Alkali metals Questions in English

(A) Elements with low ionization energy can easily excite their valence electrons to higher energy levels by absorbing heat from the Bunsen flame. When these electrons return to the ground state,they emit radiation in the visible region. Sodium,having a low ionization energy,emits light in the golden yellow part of the visible spectrum.

(B) In alkali metals,as the atomic number increases,the metallic bonding becomes weaker due to an increase in atomic size. Consequently,the melting point decreases.

(D) The ionic character of alkali metal halides increases down the group due to the increase in their electropositive nature. Therefore,the statement in option $D$ is incorrect.

(A) $Li$ is the only alkali metal that reacts directly with nitrogen to form a nitride.
$6Li + N_2 \to 2Li_3N$

(D) In alkali metals,the melting point decreases as we move down the group from $Li$ to $Cs$. Therefore,$Cs$ has the lowest melting point among the given options.

(D) When sodium dissolves in liquid ammonia,it forms ammoniated electrons and ammoniated sodium ions: $Na + (x+y)NH_3 \rightarrow [Na(NH_3)_x]^+ + [e(NH_3)_y]^-$.
The blue color and electrical conductivity are due to the presence of ammoniated electrons.
The presence of unpaired electrons makes the solution paramagnetic,not diamagnetic.
Therefore,the statement that the solution remains diamagnetic is incorrect.

(A) The photoelectric effect is inversely proportional to the ionization energy of the element.
Among the given alkali metals,$Cs$ (Cesium) has the lowest ionization energy.
Therefore,$Cs$ exhibits the maximum photoelectric effect.

(B) The reaction of $Na$ with oxygen at $300\,^oC$ produces sodium peroxide $(X)$:
$2Na + O_2 \xrightarrow{300\,^oC} Na_2O_2$ $(X)$
Sodium peroxide reacts with $CO_2$ to form sodium carbonate and oxygen $(Y)$:
$2Na_2O_2 + 2CO_2 \rightarrow 2Na_2CO_3 + O_2$ $(Y)$
Therefore,$Y$ is $O_2$.

(B) The hydroxides of alkali metals are highly soluble in water. $K^+$ is an alkali metal ion,therefore $KOH$ is highly soluble in water.

(D) The solubility of alkali metal bicarbonates increases down the group due to the increase in the size of the cation,which decreases the lattice energy more significantly than the hydration energy.
$CaCO_3$ (Calcium carbonate) is practically insoluble in water.
Among the bicarbonates,$NaHCO_3$ is less soluble than $KHCO_3$ because the smaller $Na^+$ ion has a higher hydration energy but also a higher lattice energy compared to $K^+$.
Therefore,the correct order of increasing solubility is $CaCO_3 < NaHCO_3 < KHCO_3$.

(B) Lithium shows a diagonal relationship with magnesium because their polarizing power is approximately the same,which means their charge-to-size ratio is approximately the same.

(D) When sodium metal dissolves in liquid ammonia,it undergoes ionization to form $Na^+$ ions and solvated electrons,represented as $[Na(NH_3)_x]^+$ and $[e(NH_3)_y]^-$.
These solvated electrons are responsible for the blue color,high electrical conductivity,and strong reducing property of the solution.

(A) The aqueous solutions of chlorides of Group $1$ elements (alkali metals) are neutral because they are salts of strong bases and strong acids.
For example,$NaCl$ in water dissociates into $Na^+$ and $Cl^-$,which do not undergo hydrolysis,resulting in a neutral solution $(pH = 7)$.

(B) Alkali metals are highly electropositive and their oxides are basic in nature. Sodium $(Na)$ is a typical alkali metal and its oxide $(Na_2O)$ is strongly basic,not amphoteric. Therefore,the statement that sodium is amphoteric is incorrect.

(D) When $Zn$ reacts with caustic soda $(NaOH)$,it produces sodium zincate and hydrogen gas,which is a combustible gas.
$2NaOH + Zn \to Na_2ZnO_2 + H_2(g)$

(A) As we move down the group in the periodic table,the basic character of alkali metal oxides increases.
Since $Cs$ is at the bottom of the alkali metal group,its oxide $(Cs_2O)$ is strongly basic in nature.

(A) The small size of the $Li$ atom and the $Li^+$ ion is the primary reason why lithium differs from other alkali metals. This small size leads to high polarizing power and anomalous properties compared to the rest of the group.

(A) $2Na + 2H_2O \to 2NaOH(aq) + H_2(g)$
$(A) \quad \quad \quad \quad \quad \quad \quad (C) \quad \quad \quad (B)$
$2NaOH + Zn \to Na_2ZnO_2 + H_2(g)$
$(C) \quad \quad \quad (D) \quad \quad \quad \quad \quad \quad \quad \quad (B)$
$Na$ $(A)$ gives a golden yellow flame in the Bunsen flame test.

(B) When sodium reacts with an excess of oxygen or air,it forms sodium peroxide $(Na_2O_2)$.
The balanced chemical equation is:
$2Na + O_2 \to Na_2O_2$

(B) The reaction of sodium metal with $Al_2O_3$ at high temperature is: $6Na + Al_2O_3 \to 3Na_2O + 2Al$. Here,$X$ is $Na_2O$.
Next,$Na_2O$ reacts with $CO_2$ to form sodium carbonate: $Na_2O + CO_2 \to Na_2CO_3$. Thus,$Y$ is $Na_2CO_3$.

(D) Washing soda is sodium carbonate decahydrate,$Na_2CO_3 \cdot 10H_2O$. When heated,it loses its water of crystallization. The reaction is: $Na_2CO_3 \cdot 10H_2O$ $\xrightarrow{\Delta} Na_2CO_3 \cdot H_2O + 9H_2O$ $\xrightarrow{\Delta} Na_2CO_3 + H_2O$. Thus,water vapor is released.

(D) Potassium is a highly reactive alkali metal that reacts vigorously with air and moisture. Therefore,it is stored in $Kerosene$ to prevent contact with air and moisture.

(C) Sodium metal reacts with alcohol to release hydrogen gas,therefore it cannot be stored in alcohol.
$2R-OH + 2Na \to 2R-ONa + H_2 \uparrow$

(B) In a soda-acid fire extinguisher,the reaction occurs between sodium bicarbonate $(NaHCO_3)$ and sulfuric acid $(H_2SO_4)$. The chemical equation is: $2NaHCO_3 + H_2SO_4 \to Na_2SO_4 + 2CO_2 + 2H_2O$. The produced $CO_2$ gas helps in extinguishing the fire.

(D) $Na_2CO_3$ is thermally stable and does not decompose upon heating. Therefore,no gas is released.

(A) The process of caustification is used for the industrial preparation of $NaOH$ (caustic soda) by reacting sodium carbonate with slaked lime $(Ca(OH)_2)$.

(C) Baking powder is a mixture of baking soda $(NaHCO_3)$ and a mild edible acid such as tartaric acid. Therefore,$NaHCO_3$ is a component of baking powder.

(B) $Na$ metal is used for drying benzene because it reacts with water present as an impurity in benzene. $Na$ cannot be used for drying alcohols or ammonia because it reacts with them.

(A) Alkali metals are highly electropositive and react with hydrogen to form ionic (salt-like) hydrides of the general formula $MH$ (where $M$ is an alkali metal).
Therefore,the correct statement is that they form salt-like hydrides.

(D) Sodium is an alkali metal with a low ionization enthalpy,which makes it a strong reducing agent. Therefore,it loses electrons easily,meaning it can be easily oxidized $(Na \rightarrow Na^+ + e^-)$.

(B) The stability of alkali metal halides is directly related to their lattice energy and enthalpy of formation.
As the size of the alkali metal cation increases,the lattice energy decreases,leading to lower stability.
Therefore,the stability order is $LiCl > NaCl > KCl > CsCl$.
However,looking at the provided options,the question asks for the order of stability. Based on standard chemical trends,the stability decreases as we move down the group.
Thus,the correct order is $CsCl < KCl < NaCl < LiCl$.

(D) The reactivity of alkali metals with water increases as we move down the group in the periodic table. The order of reactivity is $Li < Na < K < Rb < Cs$. Therefore,$Rb$ (Rubidium) has a higher reaction rate with water compared to $Li$,$Na$,and $K$.

(D) The Solvay process involves the reaction of ammonia with carbon dioxide and water to form ammonium bicarbonate,which then reacts with sodium chloride to produce sodium bicarbonate. The reaction $2NaOH + CO_2 \to Na_2CO_3 + H_2O$ is not part of the Solvay process,as sodium hydroxide is not used as a starting material in this industrial method.

(B) Alkali metals have low ionization energy,which allows them to lose electrons easily.
This process of losing electrons is called oxidation.
Since they undergo oxidation easily,they act as strong reducing agents.

(C) In the Solvay process,the overall reaction is $2NaCl + CaCO_3 \rightarrow Na_2CO_3 + CaCl_2$.
Ammonia $(NH_3)$ is recovered and recycled,but $CaCl_2$ is obtained as a major by-product.
Therefore,the correct by-product is $CaCl_2$.

(D) $NaOH$ is commercially produced by the electrolysis of brine solution using the $Castner-Kellner$ cell process.

(B) Sodium metal is a strong reducing agent and it reduces $CO_2$ to carbon: $4Na + 3CO_2 \to 2Na_2CO_3 + C$.
Option $A$ is incorrect because $Downs$ process uses a mixture of $NaCl$ and $CaCl_2$ to lower the melting point.
Option $C$ is incorrect because $Mg$ reacts with hot water or steam,not cold water.
Option $D$ is incorrect because Magnalium is an alloy of $Mg$ and $Al$.

(B) When $CO_2$ is passed through an aqueous solution of sodium carbonate $(Na_2CO_3)$,it reacts with water and the carbonate to form sodium bicarbonate $(NaHCO_3)$.
The chemical equation is: $Na_2CO_3(aq) + CO_2(g) + H_2O(l) \to 2NaHCO_3(aq)$.

(B) The solubility of alkali hydroxides increases down the group because the lattice energy decreases more rapidly than the hydration energy. Thus,the order is $LiOH < NaOH < KOH < RbOH < CsOH$.
The solubility of alkali carbonates increases down the group because the lattice energy decreases more rapidly than the hydration energy. Thus,the order is $Li_2CO_3 < Na_2CO_3 < K_2CO_3 < Rb_2CO_3 < Cs_2CO_3$.
Comparing the given options,option $B$ correctly describes the trend for alkali carbonates.

(C) When alkali metals dissolve in liquid ammonia,they undergo ionization to form ammoniated cations and ammoniated electrons:
$M + (x+y)NH_3 \rightarrow [M(NH_3)_x]^+ + [e(NH_3)_y]^-$.
The blue color of the solution is due to the excitation of ammoniated electrons to higher energy levels,which absorb light in the red region of the visible spectrum.

(D) The oxides $(M_2O)$ and peroxides $(M_2O_2)$ of alkali metals have all electrons paired,making them diamagnetic and colorless.
Superoxides $(MO_2)$ contain the superoxide ion $(O_2^-)$,which has one unpaired electron,making them paramagnetic.
$Li$ forms only the oxide,and $Na$ forms the oxide and peroxide,but neither forms a stable superoxide under normal conditions.
Therefore,all the given statements are correct.

(C) Sodium metal conducts electricity because it contains free and mobile electrons in its metallic lattice.

(B) The reducing power of alkali metals depends on their standard oxidation potential,which is related to the ease of losing an electron (low ionization enthalpy) and high hydration energy.
In the gas phase,ionization enthalpy decreases down the group $(Li > Na > K > Rb > Cs)$.
However,in aqueous solution,the standard electrode potential $(E^\circ)$ is determined by the sum of sublimation,ionization,and hydration energies.
Lithium $(Li)$ is the strongest reducing agent in aqueous solution due to its very high hydration energy.
Among the given options $(Na, K, Rb, Cs)$,the reducing power increases as we go down the group because the ionization enthalpy decreases significantly.
Therefore,$Cs$ (Cesium) is the strongest reducing agent among the given options.

(D) Due to the high reactivity of alkali metals,they are generally extracted by the electrolytic method of their fused salts.

(C) Oxides of alkali metals are basic in nature.
$Na_2O$ is an oxide of an alkali metal,therefore it is basic.
$CO_2$,$SiO_2$,and $SO_2$ are acidic oxides.

(B) Among the given metals,$K$ (Potassium) is the most reactive metal.
Reactivity in group $1$ increases down the group as the ionization enthalpy decreases.
Therefore,the order of reactivity is $Na < K < Rb$.

(A) Lithium is the only alkali metal that reacts directly with nitrogen at room temperature to form lithium nitride $(Li_3N)$.
The chemical reaction is:
$6Li + N_2 \to 2Li_3N$

(B) The solubility of alkali metal hydroxides increases down the group as the lattice energy decreases more rapidly than the hydration energy. Therefore,$CsOH$ has the highest solubility product.

(B) The reducing power of alkali metals is determined by their standard electrode potential $(E^\circ)$.
$Li$ has the most negative $E^\circ$ value,making it the strongest reducing agent in aqueous solution due to its high hydration energy.
Among the given options $(Li, Na, K, Rb)$,the reducing power generally increases down the group due to the decrease in ionization enthalpy,but in aqueous solution,$Li$ is the strongest. However,comparing the standard reduction potentials $(E^\circ)$ for $M^+/M$,the values are: $Li^+/Li = -3.04 \ V$,$Rb^+/Rb = -2.93 \ V$,$K^+/K = -2.93 \ V$,$Na^+/Na = -2.71 \ V$.
Since $Na$ has the least negative reduction potential,it acts as the weakest reducing agent among the given options.

(A) In the electrochemical series (which is based on oxidation potential),alkali metals have a higher oxidation potential than hydrogen,meaning they are placed below hydrogen in the reduction potential series (or above it in the oxidation potential series).
Because they have a high tendency to lose electrons (high oxidation potential),they can easily displace $H^+$ ions from water to release $H_2$ gas and form metal hydroxides (bases).