NCERT Solutions for Class 11 Chemistry The s – Block Elements [Old Book]

NCERT
SOLUTIONS
CLASS - 11th

aglase .co

Class : 11th
Subject : Chemistry
Chapter : 10
Chapter Name : The S-Block Elements

Q10.1 What are the common physical and chemical features of alkali metals ?

Answer. 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 ames. This is because the
heat from the ame 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 + 2H2 O ⟶ 2M + 2OH + 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 ) ] + [M(NH3 ) ]
x 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.
+ − 1
M + e + NH3(l) ⟶ MNH(cm) + H2(g)
(cm) 2

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

Page : 313 , Block Name : Exercise

Q10.2 Discuss the general characteristics and gradation in properties of alkaline earth metals.

Answer. General characteristics of alkaline earth metals are as follows.
(i) The general electronic con guration of alkaline earth metals is [noble gas] ns .
2

(ii) These metals lose two electrons to acquire the nearest noble gas con guration. 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 rst 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 ames.
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 ame.
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 Be N . 3 2

(b) Mg, being more electropositive, burns in air with a dazzling sparkle to form MgO and
Mg N . 2
3

(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, I)

(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(x) ↑

(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.
−
+2
M + (x − y)NH3 ⟶ [M(NH3 ) ] + 2[e(NH3 ) ]
x y

Page : 313 , Block Name : Exercise

Q10.3 Why are alkali metals not found in nature ?

Answer. 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.

Page : 313 , Block Name : Exercise

Q10.4 Find out the oxidation state of sodium in Na O .2 2

Answer. 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 state of sodium is +1.

Page : 313 , Block Name : Exercise

Q10.5 Explain why is sodium less reactive than potassium.

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

Page : 313 , Block Name : Exercise

Q10.6 Compare the alkali metals and alkaline earth metals with respect to
(i) ionisation enthalpy
(ii) basicity of oxides and
(iii) solubility of hydroxides.

Answer.

Page : 313 , Block Name : Exercise

Q10.7 In what ways lithium shows similarities to magnesium in its chemical behaviour?

Answer. 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.
heat
2LiOH ⟶ Li2 O + H2 O

heat
Mg(OH)2 ⟶ MgO + H2 O

(iii) Both Li and Mg react with N2 to form nitrides.
heat
2LiOH ⟶ Li2 O + H2 O

heat
Mg(OH)2 ⟶ MgO + H2 O

(iv) Neither Li nor Mg form peroxides or superoxides.
(v) The carbonates of both are covalent in nature. Also, these decompose on heating.
heat
Li2 CO3 ⟶ Li2 O + CO2
heat

MgCO3 ⟶ MgO + CO2

(vi) Li and Mg do not form solid bicarbonates.
(vii) Both LiCl and MgCl are soluble in ethanol owing to their covalent nature.
2

(viii) Both LiCl and MgCl are deliquescent in nature. They crystallize from aqueous solutions
2

as hydrates, for example, LiCl ⋅ 2H O and MgCl ⋅ 8H O
2 2 2

Page : 313 , Block Name : Exercise

Q10.8 Explain why can alkali and alkaline earth metals not be obtained by chemical reduction
methods?

Answer. 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.

Page : 313 , Block Name : Exercise

Q10.9 Why are potassium and caesium, rather than lithium used in photoelectric cells?

Answer. 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.

Page : 313 , Block Name : Exercise

Q10.10 When an alkali metal dissolves in liquid ammonia the solution can acquire different
colours. Explain the reasons for this type of colour change.

Answer. When an alkali metal is dissolved in liquid ammonia, it results in the formation of a
deep blue coloured solution.
+ −1
M + (x + y)NH3 ⟶ M (NH3 ) + e (NH3 )
x 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.

Page : 313 , Block Name : Exercise

Q10.11 Beryllium and magnesium do not give colour to ame whereas other alkaline earth
metals do so. Why ?

Answer. 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 ame is seen.

Page : 313 , Block Name : Exercise

Q10.12 Discuss the various reactions that occur in the Solvay process.

Answer. 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 + H2 O + CO2 ⟶ (NH4 ) CO3
2

This ammoniated brine is ltered 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 → NH4CO3NaCl + NH4HCO3 → NaHCO3 + NH4Cl

Step 3: The solution containing crystals of NaHCO3 is ltered to obtain NaHCO3.
Step 4: NaHCO is heated strongly to convert it into NaHCO .
3 3

2NaHCO3 ⟶ Na2 CO3 + CO2 + H2 O

Step 5: To recover ammonia, the ltrate (after removing NaHCO ) is mixed with Ca(OH)
3 2

and heated.
Ca(OH)2 + 2NH4Cl → 2NH3 + 2H2O + CaCl2

The overall reaction taking place in Solvay process is
2NaCl + CaCO3 ⟶ Na2 CO3 + CaCl2

Page : 313 , Block Name : Exercise

Q10.13 Potassium carbonate cannot be prepared by Solvay process. Why ?

Answer. 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.

Page : 313 , Block Name : Exercise

Q10.14 Why isLi CO decomposed at a lower temperature whereas Na CO at higher
2 3 2 3

temperature?

Answer. 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.
Δ
Li2 CO3 ⟶ Li2 O + CO2

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

carbonate decomposes at a high temperature.

Page : 313 , Block Name : Exercise

Q10.15 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.

Answer. (i) Nitrates
Thermal stability
Nitrates of alkali metals, except LiNO , decompose on strong heating to form nitrites.
3

2KNO3(s) ⟶ 2KNO2(s) + O2(x)

LiNO3, on decomposition, gives oxide.
Δ
2LiNO3(s) ⟶ Li2 O(s) + 2NO2(x) + O2(x)

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

2Ca(NO3 ) ⟶ 2CaO(s) + 4NO2(g) + O2(g)
(s)

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.
Δ
Na2 CO3 ⟶ No effect

Δ
Li2 CO3 ⟶ Li2 O + CO2

Δ
MgCO ⟶ MgO + CO2
3

Solubility
Carbonates of alkali metals are soluble in water with the exception of Li CO . Also, the
2 3

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.
BeSO Fairly soluble
4

MgSO Soluble
4

CaSO Sparingly soluble
4

SrSO Insoluble
4

BaSO Insoluble
4

In other words, while moving down the alkaline earth metals, the solubility of their sulphates
decreases.

Page : 314 , Block Name : Exercise

Q10.16 Starting with sodium chloride how would you proceed to prepare
(i) sodium metal
(ii) sodium hydroxide
(iii) sodium peroxide
(iv) sodium carbonate ?

Answer. (a) Sodium can be extracted from sodium chloride by Downs process.
This process involves the electrolysis of fused NaCl (40%) andCaCl (60 %) at a temperature
2

of 1123 K in Downs cell.

+ −
NaCl − Electrolysis → Na + Cl

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
+ −
At Cathode: Na + e ⟶ Na
− −
At Anode: Cl + e ⟶ Cl

Cl + Cl ⟶ Cl2

Page : 314 , Block Name : Exercise

Q10.17 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 ?

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

2Mg + O2 ⟶ 2MgO

Buming
3Mg + N2 Mg3 N2
⟶

(ii) Quick lime (CaO) combines with silica (SiO2) to form slag.
heat
CaO + SiO2 ⟶ CaSiO3

(iii) When chloride is added to slaked lime, it gives bleaching powder.
A
Ca(OH)2 + Cl2 ⟶ CaOCl2 + H2 O

Bleaching

powder

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

2Ca(NO3 ) ⟶ 2CaO(s) + 4NO2(g) + O2(g)
2(s)

Page : 314 , Block Name : Exercise

Q10.18 Describe two important uses of each of the following :
(i) caustic soda
(ii) sodium carbonate
(iii) quicklime.

Answer. (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.

Page : 314 , Block Name : Exercise

Q10.19 Draw the structure of
(i) BeCl (vapour)
2

(ii) BeCl (solid).
2

Answer. (i)In the vapour state, BeCl exists as a monomer with a linear structure.
2

(ii) Structure of BeCl2 (solid)
BeCl exists as a polymer in condensed (solid) phase.
2

Page : 314 , Block Name : Exercise

Q10.20 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.

Answer. 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.

Page : 314 , Block Name : Exercise

Q10.21 Describe the importance of the following :
(i) limestone
(ii) cement
(iii) plaster of paris.

Answer. (i) Chemically, limestone is CaCO . 3

Importance of limestone
(a) It is used in the preparation of lime and cement.
(b) It is used as a ux 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 2CaSO ⋅ H O
4 2

Importance of plaster of Paris
(a) It is used in surgical bandages.
(b) It is also used for making casts and moulds.

Page : 314 , Block Name : Exercise

Q10.22 Why are lithium salts commonly hydrated and those of the other alkali ions usually
anhydrous?

Answer. 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.3H O are commonly hydrated. As the size of the ions increases, their
2

polarizing power decreases. Hence, other alkali metal ions usually form anhydrous salts.

Page : 314 , Block Name : Exercise

Q10.23 Why is LiF almost insoluble in water whereas LiCl soluble not only in water but also in
acetone ?

Answer. 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 uoride 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 uoride 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.

Page : 314 , Block Name : Exercise

Q10.24 Explain the signi cance of sodium, potassium, magnesium and calcium in biological
uids.

Answer. Importance of sodium, potassium, magnesium, and calcium in biological uids:
(i) Sodium (Na):
Sodium ions are found primarily in the blood plasma. They are also found in the interstitial
uids surrounding the cells.
(a) Sodium ions help in the transmission of nerve signals.
(b) They help in regulating the ow 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 uids.
(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.

Page : 314 , Block Name : Exercise

Q10.25 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 ?

Answer. (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) + 2H2 O(t) ⟶ 2NaOH(αq) + 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) ⟶ Na2 O2(s)

(iii) When sodium peroxide is dissolved in water, it is readily hydrolysed to form sodium
hydroxide and water. The chemical equation involved in the reaction is:
2Na(s) + O2(s) ⟶ Na2 O2(s)

Page : 314 , Block Name : Exercise

Q10.26 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 M (aq) + 2e ⟶ M(s) (where M = Ca, )
∘ 2+ −

Answer. (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

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
+

N
3−
ion. Hence, the lattice energy released is very high. This energy also overcomes the high
amount of energy required for the formation of theN ion. 3−

(c) Electrode potential (E°) of any M /M electrode depends upon three factors:
2+

(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.

Page : 314 , Block Name : Exercise

Q10.27 State as to why
(a) a solution of Na CO is alkaline ?
2 3

(b) alkali metals are prepared by electrolysis of their fused chlorides ?
(c) sodium is found to be more useful than potassium ?

Answer. (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.
Na2 CO3 + H2 O ⟶ 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 dif culties, alkali metals are usually prepared by the electrolysis of
their fused chlorides.
(c) Blood plasma and the interstitial uids surrounding the cells are the regions where sodium
ions are primarily found. Potassium ions are located within the cell uids. Sodium ions are
involved in the transmission of nerve signals, in regulating the ow 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.

Page : 314 , Block Name : Exercise

Q10.28 Write balanced equations for reactions between
(a) Na2 O2 and water

(b) KO2 and water

(c) Na2 O and CO2

Answer. (a) The balanced chemical equation for the reaction between Na2O2 and water is:
2Na2 O2(s) + 2H2 O(l) ⟶ 4NaOH(aq) + O(2)

(b) The balanced chemical equation for the reaction between KO2 and water is:
2KO2(s) + 2H2 O(l) ⟶ 2KOH(aq) + H2 O2(aq) + O2(g)

(c) The balanced chemical equation for the reaction between Na2O and CO2 is:
Na2 O(s) + CO2(g) ⟶ Na2 CO3

Page : 314 , Block Name : Exercise

Q10.29 How would you explain the following observations?
(i) BeO is almost insoluble but BeSO is soluble in water,
4

(ii) BaO is soluble but BeSO is insoluble in water,
4

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

Answer. (i) BeO is almost insoluble in water and BeSO is soluble in water.Be is a small
4
2+

cation with a high polarising power and O is a small anion. The size compatibility of Be
2− 2+

and O is high. Therefore, the lattice energy released during their formation is also very
2−

high. When BeO is dissolved in water, the hydration energy of its ions is not suf cient to
overcome the high lattice energy. Therefore, BeO is insoluble in water. On the other hand,
ion is a large anion. Hence, Be can easily polarised SO ions, making BeSO
2− 2+ 2−
SO 4
4 4

unstable. Thus, the lattice energy of BeSO is not very high and so it is soluble in water.
4

(ii) BaO is soluble in water, but BeSO is not. Ba is a large cation and O is a small anion.
2+ 2−
4

The size compatibility of Be and O is not high. As a result, BaO is unstable. The lattice
2+ 2−

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 BeSO , Be and SO
4
2+ 2−

4

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.

Page : 314 , Block Name : Exercise

Q10.30 Which of the alkali metal is having least melting point ?
(a) Na
(b) K

(c) Rb
(d) Cs

Answer. 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.

Page : 314 , Block Name : Exercise

Q10.31 Which one of the following alkali metals gives hydrated salts ?
(a) Li
(b) Na
(c) K
(d) Cs

Answer. 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 polarising
power. Hence, it attracts water molecules more strongly than the other alkali metals. As a
result, it forms hydrated salts such as LiCl ⋅ 2H O .. The other alkali metals are larger than Li
2

and have weaker charge densities. Hence, they usually do not form hydrated salts.

Page : 314 , Block Name : Exercise

Q10.32 Which one of the alkaline earth metal carbonates is thermally the most stable ?
(a)MgCO (b) CaCO3 (c)SrCO3 (d)BaCO3
3

Answer. 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:
MgCO < CaCO3 < SrCO3 < BaCO3
3

Page : 314 , Block Name : Exercise