TTHHEE pp--BBLLOOCCKK EELLEEMMEENNTTSS.. Group-15 elements. N, P, As, Sb, Bi General characteristics:- El.confgn: ns2 np3

Extra stable due to half-filled p-confgn. B’coz of the same reason, they have higher ionisation enthalpies than gr-14 elements(Carbon family)

There is a very small in size from As to Bi due to completely filled d- and f- orbitals in heavier memebers (they have poor shielding effect)

PPhhyyssiiccaall pprrooppeerrttiieess MMeettaalllliicc cchhaarraacctteerr iinnccrreeaasseess ddoowwnn tthhee ggrroouupp BBii iiss aa ssttrroonngg mmeettaall dduuee ttoo hhiigghheerr II..EE.. bb’’ccoozz ooff iittss ssmmaallll ssiizzee EExxcceepptt nniittrrooggeenn,, aallll sshhooww aalllloottrrooppyy.. CChheemmiiccaall PPrrooppeerrttiieess

OOxxiiddaattiioonn ssttaatteess == --33,, ++33 aanndd ++55.. TThhee tteennddeennccyy ttoo sshhooww --33 ssttaattee ddeeccrreeaasseess aass wwee ggoowwnn dduuee ttoo iinnccrreeaassee iinn ssiizzee aanndd mmeettaalllliicc cchhaarraacctteerr TThhee ssttaabbiilliittyy ooff ++33 ssttaattee iinnccrreeaasseess ddoowwnn tthhee ggrroouupp wwhheerreeaass tthhaatt ooff ++55 ssttaattee ddeeccrreeaasseess dduuee ttoo IINNEERRTT--PPAAIIRR eeffffeecctt BBii sshhoowwss ++55 oonnllyy wwiitthh fflluuoorriinnee ((BBiiFF55)) dduuee ttoo hhiigghh ppoollaarriizziinngg ppoowweerr ooff fflluuoorriinnee..

Conseqences of inert-pair effect Nitrogen compounds disproportionate in acid solutions.

3HNO2 HNO3 + H2O + 2NO Similarly, in case of phosphorus nearly all intermediate oxidation states disproportionate

into +5 and –3 both in alkali and acid. But the +3 oxidation state in case of arsenic, antimony and bismuth becomes increasingly

stable and do not ungergo disproportionation. Anomalous properties of Nitrogen It can show a maximum covalency of four only whereas others show 5 and 6. for

erxample:-PCl5 and PF6-

This is due to the absence of vacant d-orbitals in N-atom It forms strong pπ-pπ multiple bonds due to its small size. This is why it exists as a

diatomic molecule with a triple bond between two N-atoms. Others do not form such bonds b’coz their atomic orbitals are so large and diffused that they

can not have effective overlapping.

AAnnoommaalloouuss pprrooppeerrttiieess ooff NNiittrrooggeenn TThhee bboonndd eenntthhaallppyy ooff nniittrrooggeenn mmoolleeccuullee iiss vveerryy hhiigghh dduuee ttoo tthhee pprreesseennccee ooff ttrriippllee bboonndd

bbeettwweeeenn tthhee ttwwoo aattoommss.. OOtthheerr eelleemmeennttss ffoorrmm oonnllyy ssiinnggllee bboonnddss wwiitthh tthheeiirr oowwnn aattoommss aanndd ootthheerr aattoommss ttoooo.. TThhee NN--NN iiss wweeaakkeerr tthhaann PP--PP bb’’ccoozz ooff hhiigghh iinntteerreelleeccttrroonniicc rreeppuullssiioonnss ooff tthhee nnoonn--bboonnddiinngg

eelleeccttrroonnss iinn NN--NN dduuee ttoo tthhee ssmmaallll ssiizzee ooff NN--aattoomm .. TThhiiss iiss wwhhyy NN ddooeess nnoott ccaatteennaattee..

RReeaaccttiivviittyy ttoowwaarrddss hhyyddrrooggeenn::-- AAllll tthhee eelleemmeennttss ooff GGrroouupp 1155 ffoorrmm hhyyddrriiddeess ooff tthhee ttyyppee EE HH33 wwhheerree EE == NN,, PP,, AAss,, SSbb oorr BBii.. TThhee ssttaabbiilliittyy ooff hhyyddrriiddeess ddeeccrreeaasseess ffrroomm NNHH33 ttoo BBiiHH33 dduuee ttoo ddeeccrreeaassee iinn tthheeiirr bboonndd

ddiissssoocciiaattiioonn eenntthhaallppyy.. CCoonnsseeqquueennttllyy,, tthhee rreedduucciinngg cchhaarraacctteerr ooff tthhee hhyyddrriiddeess iinnccrreeaasseess.. AAmmmmoonniiaa iiss oonnllyy aa mmiilldd rreedduucciinngg aaggeenntt wwhhiillee BBiiHH33 iiss tthhee ssttrroonnggeesstt rreedduucciinngg aaggeenntt aammoonnggsstt

aallll tthhee hhyyddrriiddeess.. BBaassiicciittyy aallssoo ddeeccrreeaasseess iinn tthhee oorrddeerr FFaaccttss ooff tthhee hhyyddrriiddeess NH3 > PH3 > AsH3 > SbH3 > BiH3. Stability decreases b’coz bde decreases Basicity decreases b’coz electron density around the central atom decreases due to increase in

size Reducing nature increases b’coz bde decreases BDE decreases b’coz size of E increases RReeaaccttiivviittyy ttoowwaarrddss ooxxyyggeenn AAllll tthheessee eelleemmeennttss ffoorrmm ttwwoo ttyyppeess ooff ooxxiiddeess:: EE22OO33 aanndd EE22OO55.. TThhee ooxxiiddee iinn tthhee hhiigghheerr ooxxiiddaattiioonn ssttaattee ooff tthhee eelleemmeenntt iiss mmoorree aacciiddiicc tthhaann tthhaatt ooff lloowweerr

ooxxiiddaattiioonn ssttaattee.. TThheeiirr aacciiddiicc cchhaarraacctteerr ddeeccrreeaasseess ddoowwnn tthhee ggrroouupp.. TThhee ooxxiiddeess ooff tthhee ttyyppee EE22OO33 ooff nniittrrooggeenn aanndd pphhoosspphhoorruuss aarree ppuurreellyy aacciiddiicc,, tthhaatt ooff aarrsseenniicc aanndd aannttiimmoonnyy aammpphhootteerriicc aanndd tthhoossee ooff bbiissmmuutthh((BBii22OO33))))iiss pprreeddoommiinnaannttllyy bbaassiicc..

RReeaaccttiivviittyy ttoowwaarrddss hhaallooggeennss •• TThheessee eelleemmeennttss rreeaacctt ttoo ffoorrmm ttwwoo sseerriieess ooff hhaalliiddeess:: EEXX33 aanndd EEXX55.. •• NNiittrrooggeenn ddooeess nnoott ffoorrmm ppeennttaahhaalliiddee dduuee ttoo nnoonn--aavvaaiillaabbiilliittyy ooff tthhee dd oorrbbiittaallss iinn iittss

vvaalleennccee sshheellll.. •• PPeennttaahhaalliiddeess aarree mmoorree ccoovvaalleenntt tthhaann ttrriihhaalliiddeess.. •• IInn ccaassee ooff nniittrrooggeenn,, oonnllyy NNFF33 iiss kknnoowwnn ttoo bbee ssttaabbllee dduuee ttoo hhiigghh ppooaallrriizziinngg nnaattuurree ooff FF--aattoomm •• TTrriihhaalliiddeess eexxcceepptt BBiiFF33 aarree pprreeddoommiinnaannttllyy ccoovvaalleenntt iinn nnaattuurree.. BBiiFF33 iiss iioonniicc dduuee ttoo hhiigghh

mmaattaalllliicc cchhaarraacctteerr ooff BBiissmmuutthh..

RReeaaccttiivviittyy ttoowwaarrddss mmeettaallss

AAllll tthheessee eelleemmeennttss rreeaacctt wwiitthh mmeettaallss ttoo ffoorrmm tthheeiirr bbiinnaarryy ccoommppoouunnddss eexxhhiibbiittiinngg ––33 ooxxiiddaattiioonn ssttaattee CCaa33NN22 ((ccaallcciiuumm nniittrriiddee)) CCaa33PP22 ((ccaallcciiuumm pphhoosspphhiiddee)),,

?????? QQUUEESSTTIIOONN ?????? 11..PPHH33 hhaass lloowweerr bbooiilliinngg ppooiinntt tthhaann NNHH33.. WWhhyy?? UUnnlliikkee NNHH33,, PPHH33 mmoolleeccuulleess ddoo nnoott ffoorrmm iinntteerrmmoolleeccuullaarr hhyyddrrooggeenn bboonnddiinngg iinn lliiqquuiidd ssttaattee..

TThhaatt iiss wwhhyy tthhee bbooiilliinngg ppooiinntt ooff PPHH33 iiss lloowweerr tthhaann NNHH33.. 22.. NNHH33 iiss ssoolluubbee iinn wwaatteerr wwhheerreeaass PPHH33 iiss iinnssoolluubbllee..WWhhyy?? DINITROGEN (N2) PROPERTIES OF N2 * At higher temperatures, it directly combines with some metals to form predominantly ionic

nitrides and with non-metals, covalent nitrides. 6Li + N2 2Li3N 3Mg + N2 Mg3N2 * Dinitrogen combines with dioxygen only at very high temperature (at about 2000 K) to

form nitric oxide, NO. N2 + O2(g) 2NO(g) UUSSEESS OOFF NN22 iinn tthhee mmaannuuffaaccttuurree ooff aammmmoonniiaa iinn tthhee mmaannuuffaaccttuurree ooff cchheemmiiccaallss ccoonnttaaiinniinngg nniittrrooggeenn,, ((ee..gg..,, ccaallcciiuumm ccyyaannaammiiddee)).. IItt aallssoo ffiinnddss uussee wwhheerree aann iinneerrtt aattmmoosspphheerree iiss rreeqquuiirreedd ((ee..gg..,, iinn iirroonn aanndd sstteeeell iinndduussttrryy,, iinneerrtt

ddiilluueenntt ffoorr rreeaaccttiivvee cchheemmiiccaallss)).. LLiiqquuiidd ddiinniittrrooggeenn iiss uusseedd aass aa rreeffrriiggeerraanntt ttoo pprreesseerrvvee bbiioollooggiiccaall mmaatteerriiaallss,, ffoooodd iitteemmss aanndd

iinn ccrryyoossuurrggeerryy..

EEvvaalluuaattiioonn QQuueessttiioonnss QQ.. WWrriittee tthhee rreeaaccttiioonn ooff tthheerrmmaall ddeeccoommppoossiittiioonn ooff ssooddiiuumm aazziiddee.. AAnnss::--TThheerrmmaall ddeeccoommppoossiittiioonn ooff ssooddiiuumm aazziiddee ggiivveess ddiinniittrrooggeenn ggaass.. 22NNaaNN33 22NNaa ++ 33NN22 QQ.. WWhhyy iiss NN22 lleessss rreeaaccttiivvee aatt rroooomm tteemmppeeaarraattuurree?? AAMMMMOONNIIAA OOnn aa ssmmaallll ssccaallee aammmmoonniiaa iiss oobbttaaiinneedd ffrroomm aammmmoonniiuumm ssaallttss wwhhiicchh ddeeccoommppoossee wwhheenn

ttrreeaatteedd wwiitthh ccaauussttiicc ssooddaa oorr lliimmee..

22NNHH44CCll ++ CCaa((OOHH))22 22NNHH33 ++ 22HH22OO ++ CCaaCCll22 ((NNHH44))22 SSOO44 ++ 22NNaaOOHH 22NNHH33 ++ 22HH22OO ++ NNaa22SSOO44 OOnn aa llaarrggee ssccaallee,, aammmmoonniiaa iiss mmaannuuffaaccttuurreedd bbyy HHaabbeerr’’ss pprroocceessss.. NN22((gg)) ++ 33HH22((gg)) ÖÖ 22NNHH33((gg));; EEnntthhaallppyy cchhaannggee == –– 4466..11 kkJJ mmooll––11

CCoonnddiittiioonnss ffoorr bbeetttteerr yyiieelldd aarree::-- IInn aaccccoorrddaannccee wwiitthh LLee CChhaatteelliieerr’’ss pprriinncciippllee,, hhiigghh pprreessssuurree wwoouulldd ffaavvoouurr tthhee ffoorrmmaattiioonn ooff

aammmmoonniiaa.. TThhee ooppttiimmuumm ccoonnddiittiioonnss ffoorr tthhee pprroodduuccttiioonn ooff aammmmoonniiaa aarree aa pprreessssuurree ooff 220000 aattmm,, aa

tteemmppeerraattuurree ooff ~~ 770000 KK aanndd tthhee uussee ooff aa ccaattaallyysstt ssuucchh aass iirroonn ooxxiiddee wwiitthh ssmmaallll aammoouunnttss ooff KK22OO aanndd AAll22OO33 ttoo iinnccrreeaassee tthhee rraattee ooff aattttaaiinnmmeenntt ooff eeqquuiilliibbrriiuumm..

Structure of ammonia The ammonia molecule is trigonal pyramidal with the nitrogen atom at the apex. It has three bond pairs and one lone pair of electrons as shown in the structure.

Properties of ammonia * Ammonia gas is highly soluble in water as it forms intermolecular H-bonds with water

molecules. * It is a Lewis base as it can donate the lone pair present on the N-atom. * It donates the electron pair forming linkages with metal ions and therefore form complex

compounds. This finds applications in detection of metal ions such as Cu2+, Ag+: Cu2+ (aq) + 4 NH3(aq) [Cu(NH3)4]2+(aq) Deep blue

AgCl(s) + 2NH3(aq) [Ag(NH3)2]Cl(aq) White ppt Excess Soluble

OXIDES OF NITROGEN Oxides of Nitrogen possess planar structure due to the ability of N to form multiple bonds.

(Oxides of P have cage like structures since P can not form strong multiple bonds)

STRUCTURE-

Evaluation Questions What is the covalence of N in (a)N2O5 and (b) N2O4? Why does NO2 dimerize? Can NO dimerize? Why is the precipitate of AgCl soluble in excess of ammonia? Why is ammonia a Lewis base?

NITRIC ACID Large Scale Manufacure:- Ostwald process. Based on catalytic oxidation of ammonia.

Pt /Rh 4NH3 g + 5O2 g 4NO g + 6H2 O g (from air) 500K; 9 bar

Nitric oxide thus formed combines with oxygen giving NO2. 2NO g + O2 g 2NO2 g Nitrogen dioxide so formed, dissolves in water to give HNO3. 3NO2 g + H2 O 2HNO3 aq + NO g

NO thus formed is recycled and the aqueous HNO3 can be concentrated by distillation upto ~ 68% by mass. Further concentration to 98% can be achieved by dehydration with

concentrated H2SO4.

Lab. Preparation of Nitric acid:- In lab, nitric acid is prepared by heating potassium nitrate or sodium nitrate with conc.

sulphuric acid in a glass resort. NaNO3 + H2SO4 NaHSO4 + HNO3

Shape of Nitric acid molecule In the gaseous state, HNO3 exists as a planar molecule with the structure as shown.

H O

O N

O Properties of Nitric acid In aqueous solution, nitric acid behaves as a strong acid giving hydronium and nitrate ions.

HNO3(aq) + H2O(l) H3O+(aq) + NO3-(aq)

Concentrated nitric acid is a strong oxidising agent and attacks most metals except noble metals such as gold and platinum. The products of oxidation depend upon the concentration of the acid, temperature and the nature of the material undergoing oxidation.

3Cu + 8 HNO3(dilute) 3Cu(NO3)2 + 2NO + 4H2O Cu + 4HNO3(conc.) Cu(NO3)2 + 2NO2 + 2H2O Zinc reacts with dilute nitric acid to give N2O and with concentrated acid to give NO2. 4Zn + 10HNO3(dilute) �¨ 4 Zn (NO3)2 + 5H2O + N2O Zn + 4HNO3(conc.) �¨ Zn (NO3)2 + 2H2O + 2NO2 Some metals (e.g., Cr, Al) do not dissolve in concentrated nitric acid because of the

formation of a passive film of oxide on the surface. Concentrated nitric acid also oxidises non–metals and their compounds. Iodine is oxidised

to iodic acid, carbon to carbon dioxide, sulphur to H2SO4, and phosphorus to phosphoric acid.

Properties of nitric acid-Contd.. Concentrated nitric acid also oxidises non–metals and their compounds. Iodine is oxidised

to iodic acid, carbon to carbon dioxide, sulphur to H2SO4, and phosphorus to phosphoric acid.

I2 +10HNO3 2HIO3 + 10 NO2 + H2O

C + 4HNO3 CO2 + 2H2O + 4NO2 S8 + 48HNO3(conc.) 8H2SO4 +48NO2 + 16H2O P4 + 20HNO3(conc.) 4H3PO4 +20 NO2 + 4H2O

Chemistry of Brown ring test:-

Brown Ring Test: The brown ring test for nitrates depends on the ability of Fe2+ to reduce nitrates to nitric oxide, which reacts with Fe2+ to form a brown coloured complex.

The test is usually carried out by adding dilute ferrous sulphate solution to an aqueous solution containing nitrate ion, and then carefully adding concentrated sulphuric acid along the sides of the test tube. A brown ring at the interface between the solution and sulphuric acid layers indicate the presence of nitrate ion in the solution.

NO3 - + 3Fe2+ + 4H+ NO + 3Fe3+ + 2H2O

[Fe(H2O)6]2+ + NO [Fe (H2O)5 (NO)]2+ + H2O (brown) Uses of Nitric acid In the mfr. of ammonium nitrate for fertilizer. In the mfr. of nitrates required for making explosives and pyrotechnics. In making TNT, nitroglycerin In pickling of stainless steel, etching of glass As oxidizer in rocket fuels.

PHOSPHOROUS.

White Phosphorous:-

P4 +3NaOH + 3H2O PH3 + 3NaH2 PO2

sodium hypophosphite

Reaction with Oxygen:- P4 + 5O2 P4O10

S.No White Phosphorous Red Phosphorous

1 Poisonous Non-poisonous

2 Soluble in CS2 Insoluble in CS2

3 Glows in dark Doesn’t

Structure:- It consists of discrete P4 tetrahedron units.

Red Phosphorous:-

Red phosphorus is obtained by heating white phosphorus at 573K in an inert atmosphere for several days.

Structure:- It has a chain-like structure.

Black phosphorous has two forms alpha-black phosphorous and beta-black phosphorous.

PHOSPHINE:- (PH3)

Preparation:-

1.Phosphine is prepared by the reaction of calcium phosphide with water or dilute HCl.

Ca3P2 + 6H2O3Ca(OH)2 + 2PH3

Ca3P2 + 6HCl 3CaCl2 + 2PH3

2.In the laboratory, it is prepared by heating white phosphorus with concentrated NaOH solution in an inert atmosphere of CO2.

P4 +3NaOH + 3H2O PH3 + 3NaH2 PO2

sodium hypophosphite

When pure, it is non inflammable but becomes inflammable owing to the presence of P2H4 or P4 vapours. To purify it from the impurities, it is absorbed in HI to form phosphonium iodide (PH4I) which on treating with KOH gives off phosphine.

PH4 I + KOH KI + H2O + PH3

Properties:-

1. 3CuSO4 + 2PH3 Cu3P2 + 3H2SO4

2. 3HgCl2 + 2PH3 Hg3P2 + 6HCl

3. Phosphine is weakly basic and like ammonia, gives phosphonium compounds with acids e.g., PH3 + HBr PH4 Br

Uses of Phopsphine:- The spontaneous combustion of phosphine is technically used in Holme’s signals.

Containers containing calcium carbide and calcium phosphide are pierced and thrown in the sea when the gases evolved burn and serve as a signal.

It is also used in smoke screens.

Halides of Phosphorous:-

1. PCl3:- Preparation:-

P4 + Cl2 PCl3

P4 + 8SOCl2 4PCl3 + + 4SO2 + 2S2Cl2

Properties:-

(i)It fumes in moist air due to the formation of HCl.

PCl3 + 3H2O H3PO3 + 3HCl

(ii)It reacts with organic compounds containing –OH group such as CH3COOH, C2H5OH.

3CH3COOH + PCl3 3CH3COCl +H3PO3

3C2H5OH+ PCl3 3C2H5Cl + H3PO3

Structure:_

It has a pyramidal shape as shown, in which phosphorus is sp3 hybridised.

Phosphorous Trichloride

Cl Cl Cl

P

.. Lone pair

2.PCl5:-

Preparation:-

(i)Phosphorus pentachloride is prepared by the reaction of white phosphorus with excess of dry chlorine.

P4 + 10Cl2 4PCl5

(ii)It can also be prepared by the action of SO2Cl2 on phosphorus.

P4 + 10SO2Cl2 4PCl5 + 10SO2

Properties:-

(i)In moist air, it hydrolyses to POCl3 and finally gets converted to phosphoric acid.

PCl5 + H2O POCl3 + 2HCl

POCl3 + 3H2O H3PO4 + 3HCl

(ii)When heated, it sublimes but decomposes on stronger heating.

PCl5 PCl3 + Cl2

(iii)It reacts with organic compounds containing –OH group converting them to chloro derivatives.

3CH3COOH + PCl5 3CH3COCl +POCl3 + HCl

3C2H5OH+ PCl5 3C2H5Cl + POCl3 + HCl

(iv)Finely divided metals on heating with PCl5 give corresponding chlorides.

2Ag + PCl5 2AgCl + PCl3

Sn + 2PCl5 SnCl4 + 2PCl3

(v) In the solid state it exists as an ionic solid, [PCl4]+[PCl6]– in which the cation, [PCl4]+ is tetrahedral and the anion, [PCl6]–octahedral.

Structure:- Trigonal bipyramidal

The three equatorial P–Cl bonds are equivalent, while the two axial bonds are longer than equatorial bonds. This is due to the fact that the axial bond pairs suffer more repulsion as compared to equatorial bond pairs.

Uses:-

It is used in the synthesis of some organic compounds, e.g., C2H5Cl, CH3COCl.

OXOACIDS OF PHOSPHOROUS:-

(i)The acids in +3 oxidation state of phosphorus tend to disproportionate to higher and lower oxidation states. For example, orthophophorous acid (or phosphorous acid) on heating disproportionates to give orthophosphoric acid (or phosphoric acid) and phosphine.

4H3PO3 3H3PO4 + PH3

(ii)The acids which contain P–H bond have strong reducing properties.Thus, hypophosphorous acid is a good reducing agent as it contains two P–H bonds and reduces, for example, AgNO3 to metallic silver.

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

Group 16 Elements

1.They have ns2np4 general electronic configuration.

2.The elements of this group have lower ionisation enthalpy values compared to those of Group15 in the corresponding periods.

This is due to the fact that Group 15 elements have extra stable half-filled p orbital electronic configurations.

3.Because of the compact nature of oxygen atom, it has less negative electron gain enthalpy than sulphur.

4.Next to fluorine, oxygen has the highest electronegativity value amongst the elements.

5.Oxygen and sulphur are non-metals, selenium and tellurium metalloids, whereas polonium is a metal. Polonium is radioactive and is short lived (Half-life 13.8 days).

6.The large difference between the melting and boiling points of oxygen and sulphur may be explained on the basis of their atomicity; oxygen exists as diatomic molecule (O2) whereas sulphur exists as polyatomic molecule (S8).

Oxidation states and trends in chemical reactivity:-

1. Since electronegativity of oxygen is very high, it shows only negative oxidation state as –2 except in the case of OF2 where its oxidation state is + 2

2.The stability of + 6 oxidation state decreases down the group and stability of + 4 oxidation state increase (inert pair effect)

Anomalous behaviour of oxygen:-

1.The anomalous behaviour of oxygen, like other members of p-block

present in second period is due to its small size and high electronegativity. One typical example of effects of small size and high electronegativity is the presence of strong hydrogen bonding in H2O which is not found in H2S.

2.The absence of d orbitals in oxygen limits its covalency to four and in practice, rarely exceeds two. On the other hand, in case of other elements of the group, the valence shells can be xpanded and covalence exceeds four.

Reactivity with hydrogen:

• All the elements of Group 16 form hydrides of the type H2E (E =O, S, Se, Te, Po) • H2O H2S H2Se H2Te

BDE Decreases

Stability decreases

Acidity increases

Reducing nature increases

Reactivity with oxygen:

• All these elements form oxides of the EO2 and EO3 types where E = S, Se, Te or Po. • Ozone (O3) and sulphur dioxide (SO2) are gases while selenium dioxide (SeO2) is solid. • Reducing property of dioxide decreases from SO2 to TeO2; SO2 is reducing while TeO2 is

an oxidising agent • Both types of oxides are acidic in nature.

Reactivity towards the halogens.

• The stability of the halides decreases in the order F– > Cl– > Br– > I–. • Amongst hexahalides, hexafluorides are the only stable halides. • They have octahedral structure. • Sulphurhexafluoride, SF6 is exceptionally stable for steric reasons.

Tetrafluorides, have sp3d hybridisation and thus, have trigonal bipyramidal structures in which one of the equatorial positions is occupied by a lone pair of electrons. This geometry is also regarded as see-saw geometry.

• Dihalides have sp3 hybridisation and thus, have tetrahedral structure. • The well known monohalides are dimeric in nature. Examples are S2F2, S2Cl2, S2Br2,

Se2Cl2 and Se2Br2. • These dimeric halides undergo disproportionation as given below:

2Se2Cl2 SeCl4 + 3Se

Question:-

1.H2S is less acidic than H2Te. Why?

Due to the decrease in bond (E–H) dissociation enthalpy down the group, acidic character increases.

2. Why is H2O a liquid and H2S a gas ?

DIOXYGEN

Dioxygen can be obtained in the laboratory by the following ways:

(i) By heating oxygen containing salts such as chlorates, nitrates and permanganates.

2KClO3 2 KCl + 3 O2

MnO2

(ii) By the thermal decomposition of the oxides of metals low in the electrochemical series and higher oxides of some metals.

2Ag2O(s) 4Ag(s) + O2(g);

2Pb3O4(s) 6PbO(s) + O2(g)

2HgO(s) 2Hg(l) + O2(g) ;

2PbO2(s) 2PbO(s) + O2(g)

(iii) Hydrogen peroxide is readily decomposed into water and dioxygen by catalysts such as finely divided metals and manganese dioxide.

2H2O2(aq) 2H2O(1) + O2(g)

(iv) On large scale it can be prepared from water or air. Electrolysis of water leads to the release of hydrogen at the cathode and oxygen at the anode.

Properties

1.Molecular oxygen, O2 is unique in being paramagnetic inspite of having even number of electrons .

2.Its combination with other elements is often strongly exothermic which helps in sustaining the reaction. However, to initiate the reaction, some external heating is required as bond dissociation enthalpy of oxgyen-oxygendouble bond is high (493.4 kJ mol–1).

2.Some compounds are catalytically oxidised. For e.g.,

2SO2 + O2 2SO3

5HCl + O2 2Cl2 + 2H2O

CuCl2

Uses Of Dioxygen:- Self work

Oxides:-

A binary compound of oxygen with another element is called oxide.

Oxides can be simple (e.g., MgO, Al2O3 ) or mixed (Pb3O4, Fe3O4).

Acidic or Basic or Amphoteric or Neutral

Reactions to show that aluminium oxide is amphoteric:-

Al2O3+ 6HCl+ 9H2O 2[Al(H2O)6]Cl3 (aq)

Al2O3+ 6NaOH +3H2O Na3[Al(OH)6] (aq)

OZONE:-

Ozone is an allotropic form of oxygen. It is too reactive to remain for long in the atmosphere at sea level. At a height of about 20 kilometres, it is formed from atmospheric oxygen in the presence of sunlight. This ozone layer protects the earth’s surface from an excessive concentration of ultraviolet (UV) radiations.

Preparation:-When a slow dry stream of oxygen is passed through a silent electrical discharge, conversion of oxygen to ozone (10%) occurs. The product is known as ozonised oxygen.

3O2 2O3 ∆H = +ve

Since the formation of ozone from oxygen is an endothermic process, it is necessary to use a silent electrical discharge in its preparation to prevent its decomposition.

PROPERTIES OF OZONE:-

1.Ozone is thermodynamically unstable with respect to oxygen since its decomposition into oxygen results in the liberation of heat (∆H is negative) and an increase in entropy (∆S is positive). These two effects reinforce each other, resulting in large negative Gibbs energy change (∆G) for its conversion into oxygen.

2.Due to the ease with which it liberates atoms of nascent oxygen (O3 O2 + [O] ), it acts as a powerful oxidising agent.

For e.g., it oxidizes lead sulphide to lead sulphate and iodide ions to iodine.

PbS(s) + 4O3(g) PbSO4(s) + 4O2(g)

2I–(aq) + H2O(l) + O3(g) 2OH–(aq) + I2(s) + O2(g)

3. Estimation of ozone:-When ozone reacts with an excess of potassium iodide solution buffered with a borate buffer (pH 9.2), iodine is liberated which can be titrated against a standard solution of sodium thiosulphate. This is a quantitative method for estimating O3 gas.

4.Nitrogen oxides (particularly nitric oxide) combine very rapidly with ozone and there is, thus, the possibility that nitrogen oxides emitted from the exhaust systems of supersonic jet aeroplanes might be slowly depleting the concentration of the ozone layer in the upper atmosphere.

NO + O3 NO2 + O2

Structure of Ozone:- Angular .

Uses of Ozone: It is used as a germicide, disinfectant and for sterilising water. It is also used for bleaching oils, ivory, flour, starch, etc. It acts as an oxidising agent in the manufacture of potassium permanganate.

SULPHUR

Yellow rhombic sulphur = Monoclinic sulphur

Transition temp = 369 K

Both rhombic and monoclinic sulphur have S8 molecules.

At elevated temperatures (~1000 K), S2 is the dominant species and is paramagnetic like O2

In vapour state sulphur partly exists as S2 molecule which has two unpaired electrons in the antibonding ∏* orbitals like O2 and, hence, exhibits paramagnetism.

Preparation:-

1.Sulphur dioxide is formed together with a little (6-8%) sulphur trioxide when sulphur is burnt in air or oxygen:

S(s) + O2(g) SO2 (g)

2.In the laboratory it is readily generated by treating a sulphite with dilute sulphuric acid.

SO3 2- + 2H+(aq) H2O(l) + SO2 (g)

Properties:-

1.2NaOH + SO2 Na2SO3 + H2O

Na2SO3 + H2O + SO2 2NaHSO3

(Excess)

SO2(g) + Cl2 (g) SO2Cl2(l)

Charcoal

2. When moist, sulphur dioxide behaves as a reducing agent.

For example,

1.it converts iron(III) ions to iron(II) ions

2Fe3+ + SO2 + 2H2O 2Fe2+ + SO42- + 4H+

2. it decolourises acidified potassium permanganate(VII) solution; this reaction is a convenient test for the gas.

5SO2 + 2MnO4-+2H2O 5SO4

2- + 4H+ + 2Mn2+

Structure:- The molecule of SO2 is angular. It is a resonance hybrid of the two canonical forms:

Uses:-Sulphur dioxide is used (i) in refining petroleum and sugar (ii) in bleaching wool and silk and (iii) as an anti-chlor, disinfectant and preservative. Sulphuric acid, sodium hydrogen sulphite and calcium hydrogen sulphite (industrial chemicals) are manufactured from sulphur dioxide.

Liquid SO2 is used as a solvent to dissolve a number of organic and inorganic chemicals.

Oxoacids of Sulphur

SULPHURIC ACID:-

Sulphuric acid is manufactured by the Contact Process which involves three steps:

(i) burning of sulphur or sulphide ores in air to generate SO2.

S+ O2 SO2

(ii) conversion of SO2 to SO3 by the reaction with oxygen in the presence of a catalyst (V2O5)

2SO2 + O2 2SO3

(iii) absorption of SO3 in H2SO4 to give Oleum (H2S2O7).

SO3 + H2SO4 H2S2O7

H2S2O7 + H2O 2 H2SO4

The favourable conditions for maximum yield are a pressure of 2 bar and a temperature of 720 K.

PROPERTIES OF SULPHURIC ACID:-

1.It dissolves in water with the evolution of a large quantity of heat. Hence, care must be taken while preparing sulphuric acid solution from concentrated sulphuric acid. The concentrated acid must be added slowly into water with constant stirring.

2.The chemical reactions of sulphuric acid are as a result of the following characteristics: (a) low volatility (b) strong acidic character (c) strong affinity for water and (d) ability to act as an oxidising agent.

3. In aqueous solution, sulphuric acid ionises in two steps.

H2SO4 + H2O H3O+ + HSO4- ( Ka1 = very large )

HSO4- + H2O H3O+ + SO4

2- ( Ka 2 = very low)

The larger value of Ka1 means that H2SO4 is largely dissociated into H+ and HSO4- ions–.

Greater the value of dissociation constant (Ka), the stronger is the acid.

4.. 2 MX + H2SO4 2 HX + M2SO4 (X = F, Cl, NO3)

(M = Metal)

5. Concentrated sulphuric acid is a strong dehydrating agent.

C12H22O11 12C + 11H2O

c-H2SO4

6.Hot concentrated sulphuric acid is a moderately strong oxidizing agent.

Cu + 2 H2SO4(conc.) CuSO4 + SO2 + 2H2O

3S + 2H2SO4(conc.) 3SO2 + 2H2O

C + 2H2SO4(conc.) CO2 + 2 SO2 + 2 H2O

Uses :-

In the manufacture of fertilizers (e.g., ammonium sulphate, superphosphate). petroleum refining , manufacture of pigments, paints and dyestuff intermediates , detergent industry , metallurgical applications (e.g.,cleansing metals before enameling, electroplating and galvanising, storage batteries , in the manufacture of nitrocellulose products and, as a laboratory reagent.

QUESTIONS:-

What happens when

(i) Concentrated H2SO4 is added to calcium fluoride

(ii) SO3 is passed through water?

Answer:-

(i) It forms hydrogen fluoride

CaF2 + H2 SO4 CaSO4 + 2HF

(ii) It dissolves SO3 to give H2SO4 .

SO3 + H2O H2 SO4

GROUP 17 ELENTS- HALOGENS.

1. Outermost confioguration is ns2np5.

2. They have very high ionization enthalpy due to very small atomic size. The I.E. decreases as we go down the group.

3.They have maximum negative electron gain enthalpy in the corresponding periods . This is b’coz after gaining an electron they attain the stable noble gas configuration.

However, the negative electron gain enthalpy of fluorine is less than that of chlorine. It is due to small size of fluorine atom. As a result, there are strong interelectronic repulsions in the relatively small 2p orbitals of fluorine and thus, the incoming electron does not experience much attraction.

4. All halogens are coloured. This is due to absorption of radiations in visible region which results in the excitation of outer electrons to higher energy level. For example, F2, has yellow, Cl2 , greenish yellow, Br2, red and I2, violet colour

5.The enthalpy of dissociation of F2 is less compared to that of Cl2

A reason for this anomaly is the relatively large electron-electron repulsion among the lone pairs in F2 molecule due to smaller size of F-atom where they are much closer to each other than in case of Cl2.

Cl – Cl > Br – Br > I – I.

BDE decreases b’coz atomic size and hence bond length increase.

Cl2> Br2>F2>I2

Question?

Although electron gain enthalpy of fluorine is less negative as compared to chlorine, fluorine is a stronger oxidising agent than chlorine. Why?

Ans:- It is due to

(i) low enthalpy of dissociation of F-F bond

(ii) high hydration enthalpy of F–

CHEMICAL PROPERTIES:-

1.Fluorine shows only -1 state . due to its high electronegativity.

2. The fluorine atom has no d orbitals in its valence shell and therefore cannot expand its octet.

3. Others show -1 , +1, +3, +5 and +7.

4. The chemical reactivity decreases down the group, b’coz the electronegativity decreases.

5. The Oxidising tendency also decreases down the group whereas the reducing tendency increases.

There is a regular decrease in the first ionization energy as we go down this column. As a result, there is a regular decrease in the oxidizing strength of the halogens from fluorine to iodine.

F2 > Cl2 > Br2 > I2

oxidizing strength

This trend is mirrored by an increase in the reducing strength of the corresponding halides.

I- > Br- > Cl- > F-

reducing strength

6. A halide can oxidize any other halide having a higher atomic no.

F2+ 2X- 2F- + X2 ( X = Cl/Br/I)

Cl2 + 2X- 2Cl- + X2 ( X = Br/I)

2 I-(aq) + Br2(aq) I2(aq) + 2 Br-(aq)

2F2 + 2H2O 4HF + O2

X2 + H2O HX + HOX (X= Cl /Br)

4I - + 4H+ + O2 2I2 + 2H2O

Anomalous proerties of Fluorine:-

1.It forms only one oxoacid while other halogens form a number of oxoacids (HOF).

2.Hydrogen fluoride is a liquid (b.p. 293 K) due to strong hydrogen bonding. Other hydrogen halides are gases.

HYDRIDES:-

1.The order of boiling point is:-

HF > HI> HBr> HCl

HI has exceptionally higher bpt since it forms extensive H-bonding.

The remaining hydrides follow increase in bpt with increase in molar mass.

1. Acidic strength of these acids increases in the order: .

Reason:- BDE decreases.

H–F > H–Cl > H–Br > H–I

2. The stability of these halides decreases down the group

H–F > H–Cl > H–Br > H–I.

Reason :- BDE decreases.

OXIDES:-

1. Fluorine forms two oxides OF2 and O2F2. However, only OF2 is thermally stable at 298 K.

These oxides are essentially oxygen fluorides because of the higher electronegativity of fluorine than oxygen.

Uses:-

(i).Both OF2 and O2F2 are strong fluorinating agents.

(ii) O2F2 oxidises plutonium to PuF6 and the reaction is used in removing plutonium as PuF6 from spent nuclear fuel.

2.The thermal stability of oxides decreases in the order;

I > Cl > Br > F

Reason:- Metallic character decreases and hence ionic character decreases

3. ClO2 is used as a bleaching agent for paper pulp and textiles and in sewage-water treatment.

4.I2O5 is a very good oxidising agent and is used in the estimation of carbon monoxide.

5. The ionic character of the metal halides decreases in the order

MF >MCl > MBr > MI

Reason:- Electronegativity decreases.

If a metal exhibits more than one oxidation state, the halides in higher oxidation state will be more covalent than the one in lower oxidation state. (Fajan’s rule)

For e.g., SnCl4, PbCl4, SbCl5 and UF6 are more covalent than SnCl2, PbCl2, SbCl3 and UF4 respectively.

QUESTION?:

Fluorine exhibits only –1 oxidation state whereas other halogens exhibit + 1, + 3, + 5 and + 7 oxidation states also. Explain.

Ans:-Fluorine is the most electronegative element and cannot exhibit any positive oxidation state. Other halogens have d orbitals and therefore, can expand their octets and show + 1, + 3, + 5 and + 7 oxidation states also.

CHLORINE:-

Preparation:-

It can be prepared by any one of the following methods:

(i) By heating manganese dioxide with concentrated hydrochloric acid.

MnO2 + 4HCl MnCl2 + Cl2 + 2H2O

However, a mixture of common salt and concentrated H2SO4 is used in place of HCl.

4NaCl + MnO2 + 4H2SO4 MnCl2 + 4NaHSO4 + 2H2O + Cl2

(ii) By the action of HCl on potassium permanganate.

2KMnO4 + 16HCl 2KCl + 2MnCl2 + 8H2O + 5Cl2

Manufacture of chlorine

(i) Deacon’s process: By oxidation of hydrogen chloride gas by atmospheric oxygen in the presence of CuCl2 (catalyst) at 723 K.

4HCl + O2 2Cl2 + 2H2O

CuCl2

(ii) Electrolytic process: Chlorine is obtained by the electrolysis of brine (concentrated NaCl solution). Chlorine is liberated at anode.

Properties

1. Reaction with meatls and non-meatlas:-

2Al + 3Cl2 2AlCl3;

2Na + Cl2 2NaCl;

2Fe + 3Cl2 2FeCl3 ;

P4 + 6Cl2 4PCl3

S8 + 4Cl2 4S2Cl2

2. Reaction with hydrogen and compounds of hydrogen:-

It has great affinity for hydrogen. It reacts with compounds containing hydrogen to form HCl.

H2 + Cl2 2HCl

H2S + Cl2 2HCl + S

C10H16 + 8Cl2 16 HCl + 10C

3.With excess ammonia, chlorine gives nitrogen and ammonium chloride whereas with excess chlorine, nitrogen trichloride (explosive) is formed.

8NH3 + 3Cl2 6NH4Cl + N2;

(excess)

NH3 + 3Cl2 NCl3 + 3HCl

(excess)

4..With cold and dilute alkalis chlorine produces a mixture of chloride and hypochlorite but with hot and concentrated alkalis it gives chloride and chlorate.

2NaOH + Cl2 NaCl + NaOCl + H2O

(cold and dilute)

6 NaOH + 3Cl2 5NaCl + NaClO3 + 3H2O

(hot and conc.)

5.With dry slaked lime it gives bleaching powder.

2Ca(OH)2 + 2Cl2 Ca(OCl)2 + CaCl2 + 2H2O

The composition of bleaching powder is Ca(OCl)2.CaCl2.Ca(OH)2.2H2O.

6.Chlorine water on standing loses its yellow colour due to the formation of HCl and HOCl. Hypochlorous acid (HOCl) so formed, gives nascent oxygen which is responsible for oxidising and bleaching properties of chlorine.

(i) It oxidises ferrous to ferric, sulphite to sulphate, sulphur dioxide to sulphuric acid and iodine to iodic acid.

2FeSO4 + H2SO4 + Cl2 Fe2(SO4)3 + 2HCl

Na2SO3 + Cl2 + H2O Na2SO4 + 2HCl

SO2 + 2H2O + Cl2 H2SO4 + 2HCl

I2 + 6H2O + 5Cl2 2HIO3 + 10HCl

(ii) It is a powerful bleaching agent; bleaching action is due to oxidation and therefore it is permanent.

Cl2 + H2O 2HCl + [O]

Coloured substance + [O] Colourless substance

It bleaches vegetable or organic matter in the presence of moisture.

Bleaching effect of chlorine is permanent.

Uses of Chlorine: It is used (i) for bleaching woodpulp (required for the manufacture of paper and rayon), bleaching cotton and textiles, (ii) in the extraction of gold and platinum (iii) in the manufacture of dyes, drugs and organic compounds such as CCl4, CHCl3, DDT, refrigerants, etc. (iv) in sterilising drinking water and (v) preparation of poisonous gases such as phosgene (COCl2), tear gas (CCl3NO2), mustard gas (ClCH2CH2SCH2CH2Cl).

QUESTION:-

Write the balanced chemical equation for the reaction of Cl2 with hot and concentrated NaOH. Is this reaction a disproportionation reaction? Justify.

Ans:- 3Cl2 + 6NaOH 5NaCl + NaClO3 + 3H2O

Yes, chlorine from zero oxidation state is changed to –1 and +5 oxidation states.

Hydrogen Chloride

Preparation

In laboratory, it is prepared by heating sodium chloride with concentrated sulphuric acid.

NaCl + H2SO4 NaHSO4 + HCl

420K

NaHSO4 + NaCl Na2SO4 + HCl

823K

HCl gas can be dried by passing through concentrated sulphuric acid.

Properties

1.It reacts with NH3 and gives white fumes of NH4Cl.

NH3 + HCl NH4Cl

2.When three parts of concentrated HCl and one part of concentrated HNO3 are mixed, aqua regia is formed which is used for dissolving noble metals, e.g., gold, platinum.

Au + 4H+ + NO3- + 4Cl- AuCl4

- + NO + 2H2O

3Pt + 16H+ + 4NO3- +18Cl- PtCl62- + 4NO + 8H2O

3.Hydrochloric acid decomposes salts of weaker acids, e.g., carbonates, hydrogen carbonates, sulphites, etc

Na2CO3 + 2HCl 2NaCl + H2O + CO2

NaHCO3 + HCl NaCl + H2O + CO2

Na2SO3 + 2HCl 2NaCl + H2O + SO2

Uses of HCl:- It is used (i) in the manufacture of chlorine, NH4Cl and glucose (from corn starch), (ii) for extracting glue from bones and purifying bone black, (iii)in medicine and as a laboratory reagent.

QUESTION:-

When HCl reacts with finely powdered iron, it forms ferrous chloride and not ferric chloride. Why?

Ans:-Its reaction with iron produces H2.

Fe + 2HCl FeCl2 + H2

Liberation of hydrogen prevents the formation of ferric chloride.

Oxoacids of Halogens

INTERHALOGEN COMPOUNDS

The compounds of halogens formed amongst themselves are called interhalogen compounds.

1. The interhalogen compounds can be prepared by the direct combination or by the action of halogen on lower interhalogen compounds

Cl2 + 3F2 2ClF3

(Excess)

I2 + 3Cl2 2ICl3

(Excess)

Br2 + 5F2 2BrF5

(Excess)

2. They are all covalent in nature.

3. More reactive than halogens (except Fluorine).

This is because X–X`bond in interhalogens is weaker than X–X bond in halogens except F–F bond.

3. They are diamagnetic as they do not contain any unpaired electron.

4. XX’ + H2O HX’ + HOX. (X= smaller halogen)

5.Shapes:-

XX’3 -------------- Bent T- shaped

XX’5 --------------- Square pyramidal

XX’7 (IF7) ------- Pentagonal bipyramidal.

Shape of BrF3:- Bent T –shaped.

Uses:

1. These compounds can be used as non aqueous solvents. 2. Interhalogen compounds are very useful fluorinating agents. 3. ClF3 and BrF3 are used for the production of UF6 in the enrichment of U-235.

U(s) + 3ClF3(l) UF6(g) + 3ClF(g)

GROUP 18 ELEMENTS.

1.Group 18 consists of six elements: helium, neon, argon, krypton, xenon and radon.

2.Electronic configuration is ns2np6

3.They have very high ionization enthalpy due to stable configuration.

4..All these are gases and chemically unreactive. They form very few compounds. Because of this they are termed noble gases.

5.All the noble gases except radon occur in the atmosphere. Their atmospheric abundance in dry air is ~ 1% by volume of which argon is the major constituent.

Helium and sometimes neon are found in minerals of radioactive origin e.g., pitchblende, monazite, cleveite. The main commercial source of helium is natural gas.

Xenon and radon are the rarest elements of the group.

Radon is obtained as a decay product of Radium-226.

226 222 4

Ra Rn + He

88 86 2

Because radon is highly radioactive, it is difficult to study the chemistry of radon.

6.They have very low melting and boiling points because the interatomic interaction in these elements is weak dispersion forces. Helium has the lowest boiling point (4.2 K) of any known substance. It has an unusual property of diffusing through most commonly used laboratory materials such as rubber, glass or plastics.

7.In general, noble gases are least reactive. Their inertness to chemical reactivity is attributed to the following reasons:

(i) The noble gases except helium (1s2) have completely filled ns2np6 electronic configuration in their valence shell.

(ii) They have high ionisation enthalpy and more positive electron gain enthalpy.(since they do not have any tendency to accept an additional electron as their outermost orbits are completely filled)

8.Xenon has lower ionisation enthalpy, because of large size of xenon.

9. Xenon reacts with only oxygen and fluorine atoms because these two elements are of very high polarizing capacity.

10.The fact that the first IE of xenon is comparable with that of molecular oxygen prompted Neil Bartlett to study the chemistry of xenon compounds.(he understood this from the compound O2PtF6, which he prepared )

The first compound of xenon prepared was XePtF6

Question?

Why are the elements of Group 18 known as noble gases ?

The elements present in Group 18 have their valence shell orbitals completely filled and, therefore, react with a few elements only under certain conditions. Therefore, they are known as noble gases.

(a) Xenon-fluorine compounds

Xenon forms three binary fluorides, XeF2, XeF4 and XeF6 by the direct reaction of elements under appropriate experimental conditions.

XeF6 can also be prepared by the interaction of XeF4 and O2F2 at 143K.

XeF4 + O2 F2 XeF6 + O2

Properties:-

1.XeF2 is hydrolysed to give Xe, HF and O2.

2XeF2 (s) + 2H2O(l) 2Xe (g) + 4 HF(aq) + O2(g)

2.Xenon fluorides react with fluoride ion acceptors to form cationic species and fluoride ion donors to form fluoroanions.

XeF2 + PF5 [XeF]+ [PF6]– ;

XeF4 + SbF5 [XeF3]+ [SbF6]–

XeF6 + MF M+ [XeF7]– (M = Na, K, Rb or Cs)

(b) Xenon-oxygen compounds

Hydrolysis of XeF4 and XeF6 with water gives Xe03.

6XeF4 + 12 H2O 4Xe + 2Xe03 + 24 HF + 3 O2

XeF6 + 3 H2O XeO3 + 6 HF

Structures of xenon compounds:-

XeF4

F

F F

F

Xe

..

..

Uses:-

1.Helium is a non-inflammable and light gas. Hence, it is used in filling balloons for meteorological observations. It is also used in gas-cooled nuclear reactors. Liquid helium (b.p. 4.2 K) finds use as cryogenic agent for carrying out various experiments at low temperatures. It is used to produce and sustain powerful superconducting magnets which form an essential part of modern NMR spectrometers and Magnetic Resonance Imaging (MRI) systems for clinical diagnosis. It is used as a diluent for oxygen in modern diving apparatus because of its very low solubility in blood.

2.Neon is used in discharge tubes and fluorescent bulbs for advertisement display purposes.

3.Argon is used mainly to provide an inert atmosphere in high temperature metallurgical processes (arc welding of metals or alloys) and for filling electric bulbs.

It is also used in the laboratory for handling substances that are air-sensitive.

4.Xenon and Krypton are used in light bulbs designed for special purposes.

Important chemical equations

1. NH4CI +NaNO2N2+2H2O+NaCl

2 (NH4)2Cr2O7 N2+4H2O+Cr2O3

3. 2NH4Cl+Ca(OH)22NH3+ H2O +CaCl2

4. 2FeCl3+3NH4OH Fe(OH)3+2NH4Cl

5. Cu2++4NH3[Cu(NH3)4]2+

6. 4NH3+5O2 Pt 4NO+6H2O

500K ; 9 bar

7. 2NO+O2 2NO2

8. 3NO2+H2O2HNO3+NO

9. 3Cu+8HNO33Cu(NO3)2+2NO+H2O

10. Cu+4HNO3Cu(NO3)2+2NO2+4H2O

11. I2+10 HNO3 2HIO3+10 NO2+4H2O

12. P4+20 HNO34H3PO4+20 NO2+4H2O

13. NO3- +3Fe3++4H+ NO+ 3Fe2++2H2O

14. [Fe(H2O)6]2++NO[Fe(H2O)5NO]2++H2O

15. P4+3NaOH+3H2OPH3+3NaH2PO2

16. PCl3+3H2OH3PO3+3HCl

17. PCl5+H2O POCl3+2 HCl

18. POCl3+3H2OH3PO4+3HCl

19. Al2O3+6HCl+9H2O2[Al(H2O)6]3++6Cl-

20 Al2O3. +6NaOH+3H2O.2Na3[Al(OH)6]

21. PbS+4O3PbSO4+4O2

22. 2I- +H2O+O3 2OH-+I2+ O2

23. 2Fe3+ + SO2 + 2H2O2Fe2++ SO42- +4H+

24. 2MCl+H2SO4 HCl+M2SO4 (M= any alkali or alkaline earth metal)

Conc.H2SO4

25. C12H22O11 12C+11H20

26. C+Conc.H2SO4 CO2+2SO2+2H2O

27. MnO2+4HClMnCl2+Cl2+2H2O

28. 2KMnO4+16HCl2KCl+2MnCl2+8H2O+5Cl2

29. 2NaOH+Cl2NaCl+NaOCl+ H2O

30. 6NaOH+3Cl25NaCl+NaClO3+3 H2O

31. 2Ca(OH)2+2Cl2Ca(OCl)2+CaCl2+2H2O

32. Na2SO3+Cl2+H2ONa2SO4+2HCl

33. I2+ 6H2O +5Cl22HIO3+10 HCl

34. Cl2+3F22ClF3

35. I2+3Cl22ICl3

36. Br2+3F22BrF3

37. Xe +F2 673k ; 1bar Xe F2

38. Xe+ 2F2 873k,7bar Xe F4

39. Xe+3F2 573k,60-70bar Xe F6

40. XeF2+PF5[XeF]+[PF6]-

41. XeF6+NaFNa+[XeF7]-

42. XeF6+3H2OXeO3+6HF

HOTS

1. What is called inert pair effect? 2. As we go down a group in the p-block the stable oxidation state decreases by 2 units. Explain. 3. Why do the group 15 elements have very high I.E? 4. There is only very small increase in size from As – Bi. Why? 5. Molecular nitrogen exist as a gas whereas phosphorus is solid at room temperature. Why? 6. PCl5 exists whereas NCl5 does not why? 7. N does not show allotropy. Why? 8. N does not catenate. Why? 9. Bi shows +3 states. Why 10. Why is NF3 stable whereas NCl3 is not? 11. Why is BiF3 highly ionic? 12. Which is more covalent – BiF5 or Bif3? 13. NH3 is a Lewis base. Explain. 14. Why is NH3 a stronger base than BH3. 15. Arrange the hydrides of group 15 elements in the

(a) increasing order of basicity (b) increasing order of stability (c) Increasing order of reducing nature

16. Complete (NH4)2 Cr2O7 ∆→? 17. Molecular Nitrogen is highly unreactive. Why? 18. Explain Haber’s process. (optimum conditions and catalyst) 19. Silver chloride dissolves in aqueous ammonias. Explain. 20. Complete Cu2+

(aq) + NH3(aq) → ? 21. Shapes of oxides of nitrogen.

22. Why do NO2 and NO dimerise? 23. What is the covalence of N in (a) N2O5 (b) N2O4 ? 24. Explain Ostwald process of manufacturing HNO3. 25. Draw the shape of HNO3. Why is it monobasic? 26. Explain the chemistry of the Brown ring test. 27. Complete and balance P4 + NaOH + H2O → ? 28. Why is white phosphorus less stable and more reactive? 29. Write any two differences between white and red phosphorus. 30. Mention any two uses of phosphine. 31. Bond angle in PH4+ is higher than that in PH3. Why? 32. Why does PCl3 fumes in moist air? 33. Complete and balance (a) P4 + SOCl2 →? (b) P4 + SO2Cl2 →? (c) HgCl2 + PH3→?

(d) CuSO4 + PH3→? 34. Why does PCl5 fumes in moist air? 35. Draw the shape of (a) PH3 (b) PCl3 (c) PCl5 36. Complete and balance (a) Ag + PCl5→? (b) Sn + PCl5→? 37. PCl5 in solid state exists as an ionic solid. Explain. 38. All the five bonds in PCl5 are not equivalent. Explain. 39. Draw the structure of (a) Phosphoric Acid (b) phosphorus Acid (c) Hypo phosphorus acid

Discuss the basicity of each of them. 40. Phosphorus acid disproportionates. Write the equation 41. Hypo Phosphorus acid is a better reducing agent than phosphorus acid.Explain. 42. How is N2 prepared in lab? (ans: NH4Cl + NaNO2 ∆→ N2 + 2H20 + NaCl) 43. NH3 is a stronger base than BiF3. Why? 44. NH3 form intermolecular H-bond whereas PH3 does not. Why? 45. Write the equation for the reaction of PCl5 with heavy water. (ans: PCl5 + 4D2O→D3PO4 +5DCl) 46. Hypo phosphorus acid is a reducing agent. Explain. 47. Group 16 elements have lower I.E than group 15 elements. Why? 48. Oxygen has less negative electron gain enthalpy than S. Why? 49. Oxygen is a gas at room temperature whereas Sulphur is a sold. Why? 50. Oxygen has lower melting point and boiling point than Sulphur. Why? 51. Oxygen shows -2 oxidation state preferably. Why? 52. OF2 exists whereas OCl2 does not. Why? 53. Oxygen does not show covalency beyond 4. Why?

OR SF6 exists where as OF6 does not. Why?

54. OF2 is known as Oxyfluoride but not flourooxide. Why?

QUESTIONS WITH ANSWERS 15th GROUP 1. The Bi (V) compounds are unknown except BiF5 .Why? Due to inert pair effect, +5 oxidation state of Bi is unstable 2. Nitrogen exists as a gas in elemental state , whereas other elements of the group exist as solid .Why? Due to ability of nitrogen to form pπ- pπ multiple bond , it exists as discrete molecules Of N2 . 3. The tendency of catenation for Nitrogen is less than Phosphorous . Why? N-N bond is weaker than P-P bond due to inter electronic repulsion between non bonding pair of electrons in Nitrogen owing to it’s small size . 4. Arrange the hydrides of 15th group elements in the decreasing order of the property mentioned a. Thermal stability : NH3 > PH3>AsH3>SbH3>BiH3 b. Basic strength : NH3 > PH3>AsH3>SbH3>BiH3 c. Reducing character : BiH3 > SbH3 >AsH3>PH3> NH3 5. Arrange the oxide E2O3 of 15th group element in the decreasing order of acid strength N2O3> P2O3>As2O3>Sb2O3>Bi2O3 6. Nitrogen doesn’t form pentahalides . Why? Due to non availability of ‘d’ orbitals in Nitrogen to expand it’s covalency beyond 4. 7. NH3 has exceptionally higher boiling point compared to the other hydrides of 15th group.Why? Due to strong inter molecular association in NH3 through H- bonding 8. NH3 is most basic among the hydrides of 15th group elements . Why? Due to small size of Nitrogen the availability of lone pair electron in NH3 is more. 9. BiH3 is the strongest reducing agent among the hydrides of 15th group elements .Why? B’cause Bi-H bond is the weakest 10. The pentahalides of 15th group elements are more covalent than their trihalides .Why? The +5 oxidation state of the element in their penta halides is more polarizing than the +3 state of elements in their trihalides . 11. The thermal stability of BiH3 is the lowest among the hydrides of 15th group elements Why? Bi-H bond is the weakest bond due to it’s small bond dissociation enthalpy. 12 Dinitrogen ( N2) is inert at room temperature .Why? Due to very high bond enthalpy of N= N 13. How is Nitrogen prepared in pure form ? By thermal decomposition of Sodium/Barium azide Ba(N3)2 Ba + 3N2 2NaN3 2Na + 3N2 14. How is NH3 manufactured by Haber process N2 (g) + 3H2(g) 2NH3(g) Temperature : 700K Pressure : 200At. Catalyst : Iron oxide Promoters : K2O and Al2O3

15. AgCl is soluble in NH3 . Why ? Due to formation of a soluble complex with NH3. AgCl (s) + NH3(aq) [Cu(NH3)4] 2+ 16. How does NH3 react with Cu2+ (aq) ? Cu2+ + 4NH3 [Cu(NH3)4]2+ 17. NO2 undergoes dimerisation .Why? NO2 is an odd electron molecule . So on dimerisation it is converted to N2O4 which contains even no .of electrons . (Diagram) 18. NO being an odd electron molecule , doesn’t undergo dimerisation . Why? The unpaired electron in NO is delocalized and gets stabilized .(Diagram) 19. How is HNO3 manufactured by Ostwald’s process . Pt/Rh/ 500K ,9bar 4NH3(g) + 5O2 (g) 4NO(g) 2NO(g) + O2(g) 2NO2(g) 3NO2(g) + H2O (l) 2HNO3 (aq) + NO (g) 20. What is the Chemistry of Brown ring test ? NO3

- + 3Fe2+ + 4H+ NO + 3Fe3+ + 2H2O [Fe(H2O)6]2+ + NO [Fe(H2O)5(NO)]2+ + H2O 21. Methods of preparation of oxides of Nitrogen .(Table 7.3) 22. Reactions of con.HNO3 with Zn, Cu, I2, C, S8, P4 23. How does white P4 react with boiling NaOH ? P4 + 3NaOH + 3H2O PH3 + 3NaH2PO2 24. White Phosphorous is more reactive than red phosphorous .Why? In white phosphorous there is angular strain in P4 molecule .so it is less stable and more reactive compared to red phosphorous. 25. What are the differences between white phosphorous and red phosphorous? a. White P consists of discrete tetrahedral P4 molecules whereas red p consists of chain of P4 tetrahedra. b. White P is soluble in CS2 while red P is insoluble in CS2 . c. White P glows in darkness whereas red P doesn’t glow . d. white P is more reactive than red P 26. White P glows in darkness .Why? Due to chemiluminescence . 27. White P catches fire spontaneously .Why? Because it’s ignition temperature is lower than atmospheric temperature. 28. How are α and β Black Phosphorous prepared ? α Black P is prepared by heating red P in a sealed tube at 803k . β black P is prepared by heating white P under high pressure. 29. Impure PH3 is inflammable. Why? Due to the presence of impurities like PH3 or P4 vapours which are inflammable . 30. How is PH3 purified ? Impure PH3 is absorbed in HI to form PH4I which on treating with KOH gives off PH3. 31. How does Ca3P2 react with HCl? Ca3P2 + 6HCl 3CaCl2 + 2PH3 32. How does PH3 react with a. CuSO4 b. HgCl2

3CuSO4 + 2PH3 Cu3P2 + 3H2SO4 3HgCl2 + 2PH3 Hg3P2 + 6HCl 33. Show that PH3 is Basic . PH3 is a Lewis acid due to the presence of a lone pair electron on P atom .It reacts with acids like HI and gives salt . PH3 + HI PH4I 34. Bond angle in PH4+ is higher than that in PH3 . Why? In PH3 there are 3 bond pairs and 1 lone pair of electrons , whereas in PH4

+ there are only 4 bond pairs of electrons . Due to greater repulsion between lone pair and bond pair in PH3 as compared to bond pair –bond pair repulsion in PH4

+ , the bond angle in PH3 is less than that in PH4

+. 35. PCl3 fumes in moist air .Why? Due to formation of HCl gas with water. PCl3 + 3H2O H3PO3 + 3HCl 36. What happens when PCl5 is heated ? It decomposes to give PCl3 and Cl2 . 37. Why does PCl5 fumes in moist air Due to formation of HCl gas with water. PCl5 + 4H2O H3PO4 + 5HCl 38. All the five P-Cl bonds in PCl5 are not equivalent . Why? PCl5 has a triagonal bipyramidal structure with 3 P-Cl bonds in equatorial position and 2 P-Cl bonds in axial positions . Since axial bonds experience more repulsion from equatorial bonds ,they are slightly longer than equatorial bonds . 39. H3PO3 undergoes disproportionation .Why? Due to less stability of +3 oxidation state of P compared to it’s +5 and -3 oxidation state . 4H3PO3 3H3PO4 + PH3 40. H3PO2 is a reducing agent .Why? Due to the presence of 2 P-H bonds. Eg. 4AgNO3 + 2H2O + H3PO2 4Ag + 4HNO3 + H3PO4 41. H3PO3 is a dibasic acid though there are 3 hydrogen atoms.Why? In H3PO3 there are only 2 H atoms which are ionisable ( due to 2 P-OH bonds ).The third H atom is bonded to P , so not ionisable . 42. How Many types of salts are formed by the following acids? a) H3PO3 - 2 types (basicity is 2) b) H3PO4 -- 3 types (basicity is 3) c) H3PO2 -- 1 type (basicity is 1) 43. Solid form of PCl5 is conducting. Why? In solid form PCl5 exists as an ionic compound [PCl4]+ [PCl6]- 44. CN- is a known compound whereas CP- is not known.Why? Nitrogen can form pπ- pπ bond with carbon .whereas Phosphorous can’t form pπ- pπ with carbon. GROUP-16TH 1. There is a large difference in melting and boiling points of oxygen and sulphur.Why?

Oxygen exists as a diatomic molecule O2 whereas Sulphur exists in a puckered arrangement of S8 molecules . Therefore there is strong intermolecular force in sulphur . 2. Name two compounds in which Oxygen shows positive oxidation state . OF2(-2) and O2F2 (-1) 3. In the elemental form O2 is a gas whereas other elements of the group are solids .Why? Oxygen can form pπ- pπ multiple bond with itself and form O2 molecules . But other members of the group can’t form pπ- pπ multiple bonds with their atoms and so they form puckered arrangement of their polyatomic molecules . 4. Oxygen shows positive oxidation state in compounds with Fluorine. Why? Oxygen is less electronegative than Fluorine . 5. First ionization enthalpy of 16th group elements is less than that of 15th group elements. Why? 15th group elements have extra stable half filled P orbitals as their valence orbital, so a large amount of energy is required to remove one valence electron . 6. H2O is a liquid whereas all the other hydrides of 16th group are gases. Why? Due to strong intermolecular H-bonding in H2O 7. Arrange the Hydrides of 16th group elements in increasing order of the property mentioned below a) Thermal stability : H2Po <H2Te<H2Se<H2S<H2O b) Acid strength : H2O<H2S<H2Se<H2Te<H2PO c) Reducing property : H2O<H2S<H2Se<H2Te<H2PO 8. The acid character of hydrides of 16th group elements increases down the group.Why/ Because the bond strength of E-H bond decreases down the group due to increase in the size of the element . 9. SF6 is exceptionally stable .Why? Because Sulphur is stericaly protected by six Fluorine atoms. 10. What is the hybridization and shape of SF4 molecule ? Sp3d and the shape is see-saw . 11. How are oxides classified ? Give one example for each . a) Acidic : SO2 b) Basic : Na2O c) Amphoteric : Al2O3 d) Neutral : CO 12. Give reactions to show the amphoteric nature of Al2O3 Basic character : Al2O3 (s) + 6HCl(aq) + 9H2O(l) 2[Al(H2O06]3+(aq) Acid character : Al2O3(s) + 6NaOH(aq) + 3H2O 2 Na3[Al(OH)6](aq) 13. High concentrations of ozone can be dangerously explosive .Why? This is due to liberation of large amount of heat during it’s decomposition to form O2 . O3(g) O2(g) + (O) 14. Why does O3 act as a powerful oxidizing oxidizing agent ? Due to the ease with which it liberates atoms of nascent oxygen atom . O3(g) O2(g) + (O) 15. Give two examples to show the oxidizing property of ozone .

PbS (s) + 4O3(g) PbSO4(s) + 4O2 (g) 2I-(aq) + O3(g) + H2O (l) 2OH-(aq) + I2 (s) + O2(g) 16. How is O3 estimated quantitatively ? O3 is treated with an excess of KI solution buffered with a borate buffer , the iodine so liberated is estimated by titrating with a saturated solution of Sodium thiosulphate solution . From the amount of iodine liberated the amount of O3 can be estimated . 2I-(aq) + O3(g) + H2O (l) 2OH-(aq) + I2 (s) + O2(g) 17. How does the depletion of O3 take place in the upper atmosphere ? The Nitrogen oxides emitted by supersonic jet aeroplanes depletes the O3 layer NO (g) + O3 (g) NO2(g) + O2(g) 18. The O-O bonds in O3 are identical .Why? Due to resonance both O-O bonds get partial double bond character and are identical. 19. Sulphur in the vapour phase shows paramagnetism .Why? In the vapour form sulphur exists as a diatomic molecule which contains two unpaired electrons in the antibonding pi molecular orbital like O2 . 20. The behavior of SO2 is like CO2 .Explain . SO2 reacts with alkali solution in the similar way that CO2 reacts. 2NaOH + SO2 Na2SO3 + H2O Na2SO3 + H2O + SO2 2NaHSO3 21. How is SO2 gas detected ? So2 gas on passing through acidified KMO4 solution , the purple colour of KMnO4 is discharged . 5SO2(g0 + 2MnO4

-(aq) + 2H2O 5SO42- (aq) + 2Mn2+ + 4H+

22. Both S-O bonds in SO2 are identical. Why? Due to resonance both S-O bonds are getting partial double bond character and are identical. (Diagram) 23. Structure of oxoacids of Sulphur (From Text Book ) 24. How is H2SO4 manufactured by Contact Process ? S (s) + O2(g) SO2(g) V2O5 / 720K , 2 bar 2SO2(g) + O2 (g) 2SO3 (g) SO3(g) + H2SO4 H2S2O7 H2S2O7 + H2O 2H2SO4

Reactions of Con H2SO4 refer text book GROUP 17TH 1. Mention one source each for F, Br& I F- Fluorspar (CaF2) Br—Sea water I- Chile saltpeter. 2. Halogens have very little tendency to lose electrons .Why? Due to their high ionization enthalpy. 3. Halogens have max. negative electron gain enthalpy. Why?

Due to a) Small size b) high effective nuclear charge c) they need one electron to attain stable noble gas configuration . 4. The electron gain enthalpy of Fluorine is less negative than Chlorine .Why? Due to small size of Fluorine , there is strong inter electronic repulsion in the small 2p orbital is high. 5. All halogens are coloured .Why? Due to absorption of radiation in the visible region which results in the excitation of outer electrons to higher energy levels . 6. The bond dissociation enthalpy of F2 is smaller than that of Chlorine .Why? Due to large electron –electron repulsion among the lone pairs of electrons in Fluorine 7. Fluorine is a strong oxidizing agent than Chlorine .Why? This is due to a). Low bond enthalpy of F2. b). high hydration enthalpy of F- 8. Fluorine doesn’t show positive oxidation state.Why? Due to it’s highest electronegativity . 9. Fluorine can’t expand its covalency beyond one .Why? Due to absence of vacant ‘d’ orbital. 10. The oxidizing tendency of halogens decreases down the group. Why? Due to a) decreasing electron gain enthalpy b) decreasing hydration enthalpy of X- 11. What happens when F2 gas is passed through solutions of a) KCl b)KBr c) KI Fluorine being more reactive oxidizes the halides to the respective halogens . KCl (aq) + F2 (g) 2KF (aq) + Cl2(g) KBr (aq) + F2(g) 2KF(aq) + Br2(l) KI(aq) + F2 (g) 2Kf(aq) + I2(g) 12. How does F2 gas react with water ? F2 oxidises water to oxygen gas; F2(g) + 2H2O (l) 4HF(aq) + O2(g) B’coz, F is more electroneg than O 13. How do Chlorine and Bromine react with water ? Cl2(g) + H2O(l) HCl (aq) + HOCl (aq) Br2(l) + H2O(l) HBr(aq) + HOBr(aq) 14. What are the reasons for the anomalous behavior of Fluorine ? a) small size b) highest electronegativity c) low F-F bond dissociation enthalpy d) non availability of vacant ‘d ‘ orbitals in the valence shell. 15. Most of the compounds of Fluorine are exothermic. Why? Due to small size and strong bond formed by Fluorine with atoms of other elements . 16. Fluorine forms only one oxoacid HOF . Why? Due to absence of vacant d - orbitals in its valence shell. 17. HF is a liquid . Why? Due to strong intermolecular H-bonding 18. Arrange the hydrides of 17th group elements in the increasing order of the property mentioned . Acid strength : HF <HCl<HBr<HI Stability : HI<HBr<HCl<HF 19. The stability of the hydrides of 17th group elements decreases down the group. Why? Due to the decreasing bond strength of E-H bond .

20. What is the use of O2F2 ? Used for the removal of Pu from the spent nuclear of nuclear power plant . 21. What is the use of ClO2 ? Used in the bleaching of paper pulp and textiles and in water treatment . 22. Name the oxide of halogens used in the estimation of CO. I2O5 23. Which is more covalent SnCl4 or SnCl2 and why? SnCl4 , because the higher oxidation state of the metal is more polarizing than it’s lower oxidation state . 24. How does Cl2 reacts with a) cold dilute NaOH b) hot concentrated NaOH Cold and dilute NaOH 2NaOH + Cl2 NaCl + NaOCl + H2O Hot and concentrated NaOH 6NaOH + 3Cl2 5NaCl + NaClO3 + 3H2O 25. How is Bleaching powder manufactured ? What is it’s composition ? By passing Cl2 gas through dry slaked lime . 2Ca(OH) 2 +2Cl2 Ca(OCl)2 + CaCl2 + 2H2O Composition is Ca(OCl)2.CaCl2.Ca(OH)2.2H2O 26. Chlorine water loses its yellow colour on standing .Why? Due to the formation of HCl and HOCl . Cl2 + H2O HCl +HOCl 26. Give the reason for the bleaching and oxidizing properties of Chlorine . Moist Chlorine gives HCl and HOCl . HOCl decomposes to give HCl and nascent oxygen which is responsible for oxidizing and bleaching properties . Bleaching action is due to oxidation Cl2 + H2O 2HCl + (O) Coloured substance + (O) colourless substance . 27. Name two poisonous gases that can be prepared from Chlorine . a) Phosgene (COCl2) b) Tear gas (CCl3NO2) 28. How does moist Chlorine react with I2 ? I2 + 6H2O + 5Cl2 2HIO3 + 10HCl 29. What is aquaregia ? What is its use ? Aquqregia is a mixture of Con.HCl and Con.HNO3 in the ratio 3:1. It is used for dissolving noble metals like Gold and platinum. Au + 4H+ + NO3

- + 4Cl- AuCl4- + NO + 2H2O 30. When HCl reacts with finely powdered Iron , it forms ferrous chloride and not ferric chloride .Why? HCl reacts with Iron and produces H2 . Fe + 2HCl FeCl2 + H2 Liberation of hydrogen prevents the formation of ferric chloride. 31. Write the geometry of the following interhalogen compounds . a) ClF (XX’1) : Linear. b) ClF3 (XX’3) : Bent T – shape. c) BrF5 (XX’5) : Square pyramidal d) IF5 : Pentagonal bipyramidal .

32. Interhalogen compounds are more reactive than halogens except F2 .Why? X-X’ bond in interhalogens is weaker than X-X bond in halogens except F-F bond . 33. Name the interhalogen which is used in the enrichment of Uranium? ClF3. 34. Write the produt of Hydrolysis of the following interhalgens . a) XX’ : XX’ + H2O HX’ + HOX b) XX”3 : XX’3 + 2H2O 3HX’ + HOXO c) XX’5 : XX”5 + 3H2O 5HX’ + HOXO2 d) XX’7 : XX’7 + 4H2O 7HX’ + HOXO3 GROUP 18TH 1. The elements of group18 are known as noble gases . Why? They have completely filled valence shell . So they react with a few elements only under certain conditions . 2. 18th group elements have large positive values of electron gain enthalpy .Why? Due to their stable electronic configuration they have no tendency to accept the electron . 3. 18th group elements have very low melting and boiling points . why? Due to very weak intermolecular interaction, i.e.,weak dispersion forces . 4. Noble gases are liquefied at very low temperatures .Why? Due to weak intermolecular forces . 5. Noble gases are chemically inert .Why? Due to a) completely filled valence shell. b) High ionization enthalpy c) positive electron gain enthalpy . 6. What inspired N. Bartlett for carrying out reaction between Xe and PtF6 ? The first ionization enthalpy of molecular oxygen was almost identical with that of Xe . 7. What was the compound prepared by N. Bartlett that led to the discovery of first noble gas compound ? O2

+PtF6-

8. What was the first noble gas compound prepared ? Xe+PtF6

- 9. XeF2 is a linear molecule without bend .Why? XeF2 has 3 lone pair and 2 bond pair electrons .The 2 bond pairs are occupying the axial positions and 3 lone pairs are occupying the equatorial positions making 1200 with each other and 900 with the bond pairs . Since the bond pairs are experiencing same amount of repulsion from the lone pairs the molecule is linear without bend. 10. Give reactions to show that Xe fluorides act as both Lewis Acid and Lewis Base Lewis Acid : XeF6 + NaF M+ [XeF7]- Lewis Base : XeF2 + PF5 [XeF]+[PF6]- 11. write a balanced reaction between water and a) XeF4 b) XeF6 6XeF4 + 12H2O 4Xe + 2XeO3 + 24HF + 3O2 XeF6 + 3H2O XeO3 + 6HF 12. Why is He used in diving apparatus ? Because He gas is very light and its solubility in water is very less.

13. Why is it difficult to study the chemistry of Radon ? Because it is radioactive . 14. Give the formula and describe the structure of a noble gas species which is isostructural with a) ICl4

- b) IBr2- c) BrO3

- a) XeF4 : Square pyramidal with 2 lone pairs above and below the plane . b) XeF2 : Linear with 3 lone pairs occupying the equatorial position . c) XeO3 : Pyramidal with a lone pair electron at the apex. ******************