s-Block Elements

$\mathrm{Ca}(\mathrm{OH})_2$ 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} $

(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.

(A) Both are acidic in nature.
B(OH)₃ is orthoboric acid. It can accept a pair of electrons from Lewis bases and hence, act as Lewis acid.

H₃PO₃ is (phosphorus acid) diprotic acid as it readily ionises two protons.

(B) B(OH)₃ is acidic and Al(OH)₃ is amphoteric in nature.

(C) Both NaOH and Ca(OH)₂ are basic hydroxides.

(D) Both Be(OH)₂ and Al(OH)₃ are amphoteric hydroxides.

Hence, only B(OH)₃ and Al(OH)₃ have different nature.

The element (M) is potassium (K). It reacts with O₂ to form KO₂, which is paramagnetic in nature. All other elements form oxides or peroxides which are diamagnetic in nature.

Alkali metals have very low value of ionisation energy as compared to other metals. So, alkali metals easily get excited and impart colour to flame.

Hence, Rb is most excited and having high value of wavelength in all alkali metals.

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.

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

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.