Question

Which of the following cation cannot be identified by flame test?
A.$N{a^ + }$
B.${K^ + }$
C.$B{a^{2 + }}$
D.$M{g^{2 + }}$

Answer 1

Flame tests are used to identify the presence of a relatively small number of metal ions in a compound. Not all metal ions give flame colours. For Group 1 compounds, flame tests are usually by far the easiest way of identifying which metal you have got. For other metals, there are usually other easy methods that are more reliable - but the flame test can give a useful hint as to where to look.

Complete step by step answer:
The emission of radiation in the visible region is the reason for cations to exhibit the flame test. Cations such as $N{a^ + }$,${K^ + }$ and $B{a^{2 + }}$ give the flame test as the loosely bound electrons can excite to higher energy levels and on returning to the ground state emit the characteristic colour. The characteristic colour depends on the energy gap that exists between the ground and excited levels.
In the case of $M{g^{2 + }}$, the atomic size is abnormally small and the electron is tightly bound and so it cannot go into an excited state.
Thus, we cannot identify $M{g^{2 + }}$ with a flame test. Same is the case with beryllium.
And hence Option D is the correct answer.

Note: If you excite an atom or an ion by very strong heating, electrons get promoted from their normal unexcited state into higher orbitals. As they fall back down to lower levels (either in one go or in several steps), the energy is released as light. Each of these jumps involves a specific amount of energy being released as light energy, and each of them correspond to a particular wavelength (or frequency). As a result of all these jumps, a spectrum of lines is eventually produced, some of which will be in the visible part of the spectrum. The colour you will see will be a combination of all these individual colours.