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

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.

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 $L i$ is least $(-3.04 \mathrm{~V})$ among all alkali metals.

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

(i) $\mathrm{Li}(\mathrm{s}) \xrightarrow{\text { Sublimation }} \mathrm{Li}(\mathrm{g}) \Delta \mathrm{H}_{s}=$ Enthalpy of sublimation

(ii) $\mathrm{Li}(g) \longrightarrow \mathrm{Li}^{+}(s) \quad I E_{1}=$ lonisation enthalpy

(iii) $\mathrm{Li}^{+}(g) \longrightarrow \mathrm{Li}^{+}(\mathrm{aq}) \quad \Delta H_{n}=$ 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.

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.

$$ MCO_{3} \rightleftharpoons 1 MO+CO_{2} \quad[M=Be, Mg, Ca, Sr, Ba] $$

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

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

Answer

(d) $BaCO_{3}$ 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 $BaCO_{3}$.

Answer

(a) $BeCO_{3}$ is unstable to the extent that it is stable only in atmosphere of $CO_{2}$. $BeCO_{3}$ shows reversible reaction because stability of oxide formed is more than carbonates.

$$ BeCO_{3} \rightleftharpoons BeO+CO_{2} $$

$BeCO_{3}$ is unstable due to strong polarising effect of small $Be^{2+}$ ion on the large polarisable carbonation. Moreover, an extrastability of the oxide achieved through lattice energy by packing small cation with small oxide ion.

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.

$$ \Delta H_{\text {solution }}=\Delta H_{\text {lattice energy }}+\Delta H_{\text {hydration energy }} $$

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, $Be(OH)_2 \quad and \quad Mg(OH)_2 \text{have less negative values for }$

$\Delta H_{\text{solution} }$ hence, least basic.

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.

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.

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.

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.

$$ Be(OH)_2 +2 HCl \longrightarrow BeCl_2+2 H_2 O $$

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

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

Answer

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

The reactions involved are

$$ \begin{aligned} & NH_{3}+H_{2} O+CO_{2} \underset{\text { Ammonium bicarbonate }}{NH_{4} HCO_{3}} \\ & NaCl+NH_{4} HCO_{3} \underset{\text { Sodium bicarbonate }}{NaHCO_{3} \downarrow+NH_{4} Cl} \\ & 2 NaHCO_{3} \xrightarrow{\Delta} Na_{2} CO_{3}+H_{2} O+CO_{2} \end{aligned} $$

$NH_{3}$ is recovered from $NH_{4} HCO_{3}$ and $NH_{4} Cl$ formed during the reaction.

$\begin{aligned} & \mathrm{NH}_4 \mathrm{HCO}_3 \xrightarrow{\text { Heat }} \mathrm{NH}_3+\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2 \\ & \underset{\text { Ammonium chloride }}{2 \mathrm{NH}_4 \mathrm{Cl}}+\mathrm{Ca}(\mathrm{OH})_2 \longrightarrow \underset{\text { Ammonia }}{2 \mathrm{NH}_3}+\underset{\text { Calcium chloride }}{\mathrm{CaCl}_2}+2 \mathrm{H}_2 \mathrm{O} \\ & \end{aligned}$

Answer

(a) 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.

Answer

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

Gypsum $\left(CaSO_{4} \cdot 5 H_{2} O\right)$ 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.

Answer

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

$$ \underset{\text { Gypsum }}{2 CaSO_4 \cdot 2 H_2 O} \longrightarrow \underset{\text { Plaster of Paris }}{\left(CaSO_{4}\right)_2 \cdot \frac{1}{2} H_2 O+3 H_2 O} $$

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.

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

Answer

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

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

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.

Answer

(a) 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 $BeCl_{2}$.

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

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.

$\begin{aligned} & \mathrm{Na}_2 \mathrm{CO}_3 \cdot 10 \mathrm{H}_2 \mathrm{O} \xrightarrow{>373 \mathrm{~K}} \mathrm{Na}_2 \mathrm{CO}_3 \cdot 10 \mathrm{H}_2 \mathrm{O} \\ & \text { Washing soda } \quad\quad\quad\quad \text { Soda ash }\\ & \end{aligned}$

Answer

(b) 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.

Answer

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

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

Answer

(c) 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.

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

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

Answer

$(a, c)$

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

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

Answer

$(a, b)$

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.

Answer

$(b, c)$

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.

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

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 $Al_2 O_3$, $BeO$ 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.

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

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.

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) $\mathrm{Li}(\mathrm{s}) \xrightarrow[\text { Enthalpy }]{\text { Sublimation }} \mathrm{Li}(g)$

(ii) $\mathrm{Li}(g) \xrightarrow[\text { Enthalpy }]{\text { lonisation }} \mathrm{Li}^{+}(g)+\mathrm{e}^{-}$

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

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

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

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 $H_{2} O_{2}$

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

(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 $\mathrm{CO}_{2}$

$$ \begin{aligned} Li_{2} CO_{3} & \longrightarrow Li_{2} O+CO_{2} \\ MgCO_{3} & \longrightarrow MgO+CO_{2} \end{aligned} $$

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.

$$ \begin{gathered} Al_{2} O_{3}+2 NaOH \longrightarrow 2 NaAlO_{2}+H_{2} O \\ Al_{2} O_{3}+6 HCl \longrightarrow 2 AlCl_{3}+3 H_{2} O \end{gathered} $$

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 $Ba$.

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

$BeCO_{3}$ 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 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

$No,\left(NH_4\right)_2 CO_3$ reacts with NaCl as

$$ \left(NH_4 \right)_2 CO_3+2 NalC \rightleftharpoons Na_2 CO_3+2 NH_4 Cl $$

Because the products obtained $Na_{2} CO_{3}$ and $NH_{4} Cl$ 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 $\left(NH_4 \right)_2 CO_3$ with sodium chloride.

Answer

The Lewis structure of $\mathrm{O}_{2}^{-}$is $\quad: \mathrm{O}^{-}-\ddot{O}^{-}:$

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 $=\frac{1}{2}$

$$ \begin{aligned} \mathrm{O}_{2}^{-} & =2 x=-1 \\ x & =-\frac{1}{2} \end{aligned} $$

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. $\rightarrow(3)$

B. $\rightarrow(2)$

C. $\rightarrow(4)$

D. $\rightarrow(5)$

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].

$6 s^{2}$ 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. $\mathrm{Na}$ - 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.

Assertion and Reason

In the following questions a statement of Assertion (A) followed by a statement of Reason $(\mathrm{R})$ is given. Choose the correct option out of the choices given below in each question.

Answer

(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) $\mathrm{BeO}$ 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

$$ M+(x+y) NH_{3} \longrightarrow\left[M\left(NH_{3}\right)_{x}\right]^{+}+\left[\left(NH_3) y\right]^{-} e\right. $$

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.

$$ \underset{\text { Ammoniacal }}{M^{+}+e^{-}}+\underset{\text { Amide }}{NH_{3}} \underset{2}{MNH_{2}}+\frac{1}{2} H_{2} $$

Answer

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

$$ KO_{2}<RbO_{2}<CsO_{2} $$

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

The largest $K^{+}, Rb^{+}$and $Cs^{+}$ions permit $O_{2}^{2-}$ ion to react with $O_{2}$ forming superoxide ion $O_{2}^{-}$.

$\underset{\substack{\text { Oxide }}}{\mathrm{O}_2^{2-}} \xrightarrow{\frac{1}{2} \mathrm{O}_2} \underset{\text { Peroxide }}{\mathrm{O}^{2-}} \xrightarrow{\mathrm{O}_2} \underset{\text { Superoxide }}{2 \mathrm{O}_2^{-}}$

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 $CO_{2}$ in the solution of compound $B$ indicates that compound $B$ is lime water and compound $C$ is $CaCO_{3}$. 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{\substack{\text{Calciun} \\ \text{oxide} \\ \text{(A)} }}{CaO} + H_2o \longrightarrow \underset{\substack{\text{Lime water}\\ \text{B}}}{Ca(OH)_2}$

$ \underset{\text{(B)}}{Ca(OH)_2} + CO_2 \longrightarrow \underset{\substack{\text{Calcium carbonate} \\ \text{(Milkiness)}}}{CaCO_3} + \underset{\text{(C)}}{H_2O} $

(iii) When excess of $\mathrm{CO}_{2}$ is passed, milkiness disappears due to the formation of soluble calcium bicarbonate $(D)$.

$$ \underset{(c)}{\underset{Milkiness}{CaCO_3}} + CO_2 + H_2 O \longrightarrow \underset{(D)}{\underset{(Soluble \quad in \quad H_2 O)}{\underset{Calcium \quad bicarbonate}{Ca(HCO_3)_2}}} $$

Answer

$BeH_{2}$ can be prepared from the corresponding halides by the reduction with complex alkali metal hydrides such as lithium aluminium hydride $LiAlH_{4}$.

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

$$ 2BECl_2 + LiAlH_4 \longrightarrow 2BeH_2 + LiCl + AlCl_2 $$

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.

$$ BeO+H_{2} O \longrightarrow \underset{\substack{\text { Beryllium } \\ \text { hydroxide }}}{Be(OH)_{2}} $$

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

$$ 2 NaOH(aq)+ Be(OH)_2(~s) \longrightarrow \underset{\text { Sodium tetra hydroxidoberyllate }}{Na_2 Be(OH)_4(aq)} $$

With acids, it forms beryllium salts.

$$ Be(OH)_2 +\underset{\substack{\text { Sulphuric } \\ \text { acid }}}{H_2 SO_4} \longrightarrow \underset{\begin{array}{c} \text { Beryllium } \\ \text { sulphate } \end{array}}{BeSO_4}+2 H_2 O $$

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

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

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.