GSEB Class 11 Chemistry Solutions Chapter 10 The s-Block Elements

 

Question 1. What are the common physical and chemical features of alkali metals?
Answer: Alkali metals make up group-I of the s-block in the periodic table. These metals are identified by having only one s-electron in the outer shell of their atoms. Their properties are described below:

(b) Atomic and ionic size: Alkali metal atoms possess the biggest sizes within their respective periods in the periodic table. Their monovalent ions, \( M^+ \), are smaller than their original atoms.
(c) Ionisation enthalpy: The energies needed to remove an electron from alkali metals are quite low.
Physical Properties:
(i) Physical state: All these elements are shiny white, soft, and lightweight metals.
(ii) Density: Their densities are quite low and go up from lithium to caesium. However, potassium is lighter than sodium.
(iii) Melting and Boiling points: These points are low because the bonds in the crystal structures of the metals are weak.
(iv) Due to their low ionisation energies, these metal atoms really want to lose their \( ns^1 \) electron and create positive ions. \( M \rightarrow M^+ + e^- \)
Chemical Properties:
(i) Action of air: Alkali metals get dull in the air because an oxide forms on their surface, so they are kept in kerosene oil. \( 4Li + O_2 \rightarrow 2Li_2O \) Sodium creates sodium peroxide \( (Na_2O_2) \), and other alkali metals form superoxides \( [MO_2 \text{ where } M = K, Rb, Cs] \).
(ii) Action of water: Alkali metals react quickly and strongly with water, forming hydroxides and releasing hydrogen. \( 2M + 2H_2O \rightarrow 2MOH + H_2 \text{ [M = Li, Na, K, Rb, Cs]} \)
(iii) Action of Hydrogen: \( 2M + H_2 \rightarrow MH \text{ [M = Li, Na, K etc.] (Ionic hydride)} \)
(iv) Action with halogens: \( 2M + X_2 \rightarrow 2MX \text{ (Metal halide)} \)
(v) Solubility in liquid ammonia: All alkali metals dissolve in liquid ammonia solutions and appear blue in dilute metal-ammonia solutions. Such solutions have reducing properties.
(vi) Reducing nature: The tendency to lose electrons is a key property of these metals. Therefore, all these metals behave as strong reducing agents.
(vii) Formation of alloys: Alkali metals can form alloys with each other and with other metals. They readily dissolve in mercury, creating amalgams. This process gives off a lot of heat.
In simple words: Alkali metals are Group 1 elements with one outer electron. They are soft, light, and shiny. They react easily with air, water, hydrogen, and halogens, forming positive ions. They also dissolve in liquid ammonia.

Exam Tip: Remember to categorize the properties as 'physical' and 'chemical' and list specific examples or equations for each to score full marks.

 

Question 2. Discuss the general characteristics and gradation in properties of alkaline earth metals.
Answer:
General characteristics of the alkaline earth metals:
Group 2 of the periodic table includes beryllium, magnesium, calcium, strontium, barium, and radium. These elements (except beryllium) are known as alkaline earth metals.
(i) Atomic Properties:
(a) Electronic configuration: They are distinguished by the presence of two electrons in their outer valence shells. These two electrons are found in the s-subshell. The electronic configuration of metal M is \( - (\text{Noble gas})ns^2 \).
(b) Atomic and ionic size: The atomic and ionic radii of these metals are smaller than those of alkali metals. From top to bottom, the atomic and ionic sizes generally increase. For example: \( Be < Mg < Ca < Sr < Ba \) and \( Be^{2+} < Mg^{2+} < Ca^{2+} < Sr^{2+} < Ba^{2+} \). However, \( Mg^{2+} < Mg \), \( Ca^{2+} < Ca \), and so on.
(c) Ionisation Enthalpies: The first ionisation enthalpies of alkaline earth metals are greater than those of alkali metals because of an increase in nuclear charge. However, the second ionisation enthalpies \( (2^{nd} I.Es) \) of alkaline earth metals are smaller than those of the corresponding alkali metals. From top to bottom, ionisation enthalpies decrease with the addition of a new electron shell each time.
(ii) Physical Properties of Alkaline Earth Metals -
(a) Physical appearance: These metals are usually silvery white, shiny, and relatively soft, but they are tougher than alkali metals. Beryllium and magnesium appear somewhat greyish.
(b) Melting and boiling point: The relatively higher melting and boiling points of alkaline earth metals, when compared to corresponding alkali metals, are due to their smaller sizes and two valence electrons. The pattern, however, is not consistent. Melting points and boiling points are highest within the group.
(c) Flame colourations: Chlorides of alkaline earth metals (except beryllium and magnesium) create a specific colour when heated in a flame. This happens because electrons are easily excited to higher energy levels.
(d) Densities: Excluding magnesium and calcium, which are lighter, densities increase in the order \( Be < Sr < Ba < Ra \).
(iii) Chemical Properties Of Alkaline Earth Metals –
Alkaline earth metals are less reactive than the similar alkali metals. The reactivity of these elements goes up as you move down the group.
(i) Reactivity towards air and water: Beryllium and magnesium are not reactive to \( O_2 \) and \( H_2O \) because an oxide film forms on their surface. However, beryllium in powder form burns brightly when ignited, producing \( BeO \) and \( Be_3N_2 \). \( 2Be + O_2 \xrightarrow{ignition} 2BeO \) powdered air \( \xrightarrow{\Delta} Be_3N_2 \). Similarly, magnesium, which is more electropositive than beryllium, burns with a strong brightness in air to create \( MgO \) and \( Mg_3N_2 \). Calcium, strontium, and barium are easily attacked by air to form their oxides and nitrides.
(ii) Reactivity towards the halogens: All alkaline earth metals react with halogens at high temperatures, forming halides. \( M + X_2 \rightarrow MX_2 \text{ [X = F, Cl, Br, I]} \)
(iii) Reactivity towards hydrogen: Except for beryllium, all other alkaline earth metals combine with \( H_2 \) when heated to form hydrides. \( Ca + H_2 \rightarrow CaH_2 \)
(iv) Reactivity towards acids: Alkaline earth metals react quickly with acids, releasing \( H_2 \).
(v) Reducing nature: Like alkali metals, alkaline earth metals are powerful reducing agents, but their reducing strength is less than that of alkali metals. Generally, reducing character goes up from top to bottom.
(vi) Solutions in liquid ammonia: Similar to alkali metals, alkaline earth metals dissolve in liquid ammonia, creating deep blue-black solutions. \( M + (x + y) NH_3 \rightarrow M(NH_3)_x^{2+} + 2e^- (NH_3)_y \).
In simple words: Alkaline earth metals are Group 2 elements with two outer electrons. They are silvery, soft, and react with air, water, halogens, and acids, forming \( M^{2+} \) ions. Their atomic size and reactivity increase down the group.

Exam Tip: When discussing gradation, explicitly mention the trend (increase/decrease) and provide a reason (e.g., nuclear charge, electron shells) for each property.

 

Question 3. Why are alkali metals not found in nature?
Answer: The loosely held s-electron in the outermost shell of these metal atoms makes them extremely electropositive. This means they readily form \( M^+ \) ions. Because of this high reactivity, they are not naturally found in their free state.
In simple words: Alkali metals are very reactive because they easily lose their outer electron, forming positive ions. So, they always react with other elements and are not found alone in nature.

Exam Tip: Emphasize the 'loosely held s-electron' and 'electropositive nature' as key reasons for their high reactivity and non-existence in free form.

 

Question 4. Find out the oxidation state of sodium in \( Na_2O_2 \).
Answer: Let the oxidation number of sodium \( (Na) = x \).
It is a peroxide.
\( \implies \) Oxidation number of \( O = -1 \).
\( \implies \) For \( Na_2O_2 \), \( 2x + 2(-1) = 0 \)
\( \implies 2x - 2 = 0 \)
\( \implies 2x = 2 \)
\( \implies x = +1 \)
Therefore, the oxidation state of sodium is \( +1 \).
In simple words: In \( Na_2O_2 \), oxygen has an oxidation state of -1 because it's a peroxide. Since the compound is neutral, the two sodium atoms must balance the charge, so each sodium atom has a +1 oxidation state.

Exam Tip: Always remember that in peroxides, oxygen has an oxidation state of -1, which is different from its usual -2 in most compounds. This is a common pitfall.

 

Question 5. Explain why is sodium less reactive than potassium?
Answer: The ionisation enthalpy of sodium is greater than that of potassium. This means sodium needs more energy to lose its electrons compared to potassium. Therefore, sodium will lose electrons less readily, making it less reactive than potassium.
In simple words: Sodium holds its outer electron more tightly than potassium does. Because it's harder for sodium to lose this electron, it reacts less strongly than potassium.

Exam Tip: Relate reactivity of alkali metals directly to their ionization enthalpy: lower ionization enthalpy means higher reactivity due to easier electron loss.

 

Question 6. Compare the properties of alkaline earth metals with respect to (i) ionisation enthalpy (ii) basicity of oxides (iii) solubility of hydroxides.
Answer:
(i) Ionisation enthalpy: The first ionisation enthalpies of alkaline earth metals are greater than those of alkali metals because of an increased nuclear charge. However, the second ionisation enthalpies of alkaline earth metals are lower than those of corresponding alkali metals, for example, the first ionisation enthalpy of \( Be > Li \) and \( Mg > Na \).
(ii) Basicity of oxides: The oxides of alkaline earth metals are less basic compared to the corresponding oxides of alkali metals. This is because of their increased nuclear charge, which results in higher ionisation enthalpies.
(iii) Solubilities of hydroxides: Alkaline earth hydroxides are less soluble in water compared to alkali hydroxides.
In simple words: Alkaline earth metals have higher ionization energy and form less basic oxides and less soluble hydroxides compared to alkali metals.

Exam Tip: When comparing properties between groups, always explain the underlying atomic reasons like nuclear charge or atomic size differences.

 

Question 7. In what ways lithium shows similarities to cesium in its chemical behaviour?
Answer: Lithium resembles magnesium mostly due to the similar sizes of their atoms [\( Li = 152 \text{ pm} \), \( Mg = 160 \text{ pm} \)] and ions [\( Li^+ = 76 \text{ pm} \), \( Mg^{2+} = 72 \text{ pm} \)].
The main points of similarity are:
(i) Both \( LiOH \) and \( Mg(OH)_2 \) are weak bases.
(ii) Both form ionic nitrides in a nitrogen atmosphere.
\( 6Li + N_2 \xrightarrow{\Delta} 2Li_3N \)
\( 3Mg + N_2 \xrightarrow{\Delta} Mg_3N_2 \)
(iii) The hydroxides and carbonates of both \( Li \) and \( Mg \) break down when heated, yielding their respective oxides.
\( 2LiOH \xrightarrow{\Delta} Li_2O + H_2O \)
\( Mg(OH)_2 \xrightarrow{\Delta} MgO + H_2O \)
\( Li_2CO_3 \xrightarrow{\Delta} Li_2O + CO_2 \)
\( MgCO_3 \xrightarrow{\Delta} MgO + CO_2 \)
(iv) Both lithium and magnesium do not create solid bicarbonates.
(v) \( Li_2O \) and \( MgO \) do not react with extra oxygen to give peroxide or superoxides.
(vi) Both \( LiNO_3 \) and \( Mg(NO_3)_2 \) decompose when heated, producing nitrogen dioxide.
\( 4LiNO_3 \xrightarrow{\Delta} 2Li_2O + 4NO_2 + O_2 \)
\( 2Mg(NO_3)_2 \xrightarrow{\Delta} 2MgO + 4NO_2 + O_2 \)
(vii) The hydroxides, carbonates, and fluorides of both lithium and magnesium are only slightly soluble in water.
(viii) Because of their covalent nature, \( LiCl \) and \( MgCl_2 \) dissolve in ethanol.
(ix) Both \( LiClO_4 \) and \( Mg(ClO_4)_2 \) are very soluble in ethanol.
(x) \( LiCl \) and \( MgCl_2 \) absorb moisture from the air and crystallise from aqueous solution as hydrates \( LiCl.2H_2O \) and \( MgCl_2.2H_2O \).
In simple words: Lithium shares many chemical similarities with magnesium, mainly because their atoms and ions are almost the same size. They both form weak bases, ionic nitrides, and their compounds decompose similarly or have similar solubility.

Exam Tip: This question highlights the 'diagonal relationship' between elements in the periodic table. For Li and Mg, focus on similarities in their small size and high polarizing power leading to covalent character in some compounds.

 

Question 8. Explain why can alkali and alkaline earth metals not be obtained by chemical reduction methods?
Answer:
(i) Alkali and alkaline earth metals are naturally strong reducing agents. Therefore, they cannot be removed from their oxides and alkaline earth metals by reduction.
(ii) Alkali metals are highly electropositive, meaning they cannot be moved from the aqueous solutions of their salts by other metals.
(iii) Alkali metals cannot be separated by the electrolysis of their aqueous solution. This is because hydrogen gas is released at the cathode instead of the alkali metal.
In simple words: Alkali and alkaline earth metals are very good at reducing other substances. This means we can't use typical chemical reduction to get them, and their high reactivity with water also makes electrolysis of their solutions impossible.

Exam Tip: Remember that highly electropositive metals are strong reducing agents and therefore difficult to reduce further; chemical reduction requires a stronger reducing agent, which is not feasible for these metals.

 

Question 9. Why are potassium and caesium, rather than lithium used in photo-electric cells?
Answer: Potassium and caesium are used in photo-electric cells because they have low ionisation enthalpy and readily eject electrons when exposed to light. Lithium, on the other hand, has the highest ionisation enthalpy among alkali metals, so it cannot release electrons when exposed to light, making it unsuitable for photo-electric cells.
In simple words: Potassium and caesium release electrons easily when light hits them because they have low ionisation energy. Lithium needs more energy to release electrons, so it's not used in photo-electric cells.

Exam Tip: Connect the principle of photo-electric cells directly to ionization enthalpy: elements with lower ionization enthalpy require less energy (like light) to release electrons, making them suitable for this application.

 

Question 10. When an alkali metal dissolves in liquid ammonia, the solution acquires different colours. Explain the reasons for this type colour change.
Answer: All alkali metals dissolve in liquid ammonia, creating highly conductive deep blue solutions. \( M + (x+y)NH_3 \rightarrow M^+(NH_3)_x + e^-(NH_3)_y \)
ammoniated cation ammoniated anion
(i) When normal light shines on these ammoniated electrons, they get excited and move to higher energy levels. This happens by absorbing energy that matches the red part of the visible spectrum. As a result, the light that passes through appears blue, which gives the solution its blue colour.
(ii) Dilute solutions of alkali metals in liquid ammonia are dark blue. However, as the concentration goes above \( 3M \), the colour changes to copper-bronze. This happens because metal ion clusters form in the solution, giving it a metallic sheen.
In simple words: When alkali metals dissolve in liquid ammonia, they form ammoniated electrons. These electrons absorb red light and reflect blue, making dilute solutions blue. At higher concentrations, metal ion clusters form, changing the colour to copper-bronze with a metallic look.

Exam Tip: Differentiate between the causes of color in dilute (ammoniated electrons) and concentrated (metal ion clusters) solutions, and mention the role of light absorption.

 

Question 11. Beryllium and magnesium do not give colour to flame whereas other alkaline earth metals do so why?
Answer: Chlorides of alkaline earth metals, except for beryllium and magnesium, create characteristic flame colours. This happens because their electrons are easily excited to higher energy levels. Beryllium and magnesium, however, are small in size and hold their electrons more tightly. Their ionisation enthalpies are greater, meaning they need higher energy levels to excite their electrons. Therefore, their electrons do not get excited and do not produce any colour in the flame.
In simple words: Beryllium and magnesium atoms are small and hold their electrons very tightly, needing a lot of energy to excite them. Other alkaline earth metals have larger atoms and looser electrons, so they get excited easily and give off colours in a flame.

Exam Tip: Link the inability to produce flame color to high ionization enthalpy and small atomic size, which prevent electrons from being easily excited to higher energy levels.

 

Question 12. What are the reactions that occur in the SOLVAY process?
Answer: Sodium carbonate \( (Na_2CO_3) \) is produced using the SOLVAY process.
(i) Ammonium carbonate \( [(NH_4)_2CO_3] \) is made by passing \( CO_2 \) through a strong solution of sodium chloride \( [NaCl] \) saturated with ammonia \( (NH_3) \).
\( 2NH_3 + H_2O + CO_2 \rightarrow (NH_4)_2CO_3 \)
(ii) Ammonium carbonate then reacts further with \( CO_2 \) to create ammonium bicarbonate \( [NH_4HCO_3] \).
(iii) Ammonium bicarbonate reacts with \( NaCl \) to form sodium bicarbonate \( [NaHCO_3] \).
\( NaCl + NH_4HCO_3 \rightarrow NH_4Cl + NaHCO_3 \).
(iv) Sodium bicarbonate crystals separate. These are heated to yield sodium carbonate (washing soda).
\( 2NaHCO_3 \rightarrow Na_2CO_3 + CO_2 + H_2O \)
In this method, \( NH_3 \) is recovered when the solution containing \( NH_4Cl \) is treated with \( Ca(OH)_2 \). Calcium chloride \( (CaCl_2) \) is obtained as a secondary product.
\( 2NH_4Cl + Ca(OH)_2 \rightarrow 2NH_3 + CaCl_2 + H_2 \).
In simple words: The Solvay process makes sodium carbonate. It involves reacting ammonia, water, carbon dioxide, and salt to form ammonium bicarbonate, then sodium bicarbonate, which is heated to produce sodium carbonate and recover ammonia.

Exam Tip: To answer questions about industrial processes, list each step with its corresponding balanced chemical equation clearly. Highlighting the recycling of ammonia is also important.

 

Question 13. Potassium carbonate cannot be prepared by Solvay process why?
Answer: Potassium carbonate cannot be produced by the Solvay process because potassium bicarbonate is too soluble. It cannot be precipitated by adding ammonium bicarbonate to a concentrated solution of potassium chloride.
In simple words: The Solvay process doesn't work for potassium carbonate because potassium bicarbonate won't form a solid and precipitate out of the solution, unlike sodium bicarbonate.

Exam Tip: The key reason is the high solubility of potassium bicarbonate, which prevents its precipitation—a crucial step in the Solvay process. This highlights a limitation of the method for potassium compounds.

 

Question 14. Why is \( Li_2CO_3 \) decomposed at a lower temperature whereas \( Na_2CO_3 \) at higher temperature?
Answer: \( Li_2CO_3 \rightarrow Li_2O + CO_2 \)
\( Li_2CO_3 \) is not stable and breaks down to form \( Li_2O \). This is because the \( Li^+ \) ion is small compared to \( CO_3^{2-} \) in the \( Li_2CO_3 \) lattice. It splits apart to create a stable crystal lattice of smaller \( Li^+ \) ions and smaller \( O^{2-} \) ions in \( Li_2O \). In contrast, \( Na^+ \) ions are larger than \( Li^+ \) ions and create \( CO_3^{2-} \) ions. Therefore, \( Li_2CO_3 \) breaks down to form a stable crystal lattice of \( Li_2O \), while \( Na_2CO_3 \) is quite stable.
In simple words: Lithium carbonate breaks down easily at low temperatures because the small \( Li^+ \) ion makes its lattice unstable. Sodium carbonate is more stable and needs higher temperatures to decompose because the larger \( Na^+ \) ion forms a more stable crystal structure.

Exam Tip: Explain the difference in thermal stability using the concept of lattice energy and the polarizing power of the cation, relating small cation size (Li+) to greater polarizing power and thus lower thermal stability for carbonates.

 

Question 15. Compare the solubility and thermal stability of the following compounds of the alkali metals with those of alkaline earth metals
(a) nitrates
(b) carbonates
(c) sulphates.
Answer:
(a) Nitrates of alkali metals and alkaline earth metals.
(i) Nitrates of alkali metals are not thermally stable and break down when heated to produce \( MNO_2 \) and \( O_2 \) (except \( LiNO_3 \)). In contrast, nitrates of alkaline earth metals break down when heated to produce their oxides, nitrogen dioxide, and oxygen gas.
\( 2M(NO_3)_2 \xrightarrow{\Delta} 2MNO_2 + O_2 \)
[ \( M = Be, Mg, Ca, Sr \text{ or } Ba \) ]
e.g., \( 2Ba(NO_3)_2 \xrightarrow{\Delta} 2BaO + 4NO_2 + O_2 \)
(ii) Nitrates of alkali metals are very soluble in water, while alkaline earth metal nitrates are only slightly soluble and crystallise with six water molecules.
(b) Carbonates of alkali metals and alkaline earth metals:
(i) Carbonates of alkali metals (except \( Li \)) are quite stable, not breaking down up to \( 1273 \text{ K} \). However, carbonates of alkaline earth metals break down at different temperatures, producing their oxides and carbon dioxide.
\( CaCO_3 \xrightarrow{\Delta} CaO + CO_2 \uparrow \)
The thermal stability of alkaline earth metal carbonates goes up as you move down the group. \( BeCO_3 \) is the least stable, and \( BaCO_3 \) is the most stable.
(ii) Solubility:
All carbonates of alkali metals are generally soluble in water, and their solubility quickly increases as you go down the group. This is because their lattice energies decrease faster than their hydration energies as you move down the group. For alkaline earth metal carbonates, they are only slightly soluble in water, and their solubility decreases from beryllium to barium. For example, \( MgCO_3 \) is slightly soluble in water, but \( BaCO_3 \) is almost insoluble.
(c) Sulphates.
(i) The sulphates of alkali metals are thermally quite stable, while the sulphates of alkaline earth metals break down when heated, forming oxides and \( SO_3 \). The decomposition temperature goes up as you move down the group.
\( MSO_4 \xrightarrow{\Delta} MO + SO_3 \uparrow \)
(ii) The sulphates of alkali metals like sodium and potassium dissolve in water. Regarding the solubility of alkaline earth metal sulphates in water, \( BeSO_4 \) and \( MgSO_4 \) are highly soluble, \( CaSO_4 \) is only slightly soluble, but the sulphates of \( Sr, Ba \), and \( Ra \) are practically insoluble. Thus, the solubility of their sulphates in water goes down the group. \( BeSO_4 > MgSO_4 > CaSO_4 > SrSO_4 > BaSO_4 \).
In simple words: Alkali metal nitrates, carbonates, and sulphates are generally more stable and soluble than alkaline earth metal compounds. Alkaline earth compounds decompose at lower temperatures and have lower solubility, with stability increasing and solubility decreasing down their group.

Exam Tip: For comparison questions, create a mental table comparing each property (thermal stability, solubility) for both alkali and alkaline earth metals, noting trends within each group and differences between them. Use specific examples where appropriate.

 

Question 16. Starting with sodium chloride how would you proceed to prepare
(i) Sodium metal
(ii) Sodium hydroxide
(iii) Sodium peroxide
(iv) Sodium carbonate?
Answer:
(i) Sodium metal from sodium chloride: Sodium is produced from fused (molten) sodium chloride. Sodium chloride is mixed with \( CaCl_2 \) and \( KF \) (to reduce the melting point of \( NaCl \) to \( 850-875 \text{K} \)) and then put through electrolysis (in DOWN'S CELL), which causes the following reactions:
\( NaCl(l) \xrightarrow{\text{Electrolysis}} Na^+ + Cl^- \)
\( \implies \) At cathode: \( Na^+ + e^- \rightarrow Na \)
\( \implies \) At anode: \( Cl^- \rightarrow Cl + e^- \)
\( \implies Cl + Cl \rightarrow Cl_2 \)
Sodium, which forms at the cathode, is gathered in kerosene oil. Chlorine gas is released at the anode.
(ii) Sodium hydroxide from sodium chloride: Sodium hydroxide (caustic soda) is generally made by electrolysing brine solution ( \( NaCl \) solution in water) in a Castner Kellner cell. A mercury cathode and carbon anode are utilized. Sodium metal discharged at the cathode forms sodium amalgam. \( Cl_2 \) gas is released at the anode.
\( NaCl(aq) \rightarrow Na^+(aq) + Cl^-(aq) \)
\( \implies \) At cathode: \( Na^+ + e^- \xrightarrow{Hg} Na \text{-amalgam} \)
\( \implies \) At anode: \( Cl^- \rightarrow \frac{1}{2}Cl_2 + e^- \)
The amalgam is treated with water to yield sodium hydroxide and hydrogen gas.
\( 2Na \text{-amalgam} + 2H_2O \rightarrow 2NaOH + 2Hg + H_2 \uparrow \)
(iii) Sodium peroxide from sodium chloride: Sodium metal, obtained by the electrolysis of molten sodium chloride, is heated with \( O_2 \) at approximately \( 575 \text{ K} \). This process mainly forms sodium peroxide.
\( 2Na + O_2 \xrightarrow{573K} Na_2O_2 \)
(iv) Sodium carbonate \( (Na_2CO_3) \) from sodium chloride: Sodium carbonate is produced from an aqueous solution of \( NaCl \) using the SOLVAY PROCESS. In this process, \( CO_2 \) is passed through an \( NaCl \) solution saturated with ammonia, leading to the following reactions:
\( 2NH_3 + H_2O + CO_2 \rightarrow (NH_4)_2CO_3 \)
\( (NH_4)_2CO_3 + H_2O + CO_2 \rightarrow 2NH_4HCO_3 \)
\( NH_4HCO_3 + NaCl \rightarrow NH_4Cl + NaHCO_3 \)
\( 2NaHCO_3 \xrightarrow{Heat} Na_2CO_3 + CO_2 \uparrow + H_2O \)
In simple words: From sodium chloride, sodium metal is made by melting it and using electricity. Sodium hydroxide is made by passing electricity through salt water. Sodium peroxide is made by heating sodium metal with oxygen. Sodium carbonate is made using the Solvay process, which involves several steps with ammonia and carbon dioxide.

Exam Tip: For preparation methods, clearly state the starting materials, conditions (temperature, catalyst, electrolysis setup), and balanced chemical equations for each step. Mentioning by-products or purification helps to score well.

 

Question 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) When magnesium is burnt in air, it produces a brilliant dazzling light, and a mixture of magnesium oxide and magnesium nitride is formed.
\( Mg + \text{air} \xrightarrow{\Delta} MgO + Mg_3N_2 \).
(ii) Quick lime is treated with silica: When quick lime \( (CaO) \) is treated with silica \( (SiO_2) \) at high temperatures, calcium silicate \( (CaSiO_3) \) is formed.
\( CaO + SiO_2 \xrightarrow{\text{High temp}} CaSiO_3 \).
(iii) Chlorine reacts with slaked lime: When \( Cl_2 \) gas is passed through milk of lime (slaked lime) \( Ca(OH)_2 \), it produces calcium hypochlorite, a component of bleaching powder.
\( 2Ca(OH)_2 + 2Cl_2 \rightarrow CaCl_2 + Ca(OCl)_2 + 2H_2O \).
(iv) Calcium nitrate is heated: When calcium nitrate is heated, it breaks down to produce calcium oxide, \( NO_2 \), and \( O_2 \).
\( 2Ca(NO_3)_2 \xrightarrow{\Delta} 2CaO + 4NO_2 + O_2 \).
In simple words: When magnesium burns in air, it glows brightly and forms magnesium oxide and nitride. Quicklime reacts with silica at high temperatures to make calcium silicate. Chlorine gas with slaked lime creates bleaching powder. Calcium nitrate, when heated, breaks down into calcium oxide, nitrogen dioxide, and oxygen.

Exam Tip: For "What happens when..." questions, always provide the balanced chemical equation along with a brief description of the observation or product formed. Ensure correct conditions (e.g., heat, specific reactants) are noted.

 

Question 18. Describe two important uses of each of the following:
(i) Caustic soda
(ii) Sodium carbonate
(iii) Quicklime.
Answer:
(i) Two important uses of caustic soda:
• It is utilized in the paper, soap, rayon, and textile industries.
• It is beneficial in refining petroleum.
(ii) Two important uses of sodium carbonate:
• It is employed in making glass, soap, borax, and caustic soda.
• It is used for water softening, laundering, and cleaning.
(iii) Two important uses of quick lime:
• It is used in purifying sugar and in making dyestuffs.
• It is a crucial primary material and the most affordable form of alkali.
In simple words: Caustic soda is used in paper and soap production. Sodium carbonate helps make glass and soften water. Quicklime is used to purify sugar and as a cheap alkali.

Exam Tip: When listing uses, provide specific applications rather than general categories to demonstrate a deeper understanding of the compound's industrial relevance.

 

Question 19. Draw the structure of (i) BeCl2 (vapour) and (ii) BeCl2 (solid).
Answer:
(i) Structure of \( BeCl_2 \) (vapour): \( BeCl_2 \) has a bridged chloride structure in the vapour phase.
(ii) Structure of \( BeCl_2 \) in the solid state: \( BeCl_2 \) has a chain structure in the solid state.
In simple words: In its gas form, beryllium chloride has a bridge-like structure where chlorine atoms connect beryllium atoms. In its solid form, it creates a long chain where beryllium atoms are linked together by chlorine atoms in a repeating pattern.

Exam Tip: For structural questions, ensure you draw the correct bonding (covalent, ionic, coordinate) and geometry, labeling all atoms. Specifically for \( BeCl_2 \), highlight the difference between its monomeric, dimeric (vapor), and polymeric (solid) forms.

 

Question 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 solubility of a salt in water depends upon its:
(i) lattice energy
(ii) hydration energy.
For hydroxides and carbonates of sodium and potassium, their hydration energy is greater than their lattice energy. This is why they dissolve in water. However, for hydroxides and carbonates of magnesium and calcium, their hydration energy is less than their lattice energy, which makes them only slightly soluble.
The amount of lattice energy stays almost constant because the ion is so big that the small change in the size of the cations from \( Be^{2+} \) to \( Ba^{2+} \) does not make any difference. However, the hydration energy significantly decreases from \( Be^{2+} \) to \( Ba^{2+} \) as the size of the cation increases. Therefore, in the case of \( BaSO_4 \), its lattice energy happens to be greater than its hydration energy. Consequently, \( BaSO_4 \) is insoluble in water.
In simple words: Sodium and potassium salts dissolve easily because water can attract and surround their ions more strongly than the ions attract each other in the solid. Magnesium and calcium salts don't dissolve easily because their ions stick together more strongly in the solid than water can pull them apart.

Exam Tip: Explain solubility using the balance between lattice energy (energy needed to break the ionic solid) and hydration energy (energy released when ions are surrounded by water molecules). A higher hydration energy relative to lattice energy leads to greater solubility.

 

Question 21. Describe the importance of the following
(i) lime stone
(ii) Cement
(iii) Plaster of paris.
Answer:
(i) Importance of lime stone: Limestone is an important primary material and is the most affordable type of alkali.
(a) It is used to prepare Quick lime \( CaO \).
\( CaCO_3 \xrightarrow{1070-1270K} CaO + CO_2 \).
(b) Quick lime when combined with a limited amount of water, produces slaked lime or milk of lime \( Ca(OH)_2 \).
\( CaO + H_2O \rightarrow Ca(OH)_2 \)
(c) It is utilized as a flux to remove acidic impurities.
\( CaO + SiO_2 \rightarrow CaSiO_3 \)
\( 6CaO + P_4O_{10} \rightarrow 2Ca_3(PO_4)_2 \)
(ii) Importance of cement: Cement is an important construction material. It is used in concrete and reinforced concrete, for plastering, and in building dams, bridges, and other structures.
(iii) Importance of Plaster of Paris: Because of its remarkable property of hardening with water, Plaster of Paris is used in the building industry and in plasters. It is also used to immobilize injured body parts, such as bone fractures or sprains. Furthermore, it is used in dentistry and for making casts of statues and busts.
In simple words: Limestone is a cheap, main material for making quicklime. Cement is a key building material used in concrete, plaster, and large structures. Plaster of Paris sets when mixed with water and is used in building, medical casts for breaks, and dental work.

Exam Tip: For each substance, list at least two distinct uses and provide relevant chemical equations if the use involves a chemical reaction. This demonstrates both practical knowledge and chemical understanding.

 

Question 22. Why are lithium salts commonly hydrated and those of other alkali ions usually anhydrous?
Answer: Lithium salts are often hydrated, like \( \text{LiCl}\cdot2\text{H}_2\text{O} \), while other alkali ions are generally anhydrous. The hydration enthalpy of the \( \text{Li}^+ \) ion is highest among the alkali metal ions. Due to its very small size, the \( \text{Li}^+ \) ion gets highly hydrated, so its actual size in aqueous solution becomes the largest. The hydration enthalpy reduces as ionic size grows, following the order \( \text{Li}^+ > \text{Na}^+ > \text{K}^+ > \text{Rb}^+ > \text{Cs}^+ \). Hence, other alkali metal ions are typically anhydrous.
In simple words: Lithium ions are tiny, so they attract many water molecules (they get hydrated). This makes them appear larger in water than other alkali ions. Bigger ions attract less water, so they are usually anhydrous.

Exam Tip: For comparative questions, explicitly mention the factor (like size, charge, or hydration enthalpy) that causes the observed difference in properties.

 

Question 23. Why is LiF almost insoluble in water whereas LiCl is soluble not only in water but also in organic solvents?
Answer: \( \text{LiF} \) hardly dissolves in water [0.27g/100g \( \text{H}_2\text{O} \)] because of its strong lattice energy. Because of the \( \text{Li}^+ \) ion's high hydration energy, \( \text{LiCl} \) dissolves in water. \( \text{LiCl} \) also dissolves in acetone since it mostly has covalent character. As the anion's size gets bigger, the covalent character grows.
In simple words: \( \text{LiF} \) doesn't dissolve well in water because its particles are held together very tightly (high lattice energy). \( \text{LiCl} \) dissolves well in water because the water molecules strongly pull the \( \text{Li}^+ \) ion apart. \( \text{LiCl} \) also dissolves in things like acetone because it acts more like a covalent (shared electron) compound.

Exam Tip: When explaining solubility, remember to discuss both lattice energy and hydration energy. For solubility in organic solvents, consider the covalent character of the compound.

 

Question 24. Explain the significance of sodium, potassium, magnesium and calcium in biological fluids.
Answer:
**Sodium and Potassium:** These ions perform a crucial role in living fluids. They exist in biological fluids, and the K to Na ratio is 7:1 in animals like humans, rats, and horses. These ions gather in cells and form a concentration difference and membrane potential. Nerve impulses are created when a chemical is freed, triggering the membrane potential to change.
**Magnesium:** Magnesium forms part of chlorophyll, the green pigment in plants. Photosynthesis only happens when chlorophyll is present.
**Calcium:** Calcium exists as calcium phosphate \( \text{Ca}_3(\text{PO}_4)_2 \) in bones. Calcium and Magnesium are vital for making phosphorus-oxygen connections in biological systems for energy storage. Calcium ions manage the hydrolysis of pyrophosphate, which frees up energy. These ions carry out vital biological functions, like keeping ion balance and nerve signal transmission.
In simple words: Sodium and potassium help nerve signals work and keep cell balance. Magnesium is needed for plants to make food (photosynthesis). Calcium helps build bones and controls energy release in the body.

Exam Tip: For detailed questions on biological roles, break down the answer by each element and list their specific functions, highlighting key compounds or processes involved.

 

Question 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 sodium metal is dropped in water, hydrogen gas is released. Because of the high heat given off as the reaction releases heat, the hydrogen gas ignites.
\( \text{2Na} + \text{2H}_2\text{O} \rightarrow \text{2NaOH} + \text{H}_2 \)
(ii) When sodium metal is heated with plenty of air or oxygen, it combusts, producing a mix of sodium monoxide and sodium peroxide.
\( \text{4Na} + \text{O}_2 \rightarrow \text{2Na}_2\text{O} \)
\( \text{2Na}_2\text{O} + \text{O}_2 \rightarrow \text{2Na}_2\text{O}_2 \)
(iii) When sodium peroxide is mixed in water, it forms hydrogen peroxide.
\( \text{Na}_2\text{O}_2 + \text{2H}_2\text{O} \rightarrow \text{2NaOH} + \text{H}_2\text{O}_2 \)
In simple words: (i) Sodium in water makes hydrogen gas and a lot of heat, which can start a fire. (ii) Heating sodium in air forms a mixture of sodium monoxide and sodium peroxide. (iii) Sodium peroxide in water creates hydrogen peroxide.

Exam Tip: For chemical reactions, always write balanced equations and describe any observable phenomena such as heat release or ignition.

 

Question 26. Comment on each of the following observations:
(a) The mobilities of the alkali metal ions in aqueous solution are \( \text{Li}^+ < \text{Na}^+ < \text{K}^+ < \text{Rb}^+ < \text{Cs}^+ \)
(b) Lithium is the only alkali metal to form a nitride directly.
(c) \( \text{E}^\circ \) for \( \text{M}^{2+}(\text{aq}) + \text{2e}^- \rightarrow \text{M(s)} \) (where M = Ca, Sr, Ba) is nearly constant.
Answer:
(a) Due to its tiniest size among alkali metal ions, \( \text{Li}^+ \) has the highest level of hydration. Because of its high hydration, its movement is lowest. When the size grows from \( \text{Li}^+ \) to \( \text{Cs}^+ \), the ion size grows, hence the level of hydration reduces, and ion movement increases. Thus, mobilities of alkali metal ions in aqueous solutions follow the order: \( \text{Li}^+ < \text{Na}^+ < \text{K}^+ < \text{Rb}^+ < \text{Cs}^+ \).
(b) Lithium is the sole alkali metal that directly creates a nitride. \( \text{Li}^+ \) and \( \text{N}^{3-} \) ions are very small, so they create stable \( \text{Li}_3\text{N} \). The \( \text{Li}^+ \) ion, being the tiniest, will join with small \( \text{N}^{3-} \) ions to form a strong crystal structure.
\( \text{6Li} + \text{N}_2 \rightarrow \text{2Li}_3\text{N} \)
(c) The standard electrode potential \( \text{E}^\circ \) for \( \text{M}^{2+} + \text{2e}^- \rightarrow \text{M(s)} \) where M = Ca, Sr, Ba is almost identical. \( \text{Be}^{2+} \) and \( \text{Mg}^{2+} \) ions are tinier than \( \text{Ca}^{2+} \), \( \text{Sr}^{2+} \) or \( \text{Ba}^{2+} \). So, they lose electrons easily instead of Be or Mg. Their standard reduction potentials possess more negative values than \( \text{Be}^{2+} \) or \( \text{Mg}^{2-} \). Thus, their likelihood to lose electrons is quite similar. The \( \text{E}^\circ \) for \( \text{Ca}^{2+} \), \( \text{Sr}^{2+} \), \( \text{Ba}^{2+} \) is nearly constant (-2.89 V).
In simple words: (a) Lithium ions are tiny and attract a lot of water, making them move slowly in solution. Larger alkali ions move faster because they attract less water. (b) Lithium's small size allows it to directly combine with nitrogen to form a stable compound. (c) Calcium, strontium, and barium have very similar abilities to lose electrons, making their reduction potentials nearly constant.

Exam Tip: When explaining trends, always link them to fundamental properties like atomic/ionic size, charge density, hydration energy, or polarizing power.

 

Question 27. State as to why:
(a) a solution of \( \text{Na}_2\text{CO}_3 \) is alkaline?
(b) alkali metals are prepared by electrolysis of their fused chlorides?
(c) sodium is found to be more useful than potassium?
Answer:
(a) \( \text{Na}_2\text{CO}_3 \) solution is basic because of hydrolysis of \( \text{Na}_2\text{CO}_3 \). Sodium carbonate is the salt from a strong base \( \text{NaOH} \) and a weak acid \( \text{H}_2\text{CO}_3 \). Due to hydrolysis, the resulting solution becomes basic due to the presence of extra \( \text{OH}^- \) ions.
\( \text{Na}_2\text{CO}_3 + \text{2H}_2\text{O} \rightarrow \text{2NaOH} + \text{H}_2\text{CO}_3 \)
(b)

• Alkali metals are powerful reducing agents, so they cannot be removed by reducing their oxides or chlorides.
• Alkali metals, being very electropositive, cannot be pushed out from the watery solutions of their salts by different metals.
• Alkali metals cannot be separated by electrolyzing the watery solution of their salts since hydrogen gas is released at the cathode instead of the alkali metal itself.

(c) Sodium is more useful than potassium because:

• Sodium is employed as a reducing agent.
• It is used in Lassaigne's test for finding nitrogen, sulfur, and halogens.
• About 60% of the global output of sodium goes into making TEL \( \text{Pb}(\text{C}_2\text{H}_5)_4 \), a knock inhibitor.
• It is utilized in sodium vapor lights.

In simple words: (a) Sodium carbonate solution is basic because it reacts with water, producing excess hydroxide ions. (b) Alkali metals are so reactive that they can only be extracted by melting their salts and using electricity, not by simpler chemical reduction or from water solutions. (c) Sodium is more versatile than potassium, used in many industrial processes, tests, and even streetlights.

Exam Tip: When explaining "why" questions, provide a clear, concise reason, often involving chemical properties like electronegativity, hydrolysis, or reducing power. For uses, list specific applications.

 

Question 28. Write balanced equations for reactions between:
(a) \( \text{Na}_2\text{O}_2 \) and water
(b) \( \text{KO}_2 \) and water
Answer:
(a) \( \text{Na}_2\text{O}_2 + \text{2H}_2\text{O} \rightarrow \text{2NaOH} + \text{H}_2\text{O}_2 \)
(b) \( \text{4KO}_2 + \text{2H}_2\text{O} \rightarrow \text{4KOH} + \text{3O}_2 \)
or \( \text{2KO}_2 + \text{2H}_2\text{O} \rightarrow \text{KOH} + \text{H}_2\text{O}_2 + \text{O}_2 \)
(c) \( \text{Na}_2\text{O} + \text{CO}_2 \rightarrow \text{Na}_2\text{CO}_3 \)
In simple words: (a) Sodium peroxide with water produces sodium hydroxide and hydrogen peroxide. (b) Potassium superoxide with water can form potassium hydroxide and oxygen, or sometimes hydrogen peroxide too. (c) Sodium oxide with carbon dioxide creates sodium carbonate.

Exam Tip: Remember to balance all chemical equations correctly. For reactions with superoxides, be aware of the different possible products depending on conditions (e.g., \( \text{H}_2\text{O}_2 \) vs. \( \text{O}_2 \)).

 

Question 29. How would you explain the following observations?
(i) BeO is almost insoluble but BeSO4 is soluble in water.
(ii) BaO is soluble but BaSO4 is insoluble in water.
(iii) Lil is more soluble than Kl in ethanol.
Answer:
(i) \( \text{BeO} \) does not dissolve in water because its lattice energy is higher than its hydration energy. Conversely, \( \text{BeSO}_4 \) dissolves in water because its hydration energy exceeds the lattice energy.
(ii) In how oxides of Group II elements dissolve, both lattice and hydration energies drop down the group as cation size grows. However, lattice energy declines more sharply than the hydration energy. Hence, \( \text{BaO} \) dissolves in water as its hydration energy is greater than its lattice energy. \( \text{BaSO}_4 \), conversely, does not dissolve in water. The value of lattice energy stays nearly constant as the ion is very large, so a tiny rise in size of the cations from \( \text{Be}^{2+} \) to \( \text{Ba}^{2+} \) makes no real change. However, the hydration energy noticeably reduces from \( \text{Be}^{2+} \) to \( \text{Ba}^{2+} \) as the cation's size increases. Hence, in the case of \( \text{BaSO}_4 \), its lattice energy turns out to be greater than its hydration energy. Therefore, \( \text{BaSO}_4 \) does not dissolve in water.
(iii) \( \text{LiI} \) dissolves better in ethanol than \( \text{KI} \). In the case of \( \text{LiI} \), because of the small size of \( \text{Li}^+ \), its polarizing power is higher, and so its covalent nature is greater for the \( \text{LiI} \) molecule. Hence, it dissolves well in covalent solvents like ethanol. In the case of \( \text{KI} \), as it is an ionic compound, it does not dissolve as well in ethanol.
In simple words: (i) \( \text{BeO} \) doesn't dissolve much because its particles are held tightly, but \( \text{BeSO}_4 \) dissolves well because water pulls its particles apart easily. (ii) \( \text{BaO} \) dissolves because water can hydrate it well, but \( \text{BaSO}_4 \) doesn't because its particles are too strongly bound together. (iii) \( \text{LiI} \) dissolves better in ethanol than \( \text{KI} \) because \( \text{LiI} \) has more covalent character due to the small size of the \( \text{Li}^+ \) ion.

Exam Tip: For solubility questions, always compare lattice energy and hydration energy. For solubility in organic solvents, consider covalent character and the polarizing power of the cation.

 

Question 30. Which of the following alkali metals has the least melting point?
(a) Na
(b) K
(c) Rb
(d) Cs.
Answer: (d) Caesium is having the least melting point (= 312 K).
In simple words: Out of the choices, Caesium has the lowest temperature at which it melts.

Exam Tip: Remember that melting points generally decrease down the alkali metal group due to decreasing metallic bonding strength as atomic size increases.

 

Question 31. Which one of the following alkali metal gives hydrated salts?
(a) Li
(b) Na
(c) K
(d) Cs.
Answer: (a) Lithium is the only alkali metal which gives hydrated salts.
In simple words: Lithium is the only alkali metal that forms salts with water molecules attached (hydrated salts).

Exam Tip: Lithium's small ionic size and high charge density lead to a very high hydration energy, making it prone to forming hydrated salts.

 

Question 32. Which one of the alkaline earth metal carbonates is thermally the most stable?
(a) MgCO3
(b) CaCO3
(c) SrCO3
(d) BaCO3.
Answer: (d) BaCO3 is thermally the most stable alkaline earth metal carbonate. It decomposes only above 1633 K.
In simple words: Barium carbonate is the most heat-stable among the alkaline earth metal carbonates, meaning it needs a very high temperature to break down.

Exam Tip: For alkaline earth metal carbonates, thermal stability increases down the group as the cation's size increases, which stabilizes the larger carbonate ion more effectively.