogen bubbles for floating, the trend of reactivity down Group 1 (lower ionization enthalpy for sodium), and the exothermic nature of sodium's reaction with water leading to fire.

 

2. Write Balanced Chemical Equations For The Following.

 

Question A. CO₂ is passed into concentrated solution of NaCl, which is saturated with NH₃.
Answer:\[2\text{NH}_{3(\text{aq})} + \text{H}_2\text{O}_{(\text{l})} + \text{CO}_{2(\text{g})} \to (\text{NH}_4)_2\text{CO}_{3(\text{aq})}\] (Ammonia) (Ammonium carbonate) \[(\text{NH}_4)_2\text{CO}_{3(\text{aq})} + \text{H}_2\text{O}_{(\text{l})} + \text{CO}_{2(\text{g})} \to 2\text{NH}_4\text{HCO}_{3(\text{s})}\] (Ammonium carbonate) (Ammonium bicarbonate) \[\text{NH}_4\text{HCO}_{3(\text{aq})} + \text{NaCl}_{(\text{aq})} \to \text{NaHCO}_{3(\text{s})} + \text{NH}_4\text{Cl}_{(\text{aq})}\] (Ammonium bicarbonate) (From brine solution) (Sodium bicarbonate (precipitate))
In simple words: The process involves several steps to produce sodium bicarbonate. Ammonia, water, and carbon dioxide first react to form ammonium carbonate, which then further reacts with water and carbon dioxide to form ammonium bicarbonate. Finally, ammonium bicarbonate reacts with sodium chloride to precipitate sodium bicarbonate.

🎯 Exam Tip: Make sure to balance each individual equation and correctly label the phases (aq, l, g, s) and names of the compounds. This is a sequence of reactions, so the products of one step are reactants for the next.

 

Question B. A 50% solution of sulphuric acid is subjected to electrolyte oxidation and the product is hydrolysed.
Answer:
(a) Electrolytic oxidation: At anode: \[\text{2HSO}_4^- \xrightarrow{\text{Electrolysis}} \text{H}_2\text{S}_2\text{O}_8 + \text{2e}^-\] (Peroxydisulphuric acid)
(b) Hydrolysis: \[\text{HO-SO}_2\text{-O-O-SO}_2\text{-OH} + \text{2H}_2\text{O} \to \text{2H}_2\text{SO}_4 + \text{H}_2\text{O}_2\] (Peroxydisulphuric acid) (Hydrogen peroxide)
In simple words: First, a concentrated sulphuric acid solution undergoes electrolysis, where bisulphate ions at the anode form peroxydisulphuric acid. Then, this peroxydisulphuric acid is hydrolyzed by water, breaking down to produce sulphuric acid and hydrogen peroxide.

🎯 Exam Tip: For these multi-step reactions, identify the reactants and products at each stage, especially noting the formation of peroxydisulphuric acid as an intermediate and its subsequent hydrolysis into H₂SO₄ and H₂O₂.

 

Question C. Magnesium is heated in air.
Answer:
(a) \[\text{2Mg} + \text{O}_2 \overset{\triangle}{\longrightarrow} \text{2MgO}\] (Magnesium) (Oxygen (air)) (Magnesium oxide)
(b) \[\text{3Mg} + \text{N}_2 \overset{\triangle}{\longrightarrow} \text{Mg}_3\text{N}_2\] (Magnesium) (Nitrogen (air)) (Magnesium nitride)
In simple words: When magnesium is heated in air, it reacts with both oxygen and nitrogen present in the air. With oxygen, it forms magnesium oxide, and with nitrogen, it forms magnesium nitride.

🎯 Exam Tip: Remember that "air" implies the presence of both oxygen and nitrogen. Therefore, magnesium reacts to form two products: an oxide and a nitride. Ensure both reactions are balanced and include the heat symbol (triangle).

 

Question D. Beryllium oxide is treated separately with aqueous HCl and aqueous NaOH solutions.
Answer: Beryllium oxide (BeO) is an amphoteric oxide and thus, it reacts with both acid (HCl) as well as base (NaOH) to give the corresponding products.
(i) \[\text{BeO} + \text{2HCl} \to \text{BeCl}_2 + \text{H}_2\text{O}\] (Acid)
(ii) \[\text{BeO} + \text{2NaOH} \to \text{Na}_2\text{BeO}_2 + \text{H}_2\text{O}\] (Base)
In simple words: Beryllium oxide is amphoteric, meaning it can react with both acids and bases. With hydrochloric acid, it forms beryllium chloride and water. With sodium hydroxide, it forms sodium beryllate and water.

🎯 Exam Tip: The key concept here is the amphoteric nature of BeO. Make sure to show its reaction with both an acid (HCl) and a base (NaOH) and correctly identify the salt and water products in both cases.

 

3. Answer The Following Questions

 

Question A. Describe the diagonal relationship between Li and Mg with the help of two illustrative properties.
Answer:
(a) The relative placement of these elements with similar properties in the periodic table is across a diagonal and is called diagonal relationship.
(b) Lithium is placed in the group 1 and period 2 of the modem periodic table. It resembles with magnesium which is placed in the group 2 and period 3.
ℹ️ चित्र व्याख्या (Diagram Explanation): एक तालिका है जो आवर्त सारणी में लिथियम (Li) और सोडियम (Na) को समूह 1 में, तथा बेरिलियम (Be) और मैग्नीशियम (Mg) को समूह 2 में उनके स्थान को दर्शाती है। यह विशेष रूप से लिथियम और मैग्नीशियम के बीच विकर्ण संबंध को रेखांकित करती है, जो समान गुणों को साझा करते हैं।
(ii) Li and Mg show similarities in many of their properties. e. g.
(a) Reaction with oxygen: 1. Group 1 elements except lithium, react with oxygen present in the air to form oxides (M₂O) as well as peroxides (M₂O₂) and superoxides (MO₂) on further reaction with excess of oxygen. 2. This anomalous behaviour of lithium is due to its resemblance with magnesium as a result of diagonal relationship. 3. As group 2 elements form monoxides i.e., oxides, lithium also form monoxides. \[\text{4Li} + \text{O}_2 \overset{\triangle}{\longrightarrow} \text{2Li}_2\text{O}\] (Lithium metal) (Lithium monoxide) \[\text{Mg} + \text{O}_2 \overset{\triangle}{\longrightarrow} \text{2MgO}\] (Magnesium metal) (Magnesium monoxide (oxide))
(b) Reaction with nitrogen: 1. All the group 1 elements react only with oxygen present in the air to form oxides while group 2 elements react with both nitrogen and oxygen present in the air forming corresponding oxides and nitrides. 2. However, lithium reacts with oxygen as well as nitrogen present in the air due to its resemblance with magnesium. \[\text{6Li} + \text{N}_2 \overset{\triangle}{\longrightarrow} \text{2Li}_3\text{N}\] (Lithium) (Lithium nitride) \[\text{3Mg} + \text{N}_2 \overset{\triangle}{\longrightarrow} \text{Mg}_3\text{N}_2\] (Magnesium) (Magnesium nitride)
In simple words: Diagonal relationship describes how elements diagonally opposite in the periodic table (like Li and Mg) show similar properties due to comparable ionic sizes and polarizing powers. For instance, both Li and Mg form monoxides with oxygen and directly react with nitrogen to form nitrides, unlike other alkali metals which mostly form peroxides/superoxides or don't react directly with nitrogen.

🎯 Exam Tip: When discussing diagonal relationships, define the concept first, then provide specific chemical reactions (like with oxygen and nitrogen) as clear evidence of the property similarities between the diagonal pair.

 

Question B. Describe the industrial production of dihydrogen from steam. Also write the chemical reaction involved.
Answer: Three stages are involved in the industrial production of dihydrogen from steam.
(i) Stage 1:
(a) Reaction of steam on hydrocarbon or coke (C) at 1270 K temperature in presence of nickel catalyst gives water-gas which is a mixture of carbon monoxide and hydrogen. 1. Reaction of steam with hydrocarbon: \[\text{CH}_{4(\text{g})} + \text{H}_2\text{O}_{(\text{g})} \xrightarrow{\text{1270 K}}_{\text{Ni}} \text{CO}_{(\text{g})} + \text{3H}_{2(\text{g})}\] (Methane (Hydrocarbon)) (Steam) (Water-gas (syngas)) 2. Reaction of steam with coke or carbon (C): \[\text{C}_{(\text{s})} + \text{H}_2\text{O}_{(\text{g})} \xrightarrow{\text{1270 K}}_{\text{Ni}} \text{CO}_{(\text{g})} + \text{H}_{2(\text{g})}\]
(b) Sawdust, scrapwood, etc. can also be used in place of carbon.
(ii) Stage 2: Water-gas shift reaction: When carbon monoxide in the water-gas reacts with steam in the presence of iron chromate (FeCrO₄) as catalyst, it gets transformed into carbon dioxide. This is called water-gas shift reaction. \[\text{CO}_{(\text{g})} + \text{H}_2\text{O}_{(\text{g})} \xrightarrow{\text{673 K}}_{\text{Iron chromate catalyst}} \text{CO}_{2(\text{g})} + \text{H}_{2(\text{g})}\] (Carbon monoxide (from syngas mixture)) (Steam) (Carbon dioxide) (Hydrogen)
(iii) Stage 3: In the last stage, carbon dioxide is removed by scrubbing with sodium arsenite solution.
In simple words: Dihydrogen is industrially produced from steam in three stages. First, steam reacts with hydrocarbons or coke to form water-gas (CO + H₂). Next, in the water-gas shift reaction, carbon monoxide is converted to carbon dioxide using more steam and a catalyst, producing more hydrogen. Finally, carbon dioxide is removed to obtain pure dihydrogen.

🎯 Exam Tip: Remember the three key stages: water-gas formation (reforming), water-gas shift reaction, and CO₂ removal. Pay attention to catalysts, temperatures, and the balanced chemical equations for each step.

 

Question C. A water sample, which did not give lather with soap, was found to contain Ca(HCO₃)₂ and Mg(HCO₃)₂. Which chemical will make this water give lather with soap? Explain with the help of chemical reactions.
Answer:

• Soap does not lather in hard water due to presence of the soluble salts of calcium and magnesium in it. So, the given water sample is hard water.
• Hardness of hard water can be removed by removal of these calcium and magnesium salts.
• Sodium carbonate is used to make hard water soft as it precipitates out the soluble calcium and magnesium salts in hard water as carbonates. Thus, it will make water give lather with soap. e.g. \[\text{Ca}(\text{HCO}_3)_{2(\text{aq})} + \text{Na}_2\text{CO}_{3(\text{aq})} \to \text{CaCO}_{3(\text{s})} + \text{2NaHCO}_{3(\text{aq})}\]

In simple words: The water sample is hard because it contains soluble calcium and magnesium bicarbonates, which prevent soap from lathering. Sodium carbonate (washing soda) can soften this water by precipitating these metal ions as insoluble carbonates, thus allowing the soap to lather.

🎯 Exam Tip: Identify the cause of hardness (calcium and magnesium bicarbonates) and the specific chemical (sodium carbonate) used for softening. The key is the precipitation of insoluble carbonates, so ensure the chemical equation accurately shows this.

 

Question D. Name the isotopes of hydrogen. Write their atomic composition schematically and explain which of these is radioactive ?
Answer:
(i) Hydrogen has three isotopes i.e., hydrogen (\(^{1}\text{H}\)), deuterium (\(^{2}\text{H}\)) and tritium (\(^{3}\text{H}\)) with mass numbers 1, 2 and 3 respectively.
(ii) They all contain one proton and one electron but different number of neutrons in the nucleus.
(iii) Atomic composition of isotopes of hydrogen:

Name of the isotope	Symbol	Atomic number Z	Atomic mass number A	Neutron number N	Abundance	Stability
Hydrogen	H or \(^1\)H or H-1	1	1	0	99.98%	Stable
Deuterium	D or \(^2\)H or H-2	1	2	1	0.015%	Stable
Tritium	T or \(^3\)H or H-3	1	3	2	Trace	Radioactive

(iv) Tritium is a radioactive nuclide with half-life period 12.4 years and emits low energy β⁻ particles.
(v) Schematic representation of isotopes of hydrogen is as follows:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र हाइड्रोजन के तीन समस्थानिकों - प्रोटियम (सामान्य हाइड्रोजन), ड्यूटेरियम और ट्रिटियम - के परमाणु संघटन को दर्शाता है। प्रत्येक समस्थानिक में एक प्रोटॉन होता है, लेकिन न्यूट्रॉन की संख्या भिन्न होती है: प्रोटियम में शून्य, ड्यूटेरियम में एक, और ट्रिटियम में दो न्यूट्रॉन होते हैं, जिससे उनके द्रव्यमान संख्याएँ अलग-अलग होती हैं।
In simple words: Hydrogen has three isotopes: protium (\(^{1}\text{H}\)), deuterium (\(^{2}\text{H}\)), and tritium (\(^{3}\text{H}\)). They all have one proton but differ in the number of neutrons (0, 1, and 2 respectively). Among these, tritium is the radioactive isotope, emitting beta particles with a half-life of 12.4 years.

🎯 Exam Tip: Clearly list the three isotopes with their symbols and compositions. The table helps organize this information. Explicitly state that tritium is the radioactive one and mention its decay property.

 

4. Name The Following

 

Question A. Alkali metal with smallest atom.
Answer: Lithium (Li)
In simple words: Lithium is the alkali metal with the smallest atomic size because it is at the top of Group 1 in the periodic table, where atomic radius is smallest.

🎯 Exam Tip: Remember the periodic trends: atomic size generally decreases across a period and increases down a group. Lithium is the first element in Group 1, making it the smallest alkali metal.

 

Question B. The most abundant element in the universe.
Answer: Hydrogen (H)
In simple words: Hydrogen is the most common element in the entire universe, making up about 75% of all matter.

🎯 Exam Tip: This is a factual recall. Hydrogen's abundance is due to its simple structure and its role in stars and celestial bodies.

 

Question C. Radioactive alkali metal.
Answer: Francium (Fr)
In simple words: Francium is the only naturally occurring radioactive element among the alkali metals, located at the bottom of Group 1.

🎯 Exam Tip: Francium (Fr) is typically the heaviest and most unstable element in Group 1, making it radioactive. This is a direct recall question.

 

Question D. Ions having high concentration in cell sap.
Answer: Potassium ions (K⁺)
In simple words: Potassium ions (K⁺) are crucial for many biological processes in cells, and they are found in high concentrations within cell sap, playing a key role in maintaining turgor pressure and enzyme activity.

🎯 Exam Tip: Remember the biological importance of potassium ions, particularly their high concentration in the cytoplasm and cell sap, crucial for cell function and osmotic balance.

 

Question E. A compound having hydrogen, aluminium and lithium as its constituent elements.
Answer: Lithium aluminium hydride (LiAlH₄)
In simple words: Lithium aluminium hydride, often abbreviated as LAH, is a powerful reducing agent commonly used in organic chemistry, and it contains lithium, aluminum, and hydrogen atoms.

🎯 Exam Tip: Lithium aluminium hydride (LiAlH₄) is a common complex hydride. Recognize its formula and common usage as a reducing agent in chemistry.

 

5. Choose The Correct Option.

 

Question A. The unstable isotope of hydrogen is .....
(a) H-1
(b) H-2
(c) H-3
(d) H-4
Answer: (c) H-3
In simple words: Tritium, denoted as H-3, is the only radioactive and unstable isotope among the naturally occurring forms of hydrogen.

🎯 Exam Tip: Remember that H-1 (protium) and H-2 (deuterium) are stable, while H-3 (tritium) is radioactive. This is a fundamental concept regarding hydrogen isotopes.

 

Question B. Identify the odd one.
(a) Rb
(b) Ra
(c) Sr
(d) Be
Answer: (a) Rb
In simple words: Rubidium (Rb) is an alkali metal (Group 1), while Radium (Ra), Strontium (Sr), and Beryllium (Be) are all alkaline earth metals (Group 2). Therefore, Rb is the odd one out.

🎯 Exam Tip: To identify the odd one out, you need to classify each element by its group in the periodic table. Rb is Group 1, while Ra, Sr, and Be are all Group 2.

 

Question C. Which of the following is Lewis acid ?
(a) BaCl₂
(b) KCl
(c) BeCl₂
(d) LiCl
Answer: (c) BeCl₂
In simple words: BeCl₂ acts as a Lewis acid because beryllium has an incomplete octet in its valence shell, allowing it to accept a pair of electrons. BaCl₂, KCl, and LiCl are ionic compounds with complete octets, thus not acting as Lewis acids.

🎯 Exam Tip: Lewis acids are electron pair acceptors. Focus on elements that can accept lone pairs, often those with incomplete octets (like Be in BeCl₂) or vacant orbitals, to correctly identify them.

 

Question D. What happens when crystalline Na₂CO₃ is heated ?
(a) releases CO₂
(b) loses H₂O
(c) decomposes into NaHCO₃
(d) colour changes.
Answer: (b) loses H₂O
In simple words: When crystalline sodium carbonate (washing soda), which is a hydrate (Na₂CO₃·10H₂O), is heated, it loses its water of crystallization. It does not decompose or release CO₂ at typical heating temperatures.

🎯 Exam Tip: Remember that crystalline sodium carbonate is a decahydrate. Heating it primarily leads to the loss of water of crystallization, forming anhydrous sodium carbonate, not decomposition into CO₂ or NaHCO₃.

 

Activity:

 

1. Collect the information of preparation of dihydrogen and make a chart. 2. Find out the s block elements compounds importance/uses.
Answer:
1.

Sr. no.	Method of preparation of dihydrogen
a.	By action of dilute sulphuric acid on magnesium ribbon: High purity hydrogen gas can be prepared by the action of pure dil. H₂SO₄ on magnesium ribbon. \[\text{Mg} + \text{H}_2\text{SO}_{4(\text{dil.})} \to \text{MgSO}_4 + \text{H}_{2(\text{g})}\uparrow\]
b.	By action of water on sodium hydride: Preparation of pure hydrogen gas can also be carried out by action of water on sodium hydride. \[\text{NaH} + \text{H}_2\text{O} \to \text{NaOH} + \text{H}_{2(\text{g})}\uparrow\]
c.	By Uyeno's method: Dihydrogen can be prepared by action of KOH on scrap aluminium or silicon. \[\text{2Al} + \text{2KOH} + \text{2H}_2\text{O}\to \text{2KAlO}_2 + \text{3H}_{2(\text{g})}\uparrow\] This method is known as Uyeno's method and gives very pure hydrogen.
d.	By Bosch process: 1. In Bosch process, water gas (CO + H₂) is mixed with twice the volume of steam. 2. In the presence of promoter (like Cr₂O₃ or ThO₂), this mixture is then passed over heated catalyst (Fe₂O₃) at 773 K. 3. In this reaction, oxidation of CO to CO₂ takes place. \[\text{CO} + \text{H}_2 + \text{H}_2\text{O} \xrightarrow{\text{Fe}_2\text{O}_3+\text{Cr}_2\text{O}_3}_{\text{773K}} \text{CO}_{2(\text{g})} + \text{H}_{2(\text{g})}\] (Water gas) (Steam) 4. Carbon dioxide so formed is removed by dissolving it in water under pressure (20-25 atmospheres), thus leaving behind hydrogen which is collected.
e.	By Lane's process: When superheated steam is passed over iron filings heated to about 1023 - 1073 K, hydrogen is formed. \[\text{3Fe} + \text{4H}_2\text{O} \xrightarrow{\text{1023-1073K}}_{\text{Fe}} \text{Fe}_3\text{O}_{4(\text{s})} + \text{4H}_{2(\text{g})}\] (Steam)

2. Uses of s-block elements: Group 1 elements (alkali metals):
(a) Lithium: Lithium is widely used in batteries.
(b) Sodium:

• Liquid sodium metal is used as a coolant in fast breeder nuclear reactors.
• Sodium is also used as an important reagent in the Wurtz reaction.
• It is used in the manufacture of sodium vapour lamp.

(c) Potassium:

• Potassium has a vital role in biological system.
• Potassium chloride (KCl) is used as a fertilizer.
• Potassium hydroxide (KOH) is used in the manufacture of soft soaps and also as an excellent absorbent of carbon dioxide.
• Potassium superoxide (KO₂) is used as a source of oxygen.

(d) Caesium: Caesium is used in devising photoelectric cells. Group 2 elements (alkaline earth metals):
(a) Magnesium: Magnesium hydroxide [Mg(OH)₂] in its suspension form is used as an antacid.
(b) Calcium: Compounds of calcium such as limestone and gypsum are used as constituents of cement and mortar.
(c) Barium: BaSO₄ being insoluble in H₂O and opaque to X-rays is used as 'barium meal' to scan the X-ray of human digestive system.
In simple words: This activity covers various methods to prepare dihydrogen, including reactions with acids, hydrides, and industrial processes like Bosch and Lane's, each with specific reactants and conditions. It also details the diverse uses of s-block elements, from lithium in batteries and sodium as a coolant, to potassium in fertilizers and biological systems, and magnesium in antacids, highlighting their industrial and biological importance.

🎯 Exam Tip: For preparation methods, focus on the balanced equations and key conditions (catalyst, temperature). For uses of s-block elements, list at least one significant application for each element or its common compounds to demonstrate comprehensive knowledge.

 

Can You Recall? (Textbook Page No. 110)

 

Question 1. Which is the first element in the periodic table?
Answer: Hydrogen is the first element in the periodic table.
In simple words: The periodic table begins with hydrogen, which is the lightest and most abundant element.

🎯 Exam Tip: This is a fundamental recall question. Knowing hydrogen's position is basic for understanding periodic table organization.

 

Question 2. What are isotopes?
Answer: Many elements exist naturally as a mixture of two or more types of atoms or nuclides. These individual nuclides are called isotopes of that element. Isotopes of an element have the same atomic number (number of protons) but different atomic mass numbers due to different number of neutrons in their nuclei.
In simple words: Isotopes are different forms of the same element that have the same number of protons but a different number of neutrons, leading to different atomic masses.

🎯 Exam Tip: Define isotopes by clearly stating the fixed number of protons (same atomic number) and the varying number of neutrons (different mass number).

 

Question 3. Write the formulae of the compounds of hydrogen formed with sodium and chlorine.
Answer: Hydrogen combines with sodium to form sodium hydride (NaH) while it reacts with chlorine to form hydrogen chloride (HCl).
In simple words: Hydrogen forms sodium hydride (NaH) with sodium and hydrogen chloride (HCl) with chlorine.

🎯 Exam Tip: Remember hydrogen's ability to act as both a hydride (H⁻) with highly electropositive metals like Na and as a proton (H⁺) with highly electronegative non-metals like Cl.

 

Can You Tell? (Textbook Page No. 110)

 

Question 1. In which group should hydrogen be placed? In group 1 or group 17? Why?
Answer:

• Hydrogen contains one valence electron in its valence shell and thus, its valency is one. Therefore, hydrogen resembles alkali metals (group 1 elements) as they also contain one electron in their valence shell (alkali metals tend to lose their valence electron).
• However, hydrogen also shows similarity with halogens (group 17 elements) as their valency is also one because halogens tend to accept one electron in their valence shell.
• Due to this unique behaviour, it is difficult to assign any definite position to hydrogen in the modem periodic table.

In simple words: Hydrogen has properties resembling both Group 1 alkali metals (one valence electron, can lose it) and Group 17 halogens (needs one electron to complete its shell, can gain it). This dual nature makes its definitive placement in the periodic table challenging.

🎯 Exam Tip: Explain hydrogen's unique position by citing its electronic configuration and its ability to either lose an electron like alkali metals or gain an electron like halogens. Emphasize that it doesn't fit neatly into either group.

 

Just Think! (Textbook Page No. 112)

 

Question 1. \[\text{2Na}_{(\text{s})} + \text{H}_{2(\text{g})} \overset{\triangle}{\longrightarrow} \text{2NaH}_{(\text{s})}\]
In the above chemical reaction which element does undergo oxidation and which does undergo reduction?
Answer:
(i) Redox reaction can be described as electron transfer as shown below: \[\text{2Na}_{(\text{s})} + \text{H}_{2(\text{g})} \to \text{2Na}^+ + \text{2H}^-\]
(ii) Charge development suggests that each sodium atom loses one electron to form Na⁺ and each hydrogen atom gains one electron to form H⁻. This can be represented as follows:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र सोडियम और हाइड्रोजन के बीच इलेक्ट्रॉन स्थानांतरण को दर्शाता है। इसमें दिखाया गया है कि सोडियम (Na) इलेक्ट्रॉन खोकर Na+ आयन बनाता है (इलेक्ट्रॉन का नुकसान - ऑक्सीकरण), जबकि हाइड्रोजन (H2) इलेक्ट्रॉन प्राप्त करके H- आयन बनाता है (इलेक्ट्रॉन का लाभ - अपचयन)। यह एक रेडॉक्स अभिक्रिया को सचित्र करता है।
(iii) Na is oxidised to NaH by loss of electrons while the elemental hydrogen is reduced to NaH by gain of electrons.
In simple words: In the reaction between sodium and hydrogen to form sodium hydride, sodium (Na) loses electrons and is oxidized, while elemental hydrogen (H₂) gains electrons and is reduced. This is a redox reaction where electron transfer occurs.

🎯 Exam Tip: Clearly define oxidation as loss of electrons and reduction as gain of electrons. Show the change in oxidation states or explicit electron transfer for both reactants to justify your answer.

 

Can You Recall? (Textbook Page No. 113)

 

Question i. What is the name of the family of reactive metals having valency one?
Answer: The family of reactive metals having valency one is known as alkali metals (group 1).
In simple words: The reactive metals that typically have a valency of one (meaning they readily lose one electron) are called alkali metals, found in Group 1 of the periodic table.

🎯 Exam Tip: Remember that elements in Group 1 are called alkali metals and their characteristic valency is one, due to having a single valence electron.

 

Question ii. What is the name of the family of reactive metals having valency two?
Answer: The family of reactive metals having valency two is known as alkaline earth metals (group 2).
In simple words: The reactive metals that commonly exhibit a valency of two (meaning they readily lose two electrons) are known as alkaline earth metals, located in Group 2 of the periodic table.

🎯 Exam Tip: Identify Group 2 elements as alkaline earth metals, whose characteristic valency is two, reflecting their tendency to lose two valence electrons.