s-Block Elements

$ \begin{aligned} &\text { The reaction involved is, }\\ &\begin{array}{r} \mathrm{Mg}\left(\mathrm{HCO}_2\right)_2 \longrightarrow \mathrm{Mg}(\mathrm{OH})_2 \downarrow+2 \mathrm{CO}_2 \uparrow +\mathrm{H}_2 \mathrm{O} \end{array} \end{aligned} $

$ \mathrm{Ca}\left(\mathrm{HCO}_3\right)_2 \longrightarrow \mathrm{CaCO}_3 \downarrow+\mathrm{CO}_2+\mathrm{H}_2 \mathrm{O} $

So, the precipitate formed are $\mathrm{Mg}\left(\mathrm{OH}_2\right): \mathrm{CaCO}_3$

Option A:

Carbonates of alkaline earth metals are insoluble in water.

✅ True.

Alkaline earth metal carbonates (e.g., CaCO₃, SrCO₃, BaCO₃) are generally insoluble in water, except BeCO₃ and MgCO₃, which are slightly soluble.

Option B:

Beryllium halides are covalent in nature.

✅ True.

Beryllium has a small size and high charge density, which leads to high polarization of the halide ion → results in covalent bonding (as per Fajan’s rules).

Option C:

The superoxides of alkali metals are colourless.

❌ False.

Superoxides (e.g., KO₂, RbO₂, CsO₂) are colored — typically yellow to orange — because of the presence of the paramagnetic O₂⁻ ion. Hence, this statement is not correct.

Option D:

Alkali metal halides have high negative enthalpies of formation.

✅ True.

They are highly stable ionic compounds formed with large lattice energies and thus have large negative enthalpies of formation.

✅ Correct answer: Option C

The superoxides of alkali metals are colourless. (Incorrect statement)

Statement I and III are correct while II and IV are incorrect. The correct forms of II and IV are as follows.

II. LiCl is soluble in ethanol due to its polar nature and small size. $\mathrm{MgCl}_2$ is sparingly soluble but not completely insoluble.

IV. $\mathrm{Na}_2 \mathrm{CO}_3$ does not decompose to $\mathrm{CO}_2$ just by heating under normal conditions.

The major ingredient $(\sim 51 \%)$ in Portland cement is tricalcium silicate $\mathrm{Ca}_3 \mathrm{SiO}_5$, also called $\mathrm{C}_3 \mathrm{~S}$ ). It is mainly responsible for the early strength development of cement by rapidly reacting with water to form calcium silicate hydrate and calcium hydroxide. Its high percentage ensures quick setting and hardening of concrete structure.

The reaction of metals with air, which contains both oxygen and nitrogen, can lead to the formation of both oxides and nitrides. The metals that exhibit this dual reactivity are those with a sufficiently high affinity for nitrogen. Let's analyse the given metals.

• Lithium (Li)

$ \begin{aligned} & 4 \mathrm{Li}(s)+\mathrm{O}_2(g) \rightarrow 2 \mathrm{Li}_2 \mathrm{O}(s) \\ & 6 \mathrm{Li}(s)+\mathrm{N}_2(g) \rightarrow 2 \mathrm{Li}_3 \mathrm{~N}(s) \end{aligned} $

Lithium forms both an oxide and a nitride.

• Sodium (Na) and potassium (K)

$ 2 \mathrm{Na}(\mathrm{~s})+\mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{Na}_2 \mathrm{O}_2(\mathrm{~s}) $    (primarily peroxide)

$ \mathrm{K}(s)+\mathrm{O}_2(g) \rightarrow \mathrm{KO}_2(s) $   (primarily superoxide)

These alkali metals do not readily form nitrides when burnt in air.

• Beryllium (Be)

$ \begin{aligned} & 2 \mathrm{Be}(s)+\mathrm{O}_2(g) \rightarrow 2 \mathrm{BeO}(s) \\ & 3 \mathrm{Be}(s)+\mathrm{N}_2(g) \rightarrow \mathrm{Be}_3 \mathrm{~N}_2(s) \end{aligned} $

Beryllium forms both an oxide and a nitride.

$ \begin{aligned} &\text { Magnesium (Mg) }\\ &\begin{aligned} & 2 \mathrm{Mg}(s)+\mathrm{O}_2(g) \rightarrow 2 \mathrm{MgO}(s) \\ & 3 \mathrm{Mg}(s)+\mathrm{N}_2(g) \rightarrow \mathrm{Mg}_3 \mathrm{~N}_2(s) \end{aligned} \end{aligned} $

Magnesium forms both an oxide and a nitride.

• Strontium ( Sr )

$ \begin{aligned} & 2 \mathrm{Sr}(s)+\mathrm{O}_2(g) \rightarrow 2 \mathrm{SrO}(s) \\ & 3 \mathrm{Sr}(s)+\mathrm{N}_2(g) \rightarrow \mathrm{Sr}_3 \mathrm{~N}_2(s) \end{aligned} $

Strontium forms both an oxide and a nitride.

• Barium (Ba)

$ \mathrm{Ba}(s)+\mathrm{O}_2(g) \rightarrow \mathrm{BaO}_2(s) $ (primarily peroxide)

$ 3 \mathrm{Ba}(s)+\mathrm{N}_2(g) \rightarrow \mathrm{Ba}_3 \mathrm{~N}_2(s) $

Barium forms both an oxide (peroxide) and a nitride.

The metals from the list that form both an oxide and a nitride are lithium (Li), beryllium (Be), magnesium (Mg), strontium ( Sr ), and barium (Ba).

(a) Thermal stability of group 2 carbonates : The thermal stability of carbonates of alkaline earth metals increases down the group as the cation's polarising power decreases. The order $\mathrm{BeCO}_3<\mathrm{MgCO}_3<\mathrm{CaCO}_3<\mathrm{SrCO}_3$ is correct.

(b) Solubility of group 2 sulphates: The solubility of sulphates of alkaline earth metals decreases down the group. The order $\mathrm{BeSO}_4>\mathrm{MgSO}_4>\mathrm{CaSO}_4>\mathrm{SrSO}_4$ is correct.

(c) Thermal stability of group 1 carbonates : The thermal stability of alkali metal carbonates increases down the group as the cation's polarising power decreases. The correct order is

$ \mathrm{Li}_2 \mathrm{CO}_3<\mathrm{Na}_2 \mathrm{CO}_3<\mathrm{K}_2 \mathrm{CO}_3<\mathrm{Rb}_2 \mathrm{CO}_3 $

The given order,

$ \mathrm{Li}_2 \mathrm{CO}_3>\mathrm{Na}_2 \mathrm{CO}_3>\mathrm{K}_2 \mathrm{CO}_3>\mathrm{Rb}_2 \mathrm{CO}_3, $

is the reverse and therefore incorrect.

(d) Solubility of group 2 carbonates: The solubility of alkaline earth metal carbonates decreases down the group. The order $\mathrm{BeCO}_3>\mathrm{MgCO}_3>\mathrm{CaCO}_3>\mathrm{SrCO}_3$ is correct.

Thus, the incorrect order is (c)..

• (A) Mg(HCO₃)₂ → Mg(OH)₂↓ + 2CO₂↑

  This reaction shows the decomposition of magnesium bicarbonate upon heating (boiling) to form insoluble magnesium hydroxide and carbon dioxide. This process removes temporary hardness from water and is achieved by Boiling.

  So, (A) matches with (III) Boiling.

• (B) M²⁺ + Na₄P₆O₁₈²⁻ → [Na₂MP₆O₁₈]²⁻ + 2Na⁺

  In this reaction, hard water ions (M²⁺, typically Ca²⁺ or Mg²⁺) react with sodium hexametaphosphate (Na₄P₆O₁₈²⁻), commonly known as Calgon. The Calgon forms a soluble complex with the metal ions, thereby sequestering them and preventing them from causing hardness. This is the principle behind Calgon's method.

  So, (B) matches with (IV) Calgon's method.

• (C) Ca(HCO₃)₂ + Ca(OH)₂ → 2CaCO₃↓ + 2H₂O

  This reaction shows calcium bicarbonate reacting with a calculated amount of slaked lime (Ca(OH)₂) to precipitate insoluble calcium carbonate. This method is specifically known as Clark's method for removing temporary hardness.

  So, (C) matches with (I) Clark's method.

• (D) 2NaZ + Ca²⁺(aq) → 2Na⁺ + CaZ (Z = Zeolite)

  This reaction illustrates the exchange of calcium ions (Ca²⁺) from hard water with sodium ions (Na⁺) from sodium zeolite (NaZ). Zeolites are ion-exchange resins that replace hardness-causing ions with non-hardness-causing ions. This is the basis of the Ion exchange method (also known as the Zeolite process or Permutit process).

  So, (D) matches with (II) Ion exchange method.

Combining the matches:

A - III

B - IV

C - I

D - II

Comparing this with the given options:

Option A: A-III, B-IV, C-I, D-II (Matches our findings)

Option B: A-IV, B-II, C-I, D-III

Option C: A-III, B-IV, C-II, D-I

Option D: A-II, B-IV, C-I, D-III

The correct option is A.

The final answer is $\boxed{\text{A-III, B-IV, C-I, D-II}}$.

Statement I is correct but Statement II is not correct. It's correct form is,

Low solubility of LiF in water is due to its high lattice enthalpy and low solubility of CsI in water is due to its smaller hydration enthalpy.

The correct order of melting point of $s$-block element is,

$ \mathrm{Be}>\mathrm{Mg}>\mathrm{Li}>\mathrm{Na} $

Melting point decreases down the group. Group 2 has higher melting point than group 1 due to high charge density and strong metallic bonding.

A. KOH (Potassium hydroxide)

• KOH is a strong base that absorbs carbon dioxide ($ \mathrm{CO_2} $) forming potassium carbonate.

✅ Use: Absorbent for $ \mathrm{CO_2} $.

→ A → IV

B. Na(l) (Liquid sodium)

• Liquid sodium metal is used as a coolant in nuclear reactors because of its high thermal conductivity.

✅ Use: Coolant.

→ B → I

C. Li (Lithium)

• Lithium is used in electrochemical cells (batteries).

✅ Use: Electrochemical cells.

→ C → III

D. Mg(OH)$_2$ (Magnesium hydroxide)

• $ \mathrm{Mg(OH)_2} $ is a weak base, used as an antacid (“milk of magnesia”).

✅ Use: Antacid.

→ D → II

✅ Therefore, correct matching is:

A–IV, B–I, C–III, D–II

👉 Correct Option: B

The reaction mention is as follows,

(i) $2 \mathrm{Na}(s)+\underset{\text { (excess) }}{\mathrm{O}_2(g)} \longrightarrow \underset{(X)}{\mathrm{Na}_2 \mathrm{O}_2(s)}$

Sodium peroxide $\mathrm{O}_2^{2-}$ has all its electrons paired making it diamagnetic.

The superoxide has one unpaired electrons, making it paramagnetic.

Beryllium hydroxide is an amphoteric hydroxide. It can react with both acid and alkalis to produce salt and water.

$ \begin{array}{l} \mathrm{Be}(\mathrm{OH})_{2}+\underset{\mathrm{Acid}}{2 \mathrm{HCl}} \longrightarrow \mathrm{BeCl}_{2}+2 \mathrm{H}_{2} \mathrm{O} \\\\ \mathrm{Be}(\mathrm{OH})_{2}+2 \mathrm{NaOH} \longrightarrow \mathrm{Na}_{2} \mathrm{BeO}_{2}+2 \mathrm{H}_{2} \mathrm{O} \end{array} $

The complete reaction is as follolits

$ \begin{array}{c} \mathrm{Cs}+\underset{\text { (Excess) }}{\mathrm{O}_{2}} \longrightarrow \underset{\substack{\text { Casesium } \\\\ \text { superoxide }}}{\mathrm{CsO}_{2}} \quad \text {(Product X)} \end{array}$

$\begin{array}2 \mathrm{Na}+\mathrm{O}_{2} \longrightarrow \underset{\substack{\text { Sodium } \\\\ \text { peroxide }}}{\mathrm{Na}_{2} \mathrm{O}_{2}} \quad \text {(Product Y)} \end{array} $

Hence, $ Y $ is peroxide and $ X $ is superaxite

Statements given in I, II and IV are correct, while III and V are incorrect. Their correct forms are BeO is an amphoteric oxide. $ \mathrm{BeCO}_{3} $ is most soluble among all the carbonate of group 2 elements.

Lithium oxy salts are heat unstable and decomposes easily when heated, forming the stable oxide.

Lithium carbonate decomposes into lithium oxide and carbon dioxide when heated.

$ \mathrm{Li}_2 \mathrm{CO}_3 \longrightarrow \mathrm{Li}_2 \mathrm{O}+\mathrm{CO}_2 \uparrow $

Stability of superoxides increases with increase in size and electropositivity of metal.

Hence, the correct order is $\mathrm{KO}_2<\mathrm{RbO}_2<\mathrm{CsO}_2$.

Lithium nitrate $\left(\mathrm{LiNO}_3\right)$ decomposes to give lithium oxide $\left(\mathrm{Li}_2 \mathrm{O}\right)$.

Among the given alkali metal lithium ( $X$ ) has highest $E^{\circ}$ value while sodium ( $Y$ ) has highest $E^{\circ}$ value. It is because lithium has maximum tendency to lose electrons and hence is the strongest reducing agent. While sodium has minimum tendency to lose electrons and hence is weakest reducing agent.

Among the given options, statement given in option (a) is incorrect.

It is because the molecular formula of calgon is $\left(\mathrm{NaPO}_3\right)_6$.

The thermal decomposition of lithium nitrate $\left(\mathrm{LiNO}_3\right)$ gives $\mathrm{Li}_2 \mathrm{O}, \mathrm{O}_2$ and $\mathrm{NO}_2$. The reaction involved is as follows:

$ \underset{\substack{\text { Lithium } \\ \text { nitrate }}}{4 \mathrm{LiNO}_3} \xrightarrow{\Delta} \underset{\substack{\text { Lithium } \\ \text { oxide }}}{2 \mathrm{Li}_2 \mathrm{O}}+\underset{\substack{\text { Nitrogen } \\ \text { dioxide }}}{4 \mathrm{NO}_2 \uparrow}+\underset{\text { Oxygen }}{\mathrm{O}_2 \uparrow} $

Alkali metals and their salts impact colours to the flame because of their low ionisation enthalpies.

Alkali metals have lowest ionisation among different groups in periodic table. Therefore, both $(\mathrm{A})$ and $(\mathrm{R})$ are correct and $(\mathrm{R})$ is the correct explanation of (A).

Alkali metals have strong tendency to form superoxides due to their small sizes.

Therefore, potassium ( K ), rubidium (Rb) and Cs (cesium) form super oxides very easily.

Beryllium oxide is amphoteric in nature i.e. it, behaves both as an acid and a base. Beryllium doesn't react with hydrogen because its oxidation potential is low.

Beryllium hydride is formed when beryllium chloride $\left(\mathrm{BeCl}_2\right)$ reacts with lithium aluminium hydride $\left(\mathrm{LiAlH}_4\right)$.

Beryllium sulphate is readily soluble in water as higher hydration enthalpy of $\mathrm{Be}^{2+}$ overcomes the lattice enthalpy. Hence, statements I and III are correct while II and IV are incorrect.

$\mathrm{Ca}(\mathrm{OH})_2$ is an inorganic compound used for water softening. $\mathrm{Ca}(\mathrm{OH})_2$ is used as flocculant in water and sewage treatment.

$\therefore A$ is $\mathrm{Ca}(\mathrm{OH})_2$

The lattice energy decreases with increase in the size of halide ion. The decrease in lattice energy will cause decrease in melting point of alkali halides.

Latice energy $\propto \frac{\text { Charge of ion }}{\text { Size of ion }}$

∴ The correct order of melting point will be

$ \begin{aligned} & \text { LiF }>\mathrm{LiCl}>\mathrm{LiI} \\ & \mathrm{II}>\mathrm{I}>\mathrm{III} \end{aligned} $

Out of all given metal halides, the order of size of cation is

$ \mathrm{Mg}^{2+}<\mathrm{Ca}^{2+}<\mathrm{Sr}^{2+}<\mathrm{Ba}^{2+} $

The hydration of cation is indirectly proportional to the size of the cation. Hence, $\mathrm{MgCl}_2$ molecule/salt can accomodate more number of water molecules per molecule in their halide hydrates.

Potassium superoxide is a basic inorganic compound which undergoes hydrolysis as follows.

$ \mathrm{KO}_2+\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{~K}^{+}+\mathrm{OH}^{-}+\mathrm{O}_2+\mathrm{H}_2 \mathrm{O}_2 $

Lithium nitrate, on heating, decomposes to give lithium oxide, whereas other alkali metals nitrate decomposes to give their corresponding nitrite.

$ \begin{aligned} & 2 \mathrm{LiNO}_3 \longrightarrow 2 \mathrm{Li}_2 \mathrm{O}+4 \mathrm{NO}_2+\mathrm{O}_2 \\ & 2 \mathrm{KNO}_3 \longrightarrow 2 \mathrm{KNO}_2+\mathrm{O}_2 \end{aligned} $

All carbonates of alkaline earth metals decompose on heating to give metal oxide and carbon dioxide ( $\mathrm{CO}_2$ ). All these carbonates are white solids and the oxide that are produced are also white solid

$ \underset{\text { Metal carbonate }}{M \cdot \mathrm{CO}_3} \xrightarrow{\Delta} \underset{\text { Metal oxide }}{M \cdot \mathrm{O}}+\mathrm{CO}_2 $

∴ Option (b) is the correct answer, i.e. (I) and (II) only.

All alkali metals (Li, Na, K, Rb and Cs) form monoxides $\left(\stackrel{I}{M_2} \stackrel{-2}{\mathrm{O}}\right)$.

Except Li, other alkali metals form peroxides $\stackrel{I}{M_2}-{ }^{-1}{ }_2$ )

Except Li and Na other alkali metals form superoxides $\left(\begin{array}{ll}I & -1 / 2 \\ \mathrm{O}_2\end{array}\right)$.

So, statements I, II and III all are correct (option-d).

$\mathrm{BeCl}_2$ is covalent in nature and can be soluble in organic solvent, e.g. ethanol, because of small atomic size comparatively high electronegativity.

$\mathrm{Be}(\mathrm{OH})_2$ is amphoteric and the basic strength increases down the group. Due to small size and high ionisation enthalpy of $\mathrm{Be}, \mathrm{Be}(\mathrm{OH})_2$ is amphoteric.

Therefore, it dissolves both in acid and bases.

$ \begin{aligned} \mathrm{Be}(\mathrm{OH})_2+2 \mathrm{HCl}+ & 2 \mathrm{H}_2 \mathrm{O} \longrightarrow\left[\mathrm{Be}\left(\mathrm{H}_2 \mathrm{O}\right)_4\right] \mathrm{Cl}_2 \end{aligned} $

$ \text { or } \mathrm{Na}_2\left[\mathrm{Be}(\mathrm{OH})_4\right] \text { or } \mathrm{Be}(\mathrm{OH})_2+2 \stackrel{\ominus}{\mathrm{O}} \mathrm{H}\longrightarrow\left[\mathrm{Be}(\mathrm{OH})_4\right]^{2-} $

Hence, both reaction are feasible.