Borax bead, charcoal cavity tests: Exercises — Chemistry deep dive

Level 1 — Recognition

L1.1 A borax bead is blue/green in the oxidising flame and turns red/colourless in the reducing flame. Which cation is present?

Recall Solution

The pair "blue/green (OF) → red/colourless (RF)" is the fingerprint of copper, $C u^{2 +}$ .

OF keeps copper as $C u^{2 +}$ → blue copper metaborate $Cu(BO_{2})_{2}$ .
RF drops it to $C u^{+}$ (or $C u_{2} O$ ) → red/colourless. Answer: Copper.

L1.2 In the charcoal cavity test, a metal gives no bead but an incrustation that is yellow when hot and white when cold. Name it.

Recall Solution

"Yellow-hot, white-cold" is the unique signature of zinc ( $ZnO$ ). Zinc's oxide changes colour reversibly with temperature; zinc metal itself boils off (bp 907 °C) so no metallic bead forms. Answer: Zinc.

L1.3 Which single substance in the molten borax bead is the transparent glass that actually dissolves the metal oxide?

Recall Solution

Boric anhydride, == $B_{2} O_{3}$ ==, produced by $Na_{2} B_{4} O_{7} \to 2 NaBO_{2} + B_{2} O_{3}$ . The $NaBO_{2}$ is just the leftover; $B_{2} O_{3}$ is the reactive glassy solvent. Answer: $B_{2} O_{3}$ .

Level 2 — Application

L2.1 A bead is blue in the oxidising flame and stays blue in the reducing flame. Copper or cobalt? Explain using oxidation states.

Recall Solution

Copper: $C u^{2 +}$ (blue) is easily reduced to $C u^{+} / C u^{0}$ , so RF changes it to red/colourless.
Cobalt: $C o^{2 +}$ is hard to reduce in the bead, so it stays blue in RF. The bead stayed blue in RF → the metal resisted reduction → cobalt. Answer: Cobalt.

L2.2 Explain why the reducing flame must be used for the charcoal cavity test, whereas the borax bead is read in both flames.

Recall Solution

Charcoal cavity aims to produce free metal (a bead). Freeing metal from its oxide $MO \to M$ requires removing oxygen = reduction. Only the fuel-rich reducing flame supplies the electrons/CO to do this; an oxidising flame would re-oxidise the metal.
Borax bead aims to read a colour, and the colour depends on oxidation state. Comparing OF (higher state) with RF (lower state) is exactly what disambiguates metals, so you deliberately read both. Answer: cavity test needs reduction to get metal; bead uses both flames to compare oxidation states.

L2.3 Write the balanced reaction of boric anhydride with cobalt(II) oxide that gives the blue bead.

Recall Solution

Cobalt is $+ 2$ , so it needs two metaborate units: $B_{2} O_{3} + CoO ⟶ Co(BO_{2})_{2}$ Check atoms: left B $= 2$ , O $= 3 + 1 = 4$ , Co $= 1$ . Right B $= 2$ , O $= 4$ , Co $= 1$ . Balanced. Answer: $B_{2} O_{3} + CoO \to Co(BO_{2})_{2}$ .

Level 3 — Analysis

L3.1 A charcoal cavity gives a soft grey malleable bead (marks paper) and a crust that is yellow both hot and cold. Lead or zinc? Justify from two independent clues.

Recall Solution

Clue 1 — the bead: lead melts at 327 °C and forms a soft malleable bead; zinc gives no bead (boils away). A bead already rules out zinc. Clue 2 — the crust: $PbO$ is yellow hot and cold; $ZnO$ is yellow-hot but white-cold. The crust stayed yellow cold → lead. Both clues agree → lead. Answer: Lead.

L3.2 A borax bead is green in the oxidising flame and also green in the reducing flame. Two metals in the table share a colour in one flame — how does "green in both" narrow it to one?

Recall Solution

Look at the colour table:

$C r^{3 +}$ : green in OF and green in RF (chromium is stable in the $+ 3$ state, so the flame barely changes it).
$F e^{3 +}$ : yellow-brown in OF, but green/colourless in RF (iron drops to $F e^{2 +}$ ). "Green in both flames" = colour independent of oxidation state = chromium. Iron would have shown a colour change OF→RF. Answer: Chromium.

L3.3 After a charcoal cavity test you get a white infusible residue that glows brightly and gives no metal bead. Explain why no bead formed, and give the follow-up test that identifies the metal.

Recall Solution

No bead because the oxide (of $M g, A l, C a, B a$ ) has a very high melting point and is infusible — charcoal cannot reduce these reactive metals, and even the oxide won't melt into a bead; it just glows white-hot.
Follow-up = cobalt nitrate test: moisten the residue with $Co(NO_{3})_{2}$ solution and reheat.
- Blue ("Thénard's blue", $CoAl_{2} O_{4}$ ) → aluminium
- Pink → magnesium
- Green (rinmann's green) → zinc Answer: high-mp infusible oxide → no bead; confirm with cobalt nitrate (see Cobalt Nitrate Test).

Level 4 — Synthesis

L4.1 An unknown salt is tested. Borax bead: violet in OF, colourless in RF. Charcoal cavity: no metal bead, no coloured crust. Flame test (Flame Test and Atomic Emission): no distinctive colour. Identify the cation and explain every observation.

Recall Solution

Bead violet (OF) → $M n^{3 +}$ metaborate; colourless (RF) → reduced to $M n^{2 +}$ (colourless in bead). This pair = manganese.
Charcoal cavity: manganese oxide is not reduced to a low-melting metal bead and gives no bright crust — consistent, not contradictory.
Flame test: $M n^{2 +}$ has no strong characteristic flame emission → "no colour" is expected. All three observations point the same way. Answer: Manganese ( $M n^{2 +}$ salt).

L4.2 You must distinguish three unknown salts labelled X, Y, Z using only the borax bead. Data:

X: yellow-brown (OF) → green (RF)

Y: brown (OF) → grey (RF)

Z: blue (OF) → blue (RF)

Identify each and give the diagnostic reasoning for each colour change (or lack of it).

Recall Solution

X: yellow-brown OF = $F e^{3 +}$ ; green/colourless RF = reduced to $F e^{2 +}$ . The colour change confirms iron → ==iron ( $F e^{3 +}$ )==.
Y: brown OF, then grey RF — grey means the metaborate was reduced all the way to metal grains; this metal is nickel → ==nickel ( $N i^{2 +}$ )==.
Z: blue and unchanged in RF = a metal resistant to reduction = ==cobalt ( $C o^{2 +}$ )==. Key reasoning: the behaviour on reduction (changes / goes to metal / stays put) separates them even when OF colours look similar-ish. Answer: X = Fe, Y = Ni, Z = Co.

Level 5 — Mastery

L5.1 In the charcoal cavity test on lead sulphate, $PbSO_{4}$ is first converted to $PbCO_{3}$ by soda ash, which decomposes to $PbO$ , then reduced by carbon to $Pb$ . Starting from 3.03 g of $PbSO_{4}$ , what mass of lead metal (bead) can form at 100% conversion?

Molar masses: $PbSO_{4} = 303 g/mol$ , $Pb = 207 g/mol$ .

Recall Solution

One $Pb$ atom is conserved through every step: $PbSO_{4} \to PbCO_{3} \to PbO \to Pb$ , so mole ratio $PbSO_{4} : Pb = 1 : 1$ . $n (PbSO_{4}) = \frac{3.03}{303} = 0.0100 mol \Rightarrow n (Pb) = 0.0100 mol$ $m (Pb) = 0.0100 \times 207 = 2.07 g$ Answer: $2.07 g$ of lead.

L5.2 The reduction step is $PbO + C \to Pb + CO ↑$ . For the 0.0100 mol of $PbO$ from L5.1, what minimum mass of carbon is consumed, and what volume of CO (at STP, $22.4 L/mol$ ) is released?

Molar mass $C = 12 g/mol$ .

Recall Solution

Ratio $PbO : C : CO = 1 : 1 : 1$ .

Carbon: $0.0100 mol \times 12 = 0.120 g$ .
CO: $0.0100 mol \times 22.4 = 0.224 L = 224 mL$ . Answer: $0.120 g$ of carbon; $0.224 L$ ( $224 mL$ ) of CO.

L5.3 A borax bead is made and copper(II) oxide dissolved in it: $B_{2} O_{3} + CuO \to Cu(BO_{2})_{2}$ . If a bead contains 0.0800 g of $B_{2} O_{3}$ ( $= 69.6 g/mol$ ) and copper is in stoichiometric excess, what mass of $Cu(BO_{2})_{2}$ ( $= 149.4 g/mol$ ) forms?

Recall Solution

Ratio $B_{2} O_{3} : Cu(BO_{2})_{2} = 1 : 1$ (both carry the two boron atoms). $n (B_{2} O_{3}) = \frac{0.0800}{69.6} = 1.149 \times 1 0^{- 3} mol$ $m (Cu(BO_{2})_{2}) = 1.149 \times 1 0^{- 3} \times 149.4 = 0.1717 g$ Answer: $\approx 0.172 g$ of copper metaborate.

Recall One-line self-check before you leave

Bead ⇒ read both flames (oxidation-state colour). Cavity ⇒ reducing flame only (make the metal). White glowing residue ⇒ cobalt nitrate follow-up. Stoichiometry ⇒ conserve the metal atom in moles.