Learning Objectives:

After reading this unit student will be able to:

• understand the chemistry involved in the functional groups detection

• explain the principle in the preparation of some of the inorganic and organic compounds

• describe the chemistry involved in the volumetric analysis

• understand the chemical principles in the qualitative salt analysis

• explain the basic principles in some of the common physical chemistry experiments

Practical Chemistry

• It has mainly three branches under which practicals are performed viz., organic, inorganic and physical chemistry.

• Organic practicals involve detection, synthesis and determination of melting point and boiling point of the compounds.

• Inorganic practicals involve volumetric analysis, preparation and salt analysis.

• Physical practicals involve different types of practicals like enthalpy determination, kinetic study of reactions, pH metery based etc.

All the above stated practicals only cover some of the aspects of practical chemistry.

Detection of Extra Elements:

• $\mathrm{N},\mathrm{S}$ and halogens are treated as the extra elements in the organic compounds.

• $\mathrm{C},\mathrm{H}$ and $\mathrm{O}$ are the constituent elements of the organic compounds.

• Lassaigne’s Extract or Sodium Extract is prepared to detect the extra elements.

• In Sodium Extract, extra elements gets converted into sodium salts of ions like $\mathrm{NaCN},{\mathrm{Na}}_2\mathrm{S}$ and $\mathrm{NaX}$ (where $\mathrm{X}$ is the halogen).

Chemistry involved in the detection of extra elements

1. Test for Nitrogen $(\mathrm{N})$ :

$$\begin{array}{rl}& 6\mathrm{NaCN}+{\mathrm{FeSO}}_4⟶\underset{\text{ (sodium ferrocyanide) }}{{\mathrm{Na}}_4[\mathrm{Fe}(\mathrm{CN}{)}_6]}+{\mathrm{Na}}_2{\mathrm{SO}}_4\\ & 3{\mathrm{Na}}_4[\mathrm{Fe}(\mathrm{CN}{)}_6]+2{\mathrm{Fe}}_2{\left({\mathrm{SO}}_4\right)}_3⟶\underset{\begin{array}{c}\text{ (Ferric ferrocyanide)}\\ \text{ (Prussian blue) }\end{array}}{{\mathrm{Fe}}_4{[\mathrm{Fe}(\mathrm{CN}{)}_6]}_3}+6{\mathrm{Na}}_2{\mathrm{SO}}_4\end{array}$$

2. Test for Sulphur (S):

a) Sodium Nitroprusside Test:

$$\underset{\text{ (Sodium nitroprusside) }}{{\mathrm{Na}}_2\mathrm{S}+{\mathrm{Na}}_2[\mathrm{Fe}(\mathrm{CN}{)}_5\mathrm{NO}]}⟶\underset{\text{ (sodium sulphonitroprusside) }}{{\mathrm{Na}}_4[\mathrm{Fe}(\mathrm{CN}{)}_5\mathrm{NOS}]}$$

b) Lead Acetate Test:

$${\mathrm{Na}}_2\mathrm{S}+\left[\mathrm{Pb}{\left({\mathrm{CH}}_3\mathrm{COO}\right)}_2\right]⟶\underset{\begin{array}{c}\text{ Lead Sulphide }\\ \text{ ( Black ppt. })\end{array}}{\mathrm{PbS}\downarrow +2}{\mathrm{CH}}_3\mathrm{COONa}$$

3. Test for $\mathrm{N}$ and $\mathrm{S}$ present together

Test with ${\mathrm{FeCl}}_3$ : On addition of ${\mathrm{FeCl}}_3$ to sodium extract blood red colouration appears

$$3\mathrm{NaSCN}+{\mathrm{FeCl}}_3⟶\begin{array}{rl}& \text{ Ferric thiocyanate }\\ & \text{ (Blood red) }\end{array}$$

4. Test for Halogens $(X)$

$\mathrm{N}$ and $\mathrm{S}$ both interfere in the test of halogens, so before testing for halogens these should be completely removed from the sodium extract. Conc. ${\mathrm{HNO}}_3$ is added to sodium extract and solution is repeatedly boiled, $\mathrm{N}$ is expelled as $\mathrm{HCN}$ gas and $\mathrm{S}$ is expelled as ${\mathrm{H}}_2\mathrm{S}$ gas.

$$\begin{array}{r}\mathrm{NaCN}+\underset{\text{ (conc.) }}{{\mathrm{HNO}}_3}⟶{\mathrm{NaNO}}_3+\mathrm{HCN}\uparrow \end{array}$$

$$\begin{array}{r}{\mathrm{Na}}_2\mathrm{S}+\underset{\text{ (conc.) }}{{\mathrm{HNO}}_3}⟶{\mathrm{NaNO}}_3+{\mathrm{H}}_2\mathrm{S}\uparrow \end{array}$$

Test with ${\mathrm{AgNO}}_3$

Acidify sodium extract with dil ${\mathrm{HNO}}_3$ and add ${\mathrm{AgNO}}_3$ solution.

i) A curdy white ppt. soluble in ${\mathrm{NH}}_4\mathrm{OH}$ indicates chlorine

$$\begin{array}{rl}& \mathrm{NaCl}+{\mathrm{AgNO}}_3⟶\mathrm{AgCl}+{\mathrm{NaNO}}_3\end{array}$$

$$\mathrm{AgCl}+\underset{\text{ (excess) }}{2{\mathrm{NH}}_4\mathrm{OH}}⟶\underset{\begin{array}{c}\text{ Diamminesilver(I) chloride }\\ \text{ (Soluble complex) }\end{array}}{\left[\mathrm{Ag}{\left({\mathrm{NH}}_3\right)}_2\right]\mathrm{Cl}}+2{\mathrm{H}}_2\mathrm{O}$$

ii) A pale yellow ppt. sparingly soluble in ${\mathrm{NH}}_4\mathrm{OH}$ indicates $\mathrm{Br}$

$$\mathrm{NaBr}+{\mathrm{AgNO}}_3⟶\mathrm{AgBr}\downarrow +{\mathrm{NaNO}}_3$$

$$\mathrm{AgBr}+2{\mathrm{NH}}_4\mathrm{OH}⟶\underset{\begin{array}{c}\text{ Diamminesilver (I) bromide }\\ \text{ (sparingly soluble complex) }\end{array}}{\left[\mathrm{Ag}{\left({\mathrm{NH}}_3\right)}_2\right]\mathrm{Br}+2{\mathrm{H}}_2\mathrm{O}}$$

iii) A yellow ppt. insoluble in ${\mathrm{NH}}_4\mathrm{OH}$ indicates iodine

$$\begin{array}{rl}& \mathrm{Nal}+{\mathrm{AgNO}}_3⟶\underset{\text{ (yellow) }}{\mathrm{Agl}\downarrow }+{\mathrm{NaNO}}_3\\ & \mathrm{AgI}+{\mathrm{NH}}_4\mathrm{OH}⟶\text{ ppt. does not dissolve in }{\mathrm{NH}}_4\mathrm{OH}\end{array}$$

Organic layer Test - When 1-2 $\mathrm{mL}$ of organic solvent $({\mathrm{CHCl}}_3$ or ${\mathrm{CCl}}_4)$ is added to sodium extract and then few drops of acidified ${\mathrm{KMnO}}_4$ or 2-3 drops of conc. ${\mathrm{HNO}}_3$ is added, organic layer attains a colour on shaking.

Colourless organic layer - Chlorine

Brownish yellow organic layer - Bromine

Violet organic layer - lodine

Detection of the functional groups:

1. Hydroxyl (Alcohol and Phenol):

a) Alcohol

Cerric Ammonium Nitrate Test: Alcohols with few drops of cerric ammonium nitrate solution gives
red colour

$$\underset{\text{ Yellow }}{({\mathrm{NH}}_4{)}_2[\mathrm{Ce}({\mathrm{NO}}_3{)}_6]}+\mathrm{ROH}⟶{(\mathrm{NH}}_4{)}_2\underset{\text{ Red }}{[\mathrm{Ce}(\mathrm{OR})({\mathrm{NO}}_3{)}_5]}+{\mathrm{HNO}}_3$$

b) Phenol

i) Neutral ${\mathrm{FeCl}}_3$ Test: Aqueous solution of phenol with 1-2 drops of neutral ${\mathrm{FeCl}}_3$ solution gives purple or green colouration.

Colour will vary depending upon the organic compound containing phenol as functional group.

ii) Phthalein Test - When equal amount of phenol and phthalic anhydride with few drops of conc. ${\mathrm{H}}_2{\mathrm{SO}}_4$, heated and then poured into the beaker containing dilute $\mathrm{NaOH}$, formation of pink colour of phenolphthalein indicates the presence of phenol.

2. Carbonyl (Aldehyde or Ketone)

Carbonyl group - On heating the carbonyl compound with 2, 4-DNP reagent, a red-orange precipitate forms which indicates the presence of carbonyl group.

Test for aldehyde

Tollens’ Test or Silver Mirror Test- Aldehyde reduce the Tollens’ Reagent. Formation of silver mirror on the sides of boiling tube indicates aldehyde.

Fehling’s Test- given by aliphalic aldehydes. Aldehydes reduce the Fehling solution with a formation of brick red ppt. of cuprous oxide $\left({\mathrm{Cu}}_2\mathrm{O}\right)$.

$$\mathrm{R}-\mathrm{CHO}+2{\mathrm{Cu}}^{2+}+5{}^{\circ }\mathrm{O}\mathrm{H}⟶{\mathrm{RCOO}}^{\circ }+\underset{\text{ (Brick red ppt.) }}{{\mathrm{Cu}}_2\mathrm{O}\downarrow }+3{\mathrm{H}}_2\mathrm{O}$$

Note: Ketone does not reduce Fehling solution and ammonical ${\mathrm{AgNO}}_3$ i.e., does not give Fehling Test and Tollens’ Test.

lodoform Test: This test is given by those compounds which contain ${\mathrm{CH}}_3\mathrm{CO}$-group or

When organic compound is treated with NaOI, yellow ppt. of iodoform is obtained

$$\mathrm{R}-\mathrm{C}-{\mathrm{CH}}_3\stackrel{\mathrm{NaOI}}{\to }\underset{\text{ (Yellow ppt.) }}{{\mathrm{CHI}}_3\downarrow }+\mathrm{RCOONa}$$

3. Carboxyl Group

a) Sodium Bicarbonate Test $({\mathrm{NaHCO}}_3$ Test):

When a saturated solution of ${\mathrm{NaHCO}}_3$ is added to organic compound, brisk effervescence
indicates carboxylic group due to evolution of ${\mathrm{CO}}_2$.

$$\mathrm{RCOOH}+{\mathrm{NaHCO}}_3⟶\mathrm{RCOONa}+{\mathrm{CO}}_2\uparrow +{\mathrm{H}}_2\mathrm{O}$$

b) Litmus Test: Carboxylic acid being acidic in nature turns blue litmus red.

4. Amino Group:

Carbylamine Test/Isocyanide Test- This test is shown by both $1^{\circ }$ aliphatic and $1^{\circ }$ aromatic amines. When organic compound is gently heated with few drops of chloroform and alcoholic $\mathrm{KOH}$, foul smell of carbylamine or isocyanide is obtained.

$$\underset{\begin{array}{c}{1}^{\circ }\text{ aliphatic }\\ \text{ or }\\ 1^{\circ }\text{ aromatic amine) }\end{array}}{\mathrm{R}-{\mathrm{NH}}_2}+{\mathrm{CHCl}}_3+\underset{\text{ (alcoholic) }}{3\mathrm{KOH}}\stackrel{\mathrm{\Delta }}{\to }\underset{\begin{array}{l}\text{ Iso cyanide }\\ \text{ (foul smell) }\end{array}}{\mathrm{R}-\mathrm{NC}+3\mathrm{KCl}+{\mathrm{H}}_2\mathrm{O}}$$

Chemistry involved in the preparation of the following:

1. Acetanilide:

Acetic Anhydride is an acetyling agent. This is a preferred method for synthesis in laboratory.

2. p-nitroacetanilide:- Nitro derivatives of aniline cannot be prepared directly as this will lead to oxidation of aniline and a mixture of products will form alongwith the oxidised product of aniline.

So, reactive amino group is protected by acetylation.

Acetamide group is $0,p$-directing, so when acetanilide undergoes nitration, colourless $p$ nitroacetanilide mixed with yellow 0-product (smaller portion) is also formed.

Recrystallisation from ethanol will remove soluble 0 -nitroacetanilide.

3. Aniline Yellow- It is the first azo dye produced in 1861.

4. Iodoform: lodoform reaction is shown by all the carbonyl compounds which contain ${\mathrm{CH}}_3\mathrm{CO}$ group or alcohol which contain ${\mathrm{CH}}_3\mathrm{CH}(\mathrm{OH})$ - group.

5. Mohr’s Salt $({\mathrm{FeSO}}_4\cdot {\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4\cdot 6{\mathrm{H}}_2\mathrm{O})$

It is a double salt of ${\mathrm{FeSO}}_4$ and ${\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4$

$${\mathrm{FeSO}}_4\cdot 7{\mathrm{H}}_2\mathrm{O}+{\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4⟶{\mathrm{FeSO}}_4\cdot {\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4\cdot 6{\mathrm{H}}_2\mathrm{O}+{\mathrm{H}}_2\mathrm{O}$$

• In solution, double salts loose their identity.

• Double salt exist only in solid or crystalline state but breaks down into their constituent ions when they are dissolved in water or any other solvent.

$${\mathrm{FeSO}}_4\cdot {\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4\cdot 6{\mathrm{H}}_2\mathrm{O}⟶{\mathrm{Fe}}^{2+}(\mathrm{aq})+2{\mathrm{NH}}_4^{+}(\mathrm{aq})+2{\mathrm{SO}}_4^{2-}(\mathrm{aq})+6{\mathrm{H}}_2\mathrm{O}(\mathrm{l})$$

• It is prepared from ferrous sulphate and ammonium sulphate in the presence of dilute ${\mathrm{H}}_2{\mathrm{SO}}_4$.

• Dilute ${\mathrm{H}}_2{\mathrm{SO}}_4$ is used to prevent the conversion of ferrous sulphate to ferric sulphate and also to prevent its hydrolysis.
  
  

6. Potash Alum, ${\mathrm{K}}_2{\mathrm{SO}}_4\cdot {\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3.24{\mathrm{H}}_2\mathrm{O}$ :

It is a double salt of ${\mathrm{K}}_2{\mathrm{SO}}_4$ and ${\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3$. In the solvent, it ionizes to its constituent ions:

$$\begin{array}{rl}& {\mathrm{K}}_2{\mathrm{SO}}_4+{\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3\cdot 18{\mathrm{H}}_2\mathrm{O}+6{\mathrm{H}}_2\mathrm{O}⟶{\mathrm{K}}_2{\mathrm{SO}}_4\cdot {\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3\cdot 24{\mathrm{H}}_2\mathrm{O}\\ & {\mathrm{K}}_2{\mathrm{SO}}_4\cdot {\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3\cdot 24{\mathrm{H}}_2\mathrm{O}⟶2{\mathrm{K}}^{+}(\mathrm{aq})+2{\mathrm{Al}}^{3+}(\mathrm{aq})+4{\mathrm{SO}}_4^{2-}(\mathrm{aq})+24{\mathrm{H}}_2\mathrm{O}(\mathrm{l})\end{array}$$

• In the preparation of potash alum, concentrated ${\mathrm{H}}_2{\mathrm{SO}}_4$ is used to prevent hydrolysis of aluminium sulphate.

Chemistry involved in the titrimetric Analysis

In volumetric analysis, the strength of a unknown solution is determined by allowing the solution to react with standard solution (whose concentration is known) and volume of the solution required for complete reaction is measured.

Acid-Base Titrations

Types of acid-base titrations

1. Strong acid vs. strong base ( $\mathrm{HCl}$ vs. $\mathrm{NaOH})$

2. Weak acid vs. strong base $({\mathrm{CH}}_3\mathrm{COOH}$ vs. $\mathrm{NaOH})$

3. Strong acid vs. weak base $(\mathrm{HCl}$ vs. ${\mathrm{NH}}_4\mathrm{OH})$

• Indicator- Substance which is usually added into the solution taken in the titration flask to detect the equivalence/end point.

• Indicators which are used in the acid-base titrations are known as pH indicators or neutralisation indicators.

• An acid-base indicator is a weak organic acid (e.g., phenolphthalein) or weak organic base (e.g., methyl orange).

• They possess different colour in the acidic and basic medium which indicate the end point in the titration.

• Some commonly used indicators with their pH range are as follows:

		Colour change
Indicator	$\mathrm{pH}$ range	Acidic Medium	Basic Medium
Methyl Orange	$3.2-4.4$	Red	Yellow
Methyl red	$4.8-6.0$	Red	Yellow
Phenolphthalein	$8.2-10.0$	Colourless	Pink
Thymol blue	$8.0-9.6$	Yellow	blue

Phenolphthalein

It is a weak organic acid (HPh).

$\underset{\begin{array}{c}\text{(Benzenoid}\\ \text{structure)}\end{array}}{\mathrm{HPh}}\rightleftharpoons {\mathrm{H}}^{+}+\underset{\begin{array}{c}\text{Pink}\\ \text{(Quinonoid structure)}\end{array}}{{\mathrm{Ph}}^{-}}$

Methyl Orange, (MeOH)

It is a weak base.

$\underset{\begin{array}{c}\text{(Yellow)}\\ \text{(Benzenoid structure)}\end{array}}{\mathrm{MeOH}}\rightleftharpoons {\mathrm{Me}}^{+}+\underset{\begin{array}{c}\text{(Red)}\\ \text{(Quinonoid structure)}\end{array}}{{\mathrm{OH}}^{-}}$

Titration of strong Acid vs. Strong Base

$\mathrm{HCl}$ vs. $\mathrm{NaOH}$

• $\mathrm{pH}$ of the solution at the end point $=7$.

• Indicators which can be used are Methyl orange, methyl red and Phenolphthalein

Titration of weak acid vs. Strong Base

${\mathrm{CH}}_3\mathrm{COOH}$ vs. $\mathrm{NaOH}$

• $\mathrm{pH}$ of the solution at the end point $>7$

• Indicators which can be used are phenolphthalein or thymol blue.

Titration of strong acid vs. Weak base

$\mathrm{HCl}$ vs. ${\mathrm{NH}}_4\mathrm{OH}$

• $\mathrm{pH}$ of the solution at end point $<7$

• Indicator which can be used is methyl red.

Titration of weak acid vs. weak base

• No indicator is suitable in titration of a weak acid against weak base.

Titration of oxalic acid vs. ${\mathrm{KMnO}}_4$

• It is a redox titration

• ${\mathrm{KMnO}}_4$ act as a self-indicator

• colour change is colourless to pink.

• ${\mathrm{KMnO}}_4$ is an oxidising agent and oxalic acid $\left({\mathrm{C}}_2{\mathrm{O}}_4{\mathrm{H}}_2\right)$ is a reducing agent.

Reactions involved are:

$$\begin{array}{rl}& {\mathrm{MnO}}_4^{-}+8{\mathrm{H}}^{+}+5{\mathrm{e}}^{-}⟶{\mathrm{Mn}}^{2+}+4{\mathrm{H}}_2\mathrm{O}\\ & {\mathrm{C}}_2{\mathrm{O}}_4{\mathrm{H}}_2⟶2{\mathrm{CO}}_2+2{\mathrm{H}}^{+}+2{\mathrm{e}}^{-}\end{array}$$

Overall Reaction is:

$2{\mathrm{MnO}}_4^{-}+5{\mathrm{C}}_2{\mathrm{O}}_4{\mathrm{H}}_2+16{\mathrm{H}}^{+}⟶2{\mathrm{Mn}}^{2+}+10{\mathrm{CO}}_2+10{\mathrm{H}}^{+}+8{\mathrm{H}}_2\mathrm{O}$

• In this titration, rate of reaction of oxalic acid with ${\mathrm{KMnO}}_4$ is very slow, hence the oxalic acid solution is heated to $60-{70}^{\circ }\mathrm{C}$.

• ${\mathrm{Mn}}^{2+}$ produced in the solution catalyses the reaction.

• Temperature of the reaction should not be high because oxalic acid may decompose into ${\mathrm{CO}}_2$ and $\mathrm{CO}$.

Titration of Mohr’s salt vs. ${\mathrm{KMnO}}_4$

• It is a redox titration

• ${\mathrm{KMnO}}_4$ act as a self-indicator.

• Colour change is colourless to pink.

• ${\mathrm{KMnO}}_4$ is an oxising agent and Mohr’s salt $({\mathrm{FeSO}}_4\cdot {\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4\cdot 6{\mathrm{H}}_2\mathrm{O})$ is a reducing agent.

Reactions involved are:

$$\begin{array}{rl}& {\mathrm{MnO}}_4^{-}+8{\mathrm{H}}^{+}+5{\mathrm{e}}^{-}⟶{\mathrm{Mn}}^{2+}+4{\mathrm{H}}_2\mathrm{O}\\ & {\mathrm{Fe}}^{2+}⟶{\mathrm{Fe}}^{3+}+{\mathrm{e}}^{-}\end{array}$$

Overall Reaction is:

$${\mathrm{MnO}}_4^{-}+5{\mathrm{Fe}}^{2+}+8{\mathrm{H}}^{+}⟶{\mathrm{Mn}}^{2+}+5{\mathrm{Fe}}^{3+}+4{\mathrm{H}}_2\mathrm{O}$$

• Dilute ${\mathrm{H}}_2{\mathrm{SO}}_4$ is added while preparing standard solution of Mohr’s salt, to prevent the hydrolysis of ${\mathrm{Fe}}^{2+}$ ions to ${\mathrm{Fe}}^{3+}$ ions (brown precipitate of $\mathrm{Fe}(\mathrm{OH}{)}_3$ would be formed on hydrolysis)

• This titration is done in cold because ${\mathrm{Fe}}^{2+}$ is oxidised to ${\mathrm{Fe}}^{3+}$ ion by oxygen of air at high temperature.

Chemical principles involved in the qualitative salt analysis:

A. Identification of Acidic Radicals anions:

Most of these salts are acted upon by dilute ${\mathrm{H}}_2{\mathrm{SO}}_4$ or dilute $\mathrm{HCl}$ and concentrated ${\mathrm{H}}_2{\mathrm{SO}}_4$. In each case, a gas is liberated which is characteristic of the particular acidic radical. There are some acidic radicals which are not decomposed either by dilute ${\mathrm{H}}_2{\mathrm{SO}}_4$ or concentrated ${\mathrm{H}}_2{\mathrm{SO}}_4$. Hence, for the identification of the acidic radicals, the following scheme is followed:

Group I: This group consists of radicals which are detected by dilute ${\mathrm{H}}_2{\mathrm{SO}}_4$ or dilute $\mathrm{HCl}$. These are (i) carbonate, (ii) sulphite, (iii) sulphide, (iv) nitrite and (v) acetate.

Group II: This group consists of radicals which are detected by concentrated ${\mathrm{H}}_2{\mathrm{SO}}_4$. These are (i) chloride, (ii) bromide, (iii) iodide, (iv) nitrate and (v) oxalate.

Group III: The radicals which do not give any characteristic gas with dilute and concentrated ${\mathrm{H}}_2{\mathrm{SO}}_4$. These are (i) sulphate, (ii) phosphate, (iii) borate and (iv) fluoride.

Reactions with explanations

(i) Carbonate

The carbonates are decomposed with dilute $\mathrm{HCl}$ or dilute ${\mathrm{H}}_2{\mathrm{SO}}_4$ with the evolution of carbon dioxide gas. When this gas is passed through lime water, the lime water turns milky with the formation of calcium carbonate.

$${\mathrm{Na}}_2{\mathrm{CO}}_3+{\mathrm{H}}_2{\mathrm{SO}}_4⟶{\mathrm{Na}}_2{\mathrm{SO}}_4+{\mathrm{H}}_2\mathrm{O}+{\mathrm{CO}}_2$$

$$\underset{\text{ Lime water }}{\mathrm{Ca}(\mathrm{OH}{)}_2}+{\mathrm{CO}}_2\underset{\text{ White ppt. }}{⟶{\mathrm{CaCO}}_3}+{\mathrm{H}}_2\mathrm{O}$$

However, if the ${\mathrm{CO}}_2$ gas is passed in excess, the milky solution becomes colourless due to the formation of soluble calcium bicarbonate.

$$\underset{\text{ White ppt.Soluble }}{{\mathrm{CaCO}}_3+{\mathrm{H}}_2}\mathrm{O}+{\mathrm{CO}}_2⟶\mathrm{Ca}{\left({\mathrm{HCO}}_3\right)}_2$$

(ii) Sulphite

Sulphite ion with dilute ${\mathrm{H}}_2{\mathrm{SO}}_4$ gives out sulphur dioxide gas which possesses suffocating smell of burning sulphur. When acidified potassium dichromate paper is exposed to the gas, it attains green colour due to the formation of chromic sulphate.

$$\begin{array}{rl}& {\mathrm{Na}}_2{\mathrm{SO}}_3+{\mathrm{H}}_2{\mathrm{SO}}_4⟶{\mathrm{Na}}_2{\mathrm{SO}}_4+{\mathrm{H}}_2\mathrm{O}+{\mathrm{SO}}_2\\ & {\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7+{\mathrm{H}}_2{\mathrm{SO}}_4+3{\mathrm{SO}}_2⟶{\mathrm{K}}_2{\mathrm{SO}}_4+{\mathrm{Cr}}_2{\left({\mathrm{SO}}_4\right)}_3+{\mathrm{H}}_2\mathrm{O}\end{array}$$

(iii) Sulphide

Dilute ${\mathrm{H}}_2{\mathrm{SO}}_4$ decomposes a sulphide salt to form ${\mathrm{H}}_2\mathrm{S}$ gas which smells like rotten eggs.

$${\mathrm{Na}}_2\mathrm{S}+{\mathrm{H}}_2{\mathrm{SO}}_4⟶{\mathrm{Na}}_2{\mathrm{SO}}_4+{\mathrm{H}}_2\mathrm{S}$$

On exposure to this gas, the lead acetate paper turns black due to the formation of lead sulphide.

$$\mathrm{Pb}{\left({\mathrm{CH}}_3\mathrm{COO}\right)}_2+{\mathrm{H}}_2\mathrm{S}\underset{\text{ Black ppt. }}{\overset{⟶}{\to }}\mathrm{PbS}+2{\mathrm{CH}}_3\mathrm{COOH}$$

(iv) Nitrite

When a nitrite ion is treated with dilute ${\mathrm{H}}_2\mathrm{SO}4$, it yields a colourless nitric oxide gas which on contact with oxygen of the air becomes brown due to the formation of nitrogen dioxide.

$$\begin{array}{rl}& 2{\mathrm{NaNO}}_2+{\mathrm{H}}_2{\mathrm{SO}}_4⟶{\mathrm{Na}}_2{\mathrm{SO}}_4+2\underset{\text{ Nitrous acid }}{{\mathrm{HNO}}_2}\\ & 3{\mathrm{HNO}}_2⟶{\mathrm{H}}_2\mathrm{O}+2\mathrm{NO}+{\mathrm{HNO}}_3\\ & 2\mathrm{NO}+{\mathrm{O}}_2⟶\underset{\text{ Brown coloured gas }}{2{\mathrm{NO}}_2}\end{array}$$

(a) On passing the gas in dilute ${\mathrm{FeSO}}_4$ solution, brown coloured complex salt is formed.

$${\mathrm{FeSO}}_4\cdot 7{\mathrm{H}}_2\mathrm{O}+\mathrm{NO}⟶\underset{\text{ Brown coloured }}{\left[\mathrm{Fe}{\left({\mathrm{H}}_2\mathrm{O}\right)}_5\mathrm{NO}\right]{\mathrm{SO}}_4+2{\mathrm{H}}_2\mathrm{O}}$$

(b) When a mixture of iodide and nitrite is acted upon by dilute ${\mathrm{H}}_2{\mathrm{SO}}_4$, the iodide is decomposed giving violet vapours of iodine

$$2{\mathrm{NaNO}}_2+{\mathrm{H}}_2{\mathrm{SO}}_4⟶{\mathrm{Na}}_2{\mathrm{SO}}_4+2{\mathrm{HNO}}_2$$

$$\begin{array}{rl}& 2\mathrm{KI}+{\mathrm{H}}_2{\mathrm{SO}}_4⟶{\mathrm{K}}_2{\mathrm{SO}}_4+2\mathrm{HI}\\ & 2{\mathrm{HNO}}_2+2\mathrm{HI}⟶\underset{\text{ Violet vapours }}{2{\mathrm{H}}_2\mathrm{O}+{\mathrm{I}}_2+2\mathrm{NO}}\end{array}$$

(c) Starch iodide paper is turned blue due to the liberation of iodine form iodide by nitrous acid which gives blue colour with starch.

$$\begin{array}{rl}& 2{\mathrm{NaNO}}_2+{\mathrm{H}}_2{\mathrm{SO}}_4⟶{\mathrm{Na}}_2{\mathrm{SO}}_4+2{\mathrm{HNO}}_2\\ & 2\mathrm{KI}+2{\mathrm{CH}}_3\mathrm{COOH}+2{\mathrm{HNO}}_2⟶2{\mathrm{CH}}_3\mathrm{COOK}+2\mathrm{NO}+2{\mathrm{H}}_2\mathrm{O}+{\mathrm{I}}_2\\ & {\mathrm{I}}_2+\text{ Starch }⟶\text{ Blue colour }\end{array}$$

(v) Acetate

Acetates when heated with dilute ${\mathrm{H}}_2{\mathrm{SO}}_4$ decompose to give acetic acid vapours which possess characteristic smell of vinegar.

$$2{\mathrm{CH}}_3\mathrm{COONa}+{\mathrm{H}}_2{\mathrm{SO}}_4⟶2{\mathrm{CH}}_3\mathrm{COOH}+{\mathrm{Na}}_2{\mathrm{SO}}_4$$

(a) All acetates are soluble in water. On addition of neutral ${\mathrm{FeCl}}_3$ solution to the
solution of an acetate, blood red coluration develops due to the formation of ferric acetate.

$${\mathrm{FeCl}}_3+3{\mathrm{CH}}_3\mathrm{COONa}⟶\underset{\begin{array}{c}\text{ Blood red colour }\end{array}}{{\left({\mathrm{CH}}_3\mathrm{COO}\right)}_3\mathrm{Fe}+3\mathrm{NaCl}}$$

Note : (i) The ferric chloride solution supplied in the laboratory is always acidic containing $\mathrm{HCl}$. It is made neutral by the addition of dilute solution of ${\mathrm{NH}}_4\mathrm{OH}$ drop by drop with constant stirring till the precipitate formed does not dissolve. At this stage filter the solution. The filtrate is called neutral ferric chloride solution.

(ii) Before testing acetate in the aqueous solution of a salt or a mixture, it must be made sure that the solution does not contain the following ions which also combine with ${\mathrm{Fe}}^{3+}$ ions.

(i) ${\mathrm{CO}}_3{}^{2-}$, (ii) ${\mathrm{SO}}_3{}^{2-}$, (iii) ${\mathrm{PO}}_4^{3-}$, (iv) I-

These ions can be removed by addition of ${\mathrm{AgNO}}_3$ solution and only after the removal of these ions, the test of acetate should be performed by neutral ferric chloride solution.

(b) Acetates are also decomposed with oxalic acid and give off acetic acid.

$$2{\mathrm{CH}}_3\mathrm{COONa}+{\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{O}}_4⟶{\mathrm{Na}}_2{\mathrm{C}}_2{\mathrm{O}}_4+2{\mathrm{CH}}_3\mathrm{COOH}$$

Reactions with explanations

(i) Chloride

Colourless pungent fumes of hydrogen chloride are evolved on heating the sodium chloride with conc. ${\mathrm{H}}_2{\mathrm{SO}}_4$.

$$\mathrm{NaCl}+{\mathrm{H}}_2{\mathrm{SO}}_4⟶{\mathrm{NaHSO}}_4+\mathrm{HCl}$$

a) Yellowish-green gas of chlorine with suffocating odour is evolved when the sodium chloride mixed
with manganese dioixed is heated with conc. ${\mathrm{H}}_2{\mathrm{SO}}_4$.

$$\begin{array}{rl}& \mathrm{NaCl}+{\mathrm{H}}_2{\mathrm{SO}}_4⟶{\mathrm{NaHSO}}_4+\mathrm{HCl}\\ & {\mathrm{MnO}}_2+4\mathrm{HCl}⟶{\mathrm{MnCl}}_2+2{\mathrm{H}}_2\mathrm{O}+{\mathrm{Cl}}_2\end{array}$$

b) The gas evolved by heating chloride with sulphuric acid forms white fumes of ammonium chloride with ${\mathrm{NH}}_4\mathrm{OH}$.

$${\mathrm{NH}}_4\mathrm{OH}+\mathrm{HCl}⟶\underset{\text{ White fumes }}{{\mathrm{NH}}_4\mathrm{Cl}+{\mathrm{H}}_2\mathrm{O}}$$

c) The gas evolved by heating chloride with ${\mathrm{H}}_2{\mathrm{SO}}_4$ forms a curdy precipitate of silver chloride with silver nitrate solution.

$${\mathrm{AgNO}}_3+\mathrm{HCl}⟶\underset{\mathrm{ppt}}{\mathrm{AgCl}}+{\mathrm{HNO}}_3$$

Note: The curdy precipitate dissolve in ammonium hydroxide by forming a complex salt.

$$\mathrm{AgCl}+2{\mathrm{NH}}_4\mathrm{OH}⟶\left[\mathrm{Ag}{\left({\mathrm{NH}}_3\right)}_2\right]\mathrm{Cl}+2{\mathrm{H}}_2\mathrm{O}$$

When the solution having the silver complex is acidified with dilute nitric acid, a white precipitate of silver chloride is again formed.

$$\left[\mathrm{Ag}{\left({\mathrm{NH}}_3\right)}_2\right]\mathrm{Cl}+2{\mathrm{HNO}}_3⟶\mathrm{AgCl}+2{\mathrm{NH}}_4{\mathrm{NO}}_3$$

d) Chromyl chloride test : When sodium chloride is heated with conc. ${\mathrm{H}}_2{\mathrm{SO}}_4$ in presence of ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$, deep red vapours of chromyl chloride are evolved.

$$\begin{array}{rl}& \mathrm{NaCl}+{\mathrm{H}}_2{\mathrm{SO}}_4⟶{\mathrm{NaHSO}}_4+\mathrm{HCl}\\ & {\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7+2{\mathrm{H}}_2{\mathrm{SO}}_4⟶{\mathrm{KHSO}}_4+2{\mathrm{CrO}}_3+{\mathrm{H}}_2\mathrm{O}\\ & {\mathrm{CrO}}_3+2\mathrm{HCl}⟶\underset{\text{ Chromyl chloride }}{{\mathrm{CrO}}_2{\mathrm{Cl}}_2}+{\mathrm{H}}_2\mathrm{O}\end{array}$$

When these vapour are passed through $\mathrm{NaOH}$ solution, the solution becomes yellow due to the formation of sodium chromate.

$${\mathrm{CrO}}_2{\mathrm{Cl}}_2+4\mathrm{NaOH}⟶\underset{\text{ Yellow colour }}{{\mathrm{Na}}_2{\mathrm{CrO}}_4}+2\mathrm{NaCl}+2{\mathrm{H}}_2\mathrm{O}$$

The yellow solution is neutralised with acetic acid and on addition of lead acetate gives a yellow precipitate of lead chromate.

$${\mathrm{Na}}_2{\mathrm{CrO}}_4+\mathrm{Pb}{\left({\mathrm{CH}}_3\mathrm{COO}\right)}_2⟶\underset{\text{ Yellow ppt. }}{{\mathrm{PbCrO}}_4}+2{\mathrm{CH}}_3\mathrm{COONa}$$

Note: (i) This test is not given by the chlorides of mercury, tin, silver, lead and antimony. In such cases this test may be performed by taking the residue obtained after evaporation of sodium carbonate extract.

(ii) The chromyl chloride test is always to be performed in a dry test tube; otherwise the chromyl chloride vapours will be hydrolysed in the test tube.

$${\mathrm{CrO}}_2{\mathrm{Cl}}_2+2{\mathrm{H}}_2\mathrm{O}⟶{\mathrm{H}}_2{\mathrm{CrO}}_4+2\mathrm{HCl}$$

(iii) The test is said to be positive when all the three observations viz., orange-yellow (red) vapours of chromyl chloride, yellow solution of sodium chromate, yellow precipitate of lead chromate, are correct.

(iv) Sometimes, a white precipitate is obtained after the addition of lead acetate solution even in absence of chloride. This may be due to strong heating of the mixture with conc. ${\mathrm{H}}_2{\mathrm{SO}}_4$ when ${\mathrm{H}}_2{\mathrm{SO}}_4$ vapours are absorbed in $\mathrm{NaOH}$ solution or due to incomplete neutralisation of $\mathrm{NaOH}$ solution which reacts with lead acetate to form lead hydroxide.

(v) Bromides and iodides do not give this test.

(ii) Bromide

Reddish-brown fumes of bromine are formed when the sodium bromide is heated with conc. ${\mathrm{H}}_2{\mathrm{SO}}_4$.

$$\begin{array}{rl}& \mathrm{NaBr}+{\mathrm{H}}_2{\mathrm{SO}}_4⟶{\mathrm{NaHSO}}_4+\mathrm{HBr}\\ & 2\mathrm{HBr}+{\mathrm{H}}_2{\mathrm{SO}}_4⟶{\mathrm{Br}}_2+2{\mathrm{H}}_2\mathrm{O}+{\mathrm{SO}}_2\end{array}$$

More reddish-brown fumes of bromine are evolved when ${\mathrm{MnO}}_2$ is added.

$$\begin{array}{rl}& 2\mathrm{NaBr}+2{\mathrm{H}}_2{\mathrm{SO}}_4⟶2{\mathrm{NaHSO}}_4+2\mathrm{HBr}\\ & {\mathrm{MnO}}_2+{\mathrm{H}}_2{\mathrm{SO}}_4⟶{\mathrm{MnSO}}_4+{\mathrm{H}}_2\mathrm{O}+[0]\\ & 2\mathrm{HBr}+[0]⟶{\mathrm{H}}_2\mathrm{O}+{\mathrm{Br}}_2\\ & 2\mathrm{NaBr}+{\mathrm{MnO}}_2+3{\mathrm{H}}_2{\mathrm{SO}}_4⟶2{\mathrm{NaHSO}}_4+{\mathrm{MnSO}}_4+2{\mathrm{H}}_2\mathrm{O}+{\mathrm{Br}}_2\end{array}$$

(a) The aqueous solution of bromide gives pale yellow precipitate of silver bromide which dissolves
in excess of ${\mathrm{NH}}_4\mathrm{OH}$ forming a soluble complex. $[\mathrm{AgBr}$ is sparingly soluble in ${\mathrm{NH}}_4\mathrm{OH}$ solution.]

$$\begin{array}{rl}& \mathrm{NaBr}+{\mathrm{AgNO}}_3⟶\underset{\text{ Pale yellow ppt. }}{\mathrm{AgBr}+{\mathrm{NaNO}}_3}\\ & \mathrm{AgBr}+2{\mathrm{NH}}_4\mathrm{OH}⟶\mathrm{Ag}{\left({\mathrm{NH}}_3\right)}_2\mathrm{Br}+2{\mathrm{H}}_2\mathrm{O}\end{array}$$

(b) When the fresh salt or mixture is treated with dilute ${\mathrm{H}}_2{\mathrm{SO}}_4,{\mathrm{CHCl}}_3$ ore ${\mathrm{CCl}}_4$, and chlorine water, chlorine replaces bromine and the liberated bromine dissolves in ${\mathrm{CHCl}}_3$ or ${\mathrm{CCl}}_4$ layer giving it brown colour.

$$\begin{array}{rl}& 2\mathrm{KBr}+{\mathrm{Cl}}_2⟶2\mathrm{KCl}+{\mathrm{Br}}_2\\ & {\mathrm{Br}}_2+\text{ Chloroform }⟶\text{ Brown layer }\end{array}$$

(iii) lodide: Violet vapours of iodine are evolved on heating iodide with concentrated ${\mathrm{H}}_2{\mathrm{SO}}_4$.

$$\begin{array}{rl}& 2\mathrm{KI}+2{\mathrm{H}}_2{\mathrm{SO}}_4⟶2{\mathrm{KHSO}}_4+2\mathrm{HI}\\ & 2\mathrm{HI}+{\mathrm{H}}_2{\mathrm{SO}}_4⟶{\mathrm{I}}_2+{\mathrm{SO}}_2+2{\mathrm{H}}_2\mathrm{O}\end{array}$$

More violet vapours are evolved when ${\mathrm{MnO}}_2$ is added.

$$\begin{array}{rl}& 2\mathrm{KI}+2{\mathrm{H}}_2{\mathrm{SO}}_4⟶2{\mathrm{KHSO}}_4+2\mathrm{HI}\\ & {\mathrm{MnO}}_2+{\mathrm{H}}_2{\mathrm{SO}}_4⟶{\mathrm{MnSO}}_4+{\mathrm{H}}_2\mathrm{O}+[0]\\ & 2\mathrm{HI}+[0]⟶{\mathrm{H}}_2\mathrm{O}+{\mathrm{I}}_2\\ & 2\mathrm{KI}+{\mathrm{MnO}}_2+3{\mathrm{H}}_2{\mathrm{SO}}_4⟶2{\mathrm{KHSO}}_4+{\mathrm{MnSO}}_4+2{\mathrm{H}}_2\mathrm{O}+{\mathrm{I}}_2\end{array}$$

(a) Violet vapours with starch produce blue colour.

$${\mathrm{I}}_2+\text{ Starch }⟶\text{ Blue colour }$$

(b) Aqueous solution of the iodine gives yellow precipitate of $\mathrm{Agl}$ with silver nitrate solution which does not dissolve in ${\mathrm{NH}}_4\mathrm{OH}$.

$$\mathrm{Nal}+{\mathrm{AgNO}}_3⟶\underset{\text{ Yellow ppt. }}{\mathrm{AgI}+{\mathrm{NaNO}}_3}$$

(c) When the given substance is treated with dilute ${\mathrm{H}}_2{\mathrm{SO}}_4,{\mathrm{CHCl}}_3$ or ${\mathrm{CCl}}_4$ and chlorine water, chlorine replaces iodine which dissolves in ${\mathrm{CHCl}}_3$ or ${\mathrm{CCl}}_4$ layer giving it violet colour.

$$\begin{array}{rl}& 2\mathrm{KI}+{\mathrm{Cl}}_2⟶2\mathrm{KCI}+{\mathrm{I}}_2\\ & {\mathrm{I}}_2+{\mathrm{CHCl}}_3⟶\text{ Violet layer }\end{array}$$

Note: Excess of chlorine water should be avoided as the layer becomes colourless with the conversion of iodine into iodic acid.

$$\underline{\begin{array}{l}[{\mathrm{Cl}}_2+{\mathrm{H}}_2\mathrm{O}⟶2\mathrm{HCl}+0]\times 5\\ {\mathrm{I}}_2+{\mathrm{H}}_2\mathrm{O}+5[0]⟶2{\mathrm{HIO}}_3\\ {\mathrm{I}}_2+5{\mathrm{Cl}}_2+6{\mathrm{H}}_2\mathrm{O}⟶2{\mathrm{HIO}}_3+10\mathrm{HCl}\end{array}}$$

(vi) Nitrate:

Light brown fumes of nitrogen dioxide are evolved on heating the nitrate with concentrated ${\mathrm{H}}_2{\mathrm{SO}}_4$.

$$\begin{array}{rl}& {\mathrm{NaNO}}_3+{\mathrm{H}}_2{\mathrm{SO}}_4⟶{\mathrm{NaHSO}}_4+{\mathrm{HNO}}_3\\ & 4{\mathrm{HNO}}_3⟶2{\mathrm{H}}_2\mathrm{O}+4{\mathrm{NO}}_2+{\mathrm{O}}_2\end{array}$$

These fumes intensify when copper turnings are added.

$$\mathrm{Cu}+4{\mathrm{HNO}}_3⟶\mathrm{Cu}{\left({\mathrm{NO}}_3\right)}_2+2{\mathrm{NO}}_2+2{\mathrm{H}}_2\mathrm{O}$$

Ring Test: When the aqueous solution of the substance is treated with freshly prepared solution of ferrous sulphate and conc. ${\mathrm{H}}_2{\mathrm{SO}}_4$, a brown ring is formed on account of the formation of a complex at the junction of two liquids.

$$\begin{array}{rl}& {\mathrm{NaNO}}_3+{\mathrm{H}}_2{\mathrm{SO}}_4⟶{\mathrm{NaHSO}}_4+{\mathrm{HNO}}_3\\ & 6{\mathrm{FeSO}}_4+2{\mathrm{HNO}}_3+3{\mathrm{H}}_2{\mathrm{SO}}_4⟶⟶{\mathrm{Fe}}_2{\left({\mathrm{SO}}_4\right)}_3+4{\mathrm{H}}_2\mathrm{O}+2\mathrm{NO}\\ & \underset{\begin{array}{c}\text{ Ferrous sulphate }\end{array}}{\left[\mathrm{Fe}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]{\mathrm{SO}}_4{\mathrm{H}}_2\mathrm{O}+\mathrm{NO}}⟶\underset{\begin{array}{c}\text{ Brown ring }\end{array}}{\left[\mathrm{Fe}{\left({\mathrm{H}}_2\mathrm{O}\right)}_5{\mathrm{NO}}_2\right]{\mathrm{SO}}_4+2{\mathrm{H}}_2\mathrm{O}}\end{array}$$

Note: (a) Ring test is not reliable in presence of nitrite, bromide and iodide.

(b) The nitrates can be tested by boiling nitrate with $\mathrm{Zn}$ or $\mathrm{Al}$ in presence of
concentrated $\mathrm{NaOH}$ solution when ammonia is evolved which can be detected by the characteristic odour.

$$\begin{array}{rl}& \mathrm{Zn}+2\mathrm{NaOH}⟶{\mathrm{Na}}_2{\mathrm{ZnO}}_2+2\mathrm{H}\\ & \mathrm{Al}+\mathrm{NaOH}+{\mathrm{H}}_2\mathrm{O}⟶{\mathrm{NaAlO}}_2+3\mathrm{H}\\ & {\mathrm{NaNO}}_3+8\mathrm{H}⟶\mathrm{NaOH}+2{\mathrm{H}}_2\mathrm{O}+{\mathrm{NH}}_3\end{array}$$

(v) Oxalate:

When oxalate is heated with conc. ${\mathrm{H}}_2{\mathrm{SO}}_4$, a mixture of ${\mathrm{CO}}^{\text{and }}{\mathrm{CO}}_2$ is given off. The $\mathrm{CO}$ burns with blue flame.

$$\begin{array}{rl}& {\mathrm{Na}}_2{\mathrm{C}}_2{\mathrm{O}}_4+{\mathrm{H}}_2{\mathrm{SO}}_4⟶{\mathrm{Na}}_2{\mathrm{SO}}_4+{\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{O}}_4\\ & {\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{O}}_4+\left[{\mathrm{H}}_2{\mathrm{SO}}_4\right]⟶\mathrm{CO}+{\mathrm{CO}}_2+{\mathrm{H}}_2\mathrm{O}+\left[{\mathrm{H}}_2{\mathrm{SO}}_4\right]\end{array}$$

GROUP III

Some anions are identified by their characteristic chemical reactions. These radicals are sulphate, borate, phosphate and fluoride.

(i) Sulphate: Dissolve a little amount of the substance (salt or mixture) and add barium chloride solution. A white precipitate insoluble in conc. ${\mathrm{HNO}}_3$ is formed.

Reactions with explanations

White precipitate of barium sulphate is obtained when soluble sulphate is treated with barium chloride solution.

$$\underset{\text{ White ppt. }}{{\mathrm{Na}}_2{\mathrm{SO}}_4}+{\mathrm{BaCl}}_2⟶2\mathrm{NaCl}+{\mathrm{BaSO}}_4$$

The white precipitate is insoluble in conc. ${\mathrm{HNO}}_3$. Certain chlorides e.g., $\mathrm{NaCl}$ and ${\mathrm{BaCl}}_2$ when present in large quantities, may form a white precipitate which dissolves on dilution with water. Silver and lead, if present, maybe preciptated as silver Chloride and lead chloride by the addition of barium chloride. To avoid it, barium nitrate may be used in place of barium chloride.

(ii) Borate: To a small quantity of the substance (salt or mixture), add a few $\mathrm{mL}$ of ethyl alcohol and conc. ${\mathrm{H}}_2{\mathrm{SO}}_4$. Stir the contents with a glass rod. Heat the test tube and bring the mouth of the test tube near the flame. The formation of green edged flame indicates the presence of borate.

Reactions with explanations

When borate is heated with ethyl alcohol and ${\mathrm{H}}_2{\mathrm{SO}}_4$, ethyl borate vapours come out which burn with green edged flame.

$$\begin{array}{rl}& 2{\mathrm{Na}}_3{\mathrm{BO}}_3+3{\mathrm{H}}_2{\mathrm{SO}}_4⟶{\mathrm{Na}}_2{\mathrm{SO}}_4+2{\mathrm{H}}_3{\mathrm{BO}}_3\\ & {\mathrm{H}}_3{\mathrm{BO}}_3+3{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}⟶{\left({\mathrm{C}}_2{\mathrm{H}}_5\right)}_3{\mathrm{BO}}_3+3{\mathrm{H}}_2\mathrm{O}\end{array}$$

In place of ethyl alcohol, methyl alcohol can also be used. This test should be performed in a test tube and not in a porcelain basin because copper or barium salts, if present, will come in contact with the flame which also give green flame.

(iii) Phosphate: Take about $0.2\mathrm{g}$ of the substance in a test tube and add $2\mathrm{mL}$ conc. ${\mathrm{HNO}}_3$. Heat and add $2\mathrm{mL}$ ammonium molybdate solution. Again head, canary yellow precipitate indicates the presence of phosphate.

Reactions with explanations

The canary yellow precipitate is due to the formation of ammonium phosphomolybdate.

$${\mathrm{Ca}}_3{\left({\mathrm{PO}}_4\right)}_2+6{\mathrm{HNO}}_3⟶3\mathrm{Ca}{\left({\mathrm{NO}}_3\right)}_2+2{\mathrm{H}}_3{\mathrm{PO}}_4$$

$${\mathrm{H}}_3{\mathrm{PO}}_4+12{\left({\mathrm{NH}}_4\right)}_2{\mathrm{MoO}}_4+21{\mathrm{HNO}}_3⟶\underset{\text{ (Canary yellow ppt.) }}{{\left({\mathrm{NH}}_4\right)}_3\cdot {\mathrm{PO}}_4\cdot 12{\mathrm{MoO}}_3+21{\mathrm{NH}}_4{\mathrm{NO}}_3+12{\mathrm{H}}_2\mathrm{O}}$$

Arsenic under similar conditions also yields a yellow precipitate of ${\left({\mathrm{NH}}_4\right)}_3{\mathrm{AsO}}_4\cdot 12{\mathrm{MoO}}_3$ (ammonium arsenomolybdate). So, in presence of As, phosphate is tested in the filtrate of second group.

(a) The precipitate of ammoinum phosphomolybdate dissolves in excess of phosphate. Thus, the reagent
(ammonium molybdate) should always be added in excess.

(b) $\mathrm{HCl}$ interferes in this test. Hence, if the test of phosphate is to be performed with
the solution containing $\mathrm{HCl}$, the solution should be boiled to remove $\mathrm{HCl}$.

(c) Reducing agents such as sulphites, sulphides, etc., interfere as they reduce $\mathrm{Mo}(\mathrm{VI})$ to molybdenum blue $({\mathrm{MO}}_3{\mathrm{O}}_8\cdot {\mathrm{xH}}_2\mathrm{O})$. The solution, therefore, turns blue. In the presence of reducing agents, the solution should be boiled with ${\mathrm{HNO}}_3$ as to oxidise them before the addition of ammonium molybdate.

Tests of various acidic radicals with sodium carbonate extract:

(i) Sulphide

(a) Take sodium carbonate extract and add few drops of $\mathrm{NaOH}$ and then freshly prepared
sodium nitroprusside solution. Appearance of violet colour indicates the presence of sulphide.

$${\mathrm{Na}}_2\mathrm{S}+\underset{\text{ Sodium nitroprusside }}{{\mathrm{Na}}_2[\mathrm{Fe}(\mathrm{CN}{)}_5\mathrm{NO}]}⟶\underset{\text{ (Violet colour) }}{{\mathrm{Na}}_4[\mathrm{Fe}(\mathrm{CN}{)}_5\mathrm{NOS}]}$$

(b) To soda extract, add lead acetate solution. A black precipitate indicates the presence of sulphide.

$${\mathrm{Na}}_2\mathrm{S}+\mathrm{Pb}{\left({\mathrm{CH}}_3\mathrm{COO}\right)}_2⟶\underset{\text{ Lead sulphide (Black) }}{\mathrm{PbS}+2{\mathrm{CH}}_3\mathrm{COONa}}$$

(c) To soda extract add cadmium carbonate. A yellow precipitate indicates the presence of sulphide.

$${\mathrm{Na}}_2\mathrm{S}+{\mathrm{CdCO}}_3⟶\underset{\text{ Cadmium sulphide(Yellow) }}{\mathrm{CdS}+{\mathrm{Na}}_2{\mathrm{CO}}_3}$$

Cation Analysis or Analysis of Basic Radicals

For the analysis of cations, first step is the prepration of the original solution. In this process, the mixture is dissolved in a suitable solvent and then analysis is carried out using this original solution.

Order of solvents to prepare the original solution is as follows:

a) Distilled water (first cold and then hot)

b) Dilute $\mathrm{HCl}$ (first cold and then hot)

c) Concentrated $\mathrm{HCl}$ (first cold and then hot)

d) Dilute ${\mathrm{HNO}}_3$ (first cold and then hot)

e) Concentrated ${\mathrm{HNO}}_3$ (first cold and then hot)

f) Aqua regia $(3\mathrm{HCl}:1{\mathrm{HNO}}_3)$

If substance is insoluble in any of the above mentioned solvents, then it is treated as insoluble.

Classification of cations into various groups:

Group	Cation	Group reagent
Zero	${\mathrm{NH}}_4{}^{+}$	$\mathrm{NaOH}$ solution
1st	${\mathrm{Ag}}^{+},{\mathrm{Pb}}^{2+},{\mathrm{Hg}}_2^{2+}$	Dilute $\mathrm{HCl}$
2nd	${\mathrm{Hg}}^{2+},{\mathrm{Pb}}^{2+},{\mathrm{Bi}}^{2+},{\mathrm{Cu}}^{2+},{\mathrm{Cd}}^{2+},{\mathrm{As}}^{3+}{\mathrm{Sb}}^{3+},{\mathrm{Sn}}^{2+}$	${\mathrm{H}}_2\mathrm{S}$ gas in acidic medium
3rd	${\mathrm{Fe}}^{3+},{\mathrm{Al}}^{3+},{\mathrm{Cr}}^{3+}$	${\mathrm{NH}}_4\mathrm{OH}$ in the presence of ${\mathrm{NH}}_4\mathrm{Cl}$
4th	${\mathrm{Zn}}^{2+},{\mathrm{Mn}}^{2+},{\mathrm{Ni}}^{2+},{\mathrm{Co}}^{2+}$	${\mathrm{H}}_2\mathrm{S}$ gas in basic medium
5th	${\mathrm{Ba}}^{2+},{\mathrm{Sr}}^{2+},{\mathrm{Ca}}^{2+}$	${\left({\mathrm{NH}}_4\right)}_2{\mathrm{CO}}_3$ in the presence of ${\mathrm{NH}}_4\mathrm{Cl}$ and ${\mathrm{NH}}_4\mathrm{OH}$
6th	${\mathrm{K}}^{+},{\mathrm{Mg}}^{2+}$	No particular reagent

GROUP I

When dil. $\mathrm{HCl}$ is added to original solution, insoluble chlorides of lead, silver and mercurous mercury are precipitated.

$$\begin{array}{rl}& \mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2+2\mathrm{HCl}⟶{\mathrm{PbCl}}_2+2{\mathrm{HNO}}_3\\ & {\mathrm{AgNO}}_3+\mathrm{HCl}⟶\mathrm{AgCl}+{\mathrm{HNO}}_3\\ & {\mathrm{Hg}}_2{\left({\mathrm{NO}}_3\right)}_2+2\mathrm{HCl}⟶{\mathrm{Hg}}_2{\mathrm{Cl}}_2+2{\mathrm{HNO}}_3\end{array}$$

${\mathrm{Pb}}^{2+}$ (lead)

(i) ${\mathrm{PbCl}}_2$ is suluble in hot water and on cooling white crystals are again formed.

(ii) The solution of ${\mathrm{PbCl}}_2$ gives a yellow precipitate with potassium chromate solution which is insoluble in acetic acid but soluble in sodium hydroxide.

$$\begin{array}{rl}& {\mathrm{PbCl}}_2+{\mathrm{K}}_2{\mathrm{CrO}}_4⟶\underset{\text{ Yellow ppt. }}{{\mathrm{PbCrO}}_4\downarrow }+2\mathrm{KCl}\\ & {\mathrm{PbCrO}}_4+4\mathrm{NaOH}⟶{\mathrm{Na}}_2{\mathrm{PbO}}_2+{\mathrm{Na}}_2{\mathrm{CrO}}_4+2{\mathrm{H}}_2\mathrm{O}\end{array}$$

(iii) The solution of ${\mathrm{PbCl}}_2$ forms a yellow precipitate with potassium iodide solution

$${\mathrm{PbCl}}_2+2\mathrm{KI}⟶\underset{\text{ yellow ppt. }}{{\mathrm{PbI}}_2\downarrow }+2\mathrm{KCl}$$

(iv) White precipitate of lead sulphate is formed with dilute ${\mathrm{H}}_2{\mathrm{SO}}_4$. The precipitate is soluble in ammonium acetate.

$$\begin{array}{rl}& {\mathrm{PbCl}}_2+{\mathrm{H}}_2{\mathrm{SO}}_4⟶{\mathrm{PbSO}}_4\downarrow +2\mathrm{HCl}\\ & {\mathrm{PbSO}}_4+2{\mathrm{CH}}_3{\mathrm{COONH}}_4⟶\mathrm{Pb}{\left({\mathrm{CH}}_3\mathrm{COO}\right)}_2+{\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4\end{array}$$

${\mathrm{Ag}}^{+}$(silver)

(i) $\mathrm{AgCl}$ dissolves in ammonium hydroxide.

$$\mathrm{AgCl}+2{\mathrm{NH}}_4\mathrm{OH}⟶\underset{\text{ Diammine silver (I) chloride }}{\mathrm{Ag}{\left({\mathrm{NH}}_3\right)}_2\mathrm{Cl}+2{\mathrm{H}}_2\mathrm{O}}$$

(ii) On adding dilute ${\mathrm{HNO}}_3$ to the above solution, white precipitate is again obtained.

$$\mathrm{Ag}{\left({\mathrm{NH}}_3\right)}_2\mathrm{Cl}+2{\mathrm{HNO}}_3⟶\underset{\text{ White ppt. }}{\mathrm{AgCl}}+2{\mathrm{NH}}_4{\mathrm{NO}}_3$$

(iii) On adding $\mathrm{KI}$ to the complex solution, yellow precipitate is obtained.

$$\mathrm{Ag}{\left({\mathrm{NH}}_3\right)}_2\mathrm{Cl}+\mathrm{KI}⟶\underset{\text{ Yellow ppt. }}{\mathrm{AgI}+\mathrm{KCl}+2{\mathrm{NH}}_3}$$

${\mathrm{Hg}}_2{}^{2+}$ (mercurous)

(i) ${\mathrm{Hg}}_2{\mathrm{Cl}}_2$ turns black with ${\mathrm{NH}}_4\mathrm{OH}$.

$${\mathrm{Hg}}_2{\mathrm{Cl}}_2+2{\mathrm{NH}}_4\mathrm{OH}⟶\underset{\text{ Black }}{\mathrm{Hg}+\mathrm{Hg}\left({\mathrm{NH}}_2\right)\mathrm{Cl}}+{\mathrm{NH}}_4\mathrm{Cl}+2{\mathrm{H}}_2\mathrm{O}$$

(ii) The black residue dissolves in aqua-regia forming mercuric chloride.

$$\begin{array}{rl}& 3\mathrm{HCl}+{\mathrm{HNO}}_3⟶\mathrm{NOCl}+2{\mathrm{H}}_2\mathrm{O}+2\mathrm{Cl}\\ & 2\mathrm{Hg}\left({\mathrm{NH}}_2\right)\mathrm{Cl}+6\mathrm{Cl}⟶2{\mathrm{HgCl}}_2+4\mathrm{HCl}+{\mathrm{N}}_2\\ & \mathrm{Hg}+2\mathrm{Cl}⟶{\mathrm{HgCl}}_2\end{array}$$

(iii) The solution of ${\mathrm{HgCl}}_2$ forms white or slate-coloured precipitate with stannous chloride.

$$2{\mathrm{HgCl}}_2+{\mathrm{SnCl}}_2⟶\underset{\text{ White ppt. }}{{\mathrm{Hg}}_2{\mathrm{Cl}}_2}+{\mathrm{SnCl}}_4$$

$${\mathrm{Hg}}_2{\mathrm{Cl}}_2+{\mathrm{SnCl}}_2⟶\underset{\text{ Grey ppt. }}{2\mathrm{Hg}+{\mathrm{SnCl}}_4}$$

(iv) The solution of ${\mathrm{HgCl}}_2$ with copper turning forms a grey deposit.

$${\mathrm{HgCl}}_2+\mathrm{Cu}⟶\underset{\text{ Grey ppt. }}{\mathrm{Hg}}+{\mathrm{CuCl}}_2$$

GROUP II

When hydrogen sulphide is passed in acidified solution, the radicals of second group are precipitated as sulphides. The precipitate is treated with yellow ammonium sulphide. The sulphides of group IIB are first oxidised to higher sulphides which then dissolve to form thio-compounds.

$$\begin{array}{rl}& {\mathrm{As}}_2{\mathrm{S}}_3+2{\left({\mathrm{NH}}_4\right)}_2{\mathrm{S}}_2⟶2{\left({\mathrm{NH}}_4\right)}_2\mathrm{S}+{\mathrm{As}}_2{\mathrm{S}}_5\\ & {\mathrm{Sb}}_2{\mathrm{S}}_3+2{\left({\mathrm{NH}}_4\right)}_2{\mathrm{S}}_2⟶2{\left({\mathrm{NH}}_4\right)}_2\mathrm{S}+{\mathrm{Sb}}_2{\mathrm{S}}_5\\ & \mathrm{SnS}+{\left({\mathrm{NH}}_4\right)}_2{\mathrm{S}}_2⟶{\left({\mathrm{NH}}_4\right)}_2\mathrm{S}+{\mathrm{SnS}}_2\end{array}$$

In case, the precipitate does not dissolve in yellow ammonium sulphide, it may be either $\mathrm{HgS}$ or $\mathrm{PbS}$ or ${\mathrm{Bi}}_2{\mathrm{S}}_3$ or CuS or Cds. The precipitate is heated with dilute ${\mathrm{HNO}}_3$. Except $\mathrm{HgS}$, all other sulphides of IIA are soluble.

$$\begin{array}{rl}& 3\mathrm{PbS}+8{\mathrm{HNO}}_3⟶3\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2+2\mathrm{NO}+3\mathrm{S}+4{\mathrm{H}}_2\mathrm{O}\\ & {\mathrm{Bi}}_2{\mathrm{S}}_3+8{\mathrm{HNO}}_3⟶2\mathrm{Bi}{\left({\mathrm{NO}}_3\right)}_3+2\mathrm{NO}+3\mathrm{S}+4{\mathrm{H}}_2\mathrm{O}\\ & 3\mathrm{CuS}+8{\mathrm{HNO}}_3⟶3\mathrm{Cu}{\left({\mathrm{NO}}_3\right)}_2+2\mathrm{NO}+3\mathrm{S}+4{\mathrm{H}}_2\mathrm{O}\\ & 3\mathrm{CdS}+8{\mathrm{HNO}}_3⟶3\mathrm{Cd}{\left({\mathrm{NO}}_3\right)}_2+2\mathrm{NO}+3\mathrm{S}+4{\mathrm{H}}_2\mathrm{O}\end{array}$$

${\mathrm{Hg}}^{2+}$ (mercuric)

$\mathrm{HgS}$ is dissolved in aqua-regia.

$$3\mathrm{HgS}+2{\mathrm{HNO}}_3+6\mathrm{HCl}⟶3{\mathrm{HgCl}}_2+3\mathrm{S}+2\mathrm{NO}+4{\mathrm{H}}_2\mathrm{O}$$

The solution is divided into two parts

Part I: Stannous chloride solution reduces ${\mathrm{HgCl}}_2$ first into white ${\mathrm{Hg}}_2{\mathrm{Cl}}_2$ and then to grey metallic mercury.

Part II: Copper displaces $\mathrm{Hg}$ from ${\mathrm{HgCl}}_2$ which gets coated on copper turnings as a shining deposit.

${\mathrm{Pb}}^{2+}$ (lead)

In case the sulphide dissolves in dilute ${\mathrm{HNO}}_3$, a small part of the solution is taken. Dilute${\mathrm{H}}_2{\mathrm{SO}}_4$ is added. If lead is present, a white precipitate of lead sulphate appears.

$$\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2+{\mathrm{H}}_2{\mathrm{SO}}_4⟶\underset{\text{ While ppt. }}{{\mathrm{PbSO}}_4}+2{\mathrm{HNO}}_3$$

In absence of lead, the remaining solution is made alkaline by the addition of excess of ${\mathrm{NH}}_4\mathrm{OH}$. Bismuth forms a white precipitate of $\mathrm{Bi}(\mathrm{OH}{)}_3$, copper forms a deep blue coloured solution while cadmium forms a colourless soluble complex,

$$\text{Double subscripts: use braces to clarify}$$

${\mathrm{Bi}}^{3+}$ (bismuth)

The precipitate dissolves in dilute $\mathrm{HCl}$.

$$\mathrm{Bi}(\mathrm{OH}{)}_3+3\mathrm{HCl}⟶{\mathrm{BiCl}}_3+3{\mathrm{H}}_2\mathrm{O}$$

Part I: Addition of excess of water to ${\mathrm{BiCl}}_3$ solution gives a while precipitate due to hydrolysis.

$${\mathrm{BiCl}}_3+{\mathrm{H}}_2\mathrm{O}⟶\underset{\begin{array}{c}\text{ Bismuth oxychloride }\\ \text{ (White ppt.) }\end{array}}{\mathrm{BiOCl}+2\mathrm{HCl}}$$

Part II: The solution of ${\mathrm{BiCl}}_3$ is treated with sodium stannite solution when a black precipitate of metallic bismuth is formed.

$$2{\mathrm{BiCl}}_3+\underset{\text{ Sod. Stannite }}{3{\mathrm{Na}}_2{\mathrm{SnO}}_2}+6\mathrm{NaOH}⟶\underset{\text{ Sod. Stannate }}{3{\mathrm{Na}}_2{\mathrm{SnO}}_3}+2\mathrm{Bi}+6\mathrm{NaCl}+3{\mathrm{H}}_2\mathrm{O}$$

${\mathrm{Cu}}^{2+}$ (copper)

Blue coloured solution is acidified with acetic acid. When potassium ferricyanide is added, a chocolate coloured precipitate is formed.

$$\begin{array}{rl}& \mathrm{Cu}{\left({\mathrm{NH}}_3\right)}_4{\left({\mathrm{NO}}_3\right)}_2+4{\mathrm{CH}}_3\mathrm{COOH}⟶\mathrm{Cu}{\left({\mathrm{NO}}_3\right)}_2+4{\mathrm{CH}}_3{\mathrm{COONH}}_4\\ & 2\mathrm{Cu}{\left({\mathrm{NO}}_3\right)}_2+{\mathrm{K}}_4[\mathrm{Fe}(\mathrm{CN}{)}_6]⟶\underset{\begin{array}{c}\text{ Chocolate ppt. }\end{array}}{{\mathrm{Cu}}_2[\mathrm{Fe}(\mathrm{CN}{)}_6]}+4{\mathrm{KNO}}_3\end{array}$$

${\mathrm{Cd}}^{2+}$ (cadmium)

${\mathrm{H}}_2\mathrm{S}$ is passed through colourless solution. The appearance of yellow precipitate confirms the presence of cadmium.

$$\mathrm{Cd}{\left({\mathrm{NH}}_3\right)}_4{\left({\mathrm{NO}}_3\right)}_2+{\mathrm{H}}_2\mathrm{S}⟶\underset{\text{ Yellow ppt. }}{\mathrm{CdS}+2{\mathrm{NH}}_4}{\mathrm{NO}}_3+2{\mathrm{NH}}_3$$

GROUP IIB

In case the precipitate dissolves in yellow ammonium sulphide, the tests of the radicals arsenic, antimony and tin are performed. The sulphide is treated with concentrated hydrochloric acid. Antimony and tin sulphides dissolve while arsenic sulphide remains insoluble.

${\mathrm{As}}^{3+}$ (arsenic)

The insoluble sulphide is treated with concentrated nitric acid which is then treated with ammonium molybdate. Yellow precipitate of ammonium arsenomolybdate is formed.

$$\begin{array}{rl}& {\mathrm{As}}_2{\mathrm{S}}_5+10{\mathrm{HNO}}_3⟶\underset{\text{ Arsenic acid }}{2{\mathrm{H}}_3{\mathrm{AsO}}_4}+10{\mathrm{NO}}_2+2{\mathrm{H}}_2\mathrm{O}+5\mathrm{S}\\ & \\ & {\mathrm{H}}_3{\mathrm{AsO}}_4+12{\left({\mathrm{NH}}_4\right)}_2{\mathrm{MoO}}_4+21{\mathrm{HNO}}_3⟶\underset{\text{Yellow ppt.}}{{\left({\mathrm{NH}}_4\right)}_3{\mathrm{AsO}}_{4}12{\mathrm{MoO}}_3}+21{\mathrm{NH}}_4{\mathrm{NO}}_3+12{\mathrm{H}}_2\mathrm{O}\end{array}$$

${\mathrm{Sn}}^{2+}$ or ${\mathrm{Sn}}^{4+}$ (tin)

Solution of sulphide in concentrated $\mathrm{HCl}$ is reduced with iron fillings or granulated zinc.

$$\begin{array}{rl}& {\mathrm{SnS}}_2+4\mathrm{HCl}⟶{\mathrm{SnCl}}_4+2{\mathrm{H}}_2\mathrm{S}\\ & {\mathrm{SnCl}}_4+\mathrm{Fe}⟶{\mathrm{SnCl}}_2+{\mathrm{FeCl}}_2\end{array}$$

${\mathrm{HgCl}}_2$ solution is added to above solution which gives first a white precipitate that turns to grey.

$$\begin{array}{rl}& 2{\mathrm{HgCl}}_2+{\mathrm{SnCl}}_2⟶\underset{\text{ White ppt. }}{\mathrm{HgCl}}+{\mathrm{SnCl}}_4\\ & {\mathrm{Hg}}_2{\mathrm{Cl}}_2+{\mathrm{SnCl}}_2⟶\underset{\text{ Grey }}{2\mathrm{Hg}}+{\mathrm{SnCl}}_4\end{array}$$

${\mathrm{Sb}}^{3+}$ (antimony)

Filtrate of sulphide in concentrated $\mathrm{HCl}$ is divided into two parts:

Part I: On dilution with excess of water, a white precipitate of antimony oxychloride is obtained.

$${\mathrm{SbCl}}_3+{\mathrm{H}}_2\mathrm{O}⟶\underset{\text{ White ppt. }}{\mathrm{SbOCl}}+2\mathrm{HCl}$$

Part II: ${\mathrm{H}}_2\mathrm{S}$ is circulated. Orange precipitate is formed.

$$2{\mathrm{SbCl}}_3+3{\mathrm{H}}_2\mathrm{S}⟶\underset{\text{ Orange ppt. }}{{\mathrm{Sb}}_2{\mathrm{S}}_3+6\mathrm{HCl}}$$

GROUP III

Hydroxides are precipitated on addition of excess of ammonium hydroxide in presence of ammonium chloride.

$$\begin{array}{rl}& {\mathrm{AlCl}}_3+3{\mathrm{NH}}_4\mathrm{OH}⟶\underset{\text{ Gelatinous ppt. }}{\mathrm{Al}(\mathrm{OH}{)}_3+3}{\mathrm{NH}}_4\mathrm{Cl}\\ & {\mathrm{CrCl}}_3+3{\mathrm{NH}}_4\mathrm{OH}⟶\underset{\text{ Green ppt. }}{\mathrm{Cr}(\mathrm{OH}{)}_3}+3{\mathrm{NH}}_4\mathrm{Cl}\\ & {\mathrm{FeCl}}_3+2{\mathrm{NH}}_4\mathrm{OH}⟶\underset{\text{ Brownish red ppt. }}{\mathrm{Fe}(\mathrm{OH}{)}_3+3}{\mathrm{NH}}_4\mathrm{Cl}\end{array}$$

${\mathrm{Fe}}^{3+}$ (iron)

The brownish red precipitate dissolves in dilute $\mathrm{HCl}$. The solution is divided into two parts:

Part I: ${\mathrm{K}}_4[\mathrm{Fe}(\mathrm{CN}{)}_6]$ solution is added which forms deep blue solution or precipitate.

$$\begin{array}{rl}& \mathrm{Fe}(\mathrm{OH}{)}_3+3\mathrm{HCl}⟶{\mathrm{FeCl}}_3+3{\mathrm{H}}_2\mathrm{O}\\ & 4{\mathrm{FeCl}}_3+3{\mathrm{K}}_4[\mathrm{Fe}(\mathrm{CN}{)}_6]⟶\underset{\text{ Prussian blue }}{{\mathrm{Fe}}_4{[\mathrm{Fe}(\mathrm{CN}{)}_6]}_3}+12\mathrm{KCl}\end{array}$$

Part II: Addition of potassium thiocyanate solution gives a blood red colouration.

$${\mathrm{FeCl}}_3+3\mathrm{KCNS}⟶\underset{\begin{array}{c}\text{ Blood red colour }\end{array}}{\mathrm{Fe}(\mathrm{CNS}{)}_3+3\mathrm{KCl}}$$

${\mathrm{Cr}}^{3+}$ (chromium)

The green precipitate is fused with fusion mixture $({\mathrm{Na}}_2{\mathrm{CO}}_3+{\mathrm{KNO}}_3)$. The fused product is extracted with water or the precipitate is heated with $\mathrm{NaOH}$ and bromine water.

$$2\mathrm{Cr}(\mathrm{OH}{)}_3+3{\mathrm{KNO}}_3+2{\mathrm{Na}}_2{\mathrm{CO}}_3⟶2{\mathrm{Na}}_2{\mathrm{CrO}}_4+3{\mathrm{KNO}}_2+2{\mathrm{CO}}_2+3{\mathrm{H}}_2\mathrm{O}$$

or

$$2\mathrm{NaOH}+{\mathrm{Br}}_2⟶\mathrm{NaBrO}+\mathrm{NaBr}+{\mathrm{H}}_2\mathrm{O}$$

$$\begin{array}{rl}& \mathrm{NaBrO}⟶\mathrm{NaBr}+[0]\\ & 2\mathrm{Cr}(\mathrm{OH}{)}_3+4\mathrm{NaOH}+3[0]⟶2{\mathrm{Na}}_2{\mathrm{CrO}}_4+5{\mathrm{H}}_2\mathrm{O}\end{array}$$

The solution thus obtaind contains sodium chromate. The solution is acidified with acetic acid and treated with lead acelate solution. A yellow precipitate appears.

$${\mathrm{Na}}_2{\mathrm{CrO}}_4+\mathrm{Pb}{\left({\mathrm{CH}}_3\mathrm{COO}\right)}_2⟶\underset{\text{ Yellow ppt. }}{{\mathrm{PbCrO}}_4}+2{\mathrm{CH}}_3\mathrm{COONa}$$

${\mathrm{Al}}^{3+}$ (aluminium)

The gelatinous precipitate dissolves in $\mathrm{NaOH}$.

$$\mathrm{Al}(\mathrm{OH}{)}_3+\mathrm{NaOH}⟶\underset{\text{ Soluble }}{{\mathrm{NaAlO}}_2}+2{\mathrm{H}}_2\mathrm{O}$$

The solution is boiled with ammonium chloride when $\mathrm{Al}(\mathrm{OH}{)}_3$ is again formed.

$${\mathrm{NaAlO}}_2+{\mathrm{NH}}_4\mathrm{Cl}+{\mathrm{H}}_2\mathrm{O}⟶\mathrm{Al}(\mathrm{OH}{)}_3+\mathrm{NaCl}+{\mathrm{NH}}_3$$

GROUP IV

On passing ${\mathrm{H}}_2\mathrm{S}$ through the filtrate of the third group, sulphides of fourth group are precipitated. NiS and CoS are black and insoluble in concentrated $\mathrm{HCl}$ while MnS (buff coloured), ZnS (colourless) are soluble in conc. $\mathrm{HCl}$.

${\mathrm{Zn}}^{2+}$ (zinc)

The sulphide dissolves in $\mathrm{HCl}$.

$$\mathrm{ZnS}+2\mathrm{HCl}⟶{\mathrm{ZnCl}}_2+{\mathrm{H}}_2\mathrm{S}$$

When the solution is treated with $\mathrm{NaOH}$, first a white precipitate appears which dissolves in excess of $\mathrm{NaOH}$.

$$\begin{array}{rl}& {\mathrm{ZnCl}}_2+2\mathrm{NaOH}⟶\mathrm{Zn}(\mathrm{OH}{)}_2+2\mathrm{NaCl}\\ & \mathrm{Zn}(\mathrm{OH}{)}_2+2\mathrm{NaOH}⟶\underset{\text{(Soluble)}}{{\mathrm{NaZnO}}_2}+2{\mathrm{H}}_{2}0\end{array}$$

On passing ${\mathrm{H}}_2\mathrm{S}$, white precipitate of zinc sulphide is formed.

$${\mathrm{Na}}_2{\mathrm{ZnO}}_2+{\mathrm{H}}_2\mathrm{S}⟶\underset{\text{ White ppt. }}{\mathrm{ZnS}}+2\mathrm{NaOH}$$

${\mathrm{Mn}}^{2+}$ (manganese)

Manganese sulphide dissolves in $\mathrm{HCl}$.

$$\mathrm{MnS}+2\mathrm{HCl}⟶{\mathrm{MnCl}}_2+{\mathrm{H}}_2\mathrm{S}$$

On heating the solution with $\mathrm{NaOH}$ and ${\mathrm{Br}}_2$ water, manganese dioxide gets precipitated.

$$\begin{array}{rl}& {\mathrm{MnCl}}_2+2\mathrm{NaOH}⟶\mathrm{Mn}(\mathrm{OH}{)}_2+2\mathrm{NaCl}\\ & \mathrm{Mn}(\mathrm{OH}{)}_2+\mathrm{O}⟶{\mathrm{MnO}}_2+{\mathrm{H}}_2\mathrm{O}\end{array}$$

The precipitate is treated with excess of nitric acid and ${\mathrm{PbO}}_2$ or ${\mathrm{Pb}}_3{\mathrm{O}}_4$ (red lead). The contents are heated. The formation of permanganic acid imparts pink colour to the supernatant liquid.

$$\begin{array}{rl}& 2{\mathrm{MnO}}_2+4{\mathrm{HNO}}_3⟶2\mathrm{Mn}{\left({\mathrm{NO}}_3\right)}_2+2{\mathrm{H}}_2\mathrm{O}+{\mathrm{O}}_2\\ & 2\mathrm{Mn}{\left({\mathrm{NO}}_3\right)}_2+5{\mathrm{Pb}}_3{\mathrm{O}}_4+26{\mathrm{HNO}}_3⟶\underset{\text{Permagnetic acid (pink)}}{2{\mathrm{HMnO}}_4+15\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2}+12{\mathrm{H}}_2\mathrm{O}\end{array}$$

Note: The above test fails in presence of $\mathrm{HCl}$.

${\mathrm{Ni}}^{2+}$ (nickel) and ${\mathrm{Co}}^{2+}$ (cobalt)

The black precipitate is dissolved in aqua-regia.

$$\begin{array}{rl}& 3\mathrm{NiS}+6\mathrm{HCl}+2{\mathrm{HNO}}_3⟶3{\mathrm{NiCl}}_2+2\mathrm{NO}+3\mathrm{S}+4{\mathrm{H}}_2\mathrm{O}\\ & 3\mathrm{CoS}+6\mathrm{HCl}+2{\mathrm{HNO}}_3⟶3{\mathrm{CoCl}}_2+2\mathrm{NO}+3\mathrm{S}+4{\mathrm{H}}_2\mathrm{O}\end{array}$$

The solution is evaporated to dryness and residue extracted with dilute $\mathrm{HCl}$. It is divided into three parts:

Part I: Add ${\mathrm{NH}}_4\mathrm{OH}$ (excess) and dimethyl glyoxime. A rosy red precipitate appears.

Part II: Add ${\mathrm{CH}}_3\mathrm{COOH}$ in excess and ${\mathrm{KNO}}_2$. The appearance of yellow precipitate confirms the presence of cobalt.

$$\begin{array}{rl}& {\mathrm{KNO}}_2+{\mathrm{CH}}_3\mathrm{COOH}⟶{\mathrm{CH}}_3\mathrm{COOK}+{\mathrm{HNO}}_2\\ & {\mathrm{CoCl}}_2+2{\mathrm{KNO}}_2⟶\mathrm{Co}{\left({\mathrm{NO}}_2\right)}_2+2{\mathrm{HNO}}_2\\ & \mathrm{Co}{\left({\mathrm{NO}}_2\right)}_3+3{\mathrm{KNO}}_2⟶{\mathrm{K}}_3\left[\mathrm{Co}{\left({\mathrm{NO}}_2\right)}_6\right]\end{array}$$

GROUP V

Ammonium carbonate precipitates $V$ group radicals in the form of carbonates. These carbonates are soluble in acetic acid.

$$\begin{array}{rl}& {\mathrm{BaCO}}_3+2{\mathrm{CH}}_3\mathrm{COOH}⟶{\left({\mathrm{CH}}_3\mathrm{COO}\right)}_2\mathrm{Ba}+{\mathrm{CO}}_2+{\mathrm{H}}_2\mathrm{O}\\ & \\ & {\mathrm{SrCO}}_3+2{\mathrm{CH}}_3\mathrm{COOH}⟶{\left({\mathrm{CH}}_3\mathrm{COO}\right)}_2\mathrm{Sr}+{\mathrm{CO}}_2+{\mathrm{H}}_2\mathrm{O}\\ & \\ & {\mathrm{CaCO}}_3+2{\mathrm{CH}}_3\mathrm{COOH}⟶{\left({\mathrm{CH}}_3\mathrm{COO}\right)}_2\mathrm{Ca}+{\mathrm{CO}}_2+{\mathrm{H}}_2\mathrm{O}\end{array}$$

${\mathrm{Ba}}^{2+}$ (barium)

Barium chromate is insoluble and precipitated by the addition of potassium chromate solution.

$$\mathrm{Ba}{\left({\mathrm{CH}}_3\mathrm{COO}\right)}_2+{\mathrm{K}}_2{\mathrm{CrO}}_4⟶{\mathrm{BaCrO}}_4+2{\mathrm{CH}}_3\mathrm{COOK}$$

${\mathrm{Sr}}^{2+}$ (strontium)

Strontium sulphate is insoluble and precipitated by the addition of ammonium sulphate solution.

$$\mathrm{Sr}{\left({\mathrm{CH}}_3\mathrm{COO}\right)}_2+{\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4⟶\underset{\text{ White ppt. }}{{\mathrm{SrSO}}_4}+2{\mathrm{CH}}_3{\mathrm{COONH}}_4$$

${\mathrm{Ca}}^{2+}($ calcium $)$

Calcium oxalate is insoluble and precipitated by the addition of ammonium oxalate.

$$\mathrm{Ca}{\left({\mathrm{CH}}_3\mathrm{COO}\right)}_2+{\left({\mathrm{NH}}_4\right)}_2{\mathrm{C}}_2{\mathrm{O}}_4⟶\underset{\text{ White ppt. }}{{\mathrm{CaC}}_2{\mathrm{O}}_4}+2{\mathrm{CH}}_3{\mathrm{COONH}}_4$$

GROUP VI

In the filtrate of $\mathrm{V}$ group, some quantity of ammonium oxalate is added so as to remove $\mathrm{Ba},\mathrm{Ca}$ and $\mathrm{Sr}$ completely from the solution. The clear solution is concentrated and made alkaline with ${\mathrm{NH}}_4\mathrm{OH}$. Disodium hydrogen phosphate is now added, a white precipitate is formed.

$${\mathrm{MgCl}}_2+{\mathrm{Na}}_2{\mathrm{HPO}}_4+{\mathrm{NH}}_4\mathrm{OH}⟶\underset{\begin{array}{c}\text{ Magnesium ammonium }\\ \text{ phosphate (White ppt.) }\end{array}}{\mathrm{Mg}\left({\mathrm{NH}}_4\right){\mathrm{PO}}_4+2\mathrm{NaCl}}+{\mathrm{H}}_2\mathrm{O}$$

${\mathrm{NH}}^{4+}$ (ammonium)

The substance (salt or mixture) when heated with $\mathrm{NaOH}$ solution evolves ammonia.

$${\mathrm{NH}}_4\mathrm{Cl}+\mathrm{NaOH}⟶\mathrm{NaCl}+{\mathrm{NH}}_3+{\mathrm{H}}_2\mathrm{O}$$

When a rod dipped in $\mathrm{HCl}$ is brought on the mouth of the test-tube, white fumes of ammonium chloride are formed.

$${\mathrm{NH}}_3+\mathrm{HCl}⟶\underset{\text{ White fumes }}{{\mathrm{NH}}_4\mathrm{Cl}}$$

To the aqueous solution of ammonium salt when Nessler’s reagent is added, brown coloured precipitate is formed.

DRY TESTS

Dry tests are of great importance as these tests give clear indications of the presence of certain radicals. The following tests are performed in dry state:

(i) Flame test

(ii) Borax bead test

(iii) Microcosmic salt bead test

(i) Flame test

Some volatile salts impart characteristic colour to the non-luminous flame. The chlorides of the metals are more volatile in comparison to other salts. The metal chloride volatilises and its thermal ionisation takes place.

$$\begin{array}{rl}& \mathrm{NaCl}\rightleftharpoons {\mathrm{Na}}^{+}+{\mathrm{Cl}}^{-}\\ & {\mathrm{CaCl}}_2\rightleftharpoons {\mathrm{Ca}}^{2+}+2{\mathrm{Cl}}^{-}\end{array}$$

The cations impart a characteristic colour to the flame as these absorb energy from the flame and transmit the same as light of characteristic colour.

Procedure: The platinum wire fused in a glass rod is heated in the flame till it imparts colourless flame. This is achieved by dipping the wire in conc. $\mathrm{HCl}$ and heating it. The process is repeated till it gives a colourless flame. The tip of the wire is now dipped in conc. $\mathrm{HCl}$ and then into the substance. The tip of the wire is strongly heated in the non-luminous flame and the colour of the flame is observed by the naked eye.

	Colour of flame	Inference
1.	Golden yellow	Sodium
2.	Violet	Potassium
3.	Brick red	Calcium
4.	Crimson red	Strontium
5.	Apple green	Barium
6.	Green with a blue centre	Copper

Note: Flame test should not be performed in the presence of $\mathrm{As},\mathrm{Sb},\mathrm{Bi},\mathrm{Sn}$ and $\mathrm{Pb}$ as these radicals form alloy with platinum and hence, the wire is spoiled.

(ii) Borax bead test

On heating borax the colourless glassy bead formed consists of sodium metaborate and boric anhydride.

$${\mathrm{Na}}_2{\mathrm{B}}_4{\mathrm{O}}_7\cdot 10{\mathrm{H}}_2\mathrm{O}\stackrel{\text{ Heat }}{\to }{\mathrm{Na}}_2{\mathrm{B}}_4{\mathrm{O}}_7\stackrel{\text{ Heat }}{\to }\underset{\text{ Glassy bead }}{2{\mathrm{NaBO}}_2}+{\mathrm{B}}_2{\mathrm{O}}_3$$

On heating with a coloured salt, the glassy bead forms a coloured metaborate in oxidising flame.

$$\begin{array}{rl}& {\mathrm{CuSO}}_4⟶\mathrm{CuO}+{\mathrm{SO}}_3\\ & \mathrm{CuO}+{\mathrm{B}}_2{\mathrm{O}}_3⟶\underset{\text{ Copper metaborate (Blue) }}{\mathrm{Cu}{\left({\mathrm{BO}}_2\right)}_2}\end{array}$$

The metaborates possess different characteristic colours. The shade of the colour gives a clue regarding the presence of the radical.

However, in reducing flame the colours may be different due to different reactions. For example, copper metaborate may be reduced to colourless cuprous metaborate or to metallic copper, which appears red and opaque.

$$\begin{array}{rl}& 2\mathrm{Cu}{\left({\mathrm{BO}}_2\right)}_2+\mathrm{C}⟶2{\mathrm{CuBO}}_2+{\mathrm{B}}_2{\mathrm{O}}_3+\mathrm{CO}\\ & 2\mathrm{Cu}{\left({\mathrm{BO}}_2\right)}_2+2\mathrm{C}⟶2\mathrm{Cu}+2{\mathrm{B}}_2{\mathrm{O}}_3+2\mathrm{CO}\end{array}$$

Procedure: The free end of a platinum wire is coiled into a small loop and heated in the Bunsen flame until red hot. It is dipped in borax and gain heated, when borax swells up and then fused into a glassy bead.

The bead is moistened with water and dipped in the coloured salt again. It is now heated first in the oxdidising flame and then in the reducing flame and colours are noted in both the flames in hot and cold conditions.

(iii) Microcosmic salt bead test

This test is similar to borax bead test. When microcosmic salt is heated, a colourless transparent bead of sodium metaphosphate is formed.

$$\mathrm{Na}\left({\mathrm{NH}}_4\right){\mathrm{HPO}}_4⟶\underset{\text{ Sodium metaphosphate }}{{\mathrm{Na}}_3{\mathrm{PO}}_3+{\mathrm{NH}}_3}+{\mathrm{H}}_2\mathrm{O}$$

Sodium metaphosphate combines with metallic oxides to form orthophosphates which are usually coloured, the shade of the colour gives a clue regarding the presence of metal. Like borax bead test, colours are noted both in oxidising and reducing flames in hot and cold conditions.

Enthalpy of neutralization of strong acid and strong base

The amount of heat evolved when one gram equivalent of an acid is neutralised by one gram equivalent of a base is called enthalpy of neutralisation

Units: $\mathrm{\Delta }{\mathrm{H}}_{\text{neu }}=$ cal mol ${}^{-1}$ or $\mathrm{J}{\mathrm{mol}}^{-1}$

• For all, strong acids and strong basses,

$\mathrm{\Delta }{\mathrm{H}}_{\text{neu }}=13.70\mathrm{K}\mathrm{cal}{\mathrm{mol}}^{-1}\text{ or }57.1\mathrm{KJ}{\mathrm{mol}}^{-1}$

$$ This value is constant and given by Hess’s.

• Strong acids and strong bases are completely ionised in aqueous solutions (according to Arrhenius theory), so neutralisation process is represented as:

$$\begin{array}{rl}& \stackrel{+-}{\mathrm{HA}}+\stackrel{+-}{\mathrm{BOH}}⟶\stackrel{+-}{\mathrm{BA}}+{\mathrm{H}}_2\mathrm{O}\mathrm{\Delta }\mathrm{H}=-13.70\mathrm{K}\mathrm{cal}{\mathrm{mol}}^{-1}=-57.1\mathrm{KJ}{\mathrm{mol}}^{-1}\end{array}$$

This process is the same in all neutralisation reaction involving strong acids and strong bases, thus the heat of neutralisation is a constant value (practically).

Enthalpy of solution of ${\mathrm{CuSO}}_4$

• Formation of a solution is usually accompanied by heat changes.

• Heat of Solution: It is the head change per mole of the solute dissolved. This heat is either evolved or absorbed.

$$ Units: $\mathrm{\Delta }{\mathrm{H}}_{\text{sol }}={\mathrm{kJmol}}^{-1}$

• Heat of a solution is not a constant quantity, it depends on the amount of solute taken i.e., its value at a constant temperature will vary with concentration of the solution.

• Integral Heat of solution: It is the heat change when a known amount of solute (pure) is added to a known amount of solvent (pure) under constant temperature and pressure.

• The limiting value of integral heat of solution is known as heat of solution at infinite dilution.

• For ${\mathrm{CuSO}}_4$ : It exists as anhydrons & hydrated salt.

There are two types of heat of solution.

1. Heat of solution of anhydrous salt

2. Heat of solution of hydrous salt.

By taking the difference of the two heat of solution, heat of hydration of ${\mathrm{CuSO}}_4$ can be determined.

Enthalpy of hydration: It is the enthalpy change accompanying the hydration of one mole of an anhydrous salt by combining with specific number of moles of water.

Heat of hydration of salt $=$ Heat of solution of anhydrous salt - Heat of solution of hydrated salt.

Chemical Principles involved in the preparation of Lyophillic and Lyophobic sol:

Colloidal sols are divided into two categories:

1. Lyophillic sol- Solvent attracting

2. Lyophobic sol- Solvent repelling

• Lyophillic sols are more stable than the lyophobic sols because in lyophillic sols, particles of dispersed phase have an affinity for the particles of dispersion medium.

- Factors responsible for the stability of sols are:

1. Charge

2. Solvation of the colloidal particles by the solvent.

• Lyophillic sols are stable due to the solvation factor.

• Lyophobic sols are stable due to the charge on the colloidal particles. Charges can be positive or negative.

• Examples of lyophillic sols are Egg albumin, starch and gum.

• Examples of Iyophobic sols are freshly prepared ferric hydroxide, aluminium hyroxide and arsenic sulphide.

• Positively charged sols- hydrated ferric oxide (when ${\mathrm{FeCl}}_3$ is added to excess of hot water)

• Negatively charged sol- starch, arsenious sulphide and hydrated ferric oxide (when ${\mathrm{FeCl}}_3$ is added to $\mathrm{NaOH}$ solution).

• Lyophillic sols are directly formed by mixing and shaking the substance with a suitable liquid.

• Lyophobic sols cannot be prepared by direct mixing and shaking.

• Some methods for the colloids preparation are-

$$ a) Chemical methods

$$ b) Electrical disintegration

$$ c) Peptization.

• Sols are purified by dialysis.

Knietic study of the Reaction of iodide ion with hydrogen peroxide at room temperature

Reaction between iodide ions and hyrogen peroxide takes place in acidic medium as follows:

$$2{\mathrm{I}}^{-}(\mathrm{aq})+{\mathrm{H}}_2{\mathrm{O}}_2(\mathrm{I})+2{\mathrm{H}}^{+}(\mathrm{aq})⟶{\mathrm{I}}_2(\mathrm{g})+2{\mathrm{H}}_2\mathrm{O}(\mathrm{I});\text{ slow reaction }$$

• In this reaction mixture if calculated amount of sodium thiosulphate $\left({\mathrm{Na}}_2{\mathrm{S}}_2{\mathrm{O}}_3\right)$ is added in the presence of starch solution as an indicator.

• Liberated lodine reacts with thiosulphate ions as fast as it is formed and is reduced back to iodide ions till all the thiosulphate are oxidised to tetrathionate ions.

$${\mathrm{I}}_2(\mathrm{g})+{\mathrm{S}}_2{\mathrm{O}}_3{}^{2-}(\mathrm{aq})⟶{\mathrm{S}}_4{\mathrm{O}}_6{}^{2-}(\mathrm{aq})+2\mathrm{I}-(\mathrm{aq});\text{ fast reaction }$$

• After the complete consumption of thiosulphate ions, the concentration of iodine liberated in the reaction of hydrogen peroxide with iodide ions increases rapidly to a point where iodine forms intense blue complex with starch.

$${\mathrm{I}}_2+\text{ starch }⟶\text{ Blue complex }$$

• Time required to consume a fixed amount of the thiosulphate ions is reproducible.

• This reaction is also called clock reaction because time for the appearance of colour shows the accuracy of clock.
  
  

SOLVED EXAMPLES

Question 1.- An element ’ $A$ ’ exists as a yellow solid in standard state. It forms a volatile hydride ’ $B$ ’ which is a foul smelling gas and is extensively used in qualitative analysis of salts. When treated with oxygen. ’ $\mathrm{B}$ ’ forms an oxide ’ $\mathrm{C}$ ’ which is colourless, pungent smelling gas. This gas when passed through acidified ${\mathrm{KMnO}}_4$ solution decolourises it.

$A,B$ and $C$ are

a) $\mathrm{S},{\mathrm{H}}_2\mathrm{S}$ and ${\mathrm{SO}}_2$

b) $\mathrm{S},{\mathrm{H}}_2\mathrm{S}$ and ${\mathrm{SO}}_3$

c) $\mathrm{C},{\mathrm{CH}}_4$ and ${\mathrm{CO}}_2$

d) $\mathrm{C},{\mathrm{C}}_2{\mathrm{H}}_2$ and ${\mathrm{CO}}_2$

Show Answer

Answer:- (a)

The element ’ $A$ ’ which exists as yellow solid is Sulphur. The volatile hydride ’ $B$ ’ is ${\mathrm{H}}_2\mathrm{S}$ which has a rotten egg like smell. When treated with oxygen, the oxide formed is ${\mathrm{SO}}_2$ which has a pungent smell. The gas when passed through acidified ${\mathrm{KMnO}}_4$ decolourises it.

Question 2.- When concentrated ${\mathrm{H}}_2{\mathrm{SO}}_4$ was added into an unknown salt present in a test tube, a brown gas ’ $A$ ’ was evolved. This gas intensified when copper turnings were also added into this test tube. On cooling the gas ’ $A$ ’ changed into colourless gas ’ $B$ ‘. ’ $B$ ’ is

a) ${\mathrm{NO}}_2$

b) ${\mathrm{N}}_2{\mathrm{O}}_4$

c) ${\mathrm{HNO}}_3$

d) $\mathrm{NO}$

Show Answer

Answer:- (b)

The brown gas ’ $A$ ’ evolved is ${\mathrm{NO}}_2$. This gas exists as its dimer ${\mathrm{N}}_2{\mathrm{O}}_4$ at a lower temperature ${\mathrm{N}}_2{\mathrm{O}}_4$ is a colourless gas.

Question 3.- An orange solution ’ $P$ ’ turns yellow on adding $\mathrm{NaOH}$ solution to it. This yellow solution becomes orange again when an acid is added to it. $\mathrm{P}$ is

a) ${\mathrm{K}}_2{\mathrm{CrO}}_4$

b) ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$

c) ${\mathrm{Cr}}_2{\left({\mathrm{SO}}_4\right)}_3$

d) ${\mathrm{CrCl}}_3$

Show Answer

Answer:- (b)

When an alkali is added to an orange red solution of dichromate, a yellow solution results due to the formation of chromate. On acidifying, the colour again changes to orange red due to the reformation of dichromate.

Question 4.- Which one of the following is the most efficient electrolyte in coagulating a ${\mathrm{Fe}}_2{\mathrm{O}}_3,{\mathrm{H}}_2\mathrm{O}/{\mathrm{Fe}}^{3+}$ sol?

a) $\mathrm{KCl}$

b) ${\mathrm{AlCl}}_3$

c) ${\mathrm{K}}_4[\mathrm{Fe}(\mathrm{CN}{)}_6]$

d) ${\mathrm{MgCl}}_2$

Show Answer

Answer:- (c)

${\mathrm{Fe}}_2{\mathrm{O}}_3\cdot {\mathrm{H}}_2\mathrm{O}/{\mathrm{Fe}}^{3+}$ is positively charged sol. For coagulating this sol. a negative ion is required. According to Hardy Schulz rule, greater the valency of coagulating ion, greater is its power to bring about coagulation.

${[\mathrm{Fe}(\mathrm{CN}{)}_6]}^{4-}$ has a higher charge. Therefore, it is more effective to bring about the coagulation of positive sol.

Question 5.- Compount ’ $A$ ’ on reacting with $\mathrm{NaOH}$ and ${\mathrm{I}}_2$ gives two products as shown below:

$$\mathrm{A}\underset{{\mathrm{I}}_2}{\overset{\mathrm{NaOH}}{\to }}\mathrm{B}+{\mathrm{CH}}_3\mathrm{COONa}$$

$B$ is a yellow ppt.

$A$ is

a) Ethanol

b) Ethanal

c) Propanone

d) 2-Butanone

Show Answer

Answer:- (c)

The given reaction is lodoform reaction. It is given by those aldehydes and ketones which contain ${\mathrm{CH}}_3\mathrm{CO}$ - group. In this reaction, all the three $\mathrm{H}$-atoms of the methyl group are first replaced by halogen atoms to form either a trihaloaldehyde or trihaloketone which subsequently reacts with alkali to yield a haloform and the salt of carboxylic acid containing one carbon atom less than the starting aldehyde/ketone.

In the given question, since the sodium salt obtained is ${\mathrm{CH}}_3\mathrm{COONa}$ (2 carbon atoms), the starting ketone is propanone (3 carbon atoms).

Question 6.- A compound ’ $\mathrm{P}$ ’ $\left({\mathrm{C}}_7{\mathrm{H}}_6{\mathrm{O}}_3\right)$ gives a dark colour with neutral ${\mathrm{FeCl}}_3$ solution and a brisk effervescence with Sodium-bi-carbonate solution. ’ $P$ ’ is

a) Phenol

b) Benzoic acid

c) Hydroxy benzoic acid

d) An ester

Show Answer

Answer:- (c)

Since the given compound gives a brisk effervescence with ${\mathrm{NaHCO}}_3$, it is a carboxylic acid. Phenol does not give this reaction. The compound also gives a dark colour with neutral ${\mathrm{FeCl}}_3$, so the compound is Hydroxy benzoic acid.

PRACTICE QUESTIONS

Question 1.- A test-tube containing a nitrate and another containing a bromide and ${\mathrm{MnO}}_2$ are treated with conc. ${\mathrm{H}}_2{\mathrm{SO}}_4$. The brown fumes evolved are passed into water. The water will be coloured by

a) the nitrate

b) the bromide

c) both

d) none of the two

Show Answer

Answer:- b

Question 2.- An inorganic salt when heated evolves a coloured gas which bleaches moist litmus paper. The evolved gas is

a) ${\mathrm{NO}}_2$

b) ${\mathrm{Cl}}_2$

c) ${\mathrm{Br}}_2$

d) ${\mathrm{I}}_2$

Show Answer

Answer:- b

Question 3.- A salt is heated first with dil. ${\mathrm{H}}_2{\mathrm{SO}}_4$ and then with conc. ${\mathrm{H}}_2{\mathrm{SO}}_4$. No reaction takes place. It may be

a) nitrate

b) sulphide

c) oxalate

d) sulphate

Show Answer

Answer:- d

Question 4.- Chromyl chloride vapours are dissolved in water and acetic acid and lead acetate solution is added, then

a) the solution will remain colourless

b) the solution will become dark green

c) a yellow solution will be obtained

d) a yellow precipitate will be obtained

Show Answer

Answer:- d

Question 5.- The acidic solution of a salt produced a deep blue colour with strach iodide solution. The salt may be

a) chloride

b) nitrite

c) acetate

d) bromide

Show Answer

Answer:- b

Question 6.- When chlorine water is added to an aqueous solution of potassium halide in presence of chloroform, a violet colour is obtained. On adding more of chlorine water, the violet colour disappears, and a colourless solution is obtained. The test confirms the presence of the following in aqueous solution.

a) lodide

b) Bromide

c) Chloride

d) Iodide and bromide

Show Answer

Answer:- a

Question 7.- For the test of halides the soda extract is acidified with

a) dil. ${\mathrm{H}}_2{\mathrm{SO}}_4$

b) dil. ${\mathrm{HNO}}_3$

c) dil. $\mathrm{HCl}$

D) any of the three

Show Answer

Answer:- b

Question 8.- A salt solution is acidified with dil. $\mathrm{HCl}$ and ${\mathrm{BaCl}}_2$ solution is added. A white precipitate is formed. The salt contains

a) ${\mathrm{Cl}}^{-}$

b) ${\mathrm{Br}}^{-}$

c) ${\mathrm{NO}}_3^{-}$

d) ${\mathrm{SO}}_4{}^{2-}$

Show Answer

Answer:- d

Question 9.- Concentrated nitric acid is added before proceeding to test for group III members. This is to

a) reduce any remaining ${\mathrm{H}}_2\mathrm{S}$

b) convert ferrous ion into ferric ion

c) form nitrates which give granular precipitate

d) increase ionisation of ammonium hydroxide

Show Answer

Answer:- b

Question 10.- Which of the following can be used in place ${\mathrm{NH}}_4\mathrm{Cl}$ for the identification of the third radicals?

a) ${\mathrm{NH}}_4{\mathrm{NO}}_3$

b) ${\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4$

c) ${\left({\mathrm{NH}}_4\right)}_2{\mathrm{CO}}_3$

d) $\mathrm{NaCl}$.

Show Answer

Answer:- a

Question 11.- In fifth group, $\left({\mathrm{NH}}_4\right){\mathrm{CO}}_3$ is added to precipitate out the carbonates. We do not add ${\mathrm{Na}}_2{\mathrm{CO}}_3$ because

a) ${\mathrm{CaCO}}_3$ is soluble in ${\mathrm{Na}}_2{\mathrm{CO}}_3$

b) ${\mathrm{Na}}_2{\mathrm{CO}}_3$ increases the solubility of fifth group carbonates

c) ${\mathrm{MgCO}}_3$ will be precipitated out in fifth group

d) none.

Show Answer

Answer:- c

Question 12.- On addition of aqueous $\mathrm{NaOH}$ to a salt solution, a white gelatinous precipitate is formed, which dissolves in excess alkali. The salt solution contains

a) chromous ions

b) aluminium ions

c) barium ions

d) iron ions

Show Answer

Answer:- b

Question 13.- Identify the correct order of solubility of ${\mathrm{Na}}_2\mathrm{S}$, CuS and $\mathrm{ZnS}$ in aqueous medium

a) $\mathrm{CuS}>\mathrm{ZnS}>{\mathrm{Na}}_2\mathrm{S}$

b) $\mathrm{ZnS}>{\mathrm{Na}}_2\mathrm{S}>\mathrm{CuS}$

c) ${\mathrm{Na}}_2\mathrm{S}>\mathrm{CuS}>\mathrm{ZnS}$

d) ${\mathrm{Na}}_2\mathrm{S}>\mathrm{ZnS}>\mathrm{CuS}$

Show Answer

Answer:- d

Question 14.- Potassium chromate solution is added to an aqueous solution of a metal chloride. The precipitate thus obtained are insoluble in acetic acid. These are subjected to flame test, the colour of the flame is

a) lilac

b) apple green

c) crimson red

d) golden yellow

Show Answer

Answer:- b

Question 15.- A metal chloride solution on mixing with ${\mathrm{K}}_2{\mathrm{CrO}}_4$ solution gives a yellow ppt., insoluble in acetic acid. The metal may be-

a) mercury

b) zinc

c) silver

d) lead

Show Answer

Answer:- d

Question 16.- Lead has been placed in group 1 st and 2 nd because

a) it shows the valency one and two

b) it forms insoluble ${\mathrm{PbCl}}_2$

c) it forms lead sulphide

d) its chloride is partly soluble in water

Show Answer

Answer:- d

Question 17.- Potassium ferrocyanide is used in the detection

a) ${\mathrm{Cu}}^{2+\text{ions}}$

b) ${\mathrm{Fe}}^{3+}\int$ions

c) both (a) and (b)

d) none

Show Answer

Answer:- c

Question 18.- $\mathrm{Mg}$ is not precipitated in $\mathrm{V}$ group because

a) ${\mathrm{MgCO}}_3$ is soluble in water

b) ${\mathrm{MgCO}}_3$ is soluble in ${\mathrm{NH}}_4\mathrm{Cl}$

c) ${\mathrm{MgCO}}_3$ is soluble in ${\mathrm{NH}}_4\mathrm{OH}$

d) none

Show Answer

Answer:- b

Question 19.- The reaction $2{\mathrm{MnO}}_4{}^{2-}+{\mathrm{Cl}}_2⟶2{\mathrm{MnO}}_4^{-}+2{\mathrm{Cl}}^{-}$takes place in

a) basic medium

b) acidic medium

c) neutral medium

d) both (a) and (b)

Show Answer

Answer:- a

Question 20.- Manganese achieves its highest oxidation state in its compound

a) ${\mathrm{MnO}}_2$

b) ${\mathrm{Mn}}_2{\mathrm{O}}_4$

c) ${\mathrm{KMnO}}_4$

d) ${\mathrm{K}}_2{\mathrm{MnO}}_4$

Show Answer

Answer:- c

Question 21.- Pick out the incorrect statement

a) ${\mathrm{MnO}}_2$ dissolves in dilute $\mathrm{HCl}$ but does not form ${\mathrm{Mn}}^{4+}$

b) ${\mathrm{MnO}}_2$ oxidises hot conc. ${\mathrm{H}}_2{\mathrm{SO}}_4$ liberating oxygen

c) ${\mathrm{K}}_2{\mathrm{MnO}}_4$ is formed when ${\mathrm{MnO}}_2$ is fused with $\mathrm{KOH}$ in presence of ${\mathrm{KNO}}_3$

d) Decomposition of ${\mathrm{KMnO}}_4$ is not catalyzed by sunlight.

Show Answer

Answer:- d

Question 22.- ${\mathrm{K}}_2{\mathrm{MnO}}_4$ can be converted into ${\mathrm{KMnO}}_4$ using all of the following except

a) dil. ${\mathrm{H}}_2{\mathrm{SO}}_4$

b) ${\mathrm{Cl}}_2$

c) ${\mathrm{O}}_3$

d) $\mathrm{HCl}$

Show Answer

Answer:- d

Question 23.- A mixture of salts $({\mathrm{Na}}_2{\mathrm{SO}}_3+{\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7)$ in a test tube is treated with dil. ${\mathrm{H}}_2{\mathrm{SO}}_4$ and resulting gas is passed through lime water. Which of the following observations is correct about this test?

a) Solution in test tube becomes green and lime water turns milky

b) Solution in test tube is colourless and lime-water turns milky

c) Solution in test tube becomes green and lime water remains clear

d) Solution in test tube remains clear and lime water also remains clear.

Show Answer

Answer:- c

Question 24.- How many moles of iodine are liberated when I mole of ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ reacts with potassium iodide?

a) 1

b) 2

c) 3

d) 4

Show Answer

Answer:- c

Question 25.- When ${\mathrm{H}}_2{\mathrm{O}}_2$ is shaken with an acidified solution of ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ in presence of ether, the etheral layer turns blue due to the formation of

a) ${\mathrm{Cr}}_2{\mathrm{O}}_3$

b) ${\mathrm{CrO}}_4^{-}$

b) ${\mathrm{Cr}}_2{\left({\mathrm{SO}}_4\right)}_3$

d) ${\mathrm{CrO}}_5$

Show Answer

Answer:- d

Question 26.- Oxidation of oxalic acid by acidified ${\mathrm{KMnO}}_4$ is an example of autocatalysis, it is due to the presence of

a) ${\mathrm{SO}}_4{}^{2-}$

b) ${\mathrm{MnO}}_4$

c) ${\mathrm{Mn}}^{2+}$

d) ${\mathrm{K}}^{+}$

Show Answer

Answer:- c

Question 27.- The reaction

${\mathrm{MnO}}_4^{-}+{\mathrm{e}}^{-}\rightleftharpoons {\mathrm{MnO}}_4{}^{2-}$

takes place in

a) a basic medium

b) an acid medium

c) a neutral medium

d) both acidic and basic

Show Answer

Answer:- a

Question 28.- In the preparation of ${\mathrm{KMnO}}_4$, pyrolusite $\left({\mathrm{MnO}}_2\right)$ is first converted to potassium manganate $\left({\mathrm{K}}_2{\mathrm{MnO}}_4\right)$. In this conversion, the oxidation state of manganese changes from

a) +1 to +3

b) +2 to +4

c) +3 to +5

d) +4 to +6

Show Answer

Answer:- d

Question 29.- Deep green precipitate of $\mathrm{Cr}(\mathrm{OH}{)}_3$ gives yellow solution on addition ${\mathrm{H}}_2{\mathrm{O}}_2$ in presence of excess of base. The yellow colour is due to

a) $\mathrm{Cr}(\mathrm{OH}{)}_4^{-}$ions

b) ${\mathrm{Cr}}_2{\mathrm{O}}_7{}^{2-}$ ions

c) ${\mathrm{CrO}}_4{}^{2-}$ ions

d) ${\mathrm{CrO}}_5$

Show Answer

Answer:- c

Question 30.- ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ is preferred to ${\mathrm{Na}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ for use in volumetric analysis as a primary standard because

a) ${\mathrm{Na}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ is hygroscopic while ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ is not

b) ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ is hygroscopic while ${\mathrm{Na}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ is not

c) ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ is pure while ${\mathrm{Na}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ is impure

d) none of the above

Show Answer

Answer:- a

Question 31.- What would happen when a solution of potassium chromate is treated with an excess of dilute ${\mathrm{HNO}}_3$ ?

a) ${\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}$ and ${\mathrm{H}}_2\mathrm{O}$ are formed

b) ${\mathrm{CrO}}_4{}^{2-}$ is reduced to +3 state of $\mathrm{Cr}$

c) ${\mathrm{CrO}}_4{}_4^{2-}$ is oxidised to +7 state of $\mathrm{Cr}$

d) ${\mathrm{Cr}}^{3+}$ and ${\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}$ are formed

Show Answer

Answer:- a

Question 32.- The product of oxidation of ${\mathrm{I}}^{-}$with ${\mathrm{MnO}}_4$ in alkaline medium is

a) ${10}^{3-}$

b) ${\mathrm{I}}_2$

c) ${10}^{-}$

d) ${10}_4^{-}$

Show Answer

Answer:- a

Question 33.- The purple colour of ${\mathrm{KMnO}}_4$ is due to the transition

a) Charge transfer $(0⟶Mn)$

b) Charge transfer $(\mathrm{Mn}⟶0)$

c) $d-d$

d) $p$-d

Show Answer

Answer:- a

Question 34.- ${\mathrm{MnO}}_4$ reacts with ${\mathrm{Br}}^{-}$in alkaline $\mathrm{pH}$ to give

a) ${\mathrm{BrO}}_3^{-},{\mathrm{MnO}}_2$

b) ${\mathrm{Br}}_2,{\mathrm{MnO}}_4^{2-}$

c) ${\mathrm{Br}}_2,{\mathrm{MnO}}_2$

d) ${\mathrm{BrO}}^{-},{\mathrm{MnO}}_4{}^{2-}$

Show Answer

Answer:- a

Question 35.- A student accidently added conc. ${\mathrm{H}}_2{\mathrm{SO}}_4$ to potassium permanganate and it exploded due to the formation of an explosive which is

a) $\mathrm{MnO}$

b) ${\mathrm{Mn}}_2\mathrm{O}$

c) ${\mathrm{Mn}}_2{\mathrm{O}}_5$

d) ${\mathrm{Mn}}_2{\mathrm{O}}_7$

Show Answer

Answer:- d

Question 36.- A solution of ${\mathrm{Na}}_2{\mathrm{SO}}_3$ will be oxidized to ${\mathrm{Na}}_2{\mathrm{SO}}_3$ by ${\mathrm{KMnO}}_4$ solution in an acidic medium. ${\mathrm{KMnO}}_4$ is reduced to ${\mathrm{Mn}}^{2+}$. How many moles of ${\mathrm{Na}}_2{\mathrm{SO}}_3$ will be oxidized by one mole of ${\mathrm{KMnO}}_4$ ?

a) 5

b) 3

c) 2.5

d) 1

Show Answer

Answer:- c

Question 37.- ${\mathrm{KMnO}}_4$ gets reduced to

a) ${\mathrm{K}}_2{\mathrm{MnO}}_4$ in neutral medium

b) ${\mathrm{MnO}}_2$ in acidic medium

c) ${\mathrm{Mn}}^{2+}$ in alkaline medium

d) ${\mathrm{MnO}}_2$ in neutral medium

Show Answer

Answer:- d

CONCEPT MAP