Question

What do you understand about Basic radicals?

Answer 1

A radical is an atom, molecule or ion that has an unpaired valence electron. These unpaired electrons make radicals highly chemically reactive. Basic radicals are formed after the removal of hydroxide ions from acid.

Complete answer:
Basic radical is an ion coming from a base. It is a positively charged chemical species; thus we name it as the cation. Moreover, it is a portion of an inorganic salt. This ion forms as a result of the removal of a hydroxide ion from a base.
Analysis of basic radicals: analysis of basic radical involves following steps:
$(i)$ Preparation of aqueous solution or original solution. Transparent, aqueous solution of salt is known as the original solution and it is prepared by dissolving salt/mixture in suitable solvent. To prepare the original solution a small quantity of the substance is shaken with cold water, distilled water. If substances dissolve in water to prepare the original solution.
Test for $P{b^{2 + }}$:
White precipitate disappears on heating and reappears on cooling, yellow precipitate $(Pb{I_2})$ is obtained when $KI$ is added in hot solution.
Test for $B{i^{3 + }}$:
Will be confirmly present if yellow precipitate is obtained on adding concentrated $HN{O_3}$ followed by thiourea solution, in the white precipitations after dilution.
Test for $S{b^{3 + }}$:
The white precipitation obtained is dissolved in concentrated $HCl$ solution is diluted and ${H_2}S$ is passed in the solution. Orange precipitate confirms $S{b^{3 + }}$.
$(ii)$ Separation of basic radicals in different groups.

S.No.	Group	Basic radicals	Group reagent	Precipitated compound	Color of precipitation
$1$	Zero group	$N{H_4}^ +$	Dil. $NaOH$	$N{H_3}$ gas	$-$
$2$	$1st$ group	$A{g^ + },H{g^{2 + }},P{b^{2 + }}$	Dil. $HCl$	Chlorides of respective cations.$(i)PbC{l_2}$ $(ii)AgCl$ $(iii)H{g_2}C{l_2}$	White
$3$	$2nd$ group	$H{g^{2 + }},C{u^{2 + }},P{b^{2 + }},B{i^{3 + }}$$C{d^{2 + }},S{n^{2 + }},S{b^{3 + }}$$C{o^{2 + }}$	${H_2}S$ in the presence of Dil. $HCl$	(i)HgS (ii)CuS (iii)PbS (iv)$B{i_2}{S_3}$ (v)CdS (vi)SnS (vii)$S{b_2}{S_3}$ (viii)$A{s_2}{S_3}$ (ix)$Sn{S_2}$	BlackBlackBlackYellowYellowBrownOrangeYellowYellow
$4$	$3rd$ group	$F{e^{3 + }},A{l^{3 + }},C{r^{3 + }}$	$N{H_4}OH$ in presence of $N{H_4}Cl$	(i)$Fe{(OH)_3}$ (ii)$Al{(OH)_3}$ (iii)$Cr{(OH)_3}$	RedBrownGreen
$5$	$4th$ group	$Z{n^{2 + }},N{i^{2 + }},C{o^{2 + }},M{n^{2 + }}$	${H_2}S$ in the presence of$N{H_4}OH$	(i)ZnS (ii)NiS (iii)CoS (iv)MnS	WhiteBlackBlackBuff
$6$	$5th$ group	$C{a^{2 + }},S{r^{2 + }},B{a^{2 + }}$	${(N{H_4})_2}C{O_3}$ in the presence of $N{H_4}OH$	(i)$CaC{O_3}$ (ii)$SrC{O_3}$ (iii)$BaC{O_3}$	WhiteWhiteWhite
$7$	$6th$ group	$N{a^ + },{K^ + },M{g^{2 + }}$	N{H_4}OH$$$$ + N{a_2}HP{O_4}	$Mg(N{H_4})P{O_4}$	White

$(iii)$ Analysis of the precipitation obtained for each group.
$(iv)$ When substance is changed into the original solution, its constituents are ionized and each basic radical a specific route of tests is used.

Note:
$(i)$ Original solution may be colored but always transparent.
$(ii)$ Concentrated $HN{O_3}$, concentrated ${H_2}S{O_4}$ or aqueous regia can’t be used to prepare the original solution.
$(iii)$ If the original solution is prepared in conc. $HCl$, then its dilution is necessary before use. if not then diluted then $B{a^{2 + }}$ may precipitate as $BaC{l_2}$.
$(iv)$ original solution must be prepared in distilled water because ordinary water may contain some cations or anions.
$(v)$ If observed carefully, one may get some ideas regarding cations that may be present in the mixture.
$\bullet$ If original solution is colored then transition metal ions like – blue $C{u^{2 + }}$, green $F{e^{2 + }}$, or $C{r^{2 + }}$ will be present. If the original solution is colorless then these ions will be absent.
$\bullet$ If the original solution turns pink from blue, when water is added , then $C{o^{2 + }}$ is present.
$\bullet$ If the original solution is prepared in dil. $HCl$, then group $1(P{b^{2 + }},A{g^ + },H{g^{2 + }})$ will be absent.
$\bullet$ If white ppt, is obtained when the original solution is prepared in dil. $HCl$ Then $P{b^{2 + }}$ may be present.