d-Block Elements

INTRODUCTION

The series of elements, that are formed by filling the 3d, 4d and 5d shells of electrons, comprise the d-block elements. They are often called as transition elements because their position in the periodic table is between s-block and p-block elements. Their properties are transitional between the highly reactive metallic elements of the s-block, which form ionic compounds and elements of p-block which form covalent compounds. In s and p blocks electrons add to the last shell, in

d-block electrons are added to the penultimate shell. Typically the transition elements have an incompletely filled d level. The zinc group has d¹⁰ configuration and compounds of these elements show some differences from other transition elements. The elements make up three complete rows of ten elements and an incomplete fourth row. The position of the incomplete fourth series is discussed along with the f-block elements.

Illustration 1. Through copper, silver and fold have completely filled sets of d-orbitals yet they are considered as transition metals. Why?

Solution: These metals in their common oxidation states have incompletely filled d-orbitals e.g. Cu²⁺ has 3d⁹ and Au³⁺ has 5d⁸ configuration.

Illustration 2. Zinc, cadmium and mercury are generally not considered as transition metals. Give reasons.

Solution: These elements in their most common oxidation state of +2 have completely filled d-orbitals.

ELECTRONIC CONFIGURATION

The general electronic configuration is (n – 1)^1-10 ns^1-2 where n is the outermost shell. The number of electrons in their outermost subshell remains two while their penultimate shell of electrons expands from 8 to 18 electrons.

Electronic configuration of 3d series

Anamolous configuration of Cu and Cr

Copper and chromium have a single electron in 4s-orbital. This is due to the gain of additional stability by the atom having either half-filled (5 electrons) or completely filled (10 electrons)

d-shell.

Illustration 3. In what way is the electronic configuration of transition elements different form that of the non-transition elements?

Solution: Transition elements contain incompletely filled d-subshell i.e. their electronic configuration is (n – 1)d^1-10ns^0-2 whereas non-transition elements have no d-subshell or their subshell is completely filled and have ns^1-2np^1-2 in their outermost shell.

GENERAL CHARACTERISTICS OF TRANSITION ELEMENTS

(i) Except for mercury, which is a liquid at room temperature all other elements are solid metals exhibiting all the characteristics of a metal.

(ii) The show variable oxidation states unlike s and p block elements.

(iii) They, and some of their compounds, show catalytic properties.

(iv) Their compounds are coloured.

(v) They have great tendency to form complex compounds.

(vi) They form alloys and interstitial compounds.

Conductivity

All the transition metals are good conductors of heat and electricity. Silver is the best conductor of electricity.

Density

Because of small size of their atoms and strong metallic bonding the density and hardness of transition elements are high.

Ionization Energy

The ionization energy (IE) of transition elements are higher than those of s-block elements but lower than p-block elements. In a particular transition series, ionization energy although increases gradually as we move from left to right but this increase is not appreciable.

The increase in ionization energy is due to increase in nuclear charge, the effect of increase in nuclear charge is partly balanced by the increase in screening effect. Consequently, the increase in ionization energy along the period of d-block elements is very small.

Illustration 4. K₂PtCl₆ is a well known compound whereas corresponding Ni compound is not known. State a reason for it.

Solution: This is because Pt⁴⁺ is more stable than Ni⁴⁺ has the sum of four ionization energies of Pt is less than that of Ni.

VARIABLE OXIDATION STATES

Transition elements usually exist in several different oxidation states and the oxidation states changes in units of one, e.g. Fe²⁺ and Fe⁺³, Cu⁺¹ and Cu⁺².

Scandium can have an oxidation number of (+II) if both s electrons are used for bonding and (+III) when two s and one d electrons are involved. Similarly all the elements show variable oxidation states depending upon the number of electrons available for bonding in their s and d sub-shells.

Illustration 5. Why do transition elements show variable oxidation states?

Solution: In the transition elements, the energies of (n-1)d orbitals and ns orbitals are very close. Hence electrons from both can participate in bonding.

Illustration 6. Explain briefly how +2 state becomes more and more stable in the first half of the first row transition elements with increasing atomic number.

Solution: The sum IE₁ + IE₂ increases. As a result the standard reduction potentials (E⁰) becomes less and less negative. Hence the tendency to form ²M⁺ ion decreases. The greater stability of +2 state for Mn is due to half-filled d-subshell (d⁵), that for zinc is due to completely filled d-subshell (d¹⁰) and half that for nickel is due to highest negative enthalpy of hydration.

COMPLEXES AND THEIR PROPERTIES

The transition elements have an unparalleled tendency to form coordination compounds with the Lewis bases, which are called as ligands.

$C{O}^{3+}+6N{H}_3\to {\left[CO{\left(N{H}_3\right)}_6\right]}^{3+}$

$F{e}^{2+}+6C{N}^{-}\to {\left[Fe{\left(CN\right)}_6\right]}^{4-}$

s and p block elements form very few complexes. The reason transition elements are so good at forming complex is that they have small, highly charged ions and have vacant low energy orbitals to accept lone pairs of electrons donated by ligands.

Size of Atoms and Ions

The covalent radii of the elements decrease from left to right across a row in the transition series. This is because of the poor screening by the d electrons due to which, the nuclear charge attracts all of the electrons more strongly, hence a contraction in size occurs.

The elements in the first group in the d-block show the excepted increase (due to the addition of extra shell) in size Sc$\to$ Y $\to$ La. However in the subsequent groups there is an increase between first and second members, but hardly any increase between second and third elements. This is due to lanthanide contraction (discussed in f-block elements).

Colour

Many compounds of transition elements are coloured in contrasts to those of s and p block elements.

In compound state due to the surrounding groups (ligands), the d-orbitals of transition elements are not degenerate but split into two groups of different energy. Thus it is possible to promote electrons from one group to another group. This corresponds to fairly small amount of energy difference and so light is absorbed in visible region. Some compounds of transition metals are white, for example ZnSO₄ and TiO₂. In these compounds it is not possible to promote the electrons within the d-level.

Illustration 7. Why Zn²⁺ salts are white while Ni²⁺ salts are blue?

Solution: Zn²⁺ has completely filled d-orbitals (3d¹⁰) while Ni²⁺ has incompletely filled d-orbitals (3d⁸).

Illustration 8. Why Zn²⁺ salts are white while Cu²⁺ salts are blue?

Solution: Reason same as above.

Magnetic Properties

On the basis of behaviour in a magnetic field, substance are classified as paramagnetic, diamagnetic and ferromagnetic. Those substance which are attracted by the applied magnetic field are called paramagnetic where as those which are repelled by the magnetic field are called diamagnetic. Substances which are very strongly attracted by the applied field are called ferromagnetic.

Paramagnetism is a property due to the presence of unpaired electrons. Thus most of the transition metals are paramagnetic. As the number of unpaired electrons increases, the paramagnetic character also increases.

The magnetic moment is calculated from the following formula $\mu =\sqrt[]{n\left(n+2\right)}B.M$

Where n is the number of unpaired electrons and B. M stands for Bohr magneton.

Illustration 9. Why does Mn(II) show maximum papamagentic character amongst the bivalent ions of the first transition series?

Solution: Mn²⁺ has maximum number of unpaired electrons i.e. 3d⁵.

Catalytic Properties

Many transition metals and their compounds have catalytic properties. For e.g. V₂O₅, Fe, FeCl₃, Ni, Pd etc.

This property of transition elements is due to their variable oxidation states. In some cases the transition metals with their variable valency may form variable unstable intermediate compounds. In other cases the transition metal provides a suitable reaction surface.

NON STOICHIOMETRY

Another feature of the transition elements is that they sometimes form non stoichiometry compounds. These are compounds of indefinite structure and proportions. For example$F{e}_{0.94}O$. It is mostly due to the variable valency of transition elements. Sometimes, non stoichiometry is caused by defects in the solid structures.

ALLOY FORMATION

Alloys are homogenous solid solutions of two or more metals obtained by melting the components and then cooling the melt. These are formed by metals whose atomic radii differ by not more than 15% so that the atoms of one metal can easily take up the positions in the crystal lattice of the other. Since transition metals have similar atomic radii, they form alloys very readily.

STUDY OF SOME IMPORTANT COMPOUNDS

(I) Copper (II) Sulphate pentahydrate or blue vitriol, CuSO₄.5H₂O

Preparation

In the laboratory, it is prepared by dissolving cupric oxide, cupric hydroxide or carbonate in dilute H₂SO₄.

CuO + H₂SO₄$\to$ CuSO₄ + H₂O

Cu(OH)₂ + H₂SO₄$\to$ CuSO₄ + 2H₂O

CuCO₃ + H₂SO₄$\to$CuSO₄ + H₂O + CO₂

The solution of CuSO₄ thus obtained is concentrated and cooled when crystals of blue vitriol, CuSO₄.5H₂O separates out. Commercially it is prepared by the action of hot dilute sulphuric acid on scrap copper in the presence of air.

$2Cu+2H_{2}S{O}_4+O_2\to 2CuS{O}_4+2H_{2}O$

Properties

1. Action of heat

$CuS{O}_4.5H_{2}O\stackrel{373K}{\to }CuS{O}_4.H_{2}O\stackrel{423K}{\to }\underset{\left(white\right)}{CuS{O}_4}\underset{heating}{\overset{strong}{\to }}CuO+S{O}_3$

In CuSO₄.5H₂O, four H₂O molecules are coordinated to the central Cu²⁺ ion. While the 5^th one is attached between $S{O}_4^{2-}$ and H₂O molecule by hydrogen bonding. This fifth H₂O molecule is thus deep inside the crystal lattice and is not easily lost.

2. Action of alkalis

$CuS{O}_4+2NaOH\to Cu{\left(OH\right)}_2+N{a}_{2}S{O}_4$

With NH₄OH it forms tetraamminecopper (II) sulphate

$CuS{O}_4+4N{H}_{4}OH\to \left[Cu{\left(N{H}_3\right)}_4\right]S{O}_4+4H_{2}O$

3. Reaction with KI

$CuS{O}_4+2KI\to Cu{I}_2+K_{2}S{O}_4$

$2Cu{I}_2\to 2CuI+I_2$

The liberation of iodine in this reaction is quantitative. Therefore, this reaction is used to estimate copper volumetrically.

Uses

(i) It is used as an electrolyte in electroplating, electrotyping and refining of copper.

(ii) It is used in reservoirs and swimming pools to prevent the growth of weeds.

(iii) It is used as a fungicide under the name Bordeaux mixture, which is a mixture of CuSO₄ and slaked lime Ca(OH)₂.

(iv) Anhydrous CuSO₄ is used for detection of moisture in organic liquids such as alcohol, ether etc.

Silver Nitrate, AgNO₃

Preparation

It is prepared by the action of dilute nitric acid on silver and then evaporating the solution to crystallization.

$3Ag+4HN{O}_3\to 3AgN{O}_3+NO\uparrow +2H_{2}O$

Properties

1. Action of heat

It decomposes on heating.

$2AgN{O}_3\stackrel{723K}{\to }2AgN{O}_2\stackrel{980K}{\to }2Ag+2N{O}_2$

2. On coming into contact with organic matter like skin or clothes, it is reduced to finely – divided silver, giving a black stain.

3. Precipitation reactions

It gives precipitates with some salt solutions which help in the detection of acid radicals. Some of the precipitation reactions are:

$NaCl+AgN{O}_3\to AgCl\downarrow +NaN{O}_3$

$NaP{O}_4+3AgN{O}_3\to A{g}_{3}P{O}_4\downarrow +3NaN{O}_3$

$K_{2}Cr{O}_4+3AgN{O}_3\to AgCr{O}_4\downarrow +2KN{O}_3$

$N{a}_{2}S+2AgN{O}_3\to A{g}_{2}S\downarrow +2NaN{O}_3$

$N{a}_2{S}_2{O}_3+2AgN{O}_3\to A{g}_2{S}_2{O}_3\downarrow +2NaN{O}_3$

$N{a}_2{C}_2{O}_4+2AgN{O}_3\to A{g}_2{C}_2{O}_4\downarrow +2NaN{O}_3$

$N{a}_{3}B{O}_3+3AgN{O}_3\to A{g}_{3}B{O}_3\downarrow +3NaN{O}_3$

Uses

It is used for

1. Preparing silver halides used in photography.

2. For making inks and hair dyes.

3. In qualitative and quantitative analysis.

4. For silvering of glass, i.e. preparation of mirrors.

Halides

1. Silver halides (AgF, AgCl, AgBr & AgI)

Preparation

Silver halides are prepared by the action of sodium or potassium halide on silver nitrate solution (except for AgF)

$AgN{O}_3+NaX\to Ag{X}_{\left(s\right)}+NaN{O}_3$

Silver fluoride is prepared by the action of HF on silver (I) oxide.

$2HF+2A{g}_{2}O\to 2AgF+H_{2}O$

Properties

(i) AgCl is white solid, AgBr is a pale yellow solid and AgI is a yellow solid.

(ii) AgF is soluble in water whereas other halides are insoluble in water. AgCl dissolves in ammonia to form a complex.

$AgCl+2N{H}_{4}OH\to \left[Ag{\left(N{H}_3\right)}_2\right]Cl+2H_{2}O$

AbBr is partially soluble and AgI is insoluble in NH₄OH.

(iii) All the silver halides dissolve in potassium cyanide and Na₂S₂O₃ solution to form complexes.

$AgCl+2KCN\to K\left[Ag{\left(CN\right)}_2\right]+KCl$

$AgCl+2N{a}_2{S}_2{O}_3\to N{a}_2\left[Ag{\left(S_2{O}_3\right)}_2\right]+NaCl$

Uses

All silver halides (particularly AgBr) are photosensitive and hence are widely used in photography.

2. Mercury halides

(a) Mercury (I) chloride or mercurous chloride or calomel, Hg₂Cl₂

Preparation

(i) It can be prepared by mixing a chloride solution with a mercury (I) salt solution.

$H{g}_2{\left(N{O}_3\right)}_2+2NaCl\to \underset{calomel}{H{g}_{2}C{l}_2}\downarrow +2NaN{O}_3$

(ii) It can also prepared by heating a mixture of mercuric chloride and mercury in an iron vessel.

$HgC{l}_2+Hg\stackrel{\Delta }{\to }H{g}_{2}C{l}_2$

Properties

(i) It is a white power insoluble in water but soluble in chlorine water.

$H{g}_{2}C{l}_2+C{l}_2\to 2HgC{l}_2$

(ii) It decomposes on heating to HgCl₂

$H{g}_{2}C{l}_2\to HgC{l}_2+Hg$

(iii) On treatment with ammonia, if turns black due to the formation of finely divided mercury.

$H{g}_{2}C{l}_2+2N{H}_3\to \underset{\left(black\right)}{\mathrm{Hg}}+Hg\left(N{H}_2\right)Cl+N{H}_{4}Cl$

Uses

(i) In making standard calomel electrode and

(ii) As a purgative in medicine.

(b) Mercury (II) chloride HgCl₂

1. It is prepared by passing dry chlorine over heated mercury.

$Hg+C{l}_2\stackrel{\Delta }{\to }HgC{l}_2$

2. It is also obtained by treating HgO with HCl

$HgO+2HCl\to HgC{l}_2+H_{2}O$

3. Commercially, it is prepared by heating a mixture of HgSO₄ and NaCl in the presence of MnO₂

$HgS{O}_4+2NaCl\underset{\Delta }{\overset{Mn{O}_2}{\to }}HgC{l}_2+N{a}_{2}S{O}_4$

Properties

1. It is a white crystalline solid sparingly soluble in cold water but soluble in hot water. Its solubility can be increased by adding Cl^-.

$HgC{l}_2+2C{l}^{-}\to {\left[HgC{l}_4\right]}^{2-}$

2. It is readily soluble in organic solvents suggesting its covalent nature.

3. When treated with SnCl₂ it is reduced to mercury.

$2HgC{l}_2+SnC{l}_2\to SnC{l}_4+H{g}_{2}C{l}_2$

$H{g}_{2}C{l}_2+SnC{l}_2\to 2Hg+SnC{l}_4$

4. When Cu turnings are placed in its contact a shining grey film of mercury deposits over them.

$HgC{l}_2+Cu\to Hg+CuC{l}_2$

Uses

It is used for preserving wood and hides and for making fungicides.

(c) Mercury (II) Iodide

Preparation

It is prepared by treating HgCl₂ with KI.

$HgC{l}_2+2KI\to \underset{\left(scorletred\right)}{Hg{I}_2}+2KCl$

Properties

1. Mercuric iodide exists in two forms, i.e. red and yellow. The yellow form is stable above 400 K white the red form is stable below this temperature.

$\underset{\left(red\right)}{Hg{I}_2}\underset{\left(Yellow\right)}{Hg{I}_2}$

2. It readily dissolves in KI forming a complex

$Hg{I}_2+2KI\to K_2\left[Hg{I}_4\right]$

An alkaline solution of K₂HgI₄ is called Nessler's reagent and is used to detect the presence of NH₄⁺ with which it gives a brown precipitate due to the formation of iodide of Million's base.

Uses

It is used to prepare Nessler's reagent and for making ointments for treating skin infections.

Potassium Dichromate, K₂Cr₂O₇

Preparation

It is prepared from the ore called chromate or ferrochrome or chrome iron, FeO.Cr₂O₃. The various steps involved are

(a) Preparation of sodium chromate

$4FeO.C{r}_2{O}_3+O_2\to F{e}_2{O}_3+4C{r}_2{O}_3$

$4N{a}_{2}C{O}_3+2C{r}_2{O}_3+3O_2\to 4N{a}_{2}Cr{O}_4+4C{O}_2$

(b) Conversion of sodium chromate into sodium dichromate.

$2N{a}_{2}Cr{O}_4+H_{2}S{O}_4\to N{a}_{2}C{r}_2{O}_7+N{a}_{2}S{O}_4+H_{2}O$

(c) Conversion of sodium dichromate into potassium dichromate.

$N{a}_{2}C{r}_2{O}_7+2KCl\to K_{2}C{r}_2{O}_7+2NaCl$

Properties

It forms orange red crystals. It is moderately soluble in cold water but freely soluble in hot water.

1. Action of heat

When heated, it decomposed to its chromate

$4K_{2}C{r}_2{O}_7\stackrel{\Delta }{\to }4K_{2}Cr{O}_4+2C{r}_2{O}_3+3O_2$

2. Action of alkalis

With alkalis it is converted into chromate which on acidifying gives back dichromate.

$K_{2}C{r}_2{O}_7+2KOH\to 2K_{2}Cr{O}_4+H_{2}O$

$2K_{2}C{r}_2{O}_7+H_{2}S{O}_4\to K_{2}C{r}_2{O}_7+K_{2}S{O}_4+H_{2}O$

In dichromate solution the $C{r}_2{O}_7^{2-}$ ions are in equilibrium with $Cr{O}_4^{2-}$ ions at pH = 4.

$\underset{orangered}{C{r}_2{O}_7^{2-}}+H_{2}O\underset{yellow}{2Cr{O}_4^{2-}}+2HI$

3. Action of conc. H₂SO₄ solution

(a) In cold conditions

$K_{2}C{r}_2{O}_7+2H_{2}S{O}_4\to 2Cr{O}_3+2KHS{O}_4+H_{2}O$

(b) In hot conditions

$2K_{2}C{r}_2{O}_7+8H_{2}S{O}_4\to 2K_{2}S{O}_4+2C{r}_2{\left(S{O}_4\right)}_3+8H_{2}O+3O_2$

4. Oxidising properties

It is a powerful oxidising agent. In the presence of dil. H₂SO₄ it furnishes 3 atoms of available oxygen.

$K_{2}C{r}_2{O}_7+4H_{2}S{O}_4\to K_{2}S{O}_4+C{r}_2{\left(S{O}_4\right)}_3+4H_{2}O+3O$

Some of the oxidizing properties of K₂Cr₂O₇ are

(a) It liberates I₂ from KI

$K_{2}C{r}_2{O}_7+7H_{2}S{O}_4+6KI\to 4K_{2}S{O}_4+C{r}_2{\left(S{O}_4\right)}_3+3I_2+7H_{2}O$

(b) It oxidises ferrous salts to ferric salts

$K_{2}C{r}_2{O}_7+7H_{2}S{O}_4+6FeS{O}_4\to K_{2}S{O}_4+C{r}_2{\left(S{O}_4\right)}_3+3F{e}_2{\left(S{O}_4\right)}_3+2H_{2}O$

(c) It oxidises S^-2 to S

$K_{2}C{r}_2{O}_7+4H_{2}S{O}_4+3H_{2}S\to K_{2}S{O}_4+C{r}_2{\left(S{O}_4\right)}_3+7H_{2}O+3S$

(d) It oxidises nitrites to nitrates

$K_{2}C{r}_2{O}_7+4H_{2}S{O}_4+3NaN{O}_2\to K_{2}S{O}_4+C{r}_2{\left(S{O}_4\right)}_3+3NaN{O}_3+4H_{2}O$

(e) It oxidises SO₂ to $S{O}_4^{2-}$

$K_{2}C{r}_2{O}_7+H_{2}S{O}_4+3S{O}_2\to K_{2}S{O}_4+C{r}_2{\left(S{O}_4\right)}_3+3H_{2}O$

(f) It oxidises ethyl alcohol to acetaldehyde and acetic acid.

5. Chromyl chloride test

When heated with conc. HCl or with a chloride in the presence of sulphuric acid, reddish brown vapours of chromyl chloride are obtained.

$K_{2}C{r}_2{O}_7+4KCl+6H_{2}S{O}_4\to 2Cr{O}_{2}C{l}_2+6KHS{O}_4+3H_{2}O$

Thus reaction is used in the detection of chloride ions in qualitative analysis.

Uses

1. In volumetric analysis for the estimation of Fe²⁺ and $I^{-}$

2. In chrome tanning in leather industry.

3. In photography and in hardening gelatin film.

Potassium Permanganate, KMnO₄

Preparation on a large scale

It is prepared from the mineral pyrolusite, MnO₂. The preparation involves the following steps

(i) Conversion of MnO₂ into potassium manganate.

When finely powdered MnO₂ is fused with KOH. K₂MnO₄ is obtained.

$2Mn{O}_2+4KOH+O_2\to 2K_{2}Mn{O}_4+2H_{2}O$

(ii) Oxidation of potassium manganate into permanganate

(a) Chemical oxidation

K₂MnO₄ is oxidised to KMnO₄ by bubbling CO₂ or Cl₂ or ozone into the former.

$3K_{2}Mn{O}_4+2C{O}_2\to 2KMn{O}_4+Mn{O}_2+2K_{2}C{O}_3$

(b) Electrolytic oxidation

The manganate solution is electrolysed between iron electrodes. The oxygen evolved at anode converts manganate into permanganate.

$2K_{2}Mn{O}_4+H_{2}O+O\to 2KMn{O}_4+2KOH$

Properties

KMnO₄ exists as deep purple prisms. It is moderately soluble in water at room temperature and its solubility in water increases with temperature.

(i) Action of heat

When heated it decomposes to K₂MnO₄.

$2KMn{O}_4\to K_{2}Mn{O}_4+Mn{O}_2+O_2$

(ii) Action of conc. H_­2SO₄

With cold conc. H₂SO₄ it gives Mn₂O₇ which on warming decomposes to MnO₂.

$2KMn{O}_2+2H_{2}S{O}_4\to M{n}_2{O}_7+2KHS{O}_4+H_{2}O$

$2M{n}_2{O}_7\stackrel{\Delta }{\to }4Mn{O}_2+3O_2$

With hot Conc. H_­2SO₄ O₂ is evolved

$4KMn{O}_4+6H_{2}S{O}_4\to 2K_{2}S{O}_4+4MnS{O}_4+6H_{2}O+5O_2$

(iii) Oxidising properties

KMnO₄ is a powerful oxidizing agent. The actual oxidizing action depends upon the medium i.e. acidic, basic or neutral.

(a) In neutral solution, it acts as moderate oxidizing agent.

$2KMn{O}_4+H_{2}O\to 2KOH+2Mn{O}_2+3O$

Some oxidizing properties of KMnO₄ in neutral medium are

$2KMn{O}_4+3N{a}_2{S}_2{O}_3+H_{2}O\to 3K_{2}S{O}_4+8Mn{O}_2+3N{a}_{2}S{O}_4+2KOH$

$2KMn{O}_4+4H_{2}S\to 2MnS+S+K_{2}S{O}_4+4H_{2}O$

(b) In strong alkaline solution, it is converted into $Mn{O}_4^{2-}$

$2KMn{O}_4+2KOH\to 2K_{2}Mn{O}_4+H_{2}O+O$

Some reactions in alkaline medium are

$2KMn{O}_4+H_{2}O+KI\to 2Mn{O}_2+2KOH+KI{O}_3$

(c) In acidic medium, Mn⁺⁷ is converted into Mn⁺²

$2KMn{O}_4+3H_{2}S{O}_4\to K_{2}S{O}_4+2MnS{O}_4+3H_{2}O+5O$

Some other reactions are

(i) $2KMn{O}_4+3H_{2}S{O}_4+5H_{2}S\to K_{2}S{O}_4+2MnS{O}_4+3H_{2}O+5S$

(ii) $2KMn{O}_4+5S{O}_2+2H_{2}O\to K_{2}S{O}_4+2MnS{O}_4+2H_{2}S{O}_4$

(iii) $2KMn{O}_4+3H_{2}S{O}_4+5KN{O}_2\to K_{2}S{O}_4+2MnS{O}_4+3H_{2}O+5KN{O}_3$

(iv) $2KMn{O}_4+3H_{2}S{O}_4+5C_2{H}_2{O}_4\to K_{2}S{O}_4+2MnS{O}_4+8H_{2}O+10C{O}_2$

(v) $2KMn{O}_4+8H_{2}S{O}_4+10FeS{O}_4\to K_{2}S{O}_4+2MnS{O}_4+5F{e}_2{\left(S{O}_4\right)}_3+8H_{2}O$

(vi) $2KMn{O}_4+3H_{2}S{O}_4+10KI\to K_{2}S{O}_4+2MnS{O}_4+8H_{2}O+5I_2$

Uses

(i) It is used in volumetric analysis for the estimation of ferrous salts, oxalates, iodides and H₂O₂.

(ii) It is used as oxidizing agent in the laboratory as well as in industry.

(iii) It is also used as disinfectant and germicide.

PHOTOGRAPHY

Modern photography is an application of the chemical behaviour of the silver halides which decomposes and turn black in light.

For example:

$2AgBr\stackrel{light}{\to }2Ag+B{r}_2$

Following steps are involved in photography:

1. Preparation of sensitive plate or film

To a solution of NH₄Br containing gelatin, amm. AgNO₃ is added. Thus an emulsion of AgBr in gelatin is prepared

$N{H}_{4}Br+AgN{O}_3\to AgBr+N{H}_{4}N{O}_3$

The mixture is allowed to stand is order for AgBr particles to grow in size. This process is called ripening of emulsion. Then the emulsion is solidified. It is washed with water to expel NH₃NO₃ and then melted and applied uniformly on a glass plate or celluloid film.

2. Exposure

The plate or film is fixed in camera. For a fraction of a second the film is exposed so that an impression of the object to be photographed is made on the film. The silver bromide which is affected by light gets reduced.

$2AgBr\to 2Ag+B{r}_2$

An actual but inverted image of the object is thus formed on the plate which is not visible to the eye.

3. Developing

The exposed film is immersed in a solution of the developer which contains reducing agents like pyrogallol or quinol. This solution reacts with the exposed part of the film. After the developing process, those parts of the film are white which were originally dark in the object and those parts are black which were originally white. This is known as negative.

$\underset{quinol}{C_6{H}_4{\left(OH\right)}_2}+2AgBr\to 2Ag+\underset{quinone}{C_6{H}_4{O}_2}+2HBr$

The process of developing is carried out in a dark room.

4. Fixing the negative

Fixing is the process of making the image permanent. This is done by placing the negative in the hypo which will dissolve the undecomposed silver bromide.

$AgBr+2N{a}_2{S}_2{O}_3\to N{a}_3\left[Ag{\left(S_2{O}_3\right)}_2\right]+NaBr$

The negative plate can now be taken out of dark room into light.

5. Printing of the photograph

The positive paper is prepared in a similar manner as the negative with the only difference that it is less sensitive. Either the P.O.P (printing out paper) or bromide paper is generally used for the purpose. The POP has got a coating of AgCl and AgNO₃. The paper is placed below the negative and then exposed to light. A negative of negative plate or positive with respect to the object is obtained on the print paper.

6. Toning

In order to impart a beautiful golden colour to the photograph, it is dipped in a dilute solution of gold chloride. The process is called toning. If a shining grey tinge is desired in the photograph, it is dipped in potassium chloroplatinate, K₂PtCl₄ solution.

The silver particles on the print are replaced by gold or platinum particles resulting in the toning up of the photograph.

$AuC{l}_3+3Ag\to 3AgCl+Au$