Mix Examples of s-Block Elements Questions in English

(D) $1$. In an aqueous medium,$Li^{+}$ has the maximum hydrated radius due to its high charge density,which leads to extensive hydration. Thus,the order $Li^{+}_{(aq)} > Na^{+}_{(aq)} > K^{+}_{(aq)} > Rb^{+}_{(aq)} > Cs^{+}_{(aq)}$ is correct.
$2$. For halogens,the bond dissociation energy order is $Cl_2 > Br_2 > F_2 > I_2$. $F_2$ has a lower bond dissociation energy than $Cl_2$ due to significant lone pair-lone pair repulsion between the small fluorine atoms.
$3$. For alkaline earth metal sulfates,the solubility decreases down the group because the hydration energy decreases more rapidly than the lattice energy. Thus,$BeSO_4 > MgSO_4 > CaSO_4$ is correct.
Therefore,all the given statements are correct.

(D) $SO_4^{2-}$ is a large anion,therefore thermal stability increases with the size of the cation $(BeSO_4 < MgSO_4 < CaSO_4 < SrSO_4)$.
Basic nature increases down the group $(BeO < MgO < CaO)$,and $ZnO$ is amphoteric,making it less basic than the alkaline earth metal oxides.
$OH^-$ is a small anion,therefore solubility increases down the group $(LiOH < NaOH < KOH < RbOH)$.
Melting point trend: $NaCl > KCl > RbCl > CsCl > LiCl$ is correct due to the high covalent character of $LiCl$ resulting from its small size and high polarizing power.

(C) $1$. Thermal stability of alkali metal nitrates increases down the group as the electropositive character increases: $LiNO_3 < NaNO_3 < KNO_3$. This is correct.
$2$. Solubility of alkaline earth metal hydroxides increases down the group due to the decrease in lattice energy being more significant than the decrease in hydration energy: $Be(OH)_2 < Ca(OH)_2 < Sr(OH)_2$. This is correct.
$3$. Thermal stability of alkaline earth metal hydroxides increases down the group as the basic character of the metal increases: $Be(OH)_2 < Ca(OH)_2 < Sr(OH)_2$. Therefore,the given order $Be(OH)_2 > Ca(OH)_2 > Sr(OH)_2$ is incorrect.
$4$. Solubility of alkali metal carbonates increases down the group: $Li_2CO_3 < Na_2CO_3 < K_2CO_3$. This is correct.

(D) $1$. Solubility of alkaline earth metal hydroxides increases down the group due to the decrease in lattice energy being more significant than the decrease in hydration energy. Thus,$Ca(OH)_2 < Sr(OH)_2 < Ba(OH)_2$ is correct.
$2$. Solubility of alkali metal carbonates increases down the group as hydration energy decreases less rapidly than lattice energy. Thus,$Li_2CO_3 < Na_2CO_3 < K_2CO_3$ is correct.
$3$. Solubility of alkali metal bicarbonates increases down the group. Thus,$NaHCO_3 < KHCO_3$ is correct.
$4$. For alkaline earth metal thiosulfates,the solubility decreases as the size of the cation increases. The correct order is $CaS_2O_3 > SrS_2O_3 > BaS_2O_3$. Therefore,the given order $BaS_2O_3 < MgS_2O_3 < CaS_2O_3$ is incorrect.

(D) Thermal stability of alkaline earth metal sulfates increases down the group as the size of the cation increases,making the metal-oxygen bond stronger relative to the sulfate ion. Thus,$BeSO_4 < MgSO_4 < CaSO_4 < SrSO_4$ is correct.
$(B)$ Melting point of alkali metal halides decreases as the size of the cation increases (lattice energy decreases). Thus,$NaCl > KCl > RbCl > CsCl$ is correct.
$(C)$ Solubility of alkali metal hydroxides increases down the group due to a decrease in lattice energy being more significant than the decrease in hydration energy. Thus,$LiOH < NaOH < KOH < RbOH < CsOH$ is correct.
$(D)$ Lattice energy is inversely proportional to the inter-ionic distance. As the size of the anion increases $(F^- < Cl^- < Br^- < I^-)$,the lattice energy decreases. Thus,$KF > KCl > KBr > KI$ is correct.
Therefore,all statements $(A), (B), (C),$ and $(D)$ are correct.

(A) $(I)$ Thermal stability of alkaline earth metal sulfates increases down the group as the size of the cation increases,making the $M-O$ bond weaker and the sulfate ion more stable. Thus,$BeSO_4 < MgSO_4 < CaSO_4 < SrSO_4$ is correct.
$(II)$ Basic nature of oxides increases down the group. $BeO$ is amphoteric,while $MgO$ and $CaO$ are basic. The correct order is $BeO < MgO < CaO$. Thus,$(II)$ is incorrect.
$(III)$ Solubility of group $1$ hydroxides increases down the group due to the decrease in lattice energy being more significant than the decrease in hydration energy. The correct order is $LiOH < NaOH < KOH < RbOH$. Thus,$(III)$ is incorrect.
$(IV)$ Melting point of alkali metal halides depends on lattice energy. $LiCl$ has high covalent character due to polarization,leading to a lower melting point than expected. The correct order is $NaCl > KCl > RbCl > CsCl > LiCl$. Thus,$(IV)$ is correct.

(B) When copper is exposed to moist air containing $CO_2$,it reacts to form a green coating on its surface.
This green layer consists of basic copper carbonate,which has the chemical formula $CuCO_3 \cdot Cu(OH)_2$.

(D) $I$. $Na_2CO_3 \cdot 10H_2O$ (Washing soda) is produced by the Solvay process $(I-C)$.
$II$. $Mg(HCO_3)_2$ causes temporary hardness in water $(II-D)$.
$III$. $NaOH$ is produced by the Castner-Kellner process $(III-B)$.
$IV$. $Ca_3Al_2O_6$ (Tricalcium aluminate) is an important ingredient in Portland cement $(IV-A)$.
Therefore,the correct matching is $I-C, II-D, III-B, IV-A$.

(C) The alloy used in the construction of aircraft is $Mg-Al$ (Magnalium).
Magnalium is an alloy of aluminium with $5-30\%$ magnesium.
It is used in aircraft construction because it is lighter,stronger,and more corrosion-resistant than pure aluminium.

(A) $X$ is washing soda,which is $Na_2CO_3 \cdot 10H_2O$.
On heating,it loses $9$ molecules of water to form monohydrated compound $Y$ $(Na_2CO_3 \cdot H_2O)$.
Upon further heating above $373 \ K$,$Y$ loses the remaining water molecule to form anhydrous $Na_2CO_3$,known as soda ash $(Z)$.
Thus,$X$ is washing soda and $Z$ is soda ash.

(C) Amphoteric metals react with both acids and bases to liberate hydrogen gas.
Zinc $(Zn)$ is an amphoteric metal.
Reaction with acid: $Zn + 2HCl \rightarrow ZnCl_2 + H_2 \uparrow$
Reaction with base: $Zn + 2NaOH \rightarrow Na_2ZnO_2 + H_2 \uparrow$

(A) The hydrolysis of metal carbides yields specific hydrocarbons based on the carbon species present in the carbide lattice.
$Al_4C_3$ is a methanide,which reacts with water to produce methane: $Al_4C_3 + 12H_2O \longrightarrow 4Al(OH)_3 + 3CH_4$.
$CaC_2$ is an acetylide,which produces acetylene: $CaC_2 + 2H_2O \longrightarrow Ca(OH)_2 + C_2H_2$.
$Mg_2C_3$ is an allylide,which produces propyne: $Mg_2C_3 + 4H_2O \longrightarrow 2Mg(OH)_2 + CH_3C \equiv CH$.
$Be_2C$ is a methanide,which produces methane: $Be_2C + 4H_2O \longrightarrow 2Be(OH)_2 + CH_4$.
Therefore,the reaction in option $A$ is incorrect as it produces methane,not acetylene.

(C) The $s-$block elements are those in which the last electron enters the outermost $s-$orbital.
Their general valence shell electronic configuration is $ns^{1-2}$.
Option $(A)$ represents Copper $(Cu)$,which has the configuration $[Ar] 3d^{10} 4s^1$. Since the last electron enters the $s-$orbital,it is considered an $s-$block element in terms of valence shell configuration,although it is a transition metal.
Option $(B)$ represents Potassium $(K)$,which has the configuration $[Ar] 4s^1$. This is a typical $s-$block element.
Since both configurations end in an $s^1$ orbital,both belong to the $s-$block category based on the valence electron criteria.

(B) $2NaN_{3(s)} \xrightarrow{\Delta} 2Na_{(s)} + 3N_{2(g)}$. Here,$N_{2}$ is a diatomic gas and $Na$ is a metallic residue.
$(B)$ $Ni(CO)_{4(g)} \xrightarrow{250^{\circ}C} Ni_{(s)} + 4CO_{(g)}$. Here,$CO$ is a diatomic gas and $Ni$ is a metallic residue.
$(C)$ $2KClO_{3(s)} \xrightarrow{\Delta} 2KCl_{(s)} + 3O_{2(g)}$. Here,$KCl$ is a salt,not a metallic residue.
$(D)$ $2HgO_{(s)} \xrightarrow{\Delta} 2Hg_{(l)} + O_{2(g)}$. Here,$O_{2}$ is a diatomic gas and $Hg$ is a metallic residue.
$(E)$ $NH_{4}NO_{2(s)} \xrightarrow{\Delta} N_{2(g)} + 2H_{2}O_{(g)}$. Here,no metallic residue is left.
Therefore,reactions $(A), (B),$ and $(D)$ satisfy the condition.

(C) The correct answer is $(C)$.
Mercurous chloride is represented as $Hg_2Cl_2$.
The mercurous ion exists as a dimer,$Hg_2^{2+}$,which contains a $Hg-Hg$ covalent bond.
This is evidenced by the fact that it is diamagnetic (magnetic moment $\mu = 0$),which is only possible if the two $Hg^{+}$ ions are paired through a metal-metal bond.

(D) For alkaline earth metal carbonates $(Group \ 2)$,the solubility decreases down the group because the lattice energy decreases less rapidly than the hydration energy as the cation size increases. Thus,the order is $BeCO_3 > MgCO_3 > CaCO_3 > SrCO_3 > BaCO_3$. Statement $I$ is correct.
For alkali metal carbonates $(Group \ 1)$,the solubility increases down the group because the hydration energy decreases less rapidly than the lattice energy as the cation size increases. Thus,the order is $Li_2CO_3 < Na_2CO_3 < K_2CO_3 < Rb_2CO_3 < Cs_2CO_3$. Statements $II$ and $III$ are correct.
Statement $IV$ is incorrect as it contradicts the trend for alkali metal carbonates.
Therefore,the correct statements are $I, II, III$.

(B) The compositions of the given alloys are:
$1$. Bronze: $Cu + Sn + Zn$ (or $Al$)
$2$. Magnalium: $Al + Mg$
$3$. Brass: $Cu + Zn$
$4$. Bell metal: $Cu + Sn$
Among the given options,magnalium is the only alloy that does not contain copper $(Cu)$.
Therefore,the correct option is $B$.

(A) Soda lime is a mixture of sodium hydroxide $(NaOH)$ and calcium oxide $(CaO)$.
It is prepared by adding $CaO$ to $NaOH$ solution and then evaporating the water.

(D) Sodium bicarbonate $(NaHCO_3)$ decomposes on heating to form sodium carbonate,water,and carbon dioxide: $2NaHCO_3 \rightarrow Na_2CO_3 + H_2O + CO_2$.
Calcium carbonate $(CaCO_3)$ decomposes on heating to form calcium oxide and carbon dioxide: $CaCO_3 \rightarrow CaO + CO_2$.
Sodium carbonate $(Na_2CO_3)$ is stable to heat and does not decompose under ordinary heating conditions.
Therefore,both $(A)$ and $(C)$ decompose on heating.

(D) The hydrolysis of metal carbides produces specific hydrocarbons based on the carbon unit present in the carbide:
$1$. $Al_4C_3$ (methanide) yields $CH_4$ (methane): $Al_4C_3 + 12H_2O \to 4Al(OH)_3 + 3CH_4$.
$2$. $CaC_2$ (acetylide) yields $C_2H_2$ (acetylene): $CaC_2 + 2H_2O \to Ca(OH)_2 + C_2H_2$.
$3$. $Mg_2C_3$ (allylide) yields $CH_3C \equiv CH$ (propyne): $Mg_2C_3 + 4H_2O \to 2Mg(OH)_2 + CH_3C \equiv CH$.
$4$. $Be_2C$ (methanide) yields $CH_4$ (methane): $Be_2C + 4H_2O \to 2Be(OH)_2 + CH_4$.
Option $D$ is incorrect because $Be_2C$ produces methane,not ethane.

(C) $MgSO_4 \cdot 7H_2O$ is isomorphous with $ZnSO_4 \cdot 7H_2O$ because they share the same crystal lattice structure.
$II-A$ group cations (like $Mg^{2+}$,$Ca^{2+}$) are not precipitated by $H_2S$ in acidic medium,and $II-B$ group cations (like $Zn^{2+}$) are also not precipitated by $H_2S$ in acidic medium (they require a basic medium or specific conditions to precipitate as sulfides).
Therefore,both statements $(A)$ and $(B)$ are correct.

(D) $Be$ and $Al$ exhibit a diagonal relationship due to their similar charge-to-size ratios.
Both elements form protective oxide layers,making them resistant to acids.
$Be$ forms tetrahedral complexes such as $[Be(C_2O_4)_2]^{2-}$ and $[BeF_4]^{2-}$.
$Al$ forms octahedral complexes such as $[AlF_6]^{3-}$ and $[Al(C_2O_4)_3]^{3-}$.
Therefore,all the given statements are correct.

(B) $Al_4C_3$ is an ionic carbide,which on hydrolysis yields methane $(CH_4)$.
Therefore,it is classified as a methanide.

(A) $(I)$ The number of $S-S$ bonds in polythionic acids $H_{2}S_{n}O_{6}$ is $(n-1)$. Thus,the statement is $F$.
$(II)$ $F_{2}$ is a strong oxidizing agent and reacts with water to produce $HF, O_{2}$ and $O_{3}$. Thus,the statement is $T$.
$(III)$ $LiNO_{3}$ (crimson red) and $BaCl_{2}$ (apple green) are used in fireworks. Thus,the statement is $T$.
$(IV)$ Hydrides of $Be$ and $Mg$ are covalent and polymeric,not ionic. Thus,the statement is $F$.
The correct order is $F, T, T, F$.

(C) When $ZnCl_2 \cdot 2H_2O$ is heated,it undergoes hydrolysis due to the high charge density of the $Zn^{2+}$ ion.
The reaction proceeds as follows:
$ZnCl_2 \cdot 2H_2O \xrightarrow{\Delta} Zn(OH)Cl + HCl + H_2O$
Further heating leads to the formation of zinc oxide:
$Zn(OH)Cl \xrightarrow{\Delta} ZnO + HCl$
Thus,the final product obtained upon strong heating is $ZnO$.

(C) The metal $M$ is Zinc $(Zn)$. The reactions are as follows:
$1. 4Zn + 10HNO_3 \text{ (dilute)} \rightarrow 4Zn(NO_3)_2 \text{ (Colourless solution)} + NH_4NO_3 + 3H_2O$
$2. Zn(NO_3)_2 + 2NaOH \rightarrow Zn(OH)_2 \downarrow \text{ (White precipitate)} + 2NaNO_3$
$3. Zn(OH)_2 + 2NaOH \text{ (excess)} \rightarrow Na_2[Zn(OH)_4] \text{ (Colourless solution)}$
$4. Na_2[Zn(OH)_4] + H_2S \rightarrow ZnS \downarrow \text{ (White precipitate)} + 2NaOH + 2H_2O$

(C) When copper is exposed to moist air containing carbon dioxide,it reacts to form a green layer of basic copper carbonate,$CuCO_3 \cdot Cu(OH)_2$,which is commonly known as verdigris.
The chemical reaction is: $2Cu + H_2O + CO_2 + \frac{1}{2}O_2 \rightarrow CuCO_3 \cdot Cu(OH)_2$.

(C) $Ca(H_2PO_4)_2 \cdot H_2O + CaSO_4 \cdot 2H_2O$ is known as superphosphate of lime.
It is widely used as a phosphatic fertilizer in agriculture.

(B) The thermal decomposition of silver nitrate $(AgNO_3)$ occurs as follows:
$2AgNO_3(s) \xrightarrow{\Delta} 2Ag(s) + 2NO_2(g) + O_2(g)$
Thus,the products formed are silver metal $(Ag)$,nitrogen dioxide $(NO_2)$,and oxygen $(O_2)$.

(B) The chemical reaction between calcium phosphide and water is given by:
$Ca_3P_2 + 6H_2O \to 2PH_3 + 3Ca(OH)_2$
From the balanced chemical equation,$1 \ mol$ of $Ca_3P_2$ reacts with $6 \ mol$ of $H_2O$ to produce $2 \ mol$ of phosphine $(PH_3)$ and $3 \ mol$ of calcium hydroxide $(Ca(OH)_2)$.
Therefore,the correct answer is $2 \ mol$ of phosphine.

(B) $(i)$ Solvay process is used for the manufacture of $Na_2CO_3$ $(s)$.
$(ii)$ $NaHCO_3$ $(r)$ evolves $CO_2$ on heating: $2NaHCO_3 \rightarrow Na_2CO_3 + H_2O + CO_2$.
$(iii)$ Aqueous solutions of $Na_2O$ $(p)$,$Na_2O_2$ $(q)$,$NaHCO_3$ $(r)$,and $Na_2CO_3$ $(s)$ are basic,hence not neutral towards litmus.
$(iv)$ $Na_2O_2$ $(q)$ is used as an air purifier in submarines because it reacts with $CO_2$ to release $O_2$.

(D) The thermal stability $(T.S.)$ of oxo-salts (like carbonates,nitrates,and chlorates) of alkali and alkaline earth metals increases as the size of the cation increases.
This is because the polarizing power of the cation decreases as its size increases,which leads to greater stability of the lattice.
Mathematically,$T.S. \propto \text{Size of cation} \propto \frac{1}{\text{Polarizing power}}$.
In option $A$,the stability of alkali metal nitrates increases from $Li$ to $Cs$.
In option $B$,the stability of alkali metal chlorates increases from $Li$ to $Cs$.
In option $C$,the stability of alkaline earth metal carbonates increases from $Be$ to $Ba$.
Therefore,all the given orders are correct.

(D) Zeolites are aluminosilicates with a three-dimensional honeycomb-like structure.
$1$. They act as water softeners because they function as ion exchangers,replacing $Ca^{2+}$ and $Mg^{2+}$ ions in hard water with $Na^+$ ions.
$2$. They act as catalysts in shape-selective catalysis due to their porous structure,such as $ZSM-5$ which converts alcohols to gasoline.
$3$. They act as cation exchangers by exchanging their sodium ions for other cations.
Since zeolites exhibit all these properties,the correct answer is $D$.

(B) Holme's signals are produced by a mixture of calcium carbide $(CaC_2)$ and calcium phosphide $(Ca_3P_2)$.
When this mixture is placed in a container and thrown into the sea,it reacts with water.
Calcium carbide reacts with water to produce acetylene gas $(C_2H_2)$,and calcium phosphide reacts with water to produce phosphine gas $(PH_3)$.
The phosphine gas catches fire spontaneously upon contact with air,which in turn ignites the acetylene gas,creating a bright flame that acts as a signal to guide ships in the sea.

(C) Analysis of the given orders:
$(a)$ Lattice Energy is directly proportional to the product of charges and inversely proportional to the interionic distance. For $A_2B$ ($A^+$,$B^{2-}$),$AB$ ($A^{2+}$,$B^{2-}$),and $A_2B_3$ ($A^{3+}$,$B^{2-}$),the charge product increases as $A_2B < AB < A_2B_3$. This order is correct.
$(b)$ Thermal stability of alkali metal carbonates increases down the group due to the decrease in polarizing power of the cation. The order $Na_2CO_3 < K_2CO_3 < Rb_2CO_3 < Cs_2CO_3$ is correct.
$(c)$ For alkaline earth metal fluorides,the solubility order is $BeF_2 > MgF_2 > CaF_2 > SrF_2 > BaF_2$. The given order $MgF_2 < CaF_2 < SrF_2 < BaF_2 < BeF_2$ is incorrect because $BeF_2$ is highly soluble,while the others show a trend of increasing solubility down the group,but the placement of $BeF_2$ at the end is wrong.
$(d)$ Solubility and thermal stability of alkali metal hydroxides increase down the group. The order $LiOH < NaOH < KOH < RbOH < CsOH$ is correct.
Therefore,only $(c)$ is incorrect.

(B) mixture of $Ca_3P_2$ and $CaC_2$ is used in making Holme's signal.
When this mixture comes in contact with water,it produces phosphine $(PH_3)$ and acetylene $(C_2H_2)$ gases.
The phosphine gas catches fire spontaneously,which ignites the acetylene gas,producing a bright flame that guides ships in the sea.
This mixture is also used for creating smoke screens.

(C) $Al_2(SO_4)_3$ on heating gives $Al_2O_3$ (white).
$HgCO_3 \cdot 3Hg(OH)_2$ on heating gives $Hg$ (liquid metal).
$Cu(NO_3)_2$ on heating gives $CuO$ (black/brown coloured metal oxide).
$Ba(OH)_2$ on heating gives $BaO$ (white).
Therefore,the correct option is $C$.

(D) $1$. Solubility of alkaline earth metal fluorides ($BeF_2$ to $BaF_2$) increases down the group due to the decrease in lattice energy being more significant than the decrease in hydration energy. Thus,$BeF_2 < MgF_2 < CaF_2$ is correct.
$2$. Acidic nature of metal hydrides increases as the electropositivity of the metal decreases. Since $Li$ is the least electropositive among $Li, Na, K$,$LiH$ is the most acidic. Thus,$LiH > NaH > KH$. The given order $LiH < NaH < KH$ is incorrect.
$3$. Thermal stability of alkali metal carbonates increases down the group as the electropositive character of the metal increases. Thus,$Li_2CO_3 < Na_2CO_3 < K_2CO_3$. The given order $Li_2CO_3 > Na_2CO_3 > K_2CO_3$ is incorrect.
Since both $B$ and $C$ are incorrect,the question implies identifying the incorrect statements. However,if only one option is to be chosen,$D$ is the most appropriate answer as multiple options are incorrect.

(C) Let us analyze the thermal decomposition of each compound:
$I$. $2AgNO_3 \xrightarrow{\Delta} 2Ag + 2NO_2 + O_2$
$II$. $CaCO_3 \xrightarrow{\Delta} CaO + CO_2$ (Does not give $O_2$)
$III$. $2Pb(NO_3)_2 \xrightarrow{\Delta} 2PbO + 4NO_2 + O_2$
$IV$. $Na_2CO_3$ is thermally stable and does not decompose to give $O_2$.
Thus,compounds $I$ and $III$ release oxygen gas upon heating. However,checking the provided options,the question implies identifying which compounds release oxygen. Based on the chemical reactions,only $I$ and $III$ produce $O_2$. Since this specific combination is not explicitly listed as a single option,we re-evaluate the question context. If the question intended to ask for nitrates,the answer would be $I$ and $III$. Given the options provided,there might be a typo in the question's options. Assuming the question asks for compounds that release oxygen,the correct set is $I$ and $III$.

(C) The compound $Ca(H_2PO_4)_2 \cdot H_2O + CaSO_4$ is known as superphosphate of lime.
It is a widely used phosphatic fertilizer in agriculture.
$CaC_2$ is calcium carbide,used for ripening fruits.
$NaAlO_2$ is sodium aluminate,used in water purification and paper industry.

(A) The $s-$block elements are highly electropositive in nature,which makes them very reactive.
Due to their high reactivity,they readily react with other elements and do not occur in a free state in nature.
They are typically found in nature in the form of compounds like halides,carbonates,and sulphates.
Therefore,the Reason correctly explains the Assertion.

(D) In $CO_2$,the total number of electrons is $6 + 8 + 8 = 22$.
In the cyanamide ion $(CN_2^{2-})$,the total number of electrons is $6 + 7 + 7 + 2 = 22$. Since both have $22$ electrons,they are isoelectronic. Both $CO_2$ and $(CN_2^{2-})$ exhibit a linear structure. Thus,statement $(a)$ is correct.
$Mg_2C_3$ reacts with water to produce propyne: $Mg_2C_3 + 4H_2O \to 2Mg(OH)_2 + CH_3C \equiv CH$. Thus,statement $(b)$ is correct.
The crystal structure of $CaC_2$ is of the $NaCl$ type (distorted rock salt structure). Thus,statement $(c)$ is correct.
Therefore,all the given statements are correct.

(A) $Al(OH)_3$ is an amphoteric hydroxide. It is insoluble in weak bases like $NH_4OH$ but dissolves in strong bases like $NaOH$ due to the formation of a soluble complex,sodium meta-aluminate $(NaAlO_2)$.
The reaction is: $NaOH + Al(OH)_3 \to NaAlO_2 + 2H_2O$.
Since $NaOH$ is a strong alkali,it provides a high concentration of $OH^-$ ions,which is necessary to dissolve the amphoteric $Al(OH)_3$. Thus,the Reason correctly explains the Assertion.

(N/A)

Alkali metals	Alkaline earth metals
$I$. Ionization enthalpy: These have the lowest ionization enthalpies in their respective periods due to their large atomic sizes. They lose their only valence electron easily to attain a stable noble gas configuration.	$I$. Ionization enthalpy: Alkaline earth metals have smaller atomic sizes and higher effective nuclear charges compared to alkali metals,resulting in higher first ionization enthalpies. However,their second ionization enthalpy is lower than that of alkali metals because alkali metals achieve a stable noble gas configuration after losing one electron.
$II$. Basicity of oxides: The oxides of alkali metals are highly basic due to the highly electropositive nature of the metals,making the oxides strongly ionic and prone to dissociation in water.	$II$. Basicity of oxides: The oxides of alkaline earth metals are basic,but less so than those of alkali metals because alkaline earth metals are less electropositive.
$III$. Solubility of hydroxides: The hydroxides of alkali metals are generally more soluble in water.	$III$. Solubility of hydroxides: The hydroxides of alkaline earth metals are less soluble than those of alkali metals due to their higher lattice energies,which result from their higher charge densities.

(N/A) $(i)$ Nitrates:
Thermal stability: Alkali metal nitrates,except $LiNO_3$,decompose on strong heating to form nitrites $(2MNO_3 \longrightarrow 2MNO_2 + O_2)$. $LiNO_3$ and all alkaline earth metal nitrates decompose to form oxides $(2M(NO_3)_2 \longrightarrow 2MO + 4NO_2 + O_2)$. Thermal stability increases down both groups.
Solubility: Nitrates of both group $1$ and group $2$ metals are generally soluble in water.
$(ii)$ Carbonates:
Thermal stability: Alkali metal carbonates (except $Li_2CO_3$) are stable towards heat. $Li_2CO_3$ and all alkaline earth metal carbonates decompose on heating to form oxides and $CO_2$ $(MCO_3 \longrightarrow MO + CO_2)$.
Solubility: Alkali metal carbonates are soluble in water (except $Li_2CO_3$),and solubility increases down the group. Alkaline earth metal carbonates are generally insoluble in water.
$(iii)$ Sulphates:
Thermal stability: Sulphates of both group $1$ and group $2$ metals are generally stable towards heat.
Solubility: Alkali metal sulphates are soluble in water. For alkaline earth metals,solubility decreases down the group $(BeSO_4 > MgSO_4 > CaSO_4 > SrSO_4 > BaSO_4)$.

(N/A) The $s$-block elements of the Periodic Table are those in which the last electron enters the outermost $s$-orbital.
As the $s$-orbital can accommodate only two electrons,two groups ($1$ and $2$) belong to the $s$-block of the Periodic Table.
The general electronic configuration of $s$-block elements is [noble gas] $ns^{1}$ for alkali metals and [noble gas] $ns^{2}$ for alkaline earth metals.
Group-$1$ of the Periodic Table consists of the elements: lithium,sodium,potassium,rubidium,caesium,and francium. They are collectively known as the alkali metals. These are so called because they form hydroxides on reaction with water,which are strongly alkaline in nature.
The elements of Group-$2$ include beryllium,magnesium,calcium,strontium,barium,and radium. These elements,with the exception of beryllium,are commonly known as the alkaline earth metals. These are so called because their oxides and hydroxides are alkaline in nature and these metal oxides are found in the earth's crust.

(N/A) Among the alkali metals,$Na$ and $K$ are abundant,while $Li$,$Rb$,and $Cs$ have much lower abundances.
$Fr$ is highly radioactive; its longest-lived isotope $^{223}Fr$ has a half-life of only $21 \ minutes$.
Of the alkaline earth metals,$Ca$ and $Mg$ rank fifth and sixth in abundance,respectively,in the Earth's crust.
$Sr$ and $Ba$ have much lower abundances.
$Be$ is rare,and $Ra$ is the rarest of all,comprising only $10^{-10} \ \%$ of igneous rocks.

(N/A) Elements in which the last electron enters the $s$-orbital are known as $s$-block elements.
The elements of group $1$ (alkali metals) and group $2$ (alkaline earth metals),which have $ns^{1}$ and $ns^{2}$ outermost electronic configurations respectively,belong to the $s$-block.
Characteristics:
- They are all reactive metals with low ionization enthalpies.
- They readily lose their outermost electron$(s)$ to form $1+$ ions (alkali metals) or $2+$ ions (alkaline earth metals).
- Metallic character and reactivity increase as we move down the group.
- Due to high reactivity,they are never found in a pure state in nature.
- Compounds of $s$-block elements,with the exception of those of $Li$ and $Be$,are predominantly ionic.

(N/A) Nitrates :
$(i)$ Thermal stability: Nitrates of alkali metals,except $LiNO_{3}$,decompose on strong heating to form nitrites. $2KNO_{3(s)} \rightarrow 2KNO_{2(s)} + O_{2(g)}$. $LiNO_{3}$ on decomposition gives oxide: $4LiNO_{3(s)} \xrightarrow{\Delta} 2Li_{2}O_{(s)} + 4NO_{2(g)} + O_{2(g)}$. As we move down group-$1$ and group-$2$,the thermal stability of nitrates increases.
$(ii)$ Solubility: Nitrates of both group-$1$ and group-$2$ metals are soluble in water.
$(b)$ Carbonates :
$(i)$ Thermal stability: The carbonates of alkali metals are stable towards heat,except $Li_{2}CO_{3}$,which decomposes to form lithium oxide. The carbonates of alkaline earth metals decompose on heating to form oxide and carbon dioxide. $Na_{2}CO_{3} \xrightarrow{\Delta} \text{No reaction}$. $Li_{2}CO_{3} \xrightarrow{\Delta} Li_{2}O + CO_{2}$. $MgCO_{3} \xrightarrow{\Delta} MgO + CO_{2}$.
$(ii)$ Solubility: Carbonates of alkali metals are soluble in water with the exception of $Li_{2}CO_{3}$. Solubility increases down the group. Carbonates of alkaline earth metals are generally insoluble in water.
$(c)$ Sulphates :
$(i)$ Thermal stability: Sulphates of both group-$1$ and group-$2$ metals are stable towards heat.
$(ii)$ Solubility: Sulphates of alkali metals are soluble in water. Sulphates of alkaline earth metals show varied trends: $BeSO_{4}$ (fairly soluble),$MgSO_{4}$ (soluble),$CaSO_{4}$ (sparingly soluble),$SrSO_{4}$ (insoluble),$BaSO_{4}$ (insoluble). Solubility of alkaline earth metal sulphates decreases down the group.

(N/A)

Alkali metals	Alkaline earth metals
$I$. Ionization enthalpy: These have the lowest ionization enthalpies in their respective periods due to their large atomic sizes. They lose their only valence electron easily to attain a stable noble gas configuration.	Alkaline earth metals have smaller atomic sizes and higher effective nuclear charge compared to alkali metals,leading to higher first ionization enthalpies. However,their second ionization enthalpy is lower than that of alkali metals because alkali metals achieve a stable noble gas configuration after losing one electron.
$II$. Basicity of oxides: The oxides of alkali metals are highly basic due to the high electropositive nature of the metals,which makes the oxides highly ionic and prone to dissociation in water.	The oxides of alkaline earth metals are basic,but less so than those of alkali metals because alkaline earth metals are less electropositive.
$III$. Solubility of hydroxides: The hydroxides of alkali metals are generally more soluble in water.	The hydroxides of alkaline earth metals are less soluble than those of alkali metals due to their higher lattice energies resulting from higher charge density.