Alkali metals Questions in English

(C) $KO_2$ (Potassium superoxide) contains the superoxide ion $O_2^-$,which has one unpaired electron,making it paramagnetic in nature.
The molar mass of $KO_2$ is $39 + 32 = 71 \, g/mol$.
The percentage of $K$ in $KO_2$ is calculated as: $\frac{39}{71} \times 100 \approx 54.93\% \approx 55\%$.
When $KO_2$ reacts with sulfur $(S)$,it forms a salt $B$ (potassium sulfate,$K_2SO_4$),which reacts with $BaCl_2$ to form a white precipitate of $BaSO_4$.

(A) The Solvay process is used for the industrial production of sodium carbonate $(Na_2CO_3)$.
The raw materials consumed in this process are brine ($NaCl$ solution),limestone $(CaCO_3)$,and ammonia $(NH_3)$.
The overall chemical reaction is: $2NaCl + CaCO_3 \rightarrow Na_2CO_3 + CaCl_2$.
Although $NH_3$ is used as a catalyst/reagent,it is regenerated in the process,but $NaCl$ and $CaCO_3$ are the primary consumed raw materials.

(C) Sodium $(Na)$ burns in dry oxygen (excess) to form peroxide $(Na_2O_2)$.
Potassium $(K)$ burns in dry oxygen (excess) to form superoxide $(KO_2)$.
Since the statement claims both form peroxides,it is incorrect.
Therefore,option $C$ is the correct answer.

(C) The thermal decomposition of metal nitrates follows specific patterns based on the reactivity of the metal.
$1$. For alkali metal nitrates like $NaNO_3$,heating below $500 \ ^\circ C$ yields the metal nitrite and oxygen gas: $NaNO_3 \xrightarrow{< 500 \ ^\circ C} NaNO_2 + \frac{1}{2} O_2$.
$2$. For heavy metal nitrates like $Cu(NO_3)_2$,$Hg(NO_3)_2$,and $AgNO_3$,heating results in the formation of the metal oxide (or metal),nitrogen dioxide $(NO_2)$,and oxygen gas:
$Cu(NO_3)_2 \xrightarrow{\Delta} CuO + 2NO_2 + \frac{1}{2} O_2$
$Hg(NO_3)_2 \xrightarrow{\Delta} Hg + 2NO_2 + O_2$
$AgNO_3 \xrightarrow{\Delta} Ag + NO_2 + \frac{1}{2} O_2$
Therefore,$NaNO_3$ does not produce $NO_2$ upon heating.

(B) The reactions are as follows:
$Na_2O + H_2O \to 2NaOH$
$Na_2O \xrightarrow[400 \ ^oC]{O_2} Na_2O_2$
$2Na_2O_2 + 2H_2O \xrightarrow[25 \ ^oC]{} 4NaOH + O_2$
$Na_2O_2$ (Sodium peroxide) is used for the purification of air in confined spaces such as submarines because it reacts with $CO_2$ and $H_2O$ to release $O_2$.

(B) Only $K$ does not react with nitrogen directly,while others do.
The reactions are:
$6Li + N_2 \rightarrow 2Li_3N$
$3Mg + N_2 \rightarrow Mg_3N_2$
$3Ca_{(s)} + N_{2(g)} \rightarrow Ca_3N_{2(s)}$
Lithium is the smallest atom among alkali metals. When Lithium reacts with nitrogen gas $(N_2)$ at room temperature,it forms Lithium Nitride $(Li_3N)$,which is stable because the lattice energy released from the formation of $Li_3N$ is high enough to make the overall reaction exothermic.
As $K$ is a group $1$ element,it does not react with $N_2$.

(A) Most metal bicarbonates,except for those of alkali metals (like $NaHCO_3$,$KH_3$,etc.),do not exist in the solid state.
$LiHCO_3$ is an alkali metal bicarbonate,but it is highly unstable in the solid state and exists only in solution.
$Ca(HCO_3)_2$,$Zn(HCO_3)_2$,and $AgHCO_3$ are known to exist only in aqueous solutions and decompose upon attempting to isolate them as solids.
Therefore,$(i), (ii), (iii),$ and $(iv)$ do not exist in the solid state.

(A) When $Li$ or $Mg$ reacts with air (nitrogen and oxygen) upon heating,they form their respective nitrides ($Li_3N$ or $Mg_3N_2$) as product $A$.
These nitrides react with water to form ammonia gas $(NH_3)$ as product $B$.
When $NH_3$ reacts with $HCl$,it produces ammonium chloride $(NH_4Cl)$ which appears as white fumes.
Thus,the metal $M$ can be $Li$ or $Mg$.

(B) The formation of nitride upon burning in air is a characteristic property of $Li$ and $Mg$ due to their small size and high charge density.
$Na$ reacts with oxygen to form $Na_2O$ (oxide) and $Na_2O_2$ (peroxide) but does not form a nitride.
$Mg$ reacts with both oxygen and nitrogen present in the air to form $MgO$ (oxide) and $Mg_3N_2$ (nitride).
Therefore,$X$ could be $Na$ (which forms only oxide) and $Y$ could be $Mg$ (which forms both oxide and nitride).

(B) Statement $B$ is incorrect because $NaNO_3$ on heating decomposes to form sodium nitrite $(NaNO_2)$ and oxygen gas $(O_2)$,not sodium peroxide $(Na_2O_2)$.
The reaction is: $2NaNO_3 \rightarrow 2NaNO_2 + O_2$.
Statement $A$ is correct: Hydration energy is inversely proportional to ionic size $(HE \propto 1/r)$.
Statement $C$ is correct: Alkaline earth metals ($IIA$ group) have smaller ionic radii and higher charge than alkali metals ($IA$ group),leading to higher hydration energy for $IIA$ ions.
Statement $D$ is correct: The higher charge density of $IIA$ ions results in stronger electrostatic attraction with water molecules.

(A) Milk of magnesia is a suspension of $Mg(OH)_2$ in water,not $MgO + H_2O$. Therefore,statement $A$ is false.
Alkali metal peroxides $(M_2O_2)$ become more stable as the size of the alkali metal cation increases down the group due to the stabilization of the large peroxide ion by the large cation.
$AgF$ is soluble in water because its hydration energy is higher than its lattice energy.
Anhydrous $MgCl_2$ undergoes hydrolysis upon heating $MgCl_2 \cdot 6H_2O$ to form $Mg(OH)Cl$ and $HCl$,so it cannot be prepared by direct heating.

(A) Aluminium vessels should not be washed with materials containing washing soda $(Na_{2}CO_{3})$ because it reacts with aluminium to form soluble aluminate.
Under normal circumstances,aluminium does not react with water,as an impermeable protective layer of aluminium oxide $(Al_{2}O_{3})$ or aluminium hydroxide $(Al(OH)_{3})$ forms on its surface.
When washing soda is used,it hydrolyzes to produce sodium hydroxide $(NaOH)$. The $NaOH$ reacts with the protective layer and the aluminium metal,preventing the formation of a protective layer and dissolving the metal by forming soluble sodium aluminate,$Na[Al(OH)_{4}]$.
The reaction is as follows:
$2Al + 2NaOH + 6H_{2}O \rightarrow 2Na[Al(OH)_{4}] + 3H_{2}$

(A) When zinc is burned in the presence of air,woolly tufts form,which is known as 'Lana philosophica' in Latin. 'Lana philosophica' means philosopher's wool. The woolly tufts consist of zinc oxide $(ZnO)$.
Zinc sulphide $(ZnS)$ is known as zinc blende.
$ZnO_2$ is known as zinc peroxide.
$ZnSO_4$ is known as zinc sulphate.
We can prepare philosopher's wool by burning zinc in the presence of oxygen or by reacting zinc with steam:
$Zn + 1/2 O_2 \rightarrow ZnO$
$Zn + H_2O \text{ (steam)} \rightarrow ZnO + H_2$
Zinc oxide is used in ointments for skin diseases,as a white pigment in paints,and in rubber,plastic,cosmetic,and battery industries. Due to its wool-like texture,it is commonly known as philosopher's wool.

(D) Mercury $(Hg)$ forms amalgams with most metals,but it does not form an amalgam with iron $(Fe)$.
Because iron does not react with or dissolve in mercury,it is safe to store and transport mercury in iron containers.
Other metals like platinum $(Pt)$,tungsten $(W)$,and titanium $(Ti)$ also do not form amalgams with mercury,but iron is the most practical and commonly used material for containers.

(A) The reaction between sodium carbonate $(Na_2CO_3)$ and ferric oxide $(Fe_2O_3)$ at high temperature is a fusion reaction used in the extraction of iron or in specific chemical synthesis.
The balanced chemical equation is: $Na_2CO_3 + Fe_2O_3 \xrightarrow{\Delta} 2NaFeO_2 + CO_2$.
Here,$A$ is sodium ferrite $(NaFeO_2)$.

(C) Zinc $(II)$ ions react with sodium hydroxide to form a white precipitate of zinc hydroxide:
$Zn^{2+}(aq) + 2OH^-(aq) \rightarrow Zn(OH)_2(s)$ (white precipitate)
When excess $NaOH$ is added,the white precipitate of zinc hydroxide dissolves due to the formation of the soluble tetrahydroxozincate $(II)$ complex:
$Zn(OH)_2(s) + 2OH^-(aq) \rightarrow [Zn(OH)_4]^{2-}(aq)$

(A) The solution of sodium in liquid ammonia appears blue due to the presence of ammoniated electrons and ammoniated sodium ions.
$Na + (x + y)NH_3 \to [Na(NH_3)_x]^+ + [e(NH_3)_y]^-$
The blue color is specifically attributed to the excitation of ammoniated electrons in the visible region of the spectrum.

(B) The metal $X$ is $Mg$. The reaction with nitrogen is: $3Mg + N_2 \to Mg_3N_2$ $(Y)$.
The reaction of $Y$ $(Mg_3N_2)$ with water is: $Mg_3N_2 + 6H_2O \to 3Mg(OH)_2 + 2NH_3$ (colorless gas).
The reaction of the colorless gas $(NH_3)$ with $CuSO_4$ is: $4NH_3 + CuSO_4 \to [Cu(NH_3)_4]SO_4$ (deep blue complex).

(B) $Au$ and $Pt$ are noble metals and do not react with concentrated $H_2SO_4$.
$Pb$ does not react significantly due to the formation of an insoluble protective layer of $PbSO_4$.
$Ag$ reacts with concentrated $H_2SO_4$ as follows:
$2Ag + 2H_2SO_4 \to Ag_2SO_4 + SO_2 + 2H_2O$

(A, B, D) Sodium nitrate $(NaNO_3)$ decomposes upon heating.
At temperatures up to $500\,^oC$,it decomposes to form sodium nitrite and oxygen: $2NaNO_3 \rightarrow 2NaNO_2 + O_2$.
At temperatures above $800\,^oC$,sodium nitrite further decomposes to form sodium oxide,nitrogen,and oxygen: $2NaNO_2 \rightarrow Na_2O + N_2 + 1.5O_2$.
Therefore,the final products at temperatures above $800\,^oC$ include $N_2$,$O_2$,and $Na_2O$.

(A) Fluorosis is a condition caused by the excessive intake of fluoride. In the body,excess fluoride ions react with calcium $(Ca^{2+})$ ions to form calcium fluoride $(CaF_2)$,which leads to the hardening of bones and teeth,causing fluorosis.

(D) The solubility of alkaline earth metal sulfates $(MSO_4)$ decreases down the group because the hydration enthalpy decreases more rapidly than the lattice enthalpy as the size of the cation increases. Thus,the correct order is $MgSO_4 > CaSO_4 > SrSO_4 > BaSO_4$.
For alkali metal hydroxides $(MOH)$,the solubility increases down the group due to the decrease in lattice energy being more significant than the decrease in hydration energy. Thus,the correct order is $LiOH < NaOH < KOH$.
Comparing the given options:
Option $A$ is incorrect (it is the reverse of the actual trend).
Option $B$ is incorrect (solubility of sulfides varies based on $K_{sp}$ values).
Option $C$ is incorrect (solubility of carbonates increases down the group,but $Na_2CO_3$ is highly soluble).
Option $D$ is correct: $KOH > NaOH > LiOH$.

(B) The degree of hydration is directly proportional to the charge density or ionic potential of the ion,which is defined as the ratio of charge to ionic radius $(q/r)$.
As we move down the group from $Li^{+}$ to $Cs^{+}$,the ionic radius increases,which leads to a decrease in the charge density.
Therefore,the hydration energy and the degree of hydration decrease as the size of the ion increases.
The correct order of relative degree of hydration is: $Li^{+}_{(aq)} > Na^{+}_{(aq)} > K^{+}_{(aq)} > Rb^{+}_{(aq)} > Cs^{+}_{(aq)}$.

(A) Lithium carbonate $Li_{2}CO_{3}$ is the least thermally stable among the given alkali metal carbonates.
The thermal stability of alkali metal carbonates increases as we move down the group from $Li$ to $Cs$ because the ionic character increases and the polarizing power of the metal cation decreases.
Due to the very small size of the $Li^{+}$ ion,it has a high polarizing power. It distorts the electron cloud of the large carbonate ion $(CO_{3}^{2-})$,which weakens the $C-O$ bond and strengthens the $Li-O$ bond,facilitating the decomposition into oxide and carbon dioxide.
The decomposition reaction is: $Li_{2}CO_{3} \xrightarrow{\Delta} Li_{2}O + CO_{2}$

(A) For alkali metal carbonates,the solubility increases down the group because the lattice energy $(L.E.)$ decreases more rapidly than the hydration energy $(H.E.)$ as the size of the cation increases. Therefore,the solubility order is $Na_2CO_3 < K_2CO_3 < Rb_2CO_3$.

(C) Among the alkali metals,$Li$ reacts with oxygen to form only the monoxide $(Li_2O)$ because the $Li^+$ ion is very small and has a high charge density,which stabilizes the small $O^{2-}$ ion.
It cannot form a peroxide $(Li_2O_2)$ or superoxide $(LiO_2)$ because the larger peroxide and superoxide ions are not stable in the presence of the small $Li^+$ ion.
$Na$ forms peroxide $(Na_2O_2)$,while $Sr$ and $Ba$ (alkaline earth metals) form peroxides ($SrO_2$ and $BaO_2$ respectively).

(D) When magnesium $(Mg)$ burns in air,it reacts with both oxygen $(O_2)$ and nitrogen $(N_2)$ present in the atmosphere.
The reactions are as follows:
$2Mg(s) + O_2(g) \rightarrow 2MgO(s)$
$3Mg(s) + N_2(g) \rightarrow Mg_3N_2(s)$
Therefore,the products formed are magnesium oxide $(MgO)$ and magnesium nitride $(Mg_3N_2)$.

(C) $MgCl_2 \cdot 6H_2O \xrightarrow{\Delta} MgO + 2HCl + 5H_2O$ (Produces metal oxide)
$2K_2Cr_2O_7 \xrightarrow{\Delta} 2K_2CrO_4 + Cr_2O_3 + \frac{3}{2}O_2$ (Produces metal oxide $Cr_2O_3$)
$K_2CO_3 \xrightarrow{\Delta} \text{No decomposition, only melts}$ (Does not produce metal oxide)
$Cu(NO_3)_2 \xrightarrow{\Delta} CuO + 2NO_2 + \frac{1}{2}O_2$ (Produces metal oxide $CuO$)

(B) The thermal decomposition of metal nitrates follows specific patterns based on the reactivity of the metal.
For $P$: $AgNO_3 \xrightarrow{\Delta } Ag (\text{metallic solid}) + NO_2 \uparrow + \frac{1}{2} O_2 \uparrow$.
For $X$: $Pb(NO_3)_2 \xrightarrow{\Delta } PbO (\text{amphoteric oxide}) + 2 NO_2 \uparrow + \frac{1}{2} O_2 \uparrow$.
Thus,$P$ is $AgNO_3$ and $X$ is $Pb(NO_3)_2$.

(A) The reaction of $Zn$ with very dilute $HNO_3$ is given by:
$4Zn + 10HNO_3 (very \text{ } dil.) \rightarrow 4Zn(NO_3)_2 + NH_4NO_3 + 3H_2O$.
$Zn$ is a sufficiently active metal to reduce the nitrate ion to ammonium ion in very dilute acid conditions.
$Pb$,$Cu$,and $Au$ do not produce $NH_4NO_3$ under these specific conditions.

(B) Among the given compounds,$Ba(OH)_2$ (barium hydroxide) is a strong base and is soluble in water.
$BaSO_4$ is insoluble due to high lattice energy.
$Al(OH)_3$ is amphoteric and insoluble in water.
$CaF_2$ is sparingly soluble in water.

(B) The solubility of ionic compounds in water depends on the balance between lattice energy and hydration energy.
For alkali metal bicarbonates,the solubility increases down the group due to the decrease in lattice energy being more significant than the decrease in hydration energy.
Thus,the correct order of increasing solubility is $NaHCO_3 < KHCO_3 < RbHCO_3$.

(C) Metals that form nitrides react with moisture to release ammonia $(NH_3)$ gas,which has a characteristic pungent smell.
$1$. $Mg$ (Group $2$) forms $Mg_3N_2$,which reacts with water: $Mg_3N_2 + 6H_2O \to 3Mg(OH)_2 + 2NH_3 \uparrow$.
$2$. $Ca$ (Group $2$) forms $Ca_3N_2$,which reacts with water: $Ca_3N_2 + 6H_2O \to 3Ca(OH)_2 + 2NH_3 \uparrow$.
$3$. $Li$ (Group $1$) is the only alkali metal that forms a stable nitride $(Li_3N)$,which reacts with water: $Li_3N + 3H_2O \to 3LiOH + NH_3 \uparrow$.
$4$. $K$ (Group $1$) does not form a nitride under these conditions. Therefore,it does not produce ammonia gas upon reaction with moist air.
Thus,the correct option is $C$.

(A) The thermal stability of alkali metal bicarbonates increases as the size of the cation increases down the group.
This is because the polarizing power of the cation decreases as the ionic radius increases.
Therefore,the order of thermal stability is $NaHCO_3 < KHCO_3 < RbHCO_3 < CsHCO_3$.
Thus,$NaHCO_3$ has the lowest thermal stability.

(A) The atomic number of the element is $Z = 37$.
The electronic configuration is $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^1$.
Since the last electron enters the $5s$ orbital,the element belongs to the $s$-block.
The valence shell configuration is $5s^1$,which indicates it belongs to group $1$ (alkali metals).
Therefore,the block is $s$ and the group is $1$.

(A) The density of alkali metals generally increases down the group from $Li$ to $Cs$.
However,there is an anomaly between $Na$ and $K$.
The density of $K$ is slightly lower than that of $Na$ due to an abnormal increase in atomic volume in $K$.
Thus,the correct order of density is $Li < K < Na < Rb < Cs$.

(D) When sodium is heated in an excess of air,it primarily reacts with oxygen to form sodium peroxide $(Na_2O_2)$ and a small amount of sodium oxide $(Na_2O)$.
The reaction is as follows:
$2Na + \frac{1}{2}O_2 \rightarrow Na_2O$
$Na_2O + \frac{1}{2}O_2 \rightarrow Na_2O_2$
Sodium does not form $NaO_2$ (superoxide) under these conditions,as that is characteristic of larger alkali metals like potassium,rubidium,and cesium. While nitrogen is present in the air,sodium does not react with nitrogen to form $Na_3N$ under normal heating conditions in air.
Therefore,the products obtained are $Na_2O$ and $Na_2O_2$,which correspond to $II$ and $IV$.

(B) Among the alkali metals,only $Li$ forms a stable nitride $(Li_3N)$ by direct reaction with nitrogen.
This is due to the high lattice energy of $Li_3N$ resulting from the small size of the $Li^+$ ion.
As the size of the alkali metal cation increases down the group,the lattice energy decreases,making the nitrides of $Na, K, Rb,$ and $Cs$ unstable with respect to decomposition into their elements.

(C) $Mg + NaOH$ cannot be used to generate hydrogen gas.
The reactions leading to the generation of hydrogen gas are as follows:
$(A)$ $2Al + 2NaOH + 2H_{2}O \rightarrow 2NaAlO_{2} + 3H_{2}\uparrow$ (sodium meta aluminate)
$(B)$ $Zn + 2NaOH \rightarrow Na_{2}ZnO_{2} + H_{2}$
$(C)$ $Mg$ is not affected by alkalies. Hence,it does not react with $NaOH$.
$(D)$ $LiH + H_{2}O \rightarrow H_{2}\uparrow + LiOH$
Therefore,option $C$ is correct.

(B) The thermal stability of metal carbonates increases as the electropositive character of the metal increases.
$BeCO_3$ is highly unstable and decomposes at room temperature.
$Li_2CO_3$ is also relatively unstable compared to other alkali metal carbonates.
Among the given options,$Na_2CO_3$ is a carbonate of an alkali metal (Group $1$),which is more electropositive than the alkaline earth metals ($Be$ and $Ca$).
Therefore,$Na_2CO_3$ has the highest thermal stability and the maximum decomposition temperature.

(A) The solubility of alkali metal bicarbonates increases down the group due to the decrease in lattice energy being more significant than the decrease in hydration energy. Thus,the order is $LiHCO_3 < NaHCO_3 < KHCO_3$.
For alkaline earth metal sulfates,solubility decreases down the group as hydration energy decreases more rapidly than lattice energy,so $MgSO_4 > CaSO_4 > BaSO_4$.
For alkali metal chlorides,solubility increases down the group as $NaCl < KCl < RbCl$.
For alkaline earth metal hydroxides,solubility increases down the group as $Be(OH)_2 < Mg(OH)_2 < Ca(OH)_2$.
Therefore,the correct order is $LiHCO_3 < NaHCO_3 < KHCO_3$.

(C) $Be(OH)_2$,$Al(OH)_3$,and $Zn(OH)_2$ are amphoteric in nature and dissolve in excess $NaOH$ to form soluble complexes like $[Be(OH)_4]^{2-}$,$[Al(OH)_4]^-$,and $[Zn(OH)_4]^{2-}$.
$FeCl_3$ reacts with $NaOH$ to form $Fe(OH)_3$,which is a reddish-brown precipitate.
$Fe(OH)_3$ is basic in nature and does not dissolve in excess $NaOH$.

(D) The solubility of alkali metal bicarbonates increases down the group from $Li$ to $Cs$.
This is because the lattice energy decreases more rapidly than the hydration energy as the size of the cation increases.
Therefore,$RbHCO_3$ has the maximum solubility in water among the given options.

(B) The reaction sequence represents the thermal decomposition of sodium bicarbonate $(NaHCO_3)$:
$1$. $2NaHCO_3 \xrightarrow{\Delta} Na_2CO_3 + H_2O + CO_2(g)$
Here,$A = NaHCO_3$,$B = Na_2CO_3$,and $C = CO_2$.
$2$. $CO_2$ reacts with lime water $(Ca(OH)_2)$ to form a milky precipitate:
$Ca(OH)_2 + CO_2 \rightarrow CaCO_3(s) (D) + H_2O$
$3$. Excess $CO_2$ reacts with $CaCO_3$ to form soluble calcium bicarbonate $(E)$:
$CaCO_3 + H_2O + CO_2 \rightarrow Ca(HCO_3)_2 (E)$
Evaluating the statements:
- $A$ is $NaHCO_3$ (Correct).
- $B$ is $Na_2CO_3$,which is soluble in $H_2O$ (Incorrect,as $Na_2CO_3$ is highly soluble).
- $A$ $(NaHCO_3)$ is amphoteric,but $E$ $(Ca(HCO_3)_2)$ is also amphoteric (Correct).
- $D$ is $CaCO_3$ (Correct).
Thus,the incorrect statement is $B$.

(C) The stability of alkali metal hydrides decreases down the group as the size of the alkali metal cation increases.
As the size of the cation increases,the lattice energy of the ionic hydride decreases,making it less stable.
Among the given options,$Li^+$ has the smallest ionic radius,which results in the highest lattice energy for $LiH$.
Therefore,$LiH$ is the most stable hydride among the alkali metal hydrides.

(D) The ores of magnesium are:
$1$. Magnesite: $MgCO_3$
$2$. Dolomite: $MgCO_3 \cdot CaCO_3$
$3$. Carnallite: $KCl \cdot MgCl_2 \cdot 6H_2O$
Gypsum is a calcium ore with the formula $CaSO_4 \cdot 2H_2O$.
Therefore,Gypsum is not an ore of magnesium.

(C) In the Solvay process,$NaHCO_3$ is precipitated because it is less soluble in water and can be easily separated from the reaction mixture.
However,in the case of potassium,$KHCO_3$ is highly soluble in water.
Due to this high solubility,$KHCO_3$ does not precipitate out of the solution,making it impossible to isolate it by the Solvay process.
Therefore,the correct reason is that $KHCO_3$ is more soluble than $NaHCO_3$.

(A) The stability of oxides,peroxides,and superoxides of alkali metals depends on the size of the cation.
As the size of the alkali metal cation increases down the group,the lattice energy increases,which stabilizes the larger anions.
Therefore,the stability order for these compounds is:
$Normal \ oxide < Peroxide < Superoxide$ for larger cations,but generally,for a given metal,the stability order is $Normal \ oxide > Peroxide > Superoxide$ due to the increasing charge density and decreasing size of the anion.
However,considering the general trend of formation and stability across the group,the correct order is $Normal \ oxide > Peroxide > Superoxide$.

(D) $1$. Alkali metals dissolve in liquid $NH_3$ to form a deep blue solution,which is conducting and paramagnetic due to the presence of ammoniated electrons $(e^-(NH_3)_x)$.
$2$. The blue colour is due to the absorption of light in the red region of the visible spectrum,which corresponds to the excitation of ammoniated electrons.
$3$. As the concentration of the metal increases,the blue colour changes to bronze and the solution becomes diamagnetic because the ammoniated electrons pair up to form clusters.
$4$. Therefore,the paramagnetic character decreases with increasing concentration.
$5$. On standing,the blue solution slowly decomposes to form metal amide $(MNH_2)$ and hydrogen gas $(H_2)$,but this is not a property of all alkali metals under all conditions immediately.

(A) Zeolites are microporous,aluminosilicate minerals commonly used as commercial adsorbents and catalysts. The general chemical formula for zeolite is $Na_2Al_2Si_2O_8 \cdot xH_2O$. It is widely used in the softening of hard water by the ion-exchange process.