Borax bead, charcoal cavity tests: Question bank — Chemistry deep dive

This page tests the ideas built in the parent note: why the borax glass forms, why colour depends on which flame you use, why charcoal reduces, and the boundary cases where the tests almost fail.

Figure — Borax bead, charcoal cavity tests

Above: the two flame zones (RF inner/luminous, OF outer/hot) and, on the right, why colour appears — the metal ion sitting inside the transparent borax glass absorbs one wavelength via a d–d electron jump, so the bead shows the leftover colour (d–d colour).

True or false — justify

Borax bead colours arise from the same physics as flame test colours.

False. Flame colour (atomic emission) is light emitted by excited valence electrons; the bead colour is d–d absorption by the metal ion dissolved in the glass. Different mechanism entirely.

The borax bead test is a reliable general test for any cation.

False. Only ions with partly filled d-orbitals (transition metals) give coloured beads;

N a^{+}, K^{+}, C a^{2 +}, A l^{3 +}

etc. give colourless beads, so the test cannot identify them.

A bigger blob of salt gives a clearer, more diagnostic colour.

False. Excess salt makes the bead dark and opaque so no colour can be judged; the transparent glass is the diagnostic, so a tiny speck is correct.

The oxidising flame (OF) is the outer, hotter, oxygen-rich part of the flame.

True. The outer cone has excess air, burns hotter, and supplies oxygen, so it holds metals in their higher oxidation state.

B_{2} O_{3}

and

N a B O_{2}

are both produced when borax is strongly heated.

True. After losing water,

N a_{2} B_{4} O_{7} \to 2 N a B O_{2} + B_{2} O_{3}

; the

B_{2} O_{3}

is the transparent glass that dissolves metal oxides.

Chromium can be distinguished from copper by switching to the reducing flame (RF).

True.

C r^{3 +}

stays green in both flames, whereas

C u^{2 +}

(blue/green in OF) turns red/colourless in RF, so the flame-switch separates them.

In the charcoal cavity test you should use the oxidising flame for best results.

False. You need to remove oxygen to free the metal, which is reduction, so the fuel-rich reducing (luminous) flame is required; the OF would re-oxidise the metal.

A white residue on charcoal always means the test failed.

False. A white infusible residue that glows brightly is itself diagnostic of high-melting oxides of

M g, A l, C a, B a

; confirm with the cobalt nitrate test.

N a_{2} C O_{3}

is added in the cavity test only to make the salt melt more easily.

False. It is a flux and reducing aid — it swaps the salt into a carbonate that decomposes to oxide (see the reaction box above), which charcoal can then cleanly reduce to metal.

Spot the error

"Copper gives a red bead in the oxidising flame because

C u^{2 +}

is red."

Wrong flame:

C u^{2 +}

is blue/green in the oxidising flame (OF). The red appears in the reducing flame (RF) where it becomes

C u^{+} / C u_{2} O

"Zinc gives a grey metallic bead on charcoal like lead does."

Zinc gives no bead — it is volatile and escapes, leaving only a ZnO incrustation (yellow hot, white cold). Lead melts and stays as a bead.

"A yellow incrustation that stays yellow when cold proves zinc is present."

It proves lead (

P b O

), which stays yellow cold. Zinc's yellow oxide turns white on cooling, so the cold colour is the discriminator.

"We heat the borax to drive off

C O_{2}

before adding the salt."

Borax contains no carbonate; heating drives off water (10

H_{2} O

) causing intumescence, not

C O_{2}

"Cobalt turns colourless in the reducing flame, so a colourless RF bead means cobalt."

C o^{2 +}

stays blue in both flames. A violet-OF/colourless-RF bead points to manganese, not cobalt.

"Since charcoal supplies electrons, the metal must gain oxygen."

Gaining oxygen is oxidation. Charcoal (C, CO) removes oxygen — the metal oxide is reduced to metal.

M g

forms a shiny metal bead on charcoal like copper."

M g

has a very high melting point and forms an infusible white glowing oxide, no bead; only low-melting metals (Pb, Bi, Cu) give beads.

Why questions

Why does a fresh borax bead puff up ("intumesce") when first heated?

The 10 water molecules of crystallisation escape as steam (Step 1 above), swelling the solid before it collapses into clear glass.

Why does the metal's oxidation state control the bead colour at all?

Colour comes from d–d electron transitions (Transition Metal d-d Colour); changing oxidation state changes the d-electron count and orbital splitting, hence a different absorbed wavelength and colour.

Why does the reducing flame (RF) give a lower oxidation state?

It is fuel-rich, so unburnt carbon and CO supply electrons to the metal ion, pushing it (e.g.

C u^{2 +} \to C u^{+}

) to a lower state.

Why must you record both flames rather than one?

Some ions share a colour in one flame (Co and Cu are both blue in OF); the second flame breaks the tie, so both observations together identify the metal.

Why is

B_{2} O_{3}

, not

N a B O_{2}

, called "the glass"?

B_{2} O_{3}

is the transparent, reactive glassy solvent that dissolves metal oxides to form coloured metaborates;

N a B O_{2}

is a by-product.

Why is a blowpipe used to direct the reducing flame into the cavity?

It focuses the luminous, fuel-rich flame precisely onto the salt, maximising reduction and heat in the small cavity.

Why does the cobalt nitrate test follow a white glowing residue?

The colourless high-mp oxides (

A l, M g, Z n

) look alike, so cobalt nitrate is added to give distinctive colours — blue (Al), pink (Mg), green (Zn) — to tell them apart.

Edge cases

What does a colourless borax bead in both flames tell you?

The cation is non-transition (e.g.

N a^{+}, K^{+}, C a^{2 +}, A l^{3 +}, Z n^{2 +}

) with no d–d absorption — the borax bead simply can't identify these.

What happens if the salt is a chromate/dichromate rather than a

C r^{3 +}

salt?

In the hot bead it is still reduced/incorporated to give green

C r^{3 +}

metaborate; the bead colour reflects the final oxidation state in the glass, not the starting one.

What if the metal oxide is volatile (like

S b

Z n

) in the charcoal test?

The oxide vaporises and deposits as an incrustation around the cavity instead of forming a metal bead — the crust colour then becomes the diagnostic.

What is observed for a completely non-reducible high-mp oxide (Ca, Ba)?

No metal bead forms; you get a white infusible residue that glows intensely — confirm the specific metal with the cobalt nitrate or flame test.

What limits the borax bead test for iron in the reducing flame?

F e^{3 +}

(yellow-brown, OF) becomes green/colourless in RF, and pale beads are easily misread as "no colour", so a very small speck and careful lighting are needed.

What happens if you overheat and the bead cracks/goes opaque on cooling?

Overloaded or devitrified glass scatters light and hides the colour; you must remake the bead with a smaller speck to read the true diagnostic colour.

Recall One-line summary of the traps

Bead colour = d–d absorption tied to oxidation state (set by OF vs RF); cavity test = reduction on charcoal needing the luminous flame; "no bead / white residue" is a result, not a failure. Practise on the group salts you'll actually meet.