Alkali metals Questions in English

(D) The Solvay process is used for the manufacture of sodium carbonate $(Na_2CO_3)$.
The key steps involve the reaction of brine ($NaCl$ solution) with ammonia $(NH_3)$ and carbon dioxide $(CO_2)$ to form ammonium bicarbonate $(NH_4HCO_3)$ and sodium bicarbonate $(NaHCO_3)$.
$NH_4HCO_3$ and $NaHCO_3$ are intermediates in this process.
$Na_2C_2O_4$ (sodium oxalate) is not involved in the Solvay process.

(C) When an alkali metal dissolves in liquid ammonia,it forms a deep blue solution. This colour is due to the presence of ammoniated electrons,which absorb energy in the visible region of the spectrum. The reaction is represented as: $M + (x+y)NH_3 \rightarrow [M(NH_3)_x]^+ + [e(NH_3)_y]^-$. The species $[e(NH_3)_y]^-$ is known as the ammoniated electron.

(A) The chemical reactivity of alkali metals (group $IA$) increases as we move down the group.
This is because the ionization energy decreases down the group,making it easier for the atom to lose its valence electron.
Among the given options,$Cs$ (Cesium) is at the bottom of the group (excluding $Fr$ which is radioactive),making it the most reactive element.

(D) When alkali metals dissolve in liquid ammonia,they undergo the following reaction: $M + (x+y)NH_3 \rightarrow [M(NH_3)_x]^+ + e^-(NH_3)_y$.
The blue colour of the solution is due to the excitation of ammoniated electrons to higher energy levels by absorbing red light.
The solution is highly conducting due to the presence of both ammoniated cations $[M(NH_3)_x]^+$ and ammoniated electrons $e^-(NH_3)_y$.

(A) The reaction $Ca(OH)_2 + Na_2CO_3 \to 2NaOH + CaCO_3$ represents the causticizing process.
Here,$A = Ca(OH)_2$,$B = NaOH$,and $C = CaCO_3$.
Also,$Ca(OH)_2$ reacts with $CO_2$ to form $CaCO_3$,which makes the solution milky.
Thus,the correct option is $A$.

(A) Efflorescence is the process in which a hydrated salt loses its water of crystallization to the atmosphere upon exposure to air,resulting in the formation of a lower hydrate or an anhydrous salt.
In the given reaction,washing soda $(Na_2CO_3 \cdot 10H_2O)$ loses $9$ molecules of water when exposed to air to form sodium carbonate monohydrate $(Na_2CO_3 \cdot H_2O)$.

(B) Alkali metals have the general electronic configuration $ns^1$. When they form salts (e.g.,$M^+X^-$),the metal ion $M^+$ achieves a stable noble gas configuration $(ns^2 np^6)$.
Since all electrons in the $M^+$ ion are paired,the salts are diamagnetic.
Furthermore,because there are no unpaired electrons to undergo $d-d$ transitions or charge transfer,these salts do not absorb visible light and are therefore colourless.

(A) The process in which a hydrated salt loses its water of crystallization when exposed to the atmosphere is known as $Efflorescence$.
$Deliquescence$ refers to the property of a substance to absorb moisture from the atmosphere and become liquid.
$Hydration$ is the addition of water,while $Dehydration$ is the removal of water,but the specific term for the loss of water of crystallization to air is $Efflorescence$.

(B) Anhydrous sodium carbonate,$Na_2CO_3$,is commonly referred to as $Soda \ ash$.
$Washing \ soda$ refers to the decahydrate form,$Na_2CO_3 \cdot 10H_2O$.

(C) The reactivity of alkali metals with water increases down the group.
This is because the ionization potential $(I.P.)$ decreases as we move down the group from $Li$ to $Cs$.
Lower $I.P.$ means the valence electron is more loosely held and can be easily lost to form the $M^+$ ion.
Since sodium $(Na)$ has a lower $I.P.$ than lithium $(Li)$,it loses its electron more easily,making it more reactive towards water.

(A) In the Solvay process for the production of sodium carbonate,ammonia is recovered by treating the ammonium chloride $(NH_4Cl)$ by-product with calcium hydroxide $(Ca(OH)_2)$.
The chemical reaction is:
$2NH_4Cl + Ca(OH)_2 \rightarrow 2NH_3 + 2H_2O + CaCl_2$
From the reaction,it is clear that $CaCl_2$ (calcium chloride) is obtained as a by-product.

(D) When an alkali metal like sodium dissolves in liquid $NH_3$,it undergoes ionization to form ammoniated cations and ammoniated electrons.
The reaction is: $Na + (x+y)NH_3 \rightarrow [Na(NH_3)_x]^+ + [e(NH_3)_y]^-$.
The blue colour of the solution is due to the presence of ammoniated electrons,which absorb energy in the visible region of light,resulting in the characteristic deep blue colour.

(A) When sodium nitrate $(NaNO_3)$ is heated,it decomposes to form sodium nitrite $(NaNO_2)$ and oxygen gas $(O_2)$.
The balanced chemical equation is:
$2NaNO_3(s) \xrightarrow{\Delta} 2NaNO_2(s) + O_2(g)$

(A) In the Solvay process,the overall reaction is $2NaCl + CaCO_3 \rightarrow Na_2CO_3 + CaCl_2$.
Ammonia $(NH_3)$ is recovered by treating the byproduct $NH_4Cl$ with lime $(Ca(OH)_2)$,which is obtained from the thermal decomposition of limestone $(CaCO_3 \rightarrow CaO + CO_2)$.
The $CO_2$ produced during the calcination of limestone and the $NH_3$ recovered from the ammonium chloride are recycled back into the process.

(A) $KO_2$ (potassium superoxide) reacts with $CO_2$ to produce $O_2$ and $K_2CO_3$.
The reaction is: $4KO_2 + 2CO_2 \to 2K_2CO_3 + 3O_2$.
This property makes it useful in space and submarines to maintain oxygen levels and remove carbon dioxide.

(A) Hygroscopic substances are those that absorb moisture from the atmosphere.
$MgCl_2$,$CaCl_2$,and $LiCl$ are hygroscopic in nature due to their high charge density and small ionic size,which allows them to strongly attract water molecules.
Pure $NaCl$ is not hygroscopic. It does not absorb water from the atmosphere under normal conditions. The moisture often observed in common table salt is due to the presence of impurities like $MgCl_2$ or $CaCl_2$.

(D) When alkali metals (Group $1$) dissolve in liquid ammonia,they form deep blue solutions due to the presence of ammoniated electrons. These solutions are paramagnetic and conduct electricity.
When alkaline earth metals (Group $2$) dissolve in liquid ammonia,they also form blue solutions,but the rate of decomposition is generally faster compared to Group $1$ metals.
However,the most significant difference lies in the products obtained upon evaporation of the solvent. Group $1$ metals typically yield the original metal upon evaporation of ammonia,whereas Group $2$ metals form ammoniates of the type $[M(NH_3)_6]^{2+}$.

(B) Among alkali metals,only $Li$ reacts directly with $N_2$ at room temperature to form nitride $(Li_3N)$.
$1$. $6Li + N_2 \xrightarrow{\Delta} 2Li_3N (B)$
$2$. $Li_3N + 3H_2O \rightarrow 3LiOH (C) + NH_3 (D)$
$3$. $LiOH + HCl \rightarrow LiCl (E) + H_2O$
Thus,$A=Li$,$B=Li_3N$,$C=LiOH$,$D=NH_3$,and $E=LiCl$.

(B) Hydration energy is directly proportional to the charge of the ion and inversely proportional to its ionic radius.
For ions with different charges,the charge density is the primary factor.
$Mg^{2+}$ has a charge of $+2$ and $Na^{+}$ has a charge of $+1$.
Since the charge of $Mg^{2+}$ is higher than that of $Na^{+}$,$Mg^{2+}$ has a higher hydration energy than $Na^{+}$.

(B) The solubility of alkali metal carbonates 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,the correct order of solubility is $Li_2CO_3 < Na_2CO_3 < K_2CO_3 < Rb_2CO_3$.

(A) Among the alkali metal carbonates,$Li_2CO_3$ is thermally unstable because of the small size of the $Li^+$ ion,which has a high polarizing power.
It decomposes upon heating as follows: $Li_2CO_3 \xrightarrow{\Delta} Li_2O + CO_2$.
Other alkali metal carbonates like $Na_2CO_3$ are thermally stable due to the larger size of the alkali metal cations.

(A) This reaction sequence represents the Solvay process for the manufacture of sodium carbonate.
Step $1$: $CO_2 + NH_3 + H_2O \to NH_4HCO_3$ $(X)$
Step $2$: $NH_4HCO_3 + NaCl \to NaHCO_3$ $(Y)$ $+ NH_4Cl$
Step $3$: $2NaHCO_3 \xrightarrow{\Delta} Na_2CO_3$ $(Z)$ $+ H_2O + CO_2$
Thus,'$Z$' is $Na_2CO_3$.

(A) Amphoteric metals like $Zn$,$Be$,and $Al$ react with strong bases like $NaOH$ to produce hydrogen gas $(H_2)$ and their respective salts.
For example:
$Zn + 2NaOH \rightarrow Na_2ZnO_2 + H_2$
$2Al + 2NaOH + 6H_2O \rightarrow 2Na[Al(OH)_4] + 3H_2$
$Be + 2NaOH \rightarrow Na_2BeO_2 + H_2$
However,$Mg$ is not amphoteric and does not react with $NaOH$ to evolve $H_2$ gas.

(B) For $IA$ group hydroxides ($LiOH$ to $CsOH$),the solubility increases down the group because the lattice energy $(LE)$ decreases more rapidly than the hydration energy $(HE)$.
For $IIA$ group carbonates ($BeCO_3$ to $BaCO_3$),the solubility decreases down the group because the lattice energy $(LE)$ remains relatively constant while the hydration energy $(HE)$ decreases significantly due to the increase in the size of the cation.

(B) The solubility of alkali metal bicarbonates increases as we move 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.
$LiHCO_3$ is the least soluble,while $CsHCO_3$ is the most soluble among the given options.

(D) The thermal stability of metal carbonates generally increases as the electropositive character of the metal increases.
Alkali metal carbonates (Group $1$) are thermally very stable compared to transition metal or alkaline earth metal carbonates.
Among alkali metal carbonates,stability increases down the group as the size of the cation increases,which reduces the polarizing power of the cation on the carbonate ion.
$K_2CO_3$ is more stable than $Na_2CO_3$ because $K^+$ has a larger ionic radius than $Na^+$.
However,$Na_2CO_3$ and $K_2CO_3$ are both highly stable,but $K_2CO_3$ is generally considered more stable due to the larger size of the $K^+$ ion.

(D) When sodium is burnt in moist air,the final product obtained is $Na_2CO_3$.
In moist air,sodium reacts with oxygen to form sodium oxide,which then reacts with moisture to form sodium hydroxide,and finally,it reacts with atmospheric carbon dioxide to form sodium carbonate.
The complete reaction sequence is as follows:
$4 Na + O_2 \rightarrow 2 Na_2O$
$2 Na_2O + 2 H_2O \rightarrow 4 NaOH$
$2 NaOH + CO_2 \rightarrow Na_2CO_3 + H_2O$
Thus,the ultimate product is $Na_2CO_3$.

(A) $6Li_{(s)} + N_{2(g)} \to 2Li_3N_{(s)}$
$Li_3N_{(s)} + 3H_2O_{(l)} \to 3LiOH_{(aq)} + NH_{3(g)}$
$CuSO_{4(aq)} + 4NH_{3(g)} \to [Cu(NH_3)_4]SO_{4(aq)}$ (deep blue complex)
Thus,$M$ is $Li$ and $B$ is $NH_3$.

(C)

Compound	Nature
$KO_2$	Superoxide
$BaO_2$	Peroxide
$SiO_2$	Oxide
$CsO_2$	Superoxide

An oxide is a chemical compound that contains at least one oxygen atom and one other element in its chemical formula,where the oxygen is in the $-2$ oxidation state. $SiO_2$ (silicon dioxide) is a typical oxide,whereas $KO_2$ and $CsO_2$ are superoxides (containing the $O_2^-$ ion) and $BaO_2$ is a peroxide (containing the $O_2^{2-}$ ion).

(D) $Na_2O_2$ is the peroxide of sodium,not the superoxide.
The formula of sodium superoxide is $NaO_2$.
Therefore,the statement that $Na_2O_2$ is the superoxide of sodium is incorrect.

(D) The solubility of ionic compounds in water is determined by the balance between lattice enthalpy and hydration enthalpy.
For alkali metal fluorides,the lattice enthalpy decreases significantly as the size of the cation increases from $Li^+$ to $Cs^+$.
Since lattice enthalpy is inversely proportional to solubility,the solubility increases as the size of the alkali metal cation increases.
Therefore,the correct order of solubility is $LiF < NaF < KF < RbF$.

(A) Metal halides undergo hydrolysis in water to form their corresponding hydroxides and hydrochloric acid $(HCl)$.
The acidity of the resulting solution depends on the nature of the metal hydroxide formed.
As we move down the group in the periodic table,the basic strength of the metal hydroxides increases due to the increase in atomic size and the decrease in ionization energy,which weakens the $M-OH$ bond.
$Be(OH)_2$ is amphoteric in nature,while $Mg(OH)_2$,$Ca(OH)_2$,and $Sr(OH)_2$ are basic.
Since $BeCl_2$ produces the least basic (or amphoteric) hydroxide,the solution containing $BeCl_2$ will have the highest concentration of $H^+$ ions due to the hydrolysis of the $Be^{2+}$ ion,resulting in the lowest $pH$ value.

(A) The melting point of ionic compounds is primarily determined by their lattice energy.
Lattice energy is inversely proportional to the interionic distance $(U \propto \frac{1}{r_+ + r_-})$.
Among the given alkali metal chlorides,$LiCl$ has the smallest cation size,but it exhibits significant covalent character due to Fajan's rule,which lowers its melting point.
$NaCl$ has a high lattice energy and a stable crystal structure,resulting in the highest melting point among the options provided.

(C) The reaction of metal $M$ (specifically $Mg$) with nitrogen gas is: $3Mg + N_2 \to Mg_3N_2$ $(Y)$.
$Mg_3N_2$ $(Y)$ reacts with water to produce ammonia gas: $Mg_3N_2 + 6H_2O \to 3Mg(OH)_2 + 2NH_3 \uparrow$ (colourless gas).
When $NH_3$ is passed through $CuSO_4$ solution,it forms a deep blue complex: $CuSO_4 + 4NH_3 \to [Cu(NH_3)_4]SO_4$ (Blue complex).
Therefore,$Y$ is $Mg_3N_2$.

(C) The reactivity of alkali metals with water increases as we move down the group from $Li$ to $Cs$.
This trend is due to the increase in the electropositive character and the decrease in ionization enthalpy down the group.
The order of reactivity is $Li < Na < K < Rb < Cs$.
Therefore,$Rb$ (Rubidium) reacts most vigorously with water among the given options.

(B) Among the alkali metals,only $Li$ reacts directly with $N_2$ of the air to form lithium nitride $(Li_3N)$.
This is due to the small size and high charge density of the $Li^+$ ion,which provides high lattice energy to stabilize the nitride ion $(N^{3-})$.

(D) $Na_{(s)} + (x + y)NH_3 \to Na^{+}(NH_3)_x + e^-(NH_3)_y$
The deep blue color of the solution is due to the presence of ammoniated electrons,which absorb light in the red region of the visible spectrum.

(B) The metal $Rb$ (Rubidium) reacts with excess air to form a superoxide,$X = RbO_2$.
Hydrolysis of $RbO_2$ with water follows the reaction: $2RbO_2 + 2H_2O \to 2RbOH + H_2O_2 + O_2 \uparrow$.
Thus,the metal is $Rb$.

(C) $I.$ Potassium ions $(K^{+})$ are known to activate many enzymes,such as pyruvate kinase.
$II.$ Potassium ions participate in the oxidation of glucose to produce $ATP$ by maintaining the electrochemical gradient necessary for cellular processes.
$III.$ Potassium ions,along with sodium ions $(Na^{+})$,are essential for the transmission of nerve signals by maintaining the resting membrane potential.
Therefore,all three statements are correct.

(A) The thermal decomposition of $LiNO_3$ is given by: $2LiNO_3 \xrightarrow{\Delta} Li_2O + 2NO_2 + \frac{1}{2}O_2$.
Thus,it produces $Li_2O$ (lithium oxide),not $LiNO_2$ (lithium nitrite).
Therefore,the statement in option $A$ is $INCORRECT$.

(B) Alkali metals are highly reactive and show specific chemical properties.
$1$. Alkali metal nitrates (except $LiNO_3$) decompose on heating to form nitrites and oxygen $(2MNO_3 \rightarrow 2MNO_2 + O_2)$. $LiNO_3$ decomposes to form $Li_2O$,$NO_2$,and $O_2$.
$2$. Alkali metal chlorides (except $LiCl$) are generally not deliquescent and do not form stable hydrates. $LiCl$ is deliquescent and exists as a hydrate $(LiCl \cdot 2H_2O)$. Therefore,the statement that their chlorides are deliquescent and crystalline as hydrates is not true for most alkali metals.
$3$. Alkali metals react vigorously with water to form hydroxides and hydrogen gas $(2M + 2H_2O \rightarrow 2MOH + H_2)$.
$4$. Alkali metals react with halogens to form ionic halides $(M^{+}X^{-})$.

(C) $NaOH$ is a strong base. It reacts with amphoteric metals like $Zn$,$Al$,and $Pb$ to produce hydrogen gas and corresponding salts.
$2Al + 2NaOH + 6H_2O \rightarrow 2Na[Al(OH)_4] + 3H_2$
$Zn + 2NaOH \rightarrow Na_2ZnO_2 + H_2$
$Pb + 2NaOH + 2H_2O \rightarrow Na_2[Pb(OH)_4] + H_2$
$Cu$ is a noble metal and does not react with $NaOH$. Therefore,$NaOH$ can be safely stored or transported in a container made of $Cu$.

(C) Alkali metals are highly reactive because they have only one electron in their valence shell,which can be easily lost,causing the metal to get oxidized.
In the presence of water,they react to form the highly flammable gas $Hydrogen$ $(H_2)$.
Therefore,to prevent contact with moisture and oxygen,they are stored in $Kerosene$.

(C) The reaction $M + O_2 \to MO_2$ represents the formation of a superoxide.
Alkali metals react with oxygen to form different oxides depending on their size.
$Li$ forms the oxide $Li_2O$,$Na$ forms the peroxide $Na_2O_2$,while heavier alkali metals like $K$,$Rb$,and $Cs$ form superoxides of the type $MO_2$.
Therefore,for the reaction $M + O_2 \to MO_2$,the metal $M$ is $K$.
Hence,option $C$ is correct.

(C) $1$. $LiHCO_3$ does not exist in solid state; it is only known in solution form.
$2$. Potassium superoxide $(KO_2)$ contains the paramagnetic superoxide ion $(O_2^-)$,which has one unpaired electron,so it is paramagnetic,not diamagnetic.
$3$. $Be(OH)_2$ is amphoteric in nature. It is insoluble in water but reacts with strong bases like $NaOH$ to form beryllates,e.g.,$Na_2[Be(OH)_4]$,making it soluble.
$4$. $Li_2CO_3$ is less thermally stable than $Na_2CO_3$ because the small $Li^+$ ion polarizes the carbonate ion more effectively,leading to easier decomposition into $Li_2O$ and $CO_2$.

(C) The reaction sequence is as follows:
$1$. $2NaHCO_3 \xrightarrow{\Delta} Na_2CO_3 + H_2O + CO_2(g)$. Here $A = NaHCO_3$,$B = Na_2CO_3$,and $C = CO_2$.
$2$. $Ca(OH)_2 + CO_2 \rightarrow CaCO_3(milky) + H_2O$. Here $D = CaCO_3$.
$3$. $CaCO_3 + H_2O + CO_2 \rightarrow Ca(HCO_3)_2$. Here $E = Ca(HCO_3)_2$.
Evaluating the statements:
- $A$ is $NaHCO_3$ (Correct).
- $B$ $(Na_2CO_3)$ is a metal carbonate soluble in $H_2O$ (Correct).
- $D$ is $CaCO_3$ (Correct).
- $pH$ of $NaHCO_3$ $(A)$ is basic $(> 7)$ and $Ca(HCO_3)_2$ $(E)$ is also slightly basic $(> 7)$. Thus,the statement that $pH$ is $7$ is incorrect.

(D) Sodium carbonate $(Na_2CO_3)$ is a thermally stable compound.
It does not decompose upon heating at ordinary temperatures.
Therefore,no reaction occurs,and '$A$' is not formed as a decomposition product.
Thus,the correct option is $D$.

(C) The reactivity of alkali metals with water increases down the group.
This is because the ionization potential $(I.P.)$ decreases as we move down the group from $Li$ to $Na$.
$A$ lower $I.P.$ means the valence electron is more loosely held and can be lost more easily,making the metal more reactive.

(D) When sodium dissolves in liquid ammonia,it undergoes ionization to form $Na^{+}$ ions and electrons.
These electrons get surrounded by ammonia molecules,forming solvated electrons,which are also known as ammoniated electrons.
These solvated electrons absorb light in the visible region,which imparts a deep blue color to the solution and provides it with strong reducing properties.