s-Block Elements

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).

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.

(A) $\mathrm{BeSO}_4$ is soluble in water due to higher hydration energy of $\mathrm{Be}^{2+}$. It overcomes the lattice energy factors making $\mathrm{BeSO}_4$ soluble in water. The statement is correct.

(B) BeO is amphoteric oxide as it can react with both acid and base, thus acting as basic and acidic respectively. The given statement is thus correct.

(C) CO can be obtained by heating $\mathrm{CaCO}_3$ at $1070-1270 \mathrm{~K}$ is a incorrect statement.

$\mathrm{CaCO}_3 \xrightarrow{\Delta} \mathrm{CaO}+\mathrm{CO}_2 \uparrow$

$\therefore$ The correct statements are only A and B.

Group II elements are known to form both oxides and nitrides upon combustion in air. Therefore, the pair of elements from Group II that form both compounds are Magnesium (Mg) and Calcium (Ca).

$ \begin{aligned} & 3 \mathrm{Mg} + \mathrm{N}_2 \xrightarrow{\Delta} \mathrm{Mg}_3 \mathrm{N}_2 \\ & 3 \mathrm{Mg} + \mathrm{O}_2 \xrightarrow{\Delta} 2 \mathrm{MgO} \\ & 3 \mathrm{Ca} + \mathrm{N}_2 \xrightarrow{\Delta} \mathrm{Ca}_3 \mathrm{N}_2 \\ & 2 \mathrm{Ca} + \mathrm{O}_2 \xrightarrow{\Delta} 2 \mathrm{CaO} \end{aligned} $

Magnesium and Calcium react with nitrogen and oxygen to form magnesium nitride (Mg$_3$N$_2$), magnesium oxide (MgO), calcium nitride (Ca$_3$N$_2$), and calcium oxide (CaO), respectively.

Out of sulphates of group II, only $\mathrm{BeSO}_4$ and $\mathrm{MgSO}_4$ are soluble in water. This is because of the higher hydration enthalpies of both the ions which allow them to overcome the lattice enthalpy and making them soluble in water. Hence, both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

Sodium nitrate decomposes on heating to form sodium nitrite and oxygen gas. The complete balanced equation is as follows

$\mathrm{NaNO}_3 \longrightarrow \mathrm{NaNO}_2+\frac{1}{2} \mathrm{O}_2$

Thus, A and B in the given reaction are NaNO$_2$ and O$_2$ respectively.

Enamel is the hardest substance in human body and it covers outer surface of teeth, which is $\left[3 \mathrm{Ca}_3\left(\mathrm{PO}_4\right)_2 \cdot \mathrm{Ca}(\mathrm{OH})_2\right]$.

Li$_2$SO$_4$ is the only alkali metal salt that does not form double salt due to small size.

Sodium carbonate is stable and does not dissociate upon normal heating.

Two mole of sodium bicarbonate produce one mole of $\mathrm{CO}_2$, one mole of $\mathrm{H}_2 \mathrm{O}$ and one mole of sodium carbonate.

$\begin{aligned} & \mathrm{NaCl}+\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2+\mathrm{NH}_3 \longrightarrow \mathrm{NaHCO}_3+\mathrm{NH}_4 \mathrm{Cl} \\ & \underset{\text { Sodium bicarbonate }}{\mathrm{NaHCO}_3} \stackrel{\Delta}{\longrightarrow} \underset{\begin{array}{c}\text { Sodium } \\ \text { carbonate } \end{array}}{\mathrm{Na}_2 \mathrm{CO}_3}+\mathrm{CO}_2+\mathrm{H}_2 \mathrm{O} \\ \end{aligned}$

Hence, the ratio of $\mathrm{CO}_2$ and $\mathrm{H}_2 \mathrm{O}$ is $1: 1$.

Oxides such as $\mathrm{BaO}_2, \mathrm{Na}_2 \mathrm{O}_7$ etc. which contain peroxide linkage i.e. $-\mathrm{O}-\mathrm{O}-$ or $_2^{2-}$, on treatment with dilute $\mathrm{H}_2 \mathrm{SO}_4$ give $\mathrm{H}_2 \mathrm{O}_2$. $\mathrm{Rb}, \mathrm{Mn}, \mathrm{Al}$, do not give $\mathrm{H}_2 \mathrm{O}_2$ on treatment with dilute $\mathrm{H}_2 \mathrm{SO}_4$.

K, Rb and Cs form superoxides when they are burned in the air. As we move down the group, the size of an atom from K to Cs increases. So, lattice energy decreases and hence, the stability of superoxide also decreases.

Hence, A is true but R is false.

When borax is dissolved in water, it forms sodium hydroxide and boric acid $\left(\mathrm{H}_3 \mathrm{BO}_3\right)$ which forms a strong base. Hence, it is alkaline in nature.

$\mathrm{\mathop {N{a_2}{B_4}{O_7}}\limits_{Borax} + 7{H_2}O\buildrel {} \over \longrightarrow \mathop {2NaOH}\limits_{Sodium\,hydroxide} + \mathop {4{H_3}B{O_3}}\limits_{Boric\,acid}}$