Alkali metals Questions in English

(A) Saline hydrides,also known as ionic hydrides,are formed by the reaction of $s$-block elements (alkali metals and alkaline earth metals,except $Be$ and $Mg$) with dihydrogen at high temperatures.
$Li$ is an alkali metal and forms a saline hydride $(LiH)$.
$Be$ and $Mg$ form covalent/polymeric hydrides due to their high charge density and small size.
$Ni$ is a $d$-block element and forms metallic (interstitial) hydrides.

(A) The reactivity of metals with water depends on their position in the electrochemical series.
$K$ and $Na$ are highly reactive and react vigorously with cold water.
$Pt$ is a noble metal and does not react with water or steam.
$Fe$ (Iron) is moderately reactive; it does not react with cold water but reacts with steam at high temperatures to form iron oxide and release $H_2$ gas according to the reaction:
$3Fe(s) + 4H_2O(g) \rightarrow Fe_3O_4(s) + 4H_2(g)$.

(C) Among the given metals,$Sodium$ $(Na)$ is a highly reactive alkali metal. It reacts vigorously with cold water to form $Sodium$ $hydroxide$ $(NaOH)$ and $Hydrogen$ gas $(H_2)$. The reaction is: $2Na(s) + 2H_2O(l) \rightarrow 2NaOH(aq) + H_2(g)$. Copper,Nickel,and Silver are less reactive than hydrogen in the electrochemical series and do not react with water under normal conditions.

(A) Potash alum is a double salt with the chemical formula $K_2SO_4 \cdot Al_2(SO_4)_3 \cdot 24H_2O$.
It is commonly known as alum or fitkari.
Option $A$ represents Potash Alum.
Option $B$ is Chrome Alum.
Option $C$ is Ferric Alum.
Option $D$ is Mohr's salt.

(D) Washing soda is sodium carbonate $(Na_2CO_3)$,which in water undergoes hydrolysis to form a basic solution of sodium hydroxide $(NaOH)$.
$Al$ is an amphoteric metal that reacts with strong bases like $NaOH$ to form soluble sodium aluminate $(Na[Al(OH)_4])$ and releases hydrogen gas.
$2Al(s) + 2NaOH(aq) + 6H_2O(l) \rightarrow 2Na[Al(OH)_4](aq) + 3H_2(g)$.
Therefore,washing soda corrodes aluminium vessels.

(C) Metal carbides that yield methane $(CH_4)$ upon hydrolysis are called methanides.
$Al_4C_3 + 12H_2O \rightarrow 4Al(OH)_3 + 3CH_4$.
Similarly,$Be_2C$ also acts as a methanide.
Given the options,$Al_4C_3$ is a classic example of a methanide.

(A) When alkali metals like sodium dissolve in liquid ammonia,they undergo ionization to form solvated cations and solvated electrons:
$Na + (x+y)NH_3 \rightarrow [Na(NH_3)_x]^+ + [e(NH_3)_y]^-$.
The blue color of the solution is attributed to the presence of ammoniated (solvated) electrons,which absorb energy in the visible region of the spectrum. The solution also contains solvated sodium ions. Therefore,the blue color is due to the presence of solvated electrons and solvated sodium ions.

(D) The thermal decomposition of sodium nitrate $(NaNO_3)$ occurs in stages. At temperatures above $800\,^oC$,the reaction proceeds as follows:
$2NaNO_3 \rightarrow 2NaNO_2 + O_2$
$2NaNO_2 \rightarrow Na_2O + NO + NO_2$
Combining these,the final products at high temperatures include $Na_2O$,$N_2$,and $O_2$. However,the primary decomposition product at very high temperatures is $Na_2O$ along with the release of gases.

(A) Sodium nitrate $(NaNO_3)$ decomposes upon heating. At temperatures up to $500 \, ^oC$,it decomposes to form sodium nitrite and oxygen: $2NaNO_3 \rightarrow 2NaNO_2 + O_2$. However,at temperatures above $800 \, ^oC$,sodium nitrite further decomposes to form sodium oxide,nitrogen,and oxygen: $4NaNO_2 \rightarrow 2Na_2O + 2N_2 + 3O_2$. Thus,the final products at temperatures above $800 \, ^oC$ include $N_2$.

(A) Alkali metals have low ionization energy.
When these metals or their salts are heated in a Bunsen burner flame,the heat energy is sufficient to excite the 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 imparts a characteristic color to the flame.

(C) Sodium carbonate $(Na_2CO_3)$ is commercially produced by the $Solvay$ process.
In this process,ammonia $(NH_3)$ and carbon dioxide $(CO_2)$ are passed through a concentrated solution of sodium chloride $(NaCl)$ to form sodium bicarbonate $(NaHCO_3)$,which is then heated to obtain sodium carbonate.

(D) Alkali metals (Group $1$) have only one valence electron,resulting in weaker metallic bonding compared to alkaline earth metals (Group $2$),which have two valence electrons.
Due to this,alkali metals have lower melting and boiling points,are softer,and have larger atomic and ionic radii.
Furthermore,because of their larger size and lower effective nuclear charge,alkali metals exhibit lower ionization energy compared to alkaline earth metals.

(A) Alkali metals are located in Group $1$ of the periodic table.
They have the largest atomic size in their respective periods due to the presence of only one valence electron and weak nuclear attraction.
Therefore,the property 'Small atomic size' is incorrect for alkali metals.
They possess low ionization energy,low density,and low electronegativity as characteristic properties.

(A) When sodium salts are heated in a Bunsen burner flame,the valence electrons of the sodium atoms absorb energy from the flame and get excited to higher energy levels.
Since these excited states are unstable,the electrons return to their ground state by emitting the absorbed energy in the form of light.
For sodium,this emitted light corresponds to the yellow region of the visible spectrum (wavelength $\approx 589 \ nm$ and $589.6 \ nm$).
This phenomenon is primarily due to the low ionization energy of sodium,which allows its valence electrons to be easily excited by the thermal energy of the flame.

(B) Gunpowder is a mixture of sulfur,charcoal,and potassium nitrate $(KNO_3)$.
$KNO_3$ acts as an oxidizing agent in the mixture.
Therefore,the correct option is $B$.

(A) The photoelectric effect is the emission of electrons when electromagnetic radiation,such as light,hits a material.
For alkali metals,the ease of electron emission depends on the ionization enthalpy.
As we move down the group from $Li$ to $Cs$,the atomic size increases and the ionization enthalpy decreases.
$Cs$ (Cesium) has the lowest ionization enthalpy among the alkali metals,making it the easiest to lose an electron when exposed to light.
Therefore,$Cs$ exhibits the highest photoelectric effect.

(B) When sodium is heated in excess of air at $300\,^oC$,it forms sodium peroxide $(Na_2O_2)$ as the major product,which is $X$.
$2Na + O_2 \xrightarrow{300\,^oC} Na_2O_2$ $(X)$
Sodium peroxide reacts with $CO_2$ to form sodium carbonate $(Na_2CO_3)$ and oxygen gas $(O_2)$,which is $Y$.
$2Na_2O_2 + 2CO_2 \rightarrow 2Na_2CO_3 + O_2$ $(Y)$
Therefore,$Y$ is $O_2$.

(D) When sodium metal dissolves in liquid ammonia,it undergoes ionization: $Na + (x+y)NH_3 \rightarrow [Na(NH_3)_x]^+ + [e(NH_3)_y]^-$.
The blue color and the strong reducing property of the solution are due to the presence of ammoniated (solvated) electrons,$[e(NH_3)_y]^-$.
These solvated electrons are highly reactive and act as powerful reducing agents.

(A) Cesium $(Cs)$ is an alkali metal belonging to Group $1$ of the periodic table.
Alkali metal oxides are generally basic in nature.
As we move down the group,the electropositive character increases,which leads to an increase in the basic strength of the oxides.
Since cesium is the most electropositive alkali metal (excluding francium),its oxide,$Cs_2O$,is strongly basic.

(B) Sodium $(Na)$ is a highly reactive alkali metal that reacts vigorously with moisture and oxygen present in the air. To prevent these reactions,it is stored under kerosene oil,which acts as a protective layer and prevents contact with air and moisture.

(C) In the Solvay process,the reaction occurs as follows: $NH_3(g) + H_2O(l) + CO_2(g) \rightarrow NH_4HCO_3(aq)$.
Then,$NH_4HCO_3(aq) + NaCl(aq) \rightarrow NH_4Cl(aq) + NaHCO_3(s)$.
The precipitation of $NaHCO_3$ occurs because it has low solubility in the presence of a high concentration of $NaCl$ (brine solution) and $NH_4Cl$.

(B) Sodium metal reacts vigorously with water to produce $NaOH$ and $H_2$ gas. Because of this high reactivity,sodium cannot be used as a general drying agent for substances containing water or acidic protons. However,it is specifically used to remove traces of water from organic solvents like benzene or ether,as it reacts with the water content to form $NaOH$ and $H_2$,which can then be removed. Among the given options,drying of benzene is the standard application.

(A) . Alkali metals are highly electropositive and form ionic salts.
Due to the high charge density of the $Li^+$ ion,it is highly hydrated,which is why alkali metals (specifically $Li$) form hydrated salts (e.g.,$LiCl \cdot 2H_2O$).
Reactivity with oxygen increases from $Li$ to $Cs$ because the ionization energy decreases down the group.
Electronegativity decreases from $Li$ to $Cs$ as the atomic size increases.

(C) Potassium reacts with excess oxygen to form potassium superoxide $(KO_2)$.
In $KO_2$,the oxidation state of potassium is $+1$ and the superoxide ion is $O_2^-$,where the oxidation state of oxygen is $-1/2$.

(C) 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 filtrate containing $NH_4Cl$ with $Ca(OH)_2$: $2NH_4Cl + Ca(OH)_2 \rightarrow 2NH_3 + CaCl_2 + 2H_2O$.
Thus,$CaCl_2$ and $NH_3$ (which is recycled) are the primary by-products obtained during the process.

(D) For alkali hydroxides,the solubility increases down the group as the lattice energy decreases more rapidly than the hydration energy. Thus,the order is $LiOH < NaOH < KOH < RbOH < CsOH$.
For alkali carbonates,the solubility increases down the group because the lattice energy decreases more significantly than the hydration energy as the size of the cation increases. Thus,the order is $Li_2CO_3 < Na_2CO_3 < K_2CO_3 < Rb_2CO_3 < Cs_2CO_3$.
Since both statements provided in the options are incorrect based on the standard trends (the provided options in the prompt were reversed),the correct choice is $D$.

(C) When alkali metals $(M)$ dissolve in liquid ammonia,they undergo ionization as follows: $M + (x+y)NH_3 \rightarrow [M(NH_3)_x]^+ + [e(NH_3)_y]^-$.
The blue color of the solution is due to the presence of ammoniated electrons,which absorb energy in the visible region of light corresponding to the red part,resulting in the transmission of blue color.
Therefore,the correct option is $C$.

(C) Sodium is a metal,and metallic conduction is due to the presence of mobile (delocalized) electrons in the metallic lattice. These free electrons are responsible for the flow of electric current through the metal.

(A) Alkali metals have very high negative standard reduction potentials,which means they have very high oxidation potentials.
In the electrochemical series,they are placed much higher than hydrogen.
Due to this,they can easily lose electrons to reduce $H^+$ ions in water to $H_2$ gas,forming their respective hydroxides $(MOH)$.

(C) The anomalous properties of $Li$ compared to other group-$1$ metals include:
$1$. $Li$ is much harder and has higher melting and boiling points than other alkali metals.
$2$. $Li$ forms $Li_3N$ with $N_2$ directly,while other alkali metals do not.
$3$. $Li^+$ has a high charge density,leading to greater hydration.
Option $C$ states that $Li$ is much softer than other group-$1$ metals,which is incorrect because $Li$ is actually the hardest among them.

(D) The concentration of $Na^{+}$ ions in the human body is regulated by the kidneys. An excess of $Na^{+}$ ions leads to water retention in the blood vessels,which increases the volume of blood and consequently results in high blood pressure,also known as hypertension.

(B) As we move down Group $1$ (alkali metals),the atomic size increases due to the addition of new shells.
Due to the increase in atomic size,the valence electron becomes farther from the nucleus,and the effective nuclear charge experienced by the valence electron decreases.
Consequently,the force of attraction between the nucleus and the valence electron decreases,making it easier to remove the electron.
Therefore,the ionization energy decreases down the group.

(C) The reducing character of an element depends on its ability to lose electrons in an aqueous medium. This process involves three steps: sublimation,ionization,and hydration. The overall energy change is represented by the electrode potential $(E^\circ)$.
Lithium has the highest hydration energy due to its very small size,which compensates for its high ionization energy.
Therefore,lithium has the most negative standard electrode potential $(E^\circ = -3.04 \ V)$,making it the strongest reducing agent in aqueous solution.

(B) Metallic luster in alkali metals like sodium is primarily due to the presence of free electrons in the metallic lattice.
When light falls on the surface of the metal,these loosely bound electrons absorb the energy and oscillate,which results in the reflection of light,giving the metal its characteristic shiny appearance.

(C) $1$. The metal $M$ that reacts with $N_2$ to form a nitride of the type $M_3N$ is Lithium $(Li)$.
$2$. The reaction is: $6Li + N_2 \rightarrow 2Li_3N$ (Compound $A$).
$3$. When $Li_3N$ is treated with water,it undergoes hydrolysis to produce ammonia gas $(B)$: $Li_3N + 3H_2O \rightarrow 3LiOH + NH_3(g)$.
$4$. Ammonia $(NH_3)$ reacts with $CuSO_4$ solution to form a deep blue complex $[Cu(NH_3)_4]^{2+}$,which confirms the presence of $NH_3$.
$5$. Therefore,$A$ is $Li_3N$ and $B$ is $NH_3$.

(D) The conductivity of ions in an aqueous solution depends on their mobility. $Li^{+}$ has the smallest ionic size among alkali metals,which leads to a very high charge density. Due to this high charge density,$Li^{+}$ ions get highly hydrated in water,forming a large hydrated ion complex. This large size reduces the mobility of the ion in the solution,thereby decreasing its electrical conductivity.

(D) Magnesium is a highly reactive metal that can continue to burn in an atmosphere of $CO_2$ because it reduces $CO_2$ to carbon.
The chemical reaction is: $2Mg + CO_2 \rightarrow 2MgO + C$.
Similarly,magnesium also reacts with $N_2$ to form magnesium nitride: $3Mg + N_2 \rightarrow Mg_3N_2$.
Therefore,magnesium continues to burn in the presence of both $N_2$ and $CO_2$.

(C) Calcium cyanamide is a chemical compound with the formula $CaNCN$. It is commonly known as nitrolime and is used as a fertilizer. The cyanamide ion is $NCN^{2-}$,which balances the $Ca^{2+}$ ion.

(C) In fireworks,different metal salts are used to produce specific colors due to the excitation of electrons.
$Sr$ (Strontium) salts are responsible for producing a crimson red color.
$Ca$ (Calcium) produces a brick red color.
$Na$ (Sodium) produces a golden yellow color.
$Ba$ (Barium) produces an apple green color.
Therefore,the correct option is $C$.

(C) Radium $(Ra)$ is a radioactive element belonging to the alkaline earth metals group.
It is primarily extracted from the uranium ore known as Pitchblende $(U_3O_8)$.
Marie Curie discovered Radium by isolating it from this mineral.

(B) The thermal stability of carbonates of group $1$ and group $2$ elements increases as the size of the cation increases.
For group $2$ carbonates $(BeCO_3, MgCO_3, CaCO_3)$,the thermal stability increases down the group as the size of the cation increases: $Be^{2+} < Mg^{2+} < Ca^{2+}$.
Thus,the order for group $2$ carbonates is $BeCO_3 < MgCO_3 < CaCO_3$ $(IV < II < III)$.
Group $1$ carbonates (like $K_2CO_3$) are generally more thermally stable than group $2$ carbonates due to the lower charge density of the alkali metal cation.
Therefore,the overall increasing order of thermal stability is $BeCO_3 < MgCO_3 < CaCO_3 < K_2CO_3$,which corresponds to $IV < II < III < I$.

(A) Variable valency is typically exhibited by transition elements and some $p$-block elements due to the presence of vacant $d$-orbitals or the inert pair effect.
$Ba$ (Barium) is an alkaline earth metal belonging to group $2$.
It has a fixed valency of $+2$ because it loses its two valence electrons to achieve a stable noble gas configuration.
$Ti$,$Cu$,and $Pb$ are known to exhibit variable valencies.

(B) The metal $X$ is Magnesium $(Mg)$.
When $Mg$ is heated in a stream of nitrogen gas,it forms Magnesium nitride $(Mg_3N_2)$: $3Mg + N_2 \rightarrow Mg_3N_2$ $(Y)$.
When $Mg_3N_2$ reacts with water,it produces ammonia gas $(NH_3)$: $Mg_3N_2 + 6H_2O \rightarrow 3Mg(OH)_2 + 2NH_3 \uparrow$.
Ammonia gas reacts with $CuSO_4$ solution to form a deep blue complex $[Cu(NH_3)_4]^{2+}$,which confirms the presence of $NH_3$.

(B) The enthalpy of hydration is directly proportional to the charge density of the ion,which is defined as the ratio of charge to size $(q/r)$.
For the given ions,the charge density follows the order: $Be^{2+} > Mg^{2+} > Na^{+}$.
Since $Na^{+}$ has a lower charge $(+1)$ and a larger ionic radius compared to $Mg^{2+}$ $(+2)$,its charge density is lower.
Therefore,the enthalpy of hydration of $Na^{+}$ is lower than that of $Mg^{2+}$.

(A) Alkali metals have low ionization enthalpy,which means their valence electrons are loosely held.
When these metals 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 the ground state,they emit energy in the form of visible light,which gives a characteristic color to the flame.
Therefore,both the assertion and the reason are true,and the reason correctly explains the assertion.

(C) The reducing power of a metal in an aqueous medium is determined by its standard electrode potential $(E^\circ_{red})$,which depends on three factors: sublimation energy,ionization enthalpy,and hydration enthalpy.
Although $Li$ has a high ionization enthalpy,it has a very high hydration enthalpy due to its small size.
The combination of these factors results in the most negative standard reduction potential $(E^\circ = -3.04 \ V)$ for $Li$ among the alkali metals.
Therefore,$Li$ is the strongest reducing agent in an aqueous medium.