Beryllium is placed above magnesium in the group II. Beryllium dust, therefore, when added to soluti — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Periodic Table Position vs. Electrochemical Series Although beryllium ($\ce{Be}$) is above magnesium ($\ce{Mg}$) in Group 2, the actual reactivity in aqueous solution is determined by standard reduction potentials: $$E^\circ(\ce{Be^{2+}/Be}) = -1.85\text{ V}$$ $$E^\circ(\ce{Mg^{2+}/Mg}) = -2.37\text{ V}$$ Since $E^\circ(\ce{Mg^{2+}/Mg}) < E^\circ(\ce{Be^{2+}/Be})$, magnesium is more electropositive and reactive. Step 2 - Assessing the Feasibility of the Displacement Reaction The hypothetical reaction: $$\ce{Be(s) + Mg^{2+}(aq) -> Be^{2+}(aq) + Mg(s)}$$ $$E^\circ_{\text{cell}} = E^\circ(\ce{Mg^{2+}/Mg}) - E^\circ(\ce{Be^{2+}/Be}) = -2.37\text{ V} - (-1.85\text{ V}) = -0.52\text{ V}$$ Since $E^\circ_{\text{cell}} < 0$, the reaction is thermodynamically non-spontaneous ($\Delta G^\circ > 0$). Beryllium cannot displace magnesium. Step 3 - Explanation of Each Option * **Option (A) is correct:** No chemical reaction occurs — beryllium dust has no effect on $\ce{MgCl2}$ solution. * **Option (B) is incorrect:** Mg metal will not precipitate since Be cannot displace $\ce{Mg^{2+}}$. * **Option (C) is incorrect:** No reaction means no $\ce{MgO}$ forms. * **Option (D) is incorrect:** Be does not dissolve because this oxidation is non-spontaneous. $$\text{Correct Answer: } \boxed{\text{A}}$$