NCERT Exemplar Class 11 Chemistry Solutions Chapter 10 The s-Block Elements 2021

Q: Q 1. What are the major topics of NCERT Exemplar class 11 Chemistry Solutions Chapter 10?

A- The major focus of this chapter is upon the physical and chemical properties and anomalies of the entire s block elements. It especially discusses the general characteristics of compounds of the alkali earth metals and the anomalous behaviours of lithium and beryllium.

Q: Q2. Is this chapter carrying a huge share in the mark distribution in the exams?

A- Generally the questions asked from these chapters have theoretical, short and layered answers which do not amount to the highest mark securing problems but the information acquired from this chapter needs to be applied in many questions as well.

Q: Q 3. Is studying only from NCERT exemplar solutions for class 11 Chemistry chapter 10 enough for students?

A- Though all the important questions along with answers have been mentioned in this chapter, it is still suggested that students refer to their original textbook for the theoretical part of the preparation and then visit this chapter for quick revisions and practice.

Q: Q 4. Is there any way in which students can share this chapter with other interested students?

A- Yes students can share this chapter with anyone who is interested as a guide by making use of the option provided, which is NCERT exemplar class 11 Chemistry solutions chapter 10 pdf download.

NCERT Exemplar Class 11 Chemistry Solutions Chapter 10 The s Block Elements

Edited By Sumit Saini | Updated on Sep 10, 2022 05:47 PM IST

The study for s-block elements includes memorising a huge array of facts including element properties, symbols, physical and chemical properties and their general individual characteristics. The major focus and format of NCERT Exemplar Class 11 Chemistry solutions chapter 10 surrounds around the specific differentiating properties of the s block elements which includes various groups of categorised elements of alkali metals and its diversification, i.e. oxides and hydroxides and halides, etc. The key objective that NCERT exemplar Class 11 Chemistry chapter 10 solutions desires to accomplish is to lay down a systemic alignment and information delivery for the students to remember. It attempts to incorporate all the necessary information about the information through a brief range of questions.

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This Story also Contains

NCERT exemplar class 11 Chemistry solutions chapter 10: MCQ (Type 1)
NCERT Exemplar Class 11 Chemistry Solutions Chapter 10: MCQ (Type 2)
NCERT Exemplar Class 11 Chemistry Solutions Chapter 10: Short Answer Type
NCERT Exemplar Class 11 Chemistry Solutions Chapter 10: Matching Type
NCERT Exemplar Class 11 Chemistry Solutions Chapter 10: Assertion and Reason Type
NCERT Exemplar Class 11 Chemistry Solutions Chapter 10: Long Answer Type
Main Subtopics in Class 11 Chemistry NCERT Exemplar Solutions Chapter 10
What the Students will Learn From NCERT Exemplar Class 11 Chemistry Solutions Chapter 10
NCERT Exemplar Class 11 Chemistry Solutions Chapter-Wise
Important Topics To Cover From NCERT Exemplar Class 11 Chemistry Solutions Chapter 10 The S-Block Elements

NCERT exemplar class 11 Chemistry solutions chapter 10: MCQ (Type 1)

Question:1

The alkali metals are low melting. Which of the following alkali metal is expected to melt if the room temperature rises to $30^{\circ}C$ ?
(i) Na
(ii) K
(iii) Rb
(iv) Cs
Answer:

The answer is the option (iv) Melting point decreases down the groups in a periodic table.

Question:2

Alkali metals react with water vigorously to form hydroxides and dihydrogen. Which of the following alkali metals reacts with water least vigorously?
(i) Li
(ii) Na
(iii) K
(iv) Cs
Answer:

The answer is the option (i) Li has an extremely high hydrogen enthalpy. So, its reaction with water releases a high amount of energy, most of which is consumed in fusion, vaporization and ionization. Hence, its reaction with water is less vigorous.

Question:3

The reducing power of a metal depends on various factors. Suggest the factor which makes Li, the strongest reducing agent in aqueous solution.
(i) Sublimation enthalpy
(ii) Ionisation enthalpy
(iii) Hydration enthalpy
(iv) Electron-gain enthalpy

Answer:

The answer is the option (iii) Hydration enthalpy is a measure of the tendency of an element to lose an electron in aqueous solution. More negative the hydration enthalpy, greater is the ability to lose an electron, making it a strong reducing agent.

Question:4

Metal carbonates decompose on heating to give metal oxide and carbon dioxide. Which of the metal carbonates is most stable thermally?
$(i) MgCO_{3}$
$(ii) CaCO_3$
$(iii) SrCO_3$
$(iv) BaCO_3$
Answer:

The answer is the option $(iv) MCO_{3}\rightarrow MO+CO_{2}$

Thermal stability of $MCO_{3}$ depends on the stability of MO. If MO is stable $MCO_{3}$ is unstable and vice versa
BaO is least stable making $BaCO_{3}$ most stable.

Question:5

Which of the carbonates given below is unstable in air and is kept in $CO_2$ atmosphere to avoid decomposition.
$(i) BeCO_{3}$
$(ii) MgCO_{3}$
$(iii) CaCO_{3}$
$(iv) BaCO_{3}$
Answer:

The answer is the option (i)

Strong polarizing effect due to small size of $Be^{2+}$ makes $BeCO_3$ unstable.

Question:6

Metals form basic hydroxides. Which of the following metal hydroxide is the least basic?
$(i) Mg(OH)_{2}$
$(ii) Ca(OH)_{2}$
$(iii) Sr(OH)_{2}$
$(iv) Ba(OH)_{2}$
Answer:

The answer is the option (i) Basic character decreases down the group for hydroxides with an increase in the size of metal.

Question:7

Some of the Group 2 metal halides are covalent and soluble in organic solvents. Among the following metal halides, the one which is soluble in ethanol is
$(i) BeCl_2$
$(ii) MgCl_2$
$(iii) CaCl_2$
$(iv) SrCl_2$
Answer:

The answer is the option (i) Covalent metal halides are soluble in organic solvents like ethanol.
$BeCl_{2}$ has covalent character due to small size and high effective nuclear charge.

Question:8

The order of decreasing ionisation enthalpy in alkali metals is
$(i) Na > Li > K > Rb$
$(ii) Rb < Na < K < Li$
$(iii) Li > Na > K > Rb$
$(iv) K < Li < Na < Rb$
Answer:

The answer is the option $(iii) Li > Na > K > Rb$
Effective nuclear charge and ionization enthalpy decrease with increase in the size of the atom down the group.

Question:9

The solubility of metal halides depends on their nature, lattice enthalpy and hydration enthalpy of the individual ions. Amongst fluorides of alkali metals, the lowest solubility of LiF in water is due to
(i) Ionic nature of lithium fluoride
(ii) High lattice enthalpy
(iii) High hydration enthalpy for lithium ion
(iv) Low ionisation enthalpy of lithium atom
Answer:

The answer is the option (ii) Higher the lattice enthalpy and lower the hydration enthalpy, lower is the solubility of metal halides. In the case of LiF, high lattice enthalpy results in lower solubility.

Question:10

Amphoteric hydroxides react with both alkalies and acids. Which of the following Group 2 metal hydroxides is soluble in sodium hydroxide?
$(i) Be(OH)_{2}$
$(ii) Mg(OH)_{2}$
$(iii) Ca(OH)_{2}$
$(iv) Ba(OH)_{2}$
Answer:

The answer is the option $(i) Be(OH)_{2}$ is an amphoteric hydroxide. It reacts with base to give beryllate and water.
$Be(OH)_{2}+2NaOH \rightarrow BeCl_{2}+2H_{2}O$

Question:11

In the synthesis of sodium carbonate, the recovery of ammonia is done by treating $NH_{4}Cl$ with $Ca(OH)_{2}$ . The by-product obtained in this process is
$(i) CaCl_{2}$
$(ii) NaCl$
$(iii) NaOH$
$(iv) NaHCO_{3}$
Answer:

The answer is the option $(i) CaCl_{2}$
On treating $NH_{4}Cl$ with $Ca(OH)_{2}$ , $CaCl_{2}$ is formed as a by product
$2NH_{4} Cl+CaOH_{2} \rightarrow 2NH_{3}+2H_{2}O+CaCl_{2}$

Question:12

When sodium is dissolved in liquid ammonia, a solution of deep blue colour is obtained. The colour of the solution is due to
(i) ammoniated electron
(ii) sodium ion
(iii) sodium amide
(iv) ammoniated sodium ion
Answer:

The answer is the option (i) Alkali metals, when dissolved in liquid ammonia give deep blue colour. The colour of the solution is due to ammoniated electrons which get excited to a higher energy level by absorbing red wavelength and start transmitting blue colour.
$Na+\left (x+y \right )NH_{3}\rightarrow\left [ \left (Na\left (NH_{3} \right )_{x} \right ) \right ]^{+} e \left (NH_{3} \right )_{y}^{-}$

Question:13

By adding gypsum to cement
(i) setting time of cement becomes less.
(ii) setting time of cement increases.
(iii) colour of cement becomes light.
(iv) shining surface is obtained.
Answer:

The answer is the option (ii) Gypsum is added to cement to increase the setting time of cement to let it get hardened.

Question:14

Dead burnt plaster is
$(i)CaSO_{4}$
$(ii)CaSO_{4}. \frac{1}{2}H_{2}O$
$(iii)CaSO_{4}. H_{2}O$
$(iv)CaSO_{4}. 2H_{2}O$

Answer:

The answer is the option (i)

Plaster of Paris $CaSO_{4}.12H_{2}O$ on heating at 2000 changes to anhydrous $CaSO_{4}$ which is known as Dead burnt plaster.

Question:15

Suspension of slaked lime in water is known as
(i) lime water
(ii) quick lime
(iii) milk of lime
(iv) aqueous solution of slaked lime
Answer:

The answer is the option (iii) Slaked lime $Ca(OH)_{2}$ gives a white suspension in water called Milk of Lime. It is a highly exothermic reaction and the milk of lime is used in whitewashing.

Question:16

Which of the following elements does not form hydride by direct heating with dihydrogen?
(i) Be
(ii) Mg
(iii) Sr
(iv) Ba
Answer:

The answer is the option $(i) BeH_{2}$ can be prepared by reacting $BeCl_{2}$ with $LiAlH_{4}$
$2BeCl_{2}+LiAlH_{4}\rightarrow 2BeH_{2}+LiCl+AlCl_{3}$ .

Question:17

The formula of soda ash is
$(i) Na_{2} CO_{3}.10H_{2}O$
$(ii) Na_{2}CO_{3}.H_{2}O$
$(iii) Na_{2}CO_{3} .2H_{2}O$
$(iv) Na_{2}CO_{3}$

Answer:

The answer is the option (iv) Soda ash is formed when Washing soda $(Na_{2}CO_{3}.10H_{2}O )$ loses water of crystallization above 373K.

Question:18

A substance which gives brick red flame and breaks down on heating to give oxygen and a brown gas is
(i) Magnesium nitrate
(ii) Calcium nitrate
(iii) Barium nitrate
(iv) Strontium nitrate
Answer:

The answer is the option $(ii) Ca\left (NO_{3} \right )_{2} \rightarrow CaO+NO_{2}+O_{2}$
Ca gives the brick red colour and $NO_{2}$ is a brown gas.

Question:19

Which of the following statements is true about $Ca(OH)_2$ ?
(i) It is used in the preparation of bleaching powder
(ii) It is a light blue solid
(iii) It does not possess disinfectant property.
(iv) It is used in the manufacture of cement.
Answer:

The answer is the option $(i) 2Ca(OH)_{2}+2Cl_{2} \rightarrow CaOCl_{2}+CaCl_{2}+2H_{2}O$
$CaOCl_{2}$ is the formula for bleaching powder

Question:20

A chemical A is used for the preparation of washing soda to recover ammonia. When $CO_{2}$ is bubbled through an aqueous solution of A, the solution turns milky. It is used in white washing due to disinfectant nature. What is the chemical formula of A?
$(i) Ca (HCO_{3})_{2}$
$(ii) CaO$
$(iii) Ca(OH)_{2}$
$(iv) CaCO_{3}$
Answer:

The answer is the option (iii)
On passing $CO_{2}$ through $Ca(OH)_{2}$ , lime water, turns from colorless to milky due to formation of $CaCO_{3}$ .
$Ca(OH)_{2}+CO_{2}\rightarrow CaCO_{3}+H_{2}O$
Also in Solvay Ammonia soda process,
$2NH_{4}Cl+Ca(OH)_{2} \rightarrow 2NH_{3}+2H_{2}O+CaCl_{2}$

Question:21

Dehydration of hydrates of halides of calcium, barium and strontium i.e., $CaCl_{2}.6H_{2}O, BaCl_{2}.2H_{2}O, SrCl_{2}.2H_{2}O$ , can be achieved by heating. These become wet on keeping in air. Which of the following statements is correct about these halides?
(i) act as dehydrating agent
(ii) can absorb moisture from air
(iii) Tendency to form hydrate decreases from calcium to barium
(iv) All of the above
Answer:

The answer is the option (iv) Alkaline earth metal chlorides are hydrated and hygroscopic in nature. Thus, all of the above options are correct.

NCERT Exemplar Class 11 Chemistry Solutions Chapter 10: MCQ (Type 2)

Question:22

Metallic elements are described by their standard electrode potential, fusion enthalpy, atomic size, etc. The alkali metals are characterized by which of the following properties?
(i) High boiling point
(ii) High negative standard electrode potential
(iii) High density
(iv) Large atomic size
Answer:

The answer is the option (ii) and (iv) Alkali metals have the largest size and low boiling point and density (Since size decreases from left to right across a period). Due to less effective nuclear charge, they can easily lose the electrons from the outermost shell and have high negative standard electrode potential.

Question:23

Several sodium compounds find use in industries. Which of the following compounds are used for textile industry?
$(i) Na_{2}CO_{3}$
$(ii) NaHCO_{3}$
$(iii) NaOH$
$(iv) NaCl$
Answer:

The answer is the option (i) and (iii)
$Na_{2}CO_{3}$ is used in making soap powders, in laundry and textile industry. $NaOH$ is used in Soap industry and paper industry.

Question:24

Which of the following compounds are readily soluble in water?
$(i) BeSO_{4}$
$(ii) MgSO_{4}$
$(iii) BaSO_{4}$
$(iv) SrSO_{4}$
Answer:

The answer is the option (i) and (ii) Hydration energy decreases down the group as size increases.
Hydration energy of $Be^{+2}$ and $Mg^{+2}$ is very high. Hence their compounds are soluble.

Question:25

When Zeolite, which is hydrated sodium aluminium silicate is treated with hard water, the sodium ions are exchanged with which of the following ion(s)?
$(i) H^{+} ions$
$(ii) Mg^{2+} ions$
$(iii) Ca^{2+} ions$
$(iv) SO_{4}^{2-} ions$
Answer:

The answer is the option (ii) and (iii) Zeolite is used to remove the hardness of the water. Zeolite is a sodium aluminosilicate.
$Ca^{2+} and Mg^{2+}$ ions in hard water get exchanged with $Na^{+}$ ion of zeolite.

Question:26

Identify the correct formula of halides of alkaline earth metals from the following.
$(i) BaCl_{2}.2H_{2}O$
$(ii) BaCl_{2}.4H_{2}O$
$(iii) CaCl_{2}.6H_{2}O$
$(iv) SrCl_{2}.4H_{2}O$
Answer:

The answer is the option (i) and (iii) The extent of hydration of chlorides of alkaline earth metals decreases down the group.

Question:27

Choose the correct statements from the following.
(i) Beryllium is not readily attacked by acids because of the presence of an oxide film on the surface of the metal.
(ii) Beryllium sulphate is readily soluble in water as the greater hydration enthalpy of $Be^{2+}$ overcomes the lattice enthalpy factor.
(iii) Beryllium exhibits coordination number more than four.
(iv) Beryllium oxide is purely acidic in nature.
Answer:

The answer is the option (i) and (ii) Be has diagonal relation with Al and forms a protective film of oxide and is not readily attacked by acids.
The high hydration enthalpy of $Be^{2+}$ makes $BeSO_{4}$ Soluble in water.

Question:28

Which of the following are the correct reasons for anomalous behaviour of lithium?
(i) Exceptionally small size of its atom
(ii) Its high polarising power
(iii) It has high degree of hydration
(iv) Exceptionally low ionisation enthalpy
Answer:

The answer is the option (i) and (ii) Li has an exceedingly small size and high nuclear charge due to which it shows high polarizing power.

NCERT Exemplar Class 11 Chemistry Solutions Chapter 10: Short Answer Type

Question:29

How do you account for the strong reducing power of lithium in aqueous solution?
Answer:

Li ions, due to small size, have high enthalpy of ionization and hydration. In aqueous solution, high hydration enthalpy predominates, and Li loses electrons and has strong reducing power.
$Li (s) \rightarrow Li (g)$ (Sublimation enthalpy )
$Li(g) \rightarrow Li^{+}(g)+1e^{-}$ (Enthalpy of Ionization )
$Li^{+}(g)+H_{2}O \rightarrow Li^{+} (aq)$ (Enthalpy ofhydration)

Question:30

When heated in air, the alkali metals form various oxides. Mention the oxides formed by Li, Na and K.
Answer:

Reactivity of alkali metals with $O_{2}$ increases down the group. These elements give 3 types of oxides- normal oxide, peroxide and superoxide.
$4Li+O_{2} \rightarrow 2Li_{2}O$ Lithium oxide
$6Na+2O_{2} \rightarrow Na_{2}O_{2}$ Sodium peroxide $+2 Na_{2}O$ Sodium oxide
$8K+2O_{2} \rightarrow KO_{2}$ Superoxide $+K_{2}O$ Peroxide $+K_{2}O$ (Normal oxide)

Question:31

Complete the following reactions
$(i)O_{2}^{2-}+H_{2}O\rightarrow$
$(ii)O_{2}^{-}+H_{2}O\rightarrow$
Answer:

$(i)O_{2}^{2-}+H_{2}O\rightarrow$ $2OH^{-}+H_{2}O_{2}$
$(ii) 2O_{2}^{-}+ 2H_{2}O\rightarrow$ $2OH^{-}+H_{2}O_{2} +O_{2}$

Question:32

Lithium resembles magnesium in some of its properties. Mention two such properties and give reasons for this resemblance.
Answer:

Li and Mg have similar ionic radii due to which some of their properties resemble.
Both Li and Mg react slowly with water. Their oxides and hydroxides are much less soluble, and their hydroxides decompose on heating.
$2LiOH \rightarrow Li_{2}O+H_{2}O$ (On heating)
$2Mg(OH)_{2} \rightarrow Mg_{2}O+2H_{2}O$
Both form nitride, $Li_{3}N$ and $Mg_{3}N_{2}$ , by direct combination with nitrogen.
$6Li+N_{2}\rightarrow 2Li_{3}N$
$3Mg+N_{2}\rightarrow 2Mg_{3}N_{2}$

Question:33

Name an element from Group 2 which forms an amphoteric oxide and a water soluble sulphate.
Answer:

Beryllium from group 2 gives an amphoteric oxide BeO and water-soluble sulphate $BeSO_4$ .

Question:34

Discuss the trend of the following:
(i) Thermal stability of carbonates of Group 2 elements.
(ii) The solubility and the nature of oxides of Group 2 elements
Answer:

(i) Thermal stability increases down the group.
$BeCO_{3}<MgCO_{3}<CaCO_{3}<SrCO_{3}<BaCO_{3}$
(ii) All oxides are basic and ionic in nature except BeO, which is amphoteric and covalent. The lattice energy of oxides decreases with an increase in the size of cation. Basic nature also increases down the group. Except for BeO and MgO, all are soluble in water and produce a large amount of heat on dissolving. The insolubility of BeO and MgO can be attributed to their high lattice energy.

Question:35

Why are $BeSO_{4}$ and $MgSO_{4}$ readily soluble in water while $CaSO_{4}$ , $SrSO_{4}$ and $BaSO_{4}$ are insoluble?
Answer:

The solubility of sulphates of group 2 elements depend on their hydration energy which decreases down the group. The lattice energy of group 2 sulphates is almost the same. Very high hydration enthalpy of $Be^{2+}$ and $Mg^{2+}$ ions overcome the lattice enthalpy and their sulphates are soluble. However
In other elements of group 2 the hydration enthalpy is not high enough to overcome lattice enthalpy. So, they remain insoluble in water.

Question:36

All compounds of alkali metals are easily soluble in water but lithium compounds are more soluble in organic solvents. Explain.
Answer:

Alkali metal compounds are ionic in nature and hence soluble in water. In the case of Lithium, its small size and high polarizing power give it a covalent character making it soluble in organic solvents.

Question:37

In the Solvay process, can we obtain sodium carbonate directly by treating the solution containing $(NH_{4})_{2}CO_{3}$ with sodium chloride? Explain.
Answer:

$(NH_{4})_{2}CO_{3}$ reacts with NaCl to get $Na_{2}CO_{3}$ and $NH_{4}Cl$ . Both the products are highly soluble in water and equilibrium can not shift in forward direction. That is why $Na_{2}CO_{3}$ can not be prepared directly.

Question:38

Write Lewis strucure of $O_{2}^{-}$ ion and find out oxidation state of each oxygen atom? What is the average oxidation state of oxygen in this ion?

Answer:

Lewis structure of $O_{2}^{-}$ is $O-O^{-}$ .
Oxygen atom with no any charge has 6 electrons and Oxidation number O.N: 0.
but the other oxygen atom has 7 electrons and O.N: -1.
So, the average oxidation state is -1/2.

Question:39

Why do beryllium and magnesium not impart colour to the flame in the flame test?

Answer:

Be and Mg do not impart colour to the flame in the flame test because these 2 metals have very small atomic radii and electrons are more strongly bonded due to higher effective nuclear charge compared to other alkaline earth metals. The energy of Bunsen flame is not sufficient to excite the electrons.

Question:40

What is the structure of $BeCl_2$ molecule in gaseous and solid state?
Answer:

In the gaseous state, it exists as a monomer having a linear structure and zero dipole moment. Cl-Be-Cl
In solid state, it exists in polymeric chain structure in which each Be atom is surrounded by 4 Cl atoms, 2 through covalent bonds and other 2 through coordinate bonds to give a bridge structure.

NCERT Exemplar Class 11 Chemistry Solutions Chapter 10: Matching Type

Question:41

Match the elements given in Column I with the properties mentioned in Column II.

Column I	Column II
(i)Li	(a) Insoluble sulphate
(ii) Na	(b) Strongest monoacidic base
(iii) Ca	(c) Most negative $E^{\Theta }$ value among alkali metals
(iv)Ba	(d)Insoluble oxlate
	(e) $6s^{2}$ electronic onfiguration

Answer:

(i) – (c) Due to extremely high hydration enthalpy, $E^{\Theta }$ is most negative
(ii) – (b) Na gives NaOH (strong base) which is the strongest monoacidic base.
(iii) – (d) Calcium oxalate is insoluble due to low hydration energy
(iv) – (a,e) $Ba^{2+}$ is large in size.So its hydration energy is low and barium sulphate

Question:42

Match the compounds given in Column I with their uses mentioned in Column II.

Column I	Column II
(i) $CaCO_{3}$	(a) Dentistry, Ornamental work
(ii) $Ca(OH)_{2}$	(b) Manufacture of sodium carbonate from caustic soda
(iii) CaO	(c) Manufacture of high quality paper
(iv) $CaSO_{4}$	(d) Used in whote washing

Answer:

(i)- (c)
(ii)-(d); It is used in whitewashing due to its disinfectant nature and sparingly soluble in water.
(iii)-(b)
(iv)- (a)

Question:43

Match the elements given in Column I with the colour they impart to the flame given in Column II.

Column I	Column II
(i) Cs	(a) Apple green
(ii) Na	(b)Violet
(iii) K	(c) Brick red
(iv) Ca	(d) Yellow
(v)Sr	(e) Crimson red
(vi) Ba	(f) Red

Answer:

(i) – (f); (ii)- (d); (iii)- (b); (iv)- (c); (v)-(e ); (vi) –(a)
The heat from the flame excites the outermost orbital electron to a higher energy level. When the excited electron comes back to the ground state, there is the emission of radiation in the visible region which gives characteristic colour to Bunsen flame.

NCERT Exemplar Class 11 Chemistry Solutions Chapter 10: Assertion and Reason Type

Question:44

In the following questions a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.
Assertion (A): The carbonate of lithium decomposes easily on heating to form lithium oxide and $CO_2$ .
Reason (R) : Lithium being very small in size polarises large carbonate ion leading to the formation of more stable $Li_{2}O$ and $CO_2$ .
(i) Both A and R are correct and R is the correct explanation of A.
(ii) Both A and R are correct but R is not the correct explanation of A.
(iii) Both A and R are not correct
(iv) A is not correct but R is correct.
Answer:

The answer is the option (i) Both A and R are correct and R is the correct explanation of A. $Li_{2}CO_{3}$ is unstable and decomposes on heating. Lithium being very small in size polarises large carbonate ion leading to $Li_{2}O$ which is more stable due to higher lattice energy.

Question:45

In the following questions a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.

Assertion (A): Beryllium carbonate is kept in the atmosphere of carbon dioxide.
Reason (R) : Beryllium carbonate is unstable and decomposes to give beryllium oxide and carbon dioxide.
(i) Both A and R are correct and R is the correct explanation of A.
(ii) Both A and R are correct but R is not the correct explanation of A.
(iii) Both A and R are not correct.
(iv) A is not correct but R is correct.

Answer:

The answer is the option (i) Both A and R are correct and R is the correct explanation of A.BeO is more stable due to small size and strong polarising power of $Be^{2+}$ . Since BeO is stable and $BeCO_3$ is unstable, when kept in atmosphere of $CO_{2}$ a reversible process occurs and stability of $BeCO_3$ increases

NCERT Exemplar Class 11 Chemistry Solutions Chapter 10: Long Answer Type

Question:46

The s-block elements are characterised by their larger atomic sizes, lower ionisation enthalpies, invariable +1 oxidation state and solubilities of their oxosalts.In the light of these features describe the nature of their oxides, halides and oxosalts.
Answer:

The atoms of alkali metals have a large size due to which they readily form cations. They have +1 oxidation state, and their compounds are ionic in nature. Alkali metals give three types of oxides- Normal oxides $(M_2O)$ , Peroxides $M_{2}O_{2}$ and Super oxides $MO_{2}$ . Basic character of normal oxides increases down the group due to increase in their ionic character. Halides of alkali metals MX are also ionic except LiX which is covalent due to small size and high polarizing power. The Ionic character increases from Li to Cs. Oxosalts of alkali metals $M_{2}CO_{3}-MHCO_{3}$ , $MNO_{3}$ are solid water soluble ionic compounds. Oxosalts of lithium show different properties due to its small size and high polarizing power.

Question:47

Present a comparative account of the alkali and alkaline earth metals with respect to the following characteristics:
(i) Tendency to form ionic / covalent compounds
(ii) Nature of oxides and their solubility in water
(iii) Formation of oxosalts
(iv) Solubility of oxosalts
(v) Thermal stability of oxosalts
Answer:

(i) Compounds of alkaline earth metals are less ionic than alkali metals because of more effective nuclear charge and small size.
(ii) Oxides of alkaline earth metals are less basic than oxides of alkali metals. These oxides are water-soluble and reactions highly exothermic. Hydroxides of alkaline earth metals are less basic than hydroxides of alkali metals.
(iii) Alkaline earth metals give Oxo salts with oxoacids, but the reactivity of alkali metals is faster.
(iv) Oxo salts of alkaline earth metals are more soluble than that of alkali metals because of the smaller size of cations and higher hydration enthalpy.
(iv) Thermal stability of Oxo salts of alkali metals is higher than that of alkaline earth metals.
$Na_{2}CO_{3}$ is stable towards heat but $MgCO_{3}$ decomposes into MgO and $CO_{2}$ on heating.

Question:48

When a metal of group 1 was dissolved in liquid ammonia, the following observations were obtained:
(i) Blue solution was obtained initially.
(ii) On concentrating the solution, blue colour changed to bronze colour.
How do you account for the blue colour of the solution? Give the name of the product formed on keeping the solution for some time.
Answer:

(i) Alkali metals dissolve in liquid ammonia, and ammoniated electrons get excited to a higher level which imparts a blue colour to the solution.
$M+(x+y)NH_{3}\rightarrow \left [\left (M(NH_{3})_{x} \right ) \right ]^{+} e (NH_{3})_{y}^{-}$
(ii) In a concentrated solution, blue colour changes to bronze due to the formation of clusters of the metal ion. On keeping the solution for some time, blue solution liberates H2 gas with the formation of amide.
$M^{+}(aq)+e^{-}+NH_{3}(l)\rightarrow MNH_{2}(aq)+12H_{2}(g)$

Question:49

The stability of peroxide and superoxide of alkali metals increase as we go down the group. Explain giving reason.

Answer:

The stability of peroxides and superoxides increases as the size of metal ion increases. Stability increases as the size of cation increases. Peroxide ion and superoxide ion combine with large size of alkali metals. $O^{-2}<O_{2}^{-2}<O_{2}^{-}$
$4Li+O_{2}\rightarrow 2Li_{2}O$ Lithium oxide
$2Na+O_{2}\rightarrow Na_{2}O_{2}$ Sodium peroxide
$K+O_{2} \rightarrow KO_{2}$ Superoxide

Question:50

When water is added to compound (A) of calcium, solution of compound (B) is formed. When carbon dioxide is passed into the solution, it turns milky due to the formation of compound (C). If excess of carbon dioxide is passed into the solution milkiness disappears due to the formation of compound (D). Identify the compounds A, B, C and D. Explain why the milkiness disappears in the last step.
Answer:

Solution B which turns milky on passing $CO_2$ is lime water $Ca(OH)_{2}$ (Calcium hydroxide)and Compound C which gives milky appearance is $CaCO_{3}$ (Calcium carbonate). On passing excess of $CO_2$ milkiness disappears due to the formation of $Ca(HCO_{3})_{2}$ (Calcium bicarbonate). Compound A reacts with water to give B. A is CaO.
$CaO A+H_{2}O\rightarrow Ca(OH)_{2} (B)$
$Ca(OH)_{2}(B)+CO_{2} \rightarrow CaCO_{3} (C)+H_{2}O$
$Ca(OH)_{2}(B)+CO_{2}+H_{2}O\rightarrow Ca(HCO_{3})_{2} (D)$

Question:51

Lithium hydride can be used to prepare other useful hydrides. Beryllium hydride is one of them. Suggest a route for the preparation of beryllium hydride starting from lithium hydride. Write chemical equations involved in the proce

Answer:

Beryllium hydride cannot be prepared directly by reaction with $H_{2}$ . It is prepared by reacting with $LiAlH_{4}$ . The reactions that take place are given below:
$8LiH+Al_{2}Cl_{6} \rightarrow 2LiAlH_{4}+6LiCl$
$2BeCl_{2}+LiAlH_{4}\rightarrow 2BeH_{2}+LiCl+AlCl_{3}$

Question:52

An element of group 2 forms covalent oxide which is amphoteric in nature and dissolves in water to give an amphoteric hydroxide. Identify the element and write chemical reactions of the hydroxide of the element with an alkali and an acid.
Answer:

In group 2 only Be gives covalent oxide, which is amphoteric in nature. The rest elements in the group give ionic oxide, which is basic in nature. BeO on dissolving in water gives sparingly soluble hydroxide which reacts with acid and base to give salt.
$BeO+H_{2}O \rightarrow Be(OH)_{2}$
$Be(OH)_{2}+2OH^{-} \rightarrow \left [BeOH_{4} \right ]^{2-}$
$Be(OH)_{2}+2HCl+2H_{2}O \rightarrow \left [Be(OH)_{4} \right ] Cl_{2}$

Question:53

Ions of an element of group 1 participate in the transmission of nerve signals and transport of sugars and aminoacids into cells. This element imparts yellow colour to the flame in flame test and forms an oxide and a peroxide with oxygen. Identify the element and write chemical reaction to show the formation of its peroxide. Why does the element impart colour to the flame?
Answer:

Sodium ions impart a yellow colour to the flame in a flame test and form an oxide and a peroxide with oxygen. The element imparts colour to the flame as the electrons in it change energy levels when heated.
$4Na+O_{2}\rightarrow 2Na_{2}O$
$2Na+O_{2}\rightarrow Na_{2}O_{2}$
$2Na_{2}O+O_{2}\rightarrow 2Na_{2}O_{2}$

Students can readily take the guidance of this chapter whenever they need to by just availing the downloading option, which is through the NCERT exemplar Class 11 Chemistry solutions chapter 10 PDF download.

NCERT Exemplar Class 11 Chemistry solutions chapter 10 The S-Block Elements includes the following topics:

Main Subtopics in Class 11 Chemistry NCERT Exemplar Solutions Chapter 10

Group 1 Elements: Alkali Metals
General Characteristics Of The Compounds Of The Alkali Metals
Oxides And Hydroxides
Halides
Salts Of Oxo-acids
Anomalous Properties Of Lithium
Some Important Compounds Of Sodium
Biological Importance Of Sodium And Potassium
Group 2 Elements: Alkaline Earth Metals
General Characteristics Of Compounds Of The Alkaline Earth Metals
Anomalous Behaviour Of Beryllium
Some Important Compounds Of Calcium
Biological Importance Of Magnesium And Calcium.

What the Students will Learn From NCERT Exemplar Class 11 Chemistry Solutions Chapter 10

While going through the Class 11 Chemistry NCERT exemplar solutions chapter 10 students tend to garner information about the comprising elements in the s block of the periodic table which further include the categorisation of elements depending upon their properties and occurrence.

NCERT Exemplar Class 11 Chemistry solutions chapter 10 provides the student with an in-detail review around the different types of alkali metals and their varying set of properties along with their important compounds and the conditions required for their formation.

NCERT exemplar solutions for Class 11 Chemistry chapter 10 informs the students about the biological importance of elements like Magnesium and Calcium and discusses the anomalous properties of lithium and beryllium along with their justifications in chemistry.

This whole chapter is verbose in the terms factual intricacies but is essential to be understood in order to be able to solve practical questions regarding the s block.

NCERT Exemplar Class 11 Chemistry Solutions Chapter-Wise

Chapter 1 Some Basic Concepts of Chemistry

Chapter 2 Structure of Atom

Chapter 3 Classification of Elements and Periodicity in Properties

Chapter 4 Chemical Bonding and Molecular Structure

Chapter 5 States of Matter

Chapter 6 Thermodynamics

Chapter 7 Equilibrium

Chapter 8 Redox Reactions

Chapter 9 Hydrogen

Chapter 11 The p-Block Elements

Chapter 12 Organic Chemistry – Some Basic Principles and Techniques

Chapter 13 Hydrocarbons

Chapter 14 Environmental Chemistry

Important Topics To Cover From NCERT Exemplar Class 11 Chemistry Solutions Chapter 10 The S-Block Elements

Firstly, the students are introduced to all the components of the s block along with a basic overview of their properties and their element number and weight. Class 11 Chemistry NCERT exemplar solutions chapter 10 would make the students familiar with the basic information about the topic.

The second part of the chapter focuses entirely on the particular properties and qualities of the elements, along with their differentiating and common factors. The naturally occurring state of each element is discussed so as to spot any anomalous behaviour like in Beryllium and Lithium and the reasoning as to why they occur in their respective states.

Students are also educated upon the uses and industrial applications of some of these elements due to their composition like Magnesium and draw attention to the different compounds like in the case of Calcium.

NCERT Exemplar Class 11 Chemistry solutions chapter 10 provides an in-depth review of the s block elements and their peculiarities and anomalies.

NCERT Exemplar Class 11 Solutions

NCERT Exemplar Class 11 Biology Solutions

NCERT Exemplar Class 11 Mathematics Solutions

NCERT Exemplar Class 11 Physics Solutions

Also, Check Chapter Wise NCERT Solutions for Class 11 Chemistry

Chapter-1 - Some Basic Concepts of Chemistry

Chapter-2 - Structure of Atom

Chapter-3 - Classification of Elements and Periodicity in Properties

Chapter-4 - Chemical Bonding and Molecular Structure

Chapter-5 - States of Matter

Chapter-6 - Thermodynamics

Chapter-7 - Equilibrium

Chapter-8 - Redox Reaction

Chapter-9 - Hydrogen

Chapter-10 - The S-Block Elements

Chapter-11 - The P-Block Elements

Chapter-12 - Organic chemistry- some basic principles and techniques

Chapter-14 - Hydrocarbons

Chapter-15 - Environmental Chemistry

NCERT Exemplar Class 11 Solutions

NCERT Exemplar Class 11 Biology Solutions

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Frequently Asked Questions (FAQs)

1. Q 1. What are the major topics of NCERT Exemplar class 11 Chemistry Solutions Chapter 10?

A- The major focus of this chapter is upon the physical and chemical properties and anomalies of the entire s block elements. It especially discusses the general characteristics of compounds of the alkali earth metals and the anomalous behaviours of lithium and beryllium.

2. Q2. Is this chapter carrying a huge share in the mark distribution in the exams?

A- Generally the questions asked from these chapters have theoretical, short and layered answers which do not amount to the highest mark securing problems but the information acquired from this chapter needs to be applied in many questions as well.

3. Q 3. Is studying only from NCERT exemplar solutions for class 11 Chemistry chapter 10 enough for students?

A- Though all the important questions along with answers have been mentioned in this chapter, it is still suggested that students refer to their original textbook for the theoretical part of the preparation and then visit this chapter for quick revisions and practice.

4. Q 4. Is there any way in which students can share this chapter with other interested students?

A- Yes students can share this chapter with anyone who is interested as a guide by making use of the option provided, which is NCERT exemplar class 11 Chemistry solutions chapter 10 pdf download.

Also Read

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

NCERT Exemplar Class 11 Chemistry Solutions Chapter 10 The s Block Elements

NCERT exemplar class 11 Chemistry solutions chapter 10: MCQ (Type 1)

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Main Subtopics in Class 11 Chemistry NCERT Exemplar Solutions Chapter 10

What the Students will Learn From NCERT Exemplar Class 11 Chemistry Solutions Chapter 10

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Important Topics To Cover From NCERT Exemplar Class 11 Chemistry Solutions Chapter 10 The S-Block Elements

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