This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
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You're on a roll — certain basic radicals, particularly those of transition metals, produce two different colored beads with borax because their metal ions can exist in different oxidation states, and these states are favored under different flame conditions (oxidizing vs. reducing). Step 1: Formation of the borax bead. When borax (Na_2B_4O_7 · 10H_2O) is heated, it first loses its water of crystallization and swells. Upon further heating, it decomposes to form a transparent glassy bead consisting of sodium metaborate and boric anhydride. Na_2B_4O_7 · 10H_2O(s) heat Na_2B_4O_7(s) + 10H_2O(g) Na_2B_4O_7(s) heat 2NaBO_2(l) + B_2O_3(l) Step 2: Reaction with metal oxide. When a metal salt is introduced into the bead and heated, the metal oxide (formed from the decomposition of the salt) reacts with the boric anhydride (B_2O_3) to form a colored metal metaborate. For example, with copper(II) oxide: CuO(s) + B_2O_3(l) heat Cu(BO_2)_2(l) Step 3: Color in an oxidizing flame. In the outer, oxidizing flame, the metal ion is typically in its higher oxidation state. For copper, this is Cu^2+, which forms copper(II) metaborate. This compound gives a characteristic blue color when hot and cold. Cu(BO_2)_2(l) (blue bead in oxidizing flame) Step 4: Color in a reducing flame. In the inner, reducing flame (rich in carbon monoxide), the higher oxidation state metal ion is reduced to a lower oxidation state. For copper, Cu^2+ is reduced to Cu^+ (copper(I)) or even to metallic copper. Copper(I) metaborate (CuBO_2) is typically green, and elemental copper appears opaque red. 2Cu(BO_2)_2(l) + C(s) reducing flame 2CuBO_2(l) + B_2O_3(l) + CO(g) CuBO_2(l) (green bead in reducing flame) This change in oxidation state, driven by the flame conditions, results in two different colored beads for the same basic radical. Drop the next question.