Alkali metals Questions in English

(C) When $Na$ dissolves in liquid ammonia $(NH_{3(l)})$,it forms a deep blue coloured solution due to the presence of ammoniated electrons $(e^-(NH_3)_x)$.
The reaction is: $Na + (x+y)NH_3 \longrightarrow [Na(NH_3)_x]^+ + [e(NH_3)_y]^-$.
On standing,this solution reacts to form sodium amide $(NaNH_2)$ and hydrogen gas $(H_2)$:
$Na + NH_3 \longrightarrow NaNH_2 + \frac{1}{2} H_2$.
Thus,the solution is blue,$X$ is $NaNH_2$,and $Y$ is $\frac{1}{2} H_2$.

(D) $I$. $LiF$ has a very high lattice enthalpy due to the small size of both $Li^+$ and $F^-$ ions,which makes it sparingly soluble in water. This is a correct statement.
$II$. $LiBr$ is soluble in acetone due to its covalent character (Fajans' rule),which makes this statement incorrect as it claims it is not soluble.
$III$. $LiCl$ is soluble in pyridine due to its covalent character,which makes this statement correct.
$IV$. The melting point of alkali metal halides follows the order $MF > MCl > MBr > MI$ because the lattice energy decreases as the size of the halide ion increases. This is a correct statement.
Therefore,statements $I$,$III$,and $IV$ are correct.

(A) The hydration enthalpy of ions depends on their ionic size. Smaller ions have higher charge density and thus higher hydration enthalpy. The ionic size of alkali metal ions increases down the group as $Li^{+} < Na^{+} < K^{+} < Rb^{+} < Cs^{+}$. Therefore,the hydration enthalpy decreases as the size increases. The correct increasing order of hydration enthalpies is $Cs^{+} < Rb^{+} < K^{+} < Na^{+} < Li^{+}$.

(A) In a flame test,different alkali metals impart characteristic colours to the flame due to the excitation of electrons to higher energy levels.
$Li$ imparts a crimson red colour.
$Na$ imparts a golden yellow colour.
$K$ imparts a lilac (pale violet) colour.
$Cs$ (Caesium) imparts a blue colour to the flame.
Therefore,the correct answer is $Cs$.

(D) When alkali metals burn in the presence of oxygen,they form different oxides depending on their size and polarizing power:
$I$. Lithium,being the smallest,forms monoxide $(Li_2O)$.
$II$. Sodium,due to its size,forms peroxide $(Na_2O_2)$.
$III$. Larger alkali metals like Potassium,Rubidium,and Cesium form superoxides ($MO_2$,where $M = K, Rb, Cs$).
Therefore,all three statements are correct.

(A) In the flame test,the energy of the emitted radiation corresponds to the excitation of electrons from the ground state to higher energy levels. The energy difference $(\Delta E)$ between these levels decreases as the size of the alkali metal atom increases down the group $(Li < Na < K < Rb < Cs)$. Since energy is directly proportional to frequency $(\Delta E = h\nu)$,the frequency of the emitted radiation follows the same order as the energy difference. Therefore,the order of frequency is $Li > Na > K > Cs$.

(B) Alkali metals react with oxygen to form different types of oxides depending on their size and ionization energy.
$Li$ forms only the monoxide $(Li_2O)$.
$Na$ forms the peroxide $(Na_2O_2)$.
$K$,$Rb$,and $Cs$ form superoxides $(MO_2)$ when burnt in excess of air.
Therefore,the pair of elements that form superoxides is $K$ and $Rb$.

(C) Assertion $(A)$ is true: Alkali metals $K$,$Rb$,and $Cs$ react with excess oxygen to form superoxides of the type $MO_2$.
Reason $(R)$ is false: The stability of superoxides increases from $K$ to $Cs$ because the large size of the alkali metal cation stabilizes the large superoxide anion $(O_2^-)$ through a decrease in lattice energy,but the statement claims stability increases *due to* the decrease in lattice energy,which is a misinterpretation of the thermodynamic stability factor. More accurately,the stability increases due to the increasing size of the cation which matches the size of the large anion,reducing the lattice energy and making the formation more favorable. However,the reason provided is scientifically incorrect in its causal link.
Therefore,$(A)$ is true but $(R)$ is false.

(B) $(i)$ Superoxides contain the $O_2^-$ ion,which has one unpaired electron,making them paramagnetic. This statement is correct.
(ii) As we move down the group,the size of the metal ion increases,which decreases the lattice energy and increases the solubility/dissociation of hydroxides,thus increasing basic strength. This statement is correct.
(iii) Conductivity in aqueous solutions depends on ionic mobility. As we move down the group,the size of the hydrated ion decreases (due to less hydration),leading to higher ionic mobility and higher conductivity. Thus,conductivity increases down the group. This statement is incorrect.
(iv) The basic nature of carbonates is due to anionic hydrolysis (hydrolysis of the $CO_3^{2-}$ ion),not cationic hydrolysis. This statement is incorrect.
Therefore,only statements $(i)$ and $(ii)$ are correct.

(B) When sodium is heated in air at $300^{\circ} C$,it forms sodium peroxide $(X = Na_2O_2)$.
$2Na + O_2 \xrightarrow{300^{\circ} C} Na_2O_2$
Sodium peroxide reacts with $CO_2$ to form sodium carbonate and oxygen gas $(Y = O_2)$.
$2Na_2O_2 + 2CO_2 \rightarrow 2Na_2CO_3 + O_2$
Therefore,$Y$ is $O_2$.

(D) Sodium carbonate $(Na_2CO_3)$ is industrially prepared by the $Solvay$ process.
In this process,$CO_2$ and $NH_3$ are passed through a concentrated solution of sodium chloride (brine).
The reaction involves the formation of sodium bicarbonate $(NaHCO_3)$,which is then heated to obtain sodium carbonate.

(B) Let us analyze the reaction of each compound with $HCl$:
$1$. $Na_2CO_3 + 2HCl \rightarrow 2NaCl + H_2O + CO_2$ (Produces $1$ oxide: $CO_2$)
$2$. $NaNO_2 + HCl \rightarrow NaCl + HNO_2$ (Further decomposes to $H_2O + NO + NO_2$. Produces $2$ oxides: $NO$ and $NO_2$)
$3$. $Na_2SO_3 + 2HCl \rightarrow 2NaCl + H_2O + SO_2$ (Produces $1$ oxide: $SO_2$)
$4$. $NaHCO_3 + HCl \rightarrow NaCl + H_2O + CO_2$ (Produces $1$ oxide: $CO_2$)
Comparing the products,$NaNO_2$ produces $2$ oxides ($NO$ and $NO_2$),which is more than the others.

(C) The alkali metals react with oxygen to form different types of oxides depending on their size.
$Li$ forms only the monoxide $(Li_2O)$.
$Na$ forms the peroxide $(Na_2O_2)$.
$K, Rb,$ and $Cs$ have a strong tendency to form superoxides $(MO_2)$ due to the stabilization of the large superoxide anion $(O_2^-)$ by the large alkali metal cations.

(C) Potassium superoxide $(KO_{2})$ reacts with water to give potassium hydroxide,hydrogen peroxide,and oxygen gas.
The balanced chemical equation for the hydrolysis is:
$2KO_{2} + 2H_{2}O \rightarrow 2K^{+} + 2OH^{-} + H_{2}O_{2} + O_{2}$

(D) When sodium metal $(Na)$ reacts with water $(H_2O)$ at room temperature,it undergoes a vigorous exothermic reaction.
The chemical equation for this reaction is: $2Na(s) + 2H_2O(l) \rightarrow 2NaOH(aq) + H_2(g)$.
As shown in the equation,hydrogen gas $(H_2)$ is evolved.
Due to the highly exothermic nature of the reaction,some water also evaporates,resulting in the evolution of hydrogen gas along with water vapour.

(D) When sodium $(Na)$ metal is dissolved in liquid ammonia $(NH_3)$,it undergoes ionization to form ammoniated cations and ammoniated electrons:
$Na + (x+y)NH_3 \rightarrow [Na(NH_3)_x]^+ + [e(NH_3)_y]^-$
The blue colour of the solution is due to the excitation of the ammoniated electrons $[e(NH_3)_y]^-$ to higher energy levels,which absorb light in the visible region.

(C) The reduction of $NaOH$ to $Na$ is a highly endothermic and difficult process because $Na$ is a very strong reducing agent. Among the given options,$C$ (carbon) is capable of reducing $NaOH$ to $Na$ at high temperatures. The reaction is: $2NaOH + 2C \rightarrow 2Na + 2CO + H_2$.

(B) Let us analyze each reaction:
$1$. Reaction of fused $NaOH$ with $C$: $2NaOH + C \rightarrow Na_2CO_3 + H_2$. Hydrogen is liberated.
$2$. Reaction of $NaOH$ with sulphur: $6NaOH + 3S \rightarrow 2Na_2S + Na_2SO_3 + 3H_2O$. No hydrogen gas is liberated.
$3$. Heating concentrated $NaOH$ with $Si$: $Si + 2NaOH + H_2O \rightarrow Na_2SiO_3 + 2H_2$. Hydrogen is liberated.
$4$. Reaction of zinc with $NaOH$: $Zn + 2NaOH \rightarrow Na_2ZnO_2 + H_2$. Hydrogen is liberated.
Therefore,the correct option is $B$.

(C) The Solvay process is an industrial method used for the large-scale production of sodium carbonate $(Na_2CO_3)$,commonly known as soda ash.

(A) $1$. In option $A$,the reaction is $AlCl_3 + 3NaOH \longrightarrow Al(OH)_3(s) + 3NaCl(aq)$. This reaction produces a precipitate of aluminum hydroxide and aqueous sodium chloride; no gas is evolved.
$2$. In option $B$,white phosphorus reacts with $NaOH$ to produce phosphine gas $(PH_3)$.
$3$. In option $C$,aluminum reacts with $NaOH$ to produce sodium aluminate and hydrogen gas $(H_2)$.
$4$. In option $D$,zinc reacts with $NaOH$ to produce sodium zincate and hydrogen gas $(H_2)$.
Therefore,the reaction that does not liberate a gaseous product is $A$.

(D) In the extraction of sodium by Down's process,the electrolytic cell consists of an iron cathode and a graphite anode. The molten $NaCl$ is electrolyzed to produce sodium metal at the cathode and chlorine gas at the anode.

(C) $1$. Carbonates of alkaline earth metals $(MgCO_3, CaCO_3, SrCO_3, BaCO_3)$ are generally insoluble in water.
$2$. $Beryllium$ halides (e.g.,$BeCl_2$) are covalent due to the small size and high polarizing power of the $Be^{2+}$ ion.
$3$. Alkali metal superoxides (e.g.,$KO_2, RbO_2, CsO_2$) are paramagnetic and coloured (usually yellow or orange) due to the presence of the unpaired electron in the $\pi^* 2p$ molecular orbital.
$4$. Alkali metal halides have high negative enthalpies of formation due to high lattice energy.
Therefore,the statement that superoxides of alkali metals are colourless is incorrect.

(B) When calcium is heated in an atmosphere of $N_2$,it forms calcium nitride,$Ca_3N_2$,which is the ionic compound $A$.
$3Ca + N_2 \xrightarrow{\Delta} Ca_3N_2 (A)$
Calcium nitride reacts with water to produce calcium hydroxide and ammonia gas,$B$.
$Ca_3N_2 + 6H_2O \rightarrow 3Ca(OH)_2 + 2NH_3 (B)$
Therefore,$A$ is $Ca_3N_2$ and $B$ is $NH_3$.

(B) The reactions are as follows:
$1$. $2Mg + CO_2 \longrightarrow 2MgO + C$ ($MgO$ is formed).
$2$. $Mg + 2HNO_3 \text{ (dil.)} \longrightarrow Mg(NO_3)_2 + H_2$ ($MgO$ is not formed; $Mg(NO_3)_2$ is formed).
$3$. $2Mg + 2NO \xrightarrow{\Delta} 2MgO + N_2$ ($MgO$ is formed).
$4$. $3Mg + B_2O_3 \longrightarrow 3MgO + 2B$ ($MgO$ is formed).
Therefore,the correct option is $B$.

(C) The thermal stability of carbonates and bicarbonates determines how easily they release $CO_2$ upon heating.
$NaHCO_3$ (sodium bicarbonate) decomposes at a much lower temperature compared to the others: $2NaHCO_3 \rightarrow Na_2CO_3 + H_2O + CO_2$.
$CaCO_3$ and $Li_2CO_3$ require significantly higher temperatures for decomposition,while $Na_2CO_3$ is thermally very stable and does not decompose easily.

(D) Alkali metals $(Group \ 1)$ have only one valence electron per atom,which leads to weaker metallic bonding compared to alkaline earth metals $(Group \ 2)$,which have two valence electrons.
As a result,alkali metals are softer,have lower boiling points,and larger atomic/ionic radii than their corresponding alkaline earth metal counterparts.
Furthermore,due to their larger atomic size and lower effective nuclear charge,alkali metals have lower ionization enthalpy values compared to alkaline earth metals.

(C) The correct statement analysis is as follows:
$1$. The molecular formula of calgon is $Na_2[Na_4(PO_3)_6]$,which is correct.
$2$. Beryllium halides are covalent in nature due to the small size and high polarizing power of $Be^{2+}$ ions,making them soluble in organic solvents,which is correct.
$3$. Among alkali metals,Lithium $(Li)$ has the highest reducing property in aqueous solution due to its high hydration energy. Sodium $(Na)$ does not have the least reducing property; the trend is $Li > K > Rb > Cs > Na$. Thus,the statement is incorrect.
$4$. White metal (also known as Babbitt metal) is an alloy typically composed of tin,copper,and antimony,not Lithium. This statement is also incorrect.
However,in the context of standard competitive chemistry questions,the most widely recognized incorrect statement regarding the specific properties of alkali metals is that sodium has the least reducing property,as it contradicts the standard electrochemical series behavior in aqueous media.

(D) When $Mg$ is heated in air,it reacts with both $O_2$ and $N_2$ to form $MgO$ and $Mg_3N_2$ (which is $B$).
$2Mg + O_2 \xrightarrow{\Delta} 2MgO$
$3Mg + N_2 \xrightarrow{\Delta} Mg_3N_2 (B)$
Now,the hydrolysis of $Mg_3N_2$ occurs as follows:
$Mg_3N_2 + 6H_2O \rightarrow 3Mg(OH)_2 + 2NH_3$
Here,$Mg(OH)_2$ $(X)$ is sparingly soluble in water,and $NH_3$ $(Y)$ is a gas with a pungent odour.
Therefore,$X = Mg(OH)_2$ and $Y = NH_3$.

(A) Amphoteric metals like $Zn$ and $Al$ react with strong bases (alkaline medium) to produce hydrogen gas.
For example,$Zn + 2NaOH \rightarrow Na_2ZnO_2 + H_2$ and $2Al + 2NaOH + 2H_2O \rightarrow 2NaAlO_2 + 3H_2$.
Thus,the pair $Zn$ and $Al$ is capable of producing hydrogen in an alkaline medium.

(A) Electron is a magnesium-based alloy.
It typically consists of $Mg (95 \%)$,$Zn (4.5 \%)$,and $Cu (0.5 \%)$.
Therefore,the correct option is $A$.

(D) Carnallite is a hydrated potassium magnesium chloride mineral with the chemical formula $KCl \cdot MgCl_2 \cdot 6H_2O$.
It is an important source of potassium and magnesium.

(A) Only $KNO_{3}$ on heating does not produce $NO_{2}$. The thermal decomposition reactions are as follows:
$(a)$ $2 KNO_{3} \xrightarrow{\Delta} 2 KNO_{2} + O_{2}$
$(b)$ $2 Pb(NO_{3})_{2} \xrightarrow{\Delta} 2 PbO + 4 NO_{2} + O_{2}$
$(c)$ $2 Cu(NO_{3})_{2} \xrightarrow{\Delta} 2 CuO + 4 NO_{2} + O_{2}$
$(d)$ $2 AgNO_{3} \xrightarrow{\Delta} 2 Ag + 2 NO_{2} + O_{2}$
Alkali metal nitrates (like $KNO_{3}$) decompose to form nitrites and oxygen,whereas heavy metal nitrates decompose to form metal oxides (or metal),nitrogen dioxide,and oxygen.

(B) The thermal decomposition of metal nitrates follows specific patterns based on the reactivity of the metal.
$1$. Alkali metal nitrates like $KNO_3$ decompose to form metal nitrites and oxygen gas: $2KNO_3 \xrightarrow{\Delta} 2KNO_2 + O_2$.
$2$. Heavy metal nitrates like $AgNO_3$,$Cu(NO_3)_2$,and $Pb(NO_3)_2$ decompose to form metal oxides,nitrogen dioxide $(NO_2)$,and oxygen gas.
For example: $2Cu(NO_3)_2 \xrightarrow{\Delta} 2CuO + 4NO_2 + O_2$.
Therefore,$KNO_3$ does not produce $NO_2$ upon heating.

(C) In the Solvay process,$NaHCO_3$ is precipitated because it is sparingly soluble in water and can be easily separated by filtration.
However,in the case of potassium,the reaction is: $(NH_4)HCO_3 + KCl \longrightarrow KHCO_3(aq) + NH_4Cl(aq)$.
Since $KHCO_3$ is appreciably soluble in water,it does not precipitate out of the solution.
Therefore,it cannot be isolated from the reaction medium to be converted into $K_2CO_3$ by heating.

(D) The process described is the Solvay process for the manufacture of sodium carbonate.
Passing $CO_{2}$ $(X)$ through an ammoniacal brine solution $(NaCl + NH_{3} + H_{2}O)$ yields sodium bicarbonate $(Y)$ as a white precipitate.
$NH_{3} + CO_{2} + H_{2}O \longrightarrow NH_{4}HCO_{3}$
$NH_{4}HCO_{3} + NaCl \longrightarrow NaHCO_{3} \downarrow (Y) + NH_{4}Cl$
Upon heating,$NaHCO_{3}$ decomposes: $2NaHCO_{3} \xrightarrow{\Delta} Na_{2}CO_{3} (Z) + CO_{2} + H_{2}O$.
The weight loss percentage for $2NaHCO_{3} (2 \times 84 = 168 \ g)$ to $Na_{2}CO_{3} (106 \ g)$ is $\frac{168 - 106}{168} \times 100 \approx 36.9 \% \approx 37 \%$.
Thus,$(X) = CO_{2}$,$(Y) = NaHCO_{3}$,and $(Z) = Na_{2}CO_{3}$.