s-Block Elements

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.

The commercial name of calcium oxide (CaO) is quick lime.

The first ionization energy is the energy required to remove the most loosely held electron from an atom or a cation. It is measured in units of electron volts (eV) or kilojoules per mole (kJ/mol).

Among the given options, Scandium (Sc) has the highest first ionization energy because it has the smallest atomic radius and the highest nuclear charge. As we move from left to right across a period in the periodic table, the atomic radius decreases while the nuclear charge increases. This results in a stronger attraction between the positively charged nucleus and the negatively charged electrons in the outermost energy level, which requires more energy to remove the most loosely held electron.

The first ionization energy of the given options in increasing order is:

Na < K < Rb < Sc

Therefore, Scandium (Sc) has the highest first ionization energy among the given options.

The compound(s) formed upon the combustion of sodium metal in the excess air is sodium peroxide (Na₂O₂) and sodium oxide (Na₂O).

$ \begin{aligned} & 4 \mathrm{Na}+\mathrm{O}_2 \rightarrow 2 \mathrm{Na}_2 \mathrm{O} \\\\ & 2 \mathrm{Na}+\mathrm{O}_2 \rightarrow \mathrm{Na}_2 \mathrm{O}_2 \end{aligned} $

The product formed the reaction is used for chlorination of water bodies. The compound Na$_2$S$_2$O$_3$ gets oxidised by chlorine to form NaHSO$_3$. The balanced chemical reaction is,

Na$_2$S$_2$O$_3$ + 4Cl$_2$ + 5H$_2$O $\to$ 2NaHSO$_4$ + 8HCl

Statement-1 :

Alkali metals dissolve in liquid ammonia to give blue solutions.

✅ True

Explanation: When alkali metals (Li, Na, K, etc.) are dissolved in liquid ammonia, they form deep blue conducting solutions at low concentrations.

Reaction:

$ \mathrm{M (s) \xrightarrow{liq.\,NH_3} M^+_{(solv)} + e^-_{(solv)}} $

The blue color arises from solvated electrons ($ e^-_{(NH_3)} $), which absorb visible light.

Statement-2 :

Alkali metals in liquid ammonia give solvated species of the type $\left[\mathrm{M}\left(\mathrm{NH}_{3}\right)_n\right]^+ (\mathrm{M} = \text{alkali metals})$.

✅ True

During dissolution, the metal forms metal cations solvated by ammonia molecules, i.e., $\mathrm{M^+}$ surrounded by several $\mathrm{NH_3}$ molecules.

$ \mathrm{M^+ + e^- \rightarrow [M(NH_3)_n]^+ + e^-_{(solv)}} $

But:

The blue color (Statement-1) is not due to $[\mathrm{M(NH_3)_n}]^+$, but due to the solvated electrons ($e^-_{(NH_3)}$).

Therefore, although both statements are true, Statement-2 does not explain the blue color.

✅ Final Answer :

Option B — Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1.

The chemical reaction between $\mathrm{MgSO}_4$, $\mathrm{NH}_4 \mathrm{OH}$ and $\mathrm{Na}_2 \mathrm{HPO}_4$ is given below.

$\underset{\substack{\text { Magnesium } \\ \text { sulfate }}}{\mathrm{MgSO}_4}+\underset{\substack{\text { Ammonium } \\ \text { hydroxide }}}{\mathrm{NH}_4 \mathrm{OH}}+\underset{\substack{\text { Sodium hydrogen } \\ \text { phosphate }}}{\mathrm{Na}_2 \mathrm{HPO}_4} \rightarrow$\underset{\substack{\text { Magnesium ammonium } \\ \text { white precipitate }}}{\mathrm{Mg}\left(\mathrm{NH}_4\right) \mathrm{PO}_4 \downarrow}+\underset{\text { Sodium sulfate }}{\mathrm{Na}_2 \mathrm{SO}_4}+\mathrm{H}_2 \mathrm{O}$

The magnesium ammonium phosphate hexahydrate $\left[\mathrm{Mg}\left(\mathrm{NH}_4\right) \mathrm{PO}_4 \cdot 6 \mathrm{H}_2 \mathrm{O}\right]$ formed has white coloured precipitate.

The other products formed during the chemical reaction include sodium sulfate and water.

Ammonium chloride is used to prevent magnesium hydroxide and ammonia solutions from precipitating.