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

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Answer:

The melting point decreases down the group due to a decrease in metallic bond strength. In alkali metals, Cs is the last member of the group and thus, has the lowest melting point, i.e., 29°C. Therefore, it is expected to melt if the room temperature rises to 30°C.

Hence, the correct answer is option (iv).

Answer:

Although lithium has most negative E^Θ value, its reaction with water is less vigorous than that of sodium which has the least negative E^Θ value among the alkali metals. This behaviour of lithium is attributed to its small size and very high hydration energy. Other metals of the group react explosively with water.

Hence, the correct answer is option (i).

Answer:

Lithium has the highest hydration enthalpy which accounts for its high negative E^Θ value and its high reducing power.

Hence, the correct answer is option (iii).

Answer:

The thermal stability of the metal carbonates of group 2 depends on the stability of MO. If MO is stable, MCO₃ is thermally unstable and vice- versa. ${\mathrm{MCO}}_3\stackrel{ }{\to }\mathrm{MO}+{\mathrm{CO}}_2$

 (Thermal stability of MO decreases as the size of M increases)

Hence, the correct answer is option (iv).

Answer:

Beryllium carbonate is kept in the atmosphere of carbon dioxide as it is unstable and decomposes to give beryllium oxide and carbon dioxide.
${\mathrm{BeCO}}_3\rightleftharpoons \mathrm{BeO} + {\mathrm{CO}}_2$.
When it is stored in the atmosphere of CO₂, the concentration of CO₂ increases in the right side, making the reaction to shift towards the backward direction so that the decomposition of beryllium carbonate is prevented.

Hence, the correct answer is option (i).

Answer:

The basic character of metal hydroxides increases down the group from Mg(OH)₂ â€‹to Ba(OH)₂ due to the increase in size and decrease in ionization enthalpy. As a result, the M−OH bond becomes weaker and basicity increases.

Hence, the correct answer is option (i). 

Answer:

Beryllium halides are essentially covalent due to the smallest size of Be²⁺ ion among all the group members and therefore are soluble in organic solvents like ethanol.

Hence, the correct answer is option (i).

Answer:

The ionization enthalpies of the alkali metals decrease down the group from Li to Cs. This is because the effect of increasing size outweighs the increasing nuclear charge, and the outermost electron is very well screened from the nuclear charge.

Answer:

The lowest solubility of LiF is due to the fact that the lattice enthalpy is higher than hydration enthalpy. Therefore, the dissociation of the Li−F bond becomes difficult and so is the solubility.

Hence, the correct answer is option (ii).

Answer:

Beryllium hydroxide is amphoteric in nature as it reacts with both acid and alkali. It is soluble in sodium hydroxide. The respective reactions are shown below:
$$\begin{gathered} \mathrm{Be}{\left(\mathrm{OH}\right)}_2+2\mathrm{NaOH}\to {\mathrm{Na}}_2\left[\mathrm{Be}{\left(\mathrm{OH}\right)}_4\right] \\ \mathrm{Be}{\left(\mathrm{OH}\right)}_2+2\mathrm{HCl} +2{\mathrm{H}}_2\mathrm{O}\to \left[\mathrm{Be}{\left(\mathrm{OH}\right)}_4\right]{\mathrm{Cl}}_2 \end{gathered}$$

Hence, the correct answer is option (i).

Answer:

Sodium carbonate is synthesized by the Solvay process. The reactions involved are

${\mathrm{NH}}_3+{\mathrm{H}}_2\mathrm{O}+{\mathrm{CO}}_2\to {\mathrm{NH}}_4{\mathrm{HCO}}_3$

$$\begin{gathered} \mathrm{NaCI}+{\mathrm{NH}}_4{\mathrm{HCO}}_3\to {\mathrm{NaHCO}}_3\downarrow +{\mathrm{NH}}_4\mathrm{CI} \\ 2{\mathrm{NaHCO}}_3\stackrel{\mathrm{\Delta }}{\to }{\mathrm{Na}}_2{\mathrm{CO}}_3+{\mathrm{H}}_2\mathrm{O}+{\mathrm{CO}}_2 \end{gathered}$$

In this process NH₃ is recovered when the solution containing NH₄Cl is treated with Ca(OH)₂.

$2{\mathrm{NH}}_4\mathrm{Cl}+\mathrm{Ca}{\left(\mathrm{OH}\right)}_2\stackrel{}{\to }2{\mathrm{NH}}_3+{\mathrm{CaCl}}_2+2{\mathrm{H}}_2\mathrm{O}$

Therefore, the side product formed is calcium chloride, i.e., CaCl₂.

Hence, the correct answer is option (i).



 

Answer:

The blue colour of the solution is due to the ammoniated electron which absorbs energy in the visible region of light and thus imparts a blue colour to the solution. The reaction may be shown as follows:

$\mathrm{Na} + (x+y) {\mathrm{NH}}_3\to [\mathrm{Na}{\left({\mathrm{NH}}_3\right)}_x{]}^{+}+{\left[\mathrm{e}\right({\mathrm{NH}}_3\left)\right]}^{-}$

Hence, the correct answer is option (i).
                                                         

Answer:

The purpose of adding gypsum is only to slow down the process of setting of the cement so that it gets sufficiently hardened.

Hence, the correct answer is option (ii).

Answer:

 Plaster of Paris is prepared by heating gypsum at 120° C.

${\mathrm{CaSO}}_4.2{\mathrm{H}}_2\mathrm{O} \stackrel{{120}^{\circ }\mathrm{C}}{\to }{\mathrm{CaSO}}_4.\frac{1}{2}{\mathrm{H}}_2\mathrm{O}+\frac{3}{2}{\mathrm{H}}_2\mathrm{O}$

On heating plaster of Paris above 120°C, it forms anhydrous calcium sulphate, i.e., dead burnt plaster which has no setting property as it absorbs water very slowly.

${\mathrm{CaSO}}_4.\frac{1}{2}{\mathrm{H}}_2\mathrm{O}\stackrel{\mathrm{Above} 120°\mathrm{C} }{\to }{\mathrm{CaSO}}_4$

Hence, the correct answer is option (i).

Answer:

The suspension of slaked lime (CaO) in water is known as milk of lime i.e. Ca(OH)₂. The reaction involved is:

$\mathrm{CaO}+{\mathrm{H}}_2\mathrm{O}\to \mathrm{Ca}{\left(\mathrm{OH}\right)}_2$

Hence, the correct answer is option (iii). 







 

Answer:

All the elements except beryllium combine with hydrogen upon heating to form their hydrides, MH₂​. BeH₂​, however, can be prepared by the reaction of BeCl_2​ with LiAlH₄​. The reaction may be shown as follows:



Hence, the correct answer is option (i).

Answer:

On heating washing soda, it loses its water of crystallization to form monohydrate. Above 373 K, it becomes completely anhydrous white powder called soda ash.

$$\begin{gathered} \underset{\mathrm{washing} \mathrm{soda} }{{\mathrm{Na}}_2{\mathrm{CO}}_3.10{\mathrm{H}}_2\mathrm{O}}\stackrel{∆}{\to } {\mathrm{Na}}_2{\mathrm{CO}}_3·{\mathrm{H}}_2\mathrm{O} \\ {\mathrm{Na}}_2{\mathrm{CO}}_3·{\mathrm{H}}_2\mathrm{O}\stackrel{>373 \mathrm{K}}{\to } \underset{\mathrm{soda} \mathrm{ash}}{{\mathrm{Na}}_2{\mathrm{CO}}_3} \end{gathered}$$

Hence, the correct answer is option (iv).

Answer:

Calcium imparts brick red colour to the flame and calcium nitrate evolves O₂ and a brown gas, NO₂​ upon heating. The reaction may be shown as follows:

$2\mathrm{Ca}{\left({\mathrm{NO}}_3\right)}_2\stackrel{ \mathrm{\Delta } }{\to }2\mathrm{CaO}+{\mathrm{O}}_2+4{\mathrm{NO}}_2\uparrow (\mathrm{Brown} \mathrm{gas})$

Hence, the correct answer is option (ii).

Answer:

Ca(OH)₂ ( also called slaked lime ), is used in the preparation of bleaching powder. The reaction may be shown as follows:

$$\begin{gathered} \mathrm{Ca}{\left(\mathrm{OH}\right)}_2+2{\mathrm{Cl}}_2\to {\mathrm{CaCl}}_2+\mathrm{Ca}{\left(\mathrm{OCl}\right)}_2+2{\mathrm{H}}_2\mathrm{O} \\ \mathrm{Bleaching} \mathrm{powder} \end{gathered}$$

Hence, the correct answer is option (i).
 

Answer:

Ca(OH)₂ is used for the preparation of washing soda to recover ammonia. The reaction may be written as follows:
$2{\mathrm{NH}}_4\mathrm{Cl}+\mathrm{Ca}{\left(\mathrm{OH}\right)}_2\to 2{\mathrm{NH}}_3+{\mathrm{CaCl}}_2+2{\mathrm{H}}_2\mathrm{O}$
When CO₂ is bubbled through an aqueous solution of Ca(OH)₂, the solution turns milky. The reaction may be shown as follows:
$$\begin{gathered} {\mathrm{CO}}_2+\mathrm{Ca}{\left(\mathrm{OH}\right)}_2\to {\mathrm{CaCO}}_3+{\mathrm{H}}_2\mathrm{O} \\ \mathrm{Milky} \end{gathered}$$
Thus, the chemical formula of A is Ca(OH)₂.
Hence, the correct answer is option (iii).

Answer:

Halides of calcium, barium and strontium act as dehydrating agent as they are hygroscopic in nature. They absorb moisture and their hydrate forming tendency decreases from Ca²⁺ to Sr²⁺ due to an increase in size and decrease in polarization power.

Hence, the correct answer is option (iv).

Answer:

Alkali metals have high negative standard electrode potential and large atomic size.

Hence, the correct answers are options (ii) and (iv).

Answer:

Sodium carbonate and sodium hydroxide are used in the textile industry.

Hence, the correct answers are options (i) and (iii).
 

Answer:

The hydration enthalpies of BeSO₄ and MgSO₄ are quite high because of the small size of Be²⁺ and Mg²⁺ ions. These hydration enthalpy values are higher than their corresponding lattice enthalpies and therefore, BeSO₄ and MgSO₄ are highly soluble in water.

Hence, the correct answers are options (i) and (ii).

Answer:

Permutit or zeolites are packed in a suitable container and a slow stream of hard water is passed through this material. As a result, calcium and magnesium ions present in hard water are exchanged with sodium ions in the permutit (NaAlSiO₄). The outgoing water contains sodium salts, which do not cause hardness. 
$$\begin{gathered} {\mathrm{CaCl}}_2+2{\mathrm{NaAlSiO}}_4\to \mathrm{Ca}{\left({\mathrm{AlSiO}}_4\right)}_2+2\mathrm{NaCl} \\ {\mathrm{MgCl}}_2+2{\mathrm{NaAlSiO}}_4\to \mathrm{Mg}{\left({\mathrm{AlSiO}}_4\right)}_2+2\mathrm{NaCl} \end{gathered}$$
Hence, the correct answers are options (ii) and (iii). 

Answer:

Tendency to form halide hydrates gradually decreases down the group. The hydrates are MgCl₂.8H₂O, CaCl₂.6H₂O, SrCl₂.6H₂O, and BaCl₂.2H₂O.

Hence, the correct answers are options (i) and (iii).

Answer:

Beryllium is not readily attacked by acids because of the presence of an oxide film on the surface of the metal. Beryllium sulphate is readily soluble in water as the greater hydration enthalpy of Be²⁺ overcomes the lattice enthalpy factor. Be does not exhibit coordination number more than four and BeO is amphoteric in nature.

Hence, the correct answers are options (i) and (ii).
 

Answer:

The anomalous behavior of lithium is due to its exceptionally small size and high polarising power.

Hence, the correct answers are options (i) and (ii).

Answer:

The standard electrode potential is a measure of the tendency of an element to lose electrons in the aqueous solution. It depends upon the following factors:
(i) Sublimation enthalpy
(ii) Ionization enthalpy 
(iii) Hydration enthalpy 

Since Li has the smallest size, its enthalpy of hydration is the highest among alkali metals. Although ionization enthalpy of Li is the highest among alkali metals, it is more than compensated by the high enthalpy of hydration. Thus, Li has the most negative E^Θ value and the strongest reducing agent.

Answer:

The reactivity of alkali metals towards oxygen increases down the group as the atomic size increases. Thus, Li forms only lithium oxide (Li₂O), sodium forms mainly sodium peroxide (Na₂O₂) while potassium forms only potassium superoxide (KO₂).
This behaviour is due to the fact that larger cations stabilize larger anions due to higher lattice energies, therefore, the stability increases from oxide → peroxide → superoxide as the size of the metal cation increases down the group and the size of the anion increases from oxide → peroxide → superoxide.
The various reactions may be written as follows:
$$\begin{gathered} 4\mathrm{Li}+{\mathrm{O}}_2\stackrel{ }{\to }2{\mathrm{Li}}_2\mathrm{O} \\ 2\mathrm{Na}+{\mathrm{O}}_2\stackrel{}{\to }{\mathrm{Na}}_2{\mathrm{O}}_2 \end{gathered}$$
$\mathrm{K}+{\mathrm{O}}_2\stackrel{}{\to }{\mathrm{KO}}_2$

Answer:

The reactions may be completed as follows:

(i) ${\mathrm{O}}_2^{2-}+2{\mathrm{H}}_2\mathrm{O}\to 2{\mathrm{OH}}^{-}+{\mathrm{H}}_2{\mathrm{O}}_2$

(ii) $2{\mathrm{O}}_2^{-}+2{\mathrm{H}}_2\mathrm{O}\to 2{\mathrm{OH}}^{-}+{\mathrm{H}}_2{\mathrm{O}}_2+{\mathrm{O}}_2$

Answer:

Lithium resembles magnesium in some of its properties two of which are written as follows:

(i) Both Li and Mg are harder and lighter than other elements in their groups.
(ii) Both form ionic nitrides Li₃N and Mg₃N₂ by heating in an atmosphere of nitrogen.

The reason being their diagonal relationship which arises because of their similar sizes : atomic radii, Li = 152 pm, Mg = 160 pm; ionic radii : Li⁺ = 76 pm, Mg²⁺= 72 pm.

Answer:

Due to small size and high ionization enthalpy of Be, Be(OH)₂ is amphoteric in nature, i.e., it reacts with both acids and bases. Further, due to the small size, the hydration enthalpy of Be²⁺ ions is much higher than the lattice enthalpy of BeSO₄. As a result, BeSO₄ is highly soluble in water.

Answer:

(i) All the alkaline earth metals form carbonates (MCO₃). All these carbonates decompose on heating to give CO₂ and metal oxide. The thermal stability of these carbonates increases down the group i.e. from Be to Ba,

${\mathrm{BeCO}}_{3 }< {\mathrm{MgCO}}_3 < {\mathrm{CaCO}}_3 < {\mathrm{SrCO}}_3 < {\mathrm{BaCO}}_3$

BeCO₃ is unstable to the extent that it is stable only in an atmosphere of CO₂. It however shows reversible decomposition in a closed container as: ${\mathrm{BeCO}}_3 \rightleftharpoons \mathrm{BeO} + {\mathrm{CO}}_2$ 

Hence, more is the stability of oxide formed, less will be the stability of carbonates. The stability of oxides decreases down the group. Since beryllium oxide is highly stable, it makes BeCO₃ unstable.


(ii) All the members of group 2 form oxides of the type MO. Barium and radium form peroxides also. Except BeO, which is amphoteric, other MO oxides are basic in nature as they combine with water to form basic hydroxides as: $\mathrm{MO}+{\mathrm{H}}_2\mathrm{O}\to \mathrm{M}{\left(\mathrm{OH}\right)}_2$

The basic nature of oxides increases gradually from BeO to BaO.
BeO and MgO are insoluble in water on account of high lattice energy while the solubility of other oxides in increases gradually.

Answer:

The hydration enthalpies of BeSO₄ and MgSO₄ are quite high because of the small size of Be²⁺ and Mg²⁺ ions. These hydration enthalpy values are higher than their corresponding lattice enthalpies and therefore, BeSO₄ and MgSO₄ are highly soluble in water. However, hydration enthalpies of CaSO₄, SrSO₄, and BaSO₄ are not very high as compared to their respective lattice enthalpies and hence these are insoluble in water. 

Answer:

Due to small size, high electronegativity and high ionization enthalpy, lithium compounds have considerable covalent character (with reference to fajan's rule) while compounds of other alkali metals are ionic in nature. As a result, compounds of lithium are more soluble in organic solvents while those of other alkali metals are more soluble in water.

Answer:

No.
In the Solvay process, ammonium hydrogen carbonate is prepared from ammonium carbonate, which then reacts with sodium chloride to form sodium hydrogen carbonate. Due to the low solubility of NaHCO₃, it gets precipitated and decomposes on heating to give Na₂CO₃.
We cannot obtain sodium carbonate directly by treating the solution containing (NH₄)₂CO₃ with sodium chloride as both the products formed on reaction, i.e., Na₂CO₃ and NH₄Cl are soluble and the equilibrium will not shift in forward direction.

${\left({\mathrm{NH}}_4\right)}_2{\mathrm{CO}}_3+2\mathrm{NaCl}\rightleftharpoons {\mathrm{Na}}_2{\mathrm{CO}}_3+2{\mathrm{NH}}_4\mathrm{Cl}$

 

Answer:

Lewis structure of O₂⁻ ion is drawn below.


Oxygen atom with zero charge has 6 electrons, therefore the oxidation state is 0. When the oxygen atom has a negative charge, it has 7 electrons. Hence, the oxidation state is −1.

The average oxidation state is $\frac{0+(-1)}{2}=-\frac{1}{2}$
 

Answer:

Beryllium and magnesium atoms are smaller in size and their electrons are strongly bound to the nucleus. They need a large amount of energy for excitation of electrons to higher energy levels which is not available in the bunsen flame. So they do not impart color to the flame.

Answer:

Beryllium chloride has a chain structure in the solid state as shown below.


In the vapour phase BeCl₂ tends to form a chloro-bridged dimer as shown below.

Answer:

The matching may be done as follows:
 

Answer:

The matching may be done as follows:
 

Answer:

The matching may be done as follows:
 

Answer:

Unlike other alkali metal carbonates that are stable in heat, lithium being very small in size polarises a large CO₃^2– ion leading to the formation of more stable Li₂O and CO₂.
${\mathrm{Li}}_2{\mathrm{CO}}_3\to {\mathrm{Li}}_2\mathrm{O} + {\mathrm{CO}}_2$

Hence, the correct answer is option (i).

Answer:

BeO is more stable than BeCO₃ due to small size and high polarising power of Be²⁺. When BeCO₃ is kept in an atmosphere of CO₂, a reversible process takes place and according to Le Chatelier's principle, the reaction shifts towards reactant side as concentration of CO₂ is more on product side and thus, stability of BeCO₃ increases.
BeCO₃ â‡Œ BeO + CO₂

Hence, the correct answer is option (i).

Answer:

(i) Nature of oxides – Alkali metals form M₂O, M₂O₂ and MO₂ types of oxides. The stability of the peroxide or superoxide increases as the size of metal cation increases. This is due to the stabilization of large anions by larger cations. 

(ii) Nature of halides – Alkali metal halides have general formula MX. All halides are soluble in water. LiF is very less soluble in water due to its high lattice energy. Their melting points and boiling points follow the trend – fluoride > chloride > bromide > iodide. 

(iii) Oxosalts – Oxosalts of alkali metals are generally soluble in water and thermally stable. As electropositive character increases down the group, the stability of carbonates and bicarbonates increases.

Answer:

The comparative study may be shown as follows:

Alkali Metals:

• All alkali metals except Li form ionic compounds  because there is increased covalent character in lithium compounds due to exceptionally small size of its atom and ion, and high polarising power.
• Lithium forms monoxide, sodium forms peroxide, the other metals form superoxides. The solubility of oxides of alkali metals increases down the group.
• Oxo-acids are those in which the acidic proton is on a hydroxyl group with an oxo group attached to the same atom e.g., carbonic acid, H₂CO₃ (OC(OH)₂; sulphuric acid, H₂SO₄ (O₂S(OH)₂). The alkali metals form salts with all the oxo-acids.  
• The oxosalts are generally soluble in water.
• Carbonates of Li decompose on heating while the stability of carbonates and hydrogencarbonates of other metals increases, down the group due to increase in  electropositive character.

• All alkaline earth metals except Be form ionic compounds but less ionic than the corresponding compounds of alkali metals. This is due to increased nuclear charge and smaller size.
• BeO is amphoteric while oxides of other elements are ionic in nature. All these oxides except BeO are basic in nature and react with water to form sparingly soluble hydroxides. The solubility of oxides of Mg, Ca, Sr and Ba increase from Mg to Ba. BeO, however, is, covalent and insoluble in water.
• All alkaline earth metals form oxosalts such as carbonates, sulphates and nitrates.
• Carbonates of alkaline earth metals are insoluble in water. The solubility of carbonates and sulphates decreases down the group and the nitrates show a decreasing tendency to form hydrates with increasing size and decreasing hydration enthalpy.
• The carbonates, nitrates and sulphates of alkaline earth metals all decompose on heating but the "temperature of their decomposition increases down the group, i.e., their thermal stability increases.

Answer:

Alkali metals dissolve in liquid ammonia to give a deep blue colour. These solutions are also conducting in nature. The blue colour is obtained because of the ammoniated electron which absorbs energy in the visible region of light and thus imparts a blue colour to the solution. The reaction is given as:

M + (x+y)NH₃$\to$[M(NH₃)^_x]⁺ + [e(NH₃)_y]^-

As the concentration of the metal increases, the colour of the solution become bronze which is due to the formation of metal ion clusters.

Answer:

As the size of alkali metal ion increases, the stability of peroxides and superoxides increases. This is due to the stabilization of larger anions by larger cations.

As we move from Li⁺ to Cs⁺ the size of the cations increases.

• Li⁺ is the smallest cation with strong positive field around it. Hence it combines with small anion called oxide ion (O^2-) with strong negative field and form Lithium oxide.
• Na⁺ ion is bigger than Li⁺ ion with less positive field around it combines with peroxide ion (O₂^2-) with less negative field and form sodium peroxide.
• K⁺, Rb⁺ and Cs⁺ ions are bigger in size with less positive field around them. Hence they can stabilize the bigger superoxide ion (O₂^-) with less negative field and form superoxides.

Answer:

The appearance of milkiness on passing CO₂ in the solution of compound B indicates that compound B is lime water and compound C is CaCO₃. Since compound B is obtained by adding H₂O to compound A, therefore, compound A is quicklime, CaO. 
The reactions are shown as follows:

$$\begin{gathered} \underset{\left(\mathrm{A}\right)}{\mathrm{CaO}} + {\mathrm{H}}_2\mathrm{O}\to \underset{\mathrm{Lime} \mathrm{Water} \left(\mathrm{B}\right)}{\mathrm{Ca}{\left(\mathrm{OH}\right)}_2} \\ \underset{\left(\mathrm{B}\right)}{\mathrm{Ca}{\left(\mathrm{OH}\right)}_2}+ {\mathrm{CO}}_2 \to \underset{\mathrm{Calcium} \mathrm{Carbonate}\left(\mathrm{C}\right)}{{\mathrm{CaCO}}_3}+ {\mathrm{H}}_2\mathrm{O} \\ \underset{\left(\mathrm{C}\right)}{{\mathrm{CaCO}}_3}+{\mathrm{CO}}_2+{\mathrm{H}}_2\mathrm{O} \to \underset{\mathrm{Soluble} \mathrm{in} \mathrm{water}}{\mathrm{Ca}{\left({\mathrm{HCO}}_3\right)}_2} \end{gathered}$$

Answer:

Lithium hydride reacts with Al₂Cl₆ to form lithium aluminium hydride.
8LiH + Al₂Cl₆ → 2LiAlH₄ + 6LiCl
Lithium aluminium hydride reacts with BeCl₂ to form beryllium hydride.
2BeCl₂ + LiAlH₄ → 2BeH₂ + LiCl + AlCl₃

Answer:

Beryllium (Be) is the element of group 2 which forms covalent oxide which is amphoteric in nature and dissolves in water to give an amphoteric hydroxide. Beryllium hydroxide is amphoteric in nature as it reacts with acid and alkali both.

 

Answer:

The element is sodium (Na).

The chemical reaction to show the formation of its peroxide is as follows:
2Na + O₂ → Na₂O₂

Sodium imparts colour to the flame because 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 emission of radiation in the visible region of the spectrum.