Alkali metals Questions in English

(B) The stability of alkali metal halides is determined by their lattice energy.
As the size of the alkali metal cation increases down the group $(Li^+ < Na^+ < K^+ < Rb^+ < Cs^+)$,the lattice energy decreases because the electrostatic attraction between the cation and the chloride anion $(Cl^-)$ weakens.
Therefore,the stability of the chlorides decreases as the size of the cation increases.
The correct order of stability is $LiCl > NaCl > KCl > CsCl$.

(C) When carbon monoxide $(CO)$ is passed over solid sodium hydroxide $(NaOH)$ at a temperature of $200^\circ C$ and under high pressure,it reacts to form sodium formate $(HCOONa)$.
The chemical equation for the reaction is:
$NaOH + CO \xrightarrow{200^\circ C} HCOONa$

(B) The reaction sequence is as follows:
$3 Mg + N_2 \xrightarrow{\Delta} Mg_3N_2$ $(Y)$
$Mg_3N_2 + 6 H_2O \rightarrow 3 Mg(OH)_2 + 2 NH_3$ ($Z$,colourless gas)
$CuSO_4 + 4 NH_3 \rightarrow [Cu(NH_3)_4]SO_4$ or $CuSO_4 \cdot 4NH_3$ ($T$,deep blue colour)
Therefore,$Y = Mg_3N_2$ and $T = CuSO_4 \cdot 4NH_3$.

(D) The basic strength of hydroxides depends on the ionic character of the $M-OH$ bond.
$KOH$ and $NaOH$ are hydroxides of alkali metals,which are highly ionic and strong bases.
$Ca(OH)_2$ is a hydroxide of an alkaline earth metal,which is also a strong base.
$Zn(OH)_2$ is an amphoteric hydroxide. The $Zn^{2+}$ ion has a high charge density and small size,leading to significant covalent character in the $Zn-OH$ bond.
Therefore,$Zn(OH)_2$ is the weakest base among the given options.

(D) The aqueous solutions of lithium salts are poor conductors of electricity compared to those of other alkali metals because of the higher degree of hydration of $Li^{+}$ ions.
Due to its small size,the $Li^{+}$ ion has a very high charge density,which causes it to be heavily hydrated in aqueous solutions.
This large hydration shell increases the effective size of the ion,which significantly decreases its ionic mobility.
Since electrical conductivity in aqueous solutions depends on the mobility of ions,the reduced mobility leads to lower conductivity.
Hence,option $D$ is correct.

(C) $Sodium$ metal is highly reactive and reacts with protic solvents like alcohols to form alkoxides and hydrogen gas.
For example: $CH_3OH + Na \rightarrow CH_3ONa + \frac{1}{2} H_2$.
Therefore,it cannot be stored under alcohol. It is typically stored under kerosene oil,toluene,or benzene as they are inert towards $sodium$.

(A) $LiAlH_4$ is a complex hydride,which dissociates into $Li^+$ cation and $[AlH_4]^-$ anion.
In this structure,the metal $Al$ is a central atom within the complex $[AlH_4]^-$ anion.
Therefore,the metal $Al$ is present in the anionic part.
Hence,the correct option is $(A)$.

(A) The lattice energy is inversely proportional to the inter-ionic distance (sum of the radii of the cation and anion).
Since the fluoride ion $(F^-)$ is common to all,the lattice energy depends on the size of the alkali metal cation.
As the size of the cation increases from $Li^+$ to $Cs^+$,the inter-ionic distance increases,and the lattice energy decreases.
Therefore,$Li^+$ (being the smallest cation) forms the fluoride with the highest lattice energy,which is $LiF$.

(A) Crown ethers and cryptands are macrocyclic and macrobicyclic polyethers,respectively.
They possess a central cavity that can accommodate metal cations of specific sizes.
Due to the presence of lone pairs on the oxygen atoms,they act as ligands and form stable complexes with alkali metal cations $(M^+)$ through ion-dipole interactions.
These complexes are often referred to as host-guest complexes.

(C) The degree of hydration of ions depends on their charge density.
Smaller ions have higher charge density,which leads to stronger attraction for water molecules,resulting in a higher degree of hydration.
Among alkali metal ions,the ionic size increases down the group: $Li^{+} < Na^{+} < K^{+} < Rb^{+} < Cs^{+}$.
Therefore,the degree of hydration decreases as the ionic size increases.
The correct order of hydration is $Li^{+} > Na^{+} > K^{+} > Rb^{+} > Cs^{+}$,which is equivalent to $Cs^{+} < Rb^{+} < K^{+} < Na^{+} < Li^{+}$.

(C) The anomalous behavior of lithium is due to its extremely small size and high polarizing power.
$(1)$ Lithium is harder than other alkali metals,which are generally soft.
$(2)$ Lithium has higher melting and boiling points compared to other alkali metals.
$(3)$ Lithium has higher ionization potential and electronegativity than other alkali metals.
$(4)$ The lithium ion is the most heavily hydrated among alkali metals due to its small size.
Option $(C)$ states that lithium is softer than other alkali metals,which is incorrect as lithium is actually harder than other alkali metals. Therefore,this is not an anomalous property of lithium.

(B) The reaction is: $Na(l) + KCl(l) \rightleftharpoons NaCl(l) + K(g)$.
At $850 ^\circ C$,potassium $(K)$ has a lower boiling point $(774 ^\circ C)$ compared to sodium ($Na$,$883 ^\circ C$).
Since potassium is more volatile at this temperature,it distils off as a vapor.
According to Le Chatelier's principle,the continuous removal of the product (potassium vapor) shifts the equilibrium in the forward direction,allowing the reaction to proceed to completion.

(A) The reaction sequence is as follows:
$Ca + 2C \rightarrow CaC_2$
$CaC_2 + N_2 \xrightarrow{\Delta} Ca(CN)_2 + C$
Compound $(A)$ is $Ca(CN)_2$,which is known as calcium cyanamide (often mixed with carbon,it is called nitrolim).
It is widely used as a nitrogenous fertilizer in agriculture.
Therefore,option $A$ is correct.

(A) Magnetite is a rock mineral and one of the main iron ores,with the chemical formula $Fe_3O_4$. It does not contain $Mg$.
Magnesite is a magnesium carbonate mineral with a chemical composition of $MgCO_3$.
Asbestos is a group of minerals,and a common variety is $CaMg_3(SiO_3)_4$,which contains $Mg$.
Carnallite is an evaporite mineral,a hydrated potassium magnesium chloride with the formula $KMgCl_3 \cdot 6H_2O$.

(B) The Dow's process is an industrial method used for the extraction of magnesium metal.
In this process,magnesium is obtained by the electrolysis of fused magnesium chloride $(MgCl_2)$.
The process involves the electrolysis of molten $MgCl_2$ to produce magnesium metal at the cathode and chlorine gas at the anode.

(B) The chemical formula of Carnallite is $KCl \cdot MgCl_2 \cdot 6H_2O$.
From the formula,we can see that it contains Potassium $(K)$,Magnesium $(Mg)$,and Chlorine $(Cl)$.
It does not contain Calcium $(Ca)$.

(A) Amphoteric elements like $Zn$,$Al$,$Sn$,and $Pb$ react with $NaOH$ to produce $H_2$ gas.
$Zn + 2NaOH \rightarrow Na_2ZnO_2 + H_2$
$2Al + 2NaOH + 6H_2O \rightarrow 2Na[Al(OH)_4] + 3H_2$
$Sn + 2NaOH + H_2O \rightarrow Na_2SnO_3 + 2H_2$
$Pb + 2NaOH \rightarrow Na_2PbO_2 + H_2$
$S$ and $P$ do not produce $H_2$ gas with $NaOH$.
Therefore,the total number of such elements is $4$.

(D) $I$. Pure $NaCl$ is not hygroscopic; it becomes hygroscopic only due to impurities like $CaCl_2$ and $MgCl_2$.
$II$. Alkali metals have an electronic configuration of $ns^1$. The $(n-1)d$ subshell is not present for $Li$ $(n=2)$,and for others,it is completely filled,not vacant.
$III$. $Na_2CO_3$ is a salt of a strong base $(NaOH)$ and a weak acid $(H_2CO_3)$,making it basic. $NaHCO_3$ is an amphiprotic species,and its solution is less basic than $Na_2CO_3$.
$IV$. Alums are double salts with the general formula $M'M'''(SO_4)_2 \cdot 12H_2O$. $Li^{+}$ and $Mg^{2+}$ do not form alums because their ionic radii are too small to fit into the crystal lattice structure of alums.
Therefore,only statement $IV$ is $CORRECT$.

(A) Trona is a double salt with the formula $Na_2CO_3 \cdot NaHCO_3 \cdot 2H_2O$.
Upon heating,it undergoes thermal decomposition.
First,the water of crystallization is lost,and then the sodium bicarbonate $(NaHCO_3)$ component decomposes into sodium carbonate $(Na_2CO_3)$,water vapor $(H_2O)$,and carbon dioxide $(CO_2)$.
The overall reaction is: $2(Na_2CO_3 \cdot NaHCO_3 \cdot 2H_2O) \xrightarrow{\Delta} 3Na_2CO_3 + 5H_2O + CO_2$.
Thus,the final solid residue remaining is sodium carbonate $(Na_2CO_3)$.

(C) Among the alkali metals,only $Li$ reacts directly with $N_2$ to form $Li_3N$ due to its small size and high charge density.
Other alkali metals like $Cs$ (Cesium) do not react with nitrogen to form nitrides because the lattice energy of the resulting nitride would be insufficient to compensate for the high ionization energy and the energy required to break the $N \equiv N$ triple bond.
$Mg$ and $Al$ also form nitrides ($Mg_3N_2$ and $AlN$ respectively) upon heating with nitrogen.

(A) $(1)$ 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,$NaHCO_3 < KHCO_3 < RbHCO_3 < CsHCO_3$ is correct.
$(2)$ Boiling point order for group $15$ hydrides is $PH_3 < AsH_3 < SbH_3 < NH_3 < BiH_3$. $NH_3$ has a higher boiling point than $PH_3, AsH_3, SbH_3$ due to hydrogen bonding.
$(3)$ According to Fajan's rule,as the charge on the cation increases,polarization increases,which increases the covalent character and decreases the ionic character. Thus,$Fe_2O_3$ has more covalent character than $FeO$.
$(4)$ $IE_2$ of $Na$ $([Ne] 3s^1)$ is very high because the second electron is removed from a stable noble gas configuration,whereas $Mg$ $([Ne] 3s^2)$ has a lower $IE_2$ as it achieves a stable configuration after removing one electron. Thus,$Mg < Na$.

(C) When an alkali metal dissolves in liquid $NH_3$,it forms a blue-coloured,conducting solution due to the presence of ammoniated electrons $(e^-(NH_3)_x)$ and ammoniated metal ions $(M^+(NH_3)_y)$.
$1$. Dilute solutions are blue and paramagnetic due to solvated electrons.
$2$. Concentrated solutions become bronze-coloured and exhibit metallic lustre.
$3$. The conductivity is due to both solvated electrons and solvated metal ions,not just electrons.
$4$. The blue colour is stable in the absence of impurities,but it slowly decomposes to form metal amides $(MNH_2)$ and $H_2$ gas.
$5$. The metal can be recovered by the evaporation of ammonia.

(B) The reaction is: $2KOH + 2O_3 \to 2KO_3 + O_2 + H_2O$.
Here,$(a)$ is $KO_3$ (potassium ozonide) and $(b)$ is $O_2$ (oxygen gas).
$KO_3$ contains the ozonide ion $(O_3^-)$,which has one unpaired electron,making it paramagnetic.
$KO_3$ reacts with $HCl$ to produce $O_2$ gas: $2KO_3 + 2HCl \to 2KCl + 2O_2 + H_2O$.
$O_2$ does not react directly with $Xe$ under standard conditions.
Therefore,the statement '$(b)$ does not react directly with $Xe$' is correct.

(D) Cohesive energy is the energy required to break the metallic bonds holding atoms together in a solid lattice.
For alkali metals,the strength of metallic bonding depends on the number of valence electrons and the size of the atoms.
As we move down the group from $Li$ to $Cs$,the atomic size increases,which leads to a decrease in the strength of metallic bonding and consequently a decrease in cohesive energy.
Among the given options ($Cs$,$K$,$Rb$,$Na$),$Na$ has the smallest atomic size,resulting in the strongest metallic bonding and the highest cohesive energy.
Therefore,$Na$ has the highest value of cohesive energy.

(B) The thermal decomposition of $LiNO_3$ produces $Li_2O$,$NO_2$ (brown gas),and $O_2$.
$4LiNO_3 \xrightarrow{\Delta} 2Li_2O + 4NO_2 + O_2$.
$KNO_3$ decomposes to form $KNO_2$ and $O_2$ (colorless gas),so it does not liberate a brown gas.
$2KNO_3 \xrightarrow{\Delta} 2KNO_2 + O_2$.
Reaction of $Zn$ with conc $HNO_3$ produces $NO_2$ (brown gas).
$Zn + 4HNO_3 \rightarrow Zn(NO_3)_2 + 2NO_2 + 2H_2O$.
Addition of conc $H_2SO_4$ on $NaNO_3$ produces $HNO_3$ vapor,which decomposes to $NO_2$ (brown gas).
$NaNO_3 + H_2SO_4 \rightarrow NaHSO_4 + HNO_3$ followed by $4HNO_3 \rightarrow 4NO_2 + 2H_2O + O_2$.

(D) The magnitude of the enthalpy of formation $(\Delta H_f)$ for alkali metal halides is primarily governed by the lattice energy and the hydration energy.
For a given metal,the enthalpy of formation becomes less negative (smaller magnitude) as the size of the halide ion increases from $F^-$ to $I^-$. Thus,the order is $Fluoride > Chloride > Bromide > Iodide$.
For a given halide,the enthalpy of formation becomes less negative as the size of the alkali metal cation increases down the group. Thus,for iodides,the order is $LiI > NaI > RbI > CsI$.

(C) Amphoteric metals like $Zn$ and $Sn$ react with strong bases like $NaOH$ to produce hydrogen gas.
The chemical reactions are as follows:
$Zn + 2NaOH \rightarrow Na_2ZnO_2 + H_2 \uparrow$
$Sn + 2NaOH \rightarrow Na_2SnO_2 + H_2 \uparrow$
Thus,$Zn$ and $Sn$ are the elements that release $H_2$ gas.

(A) In the Solvay process,$Na_2CO_3$ is produced using $NaCl$,$NH_3$,and $CO_2$ as raw materials.
$NH_3$ is recovered by treating the byproduct $NH_4Cl$ with $Ca(OH)_2$.
$CO_2$ is recovered by the thermal decomposition of $CaCO_3$ (limestone).
Therefore,the products that are recycled in the process are $CO_2$ and $NH_3$.

(D) The reaction of rubidium superoxide $(RbO_2)$ with water $(H_2O)$ is given by the following equation:
$2RbO_2 + 2H_2O \rightarrow 2RbOH + H_2O_2 + O_2$
From the balanced chemical equation,the products formed are rubidium hydroxide $(RbOH)$,hydrogen peroxide $(H_2O_2)$,and oxygen gas $(O_2)$.
Therefore,$Rb_2O_2$ (rubidium peroxide) is not a product of this reaction.

(B) $1$. $Li$ is the strongest reducing agent in aqueous solution due to its high hydration energy.
$2$. The density of alkali metals generally increases down the group,but $K$ is an exception because it has a larger atomic volume than $Na$,making its density $(0.86 \ g/cm^3)$ lower than that of $Na$ $(0.97 \ g/cm^3)$. Thus,the statement that the density of $K$ is greater than $Na$ is false.
$3$. $Li^+$ has the smallest ionic radius in the gas phase,but in aqueous solution,it has the largest hydrated ionic radius due to extensive hydration.
$4$. All alkali metals dissolve in liquid ammonia to form deep blue,conducting,and paramagnetic solutions due to the presence of ammoniated electrons.

(D) The given reactions represent the Solvay process for the manufacture of sodium carbonate $(Na_2CO_3)$:
$1$. $2NH_3 + CO_2 + H_2O \to (NH_4)_2CO_3$ $(A)$
$2$. $(NH_4)_2CO_3 + H_2O + CO_2 \to 2NH_4HCO_3$ $(B)$
$3$. $NH_4HCO_3 + NaCl \to NaHCO_3$ $(C)$ + $NH_4Cl$
$4$. $2NaHCO_3 \xrightarrow{\Delta} Na_2CO_3$ $(D)$ + $H_2O + CO_2$
Comparing these with the options:
$A = (NH_4)_2CO_3$ (Correct)
$B = NH_4HCO_3$ (Option $D$ says $B = (NH_4)_2C_2O_4$,which is incorrect)
$C = NaHCO_3$ (Correct)
$D = Na_2CO_3$ (Correct)
Therefore,the incorrect statement is $B = (NH_4)_2C_2O_4$.

(C) The Solvay process is used for the manufacture of sodium carbonate $(Na_2CO_3)$.
$1$. $CaCO_3 \xrightarrow{\Delta} CaO + CO_2$ is used to produce $CO_2$ and $CaO$.
$2$. $NaCl + NH_3 + H_2O + CO_2 \to NH_4Cl + NaHCO_3$ is the main reaction where sodium bicarbonate is formed.
$3$. $Ca(OH)_2 + 2NH_4Cl \to 2NH_3 + CaCl_2 + 2H_2O$ is used to recover ammonia $(NH_3)$.
$4$. The reaction $Na_2CO_3 + CO_2 + H_2O \to 2NaHCO_3$ is not part of the Solvay process; in fact,the reverse reaction $(2NaHCO_3 \xrightarrow{\Delta} Na_2CO_3 + H_2O + CO_2)$ is used to obtain sodium carbonate from sodium bicarbonate.

(C) $1$. In reaction $(X)$,the electrolysis of aqueous $NaCl$ (brine) produces $NaOH$ $(A)$,$H_2$ gas $(B)$,and $Cl_2$ gas.
$2NaCl(aq) + 2H_2O(l) \rightarrow 2NaOH(aq) + H_2(g) + Cl_2(g)$.
Thus,$A = NaOH$ and $B = H_2$.
$2$. The reaction of $NaOH$ $(A)$ with $Al$ produces sodium aluminate $(NaAlO_2)$ and $H_2$ $(B)$: $2Al + 2NaOH + 2H_2O \rightarrow 2NaAlO_2 + 3H_2$.
$3$. In reaction $(Y)$,$NaOH$ $(A)$ reacts with $Cl_2$ at hot and concentrated conditions to produce $NaCl$ $(C)$ and $NaClO_3$ $(D)$:
$6NaOH + 3Cl_2 \rightarrow 5NaCl + NaClO_3 + 3H_2O$.
Therefore,$A = NaOH, B = H_2, C = NaCl, D = NaClO_3$.

(D) When alkali metals dissolve in liquid ammonia,they form a deep blue solution due to the presence of ammoniated electrons $(M + (x+y)NH_3 \rightarrow [M(NH_3)_x]^+ + [e(NH_3)_y]^-)$.
$1$. The blue colour is indeed due to the ammoniated electrons,which absorb energy in the visible region of the spectrum.
$2$. Upon dilution,the concentration of ammoniated electrons decreases,leading to a decrease in colour intensity.
$3$. On standing for a long time,the solution reacts to form metal amide and hydrogen gas $(2M + 2NH_3 \rightarrow 2MNH_2 + H_2)$,which causes the blue colour to disappear,making the solution colourless.
Since all statements $(A)$,$(B)$,and $(C)$ are correct,the incorrect statement is none of these.

(B) The solubility of compounds depends on the balance between lattice energy and hydration energy.
$(a)$ For $BeSO_4$ and $BaSO_4$,the hydration energy of the small $Be^{2+}$ ion is much higher than that of the $Ba^{2+}$ ion,making $BeSO_4$ more soluble.
$(b)$ For $NaCl$ and $MgCl_2$,$NaCl$ is more soluble in water due to its lower lattice energy compared to $MgCl_2$.
$(c)$ For $AgCl$ and $AgI$,$AgCl$ is more soluble because the $Cl^-$ ion is smaller and more easily hydrated than the larger,more polarizable $I^-$ ion,which leads to higher lattice energy for $AgI$.
Therefore,the correct order is $BeSO_4, NaCl, AgCl$.

(B) The reactions are as follows:
$1. \ 2 Al + 6 HCl \longrightarrow 2 AlCl_3 + 3 H_2 \uparrow$ (Gas $'P'$ is $H_2$)
$2. \ 2 Al + 2 NaOH + 2 H_2O \longrightarrow 2 NaAlO_2 + 3 H_2 \uparrow$ (Gas $'Q'$ is $H_2$)
From the reactions,we observe that both gases $'P'$ and $'Q'$ are hydrogen gas $(H_2)$.
Therefore,the statement 'Gas $'P'$ and Gas $'Q'$ are different' is incorrect.

(B) $KO_2$ (Potassium superoxide) is used in breathing apparatus for submarines and confined spaces because it reacts with $CO_2$ to produce $O_2$ and absorbs moisture.
The chemical reaction is:
$4KO_2(s) + 2CO_2(g) \to 2K_2CO_3(s) + 3O_2(g)$

(D) When potassium permanganate $(KMnO_4)$ is heated,it undergoes thermal decomposition to produce potassium manganate $(K_2MnO_4)$,manganese dioxide $(MnO_2)$,and oxygen gas $(O_2)$.
The balanced chemical equation is:
$2KMnO_4 \xrightarrow{\Delta} K_2MnO_4 + MnO_2 + O_2$

(B) In aqueous solutions,ions are hydrated. The extent of hydration is inversely proportional to the size of the ion. Smaller ions have higher charge density,leading to greater hydration and a larger hydrated radius.
Thus,the order of hydrated ionic size is $Na^{+} > K^{+} > Rb^{+} > Cs^{+}$.
Since ionic mobility is inversely proportional to the hydrated size,the order of ionic mobility is $Cs^{+} > Rb^{+} > K^{+} > Na^{+}$.

(D) In aqueous medium,alkali metal ions are hydrated. The degree of hydration is inversely proportional to the ionic size. Since $Li^{+}$ has the smallest ionic size,it has the highest degree of hydration,resulting in the largest hydrated radius and lowest ionic mobility.
Conversely,$Cs^{+}$ has the largest ionic size,the lowest degree of hydration,the smallest hydrated radius,and therefore the highest ionic mobility in aqueous solution.
In the molten phase,there is no hydration,so mobility depends on the size of the bare ion; smaller ions like $Li^{+}$ have higher mobility than larger ions like $Cs^{+}$.
In the gaseous phase,the size of the ions follows the order $Li^{+} < Na^{+} < K^{+} < Rb^{+} < Cs^{+}$. Thus,$K^{+} > Na^{+}$ is a correct statement.

(D) The $SOLVAY$ process is used for the industrial production of sodium carbonate $(Na_2CO_3)$.
The primary raw materials required for this process are:
$1$. Sodium chloride $(NaCl)$ (brine solution).
$2$. Ammonia $(NH_3)$.
$3$. Limestone $(CaCO_3)$,which provides carbon dioxide $(CO_2)$ upon heating and calcium oxide $(CaO)$ for ammonia recovery.
Thus,the correct set of raw materials is $NaCl$,$NH_3$,and $CaCO_3$.

(D) Elements that are amphoteric in nature react with strong bases like $NaOH$ to liberate $H_2$ gas.
$Be + 2NaOH \to Na_2BeO_2 + H_2$
$2Al + 2NaOH + 6H_2O \to 2Na[Al(OH)_4] + 3H_2$
$Sn + 2NaOH + H_2O \to Na_2SnO_3 + 2H_2$
Since $Be$,$Al$,and $Sn$ are all amphoteric,they all liberate $H_2$ gas upon reaction with $NaOH$.

(B) Among the alkali metal nitrates,$LiNO_3$ decomposes on heating to form lithium oxide,nitrogen dioxide,and oxygen gas.
$4LiNO_3 \to 2Li_2O + 4NO_2 + O_2$
$NO_2$ is a brown-colored gas.
Other alkali metal nitrates like $NaNO_3$,$KNO_3$,and $RbNO_3$ decompose to form their respective nitrites and oxygen gas,which does not produce a brown gas.

(A) The Solvay process is used for the industrial production of sodium carbonate $(Na_2CO_3)$.
The raw materials required for this process are ammonia $(NH_3)$,carbon dioxide $(CO_2)$,and sodium chloride $(NaCl)$.
The chemical reactions involved are:
$2NH_3 + H_2O + CO_2 \rightarrow (NH_4)_2CO_3$
$(NH_4)_2CO_3 + H_2O + CO_2 \rightarrow 2NH_4HCO_3$
$NH_4HCO_3 + NaCl \rightarrow NaHCO_3 + NH_4Cl$
$2NaHCO_3 \rightarrow Na_2CO_3 + H_2O + CO_2$
Sulfur dioxide $(SO_2)$ is not used in this process.

(A) The thermal decomposition of sodium nitrate $(NaNO_3)$ occurs as follows:
$2NaNO_3(s) \xrightarrow{\Delta} 2NaNO_2(s) + O_2(g)$
Thus,heating $NaNO_3$ produces sodium nitrite and oxygen gas $(O_2)$.

(A) The composition of Portland cement consists primarily of calcium oxide $(CaO)$,which typically accounts for $50-60\%$ of the total mass.
Other components include silicon dioxide ($SiO_2$,$20-25\%$),aluminum oxide ($Al_2O_3$,$5-10\%$),and iron oxide ($Fe_2O_3$,$1-2\%$).
Therefore,$CaO$ is the main compound by percentage.

(D) Hydration energy is directly proportional to the charge density of the ion,which is defined as the ratio of charge to size $(Charge/Size)$.
For cations,as the charge increases and the size decreases,the hydration energy increases.
Comparing $Al^{+3}$,$Mg^{+2}$,and $Na^{+}$: All are isoelectronic with the neon configuration,but $Al^{+3}$ has the highest charge and smallest size,followed by $Mg^{+2}$ and then $Na^{+}$.
Therefore,the correct order is $Al^{+3} > Mg^{+2} > Na^{+}$.

(C) In the Solvay process,$NaHCO_3$ is precipitated as a solid because it has low solubility in water.
However,$KHCO_3$ is much more soluble in water than $NaHCO_3$.
Due to this high solubility,$KHCO_3$ does not precipitate out when $CO_2$ is passed through a concentrated solution of $KCl$ and $NH_3$,making the Solvay process unsuitable for the preparation of potassium carbonate.

(C) $1$. The solubility of carbonates of alkali metals increases down the group because the lattice energy decreases more rapidly than the hydration energy as the size of the cation increases.
$2$. The melting point of alkali metal fluorides decreases down the group $(NaF > KF > RbF > CsF)$ due to the decrease in lattice energy as the size of the cation increases.
$3$. The thermal stability of nitrates of group-$1$ metals increases down the group because the electropositive character of the metal increases,making the $M-O$ bond stronger.
$4$. The solubility of bicarbonates of alkali metals increases down the group due to the decrease in lattice energy,following the order $LiHCO_3 > NaHCO_3 > KHCO_3 > RbHCO_3 > CsHCO_3$. Thus,the statement $Na^{+} < K^{+} < Rb^{+} < Cs^{+}$ is incorrect.

(C) Magnesium $(Mg)$ is a highly reactive metal.
Electrolysis of an aqueous solution of $MgCl_2$ results in the evolution of $H_2$ gas at the cathode because the reduction potential of water is higher than that of $Mg^{2+}$ ions.
Direct heating of $MgCl_2 \cdot 6H_2O$ leads to hydrolysis,producing $MgO$ $(MgCl_2 \cdot 6H_2O \xrightarrow{\Delta} MgO + 2HCl + 5H_2O)$.
Therefore,$Mg$ is obtained by the electrolysis of molten $MgCl_2$ (often mixed with $NaCl$ and $CaCl_2$ to lower the melting point and increase conductivity).