Alkali metals Questions in English

(A) Among alkali metals (group-$1$),$Cs$ is the most reactive because it has the lowest ionisation energy ($IE$,or $\Delta_i H_1$).
Note: $Fr$ is not considered as it is radioactive.

(C) $Li$ has the highest hydration enthalpy among alkali metals,which accounts for its high negative $E^{\circ}$ value and its high reducing power in aqueous solution.
Hence,the correct option is $(C)$.

(A) When alkali metals like sodium dissolve in liquid ammonia,they undergo the following reaction: $Na + (x+y)NH_3 \rightarrow [Na(NH_3)_x]^+ + [e(NH_3)_y]^-$.
The deep blue colour of the solution is primarily due to the presence of ammoniated electrons,which absorb energy in the visible region of the spectrum to promote electrons to higher energy levels.

(B) $LiNO_3$ decomposes as: $2 LiNO_3 \xrightarrow{\Delta} Li_2O + 2 NO_2 + \frac{1}{2} O_2$
$NaNO_3$ decomposes as: $2 NaNO_3 \xrightarrow{\Delta} 2 NaNO_2 + O_2$
$Ba(NO_3)_2$ decomposes as: $2 Ba(NO_3)_2 \xrightarrow{\Delta} 2 BaO + 4 NO_2 + O_2$
$Be(NO_3)_2$ decomposes as: $2 Be(NO_3)_2 \xrightarrow{\Delta} 2 BeO + 4 NO_2 + O_2$
Thus,$NaNO_3$ forms its nitrite $(NaNO_2)$ and $O_2$ upon heating,not its oxide.

(B) $1$. $\text{Hydration enthalpy of } M^{+}$: As the size of the alkali metal ion increases down the group $(Li^{+} < Na^{+} < K^{+} < Rb^{+} < Cs^{+})$,the hydration enthalpy decreases because it is inversely proportional to the ionic radius.
$2$. $\text{Ionization enthalpy of } M$: As the atomic size increases down the group,the outermost electron is further from the nucleus and more shielded,making it easier to remove. Thus,ionization enthalpy decreases.
$3$. $\text{Melting point of } M$: As the atomic size increases,the strength of the metallic bond decreases because the attraction between the nucleus and the delocalized electrons weakens. Therefore,the melting point decreases down the group.

(C) Alkali metals react with water to liberate hydrogen gas $(H_2)$,not oxygen gas $(O_2)$.
The general reaction is: $2M_{(s)} + 2H_2O_{(l)} \longrightarrow 2MOH_{(aq)} + H_{2(g)}$ (where $M$ is an alkali metal).
Therefore,the statement in option $C$ is incorrect.

(B) $I$. Incorrect: $Li^{+}$ has the highest charge density and is most highly hydrated,not $Cs^{+}$.
$II$. Correct: $Li$ is the only alkali metal that reacts directly with $N_2$ to form $Li_3N$.
$III$. Incorrect: $Li$ has the highest melting point among these alkali metals due to its small size and strong metallic bonding.
$IV$. Incorrect: $Li$ forms an oxide $(Li_2O)$,$Na$ forms a peroxide $(Na_2O_2)$,and $K, Rb, Cs$ form superoxides $(MO_2)$.

(D) Alkali metals have only one valence electron per atom,which results in weak metallic bonding.
Due to these weak interatomic metallic bonds,alkali metals are soft and possess low melting and boiling points.
Therefore,both $(A)$ and $(R)$ are true,and $(R)$ is the correct explanation of $(A)$.

(C) The electrolysis of brine solution ($NaCl$ aqueous) is the Chlor-alkali process,which produces $NaOH$ (caustic soda),$Cl_2$ gas,and $H_2$ gas. Here,'$X$' is $NaOH$.
$NaOH$ is used in the manufacture of paper,petroleum refining,and mercerising cotton fabrics.
$Na_2S_2O_3$ (sodium thiosulphate) is known as 'Antichlor',not $NaOH$.
Therefore,the correct option is $C$.

(A) In the Castner-Kellner cell,the electrolysis of brine ($NaCl$ solution) is performed.
At the anode (carbon),chloride ions are oxidized: $Cl^{-} \longrightarrow \frac{1}{2} Cl_2 + e^{-}$.
At the cathode (mercury),sodium ions are reduced to form a sodium amalgam: $Na^{+} + e^{-} + Hg \longrightarrow Na-Hg$.
The sodium amalgam then reacts with water to produce $NaOH$: $2Na-Hg + 2H_2O \longrightarrow 2NaOH + 2Hg + H_2 \uparrow$.
Thus,$A$ is $NaOH$ and $B$ is $NaCl$.
Therefore,option $(A)$ is correct.

(A) $1$. $AlCl_3 + 4NaOH \longrightarrow NaAlO_2 + 3NaCl + 2H_2O$. This reaction forms aqueous sodium meta-aluminate and does not release any gas.
$2$. $3NaOH + P_4 + 3H_2O \longrightarrow PH_3 (\text{gas}) + 3NaH_2PO_2$. This releases phosphine gas.
$3$. $2Al + 2NaOH + 2H_2O \xrightarrow{\Delta} 2NaAlO_2 + 3H_2 (\text{gas})$. This releases hydrogen gas.
$4$. $Zn + 2NaOH \xrightarrow{\Delta} Na_2ZnO_2 + H_2 (\text{gas})$. This releases hydrogen gas.
Therefore,the reaction between $AlCl_3$ and $NaOH$ does not liberate a gaseous product.

(A) In the Solvay process,the mother liquor contains $NH_4Cl$ and $NaHCO_3$. To recover ammonia $(NH_3)$,the $NH_4Cl$ solution is treated with slaked lime,which is calcium hydroxide,$Ca(OH)_2$.
The chemical reaction is: $2NH_4Cl + Ca(OH)_2 \rightarrow 2NH_3 + CaCl_2 + 2H_2O$.
Thus,the compound ' $X$ ' is $Ca(OH)_2$.

(A) $2 NaNO_3 \longrightarrow 2 NaNO_2 + O_2$ ($Na_2O$ is not formed).
$4 LiNO_3 \longrightarrow 2 Li_2O \text{ (basic)} + 4 NO_2 \text{ (acidic)} + O_2$.
$2 Be(NO_3)_2 \longrightarrow 2 BeO \text{ (amphoteric)} + 4 NO_2 \text{ (acidic)} + O_2$.
$2 Ca(NO_3)_2 \longrightarrow 2 CaO \text{ (basic)} + 4 NO_2 \text{ (acidic)} + O_2$.
$NO_2$ is an acidic oxide. $Li_2O$ and $CaO$ are basic oxides. $BeO$ is amphoteric. Thus,$Ca(NO_3)_2$ and $LiNO_3$ produce both acidic and basic oxides along with $O_2$.

(A) The flame test colors for alkali metals are as follows:
$Li$ (Lithium) gives a crimson red color.
$Na$ (Sodium) gives a golden yellow color.
$K$ (Potassium) gives a violet color.
Since $ACl$ gives a crimson red color,$A$ is $Li$.
Since $BCl$ gives a violet color,$B$ is $K$.
Therefore,$A$ and $B$ are $Li$ and $K$ respectively.

(B) In muscle contraction,calcium ions $Ca^{2+}$ play an important role by facilitating interactions between the proteins myosin and actin.
$Ca^{2+}$ ions bind to the troponin complex on the actin filament,which causes a conformational change that exposes the binding sites for the myosin heads,thereby stimulating muscle contraction.

(A) $Li^{+}$ (group-$1$) and $Mg^{2+}$ (group-$2$) exhibit a diagonal relationship in the periodic table,resulting in similar ionic radii.
Both $Li$ and $Mg$ react with oxygen to form only monoxides ($Li_2O$ and $MgO$ respectively) and react directly with nitrogen to form nitrides ($Li_3N$ and $Mg_3N_2$ respectively).
Therefore,the metals $A$ and $B$ are $Li$ and $Mg$.

(C) $Ca(OH)_2$ (calcium hydroxide) is used in the Solvay process for the preparation of washing soda $(Na_2CO_3 \cdot 10H_2O)$ to recover ammonia $(NH_3)$ from ammonium chloride $(NH_4Cl)$.
$2NH_4Cl + Ca(OH)_2 \longrightarrow 2NH_3 + CaCl_2 + 2H_2O$
When $CO_2$ is passed through an aqueous solution of $Ca(OH)_2$ (lime water),it forms $CaCO_3$,which makes the solution milky.
$Ca(OH)_2 + CO_2 \longrightarrow CaCO_3 \downarrow + H_2O$
$Ca(OH)_2$ is also used in white washing due to its disinfectant nature.
Therefore,the correct option is $C$.

(B) When magnesium is burnt in air,it reacts with both oxygen and nitrogen to form magnesium oxide $(A)$ and magnesium nitride $(B)$:
$2Mg_{(s)} + O_{2(g)} \rightarrow 2MgO_{(s)} (A)$
$3Mg_{(s)} + N_{2(g)} \rightarrow Mg_3N_{2(s)} (B)$
When magnesium nitride $(B)$ is hydrolyzed,it produces magnesium hydroxide $(C)$ and ammonia $(D)$:
$Mg_3N_{2(s)} + 6H_2O_{(l)} \rightarrow 3Mg(OH)_{2(aq)} (C) + 2NH_{3(g)} (D)$
Ammonia $(NH_3)$ is the reactant used in the manufacture of nitric acid by the Ostwald process.
Therefore,$C$ is $Mg(OH)_2$.

(C) Lithium and magnesium exhibit a diagonal relationship due to their similar ionic sizes and charge density.
$I$) Both $Li$ and $Mg$ react slowly with water compared to other alkali and alkaline earth metals. This is a correct statement.
$II$) Bicarbonates of $Li$ and $Mg$ are not known in solid form; they exist only in solution. This is an incorrect statement.
$III$) Chlorides of $Li$ $(LiCl)$ and $Mg$ $(MgCl_2)$ are covalent in nature and are soluble in ethanol. This is an incorrect statement.
$IV$) Nitrates of both $Li$ and $Mg$ decompose on heating to form their respective oxides,nitrogen dioxide,and oxygen. This is a correct statement.
Therefore,statements $I$ and $IV$ are correct.

(A) diagonal relationship exists between certain pairs of diagonally adjacent elements in the second and third periods of the periodic table.
These pairs are lithium $(Li)$ and magnesium $(Mg)$,beryllium $(Be)$ and aluminum $(Al)$,and boron $(B)$ and silicon $(Si)$.
These similarities in properties are due to the similar polarizing power and similar ionic charge-to-size ratio of the diagonally adjacent elements.
Therefore,lithium shows a diagonal relationship with magnesium $(X = Mg)$ and aluminum shows a diagonal relationship with beryllium $(Y = Be)$.

(A) Double salts are addition compounds that exist only in the solid state and dissociate into their constituent ions when dissolved in water. $Li_2SO_4$ does not form double salts (like alums) because of the exceptionally small size and high polarizing power of the $Li^+$ ion. Due to its small size,$Li^+$ cannot accommodate the coordination requirements necessary to form the stable crystal lattice structure characteristic of double salts,unlike the larger alkali metal ions such as $Na^+$,$K^+$,and $Rb^+$.

(B) The atomic number of $Cs$ (Caesium) is $55$.
The electronic configuration is $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6 6s^1$.
This can be written in terms of the nearest noble gas,Xenon ($Xe$,atomic number $54$),as $[Xe] 6s^1$.
Hence,the correct option is $B$.

(C) Zeolites are hydrated sodium alumino silicates which can be represented by the general formula $Na_2 Z$,where $Z = Al_2 Si_2 O_8 \cdot x H_2 O$.
Thus,the correct representation for '$Z$' is $Al_2 Si_2 O_8 \cdot x H_2 O$.

(A) The standard electrode potential $(E^\circ_{M^+/M})$ values for the alkali metals are as follows:
$E^\circ_{Li^+/Li} = -3.04 \ V$
$E^\circ_{K^+/K} = -2.93 \ V$
$E^\circ_{Na^+/Na} = -2.71 \ V$
Comparing the magnitude of these negative values,we have $|-3.04| > |-2.93| > |-2.71|$.
Therefore,the order of negative standard potential values is $Li > K > Na$.

(A) The standard reduction potential $E^{\circ}_{M^{+}/M}$ depends on the sublimation enthalpy,ionization enthalpy,and hydration enthalpy of the metal ion.
For alkali metals,$Li$ has the most negative standard reduction potential (lowest value) due to its very high hydration enthalpy,which compensates for its high ionization energy.
Therefore,$X = Li$.
Among the alkali metals,$Li$ has the most negative value,while $Na$ has the least negative (highest) standard reduction potential value compared to other alkali metals like $K, Rb, Cs$.
Thus,$Y = Na$.
Hence,$X$ and $Y$ are $Li$ and $Na$ respectively.

(C) In Down's process,the electrolysis of molten sodium chloride $(NaCl)$ is performed.
At the anode,oxidation of chloride ions $(Cl^{-})$ into chlorine gas $(Cl_2)$ takes place.
The reaction is: $2 Cl^{-} \longrightarrow Cl_2 + 2 e^{-}$.

(A) Fluorosis is caused by the accumulation of excess fluoride in the body. This excess fluoride reacts with calcium $(Ca)$ present in bones and teeth to form calcium fluoride $(CaF_2)$,which leads to the disease known as fluorosis.
The reaction is: $Ca + F_2 \rightarrow CaF_2$ (Fluorosis disease).

(C) Zeolite is a hydrated sodium aluminium silicate.
Its general chemical formula is $Na_2Al_2Si_2O_8 \cdot xH_2O$.
Thus,the metal ions present are $Na^{+}$ and $Al^{3+}$.

(D) White metal is an alloy of $Li$ and $Pb$.

(A) Feldspar is a group of rock-forming tectosilicate minerals that make up about $41\%$ of the Earth's continental crust. The general chemical formula for potassium feldspar (orthoclase) is $KAlSi_3O_8$.

(A) The chemical formula of carnallite is $KCl \cdot MgCl_2 \cdot 6H_2O$.
It is a double salt consisting of potassium chloride and magnesium chloride.
Therefore,the metal ions present in carnallite are $K^+$ and $Mg^{2+}$.

(A) Aluminium metal reacts with aqueous sodium hydroxide to produce sodium aluminate and hydrogen gas.
The balanced chemical equation is:
$2 Al_{(s)} + 2 NaOH_{(aq)} + 6 H_2O_{(l)} \rightarrow 2 Na[Al(OH)_4]_{(aq)} + 3 H_{2(g)}$
The evolution of hydrogen gas can be tested by bringing a burning candle near the reaction,which produces a characteristic pop sound.

(C) The reaction of aluminum $(Al)$ with an excess of aqueous sodium hydroxide $(NaOH)$ is a characteristic reaction where aluminum acts as an amphoteric metal.
The balanced chemical equation for this reaction is:
$2Al(s) + 2NaOH(aq) + 6H_2O(l) \longrightarrow 2Na^{+}[Al(OH)_4]^-(aq) + 3H_2(g)$
In this reaction,$P$ is the complex ion sodium tetrahydroxoaluminate$(III)$,$Na^{+}[Al(OH)_4]^-$,and $Q$ is hydrogen gas,$H_2$.
Therefore,the correct option is $C$.

(D) Nitrates of alkali metals (except $LiNO_3$) decompose on heating to form metal nitrites and oxygen gas,without releasing $NO_2$.
For potassium nitrate: $2 KNO_3 \xrightarrow{\Delta} 2 KNO_2 + O_2$.
Nitrates of heavy metals (like $Pb$) and $LiNO_3$ decompose to form metal oxides,$NO_2$,and $O_2$.

(C) The reaction of alkali metals with oxygen depends on the size of the metal cation.
Lithium $(Li)$ forms monoxide $(Li_2O)$.
Sodium $(Na)$ forms peroxide $(Na_2O_2)$.
Potassium $(K)$,Rubidium $(Rb)$,and Cesium $(Cs)$ form superoxides $(MO_2)$.
Therefore,$X$ is $CsO_2$ (superoxide) and $Y$ is $Na_2O_2$ (peroxide).

(B) When sodium is burnt in excess of oxygen,it forms sodium peroxide $(Na_2O_2)$,so $X = Na_2O_2$.
In $Na_2O_2$,the peroxide ion is $O_2^{2-}$,which has the electronic configuration $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^2$. Since all electrons are paired,$Na_2O_2$ is diamagnetic.
When potassium is burnt in excess of oxygen,it forms potassium superoxide $(KO_2)$,so $Y = KO_2$.
In $KO_2$,the superoxide ion is $O_2^-$,which has one unpaired electron in the $\pi^* 2p$ orbitals. Therefore,$KO_2$ is paramagnetic.
Thus,$X$ is diamagnetic and $Y$ is paramagnetic.

(B) Statement-$I$ is correct because both $LiF$ and $CsI$ exhibit low solubility in water compared to other alkali metal halides.
Statement-$II$ is incorrect because the reasons provided are swapped. The low solubility of $LiF$ is primarily due to its very high lattice enthalpy (because both $Li^+$ and $F^-$ are very small),while the low solubility of $CsI$ is due to its smaller hydration enthalpy (because both $Cs^+$ and $I^-$ are very large,leading to weak ion-dipole interactions with water).
Therefore,Statement-$I$ is correct,but Statement-$II$ is not correct.

(B) When metals are burnt in air,they react with $O_2$ to form oxides and with $N_2$ to form nitrides.
$1$. $Li$ reacts with both $O_2$ and $N_2$ to form $Li_2O$ and $Li_3N$.
$2$. $Mg$ reacts with both $O_2$ and $N_2$ to form $MgO$ and $Mg_3N_2$.
$3$. $Be$ reacts with both $O_2$ and $N_2$ to form $BeO$ and $Be_3N_2$.
$Na, K, Ba,$ and $Sr$ primarily form oxides or peroxides/superoxides but do not form stable nitrides under these conditions.
Thus,the metals that form both are $Li, Mg,$ and $Be$.
The total count is $3$.

(A) The stability of alkali metal superoxides $(MO_2)$ increases as the size of the alkali metal cation increases.
This is because the large cation stabilizes the large superoxide anion $(O_2^-)$ through lattice energy effects.
The ionic radii of the alkali metals follow the order: $K^+ < Rb^+ < Cs^+$.
Therefore,the stability order is $KO_2 < RbO_2 < CsO_2$.

(B) The correct matches are as follows:
$A$. $Li-Pb$ is used to make bearings for motor engines $(II)$.
$B$. $Be-Cu$ is used to make high strength springs $(IV)$.
$C$. $Mg-Al$ is used in aircraft construction $(I)$.
$D$. $Na-Pb$ is used to make tetraethyl lead $(III)$.
Thus,the correct matching is $A-II, B-IV, C-I, D-III$.

(A) Assertion $(A)$ is true because alkali metals have low ionization enthalpies,meaning their valence electrons are loosely held.
When these metals or their salts are heated in a flame,the energy from the flame is sufficient to excite these valence electrons to higher energy levels.
When these excited electrons return to their ground state,they emit energy in the form of visible light,which gives characteristic flame colours.
Therefore,the reason $(R)$ correctly explains the assertion $(A)$.

(D) Lithium nitrate $(LiNO_3)$ decomposes upon heating to form lithium oxide $(Li_2O)$,nitrogen dioxide $(NO_2)$,and oxygen $(O_2)$.
The balanced chemical equation is:
$4LiNO_3 \rightarrow 2Li_2O + 4NO_2 + O_2$

(B) 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 $(O_2)$.
Other alkali metal nitrates like sodium nitrate $(NaNO_3)$ decompose to form the corresponding nitrite $(NaNO_2)$ and oxygen $(O_2)$.
Therefore,the reaction $2 LiNO_3 \rightarrow 2 LiNO_2 + O_2$ is incorrect.

(B) Lithium nitrate $(LiNO_3)$ decomposes on heating to form lithium oxide $(Li_2O)$,nitrogen dioxide $(NO_2)$,and oxygen gas $(O_2)$.
The balanced chemical equation is:
$4LiNO_3(s) \xrightarrow{\Delta} 2Li_2O(s) + 4NO_2(g) + O_2(g)$

(A) The electrical conductivity of ions in an aqueous solution depends on their ionic mobility.
Ionic mobility is inversely proportional to the size of the hydrated ion.
Smaller ions have a higher charge density,which causes them to be more extensively hydrated.
Therefore,the order of hydrated ionic size is $Li^{+} > Na^{+} > K^{+} > Cs^{+}$.
Consequently,the order of ionic mobility and electrical conductivity is $Cs^{+} > K^{+} > Na^{+} > Li^{+}$.