Answer

(d) The energy binding the atoms in the crystal lattice of the alkali metals is low due to their large atomic radii and especially due to the presence of one valence electron per metal atom as compared to large number available vacant orbitals.

Hence, alkali metals have low melting and boiling points. The melting point of alkali metals decreases from $\mathrm{Li}$ to $\mathrm{Cs}$ as cohesive force decreases with increase in atomic size.

Melting point of $\mathrm{Cs}=302\mathrm{K}$ i.e., ${29}^{\circ }\mathrm{C}$.

• (a) Na: Sodium (Na) has a melting point of approximately 98°C, which is significantly higher than 30°C. Therefore, it will not melt at room temperature.

• (b) K: Potassium (K) has a melting point of approximately 63°C, which is also higher than 30°C. Hence, it will not melt at room temperature.

• (c) Rb: Rubidium (Rb) has a melting point of approximately 39°C, which is slightly higher than 30°C. Thus, it will not melt at room temperature.

Answer

(a) Li has most negative standard reduction potential due to very high enthalpy of hydration. Thus, reaction of $\mathrm{Li}$ with water will be most exothermic, but surprisingly $\mathrm{Li}$ reacts with water gently, whereas $\mathrm{Na}$ and $\mathrm{K}$ vigorously.

The explanation is in kinetics and not in thermodynamics of the reaction. No doubt, maximum energy is evolved with Li but its fusion, vaporisation and ionisation consume more energy. As a result reaction proceeds slowly.

$\mathrm{Na}$ or $\mathrm{K}$ have low melting points and molten metal spreads over water exposing a larger surface to water, making the reaction vigorous.

• (b) Na: Sodium reacts more vigorously with water than lithium because it has a lower melting point and the molten metal spreads over the water, exposing a larger surface area to the water, which accelerates the reaction.

• (c) K: Potassium reacts even more vigorously with water than sodium due to its even lower melting point and higher reactivity, which causes the molten metal to spread quickly over the water, increasing the reaction rate.

• (d) Cs: Cesium reacts the most vigorously with water among the alkali metals listed. It has the lowest melting point and highest reactivity, leading to an extremely rapid and exothermic reaction with water.

Answer

(c) Standard reduction potential $\left(E_{\mathrm{RP}}^{\circ }\right)$ is a measure of tendency of an element to lose electron in aqueous solution. Higher the negative $E_{\mathrm{RP}}^{\circ }$ greater is the ability to lose electrons.

$E_{\mathrm{RP}}^{\circ }$ depends on

(i) enthalpy of sublimation

(ii) ionisation enthalpy

(iii) enthalpy of hydration

Thus, in aqueous medium, order of reactivity of alkali metals is $\mathrm{Na}<\mathrm{K}<\mathrm{Rb}<\mathrm{Cs}<\mathrm{Li}$.

$E_{\mathrm{RP}}^{\circ }$ value of $Li$ is least $(-3.04\mathrm{V})$ among all alkali metals.

The formation of ${\mathrm{Li}}^{+}(\mathrm{aq})$ from $\mathrm{Li}$ involves following steps

(i) $\mathrm{Li}(\mathrm{s})\stackrel{\text{ Sublimation }}{\to }\mathrm{Li}(\mathrm{g})\mathrm{\Delta }{\mathrm{H}}_s=$ Enthalpy of sublimation

(ii) $\mathrm{Li}(g)⟶{\mathrm{Li}}^{+}(s)I{E}_1=$ lonisation enthalpy

(iii) ${\mathrm{Li}}^{+}(g)⟶{\mathrm{Li}}^{+}(\mathrm{aq})\mathrm{\Delta }{H}_n=$ Enthalpy of hydration

For alkali metals, enthalpies of sublimation are almost same. ${\mathrm{IE}}_1$ value of $\mathrm{Li}$ is endothermic and highest and hydration is exothermic and maximum for ${\mathrm{Li}}^{+}$.

The highly exothermic step (iii) for smallest ${\mathrm{Li}}^{+}$makes it strongest reducing agent.

• (a) Sublimation enthalpy: The enthalpy of sublimation for alkali metals is almost the same and does not significantly differentiate the reducing power of lithium from other alkali metals.

• (b) Ionisation enthalpy: Although the ionisation enthalpy of lithium is the highest among alkali metals, it is an endothermic process and does not contribute to making lithium the strongest reducing agent in aqueous solution.

• (d) Electron-gain enthalpy: Electron-gain enthalpy is not a relevant factor for the reducing power of alkali metals in aqueous solution, as it pertains to the energy change when an atom gains an electron, which is not the primary process involved in the reduction potential of lithium.

Thinking Process

All the alkaline earth metals form carbonates having general formula ${\mathrm{MCO}}_3$. These carbonates decompose on heating to form metal oxide and carbon dioxide.

$$MC{O}_3\rightleftharpoons 1MO+C{O}_2[M=Be,Mg,Ca,Sr,Ba]$$

Thermal stability of carbonates increases with increase in atomic number, i.e., on moving down the group

$$BeC{O}_3<MgC{O}_3<CaC{O}_3<SrC{O}_3<BaC{O}_3.$$

Answer

(d) $BaC{O}_3$ is thermally most stable because of the small size of resulting oxide ion. With the increase in atomic number, the size of the metal ion, the stability of the metal ion decreases and, hence that of carbonate increases (maximum in case of $BaC{O}_3$ ).

Therefore, the increasing size of cation destabilizes the oxides and hence does not favour the decomposition of heavier alkaline earth metal carbonates like $BaC{O}_3$.

• (a) ${\mathrm{MgCO}}_3$: Magnesium carbonate ${\mathrm{MgCO}}_3$ is less thermally stable compared to barium carbonate ${\mathrm{BaCO}}_3$ because magnesium has a smaller ionic radius. The smaller size of the magnesium ion leads to a higher charge density, which destabilizes the carbonate ion, making it easier to decompose upon heating.

• (b) ${\mathrm{CaCO}}_3$: Calcium carbonate ${\mathrm{CaCO}}_3$ is also less thermally stable than barium carbonate ${\mathrm{BaCO}}_3$. Although calcium has a larger ionic radius than magnesium, it is still smaller than barium. This results in a higher charge density compared to barium, which destabilizes the carbonate ion and makes it more prone to decomposition.

• (c) ${\mathrm{SrCO}}_3$: Strontium carbonate ${\mathrm{SrCO}}_3$ is less thermally stable than barium carbonate ${\mathrm{BaCO}}_3$ because strontium has a smaller ionic radius than barium. The smaller size of the strontium ion leads to a higher charge density, which destabilizes the carbonate ion, making it easier to decompose upon heating.

Answer

(a) $BeC{O}_3$ is unstable to the extent that it is stable only in atmosphere of $C{O}_2$. $BeC{O}_3$ shows reversible reaction because stability of oxide formed is more than carbonates.

$$BeC{O}_3\rightleftharpoons BeO+C{O}_2$$

$BeC{O}_3$ is unstable due to strong polarising effect of small $B{e}^{2+}$ ion on the large polarisable carbonation. Moreover, an extrastability of the oxide achieved through lattice energy by packing small cation with small oxide ion.

• (b) ${\mathrm{MgCO}}_3$: Magnesium carbonate (${\mathrm{MgCO}}_3$) is relatively stable in air and does not decompose easily. It does not require a ${\mathrm{CO}}_2$ atmosphere to remain stable.

• (c) ${\mathrm{CaCO}}_3$: Calcium carbonate (${\mathrm{CaCO}}_3$) is also stable in air and does not decompose under normal conditions. It is commonly found in nature as limestone and does not need a ${\mathrm{CO}}_2$ atmosphere for stability.

• (d) ${\mathrm{BaCO}}_3$: Barium carbonate (${\mathrm{BaCO}}_3$) is stable in air and does not decompose easily. It does not require a ${\mathrm{CO}}_2$ atmosphere to avoid decomposition.

Answer

(a) All the alkaline earth metals form hydroxides. Solubility of hydroxides of alkaline earth metals increases from Be to $Ba$. $Be(OH{)}_2$ and $Mg(OH{)}_2$ are almost insoluble.

The basic nature of hydroxides of alkaline earth metal depends on the solubility of hydroxide in water. More the solubility more the basicity. Solubility of hydroxides depends on lattice energy and hydration energy.

$$\mathrm{\Delta }{H}_{\text{solution }}=\mathrm{\Delta }{H}_{\text{lattice energy }}+\mathrm{\Delta }{H}_{\text{hydration energy }}$$

The magnitude of hydration energy remains almost same whereas lattice energy decreases down the group leading to more negative values for $\mathrm{\Delta }{H}_{\text{solution }}$ down the group.

More negative $\mathrm{\Delta }{H}_{\text{solution }}$ more is solubility of compounds.

Hence, $Be(OH{)}_{2}andMg(OH{)}_2\text{have less negative values for }$

$\mathrm{\Delta }{H}_{\text{solution}}$ hence, least basic.

• (b) $\mathrm{Ca}(\mathrm{OH}{)}_2$: Calcium hydroxide is more soluble in water compared to magnesium hydroxide, leading to a higher basicity. Therefore, it is not the least basic among the given options.

• (c) $\mathrm{Sr}(\mathrm{OH}{)}_2$: Strontium hydroxide is even more soluble in water than calcium hydroxide, resulting in a higher basicity. Thus, it is not the least basic among the given options.

• (d) $\mathrm{Ba}(\mathrm{OH}{)}_2$: Barium hydroxide is the most soluble in water among the given hydroxides, making it the most basic. Therefore, it is not the least basic among the given options.

Answer

(a) Ethanol is an organic compound i.e., of covalent character “Like dissolves like”. To dissolve in ethanol the compound should have more covalent character.

Beryllium halides have covalent character due to small size and high effective nuclear charge. Hence, ${\mathrm{BeCl}}_2$ is most covalent among all other chlorides.

• (b) ${\mathrm{MgCl}}_2$: Magnesium chloride is more ionic in nature compared to beryllium chloride due to the larger size and lower effective nuclear charge of the magnesium ion. This ionic character makes it less soluble in organic solvents like ethanol.

• (c) ${\mathrm{CaCl}}_2$: Calcium chloride is even more ionic than magnesium chloride because calcium has a larger ionic radius and lower effective nuclear charge. This increased ionic character further reduces its solubility in ethanol.

• (d) ${\mathrm{SrCl}}_2$: Strontium chloride is the most ionic among the given options due to the largest ionic radius and the lowest effective nuclear charge. This high ionic character makes it the least soluble in ethanol among the listed metal halides.

Thinking Process

Ionisation energies depend upon how strongly the valence electron is held by the nucleus. lonisation energy value will be high if electron is tightly held and if interaction between electron and nucleus is poor then ionisation energy will be low.

Answer

(c) On moving down in the group (from $\mathrm{Li}$ to $\mathrm{Cs}$ ), the ionisation energy value decreases from $\mathrm{Li}$ to $\mathrm{Cs}$, size of the atom increases and so valence electron is less tightly held. Increased screening effect from $\mathrm{Li}$ to $\mathrm{Cs}$ also makes the removal of electron easier.

• Option (a) is incorrect because it suggests that sodium (Na) has a higher ionization enthalpy than lithium (Li), which is not true. Lithium, being smaller in size, has a higher ionization enthalpy than sodium.

• Option (b) is incorrect because it suggests that rubidium (Rb) has a higher ionization enthalpy than sodium (Na) and potassium (K), which is not true. Rubidium, being larger in size, has a lower ionization enthalpy than both sodium and potassium.

• Option (d) is incorrect because it suggests that potassium (K) has a higher ionization enthalpy than lithium (Li) and sodium (Na), which is not true. Potassium, being larger in size, has a lower ionization enthalpy than both lithium and sodium.

Answer

(b) Solubilities of alkali metal halides in water can be explained in terms of lattice enthalpy and hydration enthalpy. Lower lattice enthalpies and higher hydration enthalpies favour dissolution.

Among fluorides, the order of solubility is $\mathrm{LiF}<\mathrm{NaF}<\mathrm{KF}<\mathrm{RbF}<\mathrm{CsF}$. Low solubility of $\mathrm{LiF}$ is due to very high lattice energy. On moving down in the group LiF to CsF, solubility increases because lattice energy decreases.

Except LiF, other halides of lithium are highly soluble in water.

• (a) The ionic nature of lithium fluoride does not explain its low solubility. Ionic compounds generally tend to be soluble in water due to the interaction between the ions and water molecules. The low solubility of LiF is specifically due to its high lattice enthalpy, not its ionic nature.

• (c) High hydration enthalpy for lithium ion would actually favor solubility, as higher hydration enthalpy means that the ion interacts strongly with water molecules, promoting dissolution. Therefore, this is not the reason for the low solubility of LiF.

• (d) Low ionisation enthalpy of lithium atom is not relevant to the solubility of LiF in water. Ionisation enthalpy refers to the energy required to remove an electron from an atom in the gas phase, which does not directly affect the solubility of the resulting ionic compound in water.

Answer

(a) The solubility of hydroxides of alkaline earth metals increases from $Be$ to $Ba$, in water.

$Be(OH{)}_2$ and $Mg(OH{)}_2$ are almost insoluble.

Due to high hydration enthalpy and high lattice energy $Be(OH{)}_2$ is not soluble in water. $Be(OH{)}_2$ is an amphoteric hydroxide. With acids, $Be(OH{)}_2$ is neutralised giving salts.

$$Be(OH{)}_2+2HCl⟶BeC{l}_2+2H_{2}O$$

$Be(OH{)}_2$ reacts with $NaOH$ also forming beryllate.

$$Be(OH{)}_2+2NaOH⟶N{a}_{2}Be{O}_2+2H_{2}O$$

• (b) $\mathrm{Mg}(\mathrm{OH}{)}_2$: Magnesium hydroxide is not amphoteric; it is only slightly soluble in water and does not react with sodium hydroxide to form a soluble complex.

• (c) $\mathrm{Ca}(\mathrm{OH}{)}_2$: Calcium hydroxide is not amphoteric; it is slightly soluble in water and does not react with sodium hydroxide to form a soluble complex.

• (d) $\mathrm{Ba}(\mathrm{OH}{)}_2$: Barium hydroxide is not amphoteric; it is highly soluble in water but does not react with sodium hydroxide to form a soluble complex.

Answer

(a) Sodium carbonate is synthesised by Solvay ammonia soda process.

The reactions involved are

$$\begin{array}{rl}& N{H}_3+H_{2}O+C{O}_2\to \underset{\text{ Ammonium bicarbonate }}{N{H}_{4}HC{O}_3}\\ & NaCl+N{H}_{4}HC{O}_3\to \underset{\text{ Sodium bicarbonate }}{NaHC{O}_3\downarrow +N{H}_{4}Cl}\\ & 2NaHC{O}_3\stackrel{\mathrm{\Delta }}{\to }N{a}_{2}C{O}_3+H_{2}O+C{O}_2\end{array}$$

$N{H}_3$ is recovered from $N{H}_{4}HC{O}_3$ and $N{H}_{4}Cl$ formed during the reaction.

$\begin{array}{rl}& {\mathrm{NH}}_4{\mathrm{HCO}}_3\stackrel{\text{ Heat }}{\to }{\mathrm{NH}}_3+{\mathrm{H}}_2\mathrm{O}+{\mathrm{CO}}_2\end{array}$

$\underset{\text{ Ammonium chloride }}{2{\mathrm{NH}}_4\mathrm{Cl}}+\mathrm{Ca}(\mathrm{OH}{)}_2⟶\underset{\text{ Ammonia }}{2{\mathrm{NH}}_3}+\underset{\text{ Calcium chloride }}{{\mathrm{CaCl}}_2}+2{\mathrm{H}}_2\mathrm{O}$

• (b) NaCl: Sodium chloride (NaCl) is not a by-product of the reaction between ammonium chloride (NH₄Cl) and calcium hydroxide (Ca(OH)₂). Instead, NaCl is a reactant in the Solvay process, where it reacts with ammonium bicarbonate (NH₄HCO₃) to form sodium bicarbonate (NaHCO₃) and ammonium chloride (NH₄Cl).

• (c) NaOH: Sodium hydroxide (NaOH) is not produced in the reaction between ammonium chloride (NH₄Cl) and calcium hydroxide (Ca(OH)₂). The reaction specifically produces ammonia (NH₃), calcium chloride (CaCl₂), and water (H₂O).

• (d) NaHCO₃: Sodium bicarbonate (NaHCO₃) is not a by-product of the reaction between ammonium chloride (NH₄Cl) and calcium hydroxide (Ca(OH)₂). NaHCO₃ is actually an intermediate product in the Solvay process, formed from the reaction of NaCl and NH₄HCO₃.

Answer

(a) All alkali metal dissolve in liquid ${\mathrm{NH}}_3$ giving highly conducting deep blue solution.

$\mathrm{Na}+(x+y){\mathrm{NH}}_3⟶\underset{\text{ Ammoniated cation }}{{\left[\mathrm{Na}\left({\mathrm{NH}}_3\right)x\right]}^{+}}+\underset{\text{ Ammoniated electron }}{e{\left({\mathrm{NH}}_3\right)}_y^{-}}$

When light fall on these solutions, the ammoniated electrons excite in higher energy level by absorbing red wavelengths and so transmitted light is blue.

• (b) Sodium ion is incorrect because the sodium ion itself does not contribute to the deep blue color of the solution. The color is specifically due to the presence of ammoniated electrons.

• (c) Sodium amide is incorrect because sodium amide is a different compound formed when sodium reacts with ammonia, but it does not cause the deep blue color. The deep blue color is due to the ammoniated electrons.

• (d) Ammoniated sodium ion is incorrect because the ammoniated sodium ion does not cause the deep blue color. The color is due to the ammoniated electrons, which absorb red wavelengths and transmit blue light.

Answer

(b) Raw materials for cement-limestone, clay, gypsum. Cement is a dirty greyish heavy powder containing calcium aluminates and silicates.

Gypsum $(CaS{O}_4\cdot 5H_{2}O)$ is added to the components to increase the setting time of cement so that it gets sufficiently hardened. Setting of cement is an exothermic process and involves hydration of calcium aluminates and silicates.

• (a) Setting time of cement becomes less: This is incorrect because gypsum is added to cement specifically to increase the setting time, not to decrease it. Without gypsum, the cement would set too quickly, making it difficult to work with.

• (c) Colour of cement becomes light: This is incorrect because the addition of gypsum does not significantly affect the color of the cement. The color of cement is primarily determined by the raw materials used, such as limestone and clay, and the manufacturing process.

• (d) Shining surface is obtained: This is incorrect because gypsum does not contribute to the surface finish of the cement. The surface finish of cement is influenced by factors such as the type of cement, the method of application, and the finishing techniques used, rather than the addition of gypsum.

Answer

(a) Plaster of Paris is prepared by heating gypsum at ${120}^{\circ }\mathrm{C}$.

$$\underset{\text{ Gypsum }}{2CaS{O}_4\cdot 2H_{2}O}⟶\underset{\text{ Plaster of Paris }}{{\left(CaS{O}_4\right)}_2\cdot \frac{1}{2}{H}_{2}O+3H_{2}O}$$

On heating plaster of Paris at ${200}^{\circ }\mathrm{C}$, if forms anhydrous calcium sulphate i.e., dead plaster which has no setting property as it absorbs water very slowly.

${\mathrm{CaSO}}_4\cdot 2{\mathrm{H}}_2\mathrm{O}\stackrel{{200}^{\circ }\text{C}}{\to }\underset{\text{Anhydride}}{{\mathrm{CaSO}}_4}\stackrel{{1100}^{\circ }\text{C}}{\to }\mathrm{CaO}+{\mathrm{SO}}_3$

• (b) $CaS{O}_4\cdot \frac{1}{2}{H}_{2}O$: This is Plaster of Paris, not dead burnt plaster. Plaster of Paris is formed by heating gypsum to about ${120}^{\circ }\mathrm{C}$, not ${200}^{\circ }\mathrm{C}$.

• (c) $CaS{O}_4\cdot H_{2}O$: This is not a common form of calcium sulfate used in plaster. The common forms are gypsum ($CaS{O}_4\cdot 2H_{2}O$) and Plaster of Paris ($CaS{O}_4\cdot \frac{1}{2}{H}_{2}O$).

• (d) $CaS{O}_4\cdot 2H_{2}O$: This is gypsum, the raw material used to produce Plaster of Paris. It is not dead burnt plaster, which is anhydrous calcium sulfate ($CaS{O}_4$).

Answer

(c) Calcium hydroxide is prepared by adding water to quicklime $(\mathrm{CaO})$.

$$\underset{\text{Quick lime}}{CaO(s)}+H_{2}O(l)⟶\underset{\text{ Slaked lime }}{Ca(OH{)}_2(s)}$$

It is a white amorphous powder. It is sparingly soluble in water.

So, it forms a suspension of slaked lime in water which is called milk of lime and the clear solution obtained after the suspension settles is known as lime water.

• (a) lime water: Lime water is the clear solution obtained after the suspension of slaked lime in water settles. It is not the suspension itself, but rather the clear liquid that remains after the solid particles have settled out.

• (b) quick lime: Quick lime is calcium oxide (CaO), which is a different compound from slaked lime (calcium hydroxide, Ca(OH)₂). Quick lime reacts with water to form slaked lime, but it is not the suspension of slaked lime in water.

• (d) aqueous solution of slaked lime: An aqueous solution of slaked lime refers to the dissolved portion of calcium hydroxide in water, which is lime water. It is not the suspension of slaked lime in water, which is called milk of lime.

Answer

(a) Except Be, all alkaline earth metals form hydrides $\left(M{H}_2\right)$ on directly heating with $H_2$. $Be{H}_2$ can’t be prepared by direct action of $H_2$ on $Be$. It is prepared by the action of $LiAl{H}_4$ on $BeC{l}_2$.

$$2BeC{l}_2+LiAl{H}_4⟶2Be{H}_2+LiCl+AlC{l}_3$$

• (b) Mg: Magnesium (Mg) forms magnesium hydride (MgH₂) by direct heating with dihydrogen (H₂).

• (c) Sr: Strontium (Sr) forms strontium hydride (SrH₂) by direct heating with dihydrogen (H₂).

• (d) Ba: Barium (Ba) forms barium hydride (BaH₂) by direct heating with dihydrogen (H₂).

Answer

(d) On heating washing soda, it loses its water of crystallisation. Above $373\mathrm{K}$, it becomes completely anhydrous white powder called soda ash.

$\begin{array}{rl}& {\mathrm{Na}}_2{\mathrm{CO}}_3\cdot 10{\mathrm{H}}_2\mathrm{O}\stackrel{>373\mathrm{K}}{\to }{\mathrm{Na}}_2{\mathrm{CO}}_3\cdot 10{\mathrm{H}}_2\mathrm{O}\\ & \text{ Washing soda }\text{ Soda ash }\\ & \end{array}$

• (a) $N{a}_{2}C{O}_3\cdot 10H_{2}O$: This is the formula for washing soda, not soda ash. Washing soda is the decahydrate form of sodium carbonate, whereas soda ash is the anhydrous form.

• (b) $N{a}_{2}C{O}_3\cdot 2H_{2}O$: This is the formula for sodium carbonate dihydrate, which is not commonly referred to as soda ash. Soda ash is the anhydrous form of sodium carbonate.

• (c) $N{a}_{2}C{O}_3\cdot H_{2}O$: This is the formula for sodium carbonate monohydrate, which is also not commonly referred to as soda ash. Soda ash is the anhydrous form of sodium carbonate.

Answer

(b) Calcium gives brick red coloured flame. Hence, calcium nitrate on heating decomposes into calcium oxide, with evolution of a mixture of $N{O}_2$ and $O_2$.

$$2Ca{\left(N{O}_3\right)}_2⟶2CaO+N{O}_2+O_2$$

$N{O}_2$ is brown coloured gas.

• Magnesium nitrate: Magnesium does not give a brick red flame; it typically gives a bright white flame. Additionally, upon heating, magnesium nitrate decomposes to form magnesium oxide, nitrogen dioxide, and oxygen, but the flame color does not match the description.

• Barium nitrate: Barium gives a green flame, not a brick red flame. When barium nitrate is heated, it decomposes to form barium oxide, nitrogen dioxide, and oxygen, but the flame color is incorrect.

• Strontium nitrate: Strontium gives a crimson red flame, not a brick red flame. Upon heating, strontium nitrate decomposes to form strontium oxide, nitrogen dioxide, and oxygen, but the flame color does not match the description.

Answer

(a) Calcium hydroxide is used in the manufacture of bleaching powder.

$$\underset{\begin{array}{c}\text{ Slaked }\\ \text{ lime }\end{array}}{2Ca(OH{)}_2}+2C{l}_2\stackrel{\text{ Cold }}{\to }CaC{l}_2+\underset{\begin{array}{c}\text{ Bleaching }\\ \text{ powder }\end{array}}{CaOC{l}_2}+2H_{2}O$$

• (b) Calcium hydroxide is not a light blue solid; it is a white solid.
• (c) Calcium hydroxide does possess disinfectant properties.
• (d) While calcium hydroxide is used in the construction industry, it is not a primary ingredient in the manufacture of cement. Cement primarily consists of calcium silicates and aluminates.

Answer

(c) To recover $N{H}_3$ in Solvay ammonia soda process $Ca(OH{)}_2$ is used.

$$2N{H}_{4}Cl+Ca(OH{)}_2⟶2N{H}_3+CaC{l}_2+2H_{2}O$$

On passing $C{O}_2$ through $Ca(OH{)}_2$, it turns milky due to the formation of $CaC{O}_3$.

$$Ca(OH{)}_2+C{O}_2⟶CaC{O}_3\downarrow +H_{2}O$$

$\mathrm{Ca}(\mathrm{OH}{)}_2$ is used for white washing.

• (a) $Ca{\left(HC{O}_3\right)}_2$: This compound is calcium bicarbonate, which is not used in the Solvay process for recovering ammonia. Additionally, it does not turn milky when $C{O}_2$ is bubbled through its solution.

• (b) $\mathrm{CaO}$: Calcium oxide, also known as quicklime, does not directly react with $C{O}_2$ to form a milky solution. It first needs to react with water to form $\mathrm{Ca}(\mathrm{OH}{)}_2$, which then reacts with $C{O}_2$ to form the milky suspension of ${\mathrm{CaCO}}_3$.

• (d) ${\mathrm{CaCO}}_3$: Calcium carbonate is the product formed when $C{O}_2$ is bubbled through a solution of $\mathrm{Ca}(\mathrm{OH}{)}_2$. It is not used in the Solvay process to recover ammonia and does not turn milky upon further reaction with $C{O}_2$.

Answer

(d) Chlorides of alkaline earth metals are hydrated salts. Due to their hygroscopic nature, they can be used as a dehydrating agent, to absorb moisture from air.

Extent of hydration decreases from $Mg$ to $Ba$ i.e., $MgC{l}_2\cdot 6H_{2}O,CaC{l}_2\cdot 6H_{2}O$, $BaC{l}_2\cdot 2H_{2}O,SrC{l}_2\cdot 2H_{2}O$

• (a) Act as dehydrating agent: This statement is not incorrect. The given halides can act as dehydrating agents due to their hygroscopic nature.

• (b) Can absorb moisture from air: This statement is not incorrect. The given halides can absorb moisture from the air due to their hygroscopic nature.

• (c) Tendency to form hydrate decreases from calcium to barium: This statement is incorrect. The correct order of decreasing tendency to form hydrates is from magnesium to barium, not from calcium to barium. The correct sequence is $MgC{l}_2\cdot 6H_{2}O$, $CaC{l}_2\cdot 6H_{2}O$, $SrC{l}_2\cdot 2H_{2}O$, $BaC{l}_2\cdot 2H_{2}O$.

Answer

$(b,d)$

Alkali metals are the first members in a period. Alkali metals have largest atomic radii in their period due to least effective nuclear charge.

They have low density because size is large and mass is least in a period. Alkali metals are soft metals to cut with a knife i.e., low boiling point.

Due to $n{s}^1$ electronic configuration, they lose electron easily and have high negative standard electrode potential.

• (a) High boiling point: Alkali metals have low boiling points because they are soft metals with weak metallic bonding, making it easier for them to transition to the gaseous state.

• (c) High density: Alkali metals have low density because they have large atomic sizes and relatively low atomic masses compared to other metals in the same period.

Answer

$(a,c)$

$\mathrm{NaOH}$ is used in manufacture of rayon.

$N{a}_{2}C{O}_3$ is used in manufacture of soap powders, in laundry for washing.

• ${\mathrm{NaHCO}}_3$ (Sodium bicarbonate) is not typically used in the textile industry; it is more commonly used in baking, as a leavening agent, and in fire extinguishers.
• $\mathrm{NaCl}$ (Sodium chloride) is not primarily used in the textile industry; it is more commonly used as a seasoning in food, in water softening, and in de-icing roads.

Answer

$(a,b)$

Solubility of sulphates of alkaline earth metals in water decreases from $Be$ to $Ba\cdot BeS{O}_4$ are fairly soluble while $BaS{O}_4$ is almost completely insoluble.

The decreasing solubility of $BeS{O}_4$ to $BaS{O}_4$ can be explained on the basis of decreasing hydration energy from $B{e}^{2+}$ to $B{a}^{2+}$ (as size increases). For $BeS{O}_4$ and $MgS{O}_4$, hydration energy is more than lattice energy and so they are readily soluble.

• (c) ${\mathrm{BaSO}}_4$: This compound is almost completely insoluble in water. The solubility of sulphates of alkaline earth metals decreases from Be to Ba, and for ${\mathrm{BaSO}}_4$, the lattice energy is much higher than the hydration energy, making it insoluble.

• (d) ${\mathrm{SrSO}}_4$: This compound is also not readily soluble in water. Similar to ${\mathrm{BaSO}}_4$, the solubility decreases down the group, and for ${\mathrm{SrSO}}_4$, the lattice energy is higher than the hydration energy, resulting in low solubility.

Answer

$(b,c)$

To make hard water soft, zeolite method is used. Sodium zeolite or sodium alumino silicate $(N{a}_{2}A{l}_{2}Si{O}_3\cdot x{H}_{2}O)$ is used. It has a unique property of exchanging cations such as $C{a}^2+$ and $M{g}^2+$ in hard water for sodium ion.

• (a) ${\mathrm{H}}^{+}$ ions: Zeolite does not exchange sodium ions with hydrogen ions (${\mathrm{H}}^{+}$) because the primary purpose of zeolite in water softening is to remove hardness caused by divalent cations like ${\mathrm{Ca}}^{2+}$ and ${\mathrm{Mg}}^{2+}$. Hydrogen ions do not contribute to water hardness.

• (d) ${\mathrm{SO}}_4^{2-}$ ions: Zeolite does not exchange sodium ions with sulfate ions (${\mathrm{SO}}_4^{2-}$) because sulfate ions are anions, whereas zeolite is designed to exchange cations. The process specifically targets the removal of cations like ${\mathrm{Ca}}^{2+}$ and ${\mathrm{Mg}}^{2+}$ that cause water hardness.

Answer

$(a,c)$

All the chlorides of alkaline earth metals are hydrated to different extent and extent of hydration decreases from $Mg$ to $Ba$ e.g., $MgC{l}_2\cdot 6H_{2}O,CaC{l}_2\cdot 6H_{2}O,BaC{l}_2\cdot 2H_{2}O,SrC{l}_2\cdot 2H_{2}O$.

• Option (b) $BaC{l}_2\cdot 4H_{2}O$: This option is incorrect because barium chloride typically forms a dihydrate ($BaC{l}_2\cdot 2H_{2}O$) rather than a tetrahydrate. The extent of hydration decreases from magnesium to barium, and barium chloride does not commonly form a tetrahydrate.

• Option (d) $SrC{l}_2\cdot 4H_{2}O$: This option is incorrect because strontium chloride typically forms a dihydrate ($SrC{l}_2\cdot 2H_{2}O$) rather than a tetrahydrate. Similar to barium chloride, the extent of hydration for strontium chloride is less than four water molecules.

Answer

$(a,b)$

Due to diagonal relationship, beryllium is similar to aluminium. It also forms an oxide film which is very stable on the surface of the metal.

Beryllium sulphate is soluble in water due to high hydration energy. Beryllium does not exhibit coordination number more than four. Like $A{l}_2{O}_3$, $BeO$ is amphoteric in nature.

Note: The anomalous behaviour of Be is mainly due to its very small size and partly due to its high electronegativity. These two factors increase the polarising power of ${\mathrm{Be}}^{2+}$ ions to such extent that it becomes significantly equal to the polarising power of ${\mathrm{Al}}^{3+}$ ions. Therefore, these two elements, $\mathrm{Be}$ and $\mathrm{Al}$ resemble (diagonal relationship) very much.

$$\text{ Polarising power }=\frac{\text{ lonic chgarge }}{(\text{ lonic radii }{)}^2}$$

• (c) Beryllium exhibits coordination number more than four: This statement is incorrect because beryllium typically does not exhibit a coordination number greater than four. Due to its small size and the resulting high charge density, beryllium can only accommodate four ligands around it, leading to a maximum coordination number of four.

• (d) Beryllium oxide is purely acidic in nature: This statement is incorrect because beryllium oxide (BeO) is not purely acidic; it is amphoteric in nature. This means that BeO can react with both acids and bases, similar to aluminum oxide (Al₂O₃).

Answer

$(a,b)$

Although Li exhibits, the characteristic properties of alkali metals but it dffers at same time in many of its properties from alkali metals.

The anomalous behaviour of lithium is due to extremely small size of lithium and its cation. On account of small size and high nuclear charge, lithium exerts the greatest polarising effect out of all alkali metals on negative ions.

• (c) It has high degree of hydration: While lithium does have a high degree of hydration due to its small size and high charge density, this is not a primary reason for its anomalous behavior compared to other alkali metals. The high degree of hydration is a consequence of its small size and high polarizing power, rather than a direct cause of its anomalous properties.

• (d) Exceptionally low ionisation enthalpy: Lithium does not have an exceptionally low ionization enthalpy. In fact, it has a higher ionization enthalpy compared to other alkali metals due to its small atomic size and high effective nuclear charge. This higher ionization enthalpy is not a reason for its anomalous behavior.

Answer

Strong reducing power of lithium in aqueous solution can be understood in terms of electrode potential. Electrode potential is a measure of the tendency of an element to lose electrons in the aqueous solution. It mainly depends upon the following three factors i.e.,

(i) $\mathrm{Li}(\mathrm{s})\underset{\text{ Enthalpy }}{\overset{\text{ Sublimation }}{\to }}\mathrm{Li}(g)$

(ii) $\mathrm{Li}(g)\underset{\text{ Enthalpy }}{\overset{\text{ lonisation }}{\to }}{\mathrm{Li}}^{+}(g)+{\mathrm{e}}^{-}$

(iii) ${\mathrm{Li}}^{+}(g)+\mathrm{aq}⟶{\mathrm{Li}}^{+}(\mathrm{aq})+$ enthalpy of hydration

With the small size of its ion, lithium has the highest hydration enthalpy. However, ionisation enthalpy of $\mathrm{Li}$ is highest among alkali metals but hydration enthalpy predominates over ionisation enthalpy.

Therefore, lithium is the strongest reducing agent in aqueous solution mainly because of its high enthalpy of hydration.

Answer

The reactivity of alkali metals towards oxygen increases on moving down the group with the increase in atomic size. Thus, $Li$ forms only lithium oxide $\left(L{i}_{2}O\right)$, sodium forms mainly sodium peroxide $N{a}_2{O}_2$ along with a small amount of sodium oxide while potassium forms only potassium superoxide $\left(K{O}_2\right)$.

$4Li+O_2\stackrel{\mathrm{\Delta }}{\to }2L{i}_{2}O$

$\begin{array}{r}6\mathrm{Na}+2{\mathrm{O}}_2\stackrel{\mathrm{\Delta }}{\to }\underset{\begin{array}{c}\text{ Sodium peroxide }\\ \text{ (major) }\end{array}}{{\mathrm{Na}}_2{\mathrm{O}}_2}+\underset{\begin{array}{c}\text{ Monoxide }\\ \text{ (minor) }\end{array}}{2{\mathrm{Na}}_2\mathrm{O}}\\ \mathrm{K}+{\mathrm{O}}_2\stackrel{\mathrm{\Delta }}{\to }\underset{\begin{array}{c}\text{ Potassium }\\ \text{ super oxides }\end{array}}{{\mathrm{KO}}_2}+\underset{\text{ Peroxide }}{{\mathrm{K}}_2{\mathrm{O}}_2}+\underset{\text{ Monoxide }}{{\mathrm{K}}_2(\mathrm{O})}\end{array}$

The superoxide, $O_2^{-}$ion is stable only in presence of large cations such as $K,Rb$ etc.

Answer

${\mathrm{O}}_2{}^{2-}$ represents a peroxide ion

${\mathrm{O}}_2^{-}$represents a superoxide ion

(i) Peroxide ion react with water to form $H_2{O}_2$

$$O_2^{2-}+2H_{2}O⟶2O{H}^{-}+H_2{O}_2$$

(ii) Superoxide ion react with water to form $H_2{O}_2$ and $O_2$

$$2O_2^{-}+2H_{2}O⟶2O{H}^{-}+\underset{\begin{array}{c}\text{ Hydrogen }\\ \text{ peroxide }\end{array}}{H_2{O}_2}+O_2$$

Answer

Lithium resembles with magnesium as its charge size ratio is closer to $\mathrm{Mg}$. Its resemblance with $\mathrm{Mg}$ is known as diagonal relationship.

Generally, the periodic properties show either increasing or decreasing trend along the group and vice-versa along the period which brought the diagonally situated elements to closer value.

Following characteristics can be noted

(i) Due to covalent nature, chlorides of $\mathrm{Li}$ and $\mathrm{Mg}$ are deliquescent and soluble in alcohol and pyridine.

(ii) Carbonates of $\mathrm{Li}$ and $\mathrm{Mg}$ decompose on heating and liberate ${\mathrm{CO}}_2$

$$\begin{array}{rl}L{i}_{2}C{O}_3& ⟶L{i}_{2}O+C{O}_2\\ MgC{O}_3& ⟶MgO+C{O}_2\end{array}$$

Answer

An element from group 2 which forms an amphoteric oxide and a water soluble sulphate is beryllium.

Beryllium forms oxides of formula $\mathrm{BeO}$. All other alkaline earth metal oxides are basic in nature. $\mathrm{BeO}$ is amphoteric in nature i.e., it reacts with acids and bases both.

$$\begin{array}{c}A{l}_2{O}_3+2NaOH⟶2NaAl{O}_2+H_{2}O\\ A{l}_2{O}_3+6HCl⟶2AlC{l}_3+3H_{2}O\end{array}$$

Sulphate of beryllium is a white solid which crystallises as hydrated salts $(BeS{O}_4\cdot 4H_{2}O)$.

$BeS{O}_4$ is fairly soluble in water due to highest hydration energy in the group (small size). For $BeS{O}_4$, hydration energy is more than lattice energy and so, they are readily soluble.

Answer

(i) All the alkaline earth melals form carbonates $\left(MC{O}_3\right)$. All these carbonates decompose on heating to give $C{O}_2$ and metal oxide. The thermal stability of these carbonates increases down the group i.e., from $Be$ to $Ba$.

$$BeC{O}_3<MgC{O}_3<CaC{O}_3<SC{O}_3<BaC{O}_3$$

$BeC{O}_3$ is unstable to the extent that it is stable only in atmosphere of $C{O}_2$. These carbonates however show reversible decomposition in closed container.

$$BeC{O}_3\rightleftharpoons BeO+C{O}_2$$

Hence, more is the stability of oxide formed, less will be stability of carbonates. Stability of oxides decreases down the group is beryllium oxide i.e., high stable making ${\mathrm{BeCO}}_3$ unstable.

(ii) All the alkaline earth metals form oxides of formula $MO$. The oxides are very stable due to high lattice energy and are used as refractory material.

Except $\mathrm{BeO}$ (predominantly covalent) all other oxides are ionic and their lattice energy decreases as the size of cation increases.

The oxides are basic and basic nature increases from $\mathrm{BeO}$ to $\mathrm{BaO}$ (due to increasing ionic nature).

$\underset{\text{Amphoteric }}{\underbrace{\mathrm{BeO}<}}\underset{\text{Weak basic }}{\underbrace{\mathrm{MgO}<}}\underset{\text{Strong basic }}{\underbrace{\mathrm{CaO}<\mathrm{SrO}<\mathrm{BaO}}}$

$\mathrm{BeO}$ dissolves both in acid and alkalies to give salts and is amphoteric

The oxides of the alkaline earth metals (except $\mathrm{BeO}$ and $\mathrm{MgO}$ ) dissolve in water to form basic hydroxides and evolve a large amount of heat. $\mathrm{BeO}$ and $\mathrm{MgO}$ possess high lattice energy and thus insoluble in water.

Answer

The lattice energy of alkaline earth metal sulphates is almost constant due to large size of sulphate ion. Thus, their solubility is decided by hydration energy which decreases on moving down the group.

The greater hydration enthalpies of ${\mathrm{Be}}^{2+}$ and ${\mathrm{Mg}}^{2+}$ ions overcome the lattice enthalpy factor and therefore, their sulphates are soluble in water.

However, hydration enthalpy is low for ${\mathrm{Ca}}^{2+},{\mathrm{Sr}}^{2+}$ ions and cannot overcome the lattice energy factor. Hence, these are insoluble.

Answer

Smallest size of ${\mathrm{Li}}^{+}$ion among all alkali metals and its high polarising power are the two factors which develop covalent character in the lithium compounds (Fajan’s rule). Compounds of other alkali metals are ionic in nature. So, they are soluble in water.

Since lithium compounds being relatively covalent are soluble in alcohol and other organic solvents in accordance with “like dissolve like”.

Answer

$No,{\left(N{H}_4\right)}_{2}C{O}_3$ reacts with NaCl as

$(N{H}_4{)}_{2}C{O}_3+2NaCl\leftrightarrow N{a}_{2}C{O}_3+2N{H}_{4}Cl$

Because the products obtained $N{a}_{2}C{O}_3$ and $N{H}_{4}Cl$ are highly soluble and the equilibrium will not shift in forward direction.

That’s why in the Solvay process, we cannot obtain sodium carbonate directly by treating the solution containing ${\left(N{H}_4\right)}_{2}C{O}_3$ with sodium chloride.

Answer

The Lewis structure of ${\mathrm{O}}_2^{-}$is $:{\mathrm{O}}^{-}-{\ddot{O}}^{-}:$

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Oxygen atom carrying no charge has six electrons, so its oxidation number is zero. But oxygen atom carrying -1 charge has 7 electrons, so its oxidation number is -1 .

Average oxidation number of each oxygen atom $=\frac{1}{2}$

$$\begin{array}{rl}{\mathrm{O}}_2^{-}& =2x=-1\\ x& =-\frac{1}{2}\end{array}$$

Answer

All alkaline earth metals (except $\mathrm{Be}$ and $\mathrm{Mg}$ ) impart a characteristic colour to the Bunsen flame. The different colours arise due to different energies required for electronic excitation and de-excitation.

$\mathrm{Be}$ and $\mathrm{Mg}$ atoms due to their small size, bind their electrons more strongly (because of higher effective nuclear charge). Hence, require high excitation energy and are not excited by the energy of the flame with the result that no flame colour is shown by them.

Answer

Beryllium chloride has different structures in solid and vapour state. In solid state, it exists in the form of polymeric chain structure in which each Be-atom is surrounded by four chlorine atoms having two of the chlorine atoms covalently bonded while the other two by coordinate bonds. The resulting bridge structure contains infinite chains.

Structure of ${\mathrm{BeCl}}_2$ in solid state

In vapour state, above $1200\mathrm{K}$, it exists as a monomer having linear structure and zero dipole moment. But below $1200\mathrm{K}$, it exists as dimer structure even in vapour state.

$\mathrm{Cl}-\mathrm{Be}-\mathrm{Cl}$

Above $1200\mathrm{K}$

Monomer

Answer

A. $\to (3)$

B. $\to (2)$

C. $\to (4)$

D. $\to (5)$

A. Li-Most negative $E^{-}$among alkali metals [Due to very high hydration energy the resulting $E^{\ominus }$ is most negative].

B. $\mathrm{Na}$-Strongest monoacidic base [Alkalies are more acidic than alkaline earth metals. $\mathrm{LiOH}$ has covalent character].

C. Ca-insoluble oxalate [Calciuim oxalate is insoluble in water]

D. Ba-Insoluble sulphate

[Hydration energy decreases as size of cation increases].

$6s^2$ outer electronic configuration

${}_{56}\mathrm{Ba}=1s^2,2s^2,2p^6,3s^2,3p^6,3d^{10},4s^2,4p^6,4d^{10},5s^2,5p^6,6s^2]$

Answer

A. $\to (3)$

B. $\to (4)$

C. $\to (2)$

D. $\to 1$

A. ${\mathrm{CaCO}}_3-$ Manufacture of high quality paper

B. $\mathrm{Ca}(\mathrm{OH}{)}_2$ - Used in white washing

C. $\mathrm{CaO}$ - Manufacture of sodium carbonate from caustic soda

D. ${\mathrm{CaSO}}_4-$ Dentistry, ornamental work

Answer

A. $\to (6)$

B. $\to (4)$

C. $\to (2)$

D. $\to$ (3)

E. $\to (5)$

F. $\to 1$

Elements with the characteristic flame colour are as follows

A. Cs - Blue

B. $\mathrm{Na}$ - Yellow

C. $\mathrm{K}$ - Violet

D. Ca-Brick red

E. Sr - Crimson red

F. Ba - Apple green

Flame colours are produced from the movement of the electrons in the metal ions present in the compounds. These movement of electrons (electronic excitation-de-excitation) requires energy.

Each atom has particular energy gap between ground and excited energy level therefore each of these movements involves a specific amount of energy emitted as light energy, and each corresponds to a particular colour. As we know energy gap between ground and excited state energy level increases wavelength of decreases and complemently colouer is observed as a result.

Answer

(a) The thermal stability of carbonates increases down the group. Hence, $L{i}_{2}C{O}_3$ is least stable.

Due to small size of ${\mathrm{Li}}^{+}$, strong polarising power distorts the electron cloud of ${\mathrm{CO}}_3^{2-}$ ion.

High lattice energy of $L{i}_{2}O$ than $L{i}_{2}C{O}_3$ also favours the decomposition of $L{i}_{2}C{O}_3$.

Answer

(a) $\mathrm{BeO}$ is more stable than ${\mathrm{BeCO}}_3$ due to small size and high polarising power of ${\mathrm{Be}}^{2+}$.

As $BeC{O}_3$ is unstable and $BeO$ is more stable thus, when $BeC{O}_3$ is kept in an atmosphere of $C{O}_2$, a reversible process takes place and stability of $BeC{O}_3$ increases.

$BeC{O}_3\rightleftharpoons BeO+C{O}_2$

Answer

Due to low ionisation energy and large atomic size, alkali metals form cation readily and so their compounds are ionic.

Oxides

Due to +1 oxidation state, alkali metals form normal oxides of general formula ${\mathrm{M}}_2\mathrm{O}$.

Only $\mathrm{Li}$ forms normal oxide ${\mathrm{Li}}_2\mathrm{O}$ when heated in air. Other form peroxide and superoxide.

Oxides of alkali metals are strongly basic and are soluble in water.

The basic character of oxide increases gradually from $L{i}_{2}O$ to $C{s}_{2}O$ due to increased ionic character.

Halides

Except lithium halides all other alkali metal halides are ionic. Due to high polarising power of ${\mathrm{Li}}^{+}$. Lithium halide is covalent in nature. Due to +1 oxidation states alkali metal halides have general formula MX. Low ionisation enthalpy allows formation of ionic halides.

Oxo salts

All alkali metals form solid carbonates of general formula $M_{2}C{O}_3$. Carbonates are stable except $L{i}_{2}C{O}_3$ due to high polarising capacity of $L{i}^{+}$which is unstable and decomposes.

All the alkali metals (except Li) form solid bicarbonates $MHC{O}_3$. All alkali metals form nitrates having formula $MN{O}_3$. They are colourless, water soluble, electrovalent compounds.

Answer

(a) Alkaline earth metals form compounds which are predominantly ionic but less ionic than the corresponding compounds of alkali metals due to increased nuclear charge and small size.

(b) Oxides of alkaline earth metals are less basic than corresponding oxides of alkali metals. The oxides dissolve in water to form basic hydroxides and evolve a large amount of heat. The alkaline earth metal hydroxides, are however less basic and less stable than alkali metal hydroxides.

(c) Alkaline earth metals form oxoacids as alkali metals. The formation of alkali metal oxoacids is much more faster and stronger than their corresponding alkaline earth metals due to increased nuclear charge and small size.

(d) Solubility of alkaline oxoacids is more than alkali oxoacids because alkaline earth metals have small size of cation and higher hydration energy. Salts like ${\mathrm{CaCO}}_3$ are insoluble in water.

(e) Oxosalts of alkali metals are thermally more stable than those of alkaline earth metals. As the electropositive character increases down the group, the stability of carbonate and hydrogen carbonates of alkali metal increases.

Whereas for alkaline earth metals, carbonate decomposes on heating to give carbon dioxide and oxygen.

Answer

(a) The reaction that takes place when alkali metal is dissolved in liquid ammonia is

$$M+(x+y)N{H}_3⟶{\left[M{\left(N{H}_3\right)}_x\right]}^{+}+[{(N{H}_3)y]}^{-}e$$

The blue colour of the solution is due to the presence of ammoniated electron which absorb enercy in the visible rencinn nf linht and thıs, imnart hlue colour to the solution.

(b) In concentrated solution, the blue colour changes to bronze colour due to the formation of metal ion clusters. The blue solution on keeping for some time liberate hydrogen slowly with the formation of amide.

$$M^{+}+e^{-}⟶MN{H}_2+1/2H-2$$

Answer

The stability of peroxide or superoxide increases as the size of metal ion increases i.e.,

$$K{O}_2<Rb{O}_2<Cs{O}_2$$

The reactivity of alkali metals toward oxygen to form different oxides is due to strong positive field around each alkali metal cation. $L{i}^{+}$is the smallest, it does not allow $O^{2-}$ ion to react with $O_2$ further. $N{a}^{+}$is larger than $Li$, its positive field is weaker than $L{i}^{+}$. It cannot prevent the conversion of $O^{2-}$ into $O_2^{2-}$.

The largest $K^{+},R{b}^{+}$and $C{s}^{+}$ions permit $O_2^{2-}$ ion to react with $O_2$ forming superoxide ion $O_2^{-}$.

$\underset{\begin{array}{c}\text{ Oxide }\end{array}}{{\mathrm{O}}_2^{2-}}\stackrel{\frac{1}{2}{\mathrm{O}}_2}{\to }\underset{\text{ Peroxide }}{{\mathrm{O}}^{2-}}\stackrel{{\mathrm{O}}_2}{\to }\underset{\text{ Superoxide }}{2{\mathrm{O}}_2^{-}}$

Futhermore, increased stability of the peroxide or superoxide with increase in the size of metal ion is due to the stabilisation of large anions by larger cations through lattice energy effect.

Answer

Appearance of milkiness on passing $C{O}_2$ in the solution of compound $B$ indicates that compound $B$ is lime water and compound $C$ is $CaC{O}_3$. Since, compound $B$ is obtained by adding $H_{2}O$ to compound $A$, therefore, compound $A$ is quicklime, $CaO$.

The reactions are as follows

(i)

$\underset{\begin{array}{c}\text{Calciun}\\ \text{oxide}\\ \text{(A)}\end{array}}{CaO}+H_{2}o⟶\underset{\begin{array}{c}\text{Lime water}\\ \text{B}\end{array}}{Ca(OH{)}_2}$

$\underset{\text{(B)}}{Ca(OH{)}_2}+C{O}_2⟶\underset{\begin{array}{c}\text{Calcium carbonate}\\ \text{(Milkiness)}\end{array}}{CaC{O}_3}+\underset{\text{(C)}}{H_{2}O}$

(iii) When excess of ${\mathrm{CO}}_2$ is passed, milkiness disappears due to the formation of soluble calcium bicarbonate $(D)$.

$$\underset{(c)}{\underset{Milkiness}{CaC{O}_3}}+C{O}_2+H_{2}O⟶\underset{(D)}{\underset{(Solublein{H}_{2}O)}{\underset{Calciumbicarbonate}{Ca(HC{O}_3{)}_2}}}$$

Answer

$Be{H}_2$ can be prepared from the corresponding halides by the reduction with complex alkali metal hydrides such as lithium aluminium hydride $LiAl{H}_4$.

$$8LiH+A{l}_{2}C{l}_6⟶2LiAl{H}_4+6LiCl$$

$$2BeC{l}_2+LiAl{H}_4⟶2Be{H}_2+LiCl+AlC{l}_2$$

Answer

The alkaline earth metals burn in oxygen to form monoxide MO. BeO is essentially covalent in nature, other being ionic in nature.

$\mathrm{BeO}$ is amphoteric while other oxides are basic in nature and react with water to form sparingly soluble hydroxides.

$\mathrm{BeO}$ dissolves both in acid and alkalis to give salt and is amphoteric.

$$BeO+H_{2}O⟶\underset{\begin{array}{c}\text{ Beryllium }\\ \text{ hydroxide }\end{array}}{Be(OH{)}_2}$$

$\mathrm{Be}(\mathrm{OH}{)}_2$ is an amphoteric hydroxide, dissolving in both acids and alkalies.

With alkalies it dissolves to form the tetrahydroxidoberyllate $\left(Z^{-}\right)$anion with sodium hydroxide solution.

$$2NaOH(aq)+Be(OH{)}_2(s)⟶\underset{\text{ Sodium tetra hydroxidoberyllate }}{N{a}_{2}Be(OH{)}_4(aq)}$$

With acids, it forms beryllium salts.

$$Be(OH{)}_2+\underset{\begin{array}{c}\text{ Sulphuric }\\ \text{ acid }\end{array}}{H_{2}S{O}_4}⟶\underset{\begin{array}{c}\text{ Beryllium }\\ \text{ sulphate }\end{array}}{BeS{O}_4}+2H_{2}O$$

Answer

Yellow colour flame in flame test indicates that the alkali metal must be sodium. It reacts with $O_2$ to form a mixture of sodium peroxide, $N{a}_2{O}_2$ and sodium oxide $N{a}_{2}O$.

$$\begin{array}{c}4Na+O_2\stackrel{\mathrm{\Delta }}{⟶}2N{a}_{2}O\text{ (Minor) }\\ 2N{a}_{2}O+O_2\stackrel{\mathrm{\Delta }}{⟶}2N{a}_2{O}_2\text{ (Major) }\\ 2Na+O_2\stackrel{\mathrm{\Delta }}{⟶}N{a}_2{O}_2\end{array}$$

Ionisation enthalpy of sodium is low. When sodium metal or its salt is heated in Bunsen flame, the flame energy causes an excitation of the outermost electron which on reverting back to its initial position gives out the absorbed energy as visible light and complementary colour of absorbed colour from the light radiation is seen.

That’s why sodium imparts yellow colour to the flame.