Chapter 7 The p- Block Elements

On addition of concentrated H₂SO₄ to iodide salt violet fumes comes out because HI formed during the reaction react with oxygen present in the atmosphere to form I₂ which is violet in colour.

2NaCl + H₂SO₄ Na₂SO₄ + 2HCl

HCl produced during the reaction is colourless and pungent smelling gas.

2NaI + H₂SO₄ Na₂SO₄ + 2HI 2H₂O + SO₂ +I₂

Thus the halogen acid(HI) obtained in this reaction is oxidized to form I₂.

Hence the correct option is (iii).

when H₂S is passed through an aqueous solution of salt acidified with dilute HCl

Cu²⁺ + H₂S CuS + 2H⁺

Black precipitate of copper sulphide is formed. When copper sulphide is heated with dil. HNO₃ it gives blue colour of copper nitrate( on boiling), when aqueous solution of ammonia is added to it deep blue colour of tetraammine copper(II)sulphate [Cu(NH₃)₄]² is obtained.

CuS + dil.HNO₃ Cu(NO₃)₂

Cu(NO₃)₂ + 4NH₃ [Cu(NH₃)₄]²⁺

Hence the correct option is (ii)

Cyclotrimetaphosphoric acid contains 3 double bonds, and 12 single bonds as described in the figure. a,b,c represents double bond between o and P while numbering from 1 to 12 represents single bonds.

Hence the correct option is (iii).

NOTE: single bond consist of only sigma bond and double bond consist of one sigma and one pie bond.

nitrogen, carbon, boron does not form d-p multiple bonds due to absence of vacant d orbitals.

Phosphorous can form d-p bonding because it contains p orbitals as well d orbitals.

Hence the correct option is (iii).

ISOELECTRONIC SPECIES: the elements or ions which have same number of electrons are known as isoelectronic species. They may vary in their chemical and physical properties.

ISOSTRUCTURAL SPECIES: the compounds which have same shape and hyberdisation is called isostructural species.

There are 32 electrons in and sp² hyberdisation.

In also there are 32 electrons and sp² hyberdisation.

Both have trigonal planar structure, therefore they are isolectronic as well as isostructural species.

Hence the correct option is (i).

BOND DISSOCIATION ENTHALPY: bond dissociation enthalpy is defined as the energy required to break the bonds between two atoms.

With increase in bond length, bond dissociation enthalpy decreases because bond of large length is easy to break than bond of smaller length. Larger amount of energy is required to break the small bond while very less amount of energy is required to break large bond.

Thus bond dissociation enthalpy is inversely proportional to bond length.

As we move from HF to HI the bond length increases and bond dissociation enthalpy decreases down the halides.

Hence the correct option is (i).

A reducing agent is a compound or element that losses electron.

A strong reducing agent looses electron more easily as compared to weak reducing agent.

SbH₃ acts as strongest reducing agent due to minimum value of bond dissociation enthalpy.

It requires less bond dissociation enthalpy to break bonds, therefore it can easily donate electrons to other chemical species.

NOTE: lesser will be bond dissociation enthalpy, less time is taken to break the bond between atom and more easily it can loses electrons to form bond with another atom

Therefore stronger is the reducing agent.

Hence the correct option is (iv)

As trihalides of nitrogen family act as lewis base due to presence of one lone pair of electron and have ability to donate it. In comparison to NH₃ and PH₃ size of central atom is larger in PH₃ and electron density on phosphorous decrease and less available for protonation. Therefore PH₃ is less basic than NH₃. Hence (iv) is correct statement

When white phosphorous react with NaOH and water it produce phosphine gas which is very harmful, and its solution in water also decomposes in the presence of sunlight. Hence (i) and (ii) statement sare also correct.

The incorrect statement is (iv).

Hence the correct option is (iv).

H₃PO₂ is monobasic as it consist of only one OH group.

Structure of H₃PO₂:

H₃BO₃: it does not act as proton donor but behaves as lewis acid. Boric acid does not dissociate in aqueous solution. It can be predicted from its structure that it have 3 OH group. Boric acid first accept one OH^- then it liberates one electron per molecule. Therefore it is monobasic.

H₃PO₃: it is dibasic as it have two OH group.

structure of H₃PO₄:

It has three OH group therefore it is tri basic. Oxidation state of phosphorous in H₃PO₄ is +5.

As it has three OH group, it means that it has 3 ionisable hydrogen atoms and hence form three series of salts. These are:

1.NaH₂PO₄

2. NaHPO₄

3. Na₃PO₄

Hence the correct option is (iii).

All oxyacids of phosphorous which have P-H bond acts as strong reducing agent. H₃PO₂ has two P-H bond. Therefore it acts as strong reducing agent.

Hence due to presence of one O-H bond and two P-H bond H₃PO₂ is strong reducing agent. Hence the correct option is (iii).

when lead nitrate is heated the following reaction takes place:

2Pb(NO₃)₂2PbO + 4NO₂ + O₂

Oxygen is released in the air.

Hence the correct option is (ii).

Allotropy is the property of elements to exists in two or more different forms.

Nitrogen does not show allotropy because of its small size and high electro negativity the N-N single bond is weak. Therefore ability of nitrogen to form polymeric structure or more than one structure becomes less.

Hence the correct option is (i).

maximum covalency of nitrogen is four because it cannot extend its covalency due to absence of vacant d orbitals. Nitrogen donates its one lone pair and shows 4 covalency.

Hence the correct option is (iii).

as electronegativity of nitrogen is greater than phosphorous due to small size of nitrogen therefore N-N bond is weaker than P-P bond.

Hence first statement is wrong.

PH₃ acts as ligand in the formation of coordinate compounds due to presence of lone pair. Therefore second statement is correct.

NO₂ is paramagnetic in nature due to presence of unpaired electrons. Therefore third statement is also correct.

Nitrogen cannot extend its covalency than 4 therefore fourth statement is also correct.

Hence the correct option is (i).

When freshly prepared solution of FeSO₄ is added in the solution containing NO₃ ion. It leads to the formation of brown coloured complex. This is known as brown colour test of nitrate.

+ 3Fe²⁺ + 4H⁺NO + 3Fe³⁺ + 2H₂O

[Fe(H₂O₆)]²⁺ + NO [Fe(H₂O)₅ (NO)]²⁺ + H₂O

Hence the correct option is (i).

Bismuth commonly shows +3 oxidation state instead of +5 oxidation state due to inert pair effect (recultance of s electrons to take part in bond formation)

But fluorine has small size and high electronegativity than oxygen, chlorine and sulphur.

Therefore bismuth forms BiF₅ and show +5 oxidation state.

Hence the correct option (ii).

(NH₄)₂Cr₂O₇N₂ + Cr₂O₃ + 4H₂O

Ba(N₃)₂ 3N₂ + Ba

As in both the reactions nitrogen gas is released. Hence the correct option is (i).

Two moles of NH₃ will produce 2 moles of NO on catalytic oxidation of ammonia in preparation of nitric acid.

2NH₃ + O₂ 2NO + 3H₂O

(In this reaction Pt acts as catalyst)

Hence the correct option is (i)

calculation of oxidation state of P

Consider the oxidation state of P be x.

NaH₂PO₂ : ( +1)+(+1*2)+x+(-2*2) = 0

X = +1

Hence the correct answer is (iii).

In NH⁴⁺ the central atom is sp³hyberdised. Therefore it has tetrahedral shape.

In SiCl₄ also central atom is sp³hyberdised. Therefore it has tetrahedral shape.

In SO₄² also central atom is sp³hyberdised. Therefore it has tetrahedral shape,but

SF₄ does not have tetrahedral shape because it have sp³d hyberdisation.

Thus it shows trigonal bipyramidal shape. Hence the correct answer is (iii)

peroxides are the compounds which consist of one O-O bond.

Therefore peroxides of sulphur must contain one O-O bond. Therefore it can be clearly seen from the structure of H₂SO₅ and H₂S₂O_{8 .}

Hence the correct option is (i).

H₂SO₄H₂O + SO₂

C + 2O₂CO₂

And the net reaction is obtained by multiplying reaction 1 by 2.

Addition of reaction 1 and reaction 2

C + 2H₂SO₄CO₂ + 2SO₂ + 2H₂O

The gaseous product formed during the oxidation of carbon by H₂SO₄ are:

CO₂ AND SO₂.

Hence the correct option is (iii)

when a black compound of manganese reacts with a halogen the following reaction takes place:

MnO₂ + 4HCl MnCl₂ + 2H₂O + Cl

A greenish yellow gas is produced, it means that chlorine is liberated during the reaction

When excess of Cl₂ reacts with ammonia:

NH₃ + 2Cl₂NCl₃ + 3HCl

Oxidation state of nitrogen in NH₃ can be calculated by

Let the oxidation state of N is x.

X + 3*(+1) = 0, x = -3

Oxidation state of NCl₃ can be calculated by

X + 3*(-1) = 0, x = +3

Therefore oxidation state of nitrogen varies from -3 to +3.

Hence the correct option is (i).

Neil Bartlett noticed that PtF₆ is a powerful oxidizing agent which combines with O₂ to form red ionic compound.

O₂ + PtF₆[PtF₆]^-

As the first ionization enthalpy of Xe gas is 1170KJ/MOL and that of oxygen is 1175KJ/MOL

Both have almost same value of ionization enthalpy. Diameter of O₂ and Xe are almost similar (4A’). Hence the correct option is (iii).

in solid state, PCl₅ exists as ionic solid [PCl₄]⁺[PCl₆]^- in which cation is tetrahedral and anion is octahedral.

Hence the correct option is (iv)

REDUCTION POTENIAL: reduction potential is a measure of the ease with which a molecule will accept electrons.

OXIDISING POWER: it is the ability to gain electrons.

Therefore oxidizing power is directly proportional to reduction potential.

Greater is the reduction potential, more will be its oxidizing power.

Hence the order is BrO₄^–> IO₄^–>ClO₄

Thus the correct option is (iii).

ISOELECTRONIC SPECIES: the ions or elements which have same number of electrons are called isoelectronic species.

Number of electrons in the given species;

ICl₂ : 53 + 2*17 = 87

ClO₂ ; 17 + 16 = 33

Therefore number of electrons is not same. Hence it cannot be isoelectronic pair.

BrO₂ : 35 + 2*8 + 1 = 52

BrF₂: 35 + 9*2 -1 = 52

Therefore number of electrons is same. Hence it is a isoelectronic pair.

ClO₂ : 17 + 16 = 33

BrF : 35 + 9 = 44

Therefore number of electrons is not same. Hence it cannot be isoelectronic pair.

CN^- : 6+7+1 = 14

O₃ : 8*3 = 24

Therefore number of electrons is not same. Hence it cannot be isoelectronic pair.

Hence the correct option is (ii).

6NaOH + 3Cl₂ 5NaCl + NaClO₃ +H₂O

Let x be the oxidation state of chlorine.

Oxidation state of Cl is -1 ( x + 1 =0, x= -1)

Oxidation state of chlorine in NaClO₃ is +5 ( x + 3*(-2) +1 = 0, x= 5)

Hence the correct options are (i) and (iii)

I₂ has more oxidizing power than halogen atoms because of large size of iodine it can easily accommodate electrons. Down the group size of the atom increases and oxidizing power also increases. therefore the order is F₂< Cl₂< Br₂< I₂

IONIC CHARACTER: the bond in which large electronegativity difference exists between two atoms.

Ionic character increases with increase in electronegativity of halogens

Therefore the order is MI <MBr<MCl< MF

BOND DISSOCIATION ENTHALPY: the amount of energy required to break the bond

As the bond length increases, bond dissociation enthalpy decreases.

More amount of energy is required to break the bond of small length. Bomd length of F-F is small hence it requires more energy to dissociate than Cl-Cl, Br-Br, I-I.

Hence the order of bond dissociation enthalpy is:

F₂> Cl₂> Br₂> I₂

Therefore bond dissociation enthalpy also increases.

Hence the correct options are (ii)and(iii)

it has four lone pair of electrons present at P atom.

among halogens, the radius ratio between iodine and fluorine is maximum because iodine have maximum radius while fluorine have minimum radius.

And due to highest radius ratio rule maximum numbers of atoms are present in iodine fluoride.

Interhalogen compound are more reactive than halogen compounds because bond between them is easy to break while the bond between X-X is very strong and cannot be dissociated easily. Hence (ii) is incorrect

Hence the correct options are (i), (iii), (iv)

sulphur dioxide act as bleaching agent in moist conditions because of reducing nature of SO₂

SO₂ + 2H₂O H₂SO₄ + 2H

COLOURED SUBSTANCE + H = COLOURLESS SUBSTANCE

Its dilute solution is also used as disinfectant.

It does not have linear shape. It has trigonal planar geometry and bent shape.

Hence the correct options are (i)and(iii)

1. All the three N—O bond lengths in HNO₃ are not equal. Double bonds are generally smaller in size than single bond.

2. All P—Cl bond lengths in PCl5 molecule in gaseous state are not equal. Axial bonds are larger than equatorial bonds.

3. ) P4 molecule in white phosphorous have angular strain therefore white phosphorus is very reactive.

4.) PCl is ionic in solid state in which cation is tetrahedral and anion is octahedral.

Cation [PCl₄]⁺

Anion [PCl₆]^-

As₂O₃< SiO₂< P₂O₃< SO₂ Acid strength

ACIDIC STRENGTH: acidic strength cab be described as the tendency of an acid to give proton to the solution.

Acidic strength of oxides decreases down the group and increases along period from left to right.

H₂O > H₂S > H₂Se > H₂Te in order of Thermal stability.

As the size of the bond length increases, more easily it can be dissociated and become less stable. Down the group atomic size increases and thermal stability decreases.

(ii) and (iii) are incorrect because in (ii) order of enthalpy of vapourisation is reverse

As size of N is smaller than P, therefore it is difficult to break the bond of NH₃ than PH₃ and AsH₃. So the order of enthalpy of vapourisation should be AsH₃> PH₃> NH₃ .

In (iii) electron gain enthalpy of fluorine is less because of small 2p orbital.

Hence the correct answer is (i) and (iii)

(i) S–S bond is present in H₂S₂O₆.

Therefore it can be clearly seen from the figure that there a bond between S-S.

(ii) In peroxosulphuric acid (H₂SO₅) sulphur is in +6 oxidation state.

Therefore the oxidation of S is +6 . to calculate the oxidation state of S. first consider it x

+1 + (-2) + x + (-2) + (-2) + (-1 + (-1) + 1 = 0

X = 6

(iii) In Haber’s process Fe and Mo is used as catalyst

(iv) Change in enthalpy is negative for the preparation of SO₃ by catalytic oxidation. The creation of sulphur dioxide is exothermic reaction.

Hence the correct options are (i) and (ii)

An oxidizing agent reduces itself and oxidation state of central atom decreases.

2HI + H₂SO₄→ I₂ + SO₂ + 2H₂O

Oxidation state of central atoms:

HI = -1

H₂SO₄ = -6

I₂= 0

SO₂ = -4

Cu + 2H₂SO₄→ CuSO₄ + SO₂ + 2H₂O

Oxidation state of central atom:

Cu = 0

H₂SO₄ = -6

CuSO₄ = +2

SO₂ = +4

Hence the correct options are (ii) and (iii).

1. Weak dispersion forces are present between noble gases and theses are also known as vanderwall forces of attraction. Hence (i) is true.

2. First ionization enthalpy of oxygen is 1175 KJ/ mol and first ionization enthalpy of xenon is 1170KJ/mol. Therefore there ionization enthalpies are almost same. Hence (ii) is true.

3. The oxidation state of all the elements during the hydrolysis of XeF₆ remains same, therefore it is not a redox reaction. Hence (iii) is false

4. Xenon fluorides are reactive. Hence (iv) is false.

Hence the correct answer is (i) and (ii).

Dissolution of SO₃ in water is highly exothermic in nature. This leads to the formation of mist of tiny droplets which are highly corrosive and they can even attack the lead pipelines (lead pipelines are used to cover the tower in contact process).

Thus the acid fog which is formed by the direct reaction of water with sulphur trioxide is very difficult to condense. Therefore the following reaction takes place.

Absoption of SO₃ in 98 percent of H₂SO₄ to form oleum

SO₃ + H₂SO₄H₂S₂O₇

Then dilution of oleum takes place to get H₂SO_4.

H₂S₂O₇ + H₂O 2H₂SO₄

NH₃ + O₂ 4NO + 6H₂O

In this reaction Pt gauze react as a catalyst, which is added to increase the rate of reaction.

nitrogen is more electronegative than phosphorous (electro negativity decreases with increase in atomic size and size of phosphorous is larger than nitrogen.) Ammonia forms hydrogen bonding with water therefore it is highly soluble in water, but PH₃ do not forms hydrogen bonding with water. Therefore it is not soluble in water and it escapes as a gas and forms bubbles.

NOTE: hydrogen bonding is the electrostatic force of attraction between hydrogen and electronegative atom which form covalent bond with hydrogen.

PCl₅ has trigonal bipyramidal structure. It has three equivalent equatorial bonds and two equivalent axial bonds. The size of axial bonds is greater than size of equatorial bonds to overcome repulsion, because the three equatorial bonds causes more repulsion.

Therefore two axial P-Cl bonds are longer and different from equatorial bonds.

paramagnetic: the species which contain unpaired electrons are known as paramagnetic.

Diamagnetic: the species which all paired electrons are called diamagnetic.

Nitric acid in gaseous state exists in monomer form. It consists of only one unpaired electron therefore it is paramagnetic in nature.

However in solid state it exists as dimer (N₂O₂). There is no unpaired electron in its dimer form, therefore it is diamagnetic in nature.

Chlorine has vacant d orbitals hence it can show an oxidation state of +3. Fluorine has no d orbitals, it cannot show positive oxidation state. Fluorine shows only -1 oxidation state. Therefore FCl₃ does not exists.

This can also be explained by the concept of atomic size. Size of chlorine atom is greater due to smaller size of fluorine atom it cannot accommodate three large chlorine atoms.

Hence ClF₃ exists but FCl₃ does not exist.

H₂O has higher bond angle than H₂S because as we move from oxygen to sulphur the size of the central atom increases and electronegativity decreases due to which bond pair goes away from central atom which result in decrease in bond pair bond pair repulsion and hence bond angle decreases.

fluorine is strongest oxidizing agent and it can oxidize sulphur to its maximum oxidation state +6 to form SF₆. Chlorine is not good oxidizing agent, it cannot oxidize sulphur to its maximum oxidation state. Chlorine can oxidize sulphur to only +4 oxidation state. Hence it can form SCl₄ but not SCl₆.

when chlorine first react with phosphorous and heated in a retort, it will form phosphorous trichloride PCl₃. PCl₃ further react with chlorine to form PCl₅.

Halide A is PCl₅ because it is yellowish white powder

Halide B is PCl₃ because it is colourless oily liquid.

PCl₃ + 3H₂O H₃PO₃ + 3HCl

PCl₅ undergoes a violent hydrolysis

PCl₅ + H₂O POCl₃ + 2HCl

PCl₅ + 4H₂O H₃PO₄ + 5HCl

+ 3Fe²⁺ + 4H⁺ NO + 3Fe²⁺ 2H₂O

[Fe(H₂O)₆]SO₄ + NO [Fe(H₂O)₅NO]SO₄

Thus the complex formed is brown in colour.

As the electronegativity of halogen decreases, the tendency of XO₃ group ( X = halogens) to withdraw electrons of the O-H bond towards itself decreases and hence the acid strength of the perhalic acid decreases.

Therefore the order of stability is HClO< HClO₂< HClO₃< HClO₄

ozone is thermodynamically less stable because it decompose into oxygen and this decomposition leads result in the liberation of heat, so its entropy is positive and free energy is negative. High concentration of oxygen is dangerously explosive.

P₄O₆ + 6H₂O 4H₃PO₃

4H₃PO₃ + 8NaOH Na₂HPO₃ + 8H₂0

NET REACTION:

P₄O₆ + 8NaOH 4Na2HPO₃ + 2H₂O

Moles of P₄O₆ = = 0.5

Acid formed by one mole of P₄o₆ requires = 8 mol

Acid formed by 0.005 mol of NaOH is present in 100mL solution

0.04 of NaOH is present in solution = *0.04 = 400mL

when white phosphorous react with chlorine:

P₄ + 6Cl₂ 4PCl₃

PCl₃ + 3H₂O H₃PO₃ + 4HCl ]*4

NET REACTION: BEFORE ADDING SECOND REACTION TO FIRST, MULTIPLY IT BY FOUR

P₄ +6Cl₂ +12H₂O 4H₃PO₃ +12HCl

Moles of white P= = 0.5mol

1mol of white P₄ produce HCl = 12mol

0.5 mol of white P₄ will produce HCl = 12*0.8 = 6mol

Mass of HCl = 6*36.5 = 219.0g

there are three oxoacids of nitrogen.

1. Nitric acid(HNO₃)

2. Nitrous acid (HNO₂)

3. Hyponitrous acid (H₂N₂O₁)

+ 3 oxidation is shown by HNO₂, Therefore it undergoes disproportion reaction,

To calculate oxidation state, consider the oxidation state of N is x.

3HNO₂ HNO₃ + 2NO +H₂O

Oxidation state of nitrogen in HNO₂ is +3

X + (+1) +2*(-2) = 0, x= +3

Oxidation state of nitrogen in HNO₃ is +5

X + (+1) + 3*(-2) = 0, x= 5

Oxidation state of nitrogen in NO is +2

X + (-2) = 0, x =2

Therefore +3 oxidation state changes to +5 and +2 oxidation states

P₄O₁₀ + 4HNO₃ 4HPO₃ + 2N₂O₅

White phosphorous is very reactive as compared to red phosphorous due to angular strain in white phosphorous.

Resonance structure of N₂O₅

dilute and concentrated nitric acid give different oxidation product

EXAMPLES ARE:

4Zn +10HNO₃(dil) 4Zn(NO₃)₂ + NH₄NO₃ + 3H₂O

4Zn +10HNO₃(dil) 4Zn(NO₃)₂ + N₂O + 5H₂O

4Zn +4HNO₃(conc.) Zn(NO₃)₂ + NO₂ + 2H₂O

PCl₅ reacts with silver to form white silver salt(AgCl). Which than dissolves on adding excess aqueous NH₃ to form soluble complex

PCl₅ + 2Ag 2AgCl(white ppt.) + PCl₃

Agcl + 2NH₃(aq) [Ag(NH₃)]⁺Cl^-

REACTIONS SHOWING REDUCING BEHAVIOR:

4AgNO₃ + H₃PO₂ + 2H₂O 4Ag + H₃PO₄ + 4HNO₃

2HgCl₂ + H₃PO₂ + 2H₂O 2Hg + H₃PO₄ + 4HCl

It shows reducing behavior due to possession of P-H bond

Mn₂O₇ on dissolution in water produce acidic oxide

Bi₂O₃ on dissolution in water produce basic oxide

Pb₃O₄ is mixed oxide of PbO.Pb₂O₃

N₂O is neutral oxide, do not have any charge

1. H₂SO₄ is used in lead storage batteries

2. CCl₃NO₂ is known as tear gas

3. Cl₂ have highest electron gain enthalpy because it need only one electron to compete octect. thus to attain stable configuration .

4. Oxygen family is known as chalcogen therefore sulphur is also chalcogen

1. Partial hydrolysis of XeF₆ does not change oxidation state of central atom

XeF₆ +2H₂O XeO₃ + 6HF

IN THIS REACTION OXIDATION STATE REMAINS SAME.

2. He is used in modern diving apparatus.

3. Ar is used to provide inert atmosphere for filling electrical bulbs

4. In XeF₄ central atom is in sp³d²hybridisation

electron gain enthalpy of nitrogen is less than phosphorous. N₂ is less reactive than P₄ due to high bond dissociation enthalpy. More energy is required to break the bond N-N, therefore it is difficult to break the bond, while less amount of enegy is required to break the bond of phosphorous, therefore it is easy to break the bonds.

Hence N₂is less reactive than P_4.

HNO₃ makes the iron passive because it form thin layer of oxide on the surface of iron.

correct reason for assertion is

HI cannot be prepared by the reaction of KI with concentrated H₂SO₄ because H₂SO₄ is strong oxidizing agent. It will convert HI to I₂.

Both rhombic and monoclinicsulphur exist as S8 but oxygen exists as O₂ because Oxygen forms pp – pp multiple bond due to small size and small bond length but pp– pp bonding is not possible in sulphur.

NaCl reacts with concentrated H₂SO₄ to give colourless fumes with pungent smell. But on adding MnO₂ the fumes become greenish because MnO₂oxidisesHCl to chlorine gas which is greenish yellow

SF₆ cannot be hydrolysed but SF4 can be because Six F atoms in SF₆ prevent the attack of H₂O on sulphur atom of SF₆.

the amorphous solid A is sulphur (s₈) while gas b is sulphur dioxide.

S₈ + 8O₂8SO₂

The gas SO₂ turns the lime water milky.

Roasting of sulphide ore :

2ZnS + 3o₂2ZnO + 2SO₂

Hence sulphur dioxide is produced during the roasting of sulphide ore.

The gas B decolourise the acidified KMnO₄ solution and reduces Fe³⁺ to Fe²⁺

2KMnO₄ + 5SO₂ +2H₂O K₂SO₄ + 2MnSO₄ + 2H₂SO₄

Fe(SO₄)₃ + SO₂ + 2H₂O 2FeSO₄ + 2H₂SO₄

A is NO₂ (Nitrogen dioxide)

B is N₂O₄ (solid colourless)

C is N₂O₃ (dinitrogen trioxide)

On heating lead nitrate(II) a brown gas A is produced

2Pb(NO₃)₂ 2PbO + 4NO₂ + O₂

2NO₂ N₂O₄

2NO + N₂O₄ 2N₂O₃

STRUCTURE OF N₂O₄

STRUCTURE OF N₂O₃

A = NH₄NO₂

B = N₂

C= NH₃

D= HNO₃

On heating compound A gives compound B

N₂ is a constituent of air

NH₄NO₂ N₂ + 2H₂O

When N₂ is treated with 3mol of hydrogen in the presence of catalyst gives another gas (c)

N₂ + 3H₂ 2NH₃

The gas C is basic in nature due to presence of lone pair on nitrogen atom .

Gas C on further oxidation in moist condition gives compound D (HNO₃)

HNO₃ is present in acid rain.

4NH₃ + 5O₂4NO + 6H₂O

2NO + O₂2NO₂

3NO₂ + H₂O 2HNO₃ + NO