Alkali metals Questions in English

(C) When alkali metals are burnt in excess of air,they form different types of oxides depending on their size and polarizing power.
$1$. Lithium $(Li)$ forms the normal oxide: $4Li + O_{2} \rightarrow 2Li_{2}O$.
$2$. Sodium $(Na)$ forms the peroxide: $2Na + O_{2} \rightarrow Na_{2}O_{2}$.
$3$. Potassium $(K)$ forms the superoxide: $K + O_{2} \rightarrow KO_{2}$.
Therefore,the major oxides formed are $Li_{2}O$,$Na_{2}O_{2}$,and $KO_{2}$.

(A) The metal ion described is $K^+$.
$K^+$ ions are essential for activating many enzymes,participating in the oxidation of glucose to produce $ATP$,and along with $Na^+$ ions,they play a crucial role in the transmission of nerve signals.

(C) $LiCl$ is deliquescent and crystallises from aqueous solution as a hydrate,$LiCl \cdot 2H_{2}O$.
The hydration enthalpies of alkaline earth metal ions decrease with an increase in ionic size down the group.
The correct order of hydration enthalpy is $Be^{2+} > Mg^{2+} > Ca^{2+} > Sr^{2+} > Ba^{2+}$.
$Lithium$ shows a diagonal relationship with $Magnesium$,hence their physical and chemical properties are similar.
$Lithium$ is the hardest among all alkali metals.

(C) The elements $Li$ and $Mg$ react with atmospheric nitrogen upon heating to form their respective nitrides,which upon hydrolysis yield ammonia. The reactions are as follows:
$6 Li + N_2 \rightarrow 2 Li_3N$
$Li_3N + 3 H_2O \rightarrow 3 LiOH + NH_3 \uparrow$
$3 Mg + N_2 \rightarrow Mg_3N_2$
$Mg_3N_2 + 6 H_2O \rightarrow 3 Mg(OH)_2 + 2 NH_3 \uparrow$

(D) The size of alkali metal ions increases down the group as follows: $Li ^{+} < Na ^{+} < K ^{+} < Rb ^{+} < Cs ^{+}$.
Hydration enthalpy is inversely proportional to the ionic size $(Hydration \ Enthalpy \ \propto \ \frac{1}{\text{ionic size}})$.
Therefore,the smaller the ion,the higher its hydration enthalpy.
Thus,the correct order of hydration enthalpies is $Li ^{+} > Na ^{+} > K ^{+} > Rb ^{+} > Cs ^{+}$.

(A) The flame colors of the given alkali metals are:
$Li$ (Lithium): Crimson red flame $(\lambda \approx 670 \ nm)$
$Na$ (Sodium): Golden yellow flame $(\lambda \approx 589 \ nm)$
$K$ (Potassium): Violet flame $(\lambda \approx 404 \ nm)$
$Cs$ (Cesium): Blue flame $(\lambda \approx 455 \ nm)$
Since the wavelength $(\lambda)$ is inversely proportional to the energy of the emitted light,the color with the lowest energy corresponds to the maximum wavelength.
Among the given options,the crimson red color of $Li$ has the longest wavelength.

(B) The conductivity of ions in an aqueous solution depends on their hydrated ionic size.
As the size of the gaseous ion decreases,the extent of hydration increases,which leads to a larger hydrated ionic radius.
Larger hydrated ions experience greater resistance (viscous drag) while moving through the solvent,resulting in lower ionic mobility and lower conductivity.
The order of gaseous ionic size is: $Cs^{+} > Rb^{+} > K^{+} > Na^{+}$.
Consequently,the order of hydrated ionic size is: $Cs^{+} < Rb^{+} < K^{+} < Na^{+}$.
Therefore,the order of ionic conductivity in water is: $Cs^{+} > Rb^{+} > K^{+} > Na^{+}$.

(C) In the Solvay process,$NH_{3}$ is recovered by treating the ammonium chloride $(NH_{4}Cl)$ solution with calcium hydroxide $(Ca(OH)_{2})$.
The chemical reaction is: $2NH_{4}Cl + Ca(OH)_{2} \rightarrow 2NH_{3} + CaCl_{2} + 2H_{2}O$.
Here,$CaCl_{2}$ (calcium chloride) is obtained as a by-product.

(D) Amphoteric compounds are those that can react with both acids and bases.
$BeO$ (Beryllium oxide) is amphoteric.
$Be(OH)_2$ (Beryllium hydroxide) is amphoteric.
$BaO$ (Barium oxide) is basic.
$Sr(OH)_2$ (Strontium hydroxide) is basic.
Therefore,there are $2$ amphoteric compounds in the given list.

(D) Alkali metal salts impart characteristic colors to the flame due to the excitation of electrons to higher energy levels and their subsequent return to the ground state.
The characteristic wavelengths for these metals are:
$Li$: $670.8 \text{ nm}$ (Crimson Red)
$Na$: $589.2 \text{ nm}$ (Yellow)
$Rb$: $780.0 \text{ nm}$ (Red-violet)
$Cs$: $455.5 \text{ nm}$ (Blue)
Therefore,the correct matching is: $(a)-(ii), (b)-(iv), (c)-(iii), (d)-(i)$.

(B) The photoelectric effect depends on the work function of the metal.
$Cs$ (Cesium) has the lowest ionization energy among the alkali metals,making it the most suitable for use in photoelectric cells.
While other alkali metals like $Na$ and $Rb$ can exhibit the photoelectric effect,$Cs$ is the standard choice for such applications due to its low work function.
Therefore,only $1$ metal $(Cs)$ is typically cited for this specific application in the context of this question.

(C) $1$. The standard enthalpy of formation for alkali metal bromides becomes more negative on descending the group due to the increase in the size of the metal cation and the decrease in lattice energy being offset by the ionization energy trends.
$2$. The low solubility of $CsI$ is not due to high lattice enthalpy,but rather due to the low hydration enthalpy of the large $Cs^+$ and $I^-$ ions.
$3$. Among the alkali metal halides,$LiF$ is the least soluble in water because of its very high lattice enthalpy compared to its hydration enthalpy.
$4$. The standard enthalpy of formation for $LiF$ is the most negative among alkali metal fluorides because of the very high lattice energy associated with the small $Li^+$ and $F^-$ ions.

(B) The major component of Portland cement is Tricalcium silicate $(3CaO \cdot SiO_2)$,which typically constitutes about $50-51\%$ of the composition.

(D) Lithium salts are hydrated because $Li^{+}$ has a very small size and a high charge density,which leads to a very high hydration enthalpy.
The high polarising power of $Li^{+}$ (due to its small ionic radius) causes it to attract water molecules strongly,resulting in the formation of hydrated salts like $LiCl \cdot 2H_2O.$
Therefore,both Assertion $(A)$ and Reason $(R)$ are correct,and $(R)$ is the correct explanation for $(A).$

(A) To determine the paramagnetism,we analyze the electronic configuration of the oxide ions present in each compound:
$Li_{2}O$: Contains $O^{2-}$ ion. Configuration: $1s^{2} 2s^{2} 2p^{6}$ (diamagnetic).
$CaO$: Contains $O^{2-}$ ion. Configuration: $1s^{2} 2s^{2} 2p^{6}$ (diamagnetic).
$Na_{2}O_{2}$: Contains peroxide ion $O_{2}^{2-}$. Configuration: $(\sigma 1s)^{2} (\sigma^{*} 1s)^{2} (\sigma 2s)^{2} (\sigma^{*} 2s)^{2} (\sigma 2p_{z})^{2} (\pi 2p_{x})^{2} (\pi 2p_{y})^{2} (\pi^{*} 2p_{x})^{2} (\pi^{*} 2p_{y})^{2}$ (diamagnetic).
$KO_{2}$: Contains superoxide ion $O_{2}^{-}$. Configuration: $(\sigma 1s)^{2} (\sigma^{*} 1s)^{2} (\sigma 2s)^{2} (\sigma^{*} 2s)^{2} (\sigma 2p_{z})^{2} (\pi 2p_{x})^{2} (\pi 2p_{y})^{2} (\pi^{*} 2p_{x})^{2} (\pi^{*} 2p_{y})^{1}$. Due to the presence of one unpaired electron in the $\pi^{*}$ orbital,it is paramagnetic.
$MgO$: Contains $O^{2-}$ ion (diamagnetic).
$K_{2}O$: Contains $O^{2-}$ ion (diamagnetic).
Only $KO_{2}$ is paramagnetic. Therefore,the total number of paramagnetic oxides is $1$.

(B) The reaction of alkali metals with excess oxygen leads to the formation of superoxides $(MO_{2})$.
Potassium $(K)$,Rubidium $(Rb)$,and Cesium $(Cs)$ form superoxides ($KO_{2}$,$RbO_{2}$,$CsO_{2}$) which are paramagnetic due to the presence of one unpaired electron in the $\pi^* 2p$ molecular orbital of the superoxide ion $(O_{2}^{-})$.
$KO_{2}$ is a pale yellow solid.
Sodium $(Na)$ forms a peroxide $(Na_{2}O_{2})$,which is diamagnetic and white.
Calcium $(Ca)$ and Magnesium $(Mg)$ form oxides ($CaO$,$MgO$) or peroxides $(CaO_{2})$,which are diamagnetic.
Therefore,the element $(M)$ is $K$.

(A) $Li_2CO_3$ decomposes easily on heating due to the high polarizing power of the small $Li^{+}$ cation,which imparts significant covalent character to the bond.
$(b)$ $NaHCO_3$ (sodium bicarbonate) is used in fire extinguishers because it releases $CO_2$ gas upon heating.
$(c)$ $K^{+}$ ions are the most abundant cations found within the intracellular fluid of cells.
$(d)$ $CsI$ (cesium iodide) has poor water solubility because both $Cs^{+}$ and $I^{-}$ are large ions,resulting in a low lattice energy and low hydration energy,making the dissolution process energetically unfavorable.
Therefore,the correct matching is: $(a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)$.

(B) $NaOH$ is a strong base,so it is $(ii)$ Basic.
$Be(OH)_{2}$ is amphoteric,so it is $(iii)$ Amphoteric.
$Ca(OH)_{2}$ is a strong base,so it is $(ii)$ Basic.
$B(OH)_{3}$ (or $H_{3}BO_{3}$) is a weak Lewis acid,so it is $(i)$ Acidic.
$Al(OH)_{3}$ is amphoteric,so it is $(iii)$ Amphoteric.
Therefore,the correct matching is: $(a)-(ii), (b)-(iii), (c)-(ii), (d)-(i), (e)-(iii)$.

(A) $Li$ forms an alloy with lead to make white metal bearings for motor engines.
Liquid $Na$ metal is used as a coolant in fast breeder nuclear reactors.
$K$ is used as an absorbent of $CO_{2}$ (specifically in the form of potassium superoxide,$KO_{2}$).
$Cs$ is used in making photoelectric cells due to its low ionization energy.

(B) Potassium superoxide $(KO_{2})$ reacts with carbon dioxide $(CO_{2})$ to produce potassium carbonate $(K_{2}CO_{3})$ and oxygen gas $(O_{2})$.
The balanced chemical equation is:
$2 KO_{2} + CO_{2} \rightarrow K_{2}CO_{3} + \frac{3}{2} O_{2}$
Due to this property,it is used in breathing apparatus for submarines and space travel.

(C) $Li$ is used in electrochemical cells.
$Na$ is used as a coolant in fast breeder reactors.
$KOH$ is used as an absorbent for $CO_2$.
$Cs$ is used in photoelectric cells.
Therefore,the correct matching is $(a)-(ii), (b)-(iii), (c)-(i), (d)-(iv)$.

(A) In $KO_{2}$,potassium $(K)$ is an alkali metal and always exhibits an oxidation state of $+1$.
The superoxide ion is $O_{2}^{-}$,where the oxidation state of oxygen is $-1/2$.
Therefore,the statement that the oxidation number of $K$ in $KO_{2}$ is $+4$ is incorrect.
The correct oxidation number of $K$ is $+1$.

(B) The melting points of the given metals are as follows:
$Hg$ (mercury) = $-38.8 \ ^\circ C$
$Ga$ (gallium) = $29.76 \ ^\circ C$
$Cs$ (cesium) = $28.44 \ ^\circ C$
$Ag$ (silver) = $961.78 \ ^\circ C$
Among the given options,$Ag$ has the highest melting point.

(A) Sodium hydrogencarbonate $(NaHCO_{3})$,commonly known as Baking soda,is used in certain types of fire extinguishers because it releases $CO_{2}$ gas upon heating or reacting with an acid,which helps in extinguishing the fire.

(A) The chemical formulas for the given substances are:
$1$. Baking soda: $NaHCO_3$ (contains bicarbonate anion,$HCO_3^-$).
$2$. Washing soda: $Na_2CO_3 \cdot 10H_2O$ (contains carbonate anion,$CO_3^{2-}$).
$3$. Caustic soda: $NaOH$ (contains hydroxide anion,$OH^-$).
Therefore,the carbonate anion $(CO_3^{2-})$ is present only in washing soda.

(A) The characteristic flame colors and corresponding wavelengths for alkali metals are as follows:

Metal	Wavelength $(\lambda / nm)$
$Li$	$670.8$
$Na$	$589.2$
$Rb$	$780.0$
$Cs$	$455.5$

Matching the values:
$A (Li) - I (670.8)$
$B (Na) - II (589.2)$
$C (Rb) - III (780.0)$
$D (Cs) - IV (455.5)$
Therefore,the correct match is $A-I, B-II, C-III, D-IV$.

(A) The density of alkali metals generally increases down the group as the atomic mass increases more significantly than the atomic volume. However,$K$ is an exception because of an unusual increase in its atomic size. The correct order of density is $Li < K < Na < Rb < Cs$.

(C) Assertion $A$ is true: $LiF$ is sparingly soluble in water due to its exceptionally high lattice energy compared to its hydration energy.
Reason $R$ is false: Although the $Li^{+}$ ion has the smallest ionic radius among its group members,it actually possesses the highest hydration enthalpy,not the least,because hydration enthalpy is inversely proportional to the ionic size $(\Delta H_{hyd} \propto 1/r)$.

(A) In group $1$,$Li^{+}$ has the maximum hydration enthalpy due to its smallest ionic size.
$Li$ shows a diagonal relationship with $Mg$ of group $2$.
Therefore,the element $B$ is $Mg$.

(A) $BeCl_2$ (in vapor phase) and $Al_2Cl_6$ exist as $Cl^{-}$-bridged structures. Due to their covalent nature,they are soluble in organic solvents and act as Lewis acids because of their electron-deficient nature.
$Be(OH)_2$ and $Al(OH)_3$ are amphoteric in nature. They react with excess alkali to form soluble beryllate $[Be(OH)_4]^{2-}$ and aluminate $[Al(OH)_4]^{-}$ ions,respectively.
Therefore,both Statement $I$ and Statement $II$ are true.

(A) The low solubility of $LiF$ in water is due to its very high lattice enthalpy,which outweighs the hydration enthalpy. Therefore,the statement in option $A$ is incorrect.

(B) Statement $I$ is false because an alloy of $Li$ and $Mg$ is used to make armour plates,not aircraft plates.
Statement $II$ is false because calcium ions $(Ca^{2+})$,not magnesium ions,are primarily responsible for cell-membrane integrity and neuromuscular function.

(D) Zinc reacts with an excess of aqueous alkali (such as $NaOH$) to evolve hydrogen gas and form the soluble tetrahydroxozincate$(II)$ complex ion.
The chemical equation is:
$Zn(s) + 2OH^{-}(aq) + 2H_2O(l) \rightarrow [Zn(OH)_4]^{2-}(aq) + H_2(g) \uparrow$
In this reaction,the zinc metal is oxidized to the $+2$ oxidation state,forming the complex ion $[Zn(OH)_4]^{2-}$,which is the stable species in an aqueous alkaline medium.

(C) When heated,lithium nitrate decomposes to form lithium oxide $(Li_2O)$,nitrogen dioxide $(NO_2)$,and oxygen $(O_2)$:
$4LiNO_3 \longrightarrow 2Li_2O + 4NO_2 + O_2$
In contrast,sodium nitrate decomposes to form sodium nitrite $(NaNO_2)$ and oxygen $(O_2)$:
$2NaNO_3 \longrightarrow 2NaNO_2 + O_2$
Therefore,the products are $Li_2O$ and $NaNO_2$ respectively.
Note: At very high temperatures (above $800 \ ^{\circ}C$),$NaNO_2$ can further decompose to $Na_2O$,but the standard decomposition product for $NaNO_3$ is $NaNO_2$.

(C) When alkali metals $(M)$ are dissolved in liquid $NH_{3}$,they undergo ionization to form ammoniated cations and ammoniated electrons.
The reaction is represented as: $M + (x+y)NH_{3} \rightarrow [M(NH_{3})_{x}]^{+} + [e(NH_{3})_{y}]^{-}$.
These ammoniated electrons are responsible for the blue color and high electrical conductivity of the solution.

(A) .
Among the $s$-block elements,$Be$ and $Mg$ salts do not impart any characteristic color to the flame because the excitation energy required to promote electrons to higher energy levels is very high,which cannot be provided by the heat of a Bunsen burner.

(B) .
$Na_2O + H_2O \longrightarrow 2 \, NaOH$
Since $Na_2O$ is a metal oxide,it reacts with water to form $NaOH$,which is a strong base. This reaction confirms the basic nature of $Na_2O$.

(D)
On moving down the group,as the atomic number of alkali metals increases,the number of shells increases,which leads to an increase in atomic radii.
Although the nuclear charge increases,the shielding effect of inner electrons predominates,resulting in a larger atomic size.
Thus,the correct order of atomic radii for alkali metals is $Cs > K > Na > Li$.

(B) . The standard reduction potential $(E^{\circ})$ order is $Li < Rb < Na$ (incorrect order given in statement $A$).
$B$. $CsI$ has low solubility in water due to its low lattice energy and hydration enthalpy (statement $B$ is false).
$C$. $Li_{2}CO_{3}$ is thermally unstable and decomposes into $Li_{2}O$ and $CO_{2}$ (statement $C$ is false).
$D$. Potassium in concentrated liquid ammonia forms a bronze-coloured,diamagnetic solution,whereas dilute solutions are blue and paramagnetic (statement $D$ is false).
$E$. All alkali metal hydrides are ionic solids (statement $E$ is true).
Note: Based on the standard chemical properties,only statement $E$ is strictly correct. However,if the question implies a specific set,$E$ is the only universally true statement among the choices provided.

(C) The reaction sequence is based on the Solvay process and the preparation of ammonia.
$1$. $Ca(OH)_2 + 2NH_4Cl \longrightarrow CaCl_2 + 2NH_3 + 2H_2O$
Here,$A = Ca(OH)_2$ and $B = NH_3$.
$2$. $NH_3 + H_2O + CO_2 \text{ (excess)} \longrightarrow NH_4HCO_3$
Here,$C = NH_4HCO_3$.
$3$. $NH_4HCO_3 + NaCl \longrightarrow NaHCO_3 + NH_4Cl$
Thus,the compounds are $A = Ca(OH)_2, B = NH_3, C = NH_4HCO_3$.

(C) Alkali metals and their salts impart characteristic colours to an oxidizing flame,not a reducing flame. Therefore,Assertion $A$ is incorrect.
However,alkali metals and their salts can indeed be detected using flame tests because the heat from the flame excites the valence electrons to higher energy levels,and when they return to the ground state,they emit light in the visible region. Therefore,Reason $R$ is correct.
Thus,$A$ is not correct but $R$ is correct.

(C) The reducing power of a metal is determined by its standard oxidation potential. Among the alkali metals,$Li$ has the highest hydration energy,which makes its standard electrode potential $(E^\circ)$ the most negative,making it the strongest reducing agent. Conversely,$Na$ has the least negative standard electrode potential among the given alkali metals $(K, Rb, Na, Li)$,making it the weakest reducing agent in this specific set.

(A) The hydration enthalpy of an ion depends on its charge density,which is directly proportional to the charge and inversely proportional to the ionic radius $(H_{hyd} \propto \frac{q}{r})$.
$1$. Comparing charges: Ions with higher charge $(Mg^{2+}, Ca^{2+})$ have significantly higher hydration enthalpies than monovalent ions $(K^{+}, Rb^{+}, Cs^{+})$.
$2$. Within the same group,hydration enthalpy decreases as the ionic radius increases.
$3$. For group $2$ ions: $Mg^{2+} > Ca^{2+}$ $(C > E)$.
$4$. For group $1$ ions: $K^{+} > Rb^{+} > Cs^{+}$ $(A > B > D)$.
Combining these,the overall order is $Mg^{2+} > Ca^{2+} > K^{+} > Rb^{+} > Cs^{+}$,which corresponds to $C > E > A > B > D$.

(D) $Li_2O$ contains the oxide ion $O^{2-}$,which has the electronic configuration $[He] 2s^2 2p^6$. Since all electrons are paired,it is diamagnetic.
$Na_2O_2$ contains the peroxide ion $O_2^{2-}$. The molecular orbital configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^2$. All electrons are paired,so it is diamagnetic.
$KO_2$ contains the superoxide ion $O_2^-$. The molecular orbital configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^1$. It has one unpaired electron in the $\pi^*$ orbital,making it paramagnetic.
Therefore,the correct sequence is diamagnetic,diamagnetic,and paramagnetic.

(C) The thermal decomposition of lithium nitrate $(LiNO_3)$ is given by the following equation:
$2LiNO_3 \xrightarrow{\Delta} Li_2O + 2NO_2 + \frac{1}{2}O_2$
From the given list $(Li_2O, N_2, O_2, LiNO_2, NO_2)$,the products formed are $Li_2O$,$NO_2$,and $O_2$.
Therefore,there are $3$ compounds formed.

(B) Lithium aluminium hydride $(LiAlH_4)$ is prepared by the reaction of lithium hydride $(LiH)$ with aluminium chloride ($Al_2Cl_6$ or $AlCl_3$) in an ether solvent.
The balanced chemical equation is:
$8 LiH + Al_2Cl_6 \longrightarrow 2 LiAlH_4 + 6 LiCl$

(C) Lithium nitrate $(LiNO_3)$ is unique among alkali metal nitrates because it decomposes upon heating to form lithium oxide $(Li_2O)$,nitrogen dioxide $(NO_2)$,and oxygen $(O_2)$.
The balanced chemical equation for this thermal decomposition is:
$4 LiNO_3 \stackrel{\Delta}{\longrightarrow} 2 Li_2O + 4 NO_2 + O_2$

(C) The ionization enthalpies of the given alkali metals are: $Li \approx 520 \ kJ \ mol^{-1}$,$K \approx 419 \ kJ \ mol^{-1}$,$Rb \approx 403 \ kJ \ mol^{-1}$,and $Cs \approx 376 \ kJ \ mol^{-1}$.
Among these,$Li$,$K$,and $Rb$ have ionization enthalpy greater than $400 \ kJ \ mol^{-1}$.
Regarding the formation of stable superoxides $(MO_2)$,only the larger alkali metals $K$,$Rb$,and $Cs$ form stable superoxides.
Comparing both conditions,only $K$ and $Rb$ satisfy both criteria (ionization enthalpy $> 400 \ kJ \ mol^{-1}$ and formation of stable superoxide).
Therefore,the number of such metals is $2$.

(C) The correct matches are as follows:
$A$. Slaked lime is $Ca(OH)_2$ $(II)$.
$B$. Dead burnt plaster is $CaSO_4$ $(IV)$.
$C$. Caustic soda is $NaOH$ $(I)$.
$D$. Washing soda is $Na_2CO_3 \cdot 10H_2O$ $(III)$.
Therefore,the correct matching is $A-II, B-IV, C-I, D-III$.