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

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

Ans. Physical properties of alkali metals are as follows :

(i)  All the alkali metals have one valence electron, ns1 outside the noble gas core.

(ii)  These are quite soft and can be cut easily with a knife. For example, sodium, metal can be easily cut using a knife.

(iii) Alkali metals have largest size in their period. They are light coloured and are mostly silvery white in appearance.

(iv) They have low density. The density increases down the group from Li to Cs. The only exception is K which has lower density than Na.

(v) Melting and boiling points of alkali metals are low because the metallic bonding present in alkali metals is quite weak. On moving down the group, melting point decreases with increasing atomic size.
(vi) Alkali metals have lowest ionisation energy in a period. The ionisation energy of alkali metals decreases down the group.

(vii) Alkali metals and their salts impart a characteristic colour to flames and can also display photoelectric effect. This is because the heat from the flame excites the electron present in the outermost orbital to 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.

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

Chemical properties of alkali metals are as follows :

(i) Reaction with water: The alkali metals react with water and other hydrogen containing to form their respective hydroxides and dihydrogen. On moving down the group reactivity towards water increases.

$$2\text{Na}+2\text{H}_2\text{O}\xrightarrow{}2\text{NaOH}+\text{H}_2\\2\text{Na}+2\text{HX}\xrightarrow{}2\text{NaX}+\text{H}_2(\text{X}=\text{Halogen} )$$

(ii) Reaction with oxygen: They burn vigorously in oxygen forming oxides. Lithium forms monoxide, sodium forms peroxide, the other metals form superoxides. The superoxide O2 ion is stable only in the presence of large cations.

$$4\text{Li}+\text{O}_2\xrightarrow{}2\text{Li}_{2}\text{O}(\text{oxide})\\2\text{Na}+\text{O}_{2}\xrightarrow{}\text{Na}_{2}\text{O}_{2}\text{(peroxide)}\\\text{M}+\text{O}_{2}\xrightarrow{}\text{MO}_{2}(\text{superoxide})$$

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

$$2\text{M}+\text{H}_{2}\xrightarrow{}2\text{MH}$$

(iv) Reaction with halogens: Alkali metals, except Li, react directly with halogens to form halides. All halides are ionic in nature except lithium halide which is covalent.

$$2\text{M}+\text{Cl}_{2}\xrightarrow{}2\text{MCl}$$

(M = Li,K,Rb,Cs)

Covalent character in alkali metals is as follows:

LiCl > NaCl > KCl > RbCl > CsCl

LiI > LiBr > LiCl > LiF

(v) Reducing nature: 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 because of its high hydration energy.

(vi) Reaction with Liquid Ammonia: 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 reason for blue colour is the ammoniated electrons. 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 + NH3(q) → MNHam + H2(g)

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

Ans. The general characteristics of alkaline earth metals are as follows :

(i) The general electronic configuration of alkaline earth metals is [noble gas] ns2. Since, these metals lose two electrons to acquire the nearest noble gas configuration, therefore, their oxidation state is +2.

(ii) Alkaline earth Metals are very reactive and do not found in free state but found in the form of silicates, carbonates, sulphates and phosphates in abundance in nature.

(iii) The atomic and ionic radii of the alkaline earth metals are bigger than all other elements in a period but smaller than those of the corresponding alkali metals in the same period.

(iv) These are denser and harder than corresponding alkali metals and the melting and boiling points of these metals are higher than the corresponding alkali metals.

(v) Since the alkaline earth metals have large size, their ionisation enthalpies are found comparatively low. However, their first ionisation enthalpies are higher than the corresponding group 1 metals.

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

(vii) They are highly electropositive in nature because of their low ionisation 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. Be and Mg do not impart any colour to the flame.

Ca–Brick red, Sr–Crimson red, Ba–Apple green.

(ix) The alkaline earth metals are less reactive than alkali metals and their reactivity increases on moving down the group.

Chemical properties of alkaline earth metal are as follows :

(i) Reaction with air and water:

(a) Beryllium and Magnesium react with air and forms beryllium oxide and Beryllium nitride.

$$2\text{Be}+\text{O}_{2}\xrightarrow{}2\text{BeO}\\3\text{Be}+\text{N}_{2}\xrightarrow{}\text{B}_{3}\text{N}_{2}$$

Mg being more electropositive, burns in air with a brilliant glow and forms magnesium oxide and magnesium nitride.

$$2\text{Mg}+\text{O}_{2}\xrightarrow{}2\text{MgO}\\3\text{Mg}+\text{N}_{2}\xrightarrow{}\text{Mg}_{3}\text{N}_{2}$$

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

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

(d) Beryllium hydroxide is amphoteric in nature because it can react with both acids and bases.

$$\text{Be(OH)}_{2}+2\text{OH}^{\normalsize-}\xrightarrow{}[\text{Be(OH)}_{4}]^{\normalsize-}\\\text{Be(OH)}_{2}+2\text{HCl}\xrightarrow{}\text{BeCl}_{2}+2\text{H}_{2}\text{O}$$

(ii) Reaction with halogens: All the alkaline earth metals combine with halogens at high temperature to form corresponding halides.

$$\text{M + X}_{2}\xrightarrow{}\text{MX}_{2}(\text{X=F,Cl,Br,I})\\\text{Mg}+\text{Cl}_{2}\xrightarrow{}\text{MgCl}_{2}$$

(iii) Reaction with hydrogen: All the elements except beryllium combine with hydrogen upon heating to form their hydrides, MH2.

$$\text{M + H}_{2}\xrightarrow{}\text{MH}_{2}\\\text{Mg}+\text{H}_{2}\xrightarrow{}\text{MgH}_{2}$$

BeH2 and MgH2 are electron deficit covalent compounds. CaH2 , SrH2 are of ionic nature. CaH2 is called Hydrolith.

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

$$\text{M + 2HCl}\xrightarrow{}\text{MCl}_{2}+\text{H}_{2}(\text{g})$$

(v) Reducing nature: Like alkali metals, the alkaline earth metals are strong reducing agents, but their reducing power is less than that of alkali metals. Generally reducing character increases from top to bottom.

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

$$\text{M}+(x+y)\text{NH}_{3}\xrightarrow{}[\text{M}(\text{NH}_{3})_{x}]^{+}+[(\text{NH}_{3})_{y}]^{\normalsize-}$$

Ans: The alkali earth metals are also called s-block elements because these elements have one electron each in the valence s-subshell of their atoms i.e., they have ns1 configuration. The alkali metals include lithium, sodium, potassium, rubidium, caesium and francium. The have only one electron in their valence shell which they easily lose owing to their low ionisation enthalpy. Also, they readily dissolve in water to form soluble hydroxides, which are strongly alkaline in nature. Therefore, alkali metals are highly reactive and are not found in nature in their free or native state and hence, they are not easily found in nature.

10.4. Find the oxidation state of sodium in Na2O2.

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.

10.5. Explain why is sodium less reactive than potassium?

Ans. Potassium is larger than sodium and we know that on moving down the group, the atomic size of Alkali metals increases and therefore effective nuclear charge decreases. Therefore, effective nuclear charge on potassium is less than sodium. Also, ionisation potential of sodium is less than potassium. As a result of all these factors, the removal of the outermost electron in potassium is easy as compared to sodium. Therefore, potassium is more reactive than sodium.

10.6. 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 because of their large atomic sizes. Also, they lose their only valence electron easily as they attain stablenoble gas configuration after losing it. (i) Ionization enthalpy: Their first ionization enthalpies is 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 because of their electropositive nature. Also these oxides are 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 less than alkali metals. This is because alkaline earth metals are less electropositive than alkali metals. (iii) Solubility of hydro-xides:The hydroxides of alkali metals are more soluble than those of alkaline earth metals. (iii) Solubility of hydro-xides: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 account forhigher lattice energies.

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

Ans. Lithium resembles magnesium mainly due to the similarity in sizes of their atoms [Li = 152 pm, Mg = 160 pm] and ions [Li+ = 76 pm, Mg2+ = 72 pm]. Similarities between lithium and magnesium are as follows :

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

The oxides of both Li and Mg are much less soluble in water and their hydroxides decompose upon heating.

$$2\text{LiOH}\xrightarrow{\text{heat}}\text{Li}_{2}\text{O}+\text{H}_{2}\text{O}\\\text{Mg(OH)}_{2}\xrightarrow{}\text{MgO}+\text{H}_{2}\text{O}$$

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

(iii) Both Lithium and magnesium react with nitrogen to form nitride.

$$6\text{Li}+\text{N}_{2}\xrightarrow{\Delta}2\text{Li}_{3}\text{N}\\3\text{Mg(OH)}_{2}\xrightarrow{\Delta}\text{Mg}_{3}\text{N}_{2}$$

(iv) The carbonates of both are covalent in nature. Also, these decompose on heating. Li and Mg do not form solid bicarbonates.

$$\text{Li}_{2}\text{CO}_{3}\xrightarrow{}\text{Li}_{2}\text{O}+\text{CO}_{2}\\\text{MgCO}_{3}\xrightarrow{}\text{MgO}+\text{CO}_{2}$$

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

(vii) Both LiCl and MgCl2 are deliquescent in nature. They crystallize from aqueous solutions as hydrates, for example, LiCl.2H2O and MgCl2.8H2O.

10.8. Explain why alkali and alkaline earth metals not be obtained by chemical reduction methods?

Ans. Alkali and alkaline earth metals cannot be obtained by chemical reduction methods because they are among the strongest reducing agents. In the process of chemical reduction, oxides of metals are reduced using a stronger reducing agent. Alkali and alkaline earth metals are among the strongest reducing agent and the reducing agents that are stronger than them are difficult to obtain. Therefore, they cannot be obtained by chemical reduction of their oxides.

10.9. Why are potassium and caesium, rather than lithium used in photoelectric cells?

Ans. Potassium and caesium are used in the photoelectric cell but Lithium does not. This is because it has small size and high ionisation energy which means a large amount of energy is required to lose an electron whereas in case of K and Cs, they have large size and low ionization energy. Hence, they can easily lose electrons. Thus this property makes caesium and potassium useful in photoelectric cells.

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

Ans. In solution the alkali metal atom readily loses the valence electron. Both the cation and electron combine with ammonia to form ammoniated cation and ammoniated electron as:

$$\text{M}+(x+y)\text{NH}_{3}\xrightarrow{}\underset{\text{Ammoniated}\space \text{cation}}{[\text{M}(\text{NH}_{3})_{x}]^{+}}+\underset{\text{Ammoniated}\space \text{electron}}{[e(\text{NH}_{3})_{y}]^{\normalsize-}}$$

Here M = alkali metal

Blue colour of the solution is due to Ammoniated electron. When light falls on these ammoniated electrons, they get excited to higher energy levels by absorbing energy corresponding to red region of visible light. As a result, transmitted light is blue in colour. The solution becomes blue in colour which is paramagnetic.

When the concentration increases, the ammoniated metal ion get bound by free unpaired electron known as expanded metals. As a result, the solution becomes bronze in colour which is diamagnetic.

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

Ans. Beryllium and magnesium do not give colour to flame because of their small size and high ionisation potential values. Thus, in Be and Mg, the electrons are strongly bound due to small size. Therefore, the energy required for excitation of these electrons is very high. And, when the electron comes back to its original position, the energy released does not fall in the visible region and no colour in the flame is seen. On the other hand, in case of other alkaline earth metals, when a metal is heated, its valence electrons are excited and shifted to a higher energy level. To attain stability, when these excited electron comes back to its lower energy level, they radiates energy. This energy belongs to a particular colour in the visible region. Hence, the colour is observed.

10.12. 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. The various steps that occur in the solvay process are as follows :

Step 1: Brine solution is saturated with ammonia.

$$2\text{NH}_3 + \text{H}_2\text{O} + \text{CO}_2 \xrightarrow{}(\text{NH}_4)_2\text{CO}_3$$

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.

$$2\text{NH}_3 + \text{H}_2\text{O} + \text{CO}_2 \xrightarrow{}\text{NH}_4\text{CO}_3\\\text{NaCl + NH}_{4}\text{HCO}_{3}\xrightarrow{}\text{NaHCO}_{3}+\text{NH}_{4}\text{Cl}$$

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

Step 4: NaHCO3 is heated strongly to convert it into Na2CO3.

$$2\text{NaHCO}_{3}\xrightarrow{}\text{Na}_{2}\text{CO}_{3}+\text{CO}_{2}+\text{H}_{2}\text{O}$$

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

$$\text{Ca(OH)}_{2}+2\text{NH}_{4}\text{Cl}\xrightarrow{}2\text{NH}_{3}+2\text{H}_{2}\text{O}+\text{CaCl}_{2}$$

Thus, the overall reaction taking place in Solvay process is :

$$2\text{NaCl+CaCO}_{3}\xrightarrow{}\text{Na}_{2}\text{CO}_{3}+\text{CaCl}_{2}$$

10.13. Potassium carbonate cannot be prepared by Solvay process. Why?

Ans. Potassium carbonate cannot be prepared by Solvay process because it is fairly soluble in water and does not precipitate out.

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

Ans. As we move down the group of alkali metals, electropositive character and stability of carbonates increases. But, lithium carbonate is not so stable to heat. This is because:

(i) Lithium ion, being very small in size, distorts the electron cloud on nearby oxygen atom of the large carbonate ion, leading to the formation of more stable lithium oxide. Strong polarising action of small Lithium ion distorts electron cloud on the oxygen atom of the carbonate ion. Due to this, CO bond gets weakened and Li-O becomes stronger. This results in the decomposition of lithium carbonate into lithium oxide and carbon dioxide.

(ii) Li2O due to increased lattice energy (due to smaller oxide ion) is more stable than Li2CO3 and thus favours its decomposition.

10.15. 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. Solubility
Nitrates: Nitrates of both group 1 and group 2 metals are soluble in water.

Alkali metals: Their solubility in water increases on moving down the group because lattice energy decreases more rapidly than hydration energy.

Alkaline Earth metals: Nitrates of alkaline earth metals are soluble in water. Their solubility decreases down the group because their hydration energy decreases more rapidly than lattice enthalpy.

(a) Nitrates : Thermal Stability : Nitrates of alkali metals, except LiNO3, decompose on strong heating to form nitrites.

$$2\text{KNO}_{3}\text{(s)}\xrightarrow{}2\text{KNO}_{2}\text{(s)}+\text{O}_{2}\text{(g)}$$

LiNO3, on decomposition, gives oxide.

$$2\text{LiNO}_{3}(\text{s})\xrightarrow{\Delta}\text{Li}_{2}\text{O(s)}+2\text{NO}_{2}\text(g)+\text{O}_{2}\text{(g)}$$

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

$$2\text{Ca}(\text{NO}_{3})_{2}\text{(s)}\xrightarrow{\text{Delta}}2\text{CaO(s)}+4\text{NO}_{2}\text{(g)}+\text{O}_{2}\text{(g)}$$

As we move down group 1 and group 2, the thermal stability of nitrate increases.

Solubility :

Nitrates of alkali metals are soluble in water and their solubility in water increases or moving down the group because lattice energy decreases more rapidly than hydration energy.

Nitrates of alkaline earth metals are also soluble in water but their solubility decreases down the group because their hydration energy decreases more rapidly than lattice enthalpy.

(b) Carbonates

Thermal Stability

Alkali metals : The carbonates of alkali metals are stable towards towards heat. However, carbonate of lithium, when heated, decomposes to form lithium oxide.

Alkaline earth metals : The carbonates of alkaline earth metals also decompose on heating to form oxide and carbon dioxide.

$$\text{Na}_{2}\text{CO}_{3}\xrightarrow{}\text{No effect}\\\text{Li}_{2}\text{CO}_{3}\xrightarrow{}\text{Li}_{2}\text{O}+\text{CO}_{2}\\\text{MgCO}_{3}\xrightarrow{}\text{MgO}+\text{CO}_{2}$$

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.

(c) Sulphates :

Thermal Stability :

Sulphates of alkaline earth metals decompose on heating giving the oxides and SO3, the stability of alkaline metal carbonates increases as electropositive character of metals increases.

$$\text{MSO}_{4}\xrightarrow{}\text{MO}+\text{SO}_{3}$$

(M = alkaline earth metal)

Alkali metals sulphates are stable to heat and do not decompose easily.

Solubility : Sulphates of alkali metals are soluble in water. However, sulphates of alkaline earth metals show varied trends as follows :

 Compound Oxidation number of Carbon 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.

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

Ans. (i) Sodium metal: It is manufactured by electrolysis of a fused mixture of NaCl and CaCl2 in Down’s cell at 873 K using iron cathode and graphite anode. Na is liberated at the cathode while Cl2 is evolved at anode.

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.

$$\text{NaCl}\xrightarrow{\text{Electrolysis}}\text{Na}^{\normalsize+}+\text{Cl}^{\normalsize-}\\\text{Molten}\\\text{At Cathode:}\text{Na}^{\normalsize+}+e^{\normalsize-}\xrightarrow{}\text{Na}\\\text{At Anode:}\space\text{Cl}^{\normalsize-}+e^{\normalsize-}\xrightarrow{}\text{Cl}\\\text{Cl}+\text{Cl}\xrightarrow{}\text{Cl}_{2}$$

(ii) Sodium Hydroxide: It is manufactured by electrolysis of an aqueous solution of NaCl in Castner-Keller cell using mercury cathode and carbon anode. Sodium metal which is discharged at cathode combines with mercury to form amalgam. Cl2 gas is evolved at anode. The sodium metal, which is discharged at cathode, combines with mercury to form an amalgam.

$$\text{At Cathode:}\space\text{Na}^{+}+\text{e}^{-}\xrightarrow{}\underset{\text{Sodium}}{\text{Na}}\\2\text{Na}+\text{Hg}\xrightarrow{}\underset{\text{Sodium}\space \text{amalgam}}{\text{Na-Hg}}\\\text{At Anode:}\text{Cl}^{\normalsize-}-e^{\normalsize-}\xrightarrow{}\text{Cl}\\\text{Cl}+\text{Cl}\xrightarrow{}\text{Cl}_{2}$$

Sodium amalgam thus obtained is treated with water to form sodium hydroxide and hydrogen gas.

$$2\text{Na} – \text{Hg} + 2\text{H}_2\text{O} \xrightarrow{} 2\text{NaOH} + \text{H}_2 + \text{Hg}$$

(iii) Sodium peroxide

First, NaCl is electrolysed to result in the formation of Na metal by Downs process.

This sodium metal is then heated on aluminium trays in excess of air (free of CO2) to form its peroxide.

$$4\text{Na(s) + O}_2\text{(g)}\xrightarrow{}2\text{Na}_{2}\text{O(s)}\\2\text{Na}_{2}\text{O(s)+\text{O}}_{2}\text{(g)}\xrightarrow{}2\text{Na}_{2}\text{O}_{2}\text{(s)}$$

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

$$2\text{NH}_{3}+\text{H}_{2}\text{O}+\text{CO}_{2}\xrightarrow{}(\text{NH}_{4})_{2}\text{CO}_{3}\\(\text{NH}_{4})_{2}\text{CO}_{3}+\text{H}_{2}\text{O}+\text{CO}_{2}\xrightarrow{}2\text{NH}_{4}\text{HCO}_{3}\\\text{NH}_{4}\text{HCO}_{3}+\text{NaCl}\xrightarrow{}\text{NH}_{4}\text{Cl}+\text{NaHCO}_{3}$$

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

$$2\text{NaHCO}_3\xrightarrow{}\text{Na}_{2}\text{CO}_{3}+\text{CO}_{2}+\text{H}_{2}\text{O}$$

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

$$2\text{Mg}+\text{O}_{2}\xrightarrow{\text{Burning}}2\text{MgO}\\3\text{Mg}+\text{N}_{2}\xrightarrow{\text{Buring}}\text{Mg}_{3}\text{N}_{2}$$

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

$$\text{CaO + SiO}_{2}\xrightarrow{\text{heat}}\text{CaSiO}_{3}$$

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

$$\text{Ca(OH)}_{2}+\text{Cl}_{2}\xrightarrow{\Delta}\underset{\text{Bleaching}\space \text{powder}}{\text{CaOCl}_{2}}+\text{H}_{2}\text{O}$$

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

$$2\text{Ca}(\text{NO}_{3})_{2}(s)\xrightarrow{\Delta}2\text{}\text{CaO(s)}+4\text{NO}_{2}\text{(g)}+\text{O}_{2}\text{(g)}$$

10.18. 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 the manufacture of soap, paper, artificial silk etc.

(b) It is used in the textile industries for mercerising cotton fabrics.

(ii) Uses of sodium carbonate.

(a) It is used in water softening, laundering and clearning.

(b) It is used in the manufacture of glass, soap, borax etc.

(iii) Uses of quick lime.

(a) It is used in the manufacture of dye stuffs and employed in the purification of sugar.

(b) It is used in the manufacture of sodium carbonate from caustic soda.

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

Ans. (i) In the vapour phase BeCl2 exists as chlorobridged dimer which dissociates into linear monomer at 1200 K.

(ii) In the solid state, beryllium chloride has a polymeric chain structure. In this structure, Be atom is tetrahedrally surrounded by four Cl atoms. Two of the chlorine atoms are bonded by covalent bonds while other two by coordinate bond. The polymeric structure of beryllium chloride is due to its electron deficient nature.

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. 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.

10.21. Describe the importance of the following:

(i) limestone (ii) cement (iii) plaster of paris.

Ans. (i) Importance of Limestone (CaCO3)

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

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

(c) It is used in the manufacture of quick lime.

(d) It is used as a building material in the form of marble.

(e) It is used as an antacid and as mild abrasive in toothpaste.

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

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

(c) It is used in concrete and reinforced concrete.

(iii) The Formula of Plaster of Paris is 2CaSO4.H2O.

Importance of Plaster of Paris :

(a) It is used in surgical bandages.

(b) Used for producing moulds for pottery, ceramics etc.

(c) It is also used for making casts and moulds. Also used in making statues, models and other decorative materials.

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

Ans. Hydration energy depends on charge to radius ratio. Due to smallest size among the alkali metals, lithium ion can polarises water molecules more easily than other alkali metals. As a result, water molecules get attached to lithium salts as water of crystallisation. Hence, lithium salts such as trihydrated lithium chloride (LiCl.3H2O) are commonly hydrated. As the size of the ions increases, their polarising power decreases. Therefore, other alkali metal ions are usually anhydrous.

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

Ans. LiF is insoluble in water whereas LiCl is soluble not only in water, but also in acetone . LIF is more ionic than LiCl. The solubility of a compound in water depends on the balance between lattice energy and hydration energy. The size of Fluoride ion is much smaller in size than chloride ion, therefore, the lattice energy of LiF is greater than that of LiCl. Thus, LiF is almost insoluble in water due to high lattice energy and low hydration enthalpy. Whereas in lithium chloride the lattice enthalpy is very small due to large size of chloride ions and hence its hydration enthalpy is high. More over LiCl has partial covalent and partial ionic character due to the polarisation of chloride ion by Lithium ion. Due to its low hydration energy and partial covalent and partial ionic character, LiCl is soluble in water as well as acetone.

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

Ans. Importance of sodium, potassium, magnesium, and calcium in biological fluids are as follows :

(i) Sodium:

(a) It is a major component of blood plasma.

(b) It plays an important role in transmission of nerve signals and function of heart.

(c) It plays an important role in transporting sugar and amino acids into the cells.

(d) It is important in regulating the flow of water across cell membranes

(ii) Potassium:

(a) Potassium ions help in activating many enzymes.

(b) It participate in oxidation of glucose to produce energy rich ATP molecules.

(c) It plays an important role in protein synthesis.

(d) K+ ion participates with Na+ ions in sodium potassium pump for transmission of nerve.

(iii) Magnesium (Mg):

(a) Magnesium helps in relaxing nerves and muscles.

(b) Magnesium helps in building and strengthening bones.

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

(d) Magnesium ions catalyses many enzymatic reactions. We know that energy in ATP is stored in the form of phosphate bond. All enzymes that utilise ATP in phosphate transfer require magnesium as the cofactor. Thus, energy is released by hydrolysis of phosphate bond. This reaction is also catalysed by magnesium ions.

(e) The main pigment for the absorption of light in plants is chlorophyll which contains magnesium.

(iv) Calcium (Ca)

(a) Calcium helps in the coagulation of blood

(b) Calcium also helps in maintaining homeostasis.

(c) About 99 % of body calcium is present in bones and teeth. In teeth it is found in the form of apatite [Ca3(PO4)2] and in enamel of teeth as fluorapatite [3Ca3(PO4)2CaF2].

(d) Calcium also plays important roles in neuromuscular function, inter-neuronal transmission, cell membrane integrity and blood coagulation.

10.25. 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:

$$2\text{Na(s)}+2\text{H}_{2}\text{O}(1)\xrightarrow{}2\text{NaOH(aq)}+\text{H}_{2}\text{(g)}$$

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

$$2\text{Na(s)}+\text{O}_{2}\text{(g)}\xrightarrow{}\text{Na}_{2}\text{O}_{2}\text{(s)}$$

(iii) When sodium peroxide is dissolved in water, it readily hydrolysed to form sodium hydroxide and water. The chemical equation involved in the reaction is:

$$\text{Na}_2\text{O}_2\text{(s) + 2H}_2\text{O(l)} \xrightarrow{} 2\text{NaOH(aq)} + \text{H}_2\text{O}_2\text{(aq)}$$

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 alkali metal to form a nitride directly.

(c) E° for M2+(aq) + 2e ® M(s) (where M = Ca, Sr or Ba) is nearly constant.

Ans. (a) The alkali metal ions are highly hydrated. The smaller the size of the ion, the greater is the degree of hydration. 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) Due to the diagonal relationship of Li and Mg (both have similar ionic sizes), lithium like magnesium forms a nitride on heating with nitrogen while other alkali metals do not. Lithium is the only alkali metal which forms a nitride directly Li+ and N3 ions are very small, so they form stable Li3N.

$$6\text{Li(s)}+\text{N}_{2}\text{(g)}\xrightarrow{\text{heat}}2\text{Li}_{3}\text{N(s)}$$

(c) Electrode potential (E°) of any M2+/M electrode depends upon three factors: Ionisation enthalpy, Enthalpy of hydration and 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.

10.27. 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) A solution of Na2CO3 is alkaline because when it is added to water, it hydrolyses to give sodium bicarbonate and sodium hydroxide which is a strong base. As a result, the solution becomes alkaline.

$$\text{Na}_{2}\text{CO}_{3}+\text{H}_{2}\text{O}\xrightarrow{}\text{NaHCO}_{3}+\text{NaOH}$$

(b) The discharge potentials of alkali metals are much higher than that of hydrogen and therefore, when the aqueous solution of alkali metal chlorides are subjected to electrolysis, hydrogen gas is evolved instead of alkali metal at the cathode. Therefore, to prepare alkali metals, electrolysis of their fused chlorides is carried out because melting point of anhydrous chloride is reasonably low to carry out the electrolytic reduction.

(c) Sodium ions are found primary on the outside of cells, being located in blood plasma and in the interstitial fluid which surrounds the cell whereas potassium ions are most abundant cations 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.

10.28. Write balanced equations for reactions between:

(a) Na2O2 and water

(b) KO2 and water

(c) Na2O and CO2

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

$$2\text{Na}_2\text{O}_2\text{(s)} + 2\text{H}_2\text{O(l)} \xrightarrow{} 4\text{NaOH(aq) + O}_2\text{(aq)}$$

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

$$2\text{KO}_{2}\text{(s)}+2\text{H}_{2}\text{O(1)}\xrightarrow{}2\text{KOH(aq)}+\text{H}_{2}\text{O}_{2}\text{(aq)}+\text{O}_{2}\text{(g)}$$

(c) The balanced chemical equation for the reaction between Na2O and CO2 is:

$$\text{Na}_{2}\text{O(s)}+\text{CO}_{2}\text{(g)}\xrightarrow{}\text{Na}_{2}\text{CO}_{3}$$

10.29. How would you explain the following observations?

(i) BeO is almost insoluble but BeSO4 in 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, SO42– ion is a large anion. Hence, Be2+ can easily polarise SO42– 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, whereas BaSO4 is insoluble in water. This is because 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, both Ba2+ and SO42– ions are largeized. The lattice energy released is high. Hence, it is not soluble in water.

(iii) LiI is more soluble than KI in ethanol because the lithium ion has a higher polarising power than the potassium ion because of its small size. 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.

10.30. Which of the alkali metal is having least melting point? (a) Na (b) K (c) Rb (d) Cs

Ans. In alkali metals, on moving down the group, melting point decreases with increasing atomic size. This is because as we move down in an alkali group, the atomic size increases. Also, the strength of metallic bonds decreases on moving down a group in the periodic table. This causes a decrease in the melting point. Therefore, since Cs has the largest size among the given elements and weaker metallic bond therefore, it has the least melting point.

10.31. Which one of the following alkali metals gives hydrated salts? (a) Li (b) Na (c) K (d) Cs

Ans. The smaller the size of the ion, the greater is the degree of hydration. Thus, Li+ ion gets much more hydrated than other alkali metals. Therefore, the extent of hydration decreases from Li+ to Cs+. Among the given alkali metals, Li is the smallest in size. Also, it has the highest charge density and highest polarising power. Hence, it attracts water molecules more strongly than the other alkali metals. As a result, it forms hydrated salts such as LiCl2H2O. The other alkali metals are larger than Li and have weaker charge densities. Hence, they usually do not form hydrated salts.

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

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

Ans. The order of stability of alkaline earth metal carbonates is as follows:

BeCO3 < MgCO3 < CaCO3 < SrCO3 < BaCO3

On moving down the group, the electropositive character of alkaline earth metals increases. Due to this, their thermal stability also increases. The smaller the positive ion, the higher the charge density, and the greater effect it will have on the carbonate ion. As the positive ions get larger down the group, their effect on the carbonate ions near them decreases. More heat must be supplied for the carbon dioxide to leave the metal oxide. In other words, the carbonates become more thermally stable down the group. Hence, BaCO3 is the most stable.