Question

The salt of an alkali metal gives violet colour in the flame test. Its aqueous solution gives a white precipitate with barium chloride in hydrochloric acid medium. The salt is:
A.${{\text{K}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$
B.${\text{KCl}}$
C.${\text{N}}{{\text{a}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$
D.${{\text{K}}_{\text{2}}}{\text{C}}{{\text{O}}_3}$
E.${\text{LiS}}{{\text{O}}_{\text{4}}}$

Answer 1

To answer this question, you should recall from the s-block elements to identify the cation from the colour of the flame test. To identify the anion, recall the basics of salt analysis.

Complete step by step answer:
We know the colour given by different s-block elements in the flame test are:

Alkali Metals	Flame colour
Li	Red
Na	Yellow
K	Violet
Rb	Crimson red
Cs	Blue

As the question mentions violet colour, thus, the cation should be ${{\text{K}}^ \oplus }$
Now, to find anion recall basics of salt analysis, according to which if any salt gives a white precipitate with barium chloride in ${\text{HCl}}$ medium, the anion must be ${\text{S}}{{\text{O}}_{\text{4}}}^{{\text{2 - }}}$ and the white precipitate formed is ${\text{BaS}}{{\text{O}}_{\text{4}}}$.
${{\text{K}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}{\text{ + BaC}}{{\text{l}}_{\text{2}}} \to {\text{2NaCl + BaS}}{{\text{O}}_{\text{4}}}{\text{(white ppt}}{\text{.)}}$
Combining the identified cation ${{\text{K}}^ \oplus }$ and anion ${\text{S}}{{\text{O}}_{\text{4}}}^{{\text{2 - }}}$, we can conclude that the given salt is ${{\text{K}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$.
Therefore, we can conclude that the correct answer to this question is option A.
Other important coloured salts are:

Colour of salt	Cation Present
Deep Green or purple	${\text{C}}{{\text{r}}^{{\text{3}} \oplus }}$
Whitish pink	${\text{M}}{{\text{n}}^{2 \oplus }}$
Deep red	${\text{C}}{{\text{o}}^{2 \oplus }}$
Green	${\text{F}}{{\text{e}}^{{\text{3}} \oplus }}$
Brown or yellow	${\text{F}}{{\text{e}}^{2 \oplus }}$
Dark blue	${\text{C}}{{\text{o}}^{2 \oplus }}$
Green	${\text{N}}{{\text{i}}^{2 \oplus }}$
Proper blue	${\text{C}}{{\text{u}}^{2 \oplus }}$
Green or blue	${\text{C}}{{\text{u}}^{2 \oplus }}$

So, the correct answer is Option A.

Additional Information:
We know that the majority of all the coloured compounds have transition element(s) ions in them. The colour of transition metal ions is due to the presence of unpaired electrons in it. These electrons absorb radiations of one colour and undergo transition from one level to another within the d-subshell. Due to this electronic transition, coloured light is emitted which is complementary colour of the light absorbed. Down you'll find the list of absorbed colour & the complementary colour which is given out.

Note:
We must know the colour of important cations and also remember the reactions of anions: ${\text{C}}{{\text{O}}_{\text{3}}}^{{\text{2 - }}}{\text{ , C}}{{\text{H}}_{\text{3}}}{\text{CO}}{{\text{O}}^{\text{-}}}{\text{, }}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}^{{\text{2 - }}}{\text{,P}}{{\text{O}}_{\text{4}}}^{{\text{3 - }}}{\text{, S}}{{\text{O}}_{\text{4}}}^{{\text{2 - }}}$ as they are most commonly asked in competitive examinations.
And when we have to determine the colour of salt, we should know the complementary colour scheme:
Violet Yellow
Blue Orange
Green Red
Yellow Violet
Orange Blue
Red Green