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.

The complete Solvay process mechanism is,

$ \begin{aligned} 2 \mathrm{NH}_4 \mathrm{Cl}+\underset{\text { (Reactant } X \text { ) }}{\mathrm{Ca}(\mathrm{OH})_2} \longrightarrow 2 \mathrm{NH}_3 & +\mathrm{CaCl}_2 +2 \mathrm{H}_2 \mathrm{O} \end{aligned} $

Statements given in II and III are correct, while I is incorrect.

It's correct form is,

All alkaline metals does not give hydride on heating with hydrogen.

Beryllium does not from hydride.

Statement given in $A$ and $C$ are correct, while $B$ and $D$ are incorrect, Their correct forms are

(B) Boiling point of alkali, metal decreases from Li to Cs .

(D) Li has tendency to form monoxide.

Analysis of Statement - I: Both $\mathrm{BeSO}_4$ and $\mathrm{MgSO}_4$ are readily soluble in water.

• The solubility of sulfates of alkaline earth metals generally decreases down the group.

• $\mathrm{BeSO}_4$ (Beryllium sulfate) is readily soluble in water. This is due to the very small size of the $\mathrm{Be}^{2+}$ ion, which leads to very high hydration energy. This high hydration energy more than compensates for the relatively high lattice energy, making it highly soluble.

• $\mathrm{MgSO}_4$ (Magnesium sulfate, commonly known as Epsom salt) is also readily soluble in water. Similar to beryllium, the $\mathrm{Mg}^{2+}$ ion is relatively small, leading to significant hydration energy.

• In contrast, $\mathrm{CaSO}_4$ (Calcium sulfate) is sparingly soluble, and $\mathrm{SrSO}_4$ (Strontium sulfate) and $\mathrm{BaSO}_4$ (Barium sulfate) are practically insoluble.

Therefore, Statement - I is correct.

Analysis of Statement - II: Among the nitrates of alkaline earth metals, only $\mathrm{Be}\left(\mathrm{NO}_3\right)_2$ on strong heating gives its oxide, $\mathrm{NO}_2$ and $\mathrm{O}_2$.

• Alkaline earth metal nitrates decompose on strong heating to give the metal oxide, nitrogen dioxide ($\mathrm{NO}_2$), and oxygen ($\mathrm{O}_2$). The general reaction is:

  $2 \mathrm{M}(\mathrm{NO}_3)_2(\mathrm{s}) \xrightarrow{\text{heat}} 2 \mathrm{MO}(\mathrm{s}) + 4 \mathrm{NO}_2(\mathrm{g}) + \mathrm{O}_2(\mathrm{g})$

• This decomposition pattern is observed for all alkaline earth metal nitrates.

  • $\mathrm{Be}(\mathrm{NO}_3)_2 \xrightarrow{\text{heat}} \mathrm{BeO} + 2\mathrm{NO}_2 + \frac{1}{2}\mathrm{O}_2$

  • $\mathrm{Mg}(\mathrm{NO}_3)_2 \xrightarrow{\text{heat}} \mathrm{MgO} + 2\mathrm{NO}_2 + \frac{1}{2}\mathrm{O}_2$

  • $\mathrm{Ca}(\mathrm{NO}_3)_2 \xrightarrow{\text{heat}} \mathrm{CaO} + 2\mathrm{NO}_2 + \frac{1}{2}\mathrm{O}_2$

  • And similarly for $\mathrm{Sr}(\mathrm{NO}_3)_2$ and $\mathrm{Ba}(\mathrm{NO}_3)_2$.

• The thermal stability of these nitrates generally increases down the group ($\mathrm{Be}(\mathrm{NO}_3)_2$ is the least stable, and $\mathrm{Ba}(\mathrm{NO}_3)_2$ is the most stable), meaning they decompose at different temperatures, but the decomposition products are the same for all of them.

• The statement claims that "only $\mathrm{Be}(\mathrm{NO}_3)_2$" decomposes in this manner, which is incorrect as all alkaline earth metal nitrates follow this decomposition pathway.

Therefore, Statement - II is not correct.

Conclusion:

Statement - I is correct.

Statement - II is not correct.

This matches option B.

The final answer is $\boxed{B}$

KCl does not exist as hydrate. The reason is $\mathrm{K}^{+}$ion have a weak ability to attract and bind water molecules due to their large size and low charge density.

The correct order of density is,

$ \underset{\left(2.63 \mathrm{~g} / \mathrm{cm}^3\right)}{\mathrm{Sr}}>\underset{\left(1.84 \mathrm{~g} / \mathrm{cm}^3\right)}{\mathrm{Be}}>\underset{\left(1.74 \mathrm{~g}^{\prime} / \mathrm{cm}^3\right)}{\mathrm{Mg}} >\underset{\left(1.55 \mathrm{~g} / \mathrm{cm}^3\right)}{\mathrm{Ca}}$

Statement given in I and IV are correct regarding lihium and magnesium, while statement II and III are incorrect. Their correct forms are (II) The biocarbonate of lithium and magnesium are liquid.

(III) LiCl or $\mathrm{MgCl}_2$ are soluble in organic solvent like ethanol.

All the three stated property are correctly ordered.

$\mathrm{NaO}_2<\mathrm{KO}_2<\mathrm{RbO}_2<\mathrm{CsO}_2$ - stability

$\mathrm{Mg}(\mathrm{OH})_2<\mathrm{Ca}(\mathrm{OH})_2<\mathrm{Sr}(\mathrm{OH})_2$ - basic strength

$\mathrm{MgCO}_3<\mathrm{CaCO}_3<\mathrm{SrCO}_3$ - thermal stability

Option A:

“Solubility of carbonates in water decreases down the group.”

For alkaline earth metals (Group 2):

• The solubility of carbonates (MCO₃) actually decreases down the group because lattice energy decreases more slowly than the hydration energy.

✅ This statement is correct.

Option B:

“All the sulphates are thermally stable.”

For sulphates (MSO₄) of alkaline earth metals:

• All of them are thermally stable, i.e., they do not decompose easily on heating.

✅ This statement is correct.

Option C:

“All the nitrates decompose on heating.”

For nitrates (M(NO₃)₂) of group 2:

• All nitrates decompose on heating to give oxides, nitrogen dioxide, and oxygen.

✅ This statement is correct.

Option D:

“All halides are ionic in nature.”

For halides (MX₂):

• Most are ionic, but BeCl₂ (beryllium chloride) has covalent character due to the small size and high polarizing power of Be²⁺.

❌ This statement is incorrect.

✅ Final Answer: Option D – All halides are ionic in nature (Incorrect statement).

To match List I (Alkali Metals) with List II (Emission Wavelength in nm), we need to know the characteristic emission lines of these elements. Here is the information for the typical emission wavelengths:

• Lithium (Li): 670.8 nm
• Sodium (Na): 589.2 nm
• Rubidium (Rb): 780.0 nm
• Cesium (Cs): 455.5 nm

Now let's match these with the given options:

• A. Li (III): 670.8 nm
• B. Na (I): 589.2 nm
• C. Rb (IV): 780.0 nm
• D. Cs (II): 455.5 nm

Therefore, the correct answer is:

Option B

(A)-(III), (B)-(I), (C)-(IV), (D)-(II)

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 $\mathrm{KO}_2<\mathrm{RbO}_2<\mathrm{CsO}_2$.

The key is that only those nitrates which on heating give

• an acidic oxide (NO₂)

• a basic oxide (the metal oxide)

• and O₂

will count.

(i) NaNO₃ → NaNO₂ + ½O₂

 – gives a nitrite, not a metal oxide → rejected

(ii) Ca(NO₃)₂ → CaO (basic) + 2 NO₂ (acidic) + ½ O₂ → OK

(iii) Be(NO₃)₂ → BeO (amphoteric) + 2 NO₂ (acidic) + ½ O₂ → BeO isn’t purely basic

(iv) 2 LiNO₃ → Li₂O (basic) + 2 NO₂ (acidic) + ½ O₂ → OK

Hence only (ii) and (iv) produce both an acidic oxide and a basic oxide along with O₂.

Answer: C (ii, iv)

Na is weakest reducing agent as per $E^{\circ}$ values. Reduction potential of sodium is -2.71 V . The reduction potential of an oxidised species is its tendency to gain electrons and get reduced.

Larger the negative value of reduction potential, stronger is the reducing agent.

$ E_{M^{+} / M}^{\circ}(\text { in } V) \Rightarrow $

$ \begin{array}{ccccc} \mathrm{Li} & \mathrm{Rb} & \mathrm{Cs} & \mathrm{~K} & \mathrm{Na} \\ -3.04 & -2.93 & -2.927 & -2.925 & -2.71 \end{array} $

All group 2 elements, except beryllium (Be), react with hydrogen to form hydrides when heated. The general reaction is as follows:

$ M + \mathrm{H}_2 \longrightarrow MH_2 $

Here, $ M $ represents the alkaline earth metal. Beryllium is the exception and does not form a hydride under these conditions.

Basic Nature of Hydroxides

In group 2, as we move from magnesium to barium, the metal cation becomes larger and its charge density decreases.

This causes the hydroxides to be more ionic and more soluble in water.

More soluble hydroxides produce a greater amount of hydroxide ions in solution, meaning they are more basic.

Thus, $\mathrm{Ba(OH)_2}$ is more basic than $\mathrm{Mg(OH)_2}$.

Option A is correct.

Thermal Stability of Carbonates

The general trend for the thermal decomposition of alkaline earth metal carbonates is that the decomposition temperature increases as we go down the group.

For example, $\mathrm{MgCO_3}$ decomposes at a lower temperature than $\mathrm{CaCO_3}$, and continuing this trend, $\mathrm{BaCO_3}$ (if it could be isolated cleanly) would be thermally more stable than $\mathrm{BeCO_3}$.

Therefore, to say "thermal stability decreases from $\mathrm{BeCO_3}$ to $\mathrm{BaCO_3}$" is incorrect; it actually increases.

Option B is the wrong statement.

Solubility of Sulphates

The solubility of alkaline earth metal sulphates decreases down the group.

For instance, while $\mathrm{MgSO_4}$ and even $\mathrm{BeSO_4}$ (with its slight peculiarities due to beryllium’s atypical behavior) are relatively soluble, $\mathrm{BaSO_4}$ is famous for its insolubility (used in radiography).

Thus, the trend stated in Option C is correct.

Decomposition of Nitrates

On heating, the nitrates of alkaline earth metals decompose according to a reaction like:

$\mathrm{M(NO_3)_2 \xrightarrow{\Delta} MO + 2NO_2 + \frac{1}{2}O_2}$

where M represents the metal (e.g., Mg, Ca, Ba).

This is a well-known property of these nitrates.

Option D is correct.

Conclusion:

The statement that is NOT correct about the compounds of alkaline earth metals is Option B.

Thus, the answer is Option B.

The correct order of density for the elements Beryllium (Be), Magnesium (Mg), Calcium (Ca), and Strontium (Sr) is:

$ \mathrm{Sr} > \mathrm{Be} > \mathrm{Mg} > \mathrm{Ca} $

As we move down a group in the periodic table, the size of the alkali metals increases, leading to an increase in their volume. Volume is inversely proportional to density, which means more volume generally leads to lower density.

However, in the case of Strontium (Sr) and Beryllium (Be), the increase in volume is not as significant compared to the increase in mass. This results in an unusual pattern where the density initially decreases as we go down the group but then increases.

Polarising power $\propto \frac{1}{\text { Size of cation }}$

Polarising power $\propto$ Water of crystallisation tendency.

From the given elements, size of cation order is

$ \mathrm{Ba}^{2+}>\mathrm{Ca}^{2+} \approx \mathrm{Sr}^{2+} $

Thus, tendency to form water of crystallisation is minimum for $\mathrm{Ba}^{2+}$ and same for Ca and $\mathrm{Sr}^{2+}$.

Hence, it will be

$ \begin{aligned} & \mathrm{BaCl}_2 \cdot 2 \mathrm{H}_2 \mathrm{O}, \mathrm{CaCl}_2 \cdot 6 \mathrm{H}_2 \mathrm{O}, \mathrm{SrCl}_2 \cdot 6 \mathrm{H}_2 \mathrm{O} . \\ & x=2, y=6, z=6 \end{aligned} $

The correct order of decomposition temperature for the carbonates $\mathrm{MgCO}_3 (X)$, $\mathrm{BaCO}_3 (Y)$, and $\mathrm{CaCO}_3 (Z)$ is :

$ \mathrm{BaCO}_3 > \mathrm{CaCO}_3 > \mathrm{MgCO}_3 $

This implies the order is $Y > Z > X$.

This order of decomposition can also be described as the thermal stability order. As you move down the group in the periodic table, the thermal stability of alkaline earth metal carbonates increases. This increasing stability is explained by the rise in ionic character as you go down the group.

Alkali metals are part of Group 1 on the periodic table and include lithium (Li), sodium (Na), potassium (K), rubidium (Rb), and cesium (Cs). As you move down the group, the atomic radius increases due to the addition of energy levels (shells). While you might think that the increased atomic mass would correspond to an increase in density, the increase in atomic radius generally outpaces the increase in atomic mass, leading to a decrease in density.

The densities (at room temperature) of the mentioned alkali metals are :

• Sodium (Na) : 0.97 g/cm³
• Potassium (K) : 0.86 g/cm³
• Rubidium (Rb) : 1.53 g/cm³
• Cesium (Cs) : 1.93 g/cm³

So, the order of densities is : K < Na < Rb < Cs

The alkali metals are the elements in Group 1 of the periodic table. They are lithium (Li), sodium (Na), potassium (K), rubidium (Rb), and cesium (Cs).

The melting point of these metals decreases down the group as the size of the atoms increases. This is because the metallic bonding weakens as the atomic size increases, resulting in a lower melting point.

Therefore, the alkali metal with the lowest melting point is cesium (Cs).

So, the correct option is Option D, Cs.

Let's consider each item in List-I and match it with an appropriate item from List-II:

A. K (Potassium): Potassium ions are crucial for the operation of the sodium-potassium pump in nerve cells. So, A matches with III.

B. KCl (Potassium Chloride): Potassium chloride is widely used in agriculture as a fertilizer. So, B matches with II.

C. KOH (Potassium Hydroxide): Potassium hydroxide is a strong base, but it is not typically used in a sodium-potassium pump, as a fertilizer, for thermonuclear reactions, or as a CO₂ absorbent. However, as per the given options and its closest possible use, KOH is often used to absorb CO₂ in certain applications like rebreathing gas masks. So, C matches with IV.

D. Li (Lithium): Lithium is used in thermonuclear reactions, specifically in fusion reactions. So, D matches with I.

Therefore, the correct option is Option B: A-III, B-II, C-IV, D-I.

As you move from top to bottom in a group (vertical column) on the periodic table, the metallic character increases. This is because elements at the bottom of a group have more electron shells than those at the top. These extra shells shield the outermost electrons from the pull of the nucleus, which makes it easier for these electrons to be lost, a characteristic of metals.

On the other hand, as you move from left to right across a period (horizontal row) on the periodic table, the metallic character decreases. This is because the nuclear charge increases as you move across a period, which pulls the electrons closer to the nucleus and makes them less likely to be lost.

Given this, for the elements K (Potassium), Be (Beryllium), and Ca (Calcium), the order of metallic character should be K > Ca > Be. Potassium (K) is furthest down (group 1, period 4), calcium (Ca) is above potassium (group 2, period 4), and beryllium (Be) is above calcium (group 2, period 2). So, the correct order should be Option C: K > Ca > Be.

• Washing soda is a hydrated form of sodium carbonate, with the chemical formula Na₂CO₃·10H₂O. This means that each molecule of washing soda contains 10 water molecules.


• Soda ash is the anhydrous form of sodium carbonate, with the chemical formula Na₂CO₃. This means that it does not contain any water molecules.

Soda ash is typically produced by heating washing soda to drive off the water molecules. Washing soda can be produced by dissolving soda ash in water and then recrystallizing it.

The correct answer is Option D: 10 and 0.

Lime (Calcium oxide, CaO) reacts exothermally with water to form slaked lime (Calcium hydroxide, Ca(OH)$_2$):

CaO (s) + H$_2$O (l) $\rightarrow$ Ca(OH)$_2$ (s)

This slaked lime has low solubility in water and is used for the test of carbon dioxide (CO$_2$). In this test, the carbon dioxide reacts with the calcium hydroxide to form calcium carbonate (CaCO$_3$), an insoluble white precipitate:

Ca(OH)$_2$ (aq) + CO$_2$ (g) $\rightarrow$ CaCO$_3$ (s) + H$_2$O (l)

Further reaction of calcium carbonate with carbon dioxide forms calcium bicarbonate [Ca(HCO$_3$)$_2$], a soluble compound:

CaCO$_3$ (s) + CO$_2$ (g) + H$_2$O (l) $\rightarrow$ Ca(HCO$_3$)$_2$ (aq)

Thus, 'A' in this case is slaked lime. So, the correct option is:

Option D : Slaked lime

Statement I : "Lithium and Magnesium do not form superoxide"

This statement is correct. Lithium and magnesium are smaller in size compared to other elements in their respective groups. Although lithium can form superoxide under certain conditions, in general, the smaller size of these two elements makes it harder for them to form stable superoxides.

Statement II : "The ionic radius of Li⁺ is larger than the ionic radius of Mg²⁺"

This statement is still correct. Despite both lithium and magnesium being small in size, when they lose electrons to form ions, the Mg²⁺ ion is smaller than the Li⁺ ion due to the loss of two electrons (and thus an energy level) in magnesium compared to the loss of one electron in lithium.

So, based on the information, the answer is :

Option C : Both Statement I and Statement II are correct.

Cement is manufactured by heating limestone (calcium carbonate) with small amounts of other materials, such as clay, to a high temperature in a kiln in a process known as calcination, which produces clinker. The clinker is then cooled and ground into a fine powder that we know as cement.

However, the clinker is highly reactive, and when mixed with water, it would harden immediately. This would not be practical in a construction setting because it doesn't allow enough time for the cement mix to be poured, shaped, or smoothed out.

Gypsum ($\mathrm{CaSO}_4 \cdot 2 \mathrm{H}_2 \mathrm{O}$) is added to this clinker before it's ground into cement. The role of gypsum is to extend the time it takes for the cement to set, giving the workers enough time to work with the cement before it hardens.

The gypsum reacts with the clinker during the final grinding process, controlling the rate at which the cement will set. If no gypsum was added, the cement would set as soon as water is added, making it almost impossible to use in construction.

The electrolysis of brine, which is a solution of sodium chloride (NaCl) in water, involves the migration of ions towards electrodes. When an electric current is passed through the solution, positive sodium ions ($Na^+$) move towards the cathode (negative electrode), and negative chloride ions ($Cl^-$) move towards the anode (positive electrode).

At the same time, water ($H_2O$), which also dissociates into $H^+$ and $OH^-$ ions, also contribute to the reactions at the electrodes.

At the anode, chloride ions are oxidized to chlorine gas:

$2Cl^- \rightarrow Cl_2(g) + 2e^-$

At the cathode, water is reduced to hydrogen gas and hydroxide ions, instead of sodium ions being reduced to sodium metal. This is because water is more easily reduced than sodium ions:

$2H_2O + 2e^- \rightarrow H_2(g) + 2OH^-$

The sodium ions remain in the solution and combine with the hydroxide ions to form sodium hydroxide:

$Na^+ + OH^- \rightarrow NaOH$

So the products of the electrolysis of brine are chlorine gas, hydrogen gas, and sodium hydroxide. Hydrochloric acid (HCl) is not produced during this process because the hydrogen and chloride ions do not combine in the electrolyte solution to form HCl. They are separately reduced and oxidized at the cathode and anode respectively.