Flame Tests and Quantised Energy Levels for HSC Chemistry

Two clear salt solutions can look almost identical. Put a clean wire loop carrying one into a blue Bunsen flame and it may flash bright yellow. Test the other and you might see lilac instead.

Before going further, predict this: if electrons inside the metal species could have any energy they liked, would you expect a few characteristic colours, or a smooth spread of colours?

You would expect a smooth spread. Instead, particular metal ions produce characteristic flame colours. That observation points towards one of the most important ideas in atomic structure: electron energies are quantised.

01Why a flame can reveal energy levels

Start with what the flame actually does.

When a small amount of a metal salt is placed in a hot flame, the solvent evaporates and the salt is heated strongly. Energy from the flame is transferred to the particles formed from the salt.

Some electrons absorb this energy.

An electron can then move from a lower-energy state to a higher-energy state. We call this excitation.

The higher-energy arrangement does not usually last long. The electron returns to a lower-energy state, releasing energy as electromagnetic radiation.

If that radiation has a wavelength in the visible region, we see colour.

The basic sequence is:

1. The flame supplies energy.
2. An electron absorbs energy.
3. The electron moves to a higher allowed energy level.
4. The electron returns to a lower level.
5. The energy difference is released as a photon.

The important word is allowed.

Electrons are not allowed to possess every possible energy.

02Quantised energy means “steps”, not a ramp

Picture a staircase.

You can stand on step 2 or step 3. You cannot stand at step 2.63 while somehow hovering between them.

That is a useful first model for electron energy levels. The steps represent allowed energies. Moving upstairs requires energy to be absorbed. Moving downstairs releases energy.

The analogy has an important limit. Electron energy levels are not literal shelves or physical rings that an electron sits on. Quantum mechanics describes electrons using orbitals and probability distributions. The staircase is only a model for the fact that their energies are discrete.

If you need to refresh how electrons are arranged before continuing, see Electronic Configuration for HSC Chemistry: Atoms and Ions.

Suppose an electron moves between two levels with energies \(E_{\text{high}}\) and \(E_{\text{low}}\). The energy released is

\[
\Delta E = E_{\text{high}} – E_{\text{low}}
\]

where \(\Delta E\) is the energy difference in joules, J.

That energy leaves as a photon.

The photon energy is

\[
E = hf = \frac{hc}{\lambda}
\]

where:

• \(E\) is the photon energy in joules, J
• \(h\) is Planck’s constant, \(6.626 \times 10^{-34}\ \text{J s}\)
• \(f\) is the frequency in hertz, Hz
• \(c\) is the speed of light, \(3.00 \times 10^8\ \text{m s}^{-1}\)
• \(\lambda\) is the wavelength in metres, m

So the energy gap inside the emitting species determines the photon wavelength.

A larger energy gap produces a higher-energy photon. Higher-energy visible photons have higher frequencies and shorter wavelengths, towards the blue and violet end of the visible spectrum.

A smaller gap produces a lower-energy photon with a longer wavelength, towards the red end.

That is the link:

energy levels -> energy difference -> photon energy -> wavelength -> observed colour

03Why different metal ions give different colours

Here is the key prediction.

Suppose two different metal ions had exactly the same set of electron energy levels. What would you expect from their emission?

They could produce the same set of photon energies and therefore the same wavelengths.

But different elements have different nuclear charges and different electron arrangements. Their allowed electron energies are therefore different. The gaps between their energy levels are different too.

That gives different characteristic emission wavelengths.

In a school flame test, commonly observed colours include:

Treat this table as an experimental guide, not as a set of perfectly fixed paint colours. What you actually see depends on concentration, contamination, flame conditions, and your own colour perception.

There is another useful detail. Two salts containing the same metal ion often give similar flame colours. Sodium chloride and sodium nitrate, for example, both contain \(\mathrm{Na}^{+}\).

The anion has changed, but the characteristic sodium emission remains.

That is why flame tests are mainly used as evidence about the metal component of the salt.

04One flame colour does not mean one wavelength

It is tempting to imagine a sodium flame producing one yellow wavelength and nothing else.

That is too simple.

An atom or ion can have many allowed energy levels. Electrons can undergo several different transitions, so several photon energies may be emitted.

A spectroscope separates the emitted light by wavelength. Instead of seeing one blended flame colour, you see a series of bright emission lines.

Each line corresponds to a particular photon energy and therefore to a particular energy difference.

This is stronger evidence for quantised energy levels than the naked-eye flame colour alone.

If electron energies formed a continuous range, you would expect a continuous spread of emitted energies. Discrete lines show that only particular energy differences occur.

05What is actually emitting the light?

For HSC Chemistry, it is useful to describe a flame test in terms of electrons in the metal species absorbing energy and later releasing photons as they move to lower energy levels.

There is a little more chemistry happening inside a real flame.

The solution first evaporates. The dissolved salt is converted into gaseous species, and the high temperature can produce a mixture containing atoms, ions, and other small species. The exact emitting species depends on the element and the flame conditions.

So statements such as “the \(\mathrm{Na}^{+}\) ion makes the flame yellow” are useful shorthand for identifying the metal ion originally present in the sample. They should not be taken to mean that nothing except isolated \(\mathrm{Na}^{+}\) ions exists inside the flame.

The central conclusion is unchanged: characteristic emission occurs because electrons can move only between particular allowed energy states.

06The most tempting misconception: the flame colour comes from the heat

A student might reason like this:

A hot flame glows, so perhaps sodium makes the flame hotter and potassium makes it cooler.

That sounds plausible, but it predicts the wrong mechanism.

If colour were mainly determined by the sample changing the flame temperature, then substances at similar temperatures should show similar colours.

Instead, different metal salts placed in the same flame can produce different characteristic colours.

The Bunsen flame supplies the energy. The electronic structure of the metal species determines the characteristic emitted wavelengths.

There is also a useful distinction between two kinds of light production:

• A hot object can produce a broad, continuous thermal spectrum.
• Excited atoms and ions can produce discrete emission lines.

A flame test relies on the second effect to identify metal ions.

07How to perform a useful flame-test investigation

A practical flame test looks simple, which is exactly why careless technique can ruin it.

A sensible investigation uses known metal salt solutions first, then compares an unknown sample against those observations.

A typical method is:

1. Light a Bunsen burner and adjust it to produce a blue, non-luminous flame.
2. Clean a nichrome or platinum wire loop using hydrochloric acid as directed by your laboratory procedure.
3. Place the clean loop in the flame.
4. Check that it produces no noticeable flame colour.
5. Dip the loop into one metal salt solution.
6. Place the sample into the hot region of the flame.
7. Observe and record the flame colour.
8. Clean the loop thoroughly before testing the next solution.
9. Repeat measurements so that an unusual observation is not accepted from one trial alone.

The order matters.

If you test sodium and then put the same dirty loop straight into a potassium solution, a tiny amount of sodium may remain. Sodium’s yellow emission is intense and can mask the much weaker lilac colour you were trying to observe.

This is not a tiny technical detail. It can change your conclusion.

08Evaluating the investigation properly

An evaluation should not just say “human error may occur”. That phrase tells you almost nothing.

Identify the actual weakness, explain what effect it could have on the result, and propose an improvement that targets it.

Contamination between samples

Problem: Metal ions left on the wire from a previous trial can enter the next sample or flame.

Effect: The observed colour may contain emissions from more than one metal. Sodium contamination is especially troublesome because its yellow emission can be very intense.

Improvement: Clean the loop thoroughly between samples and heat it in the flame until no characteristic colour is visible. Separate loops can also be used for different solutions.

A yellow luminous Bunsen flame

Problem: Closing the air hole produces incomplete combustion and glowing carbon particles.

Effect: The flame already appears bright yellow, making a sodium test difficult to distinguish and other colours harder to see.

Improvement: Use a blue, non-luminous flame.

Different sample quantities

Problem: One loop carries much more solution than another.

Effect: One flame colour may appear much brighter simply because more emitting material is present.

Improvement: Use the same loop and a consistent dipping method. Keep solution concentrations controlled where possible.

Colour judgement is subjective

Problem: Words such as “red”, “crimson”, and “orange-red” depend on the observer and lighting conditions.

Effect: Similar colours may be confused, especially for unknown samples.

Improvement: Compare unknowns directly with known standards under the same conditions. Better still, use a spectroscope so that wavelength positions rather than colour names can be compared.

Sodium can mask other ions

Problem: A mixture may contain a small amount of sodium alongside another ion.

Effect: The intense yellow emission can dominate what the eye sees.

Improvement: Use spectroscopic analysis to resolve separate wavelengths. In some practical settings, a cobalt-blue filter can reduce the apparent sodium yellow and make potassium’s lilac emission easier to observe, but it does not turn the flame test into a perfectly selective measurement.

09Reliability, validity, and accuracy are different

These words are often thrown into practical reports as if they mean the same thing.

They do not.

Repeating a badly contaminated flame test can make it more reliable while it remains invalid.

For example, imagine a sodium-contaminated loop gives a yellow flame three times when testing potassium. Those results are consistent. They are also misleading.

“Repeat it three times” is therefore not a complete evaluation by itself.

10What can a flame test actually prove?

A flame test can provide evidence that a particular metal ion is present.

It usually cannot prove that the ion is definitely present on the basis of colour alone.

Why not?

First, some colours are similar. Lithium and strontium both produce red colours that may be difficult to distinguish by eye.

Second, mixtures are difficult. One strong emission can mask another.

Third, contamination can create a false colour.

Fourth, the naked eye combines several wavelengths into one perceived colour.

So a strong conclusion sounds like this:

The observed yellow flame is consistent with the presence of sodium ions.

A weaker scientific conclusion would be:

The yellow flame proves the sample is sodium chloride.

The flame test has not identified the anion, and the observation alone may not uniquely identify the cation.

11Flame tests as evidence for atomic models

The important result of a flame test is not the colour chart.

It is the pattern behind the colours.

Different elements produce particular emission wavelengths. A spectroscope resolves these into discrete lines rather than a continuous rainbow.

Any useful atomic model therefore has to explain two observations:

• electrons cannot possess just any energy
• different elements have different sets of allowed energies

The Bohr model captured the idea of discrete energy levels. It was especially successful for hydrogen.

More advanced quantum models replace simple circular electron orbits with orbitals. In HSC Chemistry, spdf notation and valence electrons give you the next layer of that description.

The simple staircase model survives, but the detailed picture of the staircase becomes much richer.

12Questions and solutions

Question 1

A student tests a clean sample and observes an intense yellow flame. The possible metal ions are \(\mathrm{Na}^{+}\), \(\mathrm{K}^{+}\), and \(\mathrm{Cu}^{2+}\).

Which ion is most consistent with the observation? Explain the observation using electron energy levels.

Solution 1

The ion most consistent with the observation is \(\mathrm{Na}^{+}\).

Sodium compounds characteristically produce an intense yellow flame under typical flame-test conditions.

Energy from the flame excites electrons in sodium species to higher allowed energy states. When the electrons move back to lower-energy states, photons are emitted.

The photon energies satisfy

\[
\Delta E = hf = \frac{hc}{\lambda}
\]

Only particular energy differences are allowed, so particular wavelengths are emitted. Strong visible sodium emissions fall in the yellow region, producing the observed colour.

The flame colour is evidence consistent with sodium being present, rather than absolute proof from colour alone.

Question 2

A student tests sodium chloride, \(\ce{NaCl}\), and sodium nitrate, \(\ce{NaNO3}\). Both produce very similar yellow flames.

Another student claims that the flame test must therefore be identifying the chloride and nitrate ions.

Explain why this conclusion is incorrect.

Solution 2

The conclusion is incorrect because the common species in the two samples is the sodium ion, \(\mathrm{Na}^{+}\), not the anion.

The two compounds contain different anions:

\[
\mathrm{Cl}^{-}
\]

and

\[
\ce{NO3-}
\]

but they both contain \(\mathrm{Na}^{+}\).

Their similar characteristic flame colours therefore provide evidence that the important emission is associated with the metal species derived from sodium.

The tempting mistake is to assume that every part of a compound contributes equally to the visible flame colour. In ordinary qualitative flame tests, the characteristic colour is mainly used to identify the metal cation.

The result also cannot distinguish sodium chloride from sodium nitrate. Another test would be needed to identify the anion.

Question 3

An emission line from a heated metal sample has a wavelength of \(520\ \text{nm}\).

Calculate the energy of one photon. A second line appears at \(680\ \text{nm}\). Without calculating its exact energy, state which line corresponds to the larger electronic energy change and explain why.

Solution 3

The \(520\ \text{nm}\) photon has an energy of approximately \(3.82 \times 10^{-19}\ \text{J}\), and it represents the larger energy change.

First convert the wavelength:

\[
520\ \text{nm}
= 5.20 \times 10^{-7}\ \text{m}
\]

Use

\[
E = \frac{hc}{\lambda}
\]

and substitute:

\[
E
= \frac{(6.626 \times 10^{-34}\ \text{J s})(3.00 \times 10^8\ \text{m s}^{-1})}
{5.20 \times 10^{-7}\ \text{m}}
= 3.82 \times 10^{-19}\ \text{J}
\]

Therefore,

\[
E = 3.82 \times 10^{-19}\ \text{J per photon}
\]

The \(520\ \text{nm}\) line corresponds to a larger energy change than the \(680\ \text{nm}\) line.

Because

\[
E = \frac{hc}{\lambda}
\]

photon energy is inversely proportional to wavelength. The shorter \(520\ \text{nm}\) wavelength must therefore have the greater photon energy and correspond to the larger difference between electronic energy levels.

Question 4

A student performs flame tests in this order:

1. sodium solution
2. potassium solution
3. calcium solution

The wire loop is dipped into hydrochloric acid between samples, but the student does not heat the loop to check that it is clean. All three samples appear yellow.

The student repeats the entire procedure three times and obtains the same result each time.

Evaluate the investigation. Your answer should distinguish reliability from validity and propose a useful improvement.

Solution 4

The investigation may be repeatable, but the evidence is not valid enough to support the conclusion that all three samples contain sodium.

Getting the same yellow result three times gives some evidence of reliability, because the observation is consistent.

However, the procedure has a serious contamination problem. Simply dipping the loop into hydrochloric acid does not demonstrate that all sodium residue has been removed. Because sodium produces a very intense yellow emission, even a small amount carried into later trials could mask the expected lilac potassium flame or orange-red calcium flame.

That threatens validity because the investigation is no longer changing only the intended metal sample. Residue from the previous sample has become an uncontrolled variable.

A useful improvement is to clean the loop and then heat it in the flame before every sample. The loop should produce no characteristic flame colour before the next solution is tested.

Using separate clean loops for different samples would reduce cross-contamination further.

The important trap is assuming that repeated results must be correct. A systematic error can occur consistently, so repeating the same flawed method can produce highly consistent but misleading data.

Question 5

An unknown solution produces a strong yellow flame when viewed by eye. A student concludes:

The solution contains sodium ions and no other metal ions.

A spectroscope is then used. Strong yellow sodium lines are present, but several weaker lines also match a potassium reference spectrum.

Evaluate the student’s original conclusion and explain why the two observations are not contradictory.

Solution 5

The student’s conclusion that sodium is the only metal ion present is not justified.

The strong yellow flame is good evidence that sodium is present. However, a naked-eye flame test does not reliably show every emitting species in a mixture.

Sodium emission is extremely intense and can dominate the perceived flame colour. Potassium may therefore be present even though its lilac contribution is difficult to see.

The spectroscope gives more detailed evidence because it separates the emitted radiation according to wavelength. Instead of combining all wavelengths into one perceived colour, it reveals individual emission lines.

The observations are therefore compatible:

• the eye mainly detects the intense sodium yellow
• the spectroscope detects both sodium and weaker potassium emissions

The hidden assumption in the student’s reasoning was that “I can only see one colour” means “only one metal is present”. That assumption fails for mixtures.

Question 6

A simplified model of an emitting species contains the following allowed energy levels:

\[
E_0 = 0
\]

\[
E_1 = 2.10 \times 10^{-19}\ \text{J}
\]

\[
E_2 = 4.90 \times 10^{-19}\ \text{J}
\]

Assume electrons can make downward transitions between any pair of these levels.

Determine the three possible photon energies. Then calculate the wavelength produced by the transition from \(E_2\) to \(E_1\).

Would every possible transition necessarily produce a visible flame colour?

Solution 6

The possible photon energies are \(2.10 \times 10^{-19}\ \text{J}\), \(2.80 \times 10^{-19}\ \text{J}\), and \(4.90 \times 10^{-19}\ \text{J}\). The \(E_2\) to \(E_1\) transition produces light with a wavelength of about \(710\ \text{nm}\), and not every possible transition must be visible.

For the transition \(E_1 \to E_0\):

\[
\Delta E
= 2.10 \times 10^{-19} – 0
= 2.10 \times 10^{-19}\ \text{J}
\]

For \(E_2 \to E_1\):

\[
\Delta E
= 4.90 \times 10^{-19} – 2.10 \times 10^{-19}
= 2.80 \times 10^{-19}\ \text{J}
\]

For \(E_2 \to E_0\):

\[
\Delta E
= 4.90 \times 10^{-19} – 0
= 4.90 \times 10^{-19}\ \text{J}
\]

For the \(E_2 \to E_1\) transition,

\[
\lambda = \frac{hc}{E}
\]

so

\[
\lambda
= \frac{(6.626 \times 10^{-34}\ \text{J s})(3.00 \times 10^8\ \text{m s}^{-1})}
{2.80 \times 10^{-19}\ \text{J}}
= 7.10 \times 10^{-7}\ \text{m}
\]

Therefore,

\[
\lambda = 710\ \text{nm}
\]

which is near the red end of the visible spectrum.

Not every possible transition necessarily produces visible light. A photon may have a wavelength in the infrared or ultraviolet region, or an emission may be too weak to contribute noticeably to the observed flame colour.

This is why “no strong visible colour” does not mean “no electronic transitions occurred”. A detector that measures wavelengths beyond what the human eye can see can reveal information that a flame-colour observation misses.

13Where this idea leads next

A flame test turns an abstract statement about quantised energy into an observable prediction: particular substances emit particular wavelengths because their electrons have particular allowed energy differences.

The next question is more detailed. Why do different atoms have those particular energy levels in the first place?

That takes you from the simple Bohr-style picture of discrete levels into electron configurations, subshells, and orbitals. Once you can connect an emission spectrum to the arrangement of electrons, flame tests stop being a colour-memorisation exercise and become evidence about atomic structure.