electrovalency covalency electrovalency...

Logic of Inorganic Reactions 1

SECTION - IVALENCY/OXIDATION NUMBER/BALANCING EQUATIONS

Modern Definition of ValencyValency is the power of an atom of an element to combine with other atoms measured

by the number of electrons which an atom or radical will lose, gain or share to form a chemicalcompound.Types of Valences:

(i)Electrovalency( shown in ionic compounds eg. NaCl, K2SO

4 etc)

(ii)Covalency (shown in covalent compounds eg SO2, NH

3, H

2SO

4 etc)

ElectrovalencyRadicals:

Radical is an atom or a group of atoms which form an ion (positive or negative) by lossor gain of electrons. It is part of an ionic compound. Two radicals form a compound.Types of Radicals:

(i)Basic Radical: This forms the +ve ion or cation part of the compound. Usuallymetals e.g Na+, K+, Ca2+, Al3+, Ag+, Hg2+, Fe3+ etc. (exception: NH

4+, H+ ). This part comes

from a base.(ii)Acid Radical: This forms the –ve ion or anion part of the compound. Usually it is

made up of nonmetals. Some acid radicals contain both metal and nonmetal atoms.(i)monoatomic acid radical: Cl–, Br–, S2-, O2-, N3-, P3-, H– etc(ii)Compound acid radical : SO

42-, PO

43-, CO

32-, SO

32-, OH–, NO

2–, NO

3–

etc. (exceptions: MnO4–, CrO

42-, AlO

2– etc which contain both metal and non-metal)

Charge of Basic Radical = + valency of the radical,Charge of Acid Radical = - velency of the radical

Task: Write the ionic representations of the following radicals: Al, Fe(ous), Cu(ic), Sn(ous),NH

4, Br, O, SO

4, NO

2, MnO

4, Cr

2O

7, C, N, PO

4Definition of Electrovalency:

Electrovalency is the number of electrons lost or gained (not shared)by a radical. Thevalency tables(table 1 and 2) given later are electrovalencies.

IMPORTANT:

In fact all the radicals listed in table 1 under basic radicals are positive ions whichare formed by the loss of electron(s). Likewise all the radicals listed in table 2 under acidradicals are negative ions which are formed by the gain of electrons. The basic and acidradicals are never to be considered as neutral species. Na is a metal which is neutral.But Na as a radical is present in a compound like NaCl. Here Na is not neutral although weconventionally write as Na. Actually it is Na+. Similarly iron(Fe) is a metal which is neutral.But Fe as a radical is present in compounds in two forms i.e ferrous(eg. ferrous sulphate)in which it remains as Fe++ or Fe2+ ions and the other ferric( e.g ferric sulphate) in whichit remains as Fe+++ or Fe3+ ions. Similarly chlorine as a gas is neutral and has formula Cl

2.

But Cl as a radical(chloride) present in any compound like NaCl is actually not Cl, insteadit is Cl– (a -ve ion).

Conclusion: All basic radicals form +ve ions and all acid radical form -ve ions. Themagnitude of charge is equal to the valency of the radical.

2 Concepts in Chemistry

Covalency :

The number of electrons which an atom shares in forming a covalent bond with otheratoms in a molecule or ion is called the covalency of the atom. There is nothing calledcovalency of a group of atoms(radical) like that we learnt in electrovalency(sulfate, carbonateetc.). Covalency is determined for an atom in a covalent molecule or ion. This type ofvalencies are found in binary covalent compounds like PCl

5, NH

3, CO

2 etc. and covalent ions

like SO4

2-, NO3

- etc. The details on covalency will be discussed later.

In compound like aluminium sulphate there are two types of valencies- electrovalency betweenAl3+ and SO

42- and covalency within S and O atoms in forming SO

42-. Similarly in compound

like Na2CO3 the valency of Na and CO3 are 1+ and 2- respectively while in CO32- thevalency of C is 4 as it is sharing four of its elelctrons with the three oxygen atoms. Note thateach covalent bond is made up of two electrons contributed one each by the the two linkingatoms.

Na2CO3 Na+

CO3

2-

+electrovalency

; C

O

O O (covalency of C=4)

Note that both electrovalency and covalency are expressed by a single term calledOxidation Number(ON) or Oxidation State(OS). The only difference between valencyand ON is that there is a sign in ON(+ve and -ve) while there no sign in valency. These thingswill be made more clear in later sections.

CHEMICAL FORMULAE

Molecular Formula:

A molecular formula is a concise way of expressing information about the atoms thatconstitute a particular chemical compound. It identifies each type of element by its chemicalsymbol and identifies the number of atoms of such element to be found in each individualmolecule of the compound. The number of atoms(if greater than one) is indicated as asubscript. Molecular formula is valid for covalent molecular substances like CO

2, NH

3, H

2SO

4,

N2O

5 etc. which exist in the form of discrete molecules.

Empirical Formula:

Empirical formula of a compound is a simple expression of the relative number ofeach type of atom or ratio of the atoms of different elements present in it. Empirical formulais valid for non-molecular substances which remain as network solids. All ionic compoundslike NaCl, CaSO

4, KNO

3 etc. and a few covalent network solids like SiO

2 fall into this

category of non-molecular substances. In any ionic compound, say for example NaCl, thereis no individual NaCl molecule. Na+ and Cl- ions are arranged alternately in a repeated mannerin three dimension to form a gigantic network. In such cases molecular formula cannot beknown rather their empirical formula which gives a simple ratio of the elements present inthe compound can be known.


Crossover Rule:

While writing chemical formula of an ionic compound, valency of basic and acidradicals are crisscrossed. In other words the valency of basic radical becomes the subscriptof acid radical and vice versa. The following examples will suffice.

Al (SO4)3 2

Al2(SO4)3

Zn (CO3)2 2

ZnCO3

Note that parenthesis is used for a compound radical which contains more than oneelement such as CO

3, SO

4 etc. only when there is any subscript associated with it. In the

first example, parenthesis is used for SO4 but not in the second example for CO

3. For

monoatomic radicals containing one element such as Al, Cl, O etc parenthesis is not used evenif there is a coefficient [e.g. AlCl

3 and not Al(Cl)

3]

Question arises why does this crisscrossing of valencies are done? This is done toequalise the total positive and negative charge as the molecule is neutral i.e net charge of themolecule should be zero.

SO4

2-

SO4

2-

SO4

2-

3+Al

3+Al

Al

3

(SO4)2

Al2(SO4)3

The hollows(depressions) in the above picture of the basic radical is the site from which electronshave been lost and the mount(bulging out) portions in the acid radical are the sites at whichelectron have been accepted. There must be a complete matching between the hollows withthe mounts.In aluminium sulphate, there are two Al3+ ions, so the total +ve charge is 2 X (+3) = +6; andthat is why there are three SO

42- ions, so that the total -ve charge is 3 X (-2) = -6. In zinc

carbonate, one Zn2+ ion has +2 charge and one CO3

2- ion has -2 charge and thus the chargebalance occurs. While writing formula for such compounds in which the valencies of acid andbasic radicals are same or are simple multiples of each other, then the valencies are simplified.In zinc carbonate the valency 2 for each radical get cancelled while writing the formula. Incupric ferrocyanide, the valencies get simplified by dividing with a factor 2. Hence to conclude,empirical formula gives the simplest whole number ratio of basic and acid radicals.

Cu [Fe(CN)6]2 4 2

42

Cu [Fe(CN)6] 2Cu [Fe(CN)6]

IMPORTANT: In ionic compounds the formula we talk of are empirical formula not molecularformula.


VALENCY TABLESAND

FORMULA WRITING TIPS

The formula writing is the most vital part in the study of chemical sciences. Unlessand until you write formula correctly it is not worthwhile to proceed further. So practiseadequately to master over it. Students often neglect this part of chemistry and becomepermanently weak in the subject. You should not try to cram the valencies of different basicand acid parts(radicals) given in the tables below rather just read them and practise writingformulae of compounds for a large number of times. Check them yourself and know yourmistakes. Within a few days of regular practice and self check, these will automatically bestored in your memory. No special effort will be necessary for the purpose.

BASIC RADICALS (TABLE-I)

Valency=1 Valency=2 Valency=3 Valency=4 Valency=5

sodium(Na) magnesium aluminium(Al) stannic arsenic(As)potassium(K) (Mg) chromic (Sn) antimonicmercurous(Hg)* barium(Ba) (Cr) plumbic (Sb)cuprous(Cu) zinc(Zn) antimonous (Pb) vanadium(V)ammonium(NH

4) cupric(Cu) (Sb) platinum

aurous(Au) mercuric(Hg) auric(Au) (Pt)silver(Ag) cobaltous arsenous(As) titaniumhydrogen(H) (Co) cobaltic(Co) (Ti)lithium(Li) nickel(Ni) bismuth(Bi)rubidium(Rb) lead(Plumbous) ferric(Fe)cesium(Cs) (Pb)thallium(Tl)(ous) ferrous(Fe) manganicantimonyl(SbO) srontium(Sr) (Mn)bismuthyl(BiO) stannous(Sn) scandium(Sc)

chromous(Cr) thallium(Tl)(ic)manganous(Mn)berryllium(Be)cadmium(Cd)calcium(Ca)palladium(Pd)platinum(Pt)

* Mercurous- Hg2

2+


ACID RADICALS(TABLE-II)

Valency = 1 Valency = 2 Valency = 3 Valency = 4fluoride(F) carbonate(CO

3) nitride(N) carbide(C)

chloride(Cl) sulphate(SO4) phosphate or ferrocyanide

bromide(Br) oxide(O) orthophosphate [Fe(CN)6]

iodide(I) sulphite(SO3) (PO

4)

nitrate(NO3) sulphide(S) ferricyanide dioxide(O

2)

nitrite(NO2) thiosulphate(S

2O

3) [Fe(CN)

6] pyrophosphate

bicarbonate or manganate(MnO4) borate or (P

2O

7)

or hydrogen chromate(CrO4) orthoborate

carbonate(HCO3) dichromate(Cr

2O

7) (BO

3)

bisulphite or oxalate(C2O

4) oxychloride(OCl)

hydorgen sulphite(HSO3) zincate(ZnO

2) phospide(P)

hypochlorite(OCl or ClO) stannite(SnO2) arsenite(AsO

3)

permanganate(MnO4) stannate(SnO

3) arsenate(AsO

4)

chlorite(ClO2) plumbite(PbO

2) cobaltinitrite

bisulphate or plumbate(PbO3) [Co(NO

2)

6]

hydorgen sulphate(HSO4) berrylate(BeO

2)

chlorate(ClO3) peroxide(O

2)

bromate(BrO3) tetrathionate(S

4O

6)

iodate(IO3) hydorgen phosphite

meta aluminate (HPO3)

(AlO2) peroxydisulphate

superoxide(O2) or persulphate(S

2O

8)

hypophosphite(H2PO

2) carbide(C

2)

perchlorate(ClO4) hydrogen phosphate

perbromate(BrO4) (HPO

4)

periodate(IO4) molybdate(MoO

4)

hydride(H) dihydrogen pyroantimonatecyanide(CN) (H

2Sb

2O

7)

isocyanide(NC) silicate or metasilicatecyanate(OCN) (SiO

3)

thiocyanate(SCN) tungstate(WO4)

isocyanate(NCO) titanate(TiO3)

isothiocyanate(NCS) tetraborate(B4O

7)

acetate(CH3COO) nitroprusside[Fe(CN)

5NO]

hydroxide(OH)dihydrogen phosphate(H

2PO

4)

metaphosphate(PO3)

chromite(CrO2)

metaborate(BO2)

vanadate(VO3)

azide(N3)

bisulphide(HS)argentocynide[Ag(CN)

2]

antimonite(SbO2)

antimonate(SbO3)

bismuthate(BiO3)

perrhenate(ReO4)

ferrite(FeO2)


SELF ASSESSMENT QUESTIONS:(SAQ)

While reading the text you will find Self Assessment Question(SAQ) very frequenty.You are advised not to skip over the SAQs. SAQs are meant for giving better understandingand build up a good foundation on the particular aspect discussed. So answer SAQs ateach step before proceeding further, check with the correct answer(response) given at theend. Rectify your mistakes if any and then proceed further.

SAQ 1: Write down the formula of the following:

(i)magnesium chloride, (ii)sodium sulphate, (iii)calcium nitrate, (iv)ferric sulphide, (v)cupricoxide, (vi)potassium sulphite, (vii)sulphuric acid, (viii)aluminium phosphate, (ix)bariumperoxide, (x)ammonium cyanide

SAQ 2: Write the formula of zinc carbonate, cuprous sulphide, ammonium sulphate,potassium oxide, mercuric sulphite, ferrous sulphide, sodium carbonate, calciumnitrite, stannous chloride, nitric acid

SAQ 3: Write the name of the following (NH4)

3PO

4, FeCl

3, K

2SO

3, Mg

3N

2, Na

2O

2,

Hg2CO

3, AlP, Ca(HCO

3)

2, HgS, H

3PO

4

SAQ 4: Write the formula of the following: Practise each set at different time andcheck. your mistakes youself.

Set-I: calcium acetate, potassium manganate, magnesium hypochlorite, ferric sulphate,mercuric nitrate, ammonium bicarbonate, cuprous sulphide, aluminium bromide, zinc sulphite,nitrous acid,

Set-II: sodium chromate, magnesium chloride, potassium phosphate, calcium phosphide,stannous sulphide, plumbous(lead)carbonate, ferrous nitrite, mercurous hydrogen carbonate,ammonium chlorate, hydrobromic acid

Set-III: sodium phosphate, calcium sulphate, potassium dichromate, magnesium chlorite,stannic chloride, aluminium carbide, silver nitrite, cobalt(ous) sulphide, ferric sulphite,calcium manganate

Set-IV: magnesium bicarbonate, sodium cyanide, ferrous sulphate, chromic hydroxide,zinc thiosulphate, potassium ferrocyanide, ammonium dichromate, arsenous nitrate,cupric(copper)bicarbonate, hypochlorous acid

Set-V: strontium nitrate, potassium sulphite, cupric ferrocyanide, aluminium phosphide,lead(plumbous) acetate, cuprous carbonate, sodium bicarbonate, chromic oxide, magnesiumperchlorate, phosphoric acid(orthophosphoric acid)

Set-VI: potassium hydrogen phosphite, sodium thiosulphate, ferrous phosphate, calciumferricyanide, stannous oxide, plumbic oxide(lead dioxide), mercuric dichromate, calciumcarbide, ammonium sulphate, chloric acid

Set-VII: sodium meta-aluminate, calcium hydride, barium peroxide, manganous phosphate,potassium chromate, magnesium permanganate, chromous oxide, ferrous hydroxide,potassium zincate, perchloric acid

Set-VIII: potassium dihydrogen pyroantimonate, ammonium oxalate, sodium tetraborate,


potassium cobaltinitrite, calcium silicate, cupric pyrophosphate, sodium tetrathionate, calciumacetate, cuprous peroxydisulphate, potassium hypophosphite

SET-IX: sodium tungstate, potassium nitroprusside, permanganic acid, ammoniumisothiocyanate, manganous bromite, potassium chromite, antimonyl chloride, mercuricperiodate, ferrous vanadate, sodium plumbate,

SET-X: aluminium hydride, bismuth hypoiodite, ferrous ferricyanide, chromic isocyanide,barium dihidrogen phosphate, rubium superoxide, ammonium molybdate, arsenic chloride,cobaltic arsenite, auric perchlorate

SET-XI: potassium argentocyanide, sodium bismuthate, ammonium metaborate, potassiumstannate, bismuthyl chloride, sodium ferrite, ferric oxide, vanadium pentoxide, plumbicsulphate, potassium antimonite


IONIC EQUATIONS

You already know that a compound has two parts or radicals namely basic radical andacid radical . For example in K

2SO

4, K is the basic part and SO

4 is the acid part.

Potassium sulphate

K2SO4 K SO4(Acid part)(Basic part)

The basic part has come from the base(in this case KOH) and the acid part has come from theacid(in this case H

2SO

4). It has already been discussed before that these are ionic compounds

and these radicals are really ions, not neutral species. In the above example, they are K+ andSO

42-, not K and SO

4 as written above.

Aqueous Solution:

Many compounds which are soluble in water form free ions on dissolution. Suchcompounds are mostly are ionic in nature. For example when solid K

2SO

4 is dissolved in water

one molecule of K2SO

4 forms two potassium ions (cations), each carrying a charge of +1 and

one sulphate ion (anion) carrying a charge of 2-. The basic part forms the cation and theacid part forms the anion.

K2SO4 K SO4(Acid part)(Basic part)

2-++2

The magnitude of charge which each ion carries is equal to the valency of the part (radical)with appropriate sign. Acid part carries -ve and basic part +ve charges. The valency ofpotassium is 1 and since it is basic radical, it will carry +1 charge. The valency of sulphate(SO

4)

is 2 and since it is acid radical, it will carry 2- (or --)charge.

Solid inorganic compounds which are soluble in water dissociate into free ions. All solidinorganic compounds whether are highly soluble in water or not have one basic and one acidpart. Silver chloride(AgCl) is feebly soluble in water(practically insoluble), but it contains thebasic radical Ag and acid radical Cl. All inorganic acids and bases also contain positive andnegative ions. The following SAQ will make the idea clear.

SAQ 5: Write down the acid and basic radicals with appropriate charge shown withthe symbol/formula. Also indicate how many ions of each type will be formed fromthem on dissociation in water.

Set-I: zinc carbonate, calcium phosphate, ammonium nitrate, hydrogen sulphide, calciumacetate, sodium chlorate, sulphuric acid, aluminium carbide, ferrous nitrite

Set-II: potassium permanganate, ferric sulphate, nitric acid, magnesium carbonate,aluminium hydroxide, sodium dichromate, ammonium perchlorate, mercurous nitrate, cupricthiosulphate, hydrochloric acid


ACIDS AND BASESAcids: substances which produce H+ ions in aqueous solution are called acids. They are of thefollowing types:

(i) Monoprotic: These possess one replaceable H atom (e.g. HCl, HBr, HI, HNO3 etc.),

(ii) Diprotic: These contain two replaceable H atoms(e.g H2SO

4, H

2CO

3 etc),

(iii) Triprotic: These contain three replaceable H atoms(e.g H3PO

4)

On the basis of their strength, acids are classified into two types.

(a) Strong Acid: which ionises almost completely in aqueous solution. For example,hydrochloric acid(HCl) when put in water ionises (dissociates) almost completely formingH+ and Cl- ions and very little unionised HCl remains after dissolution.

HCl(g) H+(aq) Cl

-(aq)+

(Very large) (Very large)(Very little)

H2O

(b) Weak Acid: A weak acid though completely soluble in water like strong acid , ionisesonly slightly or feebly in aqueous solution. HCN (hydrogen cyanide) for example whenput in water completely dissolves but ionises to a very small extent. So that very little H+

and CN- exist in solution and unionised HCN molecules exist in large number.

(Very little)HCN H

+CN

-(aq) (aq)+(g)(very large) (Very little)

H2O

Examples:

Strong Acids: HCl, HBr, HI, HNO3, HClO

4, HClO

3, H

2SO

4.These are the few strong acids.

Weak Acids: HF, HCN, H2CO

3, HNO

2, H

2SO

3, H

3PO

4, H

3PO

3, H

3PO

2, HClO, etc. and almost

all organic acids such as CH3COOH(acetic acid), HCOOH(formic acid), citric acid etc.

Bases: The oxides and hydroxides of metals are commonly called bases.(a) Soluble Strong Bases: A strong base is highly soluble in water and ionises

almost completely in water producing large number of OH- ions. The solution of the soluble baseis called an ALKALI.

+NaOH(s) Na+(aq) O H-(aq) ; CaO H2O Ca(OH)2 Ca + 2 OH2+ -

+

Oxides and hydroxides of Group 1(alkali metals e.g Li, Na, K, Rb, Cs) and three heavier elementsof Group 2(alkaline earth metals -Ca, Sr and Ba) belong to this category. Strictly speaking onlythe alkali metal oxides and hydroxides are strong bases. Oxides and hydroxides of Ca, Sr and Baare less basic and less soluble than alkali metal oxides and hydroxides.

(b) Insoluble or Weak bases: They are slightly soluble and weakly ionisable in water.

Hydroxides and oxides of all metals other than those cited under soluble strong base categoryfall into this class. e.g Fe(OH)

3, MgO etc.

Exception: NH3 is highly soluble in water but forms a weak base. when NH

3 gas dissolves

in water a weak base NH4OH is formed which ionises very little in water and forms NH

4+ and

OH- ions to a very small extent.


SAQ 6 : Indicate which one are strong acids (S) and which are weak acids(W) amongthe following:

H3PO

4, H

2SO

4, HCN, CH

3COOH, HNO

3, HF, HCl, HCOOH, HBr, HNO

2, H

2CO

3, HI,

H2SO

3, HBrO, HClO

4,

Ionic Equations:

Let us first know how ionic equations are written. For that purpose let us take an acid-basereaction. You know that an acid reacts with a base to produce a salt and water.

Acid-Base Reactions:

HCl(aq) + NaOH(aq) ---------> NaCl(aq) + H2O (l)

First, all the species which are marked with the symbol (aq) are converted to their ions(cationand anion). The symbol 'aq' stands for aqueous and such species are soluble in water andundergo ionisation to produce their respective cations(basic part) and anions(acid part). Notethat the species marked with the symbol (s), (l) or (g) representing the species existing as solid,liquid and gas respectively are written as such without converting them into their ions. Thesespecies are insoluble and do not dissolve in water to produce their respective ions.

Total Ionic Equation(TIE:

H+ Cl

- Na

+ OH

-Na

+ Cl

- H2O(l)+ + + + +

When all the species associated with (aq) mark are converted to their ions we get the equationcalled Total Ionic Equation as shown in the above example. After that we find out which ionsappear common on both LHS and RHS. These are the ions which do not actually take part in thereaction and are called spectator ions. Just like when we view a cricket match in a stadium ortelevision as spectators, we do not take part in the game. It is the the players who play and areimportant persons and not we, who merely watch the game. In the same manner, the ions whichdo not take part in the reaction are not important. In the above example, Na+ and Cl- are thespectator ions. These are to be cancelled from both the sides to get a clean equation containingonly the involved ions(the players in the chemical game). This simple ionic equation is called NetIonic Equation.

Net Ionic Equation(NIE):

+H+ OH

- H2O(l)

In this example, actually the H+ ion from an acid reacts with OH- ions from a base to formundissociated water molecule. Other ions are simply not important and do not contribute to thereaction. So they have been removed from the equation.

Note that we shall use the abbreviations TIE for Total Ionic Equation and NIE for Net IonicEquation henceforth.


SAQ7:Find the net ionic equations(NIE) in case of the following

(i) Ba(OH)2(aq) + H

2SO

4(aq) --------------> BaSO

4(s) + H

2O(l)

(ii) CH3COOH(aq) + NaOH(aq) ------------> CH

3COONa(aq) + H

2O(l)

(iii) HNO3(aq) + Cu(OH)

2 (s)-----------------> Cu(NO

3)

2(aq) + H

2O(l)

How shall we know which solid species is insoluble and hence to be associated with amark (s) and which are soluble so as to be associated with a mark (aq). Of course, there is nodifficulty in detecting liquid or gaseous species for example Cl

2(g), N

2(g), NO

2(g), H

2O

2(l),

Br2(l) etc. The real trouble lies with the solid species. The following tables give the solubility

rules for some common solids at room temperature. If a solid has a minimum solubility of 0.1mole per litre of solution at room temperature, it is regarded as a soluble solid. If the solubility isless than 0.001 mole per litre at room temperature it is regarded as insoluble solid. Solubility lyingbetween 0.001 to 0.1 mole per litre comes under slightly soluble category. In the following tablesslighly soluble solids(eg. CaSO

4, PbCl

2, PbBr

2 etc.) have been kept under insoluble category. Do

not try to cram the table, rather try to refer it at the time of your need.

Solubility Rules(Table-I)(A) Soluble Category Exceptions:(Insoluble)1. Inorganic acids (ALL) _

(HCl, H2SO

4, HNO

3 etc.)

2. alkali metal (Li+, Na+, K+ , Rb+, Cs+, Fr+)and NH

4+ salts (ALL) _

3. NO3

-(nitrates) (ALL) _4. ClO

3-(chlorates), ClO

4-(perchlorates) (ALL) _

5. CH3COO-(acetates) (ALL) _

6. SO4

2-(sulphates) Pb2+, Ca2+,Sr2+, Ba2+, Hg2

2+, Hg2+, Ag+

7. Cl-(chloride),Br-(bromide)*I-(iodide)* Hg

22+(ous), Ag+, Pb2+, Cu+, Tl+

Solubility Rules(Table-II)

(B) Insoluble Category Exceptions(Soluble)1. F-(fluorides) alkali metal(Na+, K+, NH

4+,

Rb+,Cs+, Fr+), Ag+, Tl+ flourides2. O2-(oxides), OH-(hydroxides) alkali metal (Li+, Na+, K+, Rb+,Cs+,

Fr+), alkaline earth metal (onlyCa2+,Sr2+,Ba2+ , Ra2+), Tl+ & NH

4+

oxides and hydroxides3. CO

32-(carbonates), SO

32-(sulphite) alkali metal (Li+, Na+, K+, Rb+,

PO4

3-(phosphates), Cs+, Fr+), alkaline earth metalAsO

43-(arsenates) and (only Be2+) & NH

4+ salts

CrO4

2-(chromates)4. S2-(sulphides) alkali metal (Li+, Na+, K+, Rb+,

Cs+, Fr+)alkaline earth metal(Be2+, Mg2+,Ca2+,Sr2+,Ba2+,Ra2+ )&NH

4+ sulphides

5. Ag+, Hg2

2+ and Pb2+salts acetate(CH3COO-) & nitrate(NO

3-)

* HgBr2 and HgI

2 are insoluble.


SAQ 8: Indicate whether the following are soluble(S) or insoluble(I) in water.

(i)HCl (ii)NH4Cl (iii)PbSO

4 (iv)Ca(NO

3)

2 (v)Hg

2Cl

2 (vi)Na

2SO

4 (vii)BaSO

4,

(viii)K3PO

4 (ix)H

3PO

4 (x)CH

3COONa (xi)AgBr (xii)Mg(ClO

3)

2 (xiii)AgCl,

(xiv)NaClO4(xv)Bi(NO

3)

3

SAQ 9: Indicate which are insoluble(I) and which are soluble(S) in water.(i)CaF

2 (ii)Fe(OH)

3 (iii)BeCO

3 (iv)CaCO

3 (v)CaS (vi)NaOH

(vii)NH4F (viii)Ca

3 (PO

4)

2 (ix)KF (x)CuS

Note that the solubility rules given in the above tables are never to be memorised. For thebeginners, there will be always an indication in the question about species which aresolid(s), liquid(l), gas(g) or aqueous(aq.). In case such indications are absent, then youare advised to refer the solubility rules and indicate which are soluble(aq) and which arenot (s). For gases and liquids there will be no difficulty in identifying.

SAQ 10:Write down the total and net ionic equations(TIE and NIE) for the following.The equations are not to be balanced. Do not write the number of +ve and -ve ionsproduced from each molecule while writing TIE.

(i) H2SO

4(aq) + Ca(OH)

2 (aq) → CaSO

4(aq) + H

2O(l)

(ii) AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO

3(aq)

(iii) NaOH(aq) + H3PO

4(aq) → Na

3PO

4(aq) + H

2O(l)

(iv) BaCl2(aq) + K

2CrO

4(aq) → BaCrO

4(s) + KCl(aq)

(v) Pb(NO3)

2(aq.)+ HCl(aq) → PbCl

2(s) + HNO

3(aq)

SAQ 11: Write down the Net ionic equations(NIE) for the following. Don't balance.

(i) NH4OH(aq) + AlCl

3(aq) → Al(OH)

3(s) + NH

4Cl(aq)

(ii) Fe(OH)2(s)+ HNO

3(aq) → Fe(NO

3)

2 (aq) + H

2O(l)

(iii) Na2CO

3(aq) + CaCl

2(aq) → CaCO

3(s) + NaCl(aq)

(iv) Fe2(SO

4)

3(aq) + NaOH(aq) → Fe(OH)

3(s) + Na

2SO

4(aq)

(v) HBr(aq) + Ca(OH)2(aq) → CaBr

2(aq) + H

2O(l)

SAQ 12: Write down the Net Ionic Equations for the following. Don't balance the equations.

(i) Zn(s) + HCl(aq) → ZnCl2(aq) + H

2(g)

(ii) Mg(s) + H2SO

4(aq) → MgSO

4(aq) + H

2(g)

(iii) CuSO4(aq) + Zn(s) → Cu(s) + ZnSO

4(aq)

(iv) Al(s) + NaOH(aq) → NaAlO2(aq) + H

2(g)

(v) NaOH(aq) + H3PO

4(aq) → NaH

2PO

4(aq) + H

2O(l)

SAQ 13:Write the total and net ionic equations

(i)K2Cr

2O

7(aq) +H

2SO

4(aq) +FeSO

4(aq) →Cr

2(SO

4)

3(aq)+K

2SO

4(aq)+Fe

2(SO

4)

3 (aq)+H

2O(l)

(ii)Cl2(g) + NaOH(aq) → NaClO

3(aq) + NaCl(aq) + H

2O(l)

(iii)KMnO4(aq) + H

2SO

4(aq) + H

2C

2O

4(s) → K

2SO

4(aq) + MnSO

4(aq) + CO

2(g) + H

2O(l)


OXIDATION NUMBER(O.N) or OXIDATION STATE (O.S)

At this stage it is strongly felt that the students should know about an important concept,called Oxidation Number(ON) also called Oxidation State(OS). Although this concept is mostlytheoretical in nature, it is of great significance and has tremendous application. You will knowabout this very soon.

Let us imagine a molecule of H2O. In water molecule, oxygen atom is connected with

two hydrogen atoms by two covalent bonds. Each covalent bond consists of two electrons.Imagine that all the atoms in the water molecule quarrel with each other and would like to beseparated from each other such that the electron pair constituting each covalent bond will go tothat atom which is more electronegative.

OH H Thus oxygen atom gets two additional electrons from the two hydrogen atoms and

hence carries a charge of -2 over its head after the separation. Each hydrogen atom loses oneelectron and carries a charge of +1 over its head. Thus the oxidation number(ON) of oxygen is-2 and each hydrogen is +1. We can similarly analyse for a molecule of NH

3.

NHH

H In NH3, when three pairs of electrons of the three covalent bonds go to nitrogen

atom, it gets three additional electrons and hence acquire a charge of -3 and hence its oxidationnumber(ON) is -3. Each hydrogen loses one electron and acquire a charge of +1 and hence itsON is +1.

To sum up, Oxidation Number is the theoretical charge which an atom will acquire if itis hypothetically separated from all other atoms in a molecule in such a manner thatthe bond pairs of electrons are transferred to the more electronegative atom.If the two atoms are identical ( e.g -O-O- or -S-S- etc.), then the bond pair of electronsis equally distributed(one electron each) by the two atoms. Thus each atom gets backits own electron and therefore does not acquire any charge due to that.

To find ON of an atom in a molecule in a manner described above, it is necessary tohave a knowledge of the bond structure of the molecule and electronegativity of elements. Oftenit is not easy for a beginner like you to have a thorough idea on all these now. So, we have todepend on some rules for finding the ON of an atom in a molecule or an ion. It is rather better touse these rules than to write the bond structure of the molecule and find the ON by way ofanalysis as described before.

Rules for determining O.N

(1) The O.N of any element in the elemental or uncombined state is zero. Polyatomicelements such as H

2, O

2, P

4, S

8 etc. are also included in it.

SAQ 14: What is ON of the following species: Zn, Na, Cl2, P

4, Fe

(2) The O.N of element existing as simple monoatomic ion in the compound is equal tothe charge of the ion(e.g. Na+1,Cl-1). The charge of the ion is nothing but equal tovalency with appropriate sign. For basic part(radical) +ve and acid part(radical)-ve signs are used with the the respective valences.


SAQ 15: What is ON of the underlined element:

(i) CaCl2, (ii)NH

4Cl, (iii)FeSO

4, (iv)Cu

2Br

2, (v)ZnS , (vi)Sn(NO

3)

2, (vii)Mg

3N

2, (viii)HgI

2

SAQ 16: Find the ON of the underlined elements

FeSO4, Mn

3(PO

4)

2, Zn

3P

2, As

2O

3, SnCl

4, Be

2C , KMnO

4, H

2SO

4, MgO

(3) The ON of oxygen(O) is always -2 excepting in peroxides in which it is -1,superoxides in which it is -½ and OF

2 in which it is +2.

SAQ 17: Find the ON of oxygen in the following: ZnO, O2, H

2O, H

2O

2, KO

2, Al

2O

3

(4) The ON of hydrogen(H) is always +1 excepting metal hydrides in which it is -1.

SAQ 18: Find the ON of hydrogen in the following: H2O, NH

3, H

2SO

4, NaH, CaH

2, AlH

3

(5) In a neutral compound the sum of O.Ns of all the atoms of all the elements iszero.

Example: Let us take the molecule SO3. What is the ON of S in this molecule?

Let us assume the ON of S to be x. The sum of the ONs of all the atoms in the moleculeis zero as per this rule.

x + 3(-2) =0 , So x = +6. For one oxygen atom the ON is -2 and therefore for 3 oxygenatom it is 3(-2)=-6; The ON of S in SO

3 is +6.

SAQ 19: Find the ON of underlined elements.(i)SO2, (ii)HNO

3, (iii)H

2SO

4, (iv)KMnO

4,

(v)P2O

5, (vi)K

2Cr

2O

7, (vii)HClO

4

(6)The sum of ONs of all the atoms in a polyatomic ion is equal to the charge of theion, while the charge of the ion is equal to the valency of the ion with the appropriatesign(acid radical -ve and basic radical +ve)

Example: SO42-. Since the valency of sulphate radical is 2, its charge will be -2 as it is an acid

radical. Let us find the ON of S in the ion. According to the above rule the sum of the ONs of allthe atoms should be equal to the charge of the ion.

x + 4(-2) = -2 ⇒ x = +6. So the ON of S in sulphate ion is +6.NB: Conventionally while writing ON, the sign comes first followed by the numericalvalue. But when we talk of real ion the sign comes after the numerical value. For examplewhen we say ON of S in SO

42- , it is +6 but when we say the charge of the sulphate ion, it

is written as 2- or - - . This is merely a matter of convention used to distinguish betweenthe two concepts; while ON is the theoretical charge which one atom will carry, an ionhas the actual charge which the whole ion carries. Note that often students don't give +vesign in ONs and write merely 7 instead of +7. This is an error. Always a sign + or - has tobe placed before the numerical value of ON.

SO42-+6O.N Charge

Also note that the ON is to be placed just above the atom(not as superscript written for charge).


COVALENCY:

The covalency of an atom in a covalent molecule or ion is the number of electrons which theatom shares with other atoms while forming the covalent bonds. The covalency of an atom isequal to oxidation number in magnitude. There is no sign(+ve or -ve) in covalency. It is a purenumber. For example, the ON of S in SO

42- is +6 and so its covalency is 6.

SAQ 20:

(A) Find the O.N of the underlined atom(i) NH

3, (ii) KNO

3, (iii) NO

2-, (iv) NO

2, (v) N

2, (vi) N

2O

5, (vii) KMnO

4, (viii) Cr

2O

72-

(ix) S2O

32-, (x) ClO

3-, (xi) NH

4+, (xii)NO

3-, (xiii) NH

4NO

3,, (xiv) MnO

2, (xv) Cr

2(SO

4)

3,

(xvi) HClO, (xvii) SO3

2-, (xviii) HCl, (xix) SbCl3, (xx) PH

3

(B) Find the covalency of the underlined atoms in case of the following.(i) NO

3- (ii) SO

3(iii) ClO

3-

(iv) NH3

(v)N2O

5

Oxidation and Reduction (Redox Reaction)This type of reaction is called oxidation-reduction(or redox) reaction, which involves

both oxidation as well as reduction. You already know about some primitive definitions of oxidationand reduction; oxidation as the addition of oxygen or removal of hydrogen and reduction asaddition of hydrogen or removal of oxygen. These old definitions have been outdated. Now youwill be knowing the most modern definitions of oxidation and reduction.

Oxidation: A process in which ON of certain element is increased is called oxidation.This takes place by the loss of electron. Thus oxidation can also be defined as a process inwhich electron(s) is(are) lost.

Reduction: A process in which ON of certain element is decreased is calledreduction. This takes place by the gain of electron. Thus reduction can also be defined asa process in whcih electron(s) is(are) gained. Look to the following example.

+Zn HCl ZnCl2 H2+0 +2+1 0

[O]

[R]

(RA) (OA)

In this example, the ON of Zn was zero(uncombined state) in LHS and it becomes +2 in ZnCl2

in RHS. Thus the ON of Zn has increased from 0 to +2 due to loss of 2 electrons by neutral Znatom to form Zn2+ ion. This process is oxidation. We say that Zn is oxidised to ZnCl

2. On the

other hand, the ON of H in HCl is +1 in LHS and in H2 it is 0 in the RHS. Thus the ON of H has

decreased from +1 to 0 due to gain of one electron. This is the reduction process. We say thatHCl is reduced to H

2. We found also that wherever there is oxidation, there is a reduction.

Whenever there is a loss there is definitely a gain. Just like, if you lose a 100 rupee note on theroad, it is your loss, but it is the gain of a person who finds it. Like loss and gain, oxidation andreduction also go simultaneously. There cannot be an oxidation without having a reduction orvice versa.


The substance which is oxidised is called Reducing Agent(RA) and the substancewhich is reduced is called OxidisingAgent(OA). In the above reaction, Zn is oxidised, hence itis the reducing agent(RA) as it reduces H+ to H

2(ON=0). HCl is reduced, hence it is the

oxidising agent(OA) as it oxidises Zn(ON=0) to Zn2+.

SAQ 21: Show by indicating the ON which is oxidised and which reduced. Alsoindicate which is Oxidising Agent(OA) and which Reducing Agent(RA). The elementswhich have undergone changes in ON have been underlined.The equations are notbalanced. Do not also try to balance them.

(i) Na + H2O → NaOH + H

2

(ii) N2 + H

2 → NH

3

(iii) P4 + Cl

2 → PCl

3

(iv) NH3 + CuO → Cu + N

2 + H

2O

(v) H2 + I

2 → HI

(vi) Mg + H2SO

4 → MgSO

4 + H

2

(vii) N2 + O

2 → NO

(viii) Cu + HNO3 → Cu(NO

3)

2 + NO

2 + H

2O

(ix) MnO2 + HCl → MnCl

2 + H

2O + Cl

2

(x) KMnO4 + H

2SO

4 + FeSO

4 → K

2SO

4 + MnSO

4 + Fe

2(SO

4)

3 +H

2O

Broad Classification of Inorganic Reactions:All inorganic reactions belong to the following two types.(i)Redox type: Already discussed above (ii)Metathesis type(Non-redox): No changeof ON occurs in respect of any element.

(i) Redox Reactions: Change in ON takes place in this case. One is oxidised(ON increased)and the other is reduced(ON is decreased). We have already discussed this before.

ex. Zn + HCl ZnCl20 +1 +2

H2+0

[O]

[R]

(ii) Metathesis Reactions: No change in ON takes place for any element. The hundreds ofdouble replacement reactions(also called as double decomposition) that you know belongto this category. Look to the following reactions.

eg. AgNO3(aq) HCl(aq) AgCl(s) HNO3(aq)+ ++1 +1-1 -1

NaOH H2SO4 Na2SO4 H2O(l)+ ++1 +1+1 +1

In such reactions no element undergoes change in ON. You may calculate the ON of eachelement and find for yourself that no change in ON has taken place. You are advised to avoidcalculation of ON once you see a reaction belonging to double displacement reaction type. Insuch reactions, change in ON does not take place as the same acid and basic radical(ions) arepresent on both the sides although their partners are changed.


Caution: Remember that there cannot be any reaction involving only oxidation and no reductionor vice versa. In other words, the reaction cannot be an intermediate between redox andmetathesis types. The following example will explain it more clearly.

Na2O H2O NaOH H2+ ++1 +1+1 0

(wrong)

In this case only H has undergone reduction from +1 to 0 and there is no oxidation. This is notpossible. So the products have been wrongly predicted. The correct reaction is

Na2O H2O NaOH ++1 +1+1 +1

(It is a metathesis reaction, not redox)

SAQ 22: Indicate by assigning ON which reactions are redox and which metathesis. Ifyou find a reaction to be double resplacement type(metathesis), don't find the ONs ofatoms. Leave them as such. There is no change in ON in those reactions.

(i) H2 + Cl

2 --------> HCl

(ii) Zn(NO3)

2 + NH

4OH ---------> Zn(OH)

2 + NH

4NO

3

(iii) Cu + HNO3 ------> Cu(NO

3)

2 + NO

2 + H

2O

(iv) Na + H2O ---------> NaOH + H

2

(v) NaOH + HNO3 -----> NaNO

3 + H

2O

(vi) AgNO3 + HCl -----> AgCl + HNO

3

PRACTICE QUESTIONS1. Using solubility rules, predict which of the following are insoluble(I) and whcih

are soluble(S) in water.

NaNO3, KBr, MgF

2, FeCl

2, MgCO

3, BaSO

4, MgS, Na

3PO

4, Na

2CrO

4, Ba(ClO

4)

2,

Ba(OH)2, Hg

2Cl

2, NH

4F, BeCO

3, PbI

2, Ag

2SO

4, LiBr, K

2Cr

2O

7, AgCl, Mg

3(PO

4)

2

2. Write down the anions produced from the following acids in water.

HClO3, H

3PO

4, CH

3COOH, HCN, HOCl, H

2SO

4, HNO

2, HMnO

4, H

3PO

3, H

3PO

2,

HIO4, HClO

2,

3. Find the total and net ionic equations in case of the following.Balancing is not necessary.

(i) HI(aq) + KOH(aq) --------> KI(aq) + H2O(l)

(ii) NaOH(aq) + HClO4(aq) --------> NaClO

4(aq) + H

2O(l)

(iii) HNO3(aq) + NH

3(g) ----------------> NH

4NO

3 (aq)

(iv) CuSO4(aq) + H

2S(g) ---------------> CuS(s) + H

2SO

4(aq)

(v) H2SO

4(aq) + MgO(s) ------------> MgSO

4(aq) + H

2O(l)

(vi) Zn(s) + NaOH(aq) --------------> Na2ZnO

2(aq) + H

2(g)

(vii) Al(OH)3(s) + HBr(aq) ---------> AlBr

3(aq) + H

2O(l)

(viii) Mn(NO3)

2(aq) + Na

2S(aq) ---------> MnS(s) + NaNO

3(aq)

(ix) Ba(OH)2(aq) + H

2SO

4(aq) --------------> BaSO

4(s) + H

2O(l)

(x) H3PO

4(aq) + Ca(OH)

2 (aq)-----------------> Ca

3(PO

4)

2 (s)+ H

2O(l)

(xi) MnO2(s) + HCl(aq) ------> MnCl

2(aq) + Cl

2(g) + H

2O(l)

(xii) KBr(aq) + H2SO

4(aq) -------> K

2SO

4(aq) + Br

2(l) + SO

2(g) + H

2O(l)


(xiii) Cu(s) + HNO3(aq) --------->Cu(NO

3)

2 + H

2O(l) + NO(g)

(xiv) K2Cr

2O

7(aq) +H

2SO

4(aq) + KNO

2(aq) -------> K

2SO

4 + Cr

2(SO

4)

3(aq) + KNO

3(aq)+ H

2O(l)

(xv) I2(s) + Na

2S

2O

3(aq) -----> Na

2S

4O

6(aq) + NaI(aq)

(xvi) HI +HNO2 → NO + H

2O + I

2

(xvii) KOH + KMnO4 → K

2MnO

4 + O

2 + H

2O

4. Predict the products of the following neutralisation reactions and also write the total andnet ionic equations.These are the simple double replacement(ion exchange) reactionsthat you know. Refer the tables on solubility rules to know whether the salt produced issoluble(aq) or insoluble(s).(i) NaOH(aq) + HBr(aq) → _________ + _________(ii) Ca(OH)

2(aq) + HNO

3(aq) → _________ + _________

(iii) H2SO

4(aq) + KOH(aq) → _________ + _________

(iv) Ba(OH)2(aq) + HCl(aq) → _________ + _________

(v) Mg(OH)2 (s) + H

2SO

4(aq) →_________ + _________

(vi) MnO(s) + H3PO

4 (aq) → _________ + _________

(vii) Al2O

3 (s)+ H

2SO

4 (aq) → _________ + _________

(viii) Fe(OH)3 (s)+ HBr(aq) → _________ + _________

(ix) NaOH(aq) + HClO4 (aq) → _________ + _________

(x) Cr2O

3(s) + H

2SO

4(aq.) → _________ + _________

5. Assign the Oxidation Number(ON) to element underlined in the following.

NaCl, BaF2, NH

3, H

2O, NaH, SO

3, CO

2, P

2O

3, HClO, MnCl

2, As(NO

3)

3, SnSO

4,

Hg2Cl

2, H

2SO

4, BCl

3, SF

6, HClO

3 , SO

32-, S

4O

62- , NH

4+, NO

3-, NO

2-, H

2CO

3, ClO

2-,

K2Cr

2O

7, Na

2O

2, MnO

2, SnCl

4, FeBr

3, Cu

2Cl

2, NH

4NO

3, NO, N

2O, N

2, ZnO, K

2O,

Ca(OH)2, Al, Al

2(SO

4)

3, OF

2, N

2O

5, S

2O

32-, OCN-, HCN, S

2O

82-, H

2SO

5, CrO

5, KO

2,

Ni(CO)4, CaS

5, [SnS

3]2-,[Fe(CN)

6]4-,

XeOF4, NaN

3, SCN-, O

2F

2, CN-, N

3H, [Ni(CN)

4]2-, [Cr(H

2O)

4(CN)

2]+, C

3H

8, C

6H

12O

6,

BrCl3, HIO

3, H

2[SiF

6], CaNCN, NaCN, Ca(OCl)Cl, S

2O

32-

6. Indicate in the following reactions which is oxidized and which reduced. Also identifywhich is Oxidising Agent(OA) and which Reducing Agent(RA). Assign the ON of theinvolved atoms and show the Oxidation and Reduction processes by arrow mark.(i) K + H

2O → KOH + H

2

(ii) NH3 + O

2 → N

2 + H

2O

(iii) Na + NH3 → NaNH

2 + H

2

(iv) Ca + H2O → Ca(OH)

2 + H

2

(v) Al + NaOH + H2O → NaAlO

2 + H

2

(vi) Zn + H2SO

4 → ZnSO

4 + H

2

(vii) CuO + NH3 → Cu + H

2O + N

2

(viii) P4 + O

2 → P

2O

5

(ix) H2 + Cl

2 → HCl

(x) H2O

2 + H

2S → S + H

2O


(xi) FeBr3 + Cl

2 → FeCl

3 + Br

2

(xii) I2 + Na

2S

2O

3 → NaI + Na

2S

4O

6

(xiii) SnCl2 + HgCl

2 → SnCl

4 + Hg

2Cl

2

(xiv) Cu + HNO3 → Cu(NO

3)

2 + NO

2 + H

2O

(xv) KMnO4 + H

2SO

4 + FeSO

4 --------> K

2SO

4 + MnSO

4 + Fe

2(SO

4)

3 + H

2O

(xvi) K2Cr

2O

7 + H

2SO

4 + H

2S --------> K

2SO

4 + Cr

2(SO

4)

3 + S + H

2O

7. Indicate in the following reactions which is oxidized and which reduced. Also identifywhich is Oxidising Agent(OA) and which Reducing Agent(RA). Assign the ON of theinvolved atoms and show the Oxidation and Reduction processes by connecting lines.(i) Fe

2O

3 + CO → CO

2 + Fe

(ii) I2 + S

2O

32- → S

4O

62- + I-

(iii) Fe2(SO

4)

3 + H

2O

2 → FeSO

4 + O

2 + H

2SO

4

(iv) Ca(OCl)2 + KI + HCl → I

2 + CaCl

2 + H

2O + KCl

(v) PbO2 + HBr → PbBr

2 + Br

2 + H

2O

(vi) HNO3 + Zn → Zn(NO

3)

2 + NH

4NO

3 + H

2O

(vii) NH4NO

3 → N

2O + H

2O

(viii) Pb(NO3)

2 → PbO + NO

2 + O

2

(ix) P4 + H

2SO

4 → SO

2 + H

3PO

4 + H

2O

(x) Cl2 + NaOH → NaCl + NaClO

3 + H

2O

8. Show by indicating the ON which is oxidised and which reduced.(i) CuS + HNO

3 ------> Cu(NO

3)

2 + S + H

2O + NO

(ii) Zn + HNO3 ------> Zn(NO

3)

2 + H

2O + NH

4NO

3

(iii) MnO + PbO2 + HNO

3 -----> HMnO

4 + Pb(NO

3)

2 + H

2O

(iv) KMnO4 + KCl + H

2SO

4 ----------> MnSO

4 + K

2SO

4 + Cl

2 + H

2O

(v) K2C

2O

4 + K

2Cr

2O

7 + H

2SO

4 ---------> K

2SO

4 + Cr

2(SO

4)

3 + CO

2 + H

2O

(vi) HNO3 + HI -----> NO + I

2 + H

2O

(vii) NH3 + O

2 ------> NO + H

2O

(viii) CuO + NH3 -----> N

2 + Cu + H

2O

(ix) SO2 + H

2S ----> S + H

2O

(x) NaNO3 -----> NaNO

2 + O

2

(xi) KClO3 ----> KCl + O

2

(xii) Pb(NO3)

2 ------> PbO + NO

2 + O

2

(xiii) Zn + NaOH --------> Na2ZnO

2 + H

2

(xiv) KClO3 + H

2SO

4 -----> KHSO

4 + O

2 + ClO

2 + H

2O

(xv) Sn + HNO3 -----> SnO

2 + NO

2 + H

2O

(xvi) I2 + HNO

3 -----> HIO

3 + NO

2 + H

2O

(xvii) KBr + H2SO

4 -----> K

2SO

4 + Br

2 + SO

2 + H

2O

(xviii) Ca3(PO

4)

2 + SiO

2 + C ------> CaSiO

3 + P

4 + CO

(xix) Na + NH3 -----> NaNH

2 + H

2

(xx) H2O

2 + H

2S -----> S + H

2O

(xxi) Cr(OH)3 + Na

2O

2 ----------> Na

2CrO

4 + H

2O

(xxii) O3 + Hg -------> Hg

2O + O

2


(xxiii) K4[Fe(CN)

6] + O

3 + H

2O -----> K

3[Fe(CN)

6] + O

2 + KOH

(xxiv) NaBrO3 + NaBr + HCl -----> NaCl + Br

2 + H

2O

(xxv) KMnO4 ---------> K

2MnO

4 + MnO

2 + O

2

(xxvi) MnSO4 + (NH

4)

2S

2O

8 + H

2O → MnO

2 + H

2SO

4 + (NH

4)

2SO

4

(xxvii) Zn + H2SO

4 + As

2O

3 → AsH

3 + ZnSO

4 + H

2O

(xxviii) ClO2 + H

2O

2 + NaOH → NaClO

2 + O

2 + H

2O

(xxix) N2H

4 + BrO

3- → N

2 + Br - + H

2O

(xxx) CuSO4 + KI → Cu

2I

2 + K

2SO

4 + I

2

(xxxi) Fe + H2O → Fe

3O

4 + H

2

(xxxii) H2O

2 + NaOCl → NaCl + O

2 + H

2O

RESPONSE TO SAQSSAQ 1:

(i) Mg2 Cl1 = MgCl2

(ii) Na1 (SO4)2 = Na

2SO4

(iii)Ca2 (NO3)1 = Ca(NO

3)

2(iv) Fe3 S2 = Fe

2S

3

(v) Cu2 O2 = CuO (vi) K1 (SO3)2 = K

2SO

3

(vii)H2SO

4(viii) Al3 (PO

4)3 = AlPO4

(ix) Ba2 (O2)2 = BaO

2(x) (NH

4)1 (CN)1 = NH

4CN

Note that the parenthesis used for SO4 in (ii) has been removed as there is no coefficient

for it. But in (iii) the parenthesis is retained for NO3 as there is a coefficient 2 for it. Similarly

the parentheses have been removed in (vi), (viii), (ix) and (x). If you are not thorough inwriting the formula you first practise in the way shown above i.e first place the valencies atthe top (superscript) and simplify if required and then crisscross them as subscripts to writethe formula.

SAQ 2: ZnCO3, Cu

2S, (NH

4)

2SO

4, K

2O, HgSO

3, FeS, Na

2CO

3, Ca(NO

2)

2, SnCl

2,

HNO3

SAQ 3: ammonium phosphate, ferric chloride, potassium sulphite, magnesium nitride,sodium peroxide, mercurous carbonate, aluminium phosphide, calcium bicarbonate (orcalcium hydrogen carbonate), mercuric sulphide, phosphoric acid

SAQ 4: Set-I: Ca(CH3COO)

2, K

2MnO

4, Mg(OCl)

2, Fe

2(SO

4)

3, Hg(NO

3)

2, NH

4HCO

3,

Cu2S, AlBr

3, ZnSO

3, HNO

2,

Set-II: Na2CrO

4, MgCl

2, K

3PO

4, Ca

3P

2, SnS, PbCO

3, Fe(NO

2)

2, HgHCO

3, NH

4ClO

3,

HBr

Set-III: Na3PO

4, CaSO

4, K

2Cr

2O

7, Mg(ClO

2)

2, SnCl

4, Al

4C

3, AgNO

2, CoS, Fe

2(SO

3)

3,

CaMnO4


Set-IV: Mg(HCO3)

2, NaCN, FeSO

4, Cr(OH)

3, ZnS

2O

3, K

4[Fe(CN)

6], (NH

4)

2Cr

2O

7,

As(NO3)

3, Cu(HCO

3)

2, HOCl(HClO)

Set-V: Sr(NO3)

2, K

2SO

3, Cu

2[Fe(CN)

6], AlP, Pb(CH

3COO)

2, Cu

2CO

3, NaHCO

3, Cr

2O

3,

Mg(ClO4)

2, H

3PO

4

Set-VI: K2HPO

3, Na

2S

2O

3, Fe

3(PO

4)

2, Ca

3[Fe(CN)

6]

2, SnO, PbO

2, HgCr

2O

7, CaC

2,

(NH4)

2SO

4, HClO

3

Set-VII: NaAlO2, CaH

2, BaO

2, Mn

3(PO

4)

2, K

2CrO

4, Mg(MnO

4)

2, CrO, Fe(OH)

2,

K2ZnO

2, HClO

4

Set-VIII: K2H

2Sb

2O

7, (NH

4)

2C

2O

4, Na

2B

4O

7, K

3[Co(NO

2)

6], CaSiO

3, Cu

2P

2O

7,

Na2S

4O

6, Ca(CH

3COO)

2, Cu

2S

2O

8, KH

2PO

2

SET-IX: Na2WO

4, K

2[Fe(CN)

5(NO)], HMnO

4, NH

4NCS, Mn(BrO

2)

2, KCrO

2, SbOCl,

Hg(IO4)

2, Fe(VO

3)

2, Na

2PbO

3

SET-X: AlH3, Bi(OI)

3, Fe

3[Fe(CN)

6]

2, Cr(NC)

3, Ba(H

2PO

4)

2, RbO

2, (NH

4)

2MoO

4,

AsCl5, CoAsO

3, Au(ClO

4)

3

SET-XI: K[Ag(CN)2], NaBiO

3, NH

4BO

2, K

2SnO

3, BiOCl, NaFeO

2, Fe

2O

3, V

2O

5,

Pb(SO4)

2, KSbO

2

SAQ 5:

Set-I: Zn2+ and CO3

2-, 3Ca2+ & 2PO43-, NH

4+ & NO

3-, 2H+ & S2-, Ca2+ &

2CH3COO-, Na+ & ClO

3-, 2H+& SO

42-, 4Al3+ & 3C4-, Fe2+ & NO

2-

Set-II: K+ & MnO4-, 2Fe3+ & 3SO

42-, H+ & NO

3-,Mg2+ & CO

32-, Al3+ & 3OH-,

2Na+ & Cr2O

72, NH

4+ & ClO

4-, Hg+(or Hg

22+) & NO

3-(2 NO

3-) Cu2+ &

S2O

32-, H+ and Cl-

(Note that the charges of the acid and basic parts have been given to all without consideringwhether they are soluble in water or not)

SAQ 6: H3PO

4(W), H

2SO

4(S), HCN(W), CH

3COOH(W),HNO

3(S), HF(W),

HCl(S),HCOOH(W), HBr(S), HNO2(W), H

2CO

3(W),HI(S), H

2SO

3(W), HBrO(W),

HClO4(S)

SAQ7: (i) TIE: Ba2+ + OH- + H+ + SO42- ------------> BaSO

4(s) + H

2O(l)

Since BaSO4 is associated with a (s) mark, it is insoluble and hence it has not been

ionised. H2O also is associated with (l) mark, it has also not been ionised. Note that in the above

equation we should have written 2 OH- and 2 H+ in the LHS which we did not. This is becausewe are not intending to balance the equation now and as a matter of fact the equation given inthe question is not a balanced equation. Our sole purpose is to write the ions with their appropriatecharges. That is all. We don't have to bother how many of each ion is produced in the LHS andRHS at this moment. In other words, we are not interested in balancing the equation now.

NIE: Ba2+ + OH- + H+ + SO42- ------------> BaSO

4(s) + H

2O(l)

Since no ion is found to be common in both sides, no ion is a spectator. All ions are involved andhence the net ionic equation is same as total ionic equation.


(ii) TIE: CH3COO- H+ Na+ OH- CH3COO- Na+ H2O(l)+ + + + +

NIE: H+ +OH- ----------> H2O(l)

In this case, like the first example shown in the text, H+ ion from acid reacts with OH- ion frombase to form undissociated H

2O molecule. In fact this is the net ionic equation for almost all acid-

base(neutralisation) reactions.

(iii) TIE: H+ + NO3

- + Cu(OH)2(s) ----------> Cu2+ + NO

3- + H

2O(l)

In this case, NO3

- is the spectator ion which is cancelled from both the sides to get NIE.

NIE: H+ + Cu(OH)2(s) ----------> Cu2+ + H

2O(l)

SAQ 8: Refer the solubility rules to verify answer.

(i) HCl(S): All inorganic acids are soluble; (ii)NH4Cl(S): All ammonium salts are soluble

(iii) PbSO4-(I): All sulphates are soluble excepting a few such as PbSO

4, SrSO

4, BaSO

4,

HgSO4, Ag

2SO

4 which are insoluble.

(iv) Ca(NO3)

2- (S): All nitrates are soluble: (v)Hg

2Cl

2-(I): All chorides are soluble excepting

a few such as Hg2Cl

2, AgCl and PbCl

2 which are insoluble.

(vi) Na2SO

4- (S): All Na salts are soluble; (vii)BaSO

4-(I): (viii)K

3PO

4- (S): All

K salts are soluble; (ix)H3PO

4-(S): All inorganic acids are soluble;

(x) CH3COONa-(S): All Na salts and all acetates are soluble.

(xi) AgBr-(I): All bromides are soluble excepting Ag+, Hg22+ and Pb2+.

(xii) Mg(ClO3)

2- (S): All chlorates are soluble; (xiii)AgCl-(I)

(xiv) NaClO4-(S): All perchlorates are soluble.

(xv) Bi(NO3)

2 - (S): All nitrates are soluble

SAQ 9:

(i) CaF2-(I): All fluorides are insoluble excepting those of Na+, K+, NH

4+, Ag+, Tl+

which are soluble.(ii) Fe(OH)

3 -(I): All hydroxides are insoluble excepting those of Na+, K+, NH

4+, Ca2+,

Sr2+, Ba2+ which are soluble.(iii) BeCO

3-(S): All carbonates are insoluble excepting those of Na+, K+, NH

4+, Be2+

which are soluble.(iv) CaCO

3- (I)

(v) CaS-(S): All sulphides are insoluble excepting Na+, K+, NH4

+, Mg2+,Ca2+,Sr2+,Ba2+

which are soluble.(vi) NaOH-(S)(vii) NH

4F-(S)

(viii) Ca3(PO

4)

2-(I): All phosphates are insoluble excepting those of Na+, K+, NH

4+, Be2+

which are soluble.(ix) KF-(S)(x) CuS-(I)


SAQ 10:

(i) TIE: H+ + SO42- + Ca2+ + OH- ---------> Ca2+ + SO

42- + H

2O(l)

NIE: H+ + OH- -----------> H2O

(ii) TIE: Ag+ + NO3- + Na+ + Cl- -----------> AgCl(s) + Na+ + NO

3-

NIE: Ag+ + Cl- ---------> AgCl(s)(iii) TIE: Na+ + OH- + H+ + PO

43- ---------> Na+ + PO

43- + H

2O(l)

NIE: H+ + OH- ---------> H2O

(iv) TIE: Ba2+ + Cl- + K+ + CrO4

2- ----------> BaCrO4(s) + K+ + Cl-

NIE: Ba2+ + CrO4

2- -----------> BaCrO4(s)

(v) TIE: Pb2+ + NO3- + H+ + Cl- ------------> PbCl

2(s) + H+ + NO

3-

NIE: Pb2+ + Cl- -----------> PbCl2(s)

Note that in all these equations, balancing has not been done.

SAQ 11:(i) Al3+ + OH- -------> Al(OH)

3(s)

(ii) Fe(OH)2(s) + H+ --------> Fe2+ + H

2O(l)

(iii) Ca2+ + CO3

2- --------> CaCO3(s)

(iv) Fe3+ + OH- -----------> Fe(OH)3(s)

(v) H+ + OH- ---------> H2O(l)

In all the above equations the students are advised to write first the total ionic equation(TIE) andthen write net ionic equations(NIE) after cancelling the spectator ions.

SAQ 12:(i) Zn(s) + H+ ------------> Zn2+ + H

2(g)

(ii) Mg(s) + H+ -----------> Mg2+ + H2(g)

(iii) Cu2+ + Zn(s) ---------> Cu(s) + Zn2+

(iv) Al(s) + OH- ---------> AlO2

- + H2(g)

(v) OH- + H+ ----------> H2O(l)

SAQ 13:

(i) TIE: K+ + Cr2O

72- + H+ + SO

42- + Fe2+ + SO

42- ---------> Cr3+ + SO

42- + K+ + SO

42- + Fe3+ + SO

42- + H

2O(l)

NIE: Cr2O

72- +H+ + Fe2+ --------------> Cr3+ + Fe3+ + H

2O(l)

Note that we have cancelled all K+ ions and all SO42- ions from LHS and RHS without considering

whether such ions appear same number times in both sides or not. SO4

2- ions appeared twotimes in the LHS while the same appeared three times in the RHS. Even then we cancelled allof them. This is because we have not balanced the equation and had we balanced the equation,the number of K+ ions and SO

42- ions would have definitely appeared same number of times in

both the sides. Since we are interested only to write the net ionic equations, we tactfully detectedthe unimportant(spectator) ions which are not involved in the reaction and cancelled them.

(ii) TIE: Cl2(g) + Na+ + OH- ---------> Na+ + Cl- + Na+ + ClO

3- + H

2O(l)

NIE: Cl2(g) + OH- ---------> Cl- + ClO

3- + H

2O(l)

Here also we cancelled one Na+ in LHS and two Na+ in the RHS. To remind you again that wehave not done anything wrong as we merely want the net ionic equation which is not balanced.It may be remembered here that while cancelling the spectator ions from the two sides the


students should find first the ions which do not take part in the reaction. Then remove them allwithout considering how many times they appear in which side.

(iii) TIE: K+ + MnO4- + H+ + SO

42- + H

2C

2O

4(s) -------------> K+ + SO

42- + Mn2+ + SO

42- + CO

2(g) + H

2O(l)

NIE: MnO4- + H+ + H

2C

2O

4(s) ----------------> Mn2+ + CO

2(g) + H

2O(l)

Here also we have cancelled SO4

2- one time from LHS and two times from RHS. Never mind,we are not going to get a balanced ionic equation now.

SAQ 14: Zero. All of them are in the uncombined elementary state.

SAQ 15: (i) Ca = +2(Since the valency of Ca is 2), (ii) Cl= -1(since the valency of Cl is 1)

(iii) Fe = +2(since valency of ferrous iron is 2), (iv) Cu = +1(since valency of cuprous is 1)(v) S = -2(since valency of sulphide is 2), (vi) Sn = +2(since valency of stannous is 2)(vii)Mg = +2(valency of Mg is 2) and N = -3(valency of nitride is 3)

(viii) Hg =+2(valency of mercuric is 2) and I = -1(valency of iodide is 1)

Note that the basic parts have been given +ve charge and acid parts -ve charge.

SAQ 16: +2, +2, -3, +3, +4, -4, +1, +1, -2

SAQ 17: -2, 0 (uncombined state), -2, -1(since it is peroxide: O2 has a charge of -2, so for one

oxygen atom it is -1, refer the valency table), -½ (since it is superoxide: O2 has a charge of -1, so

for one oxygen atom it is -½, refer the valency table),-2

SAQ 18: +1, +1, +1, -1, -1, -1 (since the last three compounds are metallic hydrides)

It is to to be noted that ON always is assigned for one atom, for H2O, the ON of

hydrogen is +1, not +2, similarly for NH3, the ON of H is +1 not +3. But for calculation purpose

when we find the sum of ONs we take +2 and +3 respectively. We shall see this in the SAQ 19.

SAQ 19:(i) x +2(-2) =0, ⇒ x = +4(ii) +1+x +3(-2) =0, ⇒ x = +5,(iii) +2 +x +4(-2)=0, ⇒ x =+6, Note that for one H atom the ON is +1 and so for 2 H

atoms the total ON is +2.(iv) +1 + x + 4(-2) = 0 ⇒ x = +7(v) 2x + 5(-2) =0 ⇒ x=+5(note that since there are two P atoms we wrote 2x but ON

of P=x).(vi) +2 +2x + 7(-2) =0 ⇒ x= +6,(there are two K atoms and each has an ON of +1,

again there are two Cr atoms each has ON of x)(vii) +1+x+4(-2)=0 ⇒ x= +7

SAQ 20:(A)

(i) x+3=0, ⇒ x= -3 (ii) +1+x+3(-2) = 0, ⇒ x=+5, (iii)x+2(-2)=-1 ⇒ x=+3,(iv) x+2(-2)=0, ⇒ x=+4, (v) 0(uncombined state) (vi) 2x+5(-2)=0, ⇒ x=+5,(vii) +7(see SAQ 19), (viii) 2x-14=-2, ⇒ x=+6, (ix) 2x-6=-2, ⇒ x=+2, (x) x-6=-1, ⇒ x=+5(xi) x+4=+1, ⇒ x=-3, (xii) x-6=-1, ⇒ x=+5 (xiii) For NH

4NO

3, we have to find separately

for the two N atoms, once taking NH4+(already found in bit xi as -3) and and then taking NO

3-

(already found in bit xii as +5). Note that in this compound you should not find a single ON for N


by taking 2x for N, (xiv) x-4=0, ⇒ x=+4, (xv) +3, since the valency of chromic(Cr), amonoatomic ion is 3, hence its ON has to be +3, (xvi) +1+x-2 =0, ⇒ x=+1, (xvii) x-6=-2, ⇒x=+4, (xviii) -1(Cl is a monoatomic ion having valency 1, so ON is -1) (xix) +3, Sb(ous) isa monoatomic ion having valency 3, hence ON +3. (xx) x+3 =0, ⇒ x= -3(B) (i)(NO

3-) 5 (ii) (SO

3) 6 (iii)(ClO

3-) 5 (iv) (NH

3) 3 (v) (N

2O

5) 5

SAQ 21:

Student is advised to find the ON of the elements marked with asterisk by x method explainedbefore and then proceed to analyse which is oxidation and reduction.

(i) LHS: Na=0(uncombinded state), H in H2O=+1, RHS: Na in NaOH=+1 and H in

H2=0(uncombinded state)

Na H2O NaOH H2+ +0 +1 0+1

[O]

[R](OA)(RA)

In this case Na is oxidised to NaOH and hence acts as reducing agent(RA) while H2O is

reduced to H2 and hence acts as oxidising agent(OA).

(ii) LHS: N in N2 =0(uncombinded state), H in H

2=0,RHS: N in NH

3= -3, H in NH

3=+1

N2 H2 0 0 -3 +1

[O]

NH3[R]

+(RA)(OA)

N2 is reduced to NH

3 as its ON decreased from 0 to -3 and hence acts as OA while H

2 is

oxidised to NH3 as its ON increased from 0 to +1. So it acts as RA.

(iii) LHS: P in P4=0(uncombined state), Cl in Cl

2= 0, RHS: P in PCl

3=+3, Cl in PCl

3= -1

P4 Cl2 PCl3+0 0 +3 -1

[O]

[R](RA) (OA)

(iv) LHS: N in NH3=-3, Cu in CuO = +2, RHS: Cu =0(uncombined state), N in N

2=0

CuO NH3 Cu N2 H2O+ + +0 0+2 -3

[O]

[R](RA)(OA)

CuO is reduced to Cu as the ON of Cu decreased from +2 to 0 and hence acts as OA while NH3

is oxidised to N2 as the ON of N increased from -3 to 0, and hence acts as RA.

(v) LHS: H in H2=0, I in I

2=0, RHS: H in HI= +1 and I in HI= -1,

+0 0 +1 -1

H2 I2 HI

[O]

[R](RA) (OA)

,


In this case H2 is oxidised to HI hence it is the RA while I

2 is reduced to HI so that it is OA.

(vi) LHS: Mg =0, H in H2SO

4=+1, RHS: Mg in MgSO

4=+2, H in H

2=0

+ +0 0+1 +2

MgSO4 H2Mg H2SO4

[O]

[R](RA) (OA)

In this case Mg is oxidised to MgSO4 and hence is RA while H

2SO

4 is reduced to H

2 and hence

is the OA.(vii) LHS: N in N

2=0, O in O

2=0, RHS: N in NO=+2, O in NO= -2

N2 O2+0 0 +2 -2

NO

[O]

[R](RA) (OA)

(viii) LHS: Cu =0, N in HNO3 =+5, RHS: Cu in Cu(NO

3)

2=+2, N in NO

2=+4

Cu(NO3)2 NO2 H2O+ + +0 +5 +4+2

Cu HNO3

[O]

[R](RA) (OA)

(ix) LHS: Mn in MnO2= +4, Cl in HCl =-1, RHS: Mn in MnCl

2=+2, Cl in Cl

2=0

MnO2 HCl MnCl2 Cl2 H2O0

+ + ++4 +2-1

[O]

[R](RA)(OA)

In this case MnO2 is reduced to MnCl

2 as the ON of Mn decreased from +4 to +2, hence it is

OA. HCl is reduced to Cl2 as the ON of Cl increased from -1 to 0, hence HCl is RA.

(x) LHS: Mn in KMnO4=+7, Fe in FeSO

4 =+2(valency of ferrous is 2),

RHS: Mn in MnSO4=+2(valency of Manganous is 2), Fe in Fe

2(SO

4)

3=+3(valency of ferric is 3)

K2SO4 MnSO4 Fe2(SO4)3 H2O+ + + + ++7 +2 +3+2

KMnO4 H2SO4 FeSO4

[O]

[R](RA)(OA)

In this case, FeSO4 is oxidised to Fe

2(SO

4)

3 as the ON of Fe increased from +2 to +3, hence

FeSO4 is the RA. KMnO

4 is reduced to MnSO

4 as the ON of Mn decreased from +7 to +2,

hence KMnO4 acts as OA.

SAQ 22:

(i) Redox: The ON of H changes from 0 to +1 while Cl changes from 0 to -1(ii) Metathesis: It is double replacement reaction i.e partners(acid and basic radicals) have

been exchanged. In such reactions, no element undergoes any change in ON.(iii) Redox: The ON of Cu changes from 0 to +2 while of N changes from +5 to +4.(iv) Redox: The ON of Na changes from 0 to +1 while of H changes from +1 to 0.(v) Metathesis: It is also a double replacement reaction: Neutralisation reaction.(vi) Metathesis: Double replacement reaction.


ANSWERS TO PRACTICE QUESTIONS1. NaNO

3-S(All nitrates are soluble), KBr-S(All K salts are soluble), MgF

2-I(All fluorides

are insoluble excepting Na+, K+, NH4

+, Ag+, Tl+), FeCl2-S(All chlorides are soluble

excepting Ag+, Hg22+ and Pb2+), MgCO

3-S(All carbonates are insoluble excepting Na+,

K+, NH4

+, Be2+,Mg2+), BaSO4-I(All sulphates are soluble excepting Pb2+, Sr2+,

Ba2+,Hg2+,Ag+), MgS-S(All sulphides are insoluble excepting the sulphides of Na+, K+,NH

4+, Mg2+,Ca2+,Sr2+,Ba2+), Na

3PO

4-S(All Na salts are soluble), Na

2CrO

4-S(all Na

salts are soluble), Ba(ClO4)

2-S(All perchlorates are soluble), Ba(OH)

2-S(All hydroxides

are insoluble excepting those of Na+, K+, NH4+, Ca2+, Sr2+, Ba2+), Hg

2Cl

2-I(All chlorides

are soluble excepting Ag+, Hg22+ and Pb2+), NH

4F-S(All flourides are insoluble excepting

Na+, K+, NH4+, Ag+, Tl+), BeCO

3-S(All carbonates are insoluble excepting Na+, K+,

NH4+, Be2+,Mg2+), PbI

2-I(All iodides are soluble excepting those of Ag+, Hg

22+ and

Pb2+), Ag2SO

4-I(all sulphates are soluble excepting those of Pb2+, Sr2+, Ba2+,Hg2+,Ag+),

LiBr-S(all bromides are soluble excepting those of Ag+, Hg22+ and Pb2+), K

2Cr

2O

7-S(all

K salts are soluble), AgCl-I(All Chlorides are solulbe excepting Ag+, Hg2

2+ and Pb2+),Mg

3(PO

4)

2-S(all phosphates are insoluble excepting those of Na+, K+, NH

4+, Be2+,Mg2+).

2. ClO3

-(chlorate), PO4

3-(phosphate), CH3COO-(acetate), CN-(cyanide), OCl -

(hypochlorite), SO4

2-(sulphate), NO2

-(nitrite).

3. Net Ionic equations(NIE) are given below. The readers are to first write the TIE foreach.(i) H+ + OH- -------> H

2O

(ii) H+ + OH- -------> H2O

(iii) H+ + NH3(g) -------> NH

4+

(iv) Cu2+ + H2S(g) -------> CuS(s) + H+

(v) H+ + MgO(s) -------> Mg2+ + H2O(l)

(vi) Zn(s) + OH- ----------> ZnO2

2- + H2(g)

(vii) Al(OH)3(s) + H+ -------> Al3+ + H

2O(l)

(viii) Mn2+ + S2- ------> MnS(s)(ix) Ba2+ + OH- + H+ + SO

42- -------> BaSO

4(s) + H

2O(l)

(x) H+ + PO4

3- + Ca2+ + OH- -------> Ca3(PO

4)

2(s) + H

2O(l)

(xi) TIE: MnO2(s) + H+ + Cl- -------> Mn2+ + Cl- + Cl

2(g) + H

2O(l)

NIE: MnO2(s) + H+ + Cl- -------> Mn2+ + Cl

2(g) + H

2O(l)

Note that in this case Cl- has to be removed from RHS but not from LHS. If you remove Cl-

from LHS, then where from Cl2(g) is produced in RHS? So in such case, you consider Cl- to be

an unimportant species in the RHS and is to be removed from it. But Cl- is the involved speciesin LHS which produces Cl

2 in the RHS. So tactfully keep Cl- in LHS and remove from RHS.

Same type of reaction is given in the next question.

(xii) TIE: K+ + Br- + H+ + SO42- -------> K+ + SO

42- + Br

2(l) + SO

2(g) + H

2O(l)

NIE: Br- + H+ + SO4

2- ---------> Br2(l) + SO

2(g) + H

2O(l)

Here also you found that SO4

2- has been removed from RHS but not from LHS. It is SO4

2- ofLHS which has resulted SO

2 in the RHS. SO

42- is the spectator ion in the RHS and so deleted

from that side.


(xiii) TIE: Cu(s) + H+ + NO3

- --------> Cu2+ + NO3

- + NO(g) + H2O(l)

NIE: Cu(s) + H+ + NO3- ---------> Cu2+ + NO(g) + H

2O(l)

Here also NO3

- is deleted from RHS but kept in LHS for the same reason explained before.

(xiv) NIE: Cr2O

72- + H+ + NO

2- ---------> Cr3+ + NO

3- + H

2O(l)

(xv) I2(s) + S

2O

32- ---------> S

4O

62- + I-

(xv) HI(aq) +HNO2(aq)→ NO(g) + H

2O(l) + I

2(s)

TIE: H+ + I- + H+ + NO2

- ------> NO(g) + H2O(l) + I

2(s)

NIE: H+ + I- + NO2- ------> NO(g) + H

2O(l) + I

2(s)

(xvi) KOH(aq) + KMnO4(aq) → K

2MnO

4(aq)+ O

2 (g) + H

2O (l)

NIE: OH- + MnO4

- ------> MnO4

2- + O2 + H

2O

4. (i) NaBr(aq) + H2O(l); NIE: H+ + OH- ---------> H

2O

(ii) Ca(NO3)

2(aq) + H

2O; NIE: H+ + OH- ------> H

2O

(iii) K2SO

4(aq) + H

2O(l); NIE: H+ + OH- --------> H

2O

(iv) BaCl2(aq) + H

2O(l); NIE: H+ + OH- ---------> H

2O

(v) MgSO4(aq) + H

2O(l); NIE: Mg(OH)

2(s) + H+ + -------> Mg2+ + H

2O(l)

(vi) Mn3(PO

4)

2(s) + H

2O(l); NIE: MnO(s) + H+ + PO

43- -----> Mn

3(PO

4)

2(s)+ H

2O(l)

(vii) Al2(SO

4)

3(aq) + H

2O(l); NIE: Al

2O

3(s) + H+ -----> Al3+ + H

2O(l)

(viii) FeBr3(aq) + H

2O(l); NIE: Fe(OH)

3(s) + H+ -----> Fe3+ + H

2O(l)

(ix) NaClO4(aq) + H

2O(l); NIE: H+ + OH- ------> H

2O

(x) Cr2(SO

4)

3(aq) + H

2O(l); NIE: Cr

2O

3(s) + H+ ------> Cr3+ + H

2O(l)

5. The beginner is advised to calculate by x method.

NaCl: +1(valency of Na is 1), BaF2: -1(valency of F is 1), NH

3: -3, H

2O: -2, NaH:-1(ON of

H in metallic hydrides is -1), SO3: +6, CO

2: +4, P

2O

3: +3, HClO: +1, MnCl

2: +2(valency of

Manganous is 2), As(NO3)

3: +3(valency of arsenous is 3), SnSO

4:+2(valency of stannous is 2),

Hg2Cl

2: +1(valency of mercurous is 1), H

2SO

4: +6, BCl

3: +3, SF

6: +6, HClO

3: +5,

SO3

2-: +4, S4O

62-: 2.5(note that ON can be a fraction), NH

4+: -3, NO

3-: +5, NO

2-: +3,

H2CO

3: +4, ClO

2-: +3, K

2Cr

2O

7: +6, Na

2O

2: -1(the ON of oxygen in peroxide is -1),

MnO2: +4, SnCl

4: +4(valency of stannic is 4), FeBr

3: +3(valency of ferric iron is 3),

Cu2Cl

2: +1(valency of cuprous is 1), NH

4NO

3: -3 and +5(to be found out seprately, refer SAQ

20), NO: +2, N2O: +1, N

2: 0(uncombined state), ZnO: +2(valency of Zn is 2), K

2O: +1(valency

of K is 1), Ca(OH)2: +2(valency of Ca is 2), Al: 0(uncombined state), Al

2(SO

4)

3: +3(valency

of Al is 3), OF2: +2(F is -1, so O has to be +2), N

2O

5 : +5, S

2O

32-: +2

OCN- : +4(C will lose 3 electrons to N and 1 electron to O); HCN: +2(C will lose 3 electrons toN and gain 1 electron from H); S

2O

82-: +6( there are 6 normal oxide and 1 peroxide [S

2O

6(O

2)]2-

, 2x-12-2=-2 ⇒ x =+6); H2SO

5: +6 (there are 3 normal oxide and 1 peroxide[SO

3(O

2)]2-, x-6-2

=-2 ⇒ x = +6) CrO5: +6 (there are 1 normal oxide and two peroxide groups [CrO(O

2)

2], x-2-

4=0 ⇒ x =+6) KO2 : -1/2 (superoxide) Ni(CO)

4: 0(since CO is a neutral molecule) CaS

5: -2/5

(since it is a polysulphide ion S52-); [SnS

3]2-: -2 ; [Fe(CN)

4]4: +2 and +2 respectively(since CN

has -1 charge, ON of C is +2 as C loses 3 electrons to N and gains one electron due to itsnegative charge)

XeOF4: +6; NaN

3 : -1/3; SCN-: -2; O

2F

2 : +1; CN-: -3; N

3H: -1/3; [Ni(CN)

4]2-: +2 (x-4=-2,

CN is taken as -1); [Cr(H2O)

4(CN)

2]+ : +3 (x-2=+1); C

3H

8 : -8/3; C

6H

12O

6 : 0 (note that ON


concept is irrelevant for organic compounds although it can be calculated like inorganic compounds);BrCl

3 : +3( Br is less electronegative than Cl); HIO

3: +5; H

2[SiF

6]: +4; CaNCN: +4(C will

lose two electrons to each N) NaCN: +2 (the ON of C in any cyanide ion is +2)

Ca(OCl)Cl : +1 and -1 respectively from OCl- and Cl-( Note that the average ON of Cl inCaOCl

2 is 0 although in reality there are two types of Cl atoms in the molecule); S

2O

32- : +2 (but

in reality there are two types of S atoms having ON +4 and 0 respectively so that the averageON is 0. Note that in the structure of thiosulphate ion there is one S-S linkage which breakshomolytically to make one S acquire 0 state. More you can know when you study the chapterchemical bonding)

6.(i) K goes from 0(K) to +1(KOH) and is oxidised while H goes from +1(H2O) to 0(H

2) and

thus H2O is reduced. OA: H

2O and RA: K

(ii) N goes from -3(NH3) to 0(N

2), thus NH

3 is oxidised, while O goes from 0(O

2) to

-2(H2O), thus O

2 is reduced. OA: O

2 and RA: NH

3

(iii) Na goes from 0(Na) to +1(NaNH2), thus Na is oxidised, while H goes from +1(NH

3) to

0(H2), thus NH

3 is reduced. OA: NH

3, RA: Na

(iv) Ca goes from 0 to +2 while H goes from +1 to 0. Thus OA:H2O, RA:Ca

(v) Al goes from 0(Al) to +3(NaAlO2), thus Al is oxidised, while H goes from +1(NaOH) to

0(H2), thus NaOH is reduced. OA: NaOH, RA: Al

(vi) Zn goes from 0(Zn) to +2(ZnSO4), thus Zn is oxidised, while H goes from +1(H

2SO

4) to

0(H2), thus H

2SO

4 is reduced. OA: H

2SO

4, RA: Zn

(vii) Cu goes from +2(CuO) to 0(Cu), thus CuO is reduced while N goes from -3(NH3) to

0(N2), thus NH

3 is oxidised. OA: CuO, RA: NH

3

(viii) P goes from 0 to +5 while O goes from 0 to -2.OA: O2 and RA: P

4

(ix) H goes from 0 to +1 while Cl goes from 0 to -1. OA: Cl2 and RA: H

2

(x) O goes from -1(H2O

2) to -2(H

2O), thus H

2O

2 is reduced while S goes from -2(H

2S) to

0(S), thus H2S is oxidised. OA: H

2O

2, RA: H

2S

(xi) Br goes from -1 to 0 while Cl goes from 0 to -1. OA: Cl2, RA: FeBr

3

(xii) I goes from 0(I2) to -1(NaI), thus I

2 is reduced while S goes from +2(Na

2S

2O

3) to

2.5(Na2S

4O

6), thus Na

2S

2O

3 is oxidised. OA: I

2, RA: Na

2S

2O

3

(xiii) Sn goes from +2 to +4 while Hg goes from +2 to +1. OA: HgCl2, RA: SnCl

2

(xiv) Cu goes from 0(Cu) to +2[Cu(NO3)

2], thus Cu is oxidised while N goes from +5(HNO

3)

to +4(NO2), thus HNO

3 is reduced. OA: HNO

3, RA: Cu

(xv) Mn goes from +7(KMnO4) to +2(MnSO

4), thus KMnO

4 is reduced while Fe goes from

+2(FeSO4) to +3[Fe

2(SO

4)

3], thus FeSO

4 is oxidised. OA: KMnO

4, RA: FeSO

4

(xvi) Cr goes from +6(K2Cr

2O

7) to +3[Cr

2(SO

4)

3], thus K

2Cr

2O

7 is reduced, while S goes

from -2(H2S) to 0(S), thus H

2S is oxidised. OA: K

2Cr

2O

7, RA: H

2S

7. Only the changes in ON have been given. The reader is to assign the OA and RA.(i) Fe from +3 to 0 and C from +2 to +4

(ii) I from 0 to -1 and S from +2 to +2.5(iii) Fe from +3 to +2, O from -1 to 0(iv) Cl from +1 to -1, I from -1 to 0(v) Pb from +4 to +2, Br from -1 to 0


(vi) N from +5 to -3(NH4+ in NH

4NO

3), Zn from 0 to +2

(vii) N from -3(NH4

+) to +1, N from +5(NO3-) to +1(N

2O).

In this case same substance(NH4NO

3) is oxidised as well as reduced. OA and RA are not

assigned to such redox reactions. These are called disproportionation reactions. We shallstudy about this later.

(viii) N from +5 to +4, O from -2 to 0. Same as (vii):disproportionation(ix) P from 0 to +5, S from +6 to +4.(x) Cl from 0 to -1(NaCl), Cl from 0 to +5(NaClO

3). same as above:

disproportionation

8. (i) S from -2 to 0 and N from +5 to +2.(ii) Zn from 0 to +2 and N from +5 to -3 (NH

4+)

(iii) Mn from +2 to +7(HMnO4) and Pb from +4 to +2.

(iv) Mn from +7 to +2 and Cl from -1 to 0.(v) C from +3(K

2C

2O

4) to +4 and Cr from +6 to +3.

(vi) N from +5 to +2 and I from -1 to 0.(vii) N from -3 to +2 and O from 0 to -2.(viii) Cu from +2 to 0 and N from -3 to 0.(ix) S from +4 to 0 and S from -2 to 0.(x) N from +5 to +3 and O from -2 to 0.(xi) Cl from +5 to -1 and O from -2 to 0.(xii) N from +5 to +4 and O from -2 to 0.(xiii) Zn from 0 to +2 and H from +1 to 0.(xiv) Cl from +5 to +4(ClO

2) and O from -2 to 0.

(xv) Sn from 0 to +4 and N from +5 to +4.(xvi) I from 0 to +5 and N from +5 to +4.(xvii) S from +6 to +4 and Br from -1 to 0.(xviii) P from +5 to 0 and C from 0 to +2.(xix) Na from 0 to +1 and H from +1 to 0.(xx) O from -1 to -2 and S from -2 to 0.(xxi) Cr from +3 to +6 and O from -1(Na

2O

2) to -2.

(xxii) O from 0 to -2 and Hg from 0 to +1.(xxiii) Fe from +2 to +3 and O from 0 to -2.(xxiv) Br from +5 to 0 and Br from -1 to 0.(xxv) Mn from +7 to +6(K

2MnO

4) and +4(MnO

2), two reductions, O from -2 to 0.

Note that a redox reaction can have more than one oxidation and reduction reaction, but at leastone of each type of reaction has to be there.

(xxvi) Mn from +2 to +4 and O from -1(peroxydisulphate= [S2O

6(O

2)]2-) to -2

(xxvii) Zn from 0 to +2 and H from +1 to -1(xxviii) Cl from +4 to +3 and O from -1 to 0(xxix) N from -2 to 0 and Br from +5 to -1(xxx) Cu from +2 to +1 and I from -1 to 0(xxxi) Fe from 0 to +-8/3 and H from +1 to 0(xxxii) O from -1 to 0 and Cl from +1 to -1


BALANCING CHEMICAL EQUATIONS

Balancing of an equation means placing suitable coefficients before the reactants and productsso that the number of atoms of every element in the LHS is same as that in the RHS.Example: Al(OH)

3 + H

2SO

4 -------> Al

2(SO

4)

3 + H

2O (not balanced)

2Al(OH)3

+ 3H2SO

4 -------> Al

2(SO

4)

3 + 3H

2O (balanced)

The second equation is balanced, because there are same number of Al atoms, same numberof O, H and S atoms on the two sides.Metathesis reactions: You already know that a metathesis reaction is that reaction inwhich no change in ON takes place for any element. The balancing of such reaction is doneby hit and trial or the inspection method.(1) Hit and Trial Method (Inspection Method)To balance a metathesis reaction (reaction in which ONs are not changed) such as doublereplacement reaction of the type shown before in the previous section is very easy. This isdone by hit and trial (inspection) method. Mere looking at both the sides gives us clue as towhat coefficients are to be placed before the reactants and products to get a balancedequation. Answer the following SAQs.

SAQ 1: Balance the following reactions:(i) Ca(OH)

2 + H

3PO

4 ---------> Ca

3(PO

4)

2 + H

2O

(ii) ZnO + H2SO

4 -------> ZnSO

4 + H

2O

(iii) FeCl2 + H

2S -------> FeS + HCl

(iv) P2O

5 + H

2O -------> H

3PO

4(v) NH

4Cl + Ca(OH)

2--------> CaCl

2 + NH

3 + H

2O

REDOX REACTIONS:As you must have realised that balancing metathesis reactions( double replacement

type) requires no special technique. Only hit and trial method (inspection) solves the purpose.However it is often difficult to balance a redox equation by hit and trial method. It is verymuch time consuming particularly for complicated reactions and moreover there is a possibilityof arriving at a wrong balanced equation in hit and trial method. Therefore hit and trialmethod is not recommended for such reactions. For simple redox reactions, hit and trialmethod is a good choice. The following examples will illustrate the point.

Na + H2SO

4 -------> Na

2SO

4 +H

2 (not balanced)

2Na + H2SO

4 -------> Na

2SO

4 +H

2 (balanced)

N2 + H

2 --------> NH

3(not balanced)

N2 + 3H

2 --------> 2NH

3 (balanced)

P4 + Cl

2 --------> PCl

5 (not balanced)

P4 + 10Cl

2 --------> 4PCl

5 (balanced)

However it is difficult to balance other redox equations which are not simple type as shownabove. The following examples will explain this.

4NH3 + 5O

2 --------> 4NO + 6H

2O

No doubt one can balance the above equation by hit and trial method, but it is time taking.We therefore discuss a new method called Oxidation Number(ON) method to balanceredox equations.


(2) OXIDATION NUMBER(ON) METHOD:

Let us take the following examples:

(i) CuO NH3 N2 Cu H2O+ + ++2 -3 0 0

In this equation we find that the ON of Cu goes from +2 to 0. So what is the change in ONthat the oxidising agent(CuO) has undergone? . It is |+2-0|=2. This 2 is placed as thecoefficient before the reducing agent NH

3. Note that change is merely the magnitude of

difference between the two ONs.

CuO NH3 N2 Cu H2O+ + +2

Look to the first equation. The ON of N goes from -3 to 0. What is the change of ON thatthe reducing agent has undergone? It is |-3-0| =3. This 3 is placed as coefficient before theoxidising agent CuO.

CuO NH3 N2 Cu H2O+ + +23

After that, it becomes damn easy to balance by hit and trial method. We have to place 3before Cu in RHS to equalise Cu. To equalise H we have to place 3 before H

2O in RHS,and

by doing so we find that the O is automatically balanced.

CuO NH3 N2 Cu H2O+ + +23 3 3

The beauty of this method is that while balancing for the last but one element, the lastelement is automatically balanced. In the above example, we found that while balancing forH, oxygen atoms were automatically balanced. Let us look to another example.

(ii) NH3 + O

2 --------> NO + H

2O

NH3 O2 NO H2O+ +-3 0 +2 -2

(R)

(O)

In this case we found that the change in ON in the RA(NH3) is |-3-(+2)|=5, so 5 is placed as

coefficient before OA(O2)

NH3 + 5 O

2 --------> NO + H

2O

Look again to the original equation. O goes from 0 to -2. The change is |0-2|=2. This change isfor one oxygen atom. But the OA contains two O atoms(O

2), so the total change is 2X2=4.

Thus 4 is placed as coefficient before the RA(NH3). Note that in NH

3 there is one N atom, so

the net change is 5. But in O2 there are two O atoms, so the total change is not 2, it is 4.

4NH3 + 5O

2 --------> NO + H

2O

After this, it becomes extremely easy task to balance by inspection. 4 is to be placed before NOin RHS to equalise N on both sides. To equalise H, we have to place 6 before H

2O. In doing so

we find that O is automatically balanced(10 on either side).

4NH3 + 5O

2 --------> 4NO + 6H

2O

Now let us know a few rules for balancing an equation by ON method.


RULES FOR BALANCING BY ON METHOD:(i) First the ON of the atoms undergoing oxidation-reduction are found out and placed

above the respective atoms.(ii) The total change of ON in the OA is placed as the coefficient before RA and the total

change of ON in RA is placed as coefficient before OA.(iii) Then hit and trial procedure is adopted to balance the equation.

Another example is given below for a better understanding.Cl

2 + NH

3 -------> HCl + N

2

Cl2 NH3 HCl N2+ +0 -3 -1 0

( Cl2 is the OA and NH

3 is RA)

The change in ON of one Cl =|0-(-1)| =1

For two Cl atoms present in the OA, the total change in ON =2, so 2 is placed as coefficientbefore NH

3(RA).

Cl2 + 2NH

3 -------> HCl + N

2

The change in ON of in RA(NH3) = |-3-0| = 3. Since there is one N atom in the RA, the total

change of ON in the OA is also 3, so 3 is placed as coefficient before Cl2(OA)

3Cl2 + 2NH

3 -------> HCl + N

2

(Note that for the purpose of finding out the total change in ON, the number of atoms present inproducts i.e RHS is not considered. As for example we have N

2 in RHS, but we do not make the

total change 3X2=6 for N atoms to place before Cl2. Since NH

3 has one N atom, the change is

3. For that purpose, only the reactant side, i.e LHS is considered)

After that, mere hit and trial (inspection) completes the process of balancing in a second. 6 to beplaced before HCl to equalise Cl and by that we see that H is automatically balanced (6H oneither side).

3Cl2 + 2NH

3 -------> 6HCl + N

2 (Balanced equation)

SAQ 2: Balance the following by ON method.(i) KI + H

2SO

4 ---------> I

2 + K

2SO

4 + H

2S + H

2O

(ii) I2 + HNO

3 ---------> HIO

3 + NO

2 + H

2O

(iii) Cu + HNO3 ------->Cu(NO

3)

2 + NO + H

2O

(3) PARTIAL EQUATION METHOD:

Let us now learn another simple method to balance a redox reaction. This is applicable particularlywhen the oxidising agents (OA) are KMnO

4, K

2Cr

2O

7, HNO

3, H

2SO

4 etc.

Let us take the reaction

KMnO4 + H

2SO

4 + FeSO

4 --------> K

2SO

4 + MnSO

4 + Fe

2(SO

4)

3 + H

2O

It is a tough task to balance this big equation by hit and trial method. We can balance thisequation by ON method as we have just now learnt. But there is another simpler method to


balance this type of equation. This is called partial equation method. In this method, theOA(KMnO

4) is split into a few smaller fragments.

KMnO4 ---------> K

2O + MnO + [O]

KMnO4 breaks down to oxide of potassium, oxide of manganese (ous) and nascent oxygen[O].

The nascent means freshly born. That means this nascent oxygen is now born freshly to die inthe second step. Note that all the three smaller fragments that are produced from the OA will diei.e will be used up in the subsequent steps. So that we write a few steps or partial equations.Note that in each step, the equation has to be balanced by hit and trial method. Then weshall perform algebraic operation to add all the partial equations. We shall thus cancel thosespecies which are used i.e which appear in both LHS and RHS by doing algebraic manipulationsand finally get the balanced equation of the original reaction. Let us look to the following equations.(a) Reactions involving KMnO

4

Reactions involving KMnO4 and K

2Cr

2O

7 often take place in presence of an acid, preferably

H2SO

4. In the example below, FeSO

4 (ferrous sulphate) reacts with acidified KMnO

4 to produce

a mixture of potassium sulphate, manganous sulphate, ferric sulphate and water. We know thathere FeSO

4 is the reducing agent (RA) and KMnO

4 is the oxidising agent(OA).

Let us balance the following reaction by partial equation method.

(balanced equation)+++++ 8528 KMnO4 H2SO4 FeSO4 K2SO4 MnSO4 Fe2(SO4)3 H2O

5 X

2 X(Step 4)(Step 3)(Step 2)(Step 1)

+++FeSO4 H2SO4 [O} Fe2(SO4)3 H2O

+ +MnO H2SO4 MnSO4 H2O+ +K2O H2SO4 K2SO4 H2O

22 5++KMnO4 K2O MnO [O]

+++++KMnO4 H2SO4 FeSO4 K2SO4 MnSO4 Fe2(SO4)3 H2O

2

10

In this the fragments K2O and MnO produced in the first step are consumed in the second and

third step by reacting with the acid(H2SO

4) present in the medium to produce salt and water. In

the fourth step FeSO4(RA) is oxidised to Fe

2(SO

4)

3 by nascent oxygen. Remember that whenever

FeSO4 is oxidised to Fe

2(SO

4)

3, it takes the help of H

2SO

4. Otherwise the equation cannot be

balanced. Note that before adding all the step equations, we have to see that each individual stepequation is balanced. While adding the four steps, K

2O gets cancelled. To cancel 2MnO from

RHS we have to multiply the whole equation of step(3) by 2. Likewise to cancel 5[O] fromRHS, we have to multiply the whole equation of step(4) by 5. After cancellation of K

2O, 2MnO

and 5[O] from both the sides, all the species of LHS and RHS are added separately and writtenbelow. Further simplification of this equation, if required, is done like algebraic equation. If H

2O

appears on both the sides, then it is removed from one side and excess H2O is only written on the

other side. Similarly any other species, if appears both in LHS and RHS is brought to one side.

Let us take another example on KMnO4.

2


(ii)

(balanced equation)+ ++ +5 2 8 KMnO4 H2SO4 H2S K2SO4 MnSO4 S H2O

5 X

2 X(Step 4)(Step 3)(Step 2)(Step 1)

+

+

+

+ +MnO H2SO4 MnSO4 H2O+ +K2O H2SO4 K2SO4 H2O

22 5++KMnO4 K2O MnO [O]

++++ +KMnO4 H2SO4 H2S K2SO4 MnSO4 S H2O

2

H2S [O] S H2O

3 5

In this case, the first three steps are same as the first example. Only the forth step is different.The RA is H

2S in this case which is oxidised to S. You can always look at the RHS of the original

equation to know what would be the products of each step. While adding the four steps, K2O,

2MnO and 5[O] are cancelled by suitable algebraic manipulations and we get the balancedequation.

Do you think, that you could have balanced these equations by hit and trial method easily? Theanswer is NO. So to conclude, we say that a redox reaction is balanced either by ON methoddescribed before or partial equation method described just now. Note that the partial equationmethod is merely a simple technique devised to balance a complex equation in a simple andeconomical manner. The actual reaction does not take place in the successive steps as discussedbefore. There is nothing called nascent oxygen in this reaction. The nascent atom concept in thesolution is very primitive and has been outdated. The redox reactions truly take place by the lossand gain of electrons and not by way of the generation of nascent oxygen atoms. But studentsare advised to follow this technique to get the balanced equations easily.

(b) Reactions involving K2Cr

2O

7

K2Cr2O7 H2SO4 SO2 K2SO4 Cr2(SO4)3 H2SO4 H2O+ + + + +

K2Cr2O7 K2O Cr2O3 [O]+ +

K2O H2SO4 K2SO4 H2O++

Cr2O3 H2SO4 Cr2(SO4)3 H2O++

+

+

(Step 1)(Step 2)(Step 3)(Step 4)

3

33SO2 [O] SO3

SO3 H2O H2SO4

3

3 X3 X (Step 5)

K2Cr2O7 H2SO4 SO2 K2SO4 Cr2(SO4)3 H2SO4 H2O+ + + + +4

K2Cr2O7 SO2 H2SO4 K2SO4 Cr2(SO4)3 H2O

3

3+ + ++


In this case K2Cr

2O

7 is broken into three fragments namely K

2O, Cr

2O

3 and nascent oxygen.

This equation is immediately balanced. Then in step 2 and step 3, the oxides of K and Cr reactwith H

2SO

4 present in the medium to produce the respective salt and water. In step 4, SO

2(RA)

reacts with nascent oxgyen[O] to produce SO3. But SO

3 is unstable. It immediately reacts with

H2O in the step 5 to produce the corresponding acid, H

2SO

4. So we find that H

2SO

4 was used

as a reactant and it is formed also in the product. So we add the five steps by multiplying suitablecoefficients (3) to the step equations 4 and 5, to cancel 3[O] and 3SO

3 from the two sides. K

2O

and Cr2O

3 are also cancelled as such. On addition it gives an equation, which carries H

2O and

H2SO

4 on either side. So we have to simplify further to get the balanced equation. We find that

in the net balanced equation, H2SO

4 has been wiped out from RHS. Don't worry for that. At

times, it happens like that at the last step of simplification.

SAQ 3: Show which is oxidised and which reduced by assigning ONs. Also balance theequations by partial equation methods.

(i)KMnO4 + H

2SO

4 + H

2C

2O

4 -----------> K

2SO

4 + MnSO

4 + CO

2 + H

2O

(ii)K2Cr

2O

7 + H

2SO

4 + H

2O

2 ----------> K

2SO

4 + Cr

2(SO

4)

3 + O

2 + H

2O

(c) Reactions involving HNO3:

Equations involving HNO3 can also be balanced by partial equation method. Let us take an

example.

HNO3 + Cu -------> Cu(NO

3)

2 + N

2O + H

2O

We can balance this equation by Oxidation Number Method as explained before. In partialequation method, HNO

3(OA) is broken down to small fragments i.e an oxide of nitrogen (in this

case N2O), H

2O and nascent oxygen[O]. While breaking HNO

3, the student is advised to look to

the RHS to know which oxide of nitrogen is to be written.

HNO3 ------> N

2O + H

2O + [O]

Let us immediately balance it and use [O] to react with Cu(RA) in the second step to give CuO.Again since the medium is acidic(HNO

3), CuO being a basic substance reacts with the acid to

produce salt and water. The partial equations are added up and nascent oxygen atoms arecancelled to get the balanced equation.

2HNO3 ---------> N

2O + H

2O + 4[O] (i)

4X [Cu + [O] --------> CuO] (ii) (4 Multiplied to cancel 4[O])4X [CuO + 2HNO

3 -------> Cu(NO

3)

2 + H

2O](iii) (4 multiplied to cancel 4CuO)

_____________________________________4 Cu + 10HNO

3 ------> N

2O + 4Cu(NO

3)

2 + 5H

2O` (balanced equation)

Note that the oxide of nitrogen produced in reactions involving HNO3 depends on the type of

HNO3 used. The following gives a list of products formed from HNO

3.

Conc. HNO3: NO

2

Moderately concentrated HNO3 : NO

Dilute HNO3: N

2O or N

2

Very dilute HNO3: NH

3( NH

3 + HNO

3 ------> NH

4NO

3)


This means that concentrated HNO3 forms nitrogen dioxide gas(NO

2), moderately concentrated

HNO3 forms nitric oxide(NO), dilute HNO

3 forms nitrous oxide(N

2O) or nitrogen(N

2) and very

dilute HNO3 first forms NH

3 which reacts with HNO

3 to form ammonium nitrate(NH

4NO

3). In

the example given above, the nitric acid used must be dilute as the oxide of nitrogen produced isnitrous oxide(N

2O)

For the purpose of balancing, you do not have to bother all about this. You can knowwhat is the product just by looking to RHS of a given equation.

SAQ 4: Balance the following by partial equation method.(i) HNO

3 + Zn -------> NH

4NO

3 + Zn(NO

3)

2 + H

2O

(ii) HNO3 + Mg -------> Mg(NO

3)

2 + NO + H

2O

(d) Reactions involving conc. H2SO

4Reactions involving concentrated H

2SO

4 can also be balanced by partial equation method.

H2SO

4 is broken into SO

2 gas, H

2O and nascent oxgyen[O]. The process is similar to the

balancing discussed for HNO3 above. Let us take an example.

Cu + H2SO

4(conc) -------> CuSO

4 + SO

2 + H

2O

H2SO

4 ------> SO

2 + H

2O + [O] (i)

Cu + [O] -----> CuO (ii) ([O] is cancelled)CuO + H

2SO

4 ------> CuSO

4 + H

2O (iii) (CuO is cancelled)

____________________________________Cu + 2 H

2SO

4 --------> CuSO

4 + SO

2 + 2 H

2O (balanced equation)

SAQ 5: Balance the following by partial equation method.Al + H

2SO

4(conc.) ---------> Al

2(SO

4)

3 + SO

2 + H

2O

IMPORTANT: After balancing an equation by any method, check whether the balancingis correct or not by tallying two sides(LHS and RHS) element by element and atom byatom. If you find that somewhere it is not tallying, then you have done the steps somewherewrong. In such case you revise again to detect your mistake.

(4) ION-ELECTRON METHOD:

This is another interesting technique of balancing redox reactions. This is applied to the ionicequations(not molecular equations). So if a molecular equation is given to you, first you have tofind the net ionic equation from it and then balance the ionic equation by this technique. Then youcan balance the molecular equation by looking to the coefficients in the balanced ionic equation.

Do you remember how to find the net ionic equation from a molecular equation that you havestudied before?

Example-1

KMnO4(aq) + H

2SO

4(aq) + FeSO

4(aq) ------> K

2SO

4(aq) + MnSO

4(aq) + Fe

2(SO

4)

3(aq) + H

2O(l)

Net ionic equation: MnO4

- + H+ + Fe2+ -------------> Mn2+ + Fe3+ + H2O

Now the job is to balance this ionic equation. There are rules which we have to follow for doingthat. Carefully read the following.


(A) Acidic Medium

The following rules are applicable for reactions carried out in acidic medium. If you find H+ ionpresent either in LHS or RHS, then the reaction has definitely been carried out in acidic mediumand in such case, you follow this method.

MnO4

- + H+ + Fe2+ -------------> Mn2+ + Fe3+ + H2O

From this you separate the reduction and oxidation processes by taking one reactant and thecorresponding product from the net ionic equation at a time. Build up two separate step equations,one for reduction and the other for oxidation. Then add the two to get the balanced ionic equation.In the above reaction, MnO

4-(LHS) has been reduced to Mn2+(RHS). Can you not say that the

ON of Mn changes from +7 to +2 in this case.? Copy the reduction step from the originalequation as follows.Reduction step:

MnO4

- ---------> Mn2+

Rules:(i) Equalise the number of atoms other than O and H on both sides by placing

suitable coefficient.In the above example the number of Mn atoms on either side is same (one). So we will not place

any coefficient on any side.(ii) Count how many O atoms are in excess in any particular side and add same

number of H2O molecules on the opposite side.

In the above example, four O atoms are in excess in LHS, so we have to add 4 H2O

molecules in RHS.MnO

4- ---------> Mn2+ + 4H

2O

(iii) Now count how many H atoms are in excess in any particular side and add samenumber of H+ ions on the opposite side.

In this case there are 8 H atoms excess in RHS. So we have to add 8 H+ ions in LHS.MnO

4- + 8 H + ---------> Mn2+ + 4H

2O

(iv) Then we have to equalise the charge on both sides by adding electrons. Notethat while you add one electron to a particular side, charge-wise it is -1, aselectron carries one unit of negative charge.

In this case, the total charge in LHS is +7( -1 for MnO4

- and +8 for eight H+) and the total chargein RHS is +2(from one Mn2+). So we have to add 5 electrons in LHS so as to equalise the chargeon both sides to +2.

MnO4

- + 8 H + + 5e- ---------> Mn2+ + 4H2O

Thus we have completed balancing of the reduction step. Note that in the reduction step, electronsappear in LHS as reduction is the process in which electron is gained.

Let us take the oxidation step.

Oxidation Step:

In the above example Fe2+ has been oxidised to Fe3+. Let us write that.

Fe2+ --------> Fe3+


Since the number of Fe atoms is same on either side, we do not have to put any coefficient onany side. Since there is no O and H atoms on any side, the addition of H

2O and H+ ions are not

to be done. Thus rules (i), (ii) and (iii) are not applicable in this case. Only we shall apply the lastrule(iv), i.e equalisation of charge by adding electrons. In this case one electron has to be addedin the RHS to equalise the charge on both sides(+2).

Fe2+ --------> Fe3+ + e-

Thus we found that in oxidation process, electron appears in RHS as oxidation is the processin which electron is lost.

Now we have to add the two steps (reduction and oxidation) and cancel the electrons from boththe sides by doing suitable algebraic manipulations and get the balanced ionic equation.

MnO4

- + 8 H+ + 5e----------> Mn2+ + 4H2O

5X [Fe2+ -------------------> Fe3+ + e-] (5e are cancelled on both sides)_______________________________________________________________

MnO4

- + 8H+ + 5Fe2+ -------->Mn2+ + 5Fe3+ + 4 H2O (balanced equation)

Now you can check whether the balancing is correct or not by tallying two sides(LHS and RHS)element-wise, atom-wise and charge-wise. If you find that somewhere it is not tallying,then youhave done the steps somewhere wrong. In such case you revise again to detect your mistake.

Balancing of the molecular equation from balanced ionic equation:

Now we can balance the original molecular equation by looking to the coefficients of the balancedionic equation.

KMnO4 + 4 H

2SO

4 + 5 FeSO

4 → → → → → MnSO

4 + 1/2 K

2SO

4 + 5/2 Fe

2(SO

4)

3 + 4 H

2O

We needed 8 H+ ions, so we placed the coefficient 4 with H2SO

4. We needed 5 Fe3+, so we

placed the coefficient 5/2 with Fe2(SO

4)

3. In this way the whole equation is balanced. In order

to eliminate fraction in the balanced equation, the factor 2 is multiplied throughout the equation(bothLHS and RHS).

2KMnO4 + 8H

2SO

4 + 10 FeSO

4 → → → → → 2MnSO

4 + K

2SO

4 + 5 Fe

2(SO

4)

3 + 8 H

2O

Example 2 :

Let us balance the ionic equation: Cr2O

72- + H+ + NO

2- --------> Cr3+ + NO

3- + H

2O

first and after that balance the molecular equation

K2Cr

2O

7 + H

2SO

4 + KNO

2 → → → → → K

2SO

4 + Cr

2(SO

4)

3 + KNO

3 + H

2O

In this reaction, the ON of Cr changes from +6 to +3 and N from +3 to +5. Check by x methodof finding the ON. So Cr

2O

72- is reduced to Cr3+ and NO

2- has been oxidised to NO

3-.

Reduction step:

Cr2O

72- -------> Cr3+

First we have to equalise Cr atoms by placing 2 before Cr3+ in RHSCr

2O

72- -------> 2Cr3+

Since there are 7 O atoms excess in LHS, we have to add 7 H2O molecules in RHS.

Cr2O

72- -------> 2Cr3+ + 7 H

2O


Now there are 14 H atoms excess in RHS, we have to add 14 H+ ions in LHS.

Cr2O

72- + 14 H+-------> 2Cr3+ + 7 H

2O

LHS contains +12 charge(-2+14) and RHS contains +6(2X3) charge. So we have to add

6 electrons on LHS to equalise the charge on both sides(+6).

Cr2O

72- + 14 H+ + 6e--------> 2Cr3+ + 7 H

2O

Oxidation step:

NO2

- has been oxidised to NO3

-. Let us write them and apply the same set of rules for balancingthem.

NO2

- ----------> NO3

-

N atoms are same in number, so we do not have to do anything. There is one excess O atom inRHS, so we have to add one H

2O molecule in LHS.

NO2

- + H2O ----------> NO

3-

By doing so, now 2 H atoms are in excess in LHS, and so we have to add 2 H+ ions in RHS.NO

2- + H

2O ----------> NO

3- + 2 H+

Now equalisation of charge is made by adding 2 electrons in the RHS, so that the net charge is-1 on both the sides.

NO2

- + H2O ----------> NO

3- + 2 H+ + 2e-

Now we have to add the two steps and cancel the electrons.

Cr2O

72- + 14 H+ + 6e--------> 2Cr3+ + 7 H

2O

3X [NO2

- + H2O ----------> NO

3- + 2 H+ + 2e-] ( electrons get cancelled)

________________________________________________________Cr

2O

72- + 14 H+ + 3NO

2- + 3H

2O ---------->2Cr3+ + 3NO

3- + 6H+ +7H

2O

We can further simplify by taking H2O and H+ to one side.

Cr2O

72- + 8 H+ + 3NO

2- ---------->2Cr3+ + 3NO

3- +4 H


Balancing of the molecular equation from balanced ionic equation;

In the same manner as we did for the previous example, we can balance the molecular equaitonby looking to the coefficents of balanced ionic equation.

K2Cr

2O

7 + 4 H

2SO

4 + 3 KNO

2 → → → → → K

2SO

4 + Cr

2(SO

4)

3 + 3 KNO

3 + 4 H

2O

Example 3 :

Cu2O + H+ + NO

3- ----------> Cu2+ + NO + H

2O

First identity which is reduced and which is oxidised. Cu2O is oxidised to Cu2+ as the ON of Cu

increases from +1(Cu2O) to +2 in Cu2+. NO

3- is reduced to NO as the ON of N is decreased

from +5(NO3

-) to +2(NO).So let us start with oxidation step. Note that you can start from anystep you like, not necessarily always with reduction or oxidation.

Oxidation step:Cu

2O ---------> Cu2+

Equalise the Cu atoms by placing the coefficient 2 in RHS.


Cu2O ----------> 2Cu2+

Then addition of water is to be done. Since one O atom is in excess in LHS, one H2O

molecule is to be added in RHS.Cu

2O ----------> 2Cu2+ + H

2O

Now since 2 H atoms are in excess in RHS, we have to add 2H+ in LHS.Cu

2O + 2 H+ ----------> 2Cu2+ + H

2O

Finally equalisation of charge is done by adding 2 electrons in the RHS, so that charge on eitherside becomes +2.

Cu2O + 2 H+ ----------> 2Cu2+ + H

2O +2e-

Thus we found that in the oxidation step, electron has appeared in RHS.Reduction step:

NO3

- ----------> NOSince N atoms are same on either side(one), we start by adding 2H

2O molecules on RHS as

there are two excess O atoms in LHS.NO

3- ----------> NO + 2 H

2O

Now we have to add 4H+ ions in LHS as there are 4 excess H atoms in RHS.NO

3- + 4 H+ ----------> NO + 2 H

2O

Charge is made equal by adding 3 electron on LHS, so that the net charge is zero on eitherside.We find that in reduction step, electron appears in the LHS.

NO3- + 4 H+ + 3e-----------> NO + 2 H

2O

Then we have to add the two steps. Let us copy two steps again one above the other.

3X [Cu2O + 2 H+ ----------> 2Cu2+ + H

2O +2e-

2X [NO3- + 4 H+ + 3e-----------> NO + 2 H

2O] (6 electrons are cancelled)

__________________________________3Cu

2O + 14H+ + 2 NO

3- --------> 6Cu2+ + 2NO + 7 H


Balanced molecular equation: 3Cu2O+2KNO

3+7H

2SO

4---->6CuSO

4+K

2SO

4+ 2NO+7H

2O.

Let us see how we can balance a redox reaction taking place in basic(alkaline) medium. Notethat a redox reaction can take place either in acidic medium as discussed before or in inbasic(alkaline) medium.(B) Basic or Alkaline Medium:Let us take one example to understand the rules.

Br2(l) + KIO

3(aq) + KOH(aq) --------> KBr(aq) + KIO

4(aq) + H

2O(l)

Find the net ionic equation as you were doing before by cancelling the spectator ions.Net ionic equation: Br

2 + IO

3- + OH- ---------> Br- + IO

4- + H

2O

Br2( ON=0) is reduced to Br - (ON=-1), while IO

3- (ON of I=+5)is oxidiesd to IO

4-(ON of I=+7)

Br2 IO3

- OH

- Br

- IO4

- H2O+ + + +

0 +5 +7

This reaction has been carried out in alkaline medium as we find OH- in LHS. Note that inalkaline medium, OH - has to appear either in LHS or RHS.Rules:Let us take the oxidation step first and see the rules which are to be applied.


Oxidation step:

IO3- --------> IO

4-

(i) Equalise the element other than H and O by placing suitable coefficient as you didin acidic medium. In this case number of I atoms is same on both the sides.(ii) Count how many O atoms are in excess in a particular side and add same number ofH

2O molecules on the opposite side. This is same as the rule (ii) for balancing in acidic medium.

In this case one O atom is in excess in RHS, so we have to add one H2O molecule in LHS.

IO3

- + H2O --------> IO

4-

(iii) Count how many H atoms are in excess in a particular side. Add same number ofOH- ions on the same side and same number of H

2O molecules on the opposite side.

In this case, LHS contains 2 excess H atoms, so we have to add 2 OH- ions on LHS and 2 H2O

molecules on RHS.IO

3- + H

2O + 2OH- --------> IO

4- + 2 H

2O

Note that in the alkaline medium H2O is added two times, in rule (ii) and (iii).

(iv)Finally equalisation of charge is made by adding electrons to a particular side exactly in thesame way as in acidic medium.

IO3

- + H2O + 2OH- --------> IO

4- + 2 H

2O +2e-

So the charge on either side becomes -3.Let us now take the reduction step and apply the same rules to it.Reduction step:

Br2 ------->Br -

Applying the first rule, we place coefficient 2 in RHS to equalise Br atom.Br

2 -------> 2 Br -

Rules (ii) and (iii) are not applicable in this case as there is no H and O. So we straightway cometo rule(iv) i.e equalisation of charge.

Br2 + 2e- ---------> 2 Br-

Now the charge on either side becomes -2.Finally two steps are added and electrons are cancelled.

IO3

- + H2O + 2OH- --------> IO

4- + 2 H

2O +2e-

Br2 + 2e- ---------> 2 Br- (2 electrons are cancelled)

______________________________________IO

3- + Br

2 + 2OH- + H

2O ------->IO

4- + 2Br - + 2H

2O

On further simplification, we get,IO

3- + Br

2 + 2OH- ------->IO

4- + 2Br- + H

2O (balanced ionic equation)

Balancing of the molecular equation from balanced ionic equation:KIO

3 + Br

2 + 2KOH → → → → → KIO

4 + 2 KBr + H

2O

Try the following SAQ in alkaline medium.

SAQ 6: Deduce the net ionic equation and balance the ionic equation by ion-electron method.K

2CrO

4(aq) + Cu

2O(s) + H

2O(l) --------> Cu(OH)

2(s) + KCrO

2(aq) + KOH(aq)

ELECTRON BALANCE DIAGRAMME METHODThis is the alternative method of balancing by oxidation number(ON) method. Whenever theusual ON method as explained before fails while balancing an equation, this method is adopted.In this case, the procedure is similar to balancing by ion-electron method discussed just before.


By this method, we can know the coefficients of both LHS and RHS species which we cannotknow by the usual ON method discussed earlier.Example:

Cl2 + NaOH → → → → → NaCl + NaClO

3 + H

2O

(i) First assign the ON of atoms undergoing changes.(ii) Select only the atoms(not the molecule or ion) of each side with their ONs which

suffer change and develop oxidation and reduction steps as you did in ion-electron method.Here addition of H+, OH- or H

2O is not done.

Oxidation : Cl Cl+50

Reduction:0

Cl Cl -1

(iii) Balance the charge on the two sides of each equation by adding required number ofelectrons(e-) on one of the sides as we did in the last step for balancing by ion-electronmethod. Then add the two equations by cancelling the electrons from both the equation.Thus we get the electron- balance diagramme.

(electron balance diagramme)

Cl Cl+50

0

Cl Cl -1

+ 5 e-1

+ e-1X 5

6 Cl Cl + 5 Cl0 +5 -1

(iv) Now we have to balance the molecular equation by looking to the coefficients in theelectron balance diagramme. Let us develop a part balanced molecular equation fromthis.

3 Cl2

-------> NaClO3 + 5 NaCl

Then the total equation can be balanced by mere inspection.3 Cl

2 + 6 NaOH ------> NaClO

3 + 5 NaCl + 3 H

2O

SAQ 7: Balance by electron-balance diagramme technique of ON method.(i) K

2Cr

2O

7 + HCl ------> KCl + CrCl

3 + Cl

2 + H

2O

(ii) FeS2 + O

2 -----> Fe

2O

3 + SO

2

PRACTICE QUESTIONS

1. Balance the following ionic equations by ion-electron method. Wherevernecessary add H

2O molecule in the equation.

(i) NO3- + Fe2+ + H+ -------> Fe3+ + NO

2 + H

2O

(ii) NH3 + MnO

4- + OH- ---------> MnO

2 + NO

2

(iii) Fe3+ + H2O

2 -------> Fe2+ + O

2(acidic medium)

iv) Cr(OH)2 + I

2 + OH- --------> Cr(OH)

3 + I-

(v) C2O

42- + H+ + Cr

2O

72- --------> CO

2 + Cr3+ + H

2O

(vi) MnO4

- + H+ + I- ------> Mn2+ + I2 + H

2O

(vii) S2O

32- + I

2 --------> S

4O

62- + I-

(viii) OCl- + H+ + I- ---------> I2 + Cl-

(ix) Cl2 + IO

3- ----> IO

4- + Cl- (alkaline medium)


2. Balance the following by Oxidation Number method.(i)NH

3 + O

2 -----------> NO + H

2O

(ii)H2S + SO

2 ----------> S + H

2O

(iii)Fe2O

3 + CO -----------> Fe + CO

2

(iv)SnCl2 + O

2 + HCl ----------------> SnCl

4 + H

2O

(v)NaClO3 + KI + HCl -------------> NaCl + I

2 + KCl + H

2O

(vi)SbCl5 + KI ---------> SbCl

3 + I

2 + KCl

(vii)NaPO3 + BrF

3 ---------------> NaPF

6 + Br

2 + O

2

(viii)Sn + HNO3 --------------> SnO

2 + NO + H

2O

(ix)Cr2O

3 + HNO

3 + Na

2CO

3 ----------> Na

2CrO

4 + CO

2 + HNO

2

(x) Cu + HNO3 ----------> Cu(NO

3)

2 + N

2O + H

2O

(xi)Cu + HNO3 --------> Cu(NO

3)

2 + NO + H

2O

(xii)BaCrO4 + KI + HCl ----------------> BaCl

2 + I

2 + KCl + CrCl

3 + H

2O

(xiii)Ca3(PO

4)

2 + SiO

2 + C -----------> CaSiO

3 + P

4 + CO

(xiv)Zn + HNO3 -------------> Zn(NO

3)

2 + NH

4NO

3 + H

2O

(xv)K2Cr

2O

7 + KI + H

2SO

4 -----------> K

2SO

4 + Cr

2(SO

4)

3 +I

2 +H

2O

(xvi)KMnO4 + H

2SO

4 + H

2S -------------> K

2SO

4 + MnSO

4 + S + H

2O

(xvii)Cl2 + KIO

3 + KOH --------------> KCl + KIO

4 + H

2O

(xviii)N2H

4 + KBrO

3 → N

2 + KBr + H

2O

(xix)Sn + NaOH -------> Na2SnO

3 + H

2

(xx)Al + NaOH ------> NaAlO2 + H

2

3. Balance the following by partial equation method.

(i) KMnO4 + H

2SO

4 + HI ----------> K

2SO

4 + MnSO

4 + I

2 + H

2O

(ii) KMnO4 + H

2SO

4 + H

2O

2 ----------> K

2SO

4 + MnSO

4 + O

2 + H

2O

(iii) KMnO4 + H

2SO

4 + KNO

2 ----------> K

2SO

4 + MnSO

4 + KNO

3 + H

2O

(iv) KMnO4 + H

2SO

4 + H

2S ----------> K

2SO

4 + MnSO

4 + S + H

2O

(v) KMnO4 + H

2SO

4 + H

2C

2O

4 ----------> K

2SO

4 + MnSO

4 + CO

2+ H

2O

(vi) K2Cr

2O

7 + H

2SO

4 + HI ----------> K

2SO

4 + Cr

2(SO

4)

3 + I

2 + H

2O

(vii) K2Cr

2O

7 + H

2SO

4 +H

2O

2 ----------> K

2SO

4 + Cr

2(SO

4)

3 + O

2 + H

2O

viii) K2Cr

2O

7 + H

2SO

4 +KNO

2 ----------> K

2SO

4 + Cr

2(SO

4)

3 + KNO

3 + H

2O

(ix) K2Cr

2O

7 + H

2SO

4 +H

2S ----------> K

2SO

4 + Cr

2(SO

4)

3 + S + H

2O

(x) K2Cr

2O

7 + H

2SO

4 +H

2C

2O

4 ----------> K

2SO

4 + Cr

2(SO

4)

3 + CO

2 + H

2O

(xi) Cu + HNO3 ---------> Cu(NO

3)

2 + NO + H

2O

(xii) Fe + HNO3 ---------> Fe(NO

3)

2 + N

2O + H

2O

(xiii)Mg + H2SO

4 -------> MgSO

4 + SO

2 + H

2O


4. Balance the following by ion electron method.(i)S

2O

82- + Ce3+ ---------------> Ce4+ + SO

42- (acidic medium)

(ii)Cu2O + H+ + NO

3- ----------> Cu2+ + NO + H

2O

(iii)MnO4

- + SO3

2- + OH- ---------> MnO2 + SO

42- + H

2O

(iv)Mn2+ + S2O

82- -------------> MnO

4- + SO

42- (Acidic medium)

(v)Cr2O

72- + C

2H

4O + H+ ------------> Cr3+ + C

2H

4O

2 + H

2O

(vi)S + OH- ---------------> S2- + S2O

32-

(vii)ClO3- + I- + H+ -------------> Cl- + I

2 + H

2O

(viii)Ag+ + AsH3 ----------------> H

3AsO

4 + Ag (alkaline medium)

(ix)CuS + NO3

- -------------> Cu2+ + S8 + NO (acidic medium)

(x)ClO3- + SbO

2- ------------> ClO

2- + Sb(OH)

6- (basic medium)

(xi)Zn + NO3

- -------------> Zn2+ + NH3 (alkaline medium)

(xii)BrO3- + S

2O

32- ----------> Br - + S

4O

62- + H

2O(acidic medium)

(xiii)MnO4

- + SnO22- + H

2O -----------> MnO

2 + SnO

32- (basic medium)

5. Get the net ionic equation and balance by ion-electron method. From thebalanced ionic equation, balance the molecular equation.

(i)K4[Fe(CN)

6](aq) + HCl(aq) + H

2O

2(l) ---------> K

3[Fe(CN)

6](aq) +

KCl(aq) +H2O(l)

(ii)N2H

4 (l)+ K

3[Fe(CN)

6](aq)------> N

2 (g)+ K

4[Fe(CN)

6](aq) + H

2O(l)

(alkaline medium)

(iii)C2H

2 (g)+ KMnO

4(aq)------> MnO

2(s)+ K

2C

2O

4(aq)+ KOH(aq) +H

2O(l)

(iv)HgCl2(aq)+ SnCl

2(aq)------------> Hg

2Cl

2(s)+ SnCl

4(aq)

(v)NaNO3(aq) + Zn(s) + NaOH(aq) -----> NH

3(g) + Na

2ZnO

2(aq) + H

2O(l)

(vi)HgS(s) + HNO3(aq) + HCl(aq)------> S + H

2[HgCl

4](aq) + NO(g) + H

2O(l)

(vii) K2Cr

2O

7(aq) + H

2SO

4(aq)+KNO

2(aq) ----------> K

2SO

4(aq)+ Cr

2(SO

4)

3(aq)+

KNO3(aq)+ H

2O(l)

(viii) K2Cr

2O

7(aq)+ H

2SO

4(aq)+H

2S(g) ------->K

2SO

4(aq)+ Cr

2(SO

4)

3(aq)+ S + H

2O(l)

(ix) K2Cr

2O

7(aq)+ H

2SO

4(aq)+H

2C

2O

4(s)-----> K

2SO

4(aq)+ Cr

2(SO

4)

3(aq)+ CO

2(g)+ H

2O(l)

(x) Cu(s) + HNO3(aq) ---------> Cu(NO

3)

2(aq)+ NO(g) + H

2O(l)

(xi) Fe(s) + HNO3(aq)---------> Fe(NO

3)

2(aq)+ N

2O(g)+ H

2O(l)

Supplimentary Questions:

1. Balance by any one method: (either by ion-electron or ON method)(i) CuS(s) + HNO

3(aq) ------> Cu(NO

3)

2(aq) + S + H

2O(l) + NO(g)

(ii) As2S

5(s)+ HNO

3(aq) -------> H

3AsO

4(aq)+ H

2SO

4(aq)+ H

2O(l) + NO

2(g)

(iii) Zn(s) + HNO3(aq) --------> Zn(NO

3)

2(aq) + H

2O(l) + NH

4NO

3(aq)

(iv) Na2C

2O

4(aq) + KMnO

4(aq) + H

2SO

4(aq) -----> K

2SO

4(aq) +MnSO

4(aq) +

CO2(g)+ Na

2SO

4(aq) + H

2O(l)


(v) MnO + PbO2 + HNO

3 -------> HMnO

4 + Pb(NO

3)

2 + H

2O

(vi) Na2HAsO

3(aq) + KBrO

3(aq) + HCl(aq) -----> NaCl(aq) + KBr(aq) +

H3AsO

4(l)

(vii) FeS2 + O

2 ------> Fe

2O

3 + SO

2

(viii) Ca(OCl)2(aq) + KI(aq) + HCl(aq) ------> I

2(s) + CaCl

2(aq) + H

2O(l) + KCl(aq)

(ix) NaOCl(aq) + NaOH(aq) + Bi2O

3(s) ----> NaBiO

3(aq) + H

2O + NaCl(aq)

(x) Sn + HNO3 ------> SnO

2 + NO

2 + H

2O

(xi) CrI3 + KOH + Cl

2 → K

2Cr

2O

7 + KIO

4 + KCl + H

2O

(xii) Na2TeO

3 + + NaI + HCl → NaCl + Te + I

2 + H

2O

(xiii) K2Cr

2O

7 + SnCl

2 + HCl → CrCl

3 + SnCl

4 + KCl + H

2O

(xiv) K3[Fe(CN)

6] + Cr

2O

3 + KOH → K

4[Fe(CN)

6] + K

2CrO

4 + H

2O

(xv) NaHSO4 + Al + NaOH → Na

2S + Al

2O

3 + H

2O

(xvi) KI + H2SO

4 → K

2SO

4 + I

2 + H

2S + H

2O

(xvii) Cr2O

3 + Na

2CO

3 + KNO

3 → Na

2CrO

4 + CO

2 + KNO

2

(xviii) KClO3 + H

2SO

4 → KHSO

4 + ClO

2 + O

2 + H

2O

(xix) P2H

4 → PH

3 + P

4H

2

(xx) CoCl2 + KNO

2 + HCl → K

3[Co(NO

2)

6] + NO + KCl + H

2O

(xxi) MnO4

- → MnO4

2- + O2 (alkaline medium) Also balance the molecular equation

(xxii) Au + CN- + O2 → [Au(CN)

4]- + H

2O (alkaline medium). Also balance the

molecular equation

(xxiii) Zn + As2O

3 → AsH

3 + Zn2+ (acid medium) Also balance the molecular equation

(xxiv) Cl2 + IO

3- → IO

4- + Cl -(alkaline medium) Also balance the molecular equation

(xxv) Zn + NO3- → Zn2+ + NH

4+ (acidic medium) Also balance the molecular equation

(xxvi) V + NaOH + H2O → Na

3HV

6O

17 + H

2

(xxvii) MnBr2 + PbO

2 + HNO

3 → HMnO

4 + Pb(NO

3)

2 + Pb(BrO

3)

2 + H

2O

2. Complete and balance the following:

(i) Sn2+ + H+ + NO3

- -----> Sn4+ + NO

(ii) Cr2O

72- + I- ----> I

2 + Cr3+ (acidic medium)

(iii) Zn + H+ + NO3- -----> Zn2+ + NH

4+

(iv) S2O

32- + Ag+ -------> Ag + S

4O

62-

(v) P4 + OH- ----> PH

3 + H

2PO

2-

(vi) Br2 + IO

3- ----> Br - + IO

4- (basic medium)

(vii) MnO4

- + Sn2+ ----> Mn2+ + Sn4+ (acidic medium)

(viii) S2O

82- + I- ----> I

3- + SO

2 (acidic medium)

(ix) CrO4

2- + Cu2O ----> Cu(OH)

2 + [Cr(OH)

4]- (alkaline medium)


RESPONSE TO SAQs

SAQ 1: Balance the following reactions:(i) 3Ca(OH)

2 + 2H

3PO

4 ---------> Ca

3(PO

4)

2 + 6 H

2O

(ii) ZnO + H2SO

4 -------> ZnSO

4 + H

2O

(iii) FeCl2 + H

2S -------> FeS + 2HCl

(iv) P2O

5 + 3H

2O -------> 2H

3PO

4

(v) 2NH4Cl + Ca(OH)

2--------> CaCl

2 + 2NH

3 + 2H

2O

SAQ 2:

(i) 0-1

KI H2SO4 I2 K2SO4 H2S H2O+ + + +-2+6

(KI is RA and H2SO

4 is OA)

The change in ON in RA = |-1-0| = 1, so 1 is the coefficient before OA(H2SO

4). 1 is not written

because if no other coefficient is placed, then the coefficient 1 is understood to be present.

The change in ON in OA = |+6-(-2)| = 8, so 8 is to be placed before KI(RA)

8KI + H2SO

4 ---------> I

2 + K

2SO

4 + H

2S + H

2O

Then hit and trial. To equalise I, we have to place 4 before I2 in RHS and to equalise K, we have

to place 4 before K2SO

4.

8KI + H2SO

4 ---------> 4I

2 + 4K

2SO

4 + H

2S + H

2O

Let us equalise S now. The number of S atoms in RHS is 5, so 5 is placed as coefficient beforeH

2SO

4 in LHS. Note that the coefficient 1 that was found in the first case is now revised to 5.

8KI + 5H2SO

4 ---------> 4I

2 + 4K

2SO

4 + H

2S + H

2O (Revised coefficient of H

2SO

4)

To equalise H now, we have to place 4 before H2O. Thus we see that by doing so, O is

automatically balanced(20 O in either side).

8KI + 5H2SO

4 ---------> 4I

2 + 4K

2SO

4 + H

2S + 4H


Just imagine that had you not adopted this technique, could you have balanced this equation byhit and trial method? The answer may be YES. But in that case you would have taken a verylong time to do so. Therefore you are advised not to take an attempt to balance a redox reactionby hit and trial method if it appears to be little tough. Always make use of this ON method.

(ii)0

+ + +I2 HNO3 HIO3 NO2 H2O+5 +4+5

(I2 is RA and HNO

3 is OA)

The change in ON per one I = |0-5| =5, so for two I atoms present in the RA(I2), the total change

is 10. So 10 is placed as coefficient before HNO3(OA).

The change in ON of N = |+5-(+4)| =1, and since there is one N atom in HNO3, the total change

is also 1. So 1 is to be placed as the coefficient of I2 and therefore it is left as such.

I2 + 10HNO

3 --------> HIO

3 + NO

2 + H

2O

To equalise N, we place 10 before NO2 in RHS and to equalise I, we place 2 before HIO

3 in

RHS.


I2 + 10HNO

3 --------> 2HIO

3 + 10NO

2 + H

2O

To equalise H atoms, we place 4 before H2O, and by doing so, it is seen that O is automatically

balanced(30 O on either side).

I2 + 10HNO

3 --------> 2HIO

3 + 10NO

2 + 4H


What a fantastic method!!!! Could you have balanced it by hit and trial method? The answer isagain, NO.

(iii) Cu HNO3 Cu(NO 3)2 NO H2O+ + +0 +5 +2 +2

(RA is Cu and OA is HNO3)

The change of ON in Cu(RA) is 2 and in HNO3(OA) is 3. So first 2 is placed before HNO

3 and

3 placed before Cu. Then hit and trial starts. To equalise Cu, we have to place 3 before Cu(NO3)

2.

Cu HNO3 Cu(NO3)2 NO H2O+ + +3 2 3

By doing so the total number of N atoms in RHS is 7. So we have to revise the coefficient of

HNO3 in the LHS and place 7 in place of 2. Then to equalise H, we have to place 7

2 before

H2O.

Cu HNO3 Cu(NO 3)2 NO H2O+ + +3 3772

In so doing we found that O could not be balanced. 21 O are on LHS and 22.5 on RHS. So insuch case we have to re-revise the coefficient of NO. Note that we cannot do any change in Cuand Cu(NO

3)

2. So we place 2 before NO and accordingly change HNO

3 and H

2O. For that we

revise the coefficient of HNO3 from 7 to 8 to equalise 8 N atoms on either side. Then to equalise

H we place 4 before H2O. Now we find that O atoms are automatically balanced on either side.

So the balanced equation is

3Cu HNO3 Cu(NO3)2 NO H2O+3 + +28 4 (balanced equation)

SAQ 3:(i)

3

H2OCO2[O]H2C2O4

2

KMnO4 H2SO4 H2C2O4 K2SO4 MnSO4 CO2 H2O++ + + +

KMnO4 K2O MnO [O]+ +52 2

K2O H2SO4 K2SO4 H2O++

MnO H2SO4 MnSO4 H2O++

+

+

+


2 X

5 X

KMnO4 H2SO4 H2C2O4 K2SO4 MnSO4 CO2 H2O 825 + +++

(balanced equation)

2

10

In this case also the first three steps are same as done in the example in the text. In the forth stepH

2C

2O

4(oxalic acid) reacts with nascent oxygen to form CO

2. The question is how shall you

know this? This is very simple. You merely look to the RHS, you can know the products


immediately. Can you indicate now which is oxidised and which reduced. Mn has ON +7 inKMnO

4 and +2 in MnSO

4, so KMnO

4 has been reduced. While the ON of C in H

2C

2O

4 is

+3(find it by x method)and it is +4 in CO2. So H

2C

2O

4 has been oxidised. So which is OA and

RA? You can better say that now.

(ii)

+

34 +++++K2Cr2O7 H2SO4 H2O2 K2SO4 Cr2(SO4)3 O2 H2O(Step 5)

3 X

3

3 3

3


+

+ +Cr2O3 H2SO4 Cr2(SO4)3 H2O+ +K2O H2SO4 K2SO4 H2O

++K2Cr2O7 K2O Cr2O3 [O]

+ ++++K2Cr2O7 H2SO4 H2O2 K2SO4 Cr2(SO4)3 O2 H2O

H2O2 [O] O2 H2O

7

In this case too, the product in the 4th step(H2O

2 reacting with [O]) is known just by looking at

the RHS. It is O2. The other product has to be H

2O to balance H. This is known from mere

common sense.In this case K

2Cr

2O

7 is the OA as the ON of Cr is +6 and it is reduced to +3 in Cr

2(SO

4)

3. H

2O

2

is the RA as the ON of O is -1(peroxide) and it is oxidised to 0 in O2 of RHS.

SAQ 4: (i) In this case the HNO3 used must be very dilute type as NH

4NO

3 is produced.

HNO3 + H

2O --------> NH

3 + 4[O] (i)

[NH3 + HNO

3 --------> NH

4NO

3

4X [Zn + [O] ------> ZnO] (ii) (4 multiplied to cancel [O]) 4X [ZnO + 2HNO

3 --------> Zn(NO

3)

2 + H

2O] (iii) (4 multiplied to cancel 4 ZnO)

_________________________________________4Zn + 10HNO

3 -------> 4Zn(NO

3)

2 + NH

4NO

3 + 3H


Note that in step (i) H2O appears in LHS, otherwise you cannot balance the equation,remember

this.(ii) In this case the HNO

3 used must be moderately concentrated as NO is produced.

2HNO3 ----------> 2NO + H

2O + 3[O]

3X[Mg + [O]---------> MgO ] 3X[MgO + 2HNO

3 ---> Mg(NO

3)

2 + H

2O ]

_____________________________________3Mg + 8HNO

3 -------> 3Mg(NO

3)

2 + 2NO + 4H


SAQ 5: 3X [H

2SO

4 ------> SO

2 + H

2O + [O]] (3 multiplied to cancel 3[O])

2Al + 3[O] -----> Al2O

3

Al2O

3 + 3H

2SO

4 --------> Al

2(SO

4)

3 + 3H

2O (Al

2O

3 get cancelled)

______________________________________________2Al + 6H

2SO

4 -------> Al

2(SO

4)

3 + 3SO

2 + 6H



SAQ 6:

Net ionic equation: CrO4

2- + Cu2O(s) + H

2O(l) ------> Cu(OH)

2(s) + CrO

2- + OH-

Note that this reaction has been carried out in alkaline medium as OH- is present in the RHS.Could you find which is oxidised and which reduced? K

2CrO

4 has been reduced to CrO

2- because

the ON of Cr had decreased from +6 to +3. Cu2O has been oxidised to Cu(OH)

2 as the ON of

Cu has increased from +1 to +2. So let us take the steps separtely and apply the rules.

Reduction step:CrO

42- --------> CrO

2-

Cr atoms are same, so no coefficient is to be placed in any side.Then 2 H

2O molecules are to be added on RHS as LHS has 2 excess O atoms.

CrO4

2- --------> CrO2

- + 2H2O

Then 4OH- are to be added on RHS and 4 H2O to be added on LHS as RHS has 4 excess H

atoms.CrO

42- + 4 H

2O --------> CrO

2- + 2H

2O + 4OH-

Then equalisation of charge is done by adding 3 electrons in LHS so that the charge on eitherside becomes -5.

CrO4

2- + 4 H2O + 3e---------> CrO

2- + 2H

2O + 4OH -

Oxidation step:Cu

2O --------> Cu(OH)

2

First we have to place a coefficient of 2 in RHS to equalise Cu atoms.Cu

2O --------> 2Cu(OH)

2

Since in LHS there is one O atom and in RHS there are 4 O atoms, RHS has 3 excess O atoms.So we have to add 3 H

2O molecules on LHS

Cu2O + 3 H

2O --------> 2Cu(OH)

2

Now let us find how many H atoms are excess in which side. LHS contains 2 excess H atoms.So we have to add 2 OH- on LHS and 2 H

2O molecules on RHS.

Cu2O + 3 H

2O + 2OH- --------> 2Cu(OH)

2 + 2H

2O

Then charge equalisation is made by adding 2 electrons on RHS, so that both sides have -2charge.

Cu2O + 3 H

2O + 2OH- --------> 2Cu(OH)

2 + 2H

2O + 2e-

Let us add now the two steps. 2 X [CrO

42- + 4 H

2O + 3e- --------> CrO

2- + 2H

2O + 4OH- ]

3X [Cu2O + 3 H

2O + 2OH- --------> 2Cu(OH)

2 + 2H

2O + 2e-] (6 electrons are cancelled)

______________________________________________________2CrO

42- + 17H

2O + 6OH- + 3Cu

2O --------> 2CrO

2- + 10 H

2O + 6Cu(OH)

2 +8OH-

On further simplification we get,2CrO

42- + 7H

2O + 3Cu

2O --------> 2CrO

2- + 6Cu(OH)

2 + 2 OH -

SAQ 7:

(i) K2Cr2O7 + HCl KCl + CrCl3 + Cl2 + H2O+6 +3-1 0


Cl -1

Cr + 3 e Cr +6 +3

-1

Cl + e 0

-1

Reduction

Oxidation X 3

Cr + 3 Cl Cr + 3 Cl+6 -1 +3 0


Looking to 2 Cr atoms in K2Cr

2O

7, let us multiply the above equation by 2 throughout

2Cr + 6 Cl 2Cr + 6 Cl+6 -1 +3 0

So we can develop a part balanced molecular equation,K

2Cr

2O

7 + 6 HCl ------> 2 KCl + 2 CrCl

3 + 3 Cl

2

But equation is not yet balanced and one more product is missing(H2O). Since HCl is used

both as a reducing agent(RA) and an acid to form salt, more number of Cl- ions are requiredand so the coefficient of HCl has to be changed. First for equalising O atoms on both the sides,7 is placed as the coefficient of H

2O and accordingly to balance H atoms, the coefficient of

HCl is changed to 14.K

2Cr

2O

7 + 14 HCl -----> 2 KCl + 2CrCl

3 + 3 Cl

2 + 7 H

2O

(ii)+3

FeS2 + O2 Fe2O3 + SO2+2 -2-1 0 +4

Here there are two oxidations i.e Fe2+ is oxidised to Fe3+ and S-1 is oxidised to S+4. Note thatin iron disulphide(called iron pyrite or fool's gold) Fe is in the +2 state and sulphide is adisulphide(belongs to polysulphide category) in which S is in -1 state. We get the net oxidationstep by adding the following two oxidation steps.Oxidation:

+3+2

Fe Fe + e

2S 2S + 10e

-1

-1 +4-1

Fe + 2S Fe + S + 11 e -1+2 -1 +3 +4

Note that we multiplied the 2nd oxidation step by 2 to get the ratio of Fe and S equalto 1:2 to agree with the formula of iron dilphide.

Reduction: O + 2 e O0

-1 -2

Adding the oxidaton and reduction steps by cancelling 22 electrons on either side, we get,

11 O + 2 Fe + 4 S 11 O + 2 Fe + 4 S0 +2 -1 -2 +3 +4


Looking to the electron balance diagramme, we can balance the molecular equation,2 FeS

2 + 11/2 O

2 ------> Fe

2O

3 + 4 SO

2

To eliminate fraction, we multiply the equation by 2 throughout4 FeS

2 + 11 O

2 ------> 2 Fe

2O

3 + 8 SO

2


ANSWERS TO PRACTICE QUESTIONS1. (i) NO

3- + Fe2+ + H+ -------> Fe3+ + NO

2 + H

2O (Acidic medium)

Reduction step: NO3

- + 2H+ + e- --------> NO2

+ H2O

Oxidation step: Fe2+ --------------> Fe3+ + e-

____________________________________Fe2+ + NO

3- + 2H+ -------> NO

2 + Fe3+ + H

2O

(ii) NH3 + MnO

4- + OH- ---------> MnO

2 + NO

2(Alkaline medium)

Reduction step: 7X[MnO4

- + 4H2O + 3e- ---------> MnO

2 + 2H

2O + 4OH- ]

Oxidation step: 3X[NH3 + 2H

2O + 7OH- ----------> NO

2 + 7H

2O + 7e-]

__________________________________________7MnO

4- + 28H

2O + 3NH

3 + 6H

2O + 21OH- ------>

7MnO2 + 14H

2O + 28OH- + 3NO

2 + 21H

2O

On simplifying we get: 7MnO4

- + 3NH3 ---------> 7MnO

2 + 3NO

2 + 7 OH- + H

2O

(iii) Fe3+ + H2O

2 -------> Fe2+ + O

2(acidic medium)

Oxidation step: H2O

2 ----------> O

2 + 2H+ + 2e Here H

2O addition is not necessary

Reduction step: 2X[Fe3+ +e -----------> Fe2+ ]___________________________H

2O

2 + 2Fe3+ ---------> O

2 + 2Fe2+ + 2H+

(iv) Cr(OH)2 + I

2 + OH- --------> Cr(OH)

3 + I- (alkaline medium as OH- is present)

Reduction step: [I2 + 2e- ---------> 2I- ]

Oxidation step: 2X [Cr(OH)2 + H

2O + OH- ---------> Cr(OH)

3 + H

2O + e-]

_____________________________________________2Cr(OH)

2 + I

2 + 2OH- -------> 2Cr(OH)

3 + 2I-

(v) C2O

42- + H+ + Cr

2O

72- --------> CO

2 + Cr3+ + H

2O (acidic medium)

Reduction step: Cr2O

72- + 14H+ + 6e- --------> 2Cr3+ + 7H

2O

Oxidation step: 3X [C2O

42- ------------> 2CO

2 + 2e-]

____________________________________Cr

2O

72- + 3C

2O

42- + 14H+ --------> 2Cr3+ + 6CO

2 + 7H

2O

(vi) MnO4

- + H+ + I- ------> Mn2+ + I2 + H

2O

Reduction step: 2X [MnO4

- + 8H+ + 5e----------> Mn2+ + 4H2O]

Oxidation step: 5X [2I- --------------> I2 + 2e-]

______________________________________2MnO

4- + 16H+ + 10I- -------> 2Mn2+ + 5I

2 + 8H

2O

(vii) S2O

32- + I

2 --------> S

4O

62- + I-

Reduction step: I2 + 2e- ---------> 2I-

Oxidation step: 2S2O

32- --------> S

4O

62- + 2e-

___________________________I

2 + 2S

2O

32- -------> 2I- + S

4O

62-

In this case neither acidic nor basic medium is required. As you see, in the steps, there is no needof H+ or OH- ions for balancing.(viii) OCl- + H+ + I- ---------> I

2 + Cl- (acidic medium)

Reduction step: OCl-+2H+ + 2e- --------> Cl- + H2O (Cl changed from +1 to -1)

Oxidation step: 2I- --------------> I2 + 2e- (I from -1 to 0)

______________________________OCl- + 2I- + 2H+---------> Cl- + I

2 + H

2O


(ix) Cl2 + IO

3- ----> IO

4- + Cl- (alkaline medium)

Reduction step: Cl2 + 2e- -----------> 2Cl-

Oxidation step: IO3

- + H2O + 2OH- ------------> IO

4- + 2H

2O + 2e-

_________________________________________Cl

2 + IO

3- + 2OH- ----------> IO

4- + 2Cl- + H

2O

2.(i) NH

3 + O

2 -----------> NO + H

2O

ON of N goes from -3 to +2. So the change in ON =|-3-(+2)| =5, and change in ON in two atomsof O= 2X|-2|=4

4NH3 + 5O

2 -----------> 4NO + 6H

2O

(ii) H2S + SO

2 ----------> S + H

2O

ON of S goes from -2 in H2S to 0 in S, so change in ON is 2. The ON of S change from +4 in

SO2 to 0 in S, so the change is 4.

4H2S + 2SO

2 ----------> 6 S +4 H

2O

This is further simplified as : 2H2S + SO

2 ----------> 3 S +2 H

2O

(iii) Fe2O

3 + CO -----------> Fe + CO

2

The ON of Fe goes from +3 to 0, so the change per atom is 3 and for 2 Fe atoms present inFe

2O

3, the total change is 2X3=6, The ON of C changes from +2 to +4. So the change is 2.

2Fe2O

3 + 6CO -----------> 4 Fe + 6CO

2

Further simplification results: Fe2O

3 + 3CO -----------> 2 Fe + 3CO

2

(iv)SnCl2 + O

2 + HCl ----------------> SnCl

4 + H

2O

ON of Sn goes from +2 to +4, so the change in ON is 2 which is placed before O2(OA). The

change in ON per one O atom is |0-2|=2 and for 2 atoms present in O2 the total change is 2X2=4,

which is to be placed before SnCl2(RA). Then hit and trial.

4SnCl2 + 2O

2 +8 HCl ----------------> 4SnCl

4 +4 H

2O

On simplification, we get2SnCl

2 + O

2 + 4HCl ----------------> 2SnCl

4 +2 H

2O

(v) NaClO3 + KI + HCl -------------> NaCl + I

2 + KCl + H

2O

The ON of Cl goes from +5(NaClO3) to -1(NaCl) and the change is +5-(-1)=6, which is to be

placed before KI(RA). The change in ON of I is |-1-0|=1, so no other coefficient is to be placedbefore NaClO

3(OA). Then hit and trial gives the balanced equation.

NaClO3 + 6KI + 6HCl -------------> NaCl +3 I

2 + 6KCl + 3H

2O

(vi) SbCl5 + KI ---------> SbCl

3 + I

2 + KCl

The change in ON of Sb is |+5-3|=2, which is to be placed before KI. The change in ON in I is|-1-0|=1, so no other coefficient is to be placed before SbCl

5. Then hit and trial.

SbCl5 + 2KI ---------> SbCl

3 + I

2 + 2KCl

(vii) NaPO3 + BrF

3 ---------------> NaPF

6 + Br

2 + O

2

The ON of Br goes from +3(BrF3) to 0(Br

2) and hence the change is 3 which is to be placed

before NaPO3(RA). The ON of O changes from -2(NaPO

3) to 0(O

2) and hence the change is

2 per atom. For 3 O atoms present, the total change is 2X3=6, which is to be placed beforeBrF

3(OA). Then hit and trial.

3NaPO3 + 6BrF

3 ---------------> 3NaPF

6 + 3Br

2 + 2

9 O2

To eliminate the fractional coefficient, the whole equation is multiplied by 2.6NaPO

3 + 12BrF

3 --------------->6 NaPF

6 + 6Br

2 + 9O

2


Now we can simplify by dividing the equation by 3 througout.2NaPO

3 + 4BrF

3 --------------->2 NaPF

6 + 2Br

2 + 3O

2

(viii) Sn + HNO3 --------------> SnO

2 + NO + H

2O

The Change in ON of Sn is |0-4|=4 and change in ON of N is |5-2|=3, so3Sn + 4HNO

3 --------------> SnO

2 + NO + H

2O

Then hit and trial gives3Sn + 4HNO

3 --------------> 3SnO

2 + 4NO + 2H


(ix) Cr2O

3 + HNO

3 + Na

2CO

3 ----------> Na

2CrO

4 + CO

2 + HNO

2

The ON of one Cr changes by |+3-6| =3, so for two Cr atoms present in Cr2O

3(OA), the total

change is 6. The change in ON of N is |+5-3|=2, so we get first,2Cr

2O

3 + 6HNO

3 + Na

2CO

3 ----------> Na

2CrO

4 + CO

2 + HNO

2

Hit and trial gives the following2Cr

2O

3 + 6HNO

3 + 4Na

2CO

3 ----------> 4Na

2CrO

4 + 4CO

2 + 6HNO

2

Further simplification givesCr

2O

3 +3 HNO

3 + 2Na

2CO

3 ----------> 2Na

2CrO

4 + 2CO

2 + 3HNO

2

(x) Cu + HNO3 ----------> Cu(NO

3)

2 + N

2O + H

2O

The change in ON of Cu =2 and N=|5-1)=4, so first we write as follows.4Cu + 2HNO

3 ----------> Cu(NO

3)

2 + N

2O + H

2O

Hit and trial gives4Cu + 2HNO

3 ----------> 4Cu(NO

3)

2 + N

2O + H

2O

But now RHS contains 10 N atoms, so we have to revise the coefficient of HNO3 in LHS and

accordingly balance the rest.4Cu + 10HNO

3 ----------> 4Cu(NO

3)

2 + N

2O + 5H

2O

(xi) Cu + HNO3 --------> Cu(NO

3)

2 + NO + H

2O

The change in ON of Cu=2 and N=3, so we first write3Cu + 2HNO

3 --------> Cu(NO

3)

2 + NO + H

2O

Hit and trial gives,3Cu + 2HNO

3 --------> 3Cu(NO

3)

2 + NO + H

2O

Since RHS now contains 7 N atoms, we have to revise the coefficient of HNO3 from 2 to 7 to

equalise the N atoms.3Cu + 7HNO

3 --------> 3Cu(NO

3)

2 + NO + H

2O

To balance H, we have to place 27

before H2O. But we find that O is not balanced by that.

3Cu + 7HNO3 --------> 3Cu(NO

3)

2 + NO + 2

7H

2O

So we have to again revise the number of N atoms by balancing 2 before NO(RHS) andaccordingly change the coefficient of HNO

3. After that H is adjusted again .

3Cu + 8HNO3 --------> 3Cu(NO

3)

2 + 2NO + 4H


(xii) BaCrO4 + KI + HCl ----------------> BaCl

2 + I

2 + KCl + CrCl

3 + H

2O

The ON of Cr changes from +6 to +3, so the change is 3. The change in ON of I is 1.So we firstwrite

BaCrO4 + 3KI + HCl ----------------> BaCl

2 + I

2 + KCl + CrCl

3 + H

2O

Further hit and trial gives

BaCrO4 + 3KI + 8HCl ----------------> BaCl

2 + 2

3 I2 + 3KCl + CrCl

3 + 4H

2O


To eliminate fraction, we multiply by 2 throughout2BaCrO

4 + 6KI + 16HCl ----------------> 2BaCl

2 + 3I

2 + 6KCl + 2CrCl

3 + 8H

2O

(xiii) Ca3(PO

4)

2 + SiO

2 + C -----------> CaSiO

3 + P

4 + CO

The ON of P change from +5 to 0 for one P atoms. For two P atoms present in Ca3(PO

4)

2, the

total change is 10. The change in ON of C is 2. So first we write2Ca

3(PO

4)

2 + SiO

2 +10 C -----------> CaSiO

3 + P

4 + CO

Hit and trial gives2Ca

3(PO

4)

2 + 6SiO

2 +10 C -----------> 6CaSiO

3 + P

4 + 10CO

(xiv) Zn + HNO3 -------------> Zn(NO

3)

2 + NH

4NO

3 + H

2O

The change of ON of Zn is 2 and the ON of N changes from +5(HNO3) to -3(NH

4+ of NH

4NO

3).

So the change in ON=|+5-(-3)|=8, so first we write,8Zn + 2HNO

3 -------------> Zn(NO

3)

2 + NH

4NO

3 + H

2O

Hit and trial results8Zn + 2HNO

3 -------------> 8Zn(NO

3)

2 + NH

4NO

3 + H

2O

RHS now contains 18 N atoms, so we have to modify the coefficient of HNO3 and accordingly

H is adjusted.8Zn + 18HNO

3 -------------> 8Zn(NO

3)

2 + NH

4NO

3 + 7H

2O

We find that O is not balanced. So again we have to modify the N atoms by first placing coefficient2 before NH

4NO

3 and then adjusting HNO

3 and then other atoms.

8Zn + 20HNO3 -------------> 8Zn(NO

3)

2 + 2NH

4NO

3 + 6H

2O

Now interestingly we find that O is balanced.(xv) K

2Cr

2O

7 + KI + H

2SO

4 -----------> K

2SO

4 + Cr

2(SO

4)

3 +I

2 +H

2O

The ON of Cr changes from +6 to +3, so the change is 3 per atom. But for 2 Cr atoms prsent inK

2Cr

2O

7(OA), the total change in ON=6. The change in ON of I is 1. So first we write,K

2Cr

2O

7 + 6KI + H

2SO

4 -----------> K

2SO

4 + Cr

2(SO

4)

3 +I

2 +H

2O

Further hit and trial givesK

2Cr

2O

7 + 6KI + 7H

2SO

4 -----------> 4K

2SO

4 + Cr

2(SO

4)

3 +3I

2 + 7H

2O(balanced)

(xvi) KMnO4 + H

2SO

4 + H

2S -------------> K

2SO

4 + MnSO

4 + S + H

2O

The change in ON of Mn =|7-2|=5, and that of S=|-2-0|=2, so first we write,2KMnO

4 + H

2SO

4 + 5H

2S -------------> K

2SO

4 + MnSO

4 + S + H

2O,

Further hit and trial gives,2KMnO

4 + 3H

2SO

4 + 5H

2S -------------> K

2SO

4 + 2MnSO

4 + 5S + 8H

2O

(xvii) Cl2 + KIO

3 + KOH --------------> KCl + KIO

4 + H

2O

The change in ON of Cl is 1 for one Cl atom. So for 2 Cl atoms present in the OA(Cl2), the total

change is 2. The change in ON of I=|5-7|=2, so first we write,2Cl

2 + 2KIO

3 + KOH --------------> KCl + KIO

4 + H

2O,

Hit and trial gives,2Cl

2 + 2KIO

3 + 4KOH --------------> 4KCl + 2KIO

4 + 2H

2O

On simplification, it gives,Cl

2 + KIO

3 + 2KOH --------------> 2KCl + KIO

4 + H

2O

(xviii) N2H

4 + KBrO

3 → N

2 + KBr + H

2O

N goes from -2 to 0, so for 2 N atoms the total change of ON is 4. Br goes from +5 to-1 and so the change is 6.

6N2H

4 + 4KBrO

3 -----> 6N

2 + 4KBr + 12H

2O


(xix)Sn + NaOH -------> Na2SnO

3 + H

2

Sn goes from 0 to +4, so the change is 4. H goes from +1 to 0, so the change is 1.H

2O has to be added in LHS for balancing. By adding one H

2O molecule there will be

4H atoms in LHS to account for the total change of ON in H to balance with the total change inON in Sn(4).

Sn + 2NaOH + H2O -------> Na

2SnO

3 + 2H

2

(xx)Al + NaOH ------> NaAlO2 + H

2

Al goes from 0 to +3 while H goes from +1 to 0. H2O molecule is to be added in LHS for

balancing. By adding one H2O molecule there will be 3 H atoms in LHS to account for the total

change of ON in H to balance with the total change in ON in Al(3).Al + NaOH + H

2O ---------> NaAlO

2 + 3/2 H

2

2Al + 2NaOH + 2H2O ----------> 2NaAlO

2 + 3H

2

3.(i)

KMnO4 K2O MnO [O]++2 2 5K2O H2SO4 K2SO4 H2O

+

+ +

MnO H2SO4 MnSO4 H2O+ +

HI [O] I2 H2O2 +5 X2 X

KMnO4 H2SO4 HI K2SO4 MnSO4 I2 H2O+ + + + +3 10 2 5 82

(ii)

2

KMnO4 K2O MnO [O]++2 2 5K2O H2SO4 K2SO4 H2O

+

+ +

MnO H2SO4 MnSO4 H2O+ +

+ + +

+

+3 25 8

H2O2 [O] H2O O2

2 X5 X

KMnO4 H2SO4 H2O2 K2SO4 MnSO4 O2 H2O+5

(iii)

5K2SO4 MnSO4 KNO3KMnO4 H2SO4 KNO2

KNO3KNO25 X2 X

[O]

5 23 ++++

++ MnSO4 H2OMnO H2SO4

++

+

K2SO4 H2OK2O H2SO4

522 + +K2O MnO [O]KMnO4

2 + 3H2O

(iv) Try this by partial equation method in the similar manner as shown above.2KMnO

4 + 3H

2SO

4 + 5H

2S -------> K

2SO

4 + 2MnSO

4 + 5S + 8H

2O

(v) 2KMnO4 + 3H

2SO

4 + 5H

2C

2O

4 ----> K

2SO

4 + 2MnSO

4 + 10CO

2 + 8H

2O

(vi)

7364 +++++ K2SO4 Cr2(SO4)3 I2 H2OK2Cr2O7 H2SO4 HI

3 X ++2 I2 H2OHI [O]

+ +

33 ++ Cr2(SO4)3 H2OCr2O3 H2SO4

K2SO4 H2OK2O H2SO4

++ 3K2O Cr2O3 [O]K2Cr2O7


(vii) Try this by partial equation method of your own and check with the answer.

K2Cr

2O

7 + 4H

2SO

4 + 3H

2O

2 -------> K

2SO

4 + Cr

2(SO

4)

3 + 3O

2 + 7H

2O

(viii) K2Cr

2O

7 + 4H

2SO

4 + 3KNO

2 ------> K

2SO

4 + Cr

2(SO

4)

3 + 3KNO

3 + 4H

2O

(ix) K2Cr

2O

7 + 4H

2SO

4 + 3H

2S --------> K

2SO

4 + Cr

2(SO

4)

3 + 3S + 7H

2O

(x) K2Cr

2O

7 + 4H

2SO

4 + 3H

2C

2O

4-------> K

2SO

4 + Cr

2(SO

4)

3 + 6CO

2 + 7H

2O

(xi)

8

HNO3 NO H2O [O]2 + +2 3

Cu [O] CuO+

CuO HNO3 Cu(NO3)2 H2O+ 2 +3 X3 X

Cu HNO3 Cu(NO3)2 NO H2O+ + +2 433

(xii) Try yourself and check. 4Fe + 10 HNO3 ---------> 4Fe(NO

3)

2 + N

2O + 5H

2O

(xiii)

22 +++ MgSO4 SO2 H2OMg H2SO4

++

+

++

MgSO4 H2OMgO H2SO4

MgOMg [O]

SO2 H2O [O]H2SO4

Note that whenever you find a reaction involving conc. H2SO

4, the first step is the breaking

down of H2SO

4 giving SO

2, H

2O and nascent oxygen atom.

4.(i) Reduction step: S2O

82- + 2e ----------> 2SO

42- (O changes from -1 to -2)

Oxidation step: 2X[Ce3+ ----------------> Ce4+ +e]_________________________ S

2O

82- +2Ce3+ ------> 2SO

42- + 2Ce4+

(ii) Reduction step: 2X[NO3

- + 4H+ + 3e---------> NO + 2H2O]

Oxidation step: 3X[ Cu2O + 2H+---------------------> 2Cu2+ + H

2O +2e]

________________________________________3Cu

2O + 14H+ +2NO

3- --------> 6Cu2+ + 2NO + 7H

2O

(iii) Reduction step: 2X[MnO4

- +4H2O +3e----------> MnO

2 + 2H

2O + 4OH-]

3 X[SO3

2-+H2O+2OH- ----------------> SO

42- + 2H

2O+2e]

______________________________________________2MnO

4- + H

2O + 3SO

32- --------> 2MnO

2 + 2OH- + 3SO

42-

(iv) Reduction step: 5X[S2O

82- + 2e---------------> 2SO

42-]

Oxidation step: 2X[ Mn2+ +4H2O -------------------> MnO

4- + 8H+ +5e]

__________________________________________5S

2O

82- + 2Mn2+ + 8H

2O ----------> 10SO

42- + 2MnO

4- + 16H+

(v) Reduction step: Cr2O

72- +14H+ + 6e---------> 2Cr3+ + 7H

2O

Oxidation step: 3X[ C2H

4O +H

2O------------------> C

2H

4O

2 + 2H+ +2e]

__________________________________________Cr

2O

72- + 3C

2H

4O+8H+ ------> 2Cr3+ + 3C

2H

4O

2 + 4H

2O


(vi) Reduction step: 2X[S + 2e --------> S2-]Oxidation step: 2S+3H

2O +6OH- ----------> S

2O

32- + 6H

2O +4e

______________________________________4S + 6OH- -------> 2S2- +S

2O

32- + 3H

2O

(vii) Reduction step: ClO3

- + 6H+ +6e--------------> Cl- + 3H2O

Oxidation step: 3X[ 2I- ---------------------------> I2 + 2e]

_______________________________________ClO

3- + 6H+ + 6I- ----------> Cl- + 3I

2 + 3H

2O

(viii) Reduction step: 8X[Ag+ + e -----------> Ag]Oxidation step: AsH

3 +4H

2O+8OH- -----------> H

3AsO

4 + 8H

2O +8e

_________________________________________8Ag+ + AsH

3 + 8OH- ---------> H

3AsO

4 + 8Ag + 4H

2O

(ix) Reduction step: 16X [NO3- +4H+ +3e--------> NO + 2H

2O]

Oxidation step: 3X[ 8CuS ---------> 8Cu2+ + S8 + 16e]

_______________________________24CuS + 16NO

3- + 64H+ -------> 16NO + 24Cu2+ + 3S

8 + 32H

2O

(x) Reduction step: ClO3

- + 2H2O +2e-------> ClO

2- + H

2O + 2OH-

Oxidation step: SbO2

- + 4H2O +2OH- ---------> Sb(OH)

6- + 2H

2O + 2e

____________________________________________ClO

3- + SbO

2- +3H

2O -------> ClO

2- + Sb(OH)

6-

(xi) Reduction step: NO3

- + 9H2O +8e--------> NH

3 + 3H

2O + 9OH-

Oxidation step: 4X [Zn ------------------------Zn2+ + 2e]_________________________________________4Zn + NO

3- + 6H

2O --------> 4Zn2+ + NH

3 + 9OH-

(xii) Reduction step: BrO3

- +6H+ +6e-------> Br - +3H2O

Oxidation step: 3X [2S2O

32- --------------> S

4O

62- + 2e]

_______________________________BrO

3- + 6S

2O

32- + 6H+ ----------> Br - + 3S

4O

62- + 3H

2O

(xiii) Reduction step: 2X[MnO4- + 4H

2O + 3e---------------> MnO

2 + 2H

2O + 4OH-]

Oxidation step: 3X[ SnO2

2- + H2O + 2OH- -----------------> SnO

32- +2H

2O + 2e]

______________________________________________2MnO

4- + 3SnO

22- + H

2O ----------> 2MnO

2 + 3SnO

32- +2OH-

5. (i) [Fe(CN)6]4- + H+ + H

2O

2 ---------> [Fe(CN)

6]3- + H

2O

Reduction step: H2O

2 + 2H+ +2e-------> H

2O + H

2O

Oxidation step: 2X{[Fe(CN)6]4- ----------->[Fe(CN)

6]3- +e-}

____________________________________2[Fe(CN)

6]4- + H

2O

2 + 2H+ ------> 2[Fe(CN)

6]3- + 2H

2O

Looking to the coefficients of the balanced ionic equation, the original molecular equationis blananced. See this.

2K4[Fe(CN)

6](aq) + 2HCl(aq) + H

2O

2(l) ---------> 2K

3[Fe(CN)

6](aq) +

2KCl(aq) +2H2O(l)


(ii) N2H

4 + [Fe(CN)

6]3- ----------> N

2 + [Fe(CN)

6]4- + H

2O (alkaline medium)

Reduction step: 4X{[Fe(CN)6]

3- + e ----------->[Fe(CN)6]4-}

Oxidation step: N2H

4 + 4OH- -----------> N

2 + 4H

2O + 4e

________________________________4Fe(CN)

6]3- + N

2H

4 +4OH- -------> 4Fe(CN)

6]4- + 4H

2O + N

2

N2H

4 (l)+ 4K

3[Fe(CN)

6](aq)+4KOH------> N

2 (g)+ 4K

4[Fe(CN)

6](aq) + +4H

2O

(iii) C2H

2 + MnO

4- ---------> MnO

2 + C

2O

42- + OH - + H

2O (alkaline medium)

Reduction step: 8X[MnO4- + 4H

2O + 3e-----------> MnO

2 + 2H

2O + 4OH -]

Oxidation step: 3X [C2H

2 + 4H

2O +10OH - ------------------> C

2O

42- + 10H

2O + 8e]

________________________________________________8MnO

4- +3C

2H

2 ------------> 8MnO

2 + 3C

2O

42- + 2H

2O + 2OH-

3C2H

2 (g)+ 8KMnO

4(aq)------> 8MnO

2(s)+ 3K

2C

2O

4(aq)+ 2KOH(aq) +2H

2O(l)

(iv) Hg2+ + C l- + Sn2+ ---------> Hg2Cl

2 + Sn4+

Reduction step: 2Hg2+ + 2Cl - + 2e-------> Hg2Cl

2

Oxidation step: Sn2+ -----------------> Sn4+ + 2e___________________________2Hg2+ + 2Cl - + Sn2+ -------> Hg

2Cl

2 + Sn4+

2HgCl2(aq)+ SnCl

2(aq)------------> Hg

2Cl

2(s)+ SnCl

4(aq)

(v) NO3- + Zn + OH- ---------> NH

3 + ZnO

22- +H

2O

Reduction step: NO3- +9H

2O +8e------------> NH

3 + 3H

2O + 9OH-

Oxidation step: 4X [Zn + 2H2O + 4OH-------> ZnO

22- + 4H

2O + 2e]

_______________________________________4Zn + NO

3- + 7OH- ----------> NH

3 + 4ZnO

22- + 2H

2O

NaNO3(aq) + 4Zn(s) + 7NaOH(aq) -----> NH

3(g) + 4Na

2ZnO

2(aq) + 2H

2O(l)

(vi) HgS + NO3- + H+ + Cl - -----------> S + [HgCl

4]2- + NO + H

2O

Reduction step: 2X[NO3- + 4H+ +3e---------> NO + 2H

2O]

Oxidation step: 3X{HgS + 4Cl - ----------> S + [HgCl4]2- + 2e}

__________________________________3HgS + 2NO

3- + 8H+ + 12Cl - ----> 2NO + 3S + 3[HgCl

4]2- +4H

2O

3HgS(s) + 2HNO3(aq) + 12HCl(aq)------> 3S + 3H

2[HgCl

4](aq) + 2NO(g) + 4H

2O(l)

(vii) Cr2O

72- + 8H+ + 3NO

2- ---------> 2Cr3+ + 3NO

3- + 4H

2O

K2Cr

2O

7 + 4H

2SO

4+3KNO

2 ----------> K

2SO

4+ Cr

2(SO

4)

3+ 3KNO

3+ 4H

2O

(viii) Cr2O

72- + 8H+ + 3H

2S --------> 2Cr3+ + 3S + 7H

2O

K2Cr

2O

7 + 4H

2SO

4 +3H

2S ----------> K

2SO

4 + Cr

2(SO

4)

3 + 3S + 7H

2O

(ix) Cr2O

72- + 8H+ + 3H

2C

2O

4 --------> 2Cr3+ + 6CO

2 + 7H

2O

K2Cr

2O

7 + 4H

2SO

4 +3H

2C

2O

4 ----------> K

2SO

4 + Cr

2(SO

4)

3 + 6CO

2 + 7H

2O

(x) 3Cu + 2NO3

- + 8 H+ ---------> 3Cu2+ + 2NO + 4H2O

3Cu + 8HNO3 ---------> 3Cu(NO

3)

2 + 2NO + 4H

2O

(xi) 4Fe + 2NO3

- + 10H+ -----------> 4Fe2+ + N2O + 5H

2O

4Fe + 10HNO3 ---------> 4Fe(NO

3)

2 + N

2O + 5H

2O


Supplimentary Questions:

1. Do these preferably by ON method and check with the answers given below.(i) 3CuS + 8HNO

3 ------> 3Cu(NO

3)

2 + 3S + 4H

2O + 2NO

(ii) As2S

5+ 40HNO

3-------> 2H

3AsO

4+ 5H

2SO

4+ 12H

2O + 40NO

2

(iii) 8Zn + 20HNO3 -----> 8Zn(NO

3)

2 + 6H

2O+ 2NH

4NO

3

(iv) 5Na2C

2O

4 + 2KMnO

4 + 8H

2SO

4 -----> K

2SO

4 +2MnSO

4 +10CO

2+ 5Na

2SO

4 + 8H

2O

(v) 2MnO + 5PbO2 + 10HNO

3 -----> 2HMnO

4 + 5Pb(NO

3)

2 + 4H

2O

(vi) 6Na2HAsO

3 + 2KBrO

3 + 12HCl -----> 12NaCl + 2KBr + 6H

3AsO

4

(vii) 4FeS2 + 11O

2 ------> 2Fe

2O

3 + 8SO

2

(viii) Ca(OCl)2 + 4KI + 4HCl ----> 2I

2 + CaCl

2 + 2H

2O + 4KCl

(ix) 2NaOCl + 2NaOH + Bi2O

3 ----> 2NaBiO

3 + H

2O + 2NaCl

(x) Sn + 4HNO3 ----> SnO

2 + 4NO

2 + 2H

2O

(xi) CrI3 KOH Cl2 K2Cr2O7 KIO4 KCl H2O+ + + + ++3 +6 -1+70-1

There are two oxidations and one reduction in this reaction. Let us balance it by electron-balance diagramme method.

Cr Cr+3 +6

2 2 + 6 e-

I I-1 +7

6 6 + 48 e-

Cr + 6 I 2 Cr + 6 I + 54 e-+3 -1 +6 +7

2

oxidation:

+7+60 -1-1+3

2 Cr + 6 I + 54 Cl2 Cr + 6 I + 54 Cl

reduction: X 54-10

ClCl + e-

2CrI3 + 27Cl

2 + 62KOH -------> K

2Cr

2O

7 + 6KIO

4 + 54KCl + 31H

2O

(xii) Na2TeO

3 + + 4NaI + 6HCl → 6NaCl + Te + 2 I

2 + 3H

2O

(xiii) K2Cr

2O

7 + 3SnCl

2 + 14HCl → 2CrCl

3 + 3SnCl

4 + 2KCl + 7H

2O

(xiv) 6K3[Fe(CN)

6] + Cr

2O

3 + 10KOH → 6K

4[Fe(CN)

6] + 2K

2CrO

4 + 5H

2O

(xv) 3NaHSO4 + 8Al + 3NaOH → 3Na

2S + 4Al

2O

3 + 3H

2O

(xvi) 8KI + 5H2SO

4 → 4K

2SO

4 + 4I

2 + H

2S + 4H

2O

(xvii) Cr2O

3 + 2Na

2CO

3 + 3KNO

3 → 2Na

2CrO

4 + 2CO

2 + 3KNO

2

(xviii) 4KClO3 + 4H

2SO

4 → 4KHSO

4 + 4ClO

2 + O

2 + 2H

2O

(xix) 5P2H

4 → 6PH

3 + P

4H

2

(xx) CoCl2 + 7KNO

2 + 2HCl → K

3[Co(NO

2)

6] + NO + 4KCl + H

2O


(xxi) 4OH- + 4MnO4

- → 4MnO42- + O

2 + 2H

2O

4KOH + 4KMnO4 ----> 4K

2MnO

4 + O

2 + 2H

2O

(xxii) 4Au + 16CN- + + 6 H2O + 3O

2 → 4[Au(CN)

4]- + 12OH-

4Au + 16KCN + 6 H2O + 3 O

2 -----> 4K[Au(CN)

2] + 12KOH

(xxiii) 6Zn + 12H+ + As2O

3 → 2AsH

3 + 3H

2O + 6Zn2+

6Zn + 12HCl + As2O

3 ------> 2AsH

3 + 3H

2O + 6ZnCl

2

(xxiv) Cl2 + 2OH- + IO

3- → IO

4- + 2Cl- + H

2O

Cl2 + 2KOH + KIO

3 ------> KIO

4 + 2KCl + H

2O

(xxv) 4Zn + 10 H+ + NO3- → 4Zn2+ + NH

4+ + 3H

2O

4Zn + 10HNO3 ------> 4Zn(NO

3)

2 + NH

4NO

3 + 3H

2O

(xxvi) 6V + 3OH- + 14H2O → HV

6O

173- + 15H

2

6V + 3NaOH + 14HO ------> Na3HV

6O

17 + 15H

2

(xxvii) There are two oxidations and one reduction in this case.

Mn+2

Mn+7

+ 5 e

2 Br 2 Br + 12 e+5-1

Mn+2

2 Br-1

+ Mn+7

2 Br + 17 e+5

+ X 2

Oxidation:

Pb+4

+ 2 e Pb+2

X 17

Mn+2

2 4 Br-1

Pb+4

+ 17+ Mn+7

2 4 Br + 17 Pb+5 +2

Reduction:

+

Balanced equation:2 MnBr

2 + 17 PbO

2 + 30 HNO

3 → 2 HMnO

4 + 2 Pb(BrO

3)

2 + 15 Pb(NO

3)

2 + 14 H

2O

2. Do yourself and check with the following answers.(i) 3Sn2+ + 8H+ + 2NO

3- -----> 3Sn4+ + 2NO + 4H

2O

(ii) Cr2O

72- + 6I- 14H+ ----> 3I

2 + 2Cr3+ + 7H

2O

(iii) 4Zn + 10H+ + NO3- -----> 4Zn2+ + NH

4+ + 3H

2O

(iv) 2S2O

32- + 2Ag+ -------> 2Ag + S

4O

62-

(v) 4P4 + 12OH- + 12H

2O----> 4PH

3 + 12H

2PO

2-

(vi) Br2 + IO

3- + 2OH- ----> 2Br - + IO

4- + H

2O

(vii) 2MnO4

- + 5Sn2+ + 16H+ ----> 2Mn2+ + 5Sn4+ + 8H2O

(viii) S2O

82- + 9I- + 8H+ ----> 3I

3- + 2SO

2 + 4H

2O

(ix) 2CrO4

2- + 3Cu2O +11H

2O----> 6Cu(OH)

2 + 2Cr(OH)

4- + 2OH-

electrovalency covalency electrovalency...

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