Systematic Analysis of Anionic Radicals

The white precipitate formed in the reaction of lead acetate and sodium sulphate is _____. 

The colour of the precipitate formed during the reaction between lead acetate and sulphate ion is:

Write a reaction between lead acetate and sodium sulphate.

In the detection of II group acid radical, the salt containing chloride is treated with concentrated sulphuric acid, the colourless gas is liberated. The name of the gas is

A gas having a pungent odour and a pale yellow precipitate are formed on heating dilute hydrochloric acid with an aqueous solution containing :

[X] + H₂SO₄ → [Y] a colourless gas with an irritating smell

$\left[\mathrm{Y}\right]+{\mathrm{K}}_2{\mathrm{C}\mathrm{r}}_2{\mathrm{O}}_7+{\mathrm{H}}_2{\mathrm{S}\mathrm{O}}_4\to \mathrm{g}\mathrm{r}\mathrm{e}\mathrm{e}\mathrm{n}\mathrm{ }\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}$

$\left[\mathrm{X}\right]$ and $\left[\mathrm{Y}\right]$ are, respectively -

A gas having a pungent odour and a pale yellow precipitate are formed on heating dilute hydrochloric acid with an aqueous solution containing :

Sodium sulphate can be distinguished from sodium sulphite by using

A white sodium salt dissolves in water to give a solution which is neutral to litmus. When silver nitrate solution is added to the solution, a white precipitate is obtained which does not dissolve in dilute ${\mathrm{HNO}}_3$. The anion is

Sulphate ion gives white precipitate with lead acetate.

Brown ring test is used for detection of:

The salt in solution which gives a pale green precipitate with sodium hydroxide solution and a white precipitate with barium chloride solution is 

Sodium sulphate can be distinguished from sodium sulphite by using

Name the reagent used to test the presence of chloride ions in hydrochloric acid. 

When a mixture of $\mathrm{NaCl},{\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ and concentrated ${\mathrm{H}}_2{\mathrm{SO}}_4$ is heated in a dry test tube, a red vapour $\left(\mathrm{X}\right)$ is evolved. This vapour $\left(\mathrm{X}\right)$ turns an aqueous solution of $\mathrm{NaOH}$ yellow due to the formation of $\mathrm{Y}. \mathrm{X}$ and $\mathrm{Y},$ respectively, are:

Which of them gives Chromyl Chloride test with ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7\left(\mathrm{s}\right)$ and conc. ${\mathrm{H}}_2{\mathrm{SO}}_4$ ?

Out of ${\mathrm{Cu}}^{+2}, {\mathrm{Ni}}^{+2}, {\mathrm{Co}}^{+2}$ and ${\mathrm{Mn}}^{+2}$, which ion is precipitated when ${\mathrm{H}}_2\mathrm{S}$ is passed through it in the presence of dilute $\mathrm{HCI}$?

When a mixture of $\mathrm{NaCl}$ and ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ is gently warmed with concentrated ${\mathrm{H}}_2{\mathrm{SO}}_4$:

(i) An orange-red vapour is evolved.

(ii) The vapour formed, when passed into $\mathrm{NaOH}$ solution, gives yellow solution of ${\mathrm{Na}}_2{\mathrm{CrO}}_4$.

(iii) Chlorine gas is not evolved at all.

(iv) Chromyl chloride gas is formed.

(v) ${\mathrm{SnCl}}_4$ used instead of $\mathrm{NaCl}$ also gives the same observations.

Total number of anion(s) which give white precipitate with aqueous solution of ${\mathrm{AgNO}}_3$is/are:

${\mathrm{S}}^{2-}, {\mathrm{HSO}}_3^{\mathrm{\Theta }}, {\mathrm{SO}}_3^{2-}, {\mathrm{S}}_2{\mathrm{O}}_3^{2-}, {\mathrm{CH}}_3{\mathrm{COO}}^{\ominus }, {\mathrm{Cl}}^{\ominus }, {\mathrm{I}}^{\ominus }, {\mathrm{BO}}_3^{3-}$ 

Chromyl chloride test doesn't gives by ${\mathrm{Cl}}^{-}$ of