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Write the chemical equation for burning of lanthanides in oxygen.

Last updated date: 22nd Jul 2024
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Hint: The lanthanides are by and large viewed as elements with atomic number $58 - 71$. All the elements are transition metals in which $4f$ sublevel is being filled.

Complete answer:
The chemical equation of lanthanides with oxygen
$4Ln + 3{O_2} \to 2L{n_2}{O_3}$
There is one exception. Among all the lanthanides cerium is the only element which forms $Ce{O_2}$ instead of $C{e_2}{O_3}$.
Lanthanides have fundamentally the same properties and are hard to isolate from each other, they were not helpful for essential examination and application, and henceforth they were viewed as rare elements. Since the liquid dissolvable extraction technique utilizing tributylphosphine oxide opened up, lanthanide elements have been promptly accessible and generally utilized synthetic examination as well as materials in composites, impetuses, lasers, cathode-beam, and so on.
The color of lanthanides buildings start for the most part from charge move interactions between the metal and ligand. $f \to f$ transitions are symmetry forbidden, which is also valid for transition metals. Nonetheless, changing metals can utilize vibronic coupling to disrupt this rule. The valence orbitals in lanthanides are for the most part non-bonding and as such minimal powerful vibronic coupling takes, henceforth the spectra from $f \to f$ changes are a lot more smaller and vulnerable than those from $d \to d$transition. Overall this makes the color of lanthanides complexes far fainter than those of transition metal complexes.

Cerium is possible to separate more easily from the rest of the lanthanides because of its unique ability to be oxidized to the $ + 4$ state, therefore it is considered as the one of the rare-earth metal, Cerium $ + 4$ oxidation state does not oxidize water.