NCERT Solutions Class 11 Chemistry Chapter 10

The periodic table is divided into four sections. These four blocks are known as the s, p, d, and f-blocks, and they are divided based on  how full each shell (with electrons) is. The s-orbital is the very first orbital. When the last electron in an element enters the s-orbital, it is referred to as an s-block element.

The s-Block elements are discussed in Chemistry Chapter 10 Class 11. The two groups of the periodic table – groups 1 and 2 – are known as alkali metals and alkaline earth metals, respectively. These two groups are part of the s-block element family, which consists of 13 elements.

NCERT Solutions for Class 11 Chemistry Chapter 10 – The s-Block Elements

Chapter 10 teaches everything students should know about the elements in the modern periodic table. Alkali metals, hydroxides, and oxides’ physical, chemical, and general properties are covered. It also explains all structural and functional reactions, as well as the characteristics and applications of the s-block elements. This chapter will also discuss some important trends such as atomic radius, diagonal relationship, and so on.

Students can understand all of the important concepts and practise questions well enough before their examination by consulting the NCERT solutions for class 11. These Class 11 Chemistry Chapter 10 NCERT Solutions aid students in their preparation for the final exam as well as for competitive exams such as BITSAT, JEE, NEET, and others.

Access NCERT Solutions for Class 11 Chemistry Chapter – 10 The s-Block Elements

You can accessNCERT solutions for Class 11 Chemistry Chapter 10 on Extramarks website.

NCERT Exercise Solutions

The NCERT Solutions Class 11 Chemistry Chapter 10 by Extramarks are written in accordance with the guidelines given by CBSE. The answers to all of the exercise questions in the chapter s-block elements are there to help students review the entire syllabus and improve their grades.

Download NCERT Solutions for Class 11 Chemistry s Block Elements 

After studying  this chapter, students will be able to explain the general properties of alkali compounds as well as their metals. They will also study the general characteristics of alkali metals and their compounds, as well as their properties, manufacturing descriptions, and applications in industry.

On Extramarks, you’ll find the most accurate and detailed NCERT Solutions Class 11 Chemistry Chapter 10. Our chemistry subject expert solves these s-block element solutions. Furthermore, the solutions for this chapter are in accordance with CBSE guidelines.

Topic Wise Solutions of NCERT Solutions of S Block Elements Class 11th

There are two groups in the s-block (1&2). Important concepts of s-block elements are discussed in Ch 10 Chemistry Class 11 NCERT Solutions. Some of the concepts are: properties of group 1 and group 2 elements, their electronic configuration, their occurrence, general characteristics of alkali metal and alkaline earth metal compounds, anomalous behaviour of lithium and beryllium, important sodium and calcium compounds, and finally the biological importance of sodium and potassium. Students can find the detailed description of the chapter here:

Section Number Section Title
10.1 Group 1 Elements: Alkali Metals
10.2 General Characteristics of the Compounds of the Alkali Metals
10.2.1 Oxides and Hydroxides
10.2.2 Halides
10.2.3 Salts of oxo-Acids
10.3 Anomalous Properties of Lithium
10.4 Some Important Compounds of Sodium
10.5 Biological Importance of Sodium and Potassium
10.6 Group 2 Elements: Alkaline Earth Metals
10.7 General Characteristics of Compounds of the Alkaline Earth Metals
10.8 Anomalous Behaviour of Beryllium
10.9 Some Important Compounds of Calcium
10.10 Biological Importance of Magnesium and Calcium

10.1 Group 1 Elements: Alkali Metals

With increasing atomic number, the physical and chemical properties of alkali metals follow a predictable pattern. The electronic configuration, atomic and ionic radii, ionisation enthalpy, hydration enthalpy, physical and chemical properties, and applications of alkali metals are discussed in this topic.

10.2 General Characteristics of the Compounds of the Alkali Metals

The  most common compounds of Alkali metals are all ionic in nature. Some of their compounds’ general characteristics are discussed here.

10.2.1 Oxides and Hydroxides

All of the elements in group 2 produce metallic oxide when they are burned with oxygen gas. Alkaline earth metal oxides have lower basicity. When the electropositive character of an oxide increases, it becomes more basic. 

The tendency of alkaline earth metals to form hydroxides decreases as one moves down the group. All hydroxides are basic in nature, and their basicity decreases as they progress through the group.

10.2.2 Halides

Halides are two-part binary compounds in which one part is an element and the other is a halogen atom. This section also examines its reactions, precipitation, examples, types, and applications.

10.2.3 Salts of oxo-Acids

Oxoacid salts are compounds in which a highly electropositive alkali metal reacts with oxoacids to form salts. They are generally water-soluble and heat resistant. Definitions, formulae, and examples will be used to further explain these concepts in this section.

10.3 Anomalous Properties of Lithium

Lithium’s peculiar behaviour is due to two factors: (i) its atom and ion’s extremely small size, and (ii) its high polarising power (charge/ radius ratio). The differences between lithium and alkaline metals, as well as the similarities between lithium and magnesium, are discussed in this section.

10.4 Some Important Compounds of Sodium

Sodium bicarbonate, sodium carbonate, sodium chloride, and sodium hydroxide are all important industrial compounds of sodium. The large-scale production of these elements, as well as their applications, are discussed in this topic.

10.5 Biological Importance of Sodium and Potassium

The biological significance of sodium and potassium ions is explained, including how they help regulate the flow of water across cell membranes and how they activate many enzymes.

10.6 Group 2 Elements: Alkaline Earth Metals

Barium, Beryllium, Calcium, Magnesium, Radium, and Strontium make group 2. The electronic configuration, atomic and ionic radii, ionisation enthalpy, hydration enthalpy, physical and chemical properties, and applications of alkali earth metals are discussed in this topic. It also goes over the anomalous properties of Beryllium and important calcium compounds in depth.

10.7 General Characteristics of Compounds of the Alkaline Earth Metals

The chemistry of alkaline earth metals is very similar to alkali metals. In the case of alkaline earth metals, however, some differences arise due to smaller atomic and ionic sizes and higher cationic charges. Their oxides and hydroxides are less basic than the oxides and hydroxides of alkali metals. Other chemical characteristics are also discussed.

10.8 Anomalous Behaviour of Beryllium

Beryllium, the first element in group 2, has several properties that set it apart from the other elements in its group, resulting in unusual behaviour. Students will investigate the factors that cause Beryllium’s unusual behaviour.

10.9 Some Important Compounds of Calcium

Calcium oxide (lime), calcium hydroxide (slaked lime), calcium sulphate (Plaster of Paris), calcium carbonate (limestone), and cement are all industrially important calcium compounds. This section also includes calcium-based construction and industrial materials.

10.10 Biological Importance of Magnesium and Calcium

All enzymes that use ATP require magnesium as a cofactor. Magnesium can be found in green plants’ chlorophyll pigment. Students will explore further the biological importance of Calcium and Magnesium. 

There are 32 questions in this chapter’s exercise. This NCERT Solutions Class 11 Chemistry has been curated by our chemistry experts and comprises solutions to all the questions discussed in s-block elements chapter.

Short Questions

The first section of the Class 11 Chemistry Chapter 10 NCERT Solutions contains short questions and answers such as: 

  • What are the physical and chemical properties of alkali metals?
  • Why is alkali not found in nature?
  • Why is potassium more reactive than sodium?

Very Short Questions

This section of the Ch 10 Chemistry Class 11 NCERT Solutions contains a number of questions with very short answers. Some of the questions are:

  • Which compounds are used in the Solvay process for washing soda production?
  • Which electrolyte is added in Castner’s process for sodium production?

Long Questions

The NCERT Solutions Class 11 Chemistry Chapter 10 also include detailed solutions to a number of lengthy questions, the majority of which are conceptual. For example:

  • Why is lithium the best reducing agent, particularly in aqueous solutions?
  • How to determine the oxidation state of alkali metals?

Multiple Choice Questions

The final section of this NCERT solutions Class 11 includes a number of multiple-choice questions and answers. 

Why Should Students Choose NCERT Solution of s-block Elements Class 11?

Students can download and access Class 11 Chemistry Chapter 10 NCERT Solutions as per their convenience. The solutions are prepared by subject-matter experts,after extensive research. Here are some of the reasons why students should choose NCERT solutions of s-block elements Class 11 by Extramarks:

  • NCERT Solutions Class 11 Chemistry Chapter 10 are written in a comprehensive manner, and simple language, making it easier for students to understand the concepts.
  • Due to the detailed nature of solutions, students can refer to these for revision purpose.
  • Each answer includes the necessary equations and diagrams to keep students engaged in the topic.

Related Questions

  1. Q) For photoemission, the best metal to use is:
  2. Potassium
  3. Sodium 
  4. Caesium
  5. Lithium

Ans. Caesium

  1. Q) Why does Li2CO3 decompose at a lower temperature but Na2CO3 decomposes at a higher temperature?

Ans. Li2CO3 decomposes at a lower temperature, however, Na2CO3 decomposes at a higher temperature. This is because

  • Lithium is less electropositive than Sodium
  • Na2CO3 is more stable in comparison to Li2CO3
  • Lithium carbonate is not stable to heat, whereas, Sodium carbonate is very stable to heat
  1. Q) Potassium nitrate is also known as 
  2. Mohr’s salt
  3. Indian saltpetre
  4. Gypsum
  5. Chile saltpetre

Ans. Indian saltpetre

  1. Q) Which combination of elements will have the most violent reaction?
  2. Lithium and chlorine
  3. Potassium and chlorine
  4. Lithium and iodine
  5. Potassium and iodine

Ans. Potassium and chlorine

  1. Q) It is critical that no toxic or flammable products form when dealing with a spill of metallic sodium. In the event of a spill of metallic sodium, what material should be used?
  2. Dilute Hydrochloric acid
  3. Ethanol
  4. Sand
  5. Water spray 

Ans. Sand

Q.1 What are the common physical and chemical features of alkali metals?

Ans.

Physical properties of alkali metals are as follows.

(1) They are quite soft and can be cut easily. Sodium metal can be easily cut using a knife.

(2) They are light coloured and are mostly silvery white in appearance.

(3) They have low density because of the large atomic sizes. The density increases down the group from Li to Cs. The only exception to this is K, which has lower density than Na.

(4) The metallic bonding present in alkali metals is quite weak. Therefore, they have low melting and boiling points.

(5) Alkali metals and their salts impart a characteristic colour to flames. This is because the heat from the flame excites the electron present in the outermost orbital to a high energy level. When this excited electron reverts back to the ground state, it emits excess energy as radiation that falls in the visible region.

(6) They also display photoelectric effect. When metals such as Cs and K are irradiated with light, they lose electrons.

Chemical properties of alkali metals

Alkali metals are highly reactive due to their low ionization enthalpy. As we move down the group, the reactivity increases.

(1) They react with water to form respective oxides or hydroxides. As we move down the group, the reaction becomes more and more spontaneous.

(2) They react with water to form their respective hydroxides and dihydrogens. The general reaction for the same is given as

2M + 2H2O → 2M+ + 2OHΘ + H2

(3) They react with dihydrogen to form metal hydrides. These hydrides are ionic solids and have high melting points.

2M + H2 → 2M+H

(4) Almost all alkali metals, except Li, react directly with halogens to form ionic halides.

2M + Cl2 → 2MCl

(M = Li,K,Rb,Cs)

Since Li+ ion is very small in size, it can easily distort the electron cloud around the negative halide ion. Therefore, lithium halides are covalent in nature.

(5) They are strong reducing agents. The reducing power of alkali metals increases on moving down the group. However, lithium is an exception. It is the strongest reducing agent among the alkali metals. It is because of its high hydration energy.

(6) They dissolve in liquid ammonia to form deep blue coloured solutions. These solutions are conducting in nature.

M + (x + y)NH3 → [M(NH3)x]+ + [M(NH3)y]

The ammoniated electrons cause the blue colour of the solution. These solutions are paramagnetic and if allowed to stand for some time, then they liberate hydrogen. This results in the formation of amides.

M am + + e + NH 3( t ) MNH ( am ) + 1 2 H 2 ( g ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVC0dg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeaacaGaaiaabaqaamaabaabaaGcbaGaaeytamaaDa aaleaacaqGHbGaaeyBaaqaaiaabUcaaaGccaaMe8Uaae4kaiaaysW7 caqGLbWaaWbaaSqabeaacaqGtacaaOGaaGjbVlaabUcacaaMe8Uaae OtaiaabIeadaWgaaWcbaGaae4mamaabmaabaGaaeiDaaGaayjkaiaa wMcaaaqabaGccaaMe8+aa4ajaSqaaiaaysW7aeqakiaawkziaiaays W7caqGnbGaaeOtaiaabIeadaWgaaWcbaWaaeWaaeaacaqGHbGaaeyB aaGaayjkaiaawMcaaaqabaGccaaMe8Uaae4kaiaaysW7daWcaaqaai aabgdaaeaacaqGYaaaaiaabIeadaWgaaWcbaGaaeOmaaqabaGcdaqa daqaaiaabEgaaiaawIcacaGLPaaaaaa@5D43@

In a highly concentrated solution, the blue colour changes to bronze and the solution becomes diamagnetic.

Q.2 Discuss the general characteristics and gradation in properties of alkaline earth metals.

Ans.

General characteristics of alkaline earth metals are as follows.

(i) The general electronic configuration of alkaline earth metals is [noble gas] ns2.

(ii) These metals lose two electrons to acquire the nearest noble gas configuration. Therefore, their oxidation state is +2.

(iii)These metals have atomic and ionic radii smaller than that of alkali metals. Also, when moved down the group, the effective nuclear charge decreases and this causes an increase in their atomic radii and ionic radii.

(iv)Since the alkaline earth metals have large size, their ionization enthalpies are found to be fairly low. However, their first ionization enthalpies are higher than the corresponding group 1 metals.

(v) These metals are lustrous and silvery white in appearance. They are relatively less soft as compared to alkali metals.

(vi) Atoms of alkaline earth metals are smaller than that of alkali metals. Also, they have two valence electrons forming stronger metallic bonds. These two factors cause alkaline earth metals to have high melting and boiling points as compared to alkali metals.

(vii) They are highly electropositive in nature. This is due to their low ionization enthalpies. Also, the electropositive character increases on moving down the group from Be to Ba.

(viii) Ca, Sr, and Ba impart characteristic colours to flames.

Ca – Brick red

Sr – Crimson red

Ba – Apple green

In Be and Mg, the electrons are too strongly bound to be excited. Hence, these do not impart any colour to the flame. The alkaline earth metals are less reactive than alkali metals and their reactivity increases on moving down the group. Chemical properties of alkaline earth metals are as follows.

(i) Reaction with air and water: Be and Mg are almost inert to air and water because of the formation of oxide layer on their surface.

(a) Powdered Be burns in air to form BeO and Be3N2.

(b) Mg, being more electropositive, burns in air with a dazzling sparkle to form MgO and Mg3N2.

(c) Ca, Sr, and Ba react readily with air to form respective oxides and nitrides.

(d) Ca, Ba, and Sr react vigorously even with cold water.

(ii) Alkaline earth metals react with halogens at high temperatures to form halides.

M + X2 → MX2(X = F, Cl, Br, l)

(iii) All the alkaline earth metals, except Be, react with hydrogen to form hydrides.

(iv) They react readily with acids to form salts and liberate hydrogen gas.

M + 2HCl → MCl2 + H2(g)↑

(v) They are strong reducing agents. However, their reducing power is less than that of alkali metals. As we move down the group, the reducing power increases.

(vi) Similar to alkali metals, the alkaline earth metals also dissolve in liquid ammonia to give deep blue coloured solutions.

M + (x – y)NH3 → [M(NH3)x]+2 + 2[e(NH3)y]

Q.3 Why are alkali metals not found in nature?

Ans.

Alkali metals include lithium, sodium, potassium, rubidium, cesium, and francium. These metals have only one electron in their valence shell, which they lose easily, owing to their low ionization energies. Therefore, alkali metals are highly reactive and are not found in nature in their elemental state.

Q.4 Find the oxidation state of sodium in Na 2O 2.

Ans.

Let the oxidation state of Na be x. The oxidation state of oxygen, in case of peroxides, is –1.

Therefore,

2(x) + 2(–1) = 0

2x – 2 = 0

2x = 2

x = +1

Therefore, the oxidation sate of sodium is +1.

Q.5 Explain why is sodium less reactive than potassium?

Ans.

In alkali metals, on moving down the group, the atomic size increases and the effective nuclear charge decreases. Ionisation enthalpy of potassium is less than that of sodium. Because of these factors, the outermost electron in potassium can be lost easily as compared to sodium. Hence, potassium is more reactive than sodium.

Q.6 Compare the alkali metals and alkaline earth metals with respect to (i) ionization enthalpy (ii) basicity of oxides and (iii) solubility of hydroxides.

Ans.

Alkali metals Alkaline earth metals
(i) Ionization enthalpy:

These have lowest ionization enthalpies in respective periods. This is because of their large atomic sizes. Also, they lose their only valence electron easily as they attain stable noble gas configuration after losing it.

(i) Ionization enthalpy:

Alkaline earth metals have smaller atomic size and higher effective nuclear charge as compared to alkali metals. This causes their first ionization enthalpies to be higher than that of alkali metals. However, their second ionization enthalpy is less than the corresponding alkali metals. This is because alkali metals, after losing one

electron, acquires noble gas configuration, which is very stable.

(ii) Basicity of oxides:

The oxides of alkali metals are very basic in nature. This happens due to the highly

electropositive nature of alkali metals, which makes these oxides highly ionic. Hence, they

readily dissociate in water to give hydroxide ions.

(ii) Basicity of oxides:

The oxides of alkaline earth metals are quite basic but not as basic as those of alkali metals. This is because alkaline earth metals are less electropositive than alkali metals.

(iii) Solubility of hydroxides:

The hydroxides of alkali metals are more soluble than those of alkaline earth metals.

(iii) Solubility of hydroxides:

The hydroxides of alkaline earth metals are less soluble than those of alkali metals. This is due to the high lattice

energies of alkaline earth metals. Their higher charge densities (as compared to

alkali metals) account for higher lattice energies.

Q.7 In what ways lithium shows similarities to magnesium in its chemical behaviour?

Ans.

Similarities between lithium and magnesium are as follows.

(i) Both Li and Mg react slowly with cold water.

(ii) The oxides of both Li and Mg are much less soluble in water and their hydroxides decompose at high temperature.

2LiOH heat Li 2 O+ H 2 O MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVC0dg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeaacaGaaiaabaqaamaabaabaaGcbaGaaeOmaiaabY eacaqGPbGaae4taiaabIeadaGdKaWcbaGaaeiAaiaabwgacaqGHbGa aeiDaaqabOGaayPKHaGaaeitaiaabMgadaWgaaWcbaGaaeOmaaqaba GccaqGpbGaaGjbVlaabUcacaaMe8UaaeisamaaBaaaleaacaqGYaaa beaakiaab+eaaaa@48C0@ Mg ( OH ) 2 heat MgO+ H 2 O MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVC0dg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeaacaGaaiaabaqaamaabaabaaGcbaGaaeytaiaabE gadaqadaqaaiaab+eacaqGibaacaGLOaGaayzkaaWaaSbaaSqaaiaa bkdaaeqaaOWaa4ajaSqaaiaabIgacaqGLbGaaeyyaiaabshaaeqaki aawkziaiaab2eacaqGNbGaae4taiaaysW7caqGRaGaaGjbVlaabIea daWgaaWcbaGaaeOmaaqabaGccaqGpbaaaa@4992@

(iii) Both Li and Mg .

6Li+ N 2 heat 2Li 2 N MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVC0dg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeaacaGaaiaabaqaamaabaabaaGcbaGaaeOnaiaabY eacaqGPbGaaGjbVlaabUcacaaMe8UaaeOtamaaBaaaleaacaqGYaaa beaakmaaoqcaleaacaqGObGaaeyzaiaabggacaqG0baabeGccaGLsg cacaaMe8UaaeOmaiaabYeacaqGPbWaaSbaaSqaaiaabkdaaeqaaOGa aeOtaaaa@489C@ 3Mg+ N 2 heat Mg 3 N 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVC0dg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeaacaGaaiaabaqaamaabaabaaGcbaGaae4maiaab2 eacaqGNbGaaGjbVlaabUcacaaMe8UaaeOtamaaBaaaleaacaqGYaaa beaakmaaoqcaleaacaqGObGaaeyzaiaabggacaqG0baabeGccaGLsg cacaqGnbGaae4zamaaBaaaleaacaqGZaaabeaakiaab6eadaWgaaWc baGaaeOmaaqabaaaaa@4737@

(iv) Neither Li nor Mg form peroxides or superoxides.

(v) The carbonates of both are covalent in nature. Also, these decompose

Li 2 + CO 3 heat 2Li 2 O+ CO 2 MgCO 3 heat MgO+ CO 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVC0dg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeaacaGaaiaabaqaamaabaabaaGcbaGaaeytaiaabE gacaqGdbGaae4tamaaBaaaleaacaqGZaaabeaakmaaoqcaleaacaqG ObGaaeyzaiaabggacaqG0baabeGccaGLsgcacaqGnbGaae4zaiaab+ eacaaMe8Uaae4kaiaaysW7caqGdbGaae4tamaaBaaaleaacaqGYaaa beaaaaa@47F6@

(vi) Li and Mg do not form solid bicarbonates.

(vii) Both LiCl and MgCl2 are soluble in ethanol owing to their covalent nature.

(viii) Both LiCl and MgCl2 are deliquescent in nature. They crystallize solutions as hydrates, for example,

LiCl . 2H2O and MgCl2 . 8H2O

Q.8 Explain why alkali and alkaline earth metals cannot be obtained by chemical reduction methods?

Ans.

In the process of chemical reduction, oxides of metals are reduced using a stronger reducing agent. Alkali metals and alkaline earth metals are among the strongest reducing agents and the reducing agents that are stronger than them are not available.

Therefore, they cannot be obtained by chemical reduction of their oxides.

Q.9 Why are potassium and cesium, rather than lithium used in photoelectric cells?

Ans.

All the three, lithium, potassium, and cesium, are alkali metals. Still, K and Cs are used in the photoelectric cell and not Li. This is because as compared to Cs and K, Li is smaller in size and therefore, requires high energy to lose an electron. While on the other hand, K and Cs have low ionization energy.

Hence, they can easily lose electrons. This property of K and Cs is utilized in photoelectric cells.

Q.10 When an alkali metal dissolves in liquid ammonia the solution can acquire different colours. Explain the reasons for this type of colour change.

Ans.

When an alkali metal is dissolved in liquid ammonia, it results in the formation of a deep blue coloured solution.

M + (x + y)NH3 → M+(NH3)x + e–1(NH3)y

The ammoniated electrons absorb energy corresponding to red region of visible light. Therefore, the transmitted light is blue in colour. At a higher concentration (3 M), clusters of metal ions are formed. This causes the solution to attain a copper–bronze colour and a characteristic metallic lustre.

Q.11 Beryllium and magnesium do not give colour to flame whereas other alkaline earth metals do so. Why?

Ans.

When an alkaline earth metal is heated, the valence electrons get excited to a higher energy level. When this excited electron comes back to its lower energy level, it radiates energy, which belongs to the visible region. Hence, the colour is observed. In Be and Mg, the electrons are strongly bound. The energy required to excite these electrons is very high. Therefore, when the electron reverts back to its original position, the energy released does not fall in the visible region. Hence, no colour in the flame is seen.

Q.12 Discuss the various reactions that occur in the Solvay process.

Ans.

Solvay process is used to prepare sodium carbonate. When carbon dioxide gas is bubbled through a brine solution saturated with ammonia, sodium hydrogen carbonate is formed. This sodium hydrogen carbonate is then converted to sodium carbonate.

Step 1: Brine solution is saturated with ammonia.

2NH3 + H2O + CO2 → (NH4)CO3

This ammoniated brine is filtered to remove any impurity.

Step 2: Carbon dioxide is reacted with this ammoniated brine to result in the formation of insoluble sodium hydrogen carbonate.

NH3 + H2O + CO2 → NH4HCO3

NaClNH4HCO3 → NH4HCO3 ↓ + NH4Cl

Step 3: The solution containing crystals of NaHCO3 is filtered to obtain NaHCO3.

Step 4: NaHCO3 is heated strongly to convert it into NaHCO3.

2NaHCO3 → Na2CO3 + CO2 + H2O

Step 5: To recover ammonia, the filtrate (after removing NaHCO3) is mixed with Ca(OH)2 and heated.

Ca(OH)2 + 2NH4Cl → LNH3 + 2H2O + CaCl2

The overall reaction taking place in Solvay process is

2NaCl + CaCO3 → Na2CO3 + CaCl2

Q.13 Potassium carbonate cannot be prepared by Solvay process. Why?

Ans.

Solvay process cannot be used to prepare potassium carbonate. This is because unlike sodium bicarbonate, potassium bicarbonate is fairly soluble in water and does not precipitate out.

Q.14 Why is Li2CO3 decomposed at a lower temperature whereas Na2CO3 at higher temperature?

Ans.

As we move down the alkali metal group, the electropositive character increases. This causes an increase in the stability of alkali carbonates. However, lithium carbonate is not so stable to heat. This is because lithium carbonate is covalent. Lithium ion, being very small in size, polarizes a large carbonate ion, leading to the formation of more stable lithium oxide.

Li 2 CO 3 Δ Li 2 O+ CO 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVC0dg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeaacaGaaiaabaqaamaabaabaaGcbaGaaeitaiaabM gadaWgaaWcbaGaaeOmaaqabaGccaqGdbGaae4tamaaBaaaleaacaqG ZaaabeaakiaaysW7daGdKaWcbaGaaeiLdaqabOGaayPKHaGaaGjbVl aabYeacaqGPbWaaSbaaSqaaiaabkdaaeqaaOGaae4taiaaysW7caqG RaGaaGjbVlaaboeacaqGpbWaaSbaaSqaaiaabkdaaeqaaaaa@4A54@

Therefore, lithium carbonate decomposes at a low temperature while a stable sodium carbonate decomposes at a high temperature.

Q.15 Compare the solubility and thermal stability of the following compounds of the alkali metals with those of the alkaline earth metals. (a) Nitrates (b) Carbonates (c) Sulphates.

Ans.

(i) Nitrates
Thermal stability Nitrates of alkali metals, except LiNO3, decompose on strong heating to form nitrites.
2KNO3(s) → 2KNO2(s) + O2(g)
LiNO3, on decomposition, gives oxide.

2LiNO 3( s ) Δ Li 2 O ( s ) + 2NO 2( g ) + O 2( g ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVC0dg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeaacaGaaiaabaqaamaabaabaaGcbaGaaeOmaiaabY eacaqGPbGaaeOtaiaab+eadaWgaaWcbaGaae4mamaabmaabaGaae4C aaGaayjkaiaawMcaaaqabaGccaaMe8+aa4ajaSqaaiaabs5aaeqaki aawkziaiaaysW7caqGmbGaaeyAamaaBaaaleaacaqGYaaabeaakiaa b+eadaWgaaWcbaWaaeWaaeaacaqGZbaacaGLOaGaayzkaaaabeaaki aaysW7caqGRaGaaeiiaiaaykW7caqGYaGaaeOtaiaab+eadaWgaaWc baGaaeOmamaabmaabaGaae4zaaGaayjkaiaawMcaaaqabaGccaqGGa Gaae4kaiaabccacaqGpbWaaSbaaSqaaiaabkdadaqadaqaaiaabEga aiaawIcacaGLPaaaaeqaaaaa@5955@

Similar to lithium nitrate, alkaline earth metal nitrates also decompose to give oxides.

2Ca( NO 3 ) ( s ) Δ 2CaO ( s ) + 4NO 2( g ) + O 2( g ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVC0dg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeaacaGaaiaabaqaamaabaabaaGcbaGaaeOmaiaabo eacaqGHbWaaeWaaeaacaqGobGaae4tamaaBaaaleaacaqGZaaabeaa aOGaayjkaiaawMcaamaaBaaaleaadaqadaqaaiaabohaaiaawIcaca GLPaaaaeqaaOGaaGjbVpaaoqcaleaacaqGuoaabeGccaGLsgcacaaM e8UaaeOmaiaaboeacaqGHbGaae4tamaaBaaaleaadaqadaqaaiaabo haaiaawIcacaGLPaaaaeqaaOGaaGjbVlaabUcacaqGGaGaaGPaVlaa bsdacaqGobGaae4tamaaBaaaleaacaqGYaWaaeWaaeaacaqGNbaaca GLOaGaayzkaaaabeaakiaabccacaqGRaGaaeiiaiaab+eadaWgaaWc baGaaeOmamaabmaabaGaae4zaaGaayjkaiaawMcaaaqabaaaaa@5ABE@

As we move down group 1 and group 2, the thermal stability of nitrate increases.
Solubility
Nitrates of both group 1 and group 2 metals are soluble in water.
(ii) Carbonates

Thermal stability
The carbonates of alkali metals are stable towards heat. However, carbonate of lithium, when heated, decomposes to form lithium oxide. The carbonates of alkaline earth metals also decompose on heating to form oxide and carbon dioxide.

Na 2 CO 3 Δ Noeffect Li 2 CO 3 Δ Li 2 O + CO 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVC0dg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeaacaGaaiaabaqaamaabaabaaGcbaGaaeitaiaabM gadaWgaaWcbaGaaeOmaaqabaGccaqGdbGaae4tamaaBaaaleaacaqG ZaaabeaakiaaysW7daGdKaWcbaGaaeiLdaqabOGaayPKHaGaaGjbVl aabYeacaqGPbWaaSbaaSqaaiaabkdaaeqaaOGaae4taiaaysW7caqG RaGaaeiiaiaaboeacaqGpbWaaSbaaSqaaiaabkdaaeqaaaaa@496A@ MgCO 3 Δ MgO+ CO 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVC0dg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeaacaGaaiaabaqaamaabaabaaGcbaGaaeytaiaabE gacaqGdbGaae4tamaaBaaaleaacaqGZaaabeaakiaaysW7daGdKaWc baGaaeiLdaqabOGaayPKHaGaaGjbVlaab2eacaqGNbGaae4taiaays W7caqGRaGaaGjbVlaaboeacaqGpbWaaSbaaSqaaiaabkdaaeqaaaaa @487C@

Solubility
Carbonates of alkali metals are soluble in water with the exception of Li2CO3. Also, the solubility increases as we move down the group.
Carbonates of alkaline earth metals are insoluble in water.
(iii) Sulphates
Thermal stability
Sulphates of both group 1 and group 2 metals are stable towards heat.
Solubility
Sulphates of alkali metals are soluble in water. However, sulphates of alkaline earth metals show varied trends.
BeSO4 Fairly soluble
MgSO4 Soluble
CaSO4 Sparingly soluble
SrSO4 Insoluble
BaSO4 Insoluble
In other words, while moving down the alkaline earth metals, the solubility of their sulphates decreases.

Q.16 Starting with sodium chloride how would you proceed to prepare (i) sodium metal (ii) sodium hydroxide (iii) sodium peroxide (iv) sodium carbonate?

Ans.

a. Sodium can be extracted from sodium chloride by Downs process.

This process involves the electrolysis of fused NaCl (40%) and CaCl2 (60 %) at a temperature of 1123 K in Downs cell.

Steel is the cathode and a block of graphite acts as the anode. Metallic Na and Ca are formed at cathode. Molten sodium is taken out of the cell and collected over kerosene.

NaCl Electrolysis Na + + Cl MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVC0dg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeaacaGaaiaabaqaamaabaabaaGcbaGaaeOtaiaabg gacaqGdbGaaeiBaiaabccadaGdKaWcbaGaaeyraiaabYgacaqGLbGa ae4yaiaabshacaqGYbGaae4BaiaabYgacaqG5bGaae4CaiaabMgaca qGZbaabeGccaGLsgcacaqGGaGaaeOtaiaabggadaahaaWcbeqaaiaa bUcaaaGccaqGGaGaae4kaiaabccacaqGdbGaaeiBamaaCaaaleqaba Gaae4eGaaaaaa@4E39@

Molten

At Cathode: Na+ + e → Na

At Anode: Cl + e → C

Cl + Cl → Cl2

(ii) Sodium hydroxide can be prepared by the electrolysis of sodium chloride. This is called Castner–Kellner process. In this process, the brine solution is electrolysed using a carbon anode and a mercury cathode.

The sodium metal, which is discharged at cathode, combines with mercury to form anamalgam.

Cathode: Na + + e Hg Naamalgam Anode Cl + 1 2 Cl 2 + e MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVC0dg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeaacaGaaiaabaqaamaabaabaaGceaqabeaacaqGdb GaaeyyaiaabshacaqGObGaae4BaiaabsgacaqGLbGaaeOoaiaaysW7 caaMe8UaaGjbVlaab6eacaqGHbWaaWbaaSqabeaacaqGRaaaaOGaaG jbVlaabUcacaaMe8UaaeyzamaaCaaaleqabaGaae4eGaaakmaaoqca leaacaqGibGaae4zaaqabOGaayPKHaGaaeOtaiaabggacaaMe8Uaae ifGiaaysW7caqGHbGaaeyBaiaabggacaqGSbGaae4zaiaabggacaqG TbaabaGaaeyqaiaab6gacaqGVbGaaeizaiaabwgacaaMe8UaaGjbVl aaboeacaqGSbWaaWbaaSqabeaacaqGtacaaOGaaGjbVpaaoqcaleaa aeqakiaawkziaiaabccacaqGRaGaaeiiamaalaaabaGaaeymaaqaai aabkdaaaGaae4qaiaabYgadaWgaaWcbaGaaeOmaaqabaGccaaMe8Ua ae4kaiaaysW7caqGLbWaaWbaaSqabeaacaqGtacaaaaaaa@71C3@

(iii) Sodium peroxide

First, NaCl is electrolysed to result in the formation of Na metal (Downs process).This sodium metal is then heated on aluminium trays in air (free of CO2) to form its peroxide.

2Na + O2(air) → Na2O2

(iv) Sodium carbonate is prepared by Solvay process. Sodium hydrogen carbonate is precipitated in a reaction of sodium chloride and ammonium hydrogen carbonate.

2NH3 + H2O + CO2 → (NH4)2CO3

(NH4)2CO3 + H2O + CO2 → 2NH4HCO3

These sodium hydrogen carbonate crystals are heated to give sodium carbonate.

NaHCO3 → Na2CO3 + CO2 + H2O

Q.17 What happens when (i) magnesium is burnt in air (ii) quick lime is heated with silica (iii) chlorine reacts with slaked lime (iv) calcium nitrate is heated ?

Ans.

(i) Magnesium burns in air with a dazzling light to form MgO and Mg3N2.

2Mg + O 2 Burning 2MgO MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVC0dg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeaacaGaaiaabaqaamaabaabaaGcbaGaaeOmaiaab2 eacaqGNbGaaeiiaiaabUcacaqGGaGaae4tamaaBaaaleaacaqGYaaa beaakiaaysW7daGdKaWcbaGaaeOqaiaabwhacaqGYbGaaeOBaiaabM gacaqGUbGaae4zaaqabOGaayPKHaGaaGjbVlaabkdacaqGnbGaae4z aiaab+eaaaa@4A22@ 3Mg + N 2 Burning Mg 3 N 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVC0dg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeaacaGaaiaabaqaamaabaabaaGcbaGaae4maiaab2 eacaqGNbGaaeiiaiaabUcacaqGGaGaaeOtamaaBaaaleaacaqGYaaa beaakiaaysW7daGdKaWcbaGaaeOqaiaabwhacaqGYbGaaeOBaiaabM gacaqGUbGaae4zaaqabOGaayPKHaGaaGjbVlaab2eacaqGNbWaaSba aSqaaiaabodaaeqaaOGaaeOtamaaBaaaleaacaqGYaaabeaaaaa@4B39@

(ii) Quick lime (CaO) combines with silica (SiO2) to form slag.

CaO + SiO 2 heat CaSiO 3 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVC0dg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeaacaGaaiaabaqaamaabaabaaGcbaGaae4qaiaabg gacaqGpbGaaeiiaiaabUcacaqGGaGaae4uaiaabMgacaqGpbWaaSba aSqaaiaabkdaaeqaaOGaaGjbVpaaoqcaleaacaqGObGaaeyzaiaabg gacaqG0baabeGccaGLsgcacaaMe8Uaae4qaiaabggacaqGtbGaaeyA aiaab+eadaWgaaWcbaGaae4maaqabaaaaa@4B14@

(iii) When chloride is added to slaked lime, it gives bleaching powder.

Ca ( OH ) 2 + Cl 2 Δ CaOCl 2 + H 2 O MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVC0dg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeaacaGaaiaabaqaamaabaabaaGcbaGaae4qaiaabg gadaqadaqaaiaab+eacaqGibaacaGLOaGaayzkaaWaaSbaaSqaaiaa bkdaaeqaaOGaaeiiaiaabUcacaqGGaGaae4qaiaabYgadaWgaaWcba GaaeOmaaqabaGccaaMe8+aa4ajaSqaaiaabs5aaeqakiaawkziaiaa ysW7caqGdbGaaeyyaiaab+eacaqGdbGaaeiBamaaBaaaleaacaqGYa aabeaakiaabccacaqGRaGaaeiiaiaabIeadaWgaaWcbaGaaeOmaaqa baGccaqGpbaaaa@4F58@

Bleaching

Powder

(iv)Calcium nitrate, on heating, decomposes to give calcium oxide.

2Ca ( NO 3 ) 2( s ) Δ 2CaO ( s ) + 4NO 2( g ) + O 2( g ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVC0dg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeaacaGaaiaabaqaamaabaabaaGcbaGaaeOmaiaabo eacaqGHbWaaeWaaeaacaqGobGaae4tamaaBaaaleaacaqGZaaabeaa aOGaayjkaiaawMcaamaaBaaaleaacaqGYaWaaeWaaeaacaqGZbaaca GLOaGaayzkaaaabeaakiaaysW7daGdKaWcbaGaaeiLdaqabOGaayPK HaGaaGjbVlaabkdacaqGdbGaaeyyaiaab+eadaWgaaWcbaWaaeWaae aacaqGZbaacaGLOaGaayzkaaaabeaakiaaysW7caqGRaGaaGjbVlaa bsdacaqGobGaae4tamaaBaaaleaacaqGYaWaaeWaaeaacaqGNbaaca GLOaGaayzkaaaabeaakiaaykW7caqGRaGaaGjbVlaab+eadaWgaaWc baGaaeOmamaabmaabaGaae4zaaGaayjkaiaawMcaaaqabaaaaa@5CA4@

Q.18 Describe two important uses of each of the following: (i) caustic soda (ii) sodium carbonate (iii) quicklime.

Ans.

(i) Uses of caustic soda

(a) It is used in soap industry.

(b) It is used as a reagent in laboratory.

(ii) Uses of sodium carbonate

(a) It is generally used in glass and soap industry.

(b) It is used as a water softener.

(iii) Uses of quick lime

(a) It is used as a starting material for obtaining slaked lime.

(b) It is used in the manufacture of glass and cement.

Q.19 Draw the structure of (i) BeCl2 (vapour) (ii) BeCl2 (solid).

Ans.

(a) Structure of BeCl2 (solid)

BeCl2 exists as a polymer in condensed (solid) phase.

In the vapour state, BeCl2 exists as a monomer with a linear structure.

Q.20 The hydroxides and carbonates of sodium and potassium are easily soluble in water while the corresponding salts of magnesium and calcium are sparingly soluble in water. Explain.

Ans.

The atomic size of sodium and potassium is larger than that of magnesium and calcium. Thus, the lattice energies of carbonates and hydroxides formed by calcium and magnesium are much more than those of sodium and potassium. Hence, carbonates and hydroxides of sodium and potassium dissolve readily in water whereas those of calcium and magnesium are only sparingly soluble.

Q.21 Describe the importance of the following: (i) limestone (ii) cement (iii) plaster of paris.

Ans.

(i) Chemically, limestone is CaCO3.

Importance of limestone

(a) It is used in the preparation of lime and cement.

(b) It is used as a flux during the smelting of iron ores.

(ii) Chemically, cement is a mixture of calcium silicate and calcium aluminate.

Importance of cement

(a) It is used in plastering and in construction of bridges.

(b) It is used in concrete.

(iii) Chemically, plaster of Paris is 2CaSO4.H2O.

Importance of plaster of Paris

(a) It is used in surgical bandages.

(b) It is also used for making casts and moulds.

Q.22 Why are lithium salts commonly hydrated and those of the other alkali metal ions usually anhydrous?

Ans.

Lithium is the smallest in size among the alkali metals. Hence, Li+ ion can polarize water molecules more easily than other alkali metals. As a result, water molecules get attached to lithium salts as water of crystallization. Hence, lithium salts such as trihydrated lithium chloride (LiCl.3H2O) are commonly hydrated. As the size of the ions increases, their polarizing power decreases. Hence, other alkali metal ions usually form anhydrous salts.

Q.23 Why is LiF almost insoluble in water whereas LiCl soluble not only in water but also in acetone?

Ans.

LiF is insoluble in water. On the contrary, LiCl is soluble not only in water, but also in acetone. This is mainly because of the greater ionic character of LiF as compared to LiCl.

The solubility of a compound in water depends on the balance between lattice energy and hydration energy. Since fluoride ion is much smaller in size than chloride ion, the lattice energy of LiF is greater than that of LiCl. Also, there is not much difference between the hydration energies of fluoride ion and chloride ion. Thus, the net energy change during the dissolution of LiCl in water is more exothermic than that during the dissolution of LiF in water. Hence, low lattice energy and greater covalent character are the factors making LiCl soluble not only in water, but also in acetone.

Q.24 Explain the significance of sodium, potassium, magnesium and calcium in biological fluids.

Ans.

Importance of sodium, potassium, magnesium, and calcium in biological fluids:

(i) Sodium (Na):

Sodium ions are found primarily in the blood plasma. They are also found in the interstitial fluids surrounding the cells.

(a) Sodium ions help in the transmission of nerve signals.

(b) They help in regulating the flow of water across the cell membranes.

(c) They also help in transporting sugars and amino acids into the cells.

(ii) Potassium (K):

Potassium ions are found in the highest quantity within the cell fluids.

(a) K ions help in activating many enzymes.

(b) They also participate in oxidising glucose to produce ATP.

(c) They also help in transmitting nerve signals.

(iii) Magnesium (Mg) and calcium (Ca):

Magnesium and calcium are referred to as macro-minerals. This term indicates their higher abundance in the human body system.

(a) Mg helps in relaxing nerves and muscles.

(b) Mg helps in building and strengthening bones.

(c) Mg maintains normal blood circulation in the human body system.

(d) Ca helps in the coagulation of blood

(e) Ca also helps in maintaining homeostasis.

Q.25 What happens when

(i) sodium metal is dropped in water ?

(ii) sodium metal is heated in free supply of air ?

(iii) sodium peroxide dissolves in water ?

Ans.

(i) When Na metal is dropped in water, it reacts violently to form sodium hydroxide and hydrogen gas. The chemical equation involved in the reaction is:

2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)

(ii) On being heated in air, sodium reacts vigorously with oxygen to form sodium peroxide. The chemical equation involved in the reaction is:

2Na(s) + O2(s) → Na2O2(s)

Na2O2(s) + 2H2O(l) → 2NaOH(aq) + H2O2(l)

Q.26 Comment on each of the following observations:

(a) The mobilities of the alkali metal ions in aqueous solution are Li+ < Na+ < K+ < Rb+ < Cs+

(b) Lithium is the only alkali metal to form a nitride directly.

(c) (where M=Ca,Sr or Ba) is nearly constant.

Ans.

(a) On moving down the alkali group, the ionic and atomic sizes of the metals increase. The given alkali metal ions can be arranged in the increasing order of their ionic sizes as:

Li+ <na+ < K+ < Rb+ < Cs+</na

Smaller the size of an ion, the more highly is it hydrated. Since Li+ is the smallest, it gets heavily hydrated in an aqueous solution. On the other hand, Cs+ is the largest and so it is the least hydrated. The given alkali metal ions can be arranged in the decreasing order of their hydrations as:

Li+ > Na+ > K+ > Rb+ > Cs+

Greater the mass of a hydrated ion, the lower is its ionic mobility. Therefore, hydrated Li+ is the least mobile and hydrated Cs+ is the most mobile. Thus, the given alkali metal ions can be arranged in the increasing order of their mobilities as:

Li+ < Na+ < K+ < Rb+ < Cs+

(b) Unlike the other elements of group 1, Li reacts directly with nitrogen to form lithium nitride. This is because Li+ is very small in size and so its size is the most compatible with the N3– ion. Hence, the lattice energy released is very high. This energy also overcomes the high amount of energy required for the formation of the N3– ion.

(c) Electrode potential (E°) of any M2+/M electrode depends upon three factors:

(i) Ionisation enthalpy

(ii) Enthalpy of hydration

(iii) Enthalpy of vaporisation

The combined effect of these factors is approximately the same for Ca, Sr, and Ba. Hence, their electrode potentials are nearly constant.

Q.27 State as to why

(a) a solution of Na2CO3 is alkaline ?

(b) alkali metals are prepared by electrolysis of their fused chlorides ?

(c) sodium is found to be more useful than potassium ?

Ans.

(a) When sodium carbonate is added to water, it hydrolyses to give sodium bicarbonate and sodium hydroxide (a strong base). As a result, the solution becomes alkaline.

Na2CO3 + H2O → NaHCO3 + NaOH

(b) It is not possible to prepare alkali metals by the chemical reduction of their oxides as they themselves are very strong reducing agents. They cannot be prepared by displacement reactions either (wherein one element is displaced by another). This is because these elements are highly electropositive. Neither can electrolysis of aqueous solutions be used to extract these elements. This is because the liberated metals react with water. Hence, to overcome these difficulties, alkali metals are usually prepared by the electrolysis of their fused chlorides.

(c) Blood plasma and the interstitial fluids surrounding the cells are the regions where sodium ions are primarily found. Potassium ions are located within the cell fluids. Sodium ions are involved in the transmission of nerve signals, in regulating the flow of water across the cell membranes, and in transporting sugars and amino acids into the cells. Hence, sodium is found to be more useful than potassium.

Q.28 Write balanced equations for reactions between

(a) Na2O2and water

(b) KO2 and water

(c) Na2O and CO2

Ans.

(a) The balanced chemical equation for the reaction between Na2O2 and water is:

2Na2O2(s) + 2H2O2 → 4NaOH(aq) + O2(aq)

(b) The balanced chemical equation for the reaction between KO2 and water is:

2KO2(s) + 2H2O → 2KOH(aq) + H2O2(aq)+O2(aq)

c) Na2O(s) + CO2(s) → Na2CO3

Q.29 How would you explain the following observations?

(i) BeO is almost insoluble but BeSO4 is soluble in water,

(ii) BaO is soluble but BaSO4 is insoluble in water,

(iii) LiI is more soluble than KI in ethanol.

Ans.

(i) BeO is almost insoluble in water and BeSO4 is soluble in water. Be2+ is a small cation with a high polarising power and O2– is a small anion. The size compatibility of Be2+ and O2– is high. Therefore, the lattice energy released during their formation is also very high. When BeO is dissolved in water, the hydration energy of its ions is not sufficient to overcome the high lattice energy. Therefore, BeO is insoluble in water. On the other hand

SO 4 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVC0dg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeaacaGaaiaabaqaamaabaabaaGcbaGaae4uaiaab+ eadaqhaaWcbaGaaeinaaqaaiaabobiaaaaaa@38E7@

, ion is a large anion. Hence, Be2+ can easily polarise ions, making BeSO4 unstable. Thus, the lattice energy of BeSO4 is not very high and so it is soluble in water.

(ii) BaO is soluble in water, but BaSO4 is not. Ba2+ is a large cation and O2– is a small anion. The size compatibility of Ba2+ and O2– is not high. As a result, BaO is unstable. The lattice energy released during its formation is also not very large. It can easily be overcome by the hydration energy of the ions. Therefore, BaO is soluble in water. In

BaSO4, Ba2+ and are both large-sized. The lattice energy released is high. Hence, it is not soluble in water.

(iii) LiI is more soluble than KI in ethanol. As a result of its small size, the lithium ion has a higher polarising power than the potassium ion. It polarises the electron cloud of the iodide ion to a much greater extent than the potassium ion. This causes a greater covalent character in LiI than in KI. Hence, LiI is more soluble in ethanol.

Q.30 Which of the alkali metal is having least melting point?

(a) Na (b) K (c) Rb (d) Cs

Ans.

Atomic size increases as we move down the alkali group. As a result, the binding energies of their atoms in the crystal lattice decrease. Also, the strength of metallic bonds decreases on moving down a group in the periodic table. This causes a decrease in the melting point. Among the given metals, Cs is the largest and has the least melting point.

Q.31 Which one of the following alkali metals gives hydrated salts?

(a) Li (b) Na (c) K (d) Cs

Ans.

Smaller the size of an ion, the more highly is it hydrated. Among the given alkali metals, Li is the smallest in size. Also, it has the highest charge density and highest polarizing power. Hence, it attracts water molecules more strongly than the other alkali metals. As a result, it forms hydrated salts such as LiCl2.H2O. The other alkali metals are larger than Li and have weaker charge densities. Hence, they usually do not form hydrated salts.

Q.32 Which one of the alkaline earth metal carbonates is thermally the most stable?

(a) MgCO3 (b) CaCO3 (c) SrCO3 (d) BaCO3

Ans.

Thermal stability increases with the increase in the size of the cation present in the carbonate. The increasing order of the cationic size of the given alkaline earth metals is Mg < Ca < Sr < Ba

Hence, the increasing order of the thermal stability of the given alkaline earth metal carbonates is MgCO3 < CaCO3 < SrCO3 < BaCO3

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FAQs (Frequently Asked Questions)

1. What are the elements of the s-block?

The last electron of an element’s electronic configuration is present in the s orbital, making it an s-block element. In the periodic table, they form a distinct group. These elements, as well as all of their unique physical and chemical properties, are taught to students. Their interactions with other s-block elements are investigated in-depth, and  their unique identification characteristics are studied. 

2. What are the most important topics and subtopics from the s-block elements in Chapter 10?

The chapter in Chemistry has a lot of topics and sub-topics to learn and remember, but some of these require extra attention from students. The most important topics in Chemistry Chapter 10 Class 11 are listed below –

Alkali metals (Group 1)

  • General characteristics of the compounds of the alkali metals 
  • Anomalous properties of Lithium 
  • Some important compounds of Sodium 
  • Biological importance of Sodium and Potassium 

Alkaline metals (Group 2)

  • General characteristics of compounds of the alkali metals 
  • Anomalous properties of beryllium 
  • Some important compounds of Calcium 
  • Biological importance of Magnesium and Calcium 

3. What are the benefits of using NCERT Solutions Class 11 Chemistry Chapter 10?

Extramarks has compiled a list of all the exercises and their solutions in NCERT Class 11 Chemistry Chapter 10 solutions. Referring to the solutions will assist students in gaining a better understanding of the topics covered. After studying a chapter, students frequently make the mistake of not answering questions. Students will be able to find the answers to all the questions they are having trouble answering by referring to solutions available on Extramarks.

4. What is the total number of elements in s-block elements?

We’ve already discussed how s-block elements are classified into two groups. There are 13 elements in total in the s-block, 7 from group 1 and 6 from group 2.

 Lithium (Li), sodium (Na), potassium (K), rubidium (Rb), caesium (Cs), and francium (Fr) are the elements of Group 1. (Fr). They are collectively called alkali metals

 Barium (Ba), Beryllium (Be), Calcium (Ca), Magnesium (Mg), Strontium (Sr), and Radium (Ra) are the elements of Group 2. (Ra). They are collectively called alkaline earth metals.

5. What is the importance of chapter 10 of NCERT Chemistry for class 11 students?

In the final CBSE examination, Chemistry Chapter 10 Class 11 is worth a total of three marks. In NEET, NCERT class 11 Chemistry chapter 10 is worth 2% of total marks, and at least one question from this chapter is asked every year in JEE Main.