Borax bead, charcoal cavity tests: Worked examples — Chemistry deep dive

This is the practice companion to the parent topic. Here we hunt down every kind of unknown these two tests can hand you and solve it out loud. If you have not met the flame types yet, read Oxidising vs Reducing Flame first; for the underlying colour chemistry see Transition Metal d-d Colour.

The scenario matrix

Think of an unknown salt as a point that can land in one of these "cells". A good analyst must have a rehearsed move for each cell — never a scenario you have not already seen.

Cell	Case class	What makes it tricky	Example that hits it
A	Clean single-flame borax colour	one flame is enough	Ex 1 (Cr)
B	Two ions that COLLIDE in OF	same OF colour, split by RF	Ex 2 (Co vs Cu)
C	Colour flips OF → RF	oxidation-state change	Ex 3 (Mn)
D	Degenerate / colourless bead	"no colour" is itself data	Ex 4 (Na / Ca)
E	Charcoal: low-mp metal + oxide	bead and incrustation	Ex 5 (Pb)
F	Charcoal: no bead, only crust	volatile / high-mp oxide	Ex 6 (Zn)
G	White glowing infusible residue	needs a follow-up test	Ex 7 (Al/Mg)
H	Real-world word problem	translate a scene into ions	Ex 8 (foundry slag)
I	Exam twist / contradictory clues	reject the trap reading	Ex 9 (Fe vs Cr)

(Recall: OF = oxidising flame, RF = reducing flame, defined above.)

Every cell A–I is worked below. Read the Forecast line, cover the rest, and guess before you scroll.

Worked example

Ex 1 · Cell A — a clean single-flame answer A bead is green in the oxidising flame (OF) and still green in the reducing flame (RF). Name the metal.

Forecast: which single metal ion gives the same green in both flames?

Read the OF colour. Green in OF matches $C r^{3 +}$ (chromium's metaborate is green). Why this step? OF fixes the metal's higher oxidation state, so we start from the colour that survives oxygen; that green already narrows us to the "green-in-OF" ions.
Read the RF colour. It is unchanged — green again. Why this step? If the colour had changed we'd suspect an ion whose oxidation state flips; unchanged colour means the ion is stable in both flames, which keeps $C r^{3 +}$ in play.
Match "green in both" and rule out the interferents. == $C r^{3 +}$ == keeps green across OF and RF (chromium is hard to reduce further in the bead). Why this step? Matching the pair of readings is what pins a metal, but you must actively exclude the two "green" impostors:
- $F e^{3 +}$ is yellow-brown in OF (green only appears in its RF), so a green OF rules iron out immediately.
- $N i^{2 +}$ can appear green in both flames too — a genuine interferent. To split Ni from Cr, follow up on the same salt: nickel gives a grey Ni-metal bead in the charcoal cavity (Ex 6-style) and a characteristic dimethylglyoxime confirmation, whereas chromium gives no metal bead. So a green-in-both bead means "Cr or Ni — confirm with the cavity test."

Conclusion: chromium (with nickel excluded by the charcoal cavity). ✔

Verify: cross-check against the parent table — $C r^{3 +}$ = "Green / Green"; $F e^{3 +}$ = "Yellow-brown / Green", so a green OF is impossible for iron; $N i^{2 +}$ = "Brown (OF) / grey Ni bead", separable on charcoal. Consistent.

Worked example

Ex 2 · Cell B — two ions collide in OF, split them in RF Two students each report a blue bead in the oxidising flame (OF). One insists cobalt, the other copper. Design the deciding observation.

Forecast: which single reading in the reducing flame (RF) separates them?

Accept the collision. Both $C o^{2 +}$ and $C u^{2 +}$ read blue in OF, so OF alone is ambiguous. Why this step? Recognising the collision stops you from guessing on insufficient data.
Switch to RF. Why this step? RF adds electrons; only an ion whose lower oxidation state has a different colour will change. This is a discriminating test — see Oxidising vs Reducing Flame.
Predict each ion's RF colour. Why this step? We must forecast what each candidate would show in RF, so that the single upcoming observation can only match one of them.
- $C u^{2 +} RF C u^{+} / C u^{0}$ → bead turns red / colourless.
- $C o^{2 +} RF C o^{2 +}$ (not reduced in bead) → bead stays blue.
Decide. Blue in RF ⇒ cobalt; red/colourless in RF ⇒ copper. Why this step? Each candidate's predicted RF colour (from Step 3) is distinct, so the one actual RF observation lands on exactly one metal — this is the moment the forecast collapses to a single answer.

Conclusion: the RF reading decides it — no ambiguity remains. ✔

Verify: parent table: $C o^{2 +}$ = Blue/Blue, $C u^{2 +}$ = Blue-green/Red-colourless. The two RF entries differ, so the discriminator is valid.

Worked example

Ex 3 · Cell C — the colour FLIPS between flames A bead is violet (amethyst) in OF but colourless in RF.

Forecast: a flip this dramatic means an oxidation-state change — which metal?

First, an important distinction. The parent colour table is indexed by the metal's starting cation in the salt — here $M n^{2 +}$ . But inside the molten bead the flame can change that oxidation state, so what actually colours the glass is the bead speciation, which need not equal the starting ion. Watch both.

Interpret OF violet. The starting salt supplies $M n^{2 +}$ , but the OF oxidises it up inside the glass to $M n^{3 +}$ , whose metaborate ( $M n (B O_{2})_{3}$ -type) is violet. Why this step? OF pushes Mn up; the violet you see is the colour of the oxidised bead species, not of the starting $M n^{2 +}$ .
Interpret RF colourless. RF reduces the bead species back down to $M n^{2 +}$ , and $M n^{2 +}$ in the bead is essentially colourless (its d–d transitions are spin-forbidden, hence very faint — Transition Metal d-d Colour). Why this step? The loss of colour is not "nothing happened"; it is the signature of the down-shift to the starting oxidation state.
Match the flip. Violet↔colourless across the two flames is unique to manganese.

Conclusion: manganese present. ✔

Verify: parent table row is indexed by the starting cation $M n^{2 +}$ and reads OF "Violet/amethyst" (the $M n^{3 +}$ bead species) / RF "Colourless" (the $M n^{2 +}$ bead species). The flip in bead speciation is exactly what the row encodes.

Worked example

Ex 4 · Cell D — the degenerate case: a colourless bead An unknown gives a clear, colourless bead in both flames, yet a separate flame test on the same salt glows golden-yellow. What is going on?

Forecast: "no bead colour" — does that mean the test failed?

Read the null result correctly. A colourless bead means no coloured transition-metal cation is present. Why this step? Absence of colour is data: it rules out Cu, Cr, Mn, Co, Ni, Fe.
Explain why it's colourless. Ions like $N a^{+}$ , $C a^{2 +}$ , $M g^{2 +}$ have no partly-filled d-shell, so no d–d absorption, so no visible colour in glass. Why this step? This ties the observation to structure, not luck.
Use the flame test to name it. Golden-yellow flame ⇒ sodium. Why this step? The borax bead is silent on colourless cations, so we defer identification to atomic emission (Cation Analysis - Group Salts).

Conclusion: sodium — a colourless-bead cation identified by flame colour instead. ✔

Verify: Na has electron config $[Ne] 3 s^{1}$ → no d electrons → no d–d colour, consistent with the clear bead; golden flame at ~589 nm confirms Na.

Worked example

Ex 5 · Cell E — charcoal cavity: metal bead + coloured incrustation On charcoal in the reducing flame (RF), a salt gives a soft grey bead that marks paper plus a yellow crust that stays yellow when cold.

Forecast: soft + yellow-cold crust — which metal?

Read the bead. Soft, malleable, marks paper ⇒ a low-melting-point metal. Why this step? Only metals like Pb (mp 327 °C) melt in the cavity and pool into a soft bead.
Read the crust temperature behaviour. Yellow hot and cold ⇒ $P b O$ . Why this step? This is the exact test that separates Pb from Zn — Zn's crust is yellow only when hot (see Ex 6).
Write the reduction. With $N a_{2} C O_{3}$ as flux (Fluxes in Metallurgy): $PbSO_{4} + Na_{2} CO_{3} \to PbCO_{3} + Na_{2} SO_{4}$ $PbCO_{3} Δ PbO + CO_{2} ↑, PbO + C \to Pb + CO ↑$ Why this step? Charcoal (C, CO) strips oxygen → free grey Pb bead; residual PbO forms the yellow crust.

Conclusion: lead present. ✔

In the figure below the single red-highlighted object is the metal bead — the accent colour marks the key object, not its real colour (the bead is physically grey Pb; the crust is physically yellow PbO). Read the labels, not the shading: the red disc at the centre of the cavity is the molten Pb pooling where the RF (arrow from above) reduced it, and the dashed ring around the rim is the yellow PbO incrustation carbon could not fully reduce. Note the RF arrow points down into the cavity — that geometry is why the reduction happens right where the metal collects.

Verify (mass-balance the reduction): in $PbO + C \to Pb + CO$ : Pb 1=1, O 1=1, C 1=1 — balanced. In the carbonate step, C and O balance too (checked in VERIFY).

Worked example

Ex 6 · Cell F — no bead, only a temperature-changing crust A charcoal cavity test gives no metal bead at all, but an incrustation that is yellow while hot and turns white on cooling.

Forecast: the crust changing colour on cooling is a famous fingerprint — whose?

Note the missing bead. No bead ⇒ either the metal is volatile (boils away) or its oxide is infusible. Why this step? It narrows us away from Pb/Cu-type "pooling" metals.
Read the reversible colour. Yellow-hot / white-cold is the classic zinc oxide (ZnO) thermochromic behaviour. Why this step? ZnO develops oxygen vacancies when hot (yellow) that refill on cooling (white) — nothing else in the table does this.
Conclude via the rhyme. "Zinc: yellow-hot, white-cold." Why this step? The rhyme is not decoration — it encodes the underlying chemistry of Step 2: the hot/cold colour swap is caused by reversible oxygen-vacancy formation in ZnO, so reciting the rhyme is really recalling that thermochromic mechanism and mapping it uniquely onto zinc.

Conclusion: zinc present. ✔

The figure contrasts the same cavity at two temperatures. In the left (HOT) panel the crust is red-highlighted — the accent marks it as the key object (physically it is the yellow ZnO glowing hot), and there is deliberately no metal bead below it (Zn has boiled off as vapour). In the right (COLD) panel the identical crust is white-filled with a black outline: the same ZnO after cooling. Compare the two crusts side by side — the colour flip on cooling, with no bead in either panel, is the whole diagnosis.

Verify: contrast with Ex 5 — Pb crust is yellow at both temperatures; Zn crust flips to white on cooling. The distinguishing variable is cold colour, exactly as tabulated.

Worked example

Ex 7 · Cell G — a white glowing infusible residue (needs a follow-up) A salt gives no bead and no coloured crust, only a white residue that glows brilliantly in the flame. On moistening with cobalt nitrate solution and reheating, it turns blue.

Forecast: white + glowing + blue-after-cobalt — which metal?

Read "white + glows". A white infusible residue that incandesces ⇒ oxide of a high-melting metal: $M g, A l, C a, B a$ . Why this step? These oxides melt above ~2000 °C, so they can't form a bead — they just glow.
Apply the follow-up. The Cobalt Nitrate Test resolves the ambiguity: soak in $C o (N O_{3})_{2}$ , reheat, and read the coloured mixed oxide. Why this step? The dry test alone can't separate these candidates; cobalt nitrate assigns a distinct colour.
Read the colour. Why this step? The observed colour is the deciding datum that names the metal:
- Blue (Thénard's blue, $C o A l_{2} O_{4}$ ) ⇒ aluminium.
- Pale pink / flesh ⇒ magnesium.
- Ca and Ba give no characteristic cobalt-nitrate colour (a greyish/black mass), so a distinct blue or pink actively excludes them — they would instead be confirmed by their flame colours (brick-red Ca, apple-green Ba).

Conclusion: aluminium present. ✔

Verify: blue ↔ Al is the standard cobalt-nitrate assignment; pink (Mg) was not seen, and Ca/Ba give no such colour, so Al is unambiguous.

Worked example

Ex 8 · Cell H — real-world word problem A metal-worker scrapes a grey soft crust off a soldering pool, dissolves a trace, mixes with $N a_{2} C O_{3}$ , and heats it in a charcoal cavity. He sees a malleable grey bead and a yellow crust (yellow when cold too). He also notes the original solder was a tin–lead alloy. Which component is his cavity test detecting?

Forecast: solder contains Sn and Pb — which one does this observation fingerprint?

Translate the scene into ions. Solder = Sn + Pb. The cavity test's bead + crust must match one of them. Why this step? Word problems must be reduced to the ions the test actually resolves.
Compare the two candidates' cavity signatures. Why this step? Only by contrasting what each metal would show can one observation single out the culprit.
- Pb: soft grey malleable bead + yellow crust (hot and cold).
- Sn: does not give a coloured crust (its oxide $S n O_{2}$ is white/pale, and Sn does not leave a strong yellow incrustation).
Match the observation. Malleable grey bead + yellow-cold crust ⇒ lead. Why this step? The yellow-cold crust is the decisive Pb marker (from Ex 5).

Conclusion: the cavity test is reporting the lead in the solder. ✔

Verify: the yellow-cold PbO crust matches Ex 5 exactly; tin's absence of a yellow incrustation makes Pb the only fit — self-consistent with the alloy composition.

Worked example

Ex 9 · Cell I — the exam twist (reject the trap reading) Exam claim: "The bead is green in RF, therefore the metal must be chromium." A trap answer says "obviously Cr." What extra observation must you demand before agreeing, and what if OF is yellow-brown?

Forecast: green-in-RF alone — is it sufficient?

Spot the ambiguity. Green in RF fits both $C r^{3 +}$ (green/green) and $F e^{3 +}$ (whose RF bead is green, though its OF is yellow-brown). Why this step? A single RF reading is not decisive; the parent table shows two green-in-RF ions.
Demand the OF colour. Why this step? OF splits them:
- $C r^{3 +}$ → green in OF.
- $F e^{3 +}$ → yellow-brown in OF.
Apply the given twist. OF is yellow-brown ⇒ this is iron, not chromium. The trap "obviously Cr" is wrong because it ignored the OF reading.

Conclusion: iron present — the green-RF-only argument is invalid. ✔

Verify: $F e^{3 +}$ row = "Yellow-brown / Green-colourless"; $C r^{3 +}$ row = "Green / Green". A yellow-brown OF is impossible for Cr, so Fe is forced. Trap rejected.

Recall Self-test: match cell to move

Green in both flames ::: Cr (Cell A) — but confirm Ni is absent with the charcoal cavity Both blue in OF ::: switch to RF; Co stays blue, Cu goes red/colourless (Cell B) Violet OF, colourless RF ::: Mn (Cell C) Colourless bead but yellow flame ::: Na, identified by flame not bead (Cell D) Soft bead + yellow-cold crust ::: Pb (Cell E) No bead, yellow-hot white-cold crust ::: Zn (Cell F) White glowing residue, blue after cobalt nitrate ::: Al (Cell G) Green RF alone ::: insufficient — check OF to split Fe from Cr (Cell I)