Answer:(d) Cs

Explanation:

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., $29^{\circ} \mathrm{C}$.

Now, consider the incorrect options:

(a) Na: Sodium (Na) has a melting point of approximately $ 98^{\circ}C $ , which is significantly higher than $ 30^{\circ}C $ . Therefore, it will not melt at room temperature.

(b) K: Potassium (K) has a melting point of approximately $ 63^{\circ}C $ , which is also higher than $ 30^{\circ}C $ . Hence, it will not melt at room temperature.

(c) Rb: Rubidium (Rb) has a melting point of approximately $ 39^{\circ}C $ , which is slightly higher than $ 30^{\circ}C $ . Thus, it will not melt at room temperature.

Answer:(a) $Li$

Explanation:

Both m.p. and heat of the reaction of alkali metals with $ \mathrm{H}_2 \mathrm{O}$ decreases down the group from Li to Cs .

Although the heat of reaction of Li is the highest but due to its high melting point, even the heat is not sufficient to melt the metal thereby exposing greater surface to water for reaction. As a result, Li has the least reactivity by the reactivity increases as the m.p. of the alkalimetals decrease down the group from Li to Cs .

Now, consider the incorrect options:

(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) Hydration enthalpy

Explanation:

With the small size of the $Li^+$ ion , lithium has the highest hydration enthalpy which accounts for its high negative $E^\circ$ value and its high reducing power.

Now, consider the incorrect options:

(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.

Answer:(d) $ \mathrm{BaCO}_{3}$

Explanation:

$BaCO_{3}$ 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 $BaCO_{3}$ ).

Therefore, the increasing size of cation destabilizes the oxides and hence does not favour the decomposition of heavier alkaline earth metal carbonates like $BaCO_{3}$.

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

Now, consider the incorrect options:

(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) $ \mathrm{BeCO}_{3}$

Explanation:

Metal carbonates generally decompose upon heating to form metal oxides and carbon dioxide $\left(\mathrm{CO}_2\right)$. The general reaction can be represented as:

$ \mathrm{MCO}_3 \rightarrow \mathrm{MO}+\mathrm{CO}_2 $

Here, M represents the metal.

Beryllium carbonate is known to be unstable in air. It decomposes easily into beryllium oxide (BeO) and carbon dioxide $\left(\mathrm{CO}_2\right)$.

The decomposition reaction is:

$ \mathrm{BeCO}_3 \rightarrow \mathrm{BeO}+\mathrm{CO}_2$

The carbonate ion $\left(\mathrm{CO}_3{ }^{2-}\right)$ is relatively large compared to the small beryllium ion $\left(\mathrm{Be}^{2+}\right)$, leading to instability in the compound.

Beryllium Carbonate ( $\mathbf{B e C O}_{\mathbf{3}}$ ) is the only one that is unstable in air and is kept in a $ \mathrm{CO}_2 $ atmosphere to avoid decomposition. Therefore, the answer is:

Now, consider the incorrect options:

(b) Magnesium Carbonate $(MgCO_3)$:

Magnesium carbonate is more stable and does not decompose as readily as beryllium carbonate.

(c) Calcium Carbonate ($ \mathrm{CaCO}_3$ ):

Calcium carbonate is also stable and is commonly found in nature (e.g., limestone).

(d) Barium Carbonate ($ \mathrm{BaCO}_{\mathbf{3}}$ ):

Barium carbonate is relatively stable as well and does not require special conditions to prevent decomposition.

Answer:(a) $ \mathrm{Mg}(\mathrm{OH})_{2}$

Explanation:

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.

$ \Delta H_{\text {solution }}=\Delta H_{\text {lattice energy }}+\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 $\Delta H_{\text {solution }}$ down the group.

More negative $\Delta H_{\text {solution }}$ more is solubility of compounds.

Hence, $Be(OH)_2$ and $ Mg(OH)2$ have less negative values for $\Delta H{\text{solution}}$ hence, least basic.

Now, consider the incorrect options:

(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) $ \mathrm{BeCl}_{2}$

Explanation:

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.

Now, consider the incorrect options:

(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.

Answer:(c) $ \mathrm{Li}>\mathrm{Na}>\mathrm{K}>\mathrm{Rb}$

Explanation:

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.

Hence, Ionization enthalpy decreases as the atomic size, increases, i.e., $L i>N a>K>R b$.

Now, consider the incorrect options:

(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.

(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.

(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) high lattice enthalpy

Explanation:

Solubilities of alkali metal halides in water can be explained in terms of lattice enthalpy and hydration ethalpies favour dissolution. Among fluorides, the order of solubility is LiF $<N a F<K F<R b F<C s F$. Low solubility of 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.

Now, consider the incorrect options:

(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) $ \mathrm{Be}(\mathrm{OH})_{2}$

Explanation:

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 +2 HCl \longrightarrow BeCl_2+2 H_2 O $

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

$ Be(OH)_2+2 NaOH \longrightarrow Na_2 BeO_2+2 H_2 O $

Now, consider the incorrect options:

(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) $ \mathrm{CaCl}_{2}$

Explanation:

Sodium carbonate is synthesized by the Solvay ammonia soda process. In this process the following reactions are involved. These are

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

$ \mathrm{NaCl}+\mathrm{NH}_4 \mathrm{HCO}_3 \rightarrow \mathrm{NaHCO}_3 \downarrow+\mathrm{NH}_4 \mathrm{Cl}$

On heating Sodium bicarbonate, we get sodium carbonate along with water, and carbon dioxide gas is formed.

$2 \mathrm{NaHCO}_3 \rightarrow \mathrm{Na}_2 \mathrm{CO}_3+\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2$

Again on heating ammonium carbonate, ammonia gas is evolved.

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

The ammonia regenerated in the Solvay process by reacting the by-product $ \mathrm{NH}_4 \mathrm{Cl}$ with slaked lime $ \mathrm{Ca}(\mathrm{OH})_2$

$2 \mathrm{NH}_4 \mathrm{Cl}+\mathrm{Ca}(\mathrm{OH})_2 \rightarrow 2 \mathrm{NH}_3+\mathrm{CaCl}_2+2 \mathrm{H}_2 \mathrm{O}$

Here, $ \mathrm{CaCl}_2$ is by-product.

Now, consider the incorrect options:

(b) NaCl: Sodium chloride $ (NaCl) $ is not a by-product of the reaction between ammonium chloride $ (NH_4Cl) $ and calcium hydroxide $ (Ca(OH)_2)$ . Instead, NaCl is a reactant in the Solvay process, where it reacts with ammonium bicarbonate $ (NH_4HCO_3) $ to form sodium bicarbonate $ (NaHCO_3) $ and ammonium chloride $ (NH_4Cl) $ .

(c) NaOH: Sodium hydroxide (NaOH) is not produced in the reaction between ammonium chloride $ (NH_4Cl) $ and calcium hydroxide $ (Ca(OH)_2)$ . The reaction specifically produces ammonia $ (NH_3) $ , calcium chloride $ (CaCl_2) $ , and water $ (H_2O) $.

(d) $ NaHCO_3 $: Sodium bicarbonate $ (NaHCO_3) $ is not a by-product of the reaction between ammonium chloride $ (NH_4Cl) $ and calcium hydroxide $ (Ca(OH)_2) $ . $ NaHCO_3 $ is actually an intermediate product in the Solvay process, formed from the reaction of NaCl and $ NH_4HCO_3 $ .

Answer:(a) ammoniated electron

Explanation:

All alkali metal dissolve in liquid $ \mathrm{NH}_{3}$ giving highly conducting deep blue solution.

$ \mathrm{Na}+(x+y) \mathrm{NH}_3 \longrightarrow \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.

Now, consider the incorrect options:

(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) setting time of cement increases

Explanation:

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

Gypsum 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.

Now, consider the incorrect options:

(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) $CaSO_{4}$

Explanation:

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

$ \underset{\text { Gypsum }}{2 CaSO_4 \cdot 2 H_2 O} \longrightarrow \underset{\text { Plaster of Paris }}{\left(CaSO_{4}\right) \cdot \frac{1}{2} H_2 O+3 H_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.

$ CaSO_ {4} \cdot \frac{1}{2} H_ {2} O \xrightarrow{200 ^\circ C} \underset{Anhydride}{CaSO_ {4}} \xrightarrow{1100 ^\circ \text{C}} CaO+ SO_{3} $

Now, consider the incorrect options:

(b) $CaSO_{4} \cdot \frac{1}{2} H_{2} O$: 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) $CaSO_{4} \cdot H_{2} O$: This is not a common form of calcium sulfate used in plaster. The common forms are gypsum ($CaSO_{4} \cdot 2 H_{2} O$) and Plaster of Paris ($CaSO_{4} \cdot \frac{1}{2} H_{2} O$).

(d) $CaSO_{4} \cdot 2 H_{2} O$: This is gypsum, the raw material used to produce Plaster of Paris. It is not dead burnt plaster, which is anhydrous calcium sulfate ($CaSO_{4}$).

Answer:(c) milk of lime

Explanation:

Calcium hydroxide is prepared by adding water to quicklime(CaO).

$ \mathrm{CaO}(\mathrm{s})+\mathrm{H}_2 \mathrm{O}(\mathrm{I}) \rightarrow \mathrm{Ca}(\mathrm{OH})_2(\mathrm{~s})$

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

It forms a suspension of slaked lime in water which is called milk of lime and the clear solution obtained

The aqueous solution is known as lime water and a suspension of slaked lime in water is known as milk of lime.

Now, consider the incorrect options:

(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) Be

Explanation:

Except Be, all alkaline earth metals form hydrides $\left(MH_{2}\right)$ on directly heating with $H_{2}$. $BeH_{2}$ can’t be prepared by direct action of $H_{2}$ on $Be$. It is prepared by the action of $Li AlH_{4}$ on $BeCl_{2}$.

$ 2 BeCl_{2}+LiAlH_{4} \longrightarrow 2 BeH_{2}+LiCl+AlCl_{3} $

Now, consider the incorrect options:

(b) Mg: Magnesium (Mg) forms magnesium hydride $(MgH_2)$ by direct heating with dihydrogen $(H_2)$.

(c) Sr: Strontium (Sr) forms strontium hydride $(SrH_2)$ by direct heating with dihydrogen $(H_2)$.

(d) Ba: Barium (Ba) forms barium hydride $(BaH_2)$ by direct heating with dihydrogen $(H_2)$.

Answer:(d) $Na_{2} CO_{3}$

Explanation:

Sodium carbonate is a white crystalline solid which exists as a decahydrate, $Na_2 CO_3. 10H_2 O$. This is also called washing soda. It is really soluble in water.

On heating, the decahydrate loses its water of crystallisation to form monohydrate.

Above 373k, the monohydrate becomes completely anhydrous and changes to a white powder called soda ash $(Na_2 CO_3)$.

Now, consider the incorrect options:

(a) $Na_{2}CO_{3} \cdot 10 H_{2}O$: 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) $Na_{2}CO_{3} \cdot 2 H_{2}O$: 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) $Na_{2}CO_{3} \cdot H_{2}O$: 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 nitrate

Explanation:

Calcium gives brick red coloured flame. Hence, calcium nitrate on heating decomposes into calcium oxide, with evolution of a mixture of $NO_{2}$ and $O_{2}$.

$ 2 Ca \left(NO_3 \right)_{2} \longrightarrow 2 CaO+NO_2+O_2 $

$NO_{2}$ is brown coloured gas.

Now, consider the incorrect options:

(a) 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.

(c) 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.

(d) 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) It is used in the preparation of bleaching powder

Explanation:

Calcium hydroxide is used in the manufacture of bleaching powder.

$ \underset{\substack{\text { Slaked } \\ \text { lime }}}{2 Ca(OH)_2}+2 Cl_2 \xrightarrow{\text { Cold }} CaCl_2+\underset{\substack{\text { Bleaching } \\ \text { powder }}}{(CaOCl)_2}+2 H_2 O $

Now, consider the incorrect options:

(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) $ \mathrm{Ca}(\mathrm{OH})_{2}$

Explanation:

To recover $NH_{3}$ in Solvay ammonia soda process $Ca(OH)_{2}$ is used.

$ 2 NH_4 Cl+Ca(OH)_2 \longrightarrow 2 NH_3 + CaCl_2 + 2 H_2 O $

On passing $CO_{2}$ through $Ca(OH)_2$, it turns milky due to the formation of $CaCO_3$.

$ Ca(OH)_2+CO_2 \longrightarrow CaCO_3 \downarrow + H_2 O $

$ \mathrm{Ca}(\mathrm{OH})_{2}$ is used for white washing.

Now, consider the incorrect options:

(a) $Ca \left(HCO_{3}\right)_{2}$: This compound is calcium bicarbonate, which is not used in the Solvay process for recovering ammonia. Additionally, it does not turn milky when $CO_2$ is bubbled through its solution.

(b) $ \mathrm{CaO}$: Calcium oxide, also known as quicklime, does not directly react with $CO_2$ to form a milky solution. It first needs to react with water to form $ \mathrm{Ca(OH)_2}$, which then reacts with $CO_2$ to form the milky suspension of $ \mathrm{CaCO_3}$.

(d) $ \mathrm{CaCO}_{3}$: Calcium carbonate is the product formed when $CO_2$ is bubbled through a solution of $ \mathrm{Ca(OH)_2}$. It is not used in the Solvay process to recover ammonia and does not turn milky upon further reaction with $CO_2$.

Answer:(d) All of the above

Explanation:

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., $MgCl_2 \cdot 6 H_2 O, CaCl_2 \cdot 6 H_2 O$, $BaCl_2 \cdot 2 H_2 O, SrCl_2 \cdot 2 H_2 O$

(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 $MgCl_2 \cdot 6 H_2 O$, $CaCl_2 \cdot 6 H_2 O$, $SrCl_2 \cdot 2 H_2 O$, $BaCl_2 \cdot 2 H_2 O$.

Answer:(b, d)

Explanation:

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 ns electronic configuration, they lose electron easily and have high negative standard electrode potential.

Now, consider the incorrect options:

(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)

Explanation:

$Na_{2} CO_{3}$ is used in manufacture of soap powders, in laundry for washing.

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

Now, consider the incorrect options:

(b) $ \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.

(d) $ \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)

Explanation:

Solubility of sulphates of alkaline earth metals in water decreases from $Be$ to $Ba \cdot BeSO_4$ are fairly soluble while $BaSO_4$ is almost completely insoluble.

The decreasing solubility of $BeSO_4$ to $BaSO_4$ can be explained on the basis of decreasing hydration energy from $Be^{2+}$ to $Ba^{2+}$ (as size increases). For $BeSO_4$ and $MgSO_4$, hydration energy is more than lattice energy and so they are readily soluble.

Now, consider the incorrect options:

(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)

Explanation:

To make hard water soft, zeolite method is used. Sodium zeolite or sodium alumino silicate $\left(Na_2 Al_2 SiO_3 \cdot x H_2 O\right)$ is used. It has a unique property of exchanging cations such as $Ca^{2+}$ and $Mg^{2+}$ in hard water for sodium ion.

Now, consider the incorrect options:

(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)

Explanation:

All the chlorides of alkaline earth metals are hydrated to different extent and extent of hydration decreases from $Mg$ to $Ba$ e.g., $MgCl_2 \cdot 6 H_2 O, CaCl_2 \cdot 6 H_2 O, BaCl_2 \cdot 2 H_2 O, SrCl_2 \cdot 2 H_2 O$.

Now, consider the incorrect options:

(b) $BaCl_2 \cdot 4 H_2 O$: This option is incorrect because barium chloride typically forms a dihydrate ($BaCl_2 \cdot 2 H_2 O$) rather than a tetrahydrate. The extent of hydration decreases from magnesium to barium, and barium chloride does not commonly form a tetrahydrate.

(d) $SrCl_2 \cdot 4 H_2 O$: This option is incorrect because strontium chloride typically forms a dihydrate ($SrCl_2 \cdot 2 H_2 O$) rather than a tetrahydrate. Similar to barium chloride, the extent of hydration for strontium chloride is less than four water molecules.

Answer:(a, b)

Explanation:

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 $Al_2 O_3$, $BeO$ 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 { Ionic chgarge }}{(\text { Ionic radii })^{2}} $

Now, consider the incorrect options:

(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 $(Al_2O_3)$.

Answer:(a, b)

Explanation:

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.

Now, consider the incorrect options:

(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}) \xrightarrow {}\mathrm{Li}(g) \quad \text{(Sublimation enthalpy)}$

(ii) $ \mathrm{Li}(g) \xrightarrow{} \mathrm{Li}^{+}(g)+\mathrm{e}^{-} \quad \text{( Enthalpy of ionization)}$

(iii) $ \mathrm{Li}^{+}(g)+\mathrm{H_2O} \longrightarrow \mathrm{Li}^{+}(\mathrm{aq}) \quad \text{(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 $\left(Li_2 O\right)$, sodium forms mainly sodium peroxide $(Na_2 O_2)$ along with a small amount of sodium oxide while potassium forms only potassium superoxide $\left(KO_2\right)$.

$ 4 Li+O_2 \xrightarrow\Delta 2 Li_2 O $

$ 6 \mathrm{Na}+2 \mathrm{O}_2 \xrightarrow{\Delta} \underset{\substack{\text { Sodium peroxide } \\ \text { (major) }}}{\mathrm{Na}_2 \mathrm{O}_2}+\underset{\begin{array}{c}\text { Monoxide } \\ \text { (minor) }\end{array}}{2 \mathrm{Na}_2 \mathrm{O}} $

$\mathrm{K}+\mathrm{O}_2 \xrightarrow{\Delta} \underset{\substack{\text { Potassium } \\ \text { super oxides }}}{\mathrm{KO}_2}+\underset{\text { Peroxide }}{\mathrm{K}_2 \mathrm{O}_2}+\underset{\text { Monoxide }}{\mathrm{K}_2(\mathrm{O})}$

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-}+2 H_2 O \longrightarrow 2 OH^-+H_2 O_2 $

(ii) Superoxide ion react with water to form $H_{2} O_{2}$ and $O_{2}$

$ 2 O_{2}^{-}+2 H_{2} O \longrightarrow 2 OH^- +\underset{\substack{\text { Hydrogen } \\ \text { peroxide }}}{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}$

$ Li_{2} CO_{3} \longrightarrow Li_{2} O+CO_{2} $

$ MgCO_{3} \longrightarrow MgO+CO_{2} $

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.

$ Al_{2} O_{3}+2 NaOH \longrightarrow 2 NaAlO_{2}+H_{2} O $

$ Al_{2} O_{3}+6 HCl \longrightarrow 2 AlCl_{3}+3 H_{2} O $

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

$BeSO_{4}$ is fairly soluble in water due to highest hydration energy in the group (small size). For $BeSO_{4}$, hydration energy is more than lattice energy and so, they are readily soluble.

Answer:

(i) All the alkaline earth melals form carbonates $\left(MCO_{3}\right)$. All these carbonates decompose on heating to give $CO_{2}$ and metal oxide. The thermal stability of these carbonates increases down the group i.e., from $Be$ to $Ba$.

$ BeCO_{3}<MgCO_{3}<CaCO_{3}<SrCO_{3}<BaCO_{3} $

$BeCO_{3}$ is unstable to the extent that it is stable only in atmosphere of $CO_{2}$. These carbonates however show reversible decomposition in closed container.

$ BeCO_{3} \rightleftharpoons BeO+CO_{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 $M O$. 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).

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

$ \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 hydration enthalpies of $BeSO_{4}$ and $MgSO_{4}$ are quite high because of small size of $ \mathrm{Be}^{2+}$ and $ \mathrm{Mg}^{2+}$ ions. These hydration enthalpy values are higher than their corresponding lattice enthalpies and therefore , $BeSO_{4}$ and $ \mathrm{Mg}^{2+}$ are highly soluble in water.

However, hydration enthalpies of $CaSO_{4}$, $SrSO_{4}$ and $BaSO_{4}$ are not very high as compared to there respective lattice enthalpies and hence these are insoluble in water.

Answer:

$L i^{+}$ has the smallest size among all the alkali metal ions.

Due to its smaller size it has very high polarising power(polarising power=charge/radius). So lithium compounds are covalent in nature(according to fajan’s rule). Other alkali metals are ionic compounds.

So they are readily soluble in water. But as the lithium compounds are predominantly covalent so they dissolve in organic solvents like alcohol and ether.

Answer:

No. In Solvay process, ammonium hydrogen carbonate is prepared from ammonium carbonate, which reacts with NaCl to form NaHCO .

Due to low solubility of $ \mathrm{NaHCO}_3$, it gets precipitated and decomposes on heating to give $ \mathrm{Na}_2 \mathrm{CO}_3 . \mathrm{Na}_2 \mathrm{CO}_3$ cannot be directly obtained by treating with $\left(\mathrm{NH}_4\right)_2 \mathrm{CO}_3$ and NaCl solution as both the products formed on the reaction $\left(\mathrm{Na}_2 \mathrm{CO}_3, \mathrm{NH}_4 \mathrm{Cl}\right)$ are soluble and the equilibrium will not be shifted in forward direction.

$ \left(\mathrm{NH}_4\right)_2 \mathrm{CO}_3+2 \mathrm{NaCl} \rightleftharpoons \mathrm{Na} a_2 \mathrm{CO}_3+2 \mathrm{NH}_4 \mathrm{Cl} $

Answer:

The Lewis structure $ \mathrm{O}_{2}{ }^{-}$:

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 $=\dfrac{-1}{2}$

$\dfrac {0+(-1)}{2}$= $\dfrac{-1}{2}$

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} \quad (Monomer)$

Answer:

A. $\rightarrow(3)$

B. $\rightarrow(2)$

C. $\rightarrow(4)$

D. $\rightarrow(5,1)$

A. Li- Due to very high hydration enthaply, $E^\circ $ is most negative.

B. $ \mathrm{Na}$-gives NaOH (strong base). One mole of it replaces 1 mol $H^+$ from acid therefore it is the strongest monoacidic base.

C. Hydration energy is very low therefore, calcium oxalate is insoluble.

D. Hydration energy is low as the size of $Ba^{2+}$ is large therefore barium sulphate is insoluble. $6s^2$ is valence shell electronic configuration.

$6 s^{2}$ outer electronic configuration

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

Answer:

A. $\rightarrow(3)$

B. $\rightarrow(4)$

C. $\rightarrow(2)$

D. $\rightarrow (1)$

(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) $\rightarrow(6)$

(B) $\rightarrow(4)$

(C) $\rightarrow(2)$

(D) $\rightarrow (3)$

(E) $\rightarrow(5)$

(F) $\rightarrow (1)$

Elements with the characteristic flame colour are as follows:

(A) Cs - Blue

(B) Na - Yellow

(C) 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) Both $ \mathrm{A}$ and $ \mathrm{R}$ are correct and $ \mathrm{R}$ is the correct explanation of $ \mathrm{A}$

The thermal stability of carbonates increases down the group. Hence, $Li_{2} CO_{3}$ 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 $Li_{2} O$ than $Li_{2} CO_{3}$ also favours the decomposition of $Li_{2} CO_{3}$.

Answer:(a) Both $ \mathrm{A}$ and $ \mathrm{R}$ are correct and $ \mathrm{R}$ is the correct explanation of $ \mathrm{A}$

$ \mathrm{BeO}$ is more stable than $ \mathrm{BeCO}_{3}$ due to small size and high polarising power of $ \mathrm{Be}^{2+}$.

As $BeCO_3$ is unstable and $BeO$ is more stable thus, when $BeCO_3$ is kept in an atmosphere of $CO_2$, a reversible process takes place and stability of $BeCO_3$ increases.

$BeCO_3 \rightleftharpoons BeO+CO_2$

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 $Li_2 O$ to $Cs_2 O$ 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 $M_{2} CO_{3}$. Carbonates are stable except $Li_{2} CO_{3}$ due to high polarising capacity of $Li^{+}$ which is unstable and decomposes.

All the alkali metals (except Li) form solid bicarbonates $MHCO_{3}$. All alkali metals form nitrates having formula $MNO_{3}$. 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) NH_{3} \longrightarrow\left[M\left(NH_{3}\right)_{x}\right]^{+}+\left[\left(NH_3)_y\right]^{-} \right. $

The blue colour of the solution is due to the presence of ammoniated electron which absorb energy in the visible region of light and thus, imparts blue 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^{-}}+{NH_{3}} \rightarrow {MNH_{2}}+\frac {1}{2}H_{2} $

Metal amide is formed as product.

Answer:

As we go down the group the size of alkali metal ion increases and the stability of peroxides and super oxides also increases. This is due to the stabilization of larger anions by larger cations.

As we move from $ \mathrm{Li}^{+}$ to $ \mathrm{Cs}^{+}$ the size of the cations increases. $ \mathrm{Li}^{+}$ is the smallest cation with strong positive field around $i t$. Hence it combines with small anion called oxide ion $\left(\mathrm{O}^{-2}\right)$ with strong negative field and form Lithium oxide.

$ \mathrm{Na}^{+}$ ion is bigger than $ \mathrm{Li}^{+}$ ion with less positive field around it combines with peroxide ion $\left(\mathrm{O}_2{ }^{-2}\right)$ with less negative field and form sodium peroxide.

$ \mathrm{K}^{+}, \mathrm{Rb}^{+}$ and $ \mathrm{Cs}^{+}$ ions are bigger in size with less positive field around them. Hence they can stabilize the bigger superoxide ion $\left(\mathrm{O}_2{ }^{-}\right)$ with less negative field and form super oxides.

Answer:

compound ( $B$ ) is limewater. Now, compound $(B)$ is obtained by adding water to compound $(A)$. Therfore $(A)$ in quicklime (caO).

$ \underset {(A)}{\mathrm{CaO}}+{\mathrm{H}_2} \mathrm{O} \longrightarrow \underset {(B)} {\mathrm{Ca}(\mathrm{OH}})_2 $

$ \underset {(B)}{\mathrm{Ca}(\mathrm{OH})_2}+\mathrm{CO}_2 \longrightarrow \underset {(C)} {\mathrm{CaCO}_3}+\mathrm{H}_2 \mathrm{O} $

Excess of $ \mathrm{CO}_2$ is passed

$ \mathrm{CaCO}_3+\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2 \longrightarrow \underset {(D)} {\mathrm{Ca}\left(\mathrm{HCO}_3\right)_2}$

When $ \mathrm{CO}_2$ is passed in excess, Calcium bicarbonate is formed, which is soluble in water. Hence milkiness disappears.

$ A \rightarrow \mathrm{CaO} $

$ \mathrm{B} \rightarrow \mathrm{Ca}(\mathrm{OH})_2 $

$ \mathrm{C} \rightarrow \mathrm{CaCO}_3 $

$ \mathrm{D} \rightarrow \mathrm{Ca}\left(\mathrm{HCO}_3\right)_2 $

Answer:

Beryllium hydride can be prepared from lithium hydride. $ \mathrm{BeH}_2$ cannot be prepared directly by heating powdered beryllium with dihydrogen.

Lithium hydride is first reacted with anhydrous aluminium chloride to produce lithium aluminium hydride.

$ 8 LiH+Al_{2} Cl_{6} \longrightarrow 2 LiAlH_{4}+6 LiCl $

Beryllium chloride is reacted with $ \mathrm{LixlH}_4$ to give $ \mathrm{BeH}_2$

$ 2BeCl_2 + LiAlH_4 \longrightarrow 2BeH_2 + LiCl + AlCl_3 $

Answer:

In group 2, only beryllium is amphoteric in nature which means it reacts with both acids and bases. Also, beryllium only forms covalent oxide due to the covalent nature. So, the element is beryllium.

It reacts with acid to form beryllium chloride and it reacts with base to form beryllate ion which is soluble in sodium hydroxide. The reaction is shown below.

$Be(OH)_2 + 2 OH^- \rightarrow [Be(OH_4)^{2-}]$

$Be(OH)_2 +2 HCl \rightarrow BeCl_2 + 2 H_2O$

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, $Na_2 O_2$ and sodium oxide $Na_2 O$.

$4 Na+O_2 \stackrel{\Delta}{\longrightarrow} 2 Na_2 O \text { (Minor) } $

$2 Na_2 O +O_2 \stackrel{\Delta}{\longrightarrow} 2 Na_2 O_2 \text { (Major) } $

$2 Na+O_2 \stackrel{\Delta}{\longrightarrow} Na_2 O_2 $

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.