Noesis
3.5.3 · D5Inorganic Qualitative Analysis

Question bank — Flame tests — characteristic colours

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Before we start, three ideas the traps below lean on:

  • Excitation = an electron absorbing energy and jumping to a higher level (invisible to your eye).
  • De-excitation = the electron falling back down and releasing a photon — this is the light you see.
  • ==== = the wavelength (colour) is fixed by the energy gap ; big gap → short (blue) wavelength, small gap → long (red) wavelength.

True or false — justify

Every yellow flame you see proves sodium is in the sample.
False. Sodium contamination from sweat, glass and dust makes almost anything flash yellow briefly; only a persistent yellow (several seconds) counts as real Na.
Light is emitted at the exact instant the electron absorbs flame energy.
False. Absorbing energy (excitation) is invisible; the coloured photon is emitted later, when the electron falls back down to a lower level.
Two different metals could in principle share the same flame colour.
True. Ba²⁺ and Cu²⁺ both look green because their relevant energy gaps happen to give similar visible wavelengths, which is exactly why a confirmatory wet test is needed.
A larger metal atom always gives a redder flame.
False. Colour is set by the specific level spacing via , not by atomic size, so there is no simple "bigger = redder" rule.
Using concentrated H₂SO₄ instead of HCl to make the paste gives an equally strong colour.
False. Metal sulphates are high-boiling and non-volatile, so little vaporises; HCl forms volatile chlorides that reach the flame and glow strongly.
The flame colour comes mainly from the anion (Cl⁻, SO₄²⁻, NO₃⁻) of the salt.
False. The characteristic colour comes from the metal cation's electron transitions; the anion is chosen (as Cl⁻) only to make a volatile compound.
Most transition metals give the same vivid flame colours as s-block metals.
False. Many transition/heavy metals give faint colours or transitions in the UV/IR (invisible), whereas s-block metals have small excitation energies matching visible light.
A flame test alone is enough to fully identify an unknown cation.
False. It is a screening tool; you narrow the possibilities (e.g. green → Ba or Cu) and then confirm with a precipitation test.

Spot the error

"The electron absorbs a photon from the flame and that emitted photon is what we see."
The flame supplies thermal energy for excitation; the photon we see is emitted later during de-excitation. Mixing up the two events is the error.
"Since , a bigger energy gap means a longer, redder wavelength."
is inversely proportional to , so a bigger gap gives a shorter, bluer wavelength — the trap swapped the direction.
"K⁺ is best viewed through blue cobalt glass because the glass adds a lilac tint."
The glass absorbs the masking sodium-yellow; it does not add colour. Removing yellow lets the faint natural lilac of K show through.
"We clean the wire in HCl to make it hotter so the salt burns better."
Cleaning removes old sodium residue so every sample doesn't look yellow; it is about avoiding contamination, not raising temperature.
"Copper and barium both look green, so a green flame confirms copper."
Green only narrows it to Ba²⁺ or Cu²⁺. You must confirm — Ba²⁺ gives a white BaSO₄ precipitate with dilute H₂SO₄, Cu²⁺ does not.
"We use the luminous (yellow) part of the Bunsen flame so the colour shows up brightly."
You use the non-luminous (blue) part; a luminous flame is cooler and its own yellow glow would mask the sample's true colour.

Why questions

Why does each metal produce a different colour rather than white light?
Each metal has its own set of energy levels, so its own values, so its own wavelengths — a colour "fingerprint" (see Atomic Spectra and Emission Lines).
Why do we see a discrete colour instead of a smooth rainbow?
Energy levels are quantised, so only specific gaps exist and only specific wavelengths are emitted, per Planck's Quantum Theory.
Why must the yellow of sodium be persistent before we trust it?
A brief flash usually comes from trace contamination; genuine sample sodium keeps feeding the flame, so its yellow lasts several seconds.
Why do s-block metals dominate flame-test questions?
Their loosely held outer electrons have low excitation energies, giving gaps that fall right in the visible range, producing vivid, easily seen colours.
Why does converting the salt to a chloride help?
Chlorides are more volatile (lower boiling), so more of the metal vaporises into the flame per second, giving a brighter, clearer colour.
Why is the flame test only a "screening" step and not a proof?
Different cations can share colours (Ba/Cu green; Li/Sr crimson), and contamination misleads, so a positive result must be confirmed by a specific reaction.

Edge cases

What flame colour would a salt of a metal whose only transition lies in the UV show to your eye?
Essentially no visible colour — the emitted photon is UV, so is too short for your eye to detect even though light is emitted.
What happens if two cations (say Na⁺ and K⁺) are present together, viewed with the naked eye?
The strong Na yellow swamps the faint K lilac; you must use cobalt glass to absorb the yellow and reveal potassium.
Would a completely clean wire dipped only in conc. HCl show any colour in the flame?
Ideally none (colourless) — that is the goal of cleaning; any colour left means residual contamination still needs burning off.
If the flame's energy is too low to excite the electron at all, what do you observe?
No characteristic colour, because with no excitation there is no de-excitation and therefore no emitted photon — the level gap is never crossed.
Lithium and strontium can both look crimson red — how do you tell them apart?
Not by eye alone; use a confirmatory wet test (e.g. Sr²⁺ gives a sparingly soluble sulphate, Li⁺ does not) since their colours overlap.
Is the emission-based flame test the same physics as the Photoelectric Effect?
Same link between energy and light, but opposite direction: flame test = electron falls and emits a photon; photoelectric effect = a photon ejects an electron.

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