Alkali metals Questions in English

(C) When $AgNO_3$ is heated above its melting point $(212 \, ^\circ C)$,it undergoes thermal decomposition to form silver nitrite and oxygen gas.
The balanced chemical equation is:
$2AgNO_3 \xrightarrow{\Delta, T > 212 \, ^\circ C} 2AgNO_2 + O_2$

(A) $Na^{+}$ and $K^{+}$ ions play a crucial role in regulating blood pressure. An increase in the concentration of $Na^{+}$ ions leads to an increase in blood pressure.

(A) When zinc $(Zn)$ reacts with a hot concentrated solution of caustic soda $(NaOH)$,it forms sodium zincate $(Na_2ZnO_2)$ and releases hydrogen gas $(H_2)$.
The chemical equation is:
$Zn + 2NaOH \rightarrow Na_2ZnO_2 + H_2 \uparrow$
Thus,the correct product is sodium zincate $(Na_2ZnO_2)$.

(D) $Philosopher\'s \, wool$ is the common name for zinc oxide $(ZnO)$.
It is produced by the combustion of zinc metal in air:
$2Zn(s) + O_2(g) \xrightarrow{\Delta} 2ZnO(s)$.

(C) Lithium tetrahydridoaluminate is a complex metal hydride.
Its chemical name indicates that it contains a lithium cation $(Li^+)$ and a complex anion,tetrahydridoaluminate $([AlH_4]^-)$.
Therefore,the correct molecular formula is $Li[AlH_4]$.

(A) The lightest metal is $Li$ (Lithium).
It has the lowest density among all metals and an atomic number of $3$.

(D) Baking soda is $NaHCO_3$.
Baking powder is a mixture of baking soda $(NaHCO_3)$ and a mild edible acid,such as potassium hydrogen tartrate $(KHC_4H_4O_6)$,also known as cream of tartar.
When mixed with water,the acid reacts with the baking soda to release $CO_2$ gas,which helps the dough rise.
Therefore,$KHC_4H_4O_6$ transforms baking soda into baking powder.

(A) The thermal decomposition of sodium nitrate $(NaNO_3)$ at temperatures above $800\ ^oC$ follows the reaction:
$2NaNO_3(s) \xrightarrow{>800\ ^oC} 2NaNO_2(s) + O_2(g)$
Thus,the product obtained is oxygen gas $(O_2)$.

(C) Ionic mobility is inversely proportional to the size of the hydrated ion: $\text{Ionic mobility} \propto \frac{1}{\text{size of hydrated ion}}$.
In aqueous solution,the extent of hydration depends on the charge density of the ion.
Smaller alkali metal ions have higher charge density,leading to greater hydration.
The order of hydrated ionic size is: $Li^{+}_{(aq)} > Na^{+}_{(aq)} > K^{+}_{(aq)} > Rb^{+}_{(aq)}$.
Since $Li^{+}$ has the largest hydrated size,it experiences the most drag in an electric field.
Therefore,$Li^{+}$ has the lowest ionic mobility.

(C) When nitrogen reacts with $CaC_2$ at high temperatures,calcium cyanamide $(CaCN_2)$ and carbon are formed.
The chemical equation for the reaction is:
$CaC_2 + N_2 \xrightarrow{\Delta} CaCN_2 + C$
This mixture of $CaCN_2$ and $C$ is known as nitrolim,which is used as a fertilizer.

(A) The Sodium pump is also known as a sodium-potassium pump.
This pump is an important contributor to the action potential produced by nerve cells.
The process of moving sodium and potassium ions across the cell membrane is an active transport process involving the hydrolysis of $ATP$ to provide the necessary energy.
This process is responsible for maintaining a large excess of $Na^{+}$ ions outside the cell and a large excess of $K^{+}$ ions inside the cell.

(C) When alkali metals are heated in an atmosphere of oxygen,they ignite and form oxides.
$Li$ reacts with oxygen to form the normal oxide,$Li_2O$.
$Na$ reacts with oxygen to form the peroxide,$Na_2O_2$.
$K$ and $Rb$ react with oxygen to form superoxides,$MO_2$ (where $M = K, Rb$).

(B) The ease of adsorption of hydrated alkali metal ions on an ion-exchange resin depends on the size of the hydrated ion.
Smaller alkali metal ions have a higher charge density,leading to greater hydration. Thus,the size of the hydrated ion follows the order: $Li^{+} > Na^{+} > K^{+} > Rb^{+}$.
Since the ease of adsorption is inversely proportional to the size of the hydrated ion,the order of adsorption is: $Rb^{+} < K^{+} < Na^{+} < Li^{+}$.

(C) According to Fajans' rule,the covalent character is directly proportional to the polarizability of the anion.
As the size of the anion increases,its polarizability increases,which leads to an increase in the covalent character of the bond.
The order of the size of halide ions is $F^- < Cl^- < Br^- < I^-$.
Therefore,the order of covalent character in alkali metal halides is $MI > MBr > MCl > MF$.

(A) Sodium hydroxide,$NaOH$,is a strong base. It reacts with acidic and amphoteric oxides but does not react with basic oxides.
$CaO$ is a basic oxide (alkaline earth metal oxide).
$SiO_2$ is an acidic oxide.
$BeO$ and $B_2O_3$ are amphoteric oxides.
Since $CaO$ is a basic oxide,it will not react with the base $NaOH$.

(D) In an aqueous solution,ions are hydrated. The smaller the size of the bare ion,the greater is the degree of hydration because the charge density is higher.
Thus,the degree of hydration is highest for $Li^{+}$ and lowest for $Cs^{+}$.
Due to a larger hydration shell,the hydrated ion size becomes larger for ions with smaller bare radii.
Consequently,$Cs^{+}$ has the smallest hydrated size and $Li^{+}$ has the largest hydrated size.
Since ionic mobility is inversely proportional to the size of the hydrated ion,the mobility is highest for $Cs^{+}$ and lowest for $Li^{+}$.
The correct order of ionic mobility in aqueous solution is $Cs^{+} > Rb^{+} > K^{+} > Na^{+} > Li^{+}$.

(B) The thermal stability of ionic hydrides depends on the lattice energy and the strength of the metal-hydrogen bond.
As the size of the alkali metal cation increases down the group $(Li^+ < Na^+ < K^+ < Rb^+ < Cs^+)$,the metal-hydrogen bond length increases,which leads to a decrease in the bond dissociation energy.
Consequently,the thermal stability of these hydrides decreases as the size of the cation increases.
Therefore,the correct order of thermal stability is $LiH > NaH > KH > RbH > CsH$.

(A) The thermal stability of metal carbonates depends on the electropositive nature of the metal.
As the electropositive character of the metal increases,the thermal stability of its carbonate increases.
For Group $2$ elements,the electropositive character increases down the group $(Be < Mg < Ca)$. Thus,the thermal stability order is $BeCO_3 < MgCO_3 < CaCO_3$.
Group $1$ metals are more electropositive than Group $2$ metals,so $K_2CO_3$ is more stable than the alkaline earth metal carbonates.
Therefore,the correct order is $BeCO_3 < MgCO_3 < CaCO_3 < K_2CO_3$.

(A) In an aqueous solution,alkali metal ions get hydrated. The extent of hydration depends on the size of the cation; smaller cations have higher charge density and thus higher hydration energy.
The order of hydration energy is $Li^{+} > Na^{+} > K^{+} > Rb^{+}$.
Due to higher hydration,the effective size of the hydrated $Li^{+}$ ion is the largest,which makes it move the slowest in an electric field.
Therefore,the mobility of the ions in an aqueous solution is inversely proportional to their hydrated size,resulting in the order: $Rb^{+} > K^{+} > Na^{+} > Li^{+}$.

(C) Aluminium reacts with excess $NaOH$ to form sodium tetrahydroxoaluminate$(III)$,which is a soluble complex,not $Al(OH)_3$.
The reaction is: $2 Al(s) + 2 NaOH(aq) + 6 H_2O(l) \longrightarrow 2 Na[Al(OH)_4](aq) + 3 H_2(g)$.
Therefore,the statement in option $C$ is incorrect.

(D) In aqueous solution,alkali metal ions are hydrated. The extent of hydration is inversely proportional to the ionic size. Smaller ions have higher charge density and thus get more hydrated,resulting in a larger hydrated radius.
Since ionic mobility is inversely proportional to the size of the hydrated ion,the hydrated radius order is $Li^{+} > Na^{+} > K^{+} > Rb^{+}$.
Therefore,the ionic mobility order is $Li^{+} < Na^{+} < K^{+} < Rb^{+}$.
Thus,the ionic mobility is maximum for $Rb^{+}$.

(B) When alkali metals are heated in excess of air,they form different types of oxides based on their size and polarizability:
$1$. Lithium forms the monoxide: $4Li + O_2 \longrightarrow 2Li_2O$
$2$. Sodium forms the peroxide: $2Na + O_2 \longrightarrow Na_2O_2$
$3$. Potassium forms the superoxide: $K + O_2 \longrightarrow KO_2$
Thus,the correct sequence is $Li_2O, Na_2O_2$ and $KO_2$.

(A) Lithium $(Li)$ and Magnesium $(Mg)$ exhibit a diagonal relationship due to similar ionic sizes and charge-to-size ratios.
$1$. Both form nitrides ($Li_3N$ and $Mg_3N_2$) by direct combination with nitrogen.
$2$. Both form soluble bicarbonates in solution.
$3$. Both $LiNO_3$ and $Mg(NO_3)_2$ decompose on heating to give their respective oxides,$NO_2$,and $O_2$.
$4$. Magnesium forms a basic carbonate $(4MgCO_3 \cdot Mg(OH)_2 \cdot 5H_2O)$,whereas lithium does not form a basic carbonate; it only forms a normal carbonate $(Li_2CO_3)$.
Therefore,the statement that both form basic carbonates is incorrect.

(A) . $Li$ is the hardest among alkali metals because of its small size and strong metallic bonding. This statement is $CORRECT$.
$B$. In the $Solvay$ process,$NH_3$ is recovered by treating $NH_4Cl$ with $Ca(OH)_2$,not $H_2O$.
$C$. $K_2CO_3$ is known as pearl ash,not $Na_2CO_3$.
$D$. $Be^{2+}$ and $Al^{3+}$ have a very strong tendency to form complexes like $[BeF_4]^{2-}$ and $[AlF_6]^{3-}$ due to their high charge density.

(B) The thermal decomposition of sodium oxide $(Na_2O)$ at $400 \ ^oC$ is given by the reaction:
$2Na_2O \rightleftharpoons Na_2O_2 + 2Na$
Here,$A$ is sodium peroxide $(Na_2O_2)$.
Sodium peroxide reacts with sulfur dioxide $(SO_2)$ to form sodium sulfate $(Na_2SO_4)$:
$Na_2O_2 + SO_2 \to Na_2SO_4$
Sodium peroxide reacts with carbon monoxide $(CO)$ to form sodium carbonate $(Na_2CO_3)$:
$Na_2O_2 + CO \to Na_2CO_3$
Thus,$A$ is $Na_2O_2$.

(A) Lattice energy is the energy released when gaseous ions combine to form one mole of an ionic solid.
According to Coulomb's law,the lattice energy is inversely proportional to the inter-ionic distance $(r_c + r_a)$.
As the size of the alkali metal cation increases from $Li^+$ to $Cs^+$,the inter-ionic distance increases,leading to a decrease in the lattice energy.
Therefore,the order of lattice energy for alkali metal chlorides is $LiCl > NaCl > KCl > RbCl > CsCl$.

(B) Most alkali metal carbonates are thermally stable and do not decompose upon heating.
However,$Li_2CO_3$ is an exception due to the small size of the $Li^+$ ion,which leads to high polarizing power and instability of the carbonate ion.
$Li_2CO_3$ decomposes upon heating as follows:
$Li_2CO_3 \rightarrow Li_2O + CO_2$
Other alkali metal carbonates like $Na_2CO_3$,$K_2CO_3$,and $Cs_2CO_3$ are thermally stable and do not decompose at ordinary heating temperatures.

(D) When exposed to dry air (which contains $O_2$ and $N_2$),$Li$ reacts with both $O_2$ and $N_2$ to form $Li_2O$ and $Li_3N$ respectively.
$Na$ reacts with $O_2$ to form $Na_2O$ (and $Na_2O_2$ depending on conditions),but it does not react with $N_2$ at room temperature.
Therefore,the products obtained are $Na_2O$,$Li_2O$,and $Li_3N$.

(C) When sodium $(Na)$ burns in a limited supply of oxygen $(O_2)$,it forms sodium oxide $(Na_2O)$.
The chemical reaction is: $4Na + O_2 \rightarrow 2Na_2O$.
If sodium burns in an excess of oxygen,it forms sodium peroxide $(Na_2O_2)$.

(A) $1$. Conductivity in molten state depends on the mobility of ions. As the size of the cation increases,the ionic mobility increases,leading to higher conductivity. Thus,the order $Mg^{+2} < Ca^{+2} < Sr^{+2} < Ba^{+2}$ is correct.
$2$. Hydration energy is inversely proportional to the size of the ion. The correct order for hydration energy is $Li^{+1} > Na^{+1} > K^{+1} > Rb^{+1} > Cs^{+1}$. Therefore,the order given in option $B$ is incorrect.
$3$. Reactivity of alkali metals increases down the group due to the ease of losing an electron. Thus,the order $Li < Na < K < Rb < Cs$ is correct.
$4$. Since option $B$ is incorrect,the correct answer is not 'All of them'.
$5$. Re-evaluating the options: Option $A$ is correct (conductivity increases with size). Option $C$ is correct (reactivity increases down the group). Given the structure,the question likely intended to ask for the correct statement among the choices.

(B) Heating the given compounds results in the following reactions:
$1$. $Li_2CO_3 \xrightarrow{\Delta} Li_2O + CO_2$ (Does not release $O_2$).
$2$. $2NaNO_3 \xrightarrow{\Delta} 2NaNO_2 + O_2$ (Releases $O_2$).
$3$. $K_2SO_4$ is thermally stable and does not release $O_2$ under normal heating.
$4$. $2CaSO_4 \xrightarrow{\Delta} 2CaO + 2SO_2 + O_2$ (Releases $O_2$ at very high temperatures).
Therefore,$NaNO_3$ and $CaSO_4$ release oxygen upon heating. The correct option is $B$.

(A) Hydrolysis of methanide carbides produces methane $(CH_4)$.
$Be_2C$ is a methanide carbide.
The reaction is: $Be_2C + 4 H_2O \rightarrow 2 Be(OH)_2 + CH_4$.
$CaC_2$ and $SrC_2$ are acetylides that produce ethyne $(C_2H_2)$.
$Mg_2C_3$ is an allylide that produces propyne $(C_3H_4)$.

(C) Let us analyze the reactions with water for each pair:
$1$. For $Na_2O$ and $Na$:
$Na_2O + H_2O \rightarrow 2NaOH$
$2Na + 2H_2O \rightarrow 2NaOH + H_2(g)$
The products are different ($H_2$ gas is produced in the second reaction but not the first).
$2$. For $CO_2$ and $K_2O$:
$CO_2 + H_2O \rightarrow H_2CO_3$
$K_2O + H_2O \rightarrow 2KOH$
The products are different.
$3$. For $Na$ and $NaH$:
$2Na + 2H_2O \rightarrow 2NaOH + H_2(g)$
$NaH + H_2O \rightarrow NaOH + H_2(g)$
Both reactions produce $NaOH$ and $H_2$ gas as products. Therefore,this pair produces the same products.

(B) Amphoteric hydroxides dissolve in excess of strong bases like $NaOH$ to form soluble complex salts.
$Zn(OH)_2$ is an amphoteric hydroxide.
It reacts with excess $NaOH$ as follows:
$Zn(OH)_2 + 2NaOH \rightarrow Na_2[Zn(OH)_4]$ (Sodium zincate).
$Cu(OH)_2$ and $Bi(OH)_3$ are basic in nature and do not dissolve in excess $NaOH$.

(C) The reaction of metals with dilute $HNO_3$ $(20\%)$ depends on the reactivity of the metal.
$Zn$ (Zinc) is a moderately reactive metal that reacts with $20\%\ HNO_3$ to produce $N_2O$ (nitrous oxide) gas.
The reaction is: $4Zn + 10HNO_3 \rightarrow 4Zn(NO_3)_2 + N_2O + 5H_2O$.
$Sn$ (Tin) reacts with dilute $HNO_3$ to produce $NH_4NO_3$ (ammonium nitrate),while $Pt$ (Platinum) is noble and does not react with dilute $HNO_3$.

(B) The reaction sequence is as follows:
$1$. $Na + H_2O \rightarrow NaOH (A) + \frac{1}{2} H_2$
$2$. $2NaOH (A) + CO_2 \rightarrow Na_2CO_3 (B) + H_2O$
$3$. $Na_2CO_3 (B) + SO_2 (excess) \rightarrow Na_2SO_3 (C) + CO_2$
$4$. $Na_2SO_3 (C) + SO_2 (excess) + H_2O \rightarrow 2NaHSO_3$
$5$. Upon evaporation,$2NaHSO_3$ yields $Na_2S_2O_5 (D) + H_2O$.
Thus,compound $D$ is sodium metabisulfite,$Na_2S_2O_5$.

(A) The $Cs^{+}$ ion imparts a violet colour to the Bunsen flame.
The energy of a photon is given by the equation $E = h\nu = \frac{hc}{\lambda}$.
Violet light has a shorter wavelength $(\lambda)$ compared to other visible colours like red,which implies it has a higher frequency $(\nu)$ and consequently higher energy $(E)$.
Therefore,the emission of violet light corresponds to the transition of electrons involving high energy.

(A) The reaction of element $A$ with water is: $2 Na + 2 H_2O \rightarrow 2 NaOH + H_2 (g)$.
Element $A$ is $Na$ (which imparts a golden yellow color to the Bunsen flame),gas $B$ is $H_2$,and solution $C$ is $NaOH$.
When substance $D$ $(Zn)$ reacts with $NaOH$,it produces $H_2$ gas: $Zn + 2 NaOH + 2 H_2O \rightarrow Na_2[Zn(OH)_4] + H_2$.
Also,$Zn$ reacts with dilute $H_2SO_4$ to produce $H_2$: $Zn + H_2SO_4 \rightarrow ZnSO_4 + H_2$.
Thus,$A, B, C,$ and $D$ are $Na, H_2, NaOH,$ and $Zn$ respectively.
Therefore,the correct option is $A$.

(A) The thermal stability of alkali metal carbonates increases as we move down the group from $Li$ to $Cs$.
This is because the electropositive character of the alkali metal increases down the group,which makes the $M-O$ bond stronger and the carbonate ion more stable.
$Li_2CO_3$ is the least stable among them because $Li^+$ has a very small size and high polarizing power,which polarizes the carbonate ion $(CO_3^{2-})$ and leads to its decomposition into $Li_2O$ and $CO_2$ at relatively low temperatures.
Therefore,$Li_2CO_3$ has the least thermal stability.

(D) The alkali metals exhibit different behaviors when reacting with oxygen depending on their size and ionization energy.
$Li$ primarily forms the normal oxide $(Li_2O)$.
$Na$ forms the normal oxide $(Na_2O)$ and the peroxide $(Na_2O_2)$.
$K, Rb,$ and $Cs$ form superoxides $(MO_2)$ as their stable products in excess oxygen.
However,$K$ and $Rb$ are capable of forming all three types: normal oxides $(M_2O)$,peroxides $(M_2O_2)$,and superoxides $(MO_2)$.
Therefore,$K$ and $Rb$ are the alkali metals that can form all three types of oxides.

(B) Hydration energy is directly proportional to the charge of the ion and inversely proportional to the ionic radius $(HE \propto \frac{q}{r})$.
Comparing $Mg^{2+}$ and $Na^{+}$: $Mg^{2+}$ has a higher charge $(+2)$ compared to $Na^{+}$ $(+1)$,and a smaller ionic radius.
Therefore,the hydration energy of $Mg^{2+}$ is greater than that of $Na^{+}$.

(B) $NaCl$ imparts a golden yellow color to the Bunsen flame.
This phenomenon is explained by the emission spectrum of sodium.
When sodium atoms or their salts are introduced into the flame,the valence electrons absorb thermal energy and are excited to higher energy levels.
When these excited electrons return to their ground state,they release the absorbed energy in the form of electromagnetic radiation in the visible region,specifically at a wavelength corresponding to the yellow color ($589.0 \ nm$ and $589.6 \ nm$).

(D) The solution of sodium metal in liquid ammonia is strongly reducing because reduction involves the gain of electrons.
When alkali metals like sodium dissolve in liquid ammonia,they undergo ionization to form solvated metal ions and solvated electrons:
$Na + (x+y)NH_3 \rightarrow [Na(NH_3)_x]^+ + [e(NH_3)_y]^-$.
The presence of these solvated electrons makes the solution a powerful reducing agent.
These solvated electrons are also responsible for the characteristic deep blue color and the electrical conductivity of the solution.
Therefore,option $D$ is correct.

(C) The alkali metal with the lowest density is $Li$ $(0.53 \ g \ cm^{-3})$,not $K$.
Therefore,the statement in option $C$ is incorrect.

(D) $KO_2$ (potassium superoxide) is used in oxygen cylinders for space and submarines because it reacts with exhaled $CO_2$ to release $O_2$.
The chemical reaction is:
$4 KO_2(s) + 2 CO_2(g) \rightarrow 2 K_2CO_3(s) + 3 O_2(g)$
As shown in the reaction,it absorbs $CO_2$ and produces $O_2$. Therefore,both $(A)$ and $(C)$ are correct.

(C) The reaction of $Na$ with $Al_2O_3$ at high temperature produces sodium oxide $(Na_2O)$ and aluminum metal: $6Na + Al_2O_3 \rightarrow 3Na_2O + 2Al$.
When $Na_2O$ reacts with $CO_2$ in the presence of water,it forms sodium carbonate $(Na_2CO_3)$: $Na_2O + CO_2 \rightarrow Na_2CO_3$.
Thus,$X$ is $Na_2O$ and $Y$ is $Na_2CO_3$.
Therefore,option $C$ is correct.

(D) 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 blue colour of the solution is due to the excitation of solvated electrons to higher energy levels,which absorb light in the red region of the visible spectrum.
The solution is highly conducting due to the presence of both solvated cations $[Na(NH_3)_x]^+$ and solvated electrons $[e(NH_3)_y]^-$.
Therefore,both statements $A$ and $B$ are correct.

(B) The reaction sequence is as follows:
$i) \, Ca(OH)_2 (A) + Na_2CO_3 \rightarrow 2NaOH (B) + CaCO_3 (C)$
$ii) \, Ca(OH)_2 (A) + CO_2 \rightarrow CaCO_3 (C) + H_2O$
Here,$C$ is $CaCO_3$ which forms a milky cloud (precipitate) in the solution.
Thus,$A$ is $Ca(OH)_2$ and $B$ is $NaOH$.

(D) Paramagnetism is observed in species containing unpaired electrons. Superoxides $(O_2^-)$ contain one unpaired electron in their antibonding molecular orbital,making them paramagnetic.
$A$) $KO_2$ contains the superoxide ion $(O_2^-)$,which has an unpaired electron. Thus,it is paramagnetic.
$B$) $K_2O_2$ contains the peroxide ion $(O_2^{2-})$,where all electrons are paired. Thus,it is diamagnetic.
$C$) $RbO_2$ contains the superoxide ion $(O_2^-)$,which has an unpaired electron. Thus,it is paramagnetic.
Therefore,both $KO_2$ and $RbO_2$ are paramagnetic. The correct option is $(D)$.