Alkali metals Questions in English

(D) The reaction of superoxides with water is given by: $2MO_2 + 2H_2O \rightarrow 2MOH + H_2O_2 + O_2$.
Thus,superoxides produce both hydrogen peroxide $(H_2O_2)$ and oxygen $(O_2)$ with water.
$CrO_3$ is indeed an acidic oxide.
$SnO_2$ is an amphoteric oxide as it reacts with both acids and bases.
$KO_2$ is a superoxide,not a peroxide.
Therefore,the statement '$KO_2$ is a peroxide which with $H_2O$ forms hydrogen peroxide only' is false.

(C) $1$. Thermal stability of alkali metal peroxides increases down the group due to the increase in the size of the cation,which stabilizes the large peroxide anion. Thus,$Na_2O_2 < K_2O_2 < Rb_2O_2 < Cs_2O_2$ is correct.
$2$. Thermal stability of alkali metal bicarbonates increases down the group because the lattice energy increases as the size of the cation increases. Thus,$LiHCO_3 < NaHCO_3 < KHCO_3 < RbHCO_3 < CsHCO_3$ is correct.
$3$. In $NaF, NaCl, NaBr, NaI$,the anion size increases,which leads to a decrease in lattice energy. Therefore,the melting point $(mpt)$ order should be $NaF > NaCl > NaBr > NaI$. The given order $NaF < NaCl < NaBr < NaI$ is incorrect.
$4$. According to Fajan's rule,as the charge on the cation increases $(Na^+ < Mg^{2+} < Al^{3+})$,the covalent character increases,which leads to a decrease in melting point. Thus,$NaCl > MgCl_2 > AlCl_3$ is correct.

(A) The electronic configuration of the element is $[Kr]4d^{10} 4f^{14} 5s^2 5p^6 6s^2$.
In this configuration,the last electron enters the $6s$ orbital.
According to the rules of the periodic table,if the last electron enters the $s-$orbital,the element belongs to the $s-$block.
Therefore,the element belongs to the $s-$block.

(D) Hydration energy is the amount of energy released upon solvation by water.
Hydration energy depends on the charge of the ion and the ionic radius.
Higher the charge and smaller the size,greater the hydration energy.
$H^{+}$ has the smallest size (it is essentially a bare proton) and hence has the maximum hydration energy among the four.

(C) The enthalpy of solution $(\Delta H_{sol})$ is the sum of the lattice enthalpy and the hydration enthalpy.
For alkali metal halides,as the size of the anion increases,the lattice enthalpy decreases significantly.
However,the hydration enthalpy also decreases as the anion size increases.
$F^-$ is a small anion,which results in a very high hydration enthalpy.
This high hydration energy released during the dissolution of $KF$ outweighs the lattice energy,making the overall enthalpy of solution negative (exothermic).
For other halides like $KCl$,$KBr$,and $KI$,the lattice energy is relatively lower,but the hydration energy is not high enough to make the process exothermic,resulting in a positive enthalpy of solution.

(A) The enthalpy of solution $(\Delta H_{sol})$ is the sum of lattice energy $(\Delta H_{lattice})$ and hydration energy $(\Delta H_{hyd})$.
For $LiF$,the lattice energy is very high due to the small size of both $Li^+$ and $F^-$ ions.
Since the lattice energy is significantly higher than the hydration energy,the overall enthalpy of solution for $LiF$ is positive,making it sparingly soluble in water at room temperature.

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

(B) The solubility of sulphates of alkaline earth metals decreases down the group due to the decrease in hydration energy being more significant than the decrease in lattice energy as the cationic size increases. The order is $BeSO_4 > MgSO_4 > CaSO_4 > SrSO_4 > BaSO_4$.
$Na_2SO_4$ is an alkali metal sulphate and is highly soluble in water compared to alkaline earth metal sulphates.
Therefore,the correct order of solubility is $Na_2SO_4 > BeSO_4 > MgSO_4 > BaSO_4$.

(C) The solubility of an ionic compound depends on the balance between its lattice energy and hydration energy.
As we move down the group,the size of the alkali metal cation increases.
Both the lattice energy and the hydration energy decrease as the ionic radius increases.
However,the decrease in lattice energy is more rapid than the decrease in hydration energy.
Since the lattice energy decreases more significantly,the energy required to break the crystal lattice is more easily compensated by the hydration energy,leading to an increase in solubility down the group.
Therefore,the correct option is $(C)$.

(A) The thermal stability of alkali metal hydrides decreases down the group. $LiH$ is quite stable and decomposes at much higher temperatures. $NaH$ is relatively less stable and decomposes in the range of $400-500\,^oC$.
Regarding carbonates,$Li_2CO_3$ is thermally unstable and decomposes into $Li_2O$ and $CO_2$ upon heating to $400-500\,^oC$. In contrast,$Na_2CO_3$ is highly stable and does not decompose at this temperature.
Therefore,both $NaH$ $(II)$ and $Li_2CO_3$ $(III)$ decompose in the given temperature range.

(B) The thermite reaction is a pyrotechnic composition of a metal powder and a metal oxide.
In the classic thermite reaction,a mixture of $Al$ powder and $Fe_2O_3$ is used.
The reaction is highly exothermic: $Fe_2O_3 + 2Al \rightarrow 2Fe + Al_2O_3 + \text{Heat}$.
This process is used for welding railway tracks.

(C) In the electrolysis of fused $MgCl_2$,the iron container acts as the cathode and a graphite rod acts as the anode. The magnesium metal is deposited at the cathode.

(D) Quicklime is calcium oxide $(CaO)$. When it is heated with coke $(C)$ in an electric furnace at high temperatures,it reacts to form calcium carbide $(CaC_2)$ and carbon monoxide $(CO)$.
The chemical reaction is as follows:
$CaO + 3C \rightarrow CaC_2 + CO$

(C) Among the alkali metals $(Group \ 1)$,only $Li$ reacts with atmospheric nitrogen to form a nitride $(Li_3N)$.
All alkaline earth metals $(Group \ 2)$ react with nitrogen to form nitrides.
When these nitrides react with moisture $(H_2O)$,they release ammonia $(NH_3)$ gas,which has a characteristic pungent smell.
$1. \ Mg + N_2 \to Mg_3N_2 \xrightarrow{H_2O} Mg(OH)_2 + NH_3$
$2. \ Ca + N_2 \to Ca_3N_2 \xrightarrow{H_2O} Ca(OH)_2 + NH_3$
$3. \ Li + N_2 \to Li_3N \xrightarrow{H_2O} LiOH + NH_3$
$4. \ K + N_2 \to \text{No reaction}$
Since $K$ does not form a nitride,it does not release ammonia upon exposure to moist air.

(D) As we move down the group in alkali metals,the atomic number increases,which leads to the addition of new shells.
Consequently,the atomic size increases,and the atomic radius increases.
Conversely,properties like first ionization energy,electronegativity,and hydration energy decrease as the atomic size increases.

(A) The thermal stability of alkali metal carbonates increases as we move down the group from $Li_2CO_3$ to $Cs_2CO_3$.
This is because the electropositive character of the alkali metal increases down the group,which increases the ionic character of the metal-oxygen bond.
As the ionic character increases,the lattice energy and the stability of the carbonate ion increase.
Therefore,$Cs_2CO_3$ has the highest thermal stability among the given options.

(D) The solubility of alkali metal perchlorates $(MClO_4)$ in water decreases as the size of the cation increases.
This is because the lattice energy of these salts decreases less rapidly than their hydration energy as the cation size increases.
Since $Li^+$ has the smallest ionic radius among the alkali metals,it has the highest hydration energy.
Therefore,$LiClO_4$ is the most soluble in water among the given options.

(B) When alkali metals like sodium dissolve in liquid ammonia,they undergo 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 to higher energy levels by absorbing light in the visible region.

(C) The basicity of alkali metal hydroxides depends on the ease with which the $OH^-$ ion can be released.
As we move down the group from $Li$ to $Cs$,the atomic size increases.
Due to the increase in size,the bond length between the metal cation $(M^+)$ and the hydroxide ion $(OH^-)$ increases,which decreases the bond strength.
Consequently,the $M-OH$ bond becomes easier to break,making the hydroxide more basic.
Therefore,the order of basicity is $LiOH < NaOH < KOH < RbOH < CsOH$.

(B) The metallic lustre exhibited by sodium is explained by the oscillation of loose electrons.
When light falls on the surface of sodium,the free electrons present in the metallic lattice start oscillating at their mean positions and get excited to higher energy levels.
Upon returning to their lower energy levels,these electrons emit light in all directions.
This phenomenon is responsible for the characteristic metallic lustre of sodium.

(B) When potassium is heated in an excess of oxygen,it forms potassium superoxide $(KO_2)$ as the major product.
$K + O_2 \xrightarrow{\Delta} KO_2$

(B) The reaction described is the Solvay process,where ammonia and carbon dioxide react with a concentrated solution of sodium chloride.
$NH_3 + H_2O + CO_2 \rightarrow NH_4HCO_3$
$NH_4HCO_3 + NaCl \rightarrow NaHCO_3 + NH_4Cl$
Since $NaHCO_3$ (sodium hydrogen carbonate) has low solubility in the presence of high concentrations of $NaCl$,it precipitates out as a white solid,which appears as a white cloud in the reaction mixture.

(C) When sodium metal is heated in a current of dry ammonia gas at $300-400 \ ^\circ C$,it reacts to form sodium amide $(NaNH_2)$ and hydrogen gas is evolved.
The balanced chemical equation is:
$2Na + 2NH_3 \rightarrow 2NaNH_2 + H_2$

(C) When sodium metal is heated in a current of dry ammonia gas,it reacts to form sodamide (also known as sodium amide) and releases hydrogen gas.
The chemical equation for this reaction is:
$2 Na + 2 NH_3 \rightarrow 2 NaNH_2 + H_2$

(C) Upon heating,$LiNO_3$ undergoes thermal decomposition to form lithium oxide,nitrogen dioxide,and oxygen gas.
The balanced chemical equation is:
$4LiNO_3 \rightarrow 2Li_2O + 4NO_2 + O_2$
$Li_2CO_3$ decomposes to $Li_2O$ and $CO_2$ but does not release $O_2$.
$LiOH$ and $NaOH$ are stable at moderate heating temperatures and do not release $O_2$.

(A) Sodium peroxide $(Na_2O_2)$ reacts with carbon dioxide $(CO_2)$ to produce sodium carbonate $(Na_2CO_3)$ and oxygen gas $(O_2)$.
The chemical reaction is: $2Na_2O_2(s) + 2CO_2(g) \rightarrow 2Na_2CO_3(s) + O_2(g)$.
This property makes it useful for regenerating oxygen and removing carbon dioxide in confined spaces like submarines.

(C) Washing soda is the common name for sodium carbonate decahydrate.
Its chemical formula is $Na_2CO_3 \cdot 10H_2O$.
Therefore,the correct option is $C$.

(C) When sodium metal is heated in an excess supply of air or oxygen,it primarily forms sodium peroxide $(Na_2O_2)$.
The chemical reaction is as follows:
$2Na + O_2 (\text{excess}) \rightarrow Na_2O_2$
Sodium oxide $(Na_2O)$ is formed when sodium is heated in a limited supply of air.

(D) Glauber's salt is the common name for the decahydrate form of sodium sulfate.
Its chemical formula is $Na_2SO_4 \cdot 10H_2O$.

(B) Sodium hydroxide $(NaOH)$ is produced on a large scale by the electrolysis of an aqueous solution of $NaCl$,a process known as the $Chlor-alkali$ process. The chemical reaction is: $2NaCl(aq) + 2H_2O(l) \rightarrow 2NaOH(aq) + Cl_2(g) + H_2(g)$.

(B) Sodium hydroxide $(NaOH)$ reacts with zinc $(Zn)$ to form sodium zincate $(Na_2ZnO_2)$ along with the liberation of inflammable hydrogen gas $(H_2)$.
$Zn + 2NaOH \rightarrow Na_2ZnO_2 + H_2(g)$

(B) The reducing power of a metal in an aqueous solution depends on the electrode potential,which is determined by the sum of sublimation energy,ionization energy,and hydration energy.
Although $Li$ has the highest ionization energy,it has the highest hydration energy due to its very small ionic size.
The large magnitude of hydration energy compensates for the high ionization energy,making $Li$ the strongest reducing agent in aqueous solution.

(C) . $NaHCO_3$ and $KHCO_3$ have different crystal structures; $NaHCO_3$ is monoclinic while $KHCO_3$ is monoclinic but with different parameters.
$B$. $LiNO_3$ decomposes on heating as $4LiNO_3 \rightarrow 2Li_2O + 4NO_2 + O_2$.
$C$. $Li$ imparts a crimson red colour to the flame,not just red.
$D$. $Li_2SO_4$ does not form alum because the $Li^+$ ion is too small to fit into the alum lattice structure.
Given the options,the statement in $C$ is often considered the most technically incorrect due to the specific flame colour description,but $A$ is also factually incorrect as they have different structures. However,in many standard contexts,$C$ is the intended answer as $Li$ is specifically 'crimson red'.

(B) The solubility of an ionic compound in water is determined by the balance between its lattice enthalpy and hydration enthalpy.
For a salt to be soluble,the hydration enthalpy must be sufficient to overcome the lattice enthalpy.
In the case of $LiF$,the $Li^{+}$ ion is very small,leading to a very strong electrostatic attraction with the $F^{-}$ ion.
This results in a very high lattice enthalpy for $LiF$,which is not compensated by the hydration enthalpy,making it the least soluble among alkali metal fluorides.

(C) The reducing power of a metal in an aqueous solution is determined by the standard electrode potential $(E^{\circ})$,which depends on the sum of sublimation enthalpy,ionization enthalpy,and hydration enthalpy.
Due to the very small size of the $Li^{+}$ ion,it has the highest hydration enthalpy among alkali metal ions.
This high negative value of hydration enthalpy compensates for the high ionization enthalpy of $Li$,making its standard reduction potential $(E^{\circ} = -3.04 \ V)$ the most negative.
Therefore,$Li$ acts as the strongest reducing agent in aqueous solution.

(A) The alkaline earth metals $Mg$ and $Be$ do not impart any characteristic color to the flame.
This is because the energy required to excite their electrons to higher energy levels is very high,which cannot be provided by the energy of the Bunsen flame.
Among the given options,$MgCl_2$ does not impart color to the flame.

(A) Magnesium metal reacts with nitrogen in air to form magnesium nitride,which upon hydrolysis produces ammonia gas.
$3 Mg + N_2 \xrightarrow{\Delta} Mg_3N_2$
$Mg_3N_2 + 6 H_2O \rightarrow 3 Mg(OH)_2 + 2 NH_3$

(D) Magnesium $(Mg)$ is an active metal that reacts with dilute acids to liberate hydrogen gas $(H_2)$.
$1$. Reaction with dilute $HCl$: $Mg + 2HCl \rightarrow MgCl_2 + H_2 \uparrow$
$2$. Reaction with dilute $H_2SO_4$: $Mg + H_2SO_4 \rightarrow MgSO_4 + H_2 \uparrow$
$3$. Reaction with very dilute $HNO_3$: $Mg + 2HNO_3 \rightarrow Mg(NO_3)_2 + H_2 \uparrow$
Since $Mg$ reacts with all these dilute acids to produce $H_2$ gas,the correct option is $D$.

(C) $Mg$ acts as a strong reducing agent and reduces $CO_2$ to carbon.
The chemical reaction is: $2Mg(s) + CO_2(g) \rightarrow 2MgO(s) + C(s)$.
In this reaction,$Mg$ removes oxygen from $CO_2$,thereby acting as a reducing agent.

(B) The reaction of potassium superoxide $(KO_2)$ with carbon dioxide $(CO_2)$ and water $(H_2O)$ proceeds as follows:
$4KO_2 + 2CO_2 + 2H_2O \rightarrow 4KHCO_3 + 3O_2$
In the presence of excess $CO_2$,the primary product formed is potassium bicarbonate $(KHCO_3)$ and oxygen gas $(O_2)$ is released.
Therefore,$[X] = KHCO_3$ and $[Y] = O_2$.

(B) The solubility of ionic compounds in water depends on the balance between lattice energy and hydration energy.
For alkali metal salts with small anions (like $F^-$),lattice energy dominates,and solubility increases as the cation size decreases $(LiF > NaF > KF)$.
For salts with large anions (like $HCO_3^-$,$CO_3^{2-}$,$NO_3^-$),hydration energy dominates,and solubility increases as the cation size increases.
Therefore,for bicarbonates $(HCO_3^-)$,the solubility order is $NaHCO_3 < KHCO_3 < RbHCO_3$.
Thus,option $B$ is the correct order.

(C) The enthalpy of solution,$\Delta H_{sol}$,for many ammonium salts is positive (endothermic).
However,they are highly soluble because the entropy change,$\Delta S_{sol}$,is significantly positive.
According to the Gibbs free energy equation,$\Delta G_{sol} = \Delta H_{sol} - T\Delta S_{sol}$,a large positive $\Delta S_{sol}$ makes the $\Delta G_{sol}$ value negative,which favors the dissolution process.

(D) Lithium exhibits anomalous behavior compared to other alkali metals due to its small size and high polarizing power.
$1$. $Li_2CO_3$ decomposes into $Li_2O$ and $CO_2$ upon heating,whereas other alkali metal carbonates are thermally stable: $Li_2CO_3 \xrightarrow{\Delta} Li_2O + CO_2$.
$2$. $LiCl$ is predominantly covalent in nature due to Fajan's rule,making it soluble in organic solvents like ether.
$3$. $Li$ reacts directly with $N_2$ to form $Li_3N$,while other alkali metals do not form stable nitrides under similar conditions: $6Li + N_2 \rightarrow 2Li_3N$.
Therefore,all the given statements are correct.

(D) Due to the very small ionic radius of $Li^+$,it shows anomalous behavior compared to other members of the group.
$1$. $Li_2SO_4$ has low solubility in water compared to other alkali metal sulphates.
$2$. All alkali metal sulphates except $Li_2SO_4$ form alums with trivalent metal sulphates (e.g.,$M_2SO_4 \cdot M_2'(SO_4)_3 \cdot 24H_2O$).
$3$. Alkali metal sulphates are generally stable to heat,but $Li_2SO_4$ decomposes at high temperatures,whereas others do not.
Therefore,all the given statements are correct.

(C) Alkali metals possess metallic lustre when freshly cut because they contain loosely bound valence electrons (free electrons) in their metallic lattice.
These electrons absorb incident light (photons) and re-emit them,which gives rise to the characteristic metallic lustre.

(C) $1$. $Li_2CO_3$ is sparingly soluble in water due to high lattice energy and small size of $Li^+$. $LiHCO_3$ is unstable in solid form and has not been isolated. This statement is correct.
$2$. $K_2CO_3$ cannot be prepared by the Solvay process because $KHCO_3$ is too soluble in water to precipitate out,unlike $NaHCO_3$. This statement is correct.
$3$. $Li_2CO_3$ and $MgCO_3$ are thermally unstable. They decompose on heating to form their respective oxides and $CO_2$ (e.g.,$Li_2CO_3 \xrightarrow{\Delta} Li_2O + CO_2$). This statement is incorrect.
$4$. The mineral trona has the formula $Na_2CO_3 \cdot NaHCO_3 \cdot 2H_2O$. This statement is correct.

(C) The mixture of $Na_2CO_3$ and $K_2CO_3$ is known as fusion mixture.
It melts at a lower temperature $(700^{\circ}C)$ compared to the individual melting points of $Na_2CO_3$ $(851^{\circ}C)$ and $K_2CO_3$ $(891^{\circ}C)$.

(NONE) $1$. The stability of peroxides $(O_2^{2-})$ and superoxides $(O_2^-)$ of alkali metals increases as the size of the alkali metal cation increases. This is because larger cations stabilize larger anions more effectively through lattice energy effects. Thus,statement $(a)$ and $(c)$ are correct.
$2$. $NaOH$ is highly hygroscopic and absorbs moisture from the atmosphere to form a concentrated solution,but it does not form a stable crystalline hydrated salt like $LiOH \cdot H_2O$. Thus,statement $(b)$ is correct.
$3$. The solubility of alkali metal halides depends on the balance between lattice energy and hydration energy. $LiF$ has very high lattice energy due to the small size of both $Li^+$ and $F^-$,making it sparingly soluble. $CsI$ has low lattice energy,but its solubility is limited by the very low hydration energy of the large $Cs^+$ and $I^-$ ions. However,the statement provided in $(d)$ is generally accepted in textbooks as a correct explanation for the solubility trends of these specific salts.
$4$. Upon re-evaluating the options,all statements provided are scientifically correct. If this is a multiple-choice question where one must be incorrect,there may be a typo in the source. However,based on standard chemical principles,all statements are true.

(D) Alkali metals dissolve in liquid $NH_3$ to form a deep blue solution.
The blue colour is due to the presence of ammoniated (solvated) electrons,which absorb energy in the visible region of light.
These dilute solutions are paramagnetic due to the presence of unpaired electrons.
As the concentration increases,the solution turns bronze-coloured and becomes diamagnetic due to the pairing of electrons.
Therefore,all the given statements are correct.

(D) Lithium nitrate $(LiNO_3)$ is the only alkali metal nitrate that decomposes to its oxide $(Li_2O)$ upon strong heating.
The reaction is: $4LiNO_3(s) \xrightarrow{\Delta} 2Li_2O(s) + 4NO_2(g) + O_2(g)$.
Other alkali metal nitrates like $NaNO_3$ and $KNO_3$ decompose to their respective nitrites and oxygen gas: $2MNO_3(s) \xrightarrow{\Delta} 2MNO_2(s) + O_2(g)$ (where $M = Na, K$).