# NCERT Solutions for Class 11 Chemistry Chapter 10 The S Block Elements

NCERT Solutions for Class 11 Chemistry Chapter 10 The s-Block Elements are provided here in pdf format, thereby giving better accessibility to all the students. Chapter 10 solutions PDF for Class 11 Chemistry contains answers to the questions provided in the textbook along with exemplary problems, MCQ questions from previous year question papers and CBSE sample papers. By referring to these solutions, students can have a quick revision of the entire topic in a very short interval of time. NCERT Solutions for Class 11 Chemistry Chapter 10 serves as important study material for students preparing for examinations and other assignments.

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A graphite block is the anode while steel is made the cathode. Metallic Ca and Na are formed at the cathode. Molten Na is supported by dipping in kerosene.

NaCl $overset{Electrolysis}{ ightarrow}$ Na+ + Cl

(Molten)

At Cathode: Na+ +e $ightarrow$ Na

At Anode:    Cl + e   $ightarrow$ Cl

Cl + Cl   $ightarrow$ Cl2

(ii) Sodium hydroxide

By electrolyzing a solution of sodium chloride, we can get Sodium hydroxide. This process is commonly known as Castner- Kellner process.

The process is carried out using a mercury cathode and a carbon anode.

Sodium metal, deposited at cathode forms an Amalgam by combining with Mercury.

Cathode: Na+ +e $overset{Hg}{ ightarrow}$ Na-Amalgam

2Na-Hg + 2H2O → 2NaOH + H2 + Hg

Anode:    Cl $ightarrow$ $frac{1}{2}$Cl2 + e

(iii) Sodium peroxide

After Na metal is gotten from Downs process, it is heated on Aluminium trays in presence of air (without CO2) to form Sodium peroxide.

2Na + O2(air) $ightarrow$ Na2O2

(iv) Sodium carbonate

Sodium hydrogen carbonate is obtained as a precipitate by reacting sodium chloride with ammonium hydrogen carbonate. The resultant crystals can be heated to obtain Sodium Carbonate.

2NH2 + H2O $ightarrow$ (NH4)2CO3

(NH4)2CO3 + H2O + CO2 $ightarrow$ 2NH4HCO3

2NH4HCO3 + NaCl $ightarrow$ NH4Cl + NaHCO3

The resultant crystals can be heated to obtain Sodium Carbonate.

2NaHCO3 $ightarrow$ Na2CO3+ CO2 + H2O

Q 10.17

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 10.17

(i) When Magnesium is burnt in air, it does so with a dazzling bright light resulting in the formation of Mg3N2 and MgO.

2Mg + O2 $overset{Burning}{ ightarrow}$ 2MgO

3Mg + N2 $overset{Burning}{ ightarrow}$ Mg3N2

(ii) Silica (SiO2) combines with Quick lime (CaO) resulting in formation of Slag.

CaO + SiO2 $overset{Heat}{ ightarrow}$ CaSiO3

(iii) Bleaching powder is formed when chlorine is made to react with slaked lime.

Ca(OH)2 + Cl2 $overset{Heat}{ ightarrow}$ CaOCl2 + H2O

(iv) Calcium nitrate, when heated, undergoes decomposition to form calcium oxide.

2Ca(NO3)2(s) $overset{Heat}{ ightarrow}$ 2CaO(s) + 4NO2(g) + O2(g)

Q 10.18

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

Ans 10.18

(i) Caustic soda

(a) Heavily used in soap industries.

(b) Common reagent in laboratories.

(ii) Sodium carbonate

(a) Finds uses in both soap and glass industries.

(b) Also finds use as a water softener.

(iii) Quick lime

(a)Finds use as a primary material for manufacturing slaked lime.

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

Q 10.19

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

(i) BeCl2 has a linear structure and exists as a monomer in the vapour state.

(ii) In the solid phase, BeCl2 is a polymer.

Q 10.20

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 10.20

Since the atomic sizes of magnesium and calcium are smaller than that of sodium and potassium, calcium and magnesium form carbonates and hydroxides with higher lattice energies. Thus, they are only sparingly soluble whereas those of potassium and sodium are readily soluble due to low lattice energies.

Q 10.21

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

Ans 10.21

Uses of cement:

• Bridge construction
• Plastering
• Most important ingredient in concrete

Uses of Plaster of Paris:

• Used to make casts and moulds
• Used to make surgical bandages

Uses of limestone:

• Preparation of cement and lime
• As a flux in iron ore smelting

Q 10.22

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

Ans 10.22

Since Lithium has the smallest size among all the alkali metals, it can easily polarized water molecules. Thus, smaller the size of the ion, greater is its ability to polarize water molecules.

Hence, trihydrated Lithium Chloride and other Lithium salts can be easily polarized. Due to this reason, other alkali metal ions can only form anhydrous salts.

Q 10.23

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

Ans 10.23

LiF has a greater ionic character than LiCl which disturbs the balance between hydration energy and lattice energy. This balance is crucial for the solvability of ions in solution. Due to greater covalent character and lower lattice energy, dissolution of LiCl is more exothermic in nature than that of LiF.

Q 10.24

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

Ans 10.24

Sodium (Na):

They are found in our blood plasma and the interstitial fluids around the cells. They help in

(a) Transmission of nerve signals.

(b) They regulate the flow of water across the membranes of the neighboring cells.

(c) Transport sugars and amino acids from and to cells.

Potassium (K):

They are found mostly in the cell fluids in greater quantities.

They help in

(a) Activating enzymes.

(b) Oxidising glucose to form ATP.

(c) Transmitting nerve signals.

Magnesium (Mg) and calcium (Ca):

They are also called as macro-minerals named so because of their abundance in our body. Mg helps in

(a) Relaxing nerves and muscles.

(b) Building and strengthening bones.

(c) Maintaining blood circulation in our body.

Ca helps in

(a) coagulation of blood

(b) Maintaining homeostasis.

Q 10.25

What happens when:

(i) Sodium metal is immersed in water?

(ii) Sodium metal is heated in abundance of air?

(iii) Sodium peroxide gets dissolved in water?

Ans 10.25

(i) Sodium reacts to form NaOH and H2 gas when it is dropped in water. The reaction occurs as shown below:

2Na(s) + 2H2O(l) $ightarrow$ 2NaOH(aq) + H2(g)

(ii) Sodium peroxide is formed when sodium reacts with oxygen while heating it in presence of air. The reaction proceeds as shown below:

2Na(s)  + O 2(s)  $ightarrow$ Na2O 2(s)

(iii) NaOH and water are formed as a result of hydrolysis of Sodium peroxide when it is dissolved in water.

Na2O 2(s) + 2H2O(l) $ightarrow$ 2NaOH(aq) + H2O2(aq)

Q 10.26

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 metal to form a nitride directly.

(c) E0 for M2+(aq) + 2e $ightarrow$ M(s) (where M = Ca, Sr or Ba) is nearly constant.

Ans 10.26

(a) The ionic and atomic sizes of the metals tend to increase while going down the alkali group.

The increasing order of the ionic sizes of the alkali metal ions is as shown below:

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

Smaller the size of an ion, greater is its ability to get hydrated. Li+ ion gets heavily hydrated since it is the smallest in size whereas Cs+ has the largest size and is the least hydrated ion. The alkali metal ions when arranged in the decreasing order of their hydrations is as shown below:

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

Higher the mass of a hydrated ion, the lesser is its ionic mobility. Thus, hydrated Li+ is the least mobile ion whereas hydrated Cs+ is the most mobile ion.

The ionic mobility of the alkali metal ions are in the following order:

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

(b) The only metal that can form a nitride directly is Lithium because Li+ has a smaller size and is easily compatible with the N3– ion. Thus, the lattice energy released is very high which is enough to overcome the amount of energy needed to form N3- ion.

(c) Electrode potential (E°) of any M2+/M electrode is decided by three factors:

(i) Enthalpy of hydration

(ii) Enthalpy of vaporisation

(iii) Ionisation enthalpy

The cumulative effect of these factors on Ba, Sr, and Ca is almost the same.

As a result, their electrode potentials are also same.

Q 10.27

State as to why

(a) a solution of Na2CO3 alkaline in nature?

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

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

Ans 10.27

(a) Sodium bicarbonate and sodium hydroxide are the end products when Na2CO3 is hydrolyzed. Since, the product are alkaline in nature, a solution of Na2CO3 is considered to be alkaline in nature.

(b) Chemical reduction cannot be used to prepare alkali metals since they themselves are reducing in nature. Alkali metals are highly electropositive and thus cannot be prepared by displacement reactions. Since they also react with water, these alkali metals cannot be prepared by electrolysis of their aqueous solutions. Thus, alkali metals are mostly prepared by electrolysis of their fused chlorides.

(c) Sodium ions are primarily found in the Blood plasma and the interstitial fluids around the cells whereas Potassium ions are found within the cell fluids. Sodium ions help in the transmission of nerve signals and also regulate the flow of water and transport sugars and amino acids into the cells.

Thus, Sodium is more important for our survival than potassium.

Q 10.28

Write balanced equations for reactions between:

(a) Na2O2 and water

(b) KO2 and Water

(c) Na2O and CO2

Ans 10.28

(a) 2Na2O2(s) + 2H2O(l)  $ightarrow$  4NaOH(aq) + O2(aq)

(b) 2KO2(s) + 2H2O(l)  $ightarrow$ 2KOH(aq) + H2O2(aq) + O2(aq)

(c) Na2O(s) + CO2(g) $ightarrow$ Na2CO3

Q 10.29

How would you explain the following observations?

(i) BeO is almost insoluble but BeSO4 is insoluble in water.

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

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

Ans 10.29

(i) The sizes of Be2+ and O2- are small and are highly compatible with each other. Due to this, a high amount of lattice energy is released during its formation. The hydration energy, when it is made to dissolve in water, is not enough to overcome the lattice energy. Thus, BeO is almost insoluble in water.

Whereas the size of an SO42- is large compared to Be2+ and there is lesser compatibility and lattice energy which can be easily overcome by the hydration energy. Thus, BeSO4 is easily soluble in water.

(ii) The sizes of Ba2+ and SO42- are large and are highly compatible with each other. Due to this, a high amount of lattice energy is released during its formation. The hydration energy, when it is made to dissolve in water, is not enough to overcome the lattice energy. Thus, BaSO4 is insoluble in water.

Whereas the size of an O2- is small compared to Be2+ and there is lesser compatibility and lattice energy which can be easily overcome by the hydration energy. Thus, BaO is easily soluble in water.

(iii) The lithium-ion has a smaller size and as a result of that, it has a higher polarizing capability. This enables it to polarize the electron cloud around an iodide ion thus resulting in a greater covalent character in LiI than KI. Thus, LiI is easily soluble in ethanol.

Q 10.30

Which of the following alkali metals has the least melting point?

(a) Na  (b) K  (c) Rb  (d) Cs

Ans 10.30

(d) Cs

Cs has the least melting point of the given alkali metals since it has the largest size. Due to a larger size, the binding capability of Cs is limited and the lattice energy released during the formation of its compounds is less and can be easily broken.

Q 10.31

Which one of the following alkali metals gives hydrated salts?

(a) Li  (b) Na  (c) K  (d) Cs

Ans 10.31

(a) Li

Li is capable of forming hydrated salts because of its size. Since it is smaller in size, it has a higher charge density and can easily attract water molecules around it and for hydrated salts like LiCl.2H2O. The other alkali metals have a bigger size and lesser charge density and thus aren’t capable of forming hydrated salts.

Q 10.32

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

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

Ans 10.32

(d) BaCO3

Thermal stability is directly proportional to the size of the cation i.e., larger the size of the atom, greater is its thermal stability. The biggest cation among the given compounds is Ba.

Thus, BaCO3 will be the most thermal carbonate among the given compounds followed by SrCO3, CaCO3 and MgCO3.

 Also Access NCERT Exemplar for class 11 chemistry Chapter 10 CBSE Notes for class 11 chemistry Chapter 10

Studying Chapter 10 S-Block elements will help you get acquainted with group 1 elements, their electronic configurations, alkali metals, Ionization enthalpy, atomic and ionic radii, hydration, enthalpy, physical and chemical properties use and importance. After studying this chapter students will be able to describe characters of alkali metals and their compounds, characters of alkaline earthly metals and compounds, industrial importance and production of important sodium and calcium compounds, which includes Portland cement.

CoolGyan’S provides premier study materials, notes, Sample papers, Important questions, Mock tests, textbooks, worksheets and exemplar problems for all the chapter of all the subjects of Class 11. Keep Visiting CoolGyan’S for latest CBSE sample papers, previous year questions. At CoolGyan’S we also provide CBSE notes for Classes 6 to 12.

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NCERT Solutions for Class 11 Chemistry Chapter 10 are used by students to get a proper grasp of all the concepts of the subjects and also to lay the foundation for their career or further higher studies. CoolGyan’S experts formulate these questions in an easy and understandable manner that helps students solve questions in the most efficient possible ways. We hope these solutions will help CBSE Class 11 students to build a strong foundation of the basics and secure excellent marks in their final exam.

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