Why lithium gives flame coloration?

Question

Theoretically speaking, beryllium and magnesium does not give flame test because their atoms are comparatively smaller and the valance electrons are strongly attached to the nucleus. Therefore their ionisation energies are high, so they need large amount of energy for the excitation of their valence electrons to higher energy level which is usually not available in the Bunsen flame. So they do not impart any colour to the flame.

Now, comparing the atomic radius of lithium and magnesium, both have nearly same atomic radius due to diagonal relationship.($\ce{152}$ pm and $\ce{160}$ pm respectively). Since both of them nearly have same atomic radius, why is lithium imparting flame coloration(crimson red) while magnesium is not? Both of the metals has to be supplied equal amount of energy to excite the electron which the burner is not able to supply and thus flame test should fail in this case but this is not happening.

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I don't understand despite the reasoning for beryllium not giving flame coloration is quite correct, wikipedia claims beryllium to give flame coloration(white).

Answer 1

I would love to find numbers, but alas, I have failed. If anyone can point me towards energy differences between the orbitals I am lacking, please feel free!

There is no a priori physical reason why lithium, sodium, potassium and calcium give flame colours but beryllium and magnesium would not. For all these elements (and hydrogen), the principal mechanism works in the same way: An electron is thermally excited into a higher orbital, and when it relaxes back to its ground state it releases a photon of exactly that wavelength. The flame colour is always due to gaseous atoms (not ions), so the electrons are always excited from an s-orbital.

The strongest lowest-energy excitation and relaxation should always be $n\mathrm{s} \rightarrow n\mathrm{p}$, i.e. from one shell’s s-subshell to that shell’s p-subshell. For sodium, this energy difference corresponds to $589~\mathrm{nm}$ or $2.10~\mathrm{eV}$, for lithium it is $671~\mathrm{nm}$ or $1.85~\mathrm{eV}$, for potassium $767~\mathrm{nm}$ and $1.61~\mathrm{eV}$ and for calcium $657~\mathrm{nm}$ and $1.89~\mathrm{eV}$.^[1]

We can see that the energy difference rises from potassium to calcium, even though they are neighbours and the excitation is from the same orbital to the same different one. This is because calcium’s higher nuclear charge contracts the orbitals, stabilises them and thus lowers their energy — compare with the ionisation energy which also typically rises across a period. This effect is, of course, more pronounced when going across the second or third period when compared to the fourth; thus, I would expect an even greater energy difference when going from sodium to magnesium and even greater than that from lithium to beryllium.

The best explanation I have for the colourlessness of magnesium and beryllium is that their excitations are too close to ultraviolet to be properly observed due to strong contractions of the orbitals, but without numbers that is a weak explanation at best.

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Notes:

[1]: I would be grateful for confirmation that these values I found are correct.

Answer 2

Expanding a bit on the last paragraph of Jan's answer:

The best explanation I have for the colourlessness of magnesium and beryllium is that their excitations are too close to ultraviolet to be properly observed due to strong contractions of the orbitals, but without numbers that is a weak explanation at best.

While the physics and chemistry of inductively-coupled plasmas as used in ICP-AES/ICP-OES are rather different from flame tests, given the orders-of-magnitude difference in energies involved, the usable emission lines for these various elements indirectly support Jan's supposition. The following data are drawn directly from the fantastic Interactive Periodic Table maintained by Inorganic Ventures, which lists the top few most-usable emission lines for each element; the wavelengths below are ordered so as to best highlight trends as a function of period. Lines marked with an asterisk derive from the ion as the emitting species, and thus are probably irrelevant to flame test behavior; unmarked lines are for the atom as emitting species. Bear in mind that the visible/ultraviolet transition occurs around 400 nm (per Wikipedia).

$$ \require{begingroup} \begingroup \newcommand{\nm}[0]{\ \mathrm{nm}} \begin{array}{cccc} \hline \ce{Li} & 670.784\nm & 460.286\nm & 323.261\nm \\ \ce{Na} & 589.595\nm & 588.995\nm & 330.237\nm \\ \ce{K} & 766.490\nm & 771.531\nm & 404.721\nm \\ \ce{Rb} & - & - & 420.185\nm \\ \ce{Cs} & - & - & 455.531\nm \\ \hline \ce{Be} & 313.042\nm^* & 313.107\nm^* & 234.861\nm \\ \ce{Mg} & 279.553\nm^* & 280.270\nm^* & 285.213\nm \\ \ce{Ca} & 393.366\nm^* & 396.847\nm^* & 422.673\nm \\ \ce{Sr} & 407.771\nm^* & 421.552\nm^* & 460.733\nm \\ \ce{Ba} & 230.424\nm^* & 233.527\nm^* & 455.403\nm^* \\ \hline \end{array} \endgroup $$

Unfortunately, I don't know much about the details of why certain lines are strongly emitting and some aren't--the constraints of forbidden transitions mentioned by DavePhD likely play a significant role, though, as well as complex chemical and physical phenomena occurring at the high temperatures of the plasma. Regardless, some interesting, if highly approximate, trends are present in the data.

Of relevance to this specific question, though: the alkali metals all have atomic emission lines in the visible spectrum, whereas it's only the heavier alkaline earths that have useful atomic lines juuuuust over the threshold between the near-UV and visible. While the uncertain connection between emission lines at ICP versus flame conditions must be re-emphasized, this at least provides weak corroborating support for Jan's argument.