Alkali metals Questions in English

(A) Lithium is the smallest in size among the alkali metals.
Due to its small size,the $Li^{+}$ ion has a high charge density and can polarize water molecules more effectively than other alkali metal ions.
This high polarizing power allows $LiCl$ to crystallize as a hydrate (specifically $LiCl \cdot 2H_2O$).
As the size of the alkali metal ions increases down the group,their polarizing power decreases,which is why other alkali metal chlorides typically form anhydrous salts.

(A) Baking powder is a mixture of $Sodium$ hydrogen carbonate $(NaHCO_3)$ and a mild edible acid such as $Potassium$ hydrogen tartrate (cream of tartar) or tartaric acid,along with starch to keep it dry.
$Sodium$ carbonate $(Na_2CO_3)$ is not a component of baking powder; it is a component of washing soda.
Therefore,the correct answer is $A$.

(C) The Solvay process is used for the industrial production of sodium carbonate $(Na_2CO_3)$.
The raw materials required are sodium chloride $(NaCl)$,ammonia $(NH_3)$,and limestone $(CaCO_3)$.
The chemical reactions involved are:
$2 NH_3 + H_2O + CO_2 \rightarrow (NH_4)_2CO_3$
$(NH_4)_2CO_3 + H_2O + CO_2 \rightarrow 2 NH_4HCO_3$
$NH_4HCO_3 + NaCl \rightarrow NaHCO_3 + NH_4Cl$
$2 NaHCO_3 \xrightarrow{\Delta} Na_2CO_3 + H_2O + CO_2$
Thus,the correct set of raw materials is $NaCl, NH_3, CaCO_3$.

(C) When $Zn$ reacts with an excess of $NaOH$ solution,it forms sodium zincate and releases hydrogen gas.
The reaction is:
$Zn + 2NaOH + 2H_2O \longrightarrow Na_2[Zn(OH)_4] + H_2$
This can also be represented as:
$Zn + 2NaOH \longrightarrow Na_2ZnO_2 + H_2$
Thus,the product obtained is sodium zincate $(Na_2ZnO_2)$.

(D) When magnesium burns in air,it reacts with both oxygen and nitrogen present in the atmosphere.
The chemical reactions are as follows:
$2Mg(s) + O_2(g) \rightarrow 2MgO(s)$
$3Mg(s) + N_2(g) \rightarrow Mg_3N_2(s)$
Therefore,the products formed are a mixture of magnesium oxide $(MgO)$ and magnesium nitride $(Mg_3N_2)$.

(C) In the Solvay process,$NH_4Cl$ is treated with slaked lime $(Ca(OH)_2)$ to recover ammonia $(NH_3)$.
The chemical reaction is as follows:
$2NH_4Cl + Ca(OH)_2 \rightarrow CaCl_2 + 2H_2O + 2NH_3$
Thus,$NH_3$ is recovered.

(B) Chlorophyll is a green pigment found in plants that is essential for photosynthesis.
It contains a central metal ion coordinated within a porphyrin ring.
The central metal ion in the chlorophyll molecule is $Mg^{2+}$ (Magnesium).

(C) $1$. Metals that have low ionization enthalpies are used in photoelectric cells.
$2$. The most popular metal used in photoelectric cells is Caesium $(Cs)$,which has the atomic number $55$.
$3$. Caesium belongs to group $1$ (alkali metals) and period $6$.
$4$. The electronic configuration of Caesium is $[Xe] 6s^1$.
$5$. Caesium is used widely in photocells as it can easily convert solar energy into electrical energy.
$6$. When Caesium is exposed to sunlight,it starts emitting electrons,which induces electric flow and generates electricity.
$7$. Thus,Caesium is the most preferred metal used in photoelectric cells.

(A) Lithium $(Li)$ shows a diagonal relationship with Magnesium $(Mg)$.
Due to its small ionic size and high charge density,lithium differs from other alkali metals but resembles magnesium,as the ionic size of $Li^+$ is very close to that of $Mg^{2+}$.

(A) Caesium $(Cs)$ has a very low ionization enthalpy,which makes it highly sensitive to light. Due to this property,it is widely used in the manufacturing of photoelectric cells.

(A)
Liquid sodium is used as a coolant in fast breeder nuclear reactors due to its high thermal conductivity and low melting point.

(A) The electronic configuration of Francium $(Fr)$ is $[Rn] 7s^1$.
When it loses one electron to form a $+1$ oxidation state,it becomes $Fr^+ = [Rn]$.
Since $[Rn]$ is the electronic configuration of the noble gas Radon,$Fr$ achieves a noble gas configuration in the $+1$ state.
Other elements like $Ca$,$Mg$,and $Sr$ are alkaline earth metals and typically form $+2$ oxidation states to achieve noble gas configurations.

(D) The tendency to form a metal oxide is related to the stability of the oxide formed and the reactivity of the metal.
According to the Ellingham diagram,metals that are more reactive and have a more negative Gibbs free energy of formation $(\Delta G_f^\circ)$ for their oxides have a greater tendency to form oxides.
Among the given metals ($Cr$,$Fe$,$Al$,$Ca$),$Ca$ is the most reactive alkaline earth metal.
$Ca$ forms a highly stable ionic oxide $(CaO)$ with a very large negative value of $\Delta G_f^\circ$.
Therefore,$Ca$ has the greatest tendency to form its oxide.

(C) Standard reduction potential is a measure of the tendency of an element to lose electrons in an aqueous solution.
More negative the reduction potential,higher is the ability of the element to lose electrons,and hence stronger is the reducing character.
Therefore,alkali metals behave as reducing agents.

(C) When carbon monoxide $(CO)$ is passed over solid caustic soda $(NaOH)$ at $200^{\circ} C$ and high pressure,it undergoes an addition reaction to form sodium formate $(HCOONa)$.
$CO + NaOH \xrightarrow{200^{\circ} C, \ 10 \ atm} HCOONa$

(D) Among the given nitrates,only $LiNO_3$ decomposes to give $NO_2$ on heating,whereas the nitrates of $Na, K,$ and $Rb$ decompose to form their respective nitrites and release oxygen gas.
The decomposition reaction for $LiNO_3$ is:
$4 LiNO_3 \xrightarrow{\Delta} 2 Li_2O + 4 NO_2 + O_2$
The decomposition reaction for other alkali metal nitrates $(M = Na, K, Rb)$ is:
$2 MNO_3 \xrightarrow{\Delta} 2 MNO_2 + O_2$

(D) The thermal decomposition of alkali metal nitrates follows different patterns based on the size of the cation.
$LiNO_{3}$ decomposes to form its oxide,nitrogen dioxide,and oxygen: $4LiNO_{3} \rightarrow 2Li_{2}O + 4NO_{2} + O_{2}$.
In contrast,nitrates of other alkali metals like $KNO_{3}$,$RbNO_{3}$,and $NaNO_{3}$ decompose to form nitrites and oxygen gas $(2MNO_{3} \rightarrow 2MNO_{2} + O_{2})$ and do not release $NO_{2}$ gas.
Therefore,the correct option is $D$.

(A) Amphoteric metals react with both acids and bases to liberate $H_{2}$ gas.
$Zn + 2 HCl \text{ (dil.)} \longrightarrow ZnCl_{2} + H_{2}$
$Zn + 2 NaOH + 2 H_{2}O \longrightarrow Na_{2}[Zn(OH)_{4}] + H_{2}$
$Mg$,$Ca$,and $Fe$ are not amphoteric and do not react with aqueous $NaOH$ to produce $H_{2}$.

(B) The reaction of sodium with water is highly exothermic and vigorous.
By forming an amalgam of sodium with mercury $(Na-Hg)$,the activity of sodium is reduced,which makes the reaction with water less vigorous.

(D) Potassium forms various oxides such as $K_{2}O$ (oxide),$K_{2}O_{2}$ (peroxide),and $KO_{2}$ (superoxide).
$K_{2}O_{3}$ (potassium sesquioxide) does not exist because the $O_{3}^{2-}$ ion is not a stable chemical entity.

(C) The electronic configuration $1s^{2} 2s^{2} 2p^{6} 3s^{1}$ corresponds to the element Sodium ($Na$,atomic number $11$).
Sodium is an alkali metal belonging to Group $1$ of the periodic table.
Alkali metals react with oxygen to form oxides of the general formula $M_{2}O$ (e.g.,$Na_{2}O$).
These oxides are strongly basic in nature as they react with water to form hydroxides.

(C) In general, density increases on moving downward in a group, but the density of potassium $(K)$ is lower than that of sodium $(Na)$.
This is due to an abnormal increase in atomic size from $Na$ $(186 \text{ pm})$ to $K$ $(227 \text{ pm})$.
Therefore, the correct order of density is $Li < K < Na < Rb < Cs$.
Thus, the trend given in option $C$ is incorrect.

(A) Alkali metals have low ionisation energy due to their large atomic size.
They possess the minimum value of ionisation energy in their respective periods.
Therefore,the statement that they have high ionisation energy is incorrect.

(A) Potassium superoxide $( KO_{2} )$ reacts with water to produce potassium hydroxide,hydrogen peroxide,and oxygen gas. This reaction is represented as follows:
$ 2 KO_{2} + 2 H_{2} O \rightarrow 2 KOH + H_{2} O_{2} + O_{2} $
In contrast,other metal oxides react differently:
$ Na_{2} O_{2} + 2 H_{2} O \rightarrow 2 NaOH + H_{2} O_{2} $ (Produces peroxide but no oxygen)
$ CaO + H_{2} O \rightarrow Ca(OH)_{2} $ (Produces hydroxide only)
$ Li_{2} O + H_{2} O \rightarrow 2 LiOH $ (Produces hydroxide only)

(B) The ionic character of metal hydrides depends on the electronegativity difference between the metal and hydrogen.
As we move down the group from $Li$ to $Cs$,the electronegativity of the alkali metal decreases.
Consequently,the electronegativity difference between the metal and hydrogen increases,leading to an increase in the ionic character of the metal hydride.
Therefore,the order of increasing ionic character is $LiH < NaH < KH < RbH < CsH$.
Hence,the correct option is $B$.

(D) $Li-Mg$ alloy is used in making armour plates due to its high strength-to-weight ratio.
$Cu-Be$ alloy is used in high-strength springs because of its excellent fatigue resistance and mechanical properties.
$Mg-Al$ alloy (Magnalium) is used in aircraft construction due to its lightweight and high strength properties.
Therefore,all the given pairs are correctly matched.

(D) Strongly electropositive metals form strongly basic oxides that dissolve in water to give alkalis.
$Na_2O$ contains the strongly electropositive $Na$ metal,which reacts with water to form $NaOH$,a strong base.
$Na_2O + H_2O \rightarrow 2NaOH$.

(C) $I$. The tendency to form halide hydrates decreases down the group as the size of the cation increases,which reduces the hydration energy.
$II$. As we move down the group,the size of the alkali metal cation increases,which stabilizes the large superoxide ion $(O_2^-)$ more effectively due to reduced polarizing power.
$III$. $LiF$ has a very high lattice energy due to the small size of both $Li^+$ and $F^-$ ions,which outweighs its hydration energy,leading to low solubility.
$IV$. The solubility of group-$2$ carbonates decreases down the group because the lattice energy decreases less rapidly than the hydration energy.
Therefore,statements $II$ and $III$ are correct.

(A) The strength of a reducing agent is determined by its ability to lose electrons,which is measured by its standard oxidation potential.
Greater the negative value of standard reduction potential $(E^{\circ}_{red})$,the stronger is the reducing agent.
Among the given elements,alkali metals ($Rb$ and $Na$) have more negative reduction potentials than alkaline earth metals ($Sr$ and $Mg$).
Within the alkali metal group,the reduction potential becomes more negative down the group as the ionization energy decreases.
Therefore,$Rb$ has the most negative standard reduction potential,making it the strongest reducing agent among the options provided.

(D) The reducing power of a metal is determined by its standard electrode potential $(E^{\circ})$. $A$ more negative $E^{\circ}$ value indicates a stronger reducing agent,while a less negative (or more positive) $E^{\circ}$ value indicates a weaker reducing agent.
The standard electrode potentials $(E^{\circ})$ for the given alkali metals are as follows:

Metal Ion/Metal	$E^{\circ}$ (in Volts)
$Li^{+}/Li$	$-3.05$
$Na^{+}/Na$	$-2.71$
$K^{+}/K$	$-2.93$
$Cs^{+}/Cs$	$-2.92$

Comparing the values,$Na$ has the least negative $E^{\circ}$ value $(-2.71 \ V)$. Therefore,$Na$ is the weakest reducing agent among the given options.

(A) Carnallite is a double salt. It dissociates into its constituent ions completely when dissolved in water to give $5$ ions,i.e.,$K^{+}$,$Mg^{2+}$,and $3 Cl^{-}$.
$KCl \cdot MgCl_2 \cdot 6 H_2O \longrightarrow K^{+} + Mg^{2+} + 3 Cl^{-} + 6 H_2O$

(A) Hydration enthalpy is the amount of energy released when one mole of ions is dissolved in excess of water.
It is directly proportional to the charge density of the ion.
Hydration enthalpy increases as the ionic size decreases and the ionic charge increases.
Comparing the given ions: $Na^{+}$ and $Mg^{2+}$ belong to the $3^{rd}$ period,while $K^{+}$ and $Ca^{2+}$ belong to the $4^{th}$ period.
Ions with smaller size ($Na^{+}$ and $Mg^{2+}$) have higher hydration enthalpy than those with larger size ($K^{+}$ and $Ca^{2+}$).
Between $Na^{+}$ and $Mg^{2+}$,$Mg^{2+}$ has a smaller ionic radius and a higher charge ($+2$ vs $+1$).
Therefore,$Mg^{2+}$ has the highest charge density and the highest hydration enthalpy.

(D) The correct option is $(d)$.
Among the given salts,$CaCl_2$,$LiCl$,and $BaCl_2$ have the ability to form hydrates such as $CaCl_2 \cdot 6H_2O$,$LiCl \cdot 2H_2O$,and $BaCl_2 \cdot 2H_2O$.
This is due to the relatively higher hydration enthalpies of $Ca^{2+}$,$Li^{+}$,and $Ba^{2+}$ ions compared to $Na^{+}$ and $K^{+}$.
$NaCl$ and $KCl$ do not form hydrates under normal conditions.

(A) Group $I$ elements form peroxides $(M_2O_2)$ and superoxides $(M'O_2)$.
Hydrolysis of peroxide: $M_2O_2 + 2H_2O \longrightarrow 2MOH + H_2O_2$. Here,$X = H_2O_2$.
Hydrolysis of superoxide: $2M'O_2 + 2H_2O \longrightarrow 2M'OH + H_2O_2 + O_2$. Here,$X = H_2O_2$ and $Y = O_2$.
Thus,$X = H_2O_2$ and $Y = O_2$.

(B) Hydrated sodium aluminium silicate is known as $Zeolite$.
Its general chemical formula is $Na_2Al_2Si_2O_8 \cdot xH_2O$.

(C) Nitric acid $(HNO_3)$ is a strong oxidizing agent. It usually oxidizes the $H_2$ gas produced to water and itself gets reduced to nitrogen oxides.
However,magnesium $(Mg)$ and manganese $(Mn)$ react with very dilute nitric acid to evolve hydrogen gas.
$Mg + 2HNO_3 \text{ (very dilute)} \rightarrow Mg(NO_3)_2 + H_2 \uparrow$

(D) The ability to form hydrates depends on the hydration energy of the cation and its size.
$MgCl_2$ exists as $MgCl_2 \cdot 6H_2O$.
$CaCl_2$ exists as $CaCl_2 \cdot 6H_2O$.
$LiCl$ exists as $LiCl \cdot H_2O$.
$KCl$ has a large cation size and low hydration energy,so it does not form a hydrate and crystallizes as an anhydrous salt.

(C) ) $Li^+$ has the smallest ionic size among alkali metal ions,leading to the highest charge density and thus the highest hydration enthalpy. This statement is correct.
$B$) The boiling point of alkali metals decreases from $Li$ to $Cs$ because the metallic bond strength decreases as the atomic size increases. This statement is incorrect.
$C$) The density of alkali metals generally increases down the group,but $K$ is an exception because of an unusual increase in atomic volume. Thus,the density of $K$ is less than that of $Na$ and $Rb$. This statement is correct.
$D$) $Li$ forms only the oxide $(Li_2O)$ and peroxide $(Li_2O_2)$ due to the small size of the $Li^+$ ion,which cannot stabilize the large superoxide ion $(O_2^-)$. This statement is incorrect.
Therefore,statements $A$ and $C$ are correct.

(A) The alloy of lithium and magnesium $(Li-Mg)$ is used to make armour plates.
This alloy is known for being extremely light and strong,possessing one of the lowest densities among metallic materials.

(A) The reaction of $Na_2SO_3$ with $HCl$ is: $Na_2SO_3(s) + 2HCl(aq) \longrightarrow 2NaCl(aq) + H_2O(l) + SO_2(g)$.
However,the question implies a reaction where two gaseous products are formed.
Consider $Na_2S_2O_3 + 2HCl \longrightarrow 2NaCl + H_2O + SO_2(g) + S(s)$ (not two gases).
Consider $Na_2SO_3 + 2HCl \longrightarrow 2NaCl + H_2O + SO_2(g)$.
Actually,for $Na_2SO_3$,only $SO_2$ is a gas.
Let us re-evaluate: $Na_2S + 2HCl \longrightarrow 2NaCl + H_2S(g)$. Only one gas.
Let us check $NaNO_2 + HCl \longrightarrow NaCl + HNO_2$. $HNO_2$ decomposes to $NO(g)$ and $NO_2(g)$ and $H_2O$. Thus,$NO$ and $NO_2$ are two gases.
Therefore,'$X$' is $NaNO_2$.

(B) White metal is an alloy consisting of $Li$ (Lithium) and $Pb$ (Lead).
Therefore,the correct option is $B$.

(A) $LiF$ has a very high lattice enthalpy due to its small size,which makes it less soluble in water. Thus,statement $(i)$ is correct.
The polarizing power of $Na^{+}$ is higher than that of $Cs^{+}$ due to its smaller ionic radius. Consequently,$NaCl$ has more covalent character and is less ionic than $CsCl$. Thus,statement $(ii)$ is correct.
The formation of alkali metal halides from their constituent elements is a highly exothermic process,as it involves the release of energy to form a stable crystal lattice. Thus,statement $(iii)$ is incorrect.

(B) The peroxide ion $O_2^{2-}$ has $18$ electrons and all electrons are paired,making it diamagnetic. Therefore,the statement that sodium peroxide is paramagnetic is incorrect.
Cesium forms a stable superoxide $(CsO_2)$.
Lithium chloride $(LiCl)$ is deliquescent,meaning it absorbs moisture from the atmosphere to form a hydrate.
White metal is an alloy of lithium and lead.

(A) Most metal nitrates decompose to form metal oxides,nitrogen dioxide,and oxygen gas upon heating.
However,alkali metal nitrates (except $LiNO_3$) decompose to form metal nitrites and oxygen gas.
Therefore,the thermal decomposition of sodium nitrate is represented as:
$NaNO_3 \xrightarrow{\Delta} NaNO_2 + \frac{1}{2} O_2$.
Thus,$A$ is $NaNO_2$ and $B$ is $O_2$.

(D) The thermal decomposition of alkali metal nitrates follows different pathways depending on the metal.
Lithium nitrate $(LiNO_3)$ decomposes to form its oxide,nitrogen dioxide,and oxygen gas:
$2 LiNO_3 \rightarrow Li_2 O + 2 NO_2 + \frac{1}{2} O_2$
Other alkali metal nitrates (like $NaNO_3$,$KNO_3$,etc.) decompose to form nitrites and oxygen gas:
$2 MNO_3 \rightarrow 2 MNO_2 + O_2$
Therefore,the metal '$M$' is $Li$.

(B) Potassium nitrate $(KNO_3)$ is known as Indian saltpetre.
It is a nitrogen-containing compound and is historically significant as a crude salt in India.
It is also commonly referred to as niter.

(B) Statement $A$ is correct: Alkali metals are strong reducing agents because they easily lose their valence electron $(ns^1)$ to achieve a stable noble gas configuration $(n-1)p^6$.
Statement $B$ is incorrect: $KO_2$ (potassium superoxide) contains the superoxide ion $(O_2^{-})$,which has $17$ electrons. Its molecular orbital configuration is $\sigma 1s^2, \sigma * 1s^2, \sigma 2s^2, \sigma * 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi * 2p_x^2 = \pi * 2p_y^1$. Due to the presence of one unpaired electron in the $\pi * 2p_y$ orbital,$KO_2$ is paramagnetic,not diamagnetic.
Statement $C$ is correct: Hydration enthalpy is inversely proportional to the ionic size. Since $Li^{+}$ has the smallest ionic size among alkali metals,it has the highest hydration enthalpy.
Therefore,statements $A$ and $C$ are correct.

(B) Alkali metal nitrates (except $LiNO_3$) decompose on strong heating to form metal nitrites and oxygen gas.
$2NaNO_3 \xrightarrow{\Delta} 2NaNO_2 + O_2$
Lithium nitrate $(LiNO_3)$ and alkaline earth metal nitrates (like $Mg(NO_3)_2$ and $Ca(NO_3)_2$) decompose to give metal oxides,nitrogen dioxide $(NO_2)$,and oxygen $(O_2)$.

(C) $K, Rb,$ and $Cs$ form superoxides when they are burnt in the air. \\ As we move down the group,the size of the alkali metal cation increases from $K^+$ to $Cs^+$. \\ The stability of superoxides is determined by the lattice energy; as the size of the cation increases,the lattice energy decreases,which leads to a decrease in the stability of the superoxide. \\ Therefore,the stability of superoxides decreases from $K$ to $Cs$. \\ Thus,Assertion $(A)$ is true,but Reason $(R)$ is false.

(D) Alkali metals have only one valence electron in their outermost shell.
They lose one valence electron easily due to low ionisation enthalpy and get oxidized to reduce other compounds.
Therefore,they act as strong reducing agents.