Alkali metals Questions in English

(C) Statement $I$ is correct: Lithium $(Li)$ and Magnesium $(Mg)$ have high charge density and small ionic sizes,which prevents the formation of larger superoxide ions. They typically form normal oxides ($Li_2O$ and $MgO$).
Statement $II$ is correct: The ionic radius of $Li^{+}$ is $76 \ pm$ $(0.76 \ \mathring{A})$ and the ionic radius of $Mg^{2+}$ is $72 \ pm$ $(0.72 \ \mathring{A})$. Since $76 \ pm$ > $72 \ pm$,the ionic radius of $Li^{+}$ is indeed larger than that of $Mg^{2+}$.
Therefore,both statements are correct.

(D) Assertion $A$ is true because sodium is significantly more abundant in seawater compared to potassium.
Reason $R$ is also true because potassium $(K^+)$ has a larger ionic radius than sodium $(Na^+)$ due to the presence of an additional shell.
However,the reason for the higher abundance of sodium in the ocean is not simply the size of the potassium ion,but rather the geochemical behavior of these elements. Potassium is more readily incorporated into silicate minerals (like feldspars) and clay minerals due to its size and charge,making it less likely to remain in solution in seawater compared to sodium. Therefore,$R$ is not the correct explanation for $A$.

(A) Sodium forms a stable superoxide $NaO_2$ when heated in excess of oxygen.
$(NH_4)_2BeF_4$ is a known complex compound.
$BeH_2$ is a polymeric hydride.
$PbEt_4$ (Tetraethyllead) is a well-known organometallic compound.
However,the question as stated is technically flawed because all listed compounds exist. If we consider the stability of alkali metal superoxides,$NaO_2$ is stable,but $LiO_2$ is not. Given the options,there might be a typo in the question source. Assuming the question intended to ask for a non-existent compound,none of these fit perfectly,but $NaO_2$ is often discussed in the context of stability compared to other alkali superoxides.

(D) Washing soda is the decahydrate form of sodium carbonate,represented as $Na_2CO_3 \cdot 10H_2O$,which contains $10$ molecules of water of crystallization.
Soda ash is the anhydrous form of sodium carbonate,represented as $Na_2CO_3$,which contains $0$ molecules of water of crystallization.
Therefore,the number of water molecules in washing soda and soda ash are $10$ and $0$ respectively.

(A) . $K$ is involved in the $Na^{+}/K^{+}$ pump (Sodium-Potassium pump) in biological systems. $(A-III)$
$B$. $KCl$ is widely used as a fertilizer in agriculture. $(B-II)$
$C$. $KOH$ is a strong base and is used as an absorbent for $CO_2$ gas. $(C-IV)$
$D$. $Li$ is used in thermonuclear reactions (specifically $Li-6$ isotope). $(D-I)$
Therefore,the correct matching is $A-III, B-II, C-IV, D-I$.

(D) In the alkali metal group $(Group \ 1)$,the melting point decreases as we move down the group due to the decrease in the strength of metallic bonding as the atomic size increases.
The order of melting points is $Li > Na > K > Rb > Cs$.
Therefore,$Cs$ (Cesium) has the lowest melting point.

(B) In general,moving down the group,the atomic mass increases more significantly than the atomic volume,leading to an increase in density for Group $I$ metals.
However,there is an anomaly at $K$ due to the presence of an empty $3d$ subshell,which causes a larger increase in atomic volume compared to the increase in mass.
The correct order of density for alkali metals is $Li < K < Na < Rb < Cs$.

(B) The standard enthalpy of formation $(\Delta_{f}H^0)$ for ionic compounds like sodium halides is primarily determined by the lattice energy and the hydration/sublimation energies.
As the size of the halide ion increases from $F^-$ to $I^-$,the lattice energy decreases (becomes less negative).
Consequently,the overall enthalpy of formation becomes less negative (i.e.,increases algebraically) in the order $NaF < NaCl < NaBr < NaI$.
Therefore,the correct order of $\Delta_{f}H^0$ values is $NaF < NaCl < NaBr < NaI$.

(B) $1$. All enzymes that utilize $ATP$ in phosphate transfer require $Mg$ as the cofactor.
$2$. Bone in the human body is not an inert and unchanging substance; it is continuously being solubilized and redeposited.
$3$. $Ca$ plays an important role in neuromuscular function,interneuronal transmission,cell membrane integrity,and blood coagulation.
$4$. The daily requirement of $Mg$ and $Ca$ in the human body is estimated to be $200-300 \ mg$ $(0.2-0.3 \ g)$.
Therefore,statement $B$ is correct.

(D) When an alkali metal like sodium dissolves in liquid ammonia,it forms a deep blue solution due to the presence of ammoniated electrons,$[e(NH_3)_y]^-$.
These ammoniated electrons absorb energy in the visible region of light,which imparts the blue color.
Because of the presence of unpaired electrons,the solution is paramagnetic.
The chemical reaction is: $M + (x+y) NH_3 \rightarrow [M(NH_3)_x]^+ + [e(NH_3)_y]^-$.
Therefore,the Assertion $A$ is true,but the Reason $R$ is false because the blue color is not due to the formation of amide.

(C) Statement $I$ is true because placing the $4f$ (lanthanoids) and $5f$ (actinoids) series separately prevents the periodic table from becoming unnecessarily wide and maintains the classification based on electronic configuration.
Statement $II$ is false because $S$-block elements (alkali and alkaline earth metals) are highly reactive due to their low ionization enthalpies and are therefore found in nature only in combined states (as ores,salts,etc.),not in their pure elemental form.

(D) The flame test is used to identify the presence of certain metal ions in a sample based on the characteristic color they impart to a flame.
Among the given ions,$Sr^{2+}$ (crimson red),$Ba^{2+}$ (apple green),$Ca^{2+}$ (brick red),and $Cu^{2+}$ (blue-green) show a positive flame test.
$Zn^{2+}$,$Co^{2+}$,and $Fe^{2+}$ do not give a characteristic flame test.
Therefore,the total number of metal ions that respond to the flame test is $4$.

(B) The characteristic flame emission wavelengths for alkali metals are as follows:
$Li$: $670.8 \ nm$ $(A-III)$
$Na$: $589.2 \ nm$ $(B-I)$
$Rb$: $780.0 \ nm$ $(C-IV)$
$Cs$: $455.5 \ nm$ $(D-II)$
Therefore,the correct matching is $A-III, B-I, C-IV, D-II$.

(B) Alkali metals dissolve in liquid ammonia to form a deep blue solution.
This blue color is primarily due to the presence of ammoniated (solvated) electrons,which absorb energy in the visible region of the spectrum.
$STATEMENT-1$ is True.
$STATEMENT-2$ is also True because alkali metals do form solvated cations $[M(NH_3)_n]^{+}$ and solvated electrons $[e(NH_3)_x]^{-}$.
However,the blue color is specifically attributed to the solvated electrons,not the solvated metal cations.
Therefore,$Statement-2$ is not the correct explanation for $Statement-1$.

(A) When sodium metal is heated in excess of air,it primarily forms sodium peroxide $(Na_2O_2)$.
The chemical reaction is: $2Na + O_2 (\text{excess}) \rightarrow Na_2O_2$.
Sodium oxide $(Na_2O)$ is formed when sodium is burned in a limited supply of air.
Therefore,the correct compound formed in excess air is $Na_2O_2$.

(C) The reaction between $Zn$ metal and aqueous $NaOH$ produces sodium zincate and hydrogen gas:
$Zn + 2NaOH_{(aq)} \longrightarrow Na_2ZnO_2 + H_2 \uparrow$
In other options,$Fe$ and $Cu$ react with conc. $HNO_3$ to produce nitrogen oxides ($NO_2$ or $NO$) due to the oxidizing nature of $HNO_3$,and $Au$ reacts with $NaCN$ to form a complex,not $H_2$ gas.

(B) The flame test is used to identify elements based on the characteristic color they impart to a flame:
$1$. Potassium $(K)$ imparts a violet color to the flame.
$2$. Calcium $(Ca)$ imparts a brick-red color to the flame.
$3$. Strontium $(Sr)$ imparts a crimson-red color to the flame.
$4$. Barium $(Ba)$ imparts an apple-green color to the flame.
Matching these:
$A-ii, B-i, C-iii, D-iv$.

(C) Sodium amide $(NaNH_{2})$ is prepared by passing dry ammonia gas over molten sodium at $300-400 \ ^{\circ}C$.
However,it is more commonly prepared by the reaction of sodium with liquid ammonia in the presence of a catalyst.
$Fe(NO_{3})_{3}$ (iron$(III)$ nitrate) acts as a catalyst for this reaction.
The reaction is represented as: $2Na + 2NH_{3} \xrightarrow{Fe(NO_{3})_{3}} 2NaNH_{2} + H_{2}$.

(A) Alkali metals react with liquid ammonia to form amides $(MNH_2)$ and hydrogen gas,except for lithium.
Lithium reacts with ammonia to form a complex compound known as tetraamminelithium,$[Li(NH_3)_4]$.
The chemical equation for this reaction is:
$Li + 4NH_3 \rightarrow [Li(NH_3)_4]$
Therefore,$Li$ does not form an amide under these conditions.

(C) Soft soap is manufactured using $KOH$ (potassium hydroxide),also known as caustic potash.
Potassium soaps are more soluble in water compared to sodium soaps.
In their concentrated form,these potassium-based soaps are referred to as soft soaps.
Therefore,$KOH$ is the alkali used for this purpose.

(C) The elements of group $1$ (alkali metals) have $1$ electron in their valence shell.
Among the given options,$Rb$ (Rubidium) belongs to group $1$,while $Ca$,$Ra$,and $Ba$ belong to group $2$ (alkaline earth metals).

(B) is the correct answer.
Alkali metals are the elements belonging to Group $1$ of the periodic table.
These include lithium $(Li)$,sodium $(Na)$,potassium $(K)$,rubidium $(Rb)$,caesium $(Cs)$,and francium $(Fr)$.
Among the given options,rubidium $(Rb)$ is an alkali metal.

(B) The alkali metals are the elements belonging to Group $1$ of the periodic table.
These include lithium $(Li)$,sodium $(Na)$,potassium $(K)$,rubidium $(Rb)$,caesium $(Cs)$,and francium $(Fr)$.
Among the given options,$Cs$ (caesium) is an alkali metal,while $Ba$,$Ca$,and $Sr$ are alkaline earth metals belonging to Group $2$.

(C) Cesium $(Cs)$ is the element used in photoelectric cells.
Explanation: Cesium is the preferred metal for photoelectric cells because it can easily convert sunlight into electricity.
It has a very low ionization energy and is the most electropositive element among the alkali metals.
The amount of energy required to eject electrons from a cesium surface is relatively small,only $206.5 \ kJ/mol$.

(D) $SO_3$ is an acidic oxide.
$NO$ is a neutral oxide.
$Al_2O_3$ is an amphoteric oxide.
$CaO$ is a basic oxide.
Generally,oxides of metals in Group $1$ and Group $2$ are basic,while oxides of non-metals are acidic.

(B) An amphoteric oxide is one that reacts with both acids and bases to form salt and water.
$ZnO$ is a classic example of an amphoteric oxide.
Reaction with acid: $ZnO + 2 HCl \rightarrow ZnCl_2 + H_2O$
Reaction with base: $ZnO + 2 NaOH \rightarrow Na_2ZnO_2 + H_2O$

(C) good reducing agent is a substance that readily loses electrons and has a low standard reduction potential. Among the given options,$Li$ (Lithium) has the most negative standard reduction potential $(E^\circ = -3.04 \ V)$,making it the strongest reducing agent in aqueous solution.

(B) The standard reduction potential $(E^\circ)$ is a measure of the tendency of a species to gain electrons.
Lithium $(Li)$ has the most negative standard reduction potential of approximately $-3.04 \ V$ among the alkali metals.
This is due to the very high hydration energy of the small $Li^+$ ion,which compensates for its high sublimation and ionization energy,making it the strongest reducing agent in aqueous solution.

(C) Magnesite is a mineral with the chemical formula $MgCO_3$.
It is a primary source of magnesium.
Limonite is an ore of iron,Cryolite is an ore of aluminium,and Magnetite is an ore of iron.

(C) Elements of group $1$ (alkali metals) are highly reactive. They rapidly lose their metallic luster in air due to the formation of a surface layer of oxide,peroxide,or superoxide by reacting with oxygen and moisture in the air. Among the given options,$K$ (potassium) is an alkali metal belonging to group $1$,while $Ba$,$Be$,and $Mg$ are alkaline earth metals (group $2$),which are relatively less reactive.

(A) Lithium carbonate $(Li_2CO_3)$ is thermally unstable compared to other alkali metal carbonates. Upon heating,it decomposes to form lithium oxide $(Li_2O)$ and carbon dioxide $(CO_2)$.
The balanced chemical equation is:
$Li_2CO_3 \xrightarrow{\Delta} Li_2O + CO_2$

(C) 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 reacted with excess air or oxygen.
Therefore,$K$ is the correct answer.

(C) The alkali metals are the elements of Group $1$ of the periodic table,which include Lithium $(Li)$,Sodium $(Na)$,Potassium $(K)$,Rubidium $(Rb)$,Caesium $(Cs)$,and Francium $(Fr)$.
Beryllium $(Be)$ belongs to Group $2$ of the periodic table,which are known as alkaline earth metals.
Therefore,Beryllium is not an alkali metal.

(C) Among the given options,$Na$ (Sodium) is an alkali metal. Alkali metals react with oxygen to form oxides,peroxides,or superoxides depending on their size. $Na$ reacts with excess oxygen to form sodium peroxide $(Na_2O_2)$.

(C) Alkali metals dissolve in liquid ammonia to form a deep blue coloured solution.
This colour is due to the presence of ammoniated electrons,which absorb energy in the visible region of the electromagnetic spectrum.

(D) The melting point and boiling point of alkali metals (group-$1$ elements) decrease down the group because the metallic bond strength decreases as the atomic size increases. Therefore,the statement that the melting point increases down the group is incorrect.

(B) The melting point of sodium clusters $(Na_n)$ increases as the number of atoms in the cluster increases.
Sodium clusters of $10^3$ atoms melt at $288 \ K$.
Sodium clusters of $10^4$ atoms melt at $303 \ K$.
Bulk sodium has a melting point of $371 \ K$.
Therefore,the decreasing order of melting point is: $\text{Bulk sodium} > \text{cluster of } 10^4 \text{ atoms} > \text{cluster of } 10^3 \text{ atoms}$.

(B) Group-$1$ elements (alkali metals) have a general electronic configuration of $ns^1$.
Upon losing one electron,they form unipositive ions $(M^+)$ with a stable noble gas configuration $(ns^2np^6)$,which are diamagnetic because all electrons are paired.
Therefore,the statement that compounds of unipositive ions are paramagnetic is incorrect.
They form colourless compounds in the $+1$ state because they lack unpaired electrons.
They are strong reducing agents and possess high negative values of standard reduction potential.

(D) Alkali metals are the most electropositive elements in the periodic table due to their low ionization enthalpy.
They have low density and form unipositive ions $(M^+)$ by losing their single valence electron.
Since they lose their only valence electron,the resulting ions have a noble gas configuration with no unpaired electrons,making their compounds diamagnetic.

(D) The general electronic configuration $ns^1$ corresponds to the group $1$ elements (alkali metals).
Among the given options,$Ca$,$Sr$,and $Ba$ belong to group $2$ (alkaline earth metals) with a general configuration of $ns^2$.
Francium $(Fr)$ belongs to group $1$,therefore it has the general electronic configuration $ns^1$.

(B) Cesium $(Cs)$ has the lowest ionization enthalpy among alkali metals.
Due to this,it easily loses electrons when light falls on its surface.
Therefore,it is used in the construction of photoelectric cells.

(D) The chemical formula of crystalline lithium chloride is $LiCl \cdot 2H_2O$.
Therefore,there are $2$ water molecules present in its crystalline structure.

(A) Alkali metals have a general electronic configuration of $ns^1$. When they form unipositive ions $(M^+)$,they lose their only valence electron,resulting in a stable noble gas configuration with no unpaired electrons. Therefore,their compounds are diamagnetic,not paramagnetic.
Alkali metals are indeed soft and can be cut with a knife.
They possess low density; for instance,$Li$,$Na$,and $K$ float on water.
Due to their low ionization energies,they readily lose their valence electron,making them the most electropositive elements.

(A) Potassium superoxide $(KO_{2})$ is used as a source of oxygen in submarines in emergency breathing apparatus because it has the ability to absorb carbon dioxide and release oxygen simultaneously.
$4KO_{2} + 2CO_{2} \longrightarrow 2K_{2}CO_{3} + 3O_{2} \uparrow$

(C) $Li_{2}NH$ is the chemical formula for lithium imide.
It is an inorganic compound that can be synthesized by the reaction between lithium amide and lithium hydride:
$LiNH_{2} + LiH \longrightarrow Li_{2}NH + H_{2}$
Lithium imide is a light-sensitive substance and can undergo disproportionation to form lithium nitride.

(A) When the fire extinguisher is operated by pressing the knob,the sulphuric acid reacts with the sodium bicarbonate $(NaHCO_3)$ solution.
This reaction produces a large amount of carbon dioxide $(CO_2)$ gas.
Since carbon dioxide is neither combustible nor a supporter of combustion,it effectively extinguishes the fire.
Therefore,$NaHCO_3$ is used in fire extinguishers.

(C) In the Solvay process,the recovery of ammonia is achieved by treating the ammonium chloride $(NH_4Cl)$ solution with slaked lime $(Ca(OH)_2)$.
The chemical reaction is as follows:
$2NH_4Cl + Ca(OH)_2 \rightarrow CaCl_2 + 2H_2O + 2NH_3$
As shown in the reaction,calcium chloride $(CaCl_2)$ is obtained as a by-product.

(B) Potassium superoxide $(KO_2)$ is used in breathing equipment for mountaineers and in submarines because it produces oxygen and absorbs carbon dioxide.
The chemical reaction is:
$4KO_2 + 2CO_2 \rightarrow 2K_2CO_3 + 3O_2$

(A) In the Solvay process,the preparation of sodium carbonate $(Na_2CO_3)$ involves passing $CO_{2(g)}$ through a concentrated solution of sodium chloride $(NaCl)$ and ammonia $(NH_3)$,known as ammoniated brine.
The reaction is: $NaCl + NH_3 + H_2O + CO_2 \rightarrow NaHCO_3 + NH_4Cl$.
The formed sodium bicarbonate $(NaHCO_3)$ is then heated to obtain sodium carbonate.

(B) Lithium imide is an inorganic compound with the chemical formula $Li_{2}NH$.
This white solid can be formed by a reaction between lithium amide and lithium hydride.
$LiNH_{2} + LiH \rightarrow Li_{2}NH + H_{2}$.
The product is light-sensitive and can undergo disproportionation to form lithium nitride,which is characteristically red.