CIC – 1 S-BLOCK, P-BLOCK, HYDROGEN - einstein …einsteinclasses.com/Bluetooth...

CIC – 1

Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road

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S-BLOCK, P-BLOCK, HYDROGEN

S-Block :

Q. What are the common physical and chemical features of alkali metals ?

A. Chemical features : (i) They are highly reactive metals and hence do not occur free in nature. (ii)They evolve hydrogen violently with water : 2M + H

2O 2MOH + H

2(g) (iii) They forms hydrides

with hydrogen : 2M + H2 2MH (iv) They form oxides at moderate temperature : 4M + O

2 2M

2O

(v) They combine directly with Cl2 and S : 2M + Cl

2 2MCl, 2M + S

M2S (vi) They form

hydrated ions in solution and their solution is electrically conductive. (vii) Their oxides are stronglybasic. (viii) Nearly all the alkali metal salts are soluble in water and are thermally stable

Q. Why are alkali metals not found in nature ?

A. The alkali metals do not occur free in nature. This is because they have very low ionization enthalpyand form electropositive ions

Q. Find out the oxidation state of sodium in Na2O

2.

Q. Explain, why is sodium less reactive than potassium.

A. This is because the ionization enthalpy (iH) of potassium (496 kJ mol–1) is less than that of sodium

520 kJ mol–1

Q. Why are potassium and caesium, rather than lithium used in photoelectric cells ?

A. Potassium and caesium have much much lower ionization enthalpy than that of lithium. As a result,when light falls on these metals they easily emit electrons as compared to lithium metal. Because ofthis potassium and caesium can be used in photoelectric cells as compared to lithium.

Q. Why are the elements of Group I called alkali metals ?

A. Group I elements are called alkali metals because their hydroxides are strongly alkaline

Q. Group I elements form largely univalent ions. Explain.

A. Group I elements form largely univalent ions, because after the removal of first electron, noble gasconfiguration is left, from which it is difficult to remove another electron.

Q. Why are alkali metals act as strong reducing agents ?

A. Alkali metals are strong reducing agents because they readily lose the lone valence electrons, owingto their low ionization enthalpy, low heat of automization and high heat of hydration.

Q. Why are alkali metals electrically conducting ?

A. Alkaline metals have metallic bond in their crystal lattice in which cations or kernels are held infixed positions in space, while free electrons can move under the influence of electric field.

Q. Why sodium is stored under kerosene oil ?

A. Because sodium is very reactive

Q. Name the metal which floats on water without any apparent reaction with water.

A. Lithium floats on water without any apparent reaction with water

Q. How the hydration enthalpy of alkali metals vary ?

A. The hydration enthalpies of alkali metals decrease with increase in ionic sizes.Li+ > Na+ > K+ > Rb+ > Cs+.

Q. Compare the alkali metals and alkaline earth metals with respect to (i) ionization enthalpy(ii) basicity of oxides and (iii) solubility of hydroxides.

A. Ionization enthalpy (iH) : Because of higher nuclear charge on alkaline earth metals as compared

to alkali earth metals, the iH of alkaline earth metals are higher than those of corresponding alkali

metals. (ii) Basicity of oxides : Since the electropositive character of alkali metals is higher than thatof the corresponding alkaline earth metal, therefore, the metal-hydroxyl bond formed in alkali met-als can ionize more easily, than the alkaline earth metals, hence alkali metal hydroxides are morebasic than the corresponding alkaline earth metals oxides. (iii) Solubility of hydroxides : Because ofsmaller size and higher ionic charge, the lattice enthalpies of alkaline earth metals are higher thanthose of alkali metals and hence the solubility of alkali metal hydroxides is much less than that of

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alkaline earth metal hydroxides. (As a general rule higher the lattice enthalpies lower is thesolubility.

Q. In what ways lithium shows similarities to magnesium in its chemical behaviour.

A. This is because of charge size ratio of lithium and magnesium is almost equal, i.e., rLi+ = 76 pm andrMg2+ = 72 pm. This type of similarity is also known as diagonal relationship.

Q. Explain, why can alkali and alkaline earth metals not be obtained by chemical reduction methods ?

A. Alkali metals and alkaline earth metals are strong reducing agents themselves, because of this theyare not obtained by chemical reduction process. Note : They act as strong reducing agents becausethey readily lose the valence electrons, owing to their low ionisation enthalpy, low heat of atomisationand high heat of hydration

Q. When an alkali metal dissolves in liquid ammonia the solution can acquire different colours. Explainthe reasons for this type of colour change.

A. The alkali metals dissolve in liquid ammonia to give blue to bronze colour. The blue colour is attrib-uted due to the presence of solvated electrons e.g., [e(NH

3)

4]– in dilute solution. M + (x + y)NH

3

[M(NH3)x]+ + [e(NH

3)

y]–. However, in concentrated solutions, the ammoniated metal ions are bound

by the free unpaired electrons which gives bronze colour. The blue solutions are paramagnetic, whereasthe bronze coloured solution are diamagnetic.

Q. Why are lithium salts commonly hydrated and those of the other alkali metal ions usuallyanhydrous ?

A. Lithium has the smallest size amongst all alkali metals, hence Li+ can polarise water molecules moreeasily as compared to other alkali metal ions. Because of this lithium salts are commonlyhydrated and that of other alkali salts are anhydrous

Q. Why is LiF almost insoluble in water whereas LiCl is soluble not only in water but also in acetone ?

A. In general, a salt is soluble in water when H hydration > H lattice energy i.e., LiF is almostinsoluble in water because the H hydration of LiF is much less than its H lattice energy. WhereasLiCl is soluble not only in water but also in acetone because H hydration of LiCl is much higherthan its H lattice energy

Q. Explain the significance of sodium, potassium, magnesium and calcium in biological fluids.

A. Monovalent sodium and potassium ions and divalent magnesium and calcium ions are found inlarge proportions in biological fluids. These ions perform important biological functions such asmaintenance of ion balance and nerve impulse conduction.

Q. What happens when

(i) sodium metal is dropped in water ?

(ii) sodium metal is heated in free supply of air ?

(iii) sodium peroxide dissolves in water ?

A. (i) 2Na(s) + 2H2O(l) 2NaOH (aq) + H

2(g) (ii)

quantityminor22 O(s)Na(g)O2Na(s) ,

quantitymajor222 O(s)NaO

2

1O(s)Na (iii) Na

2O

2(s) + 2H

2O(l) 2NaOH(aq) + H

2O

2(l)

Q. Comment on each of the following observations :

(a) the mobilities of alkali metal ions in aqueous solution are Li+ < Na+ < K+ < Rb+ < Cs+

(b) lithium is the only alkali metal to form a nitride directly.

(c) E0 for M2+ (aq) + 2e– M(s).

(where M = Ca, Sr or Ba) is nearly constant.

A. (a) Smaller the size of the ion larger is the hydration e.g., Li+ being smallest is highly hydrated ascompared to others and hence smallest ionic mobility. The order of size is Li+ > Na+ > K+ > Rb+.Therefore, ionic mobility increases in the order as Li+ < Na+ < K+ < Rb+. (b) Because of the diagonalrelationship of Li and Mg, Li, like Mg forms nitride, whereas other alkali metals do not give nitrides.(c) The electrode potential of any electrode of the type M2+/M depends upon three factors, i.e., (i)

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ionization enthalpy (ii) enthalpy of vaporization and (iii) enthalpy of hydration. As the combinedeffect of all the three factors is approximately the same for Ca, Ba and Sr, therefore, their electrodepotential is nearly constant.

Q. Name the alkali metals which form superoxides when heated in excess of air.

A. Potassium, rubidium and caesium forms superoxides of the type (KO2, RbO

2) and CsO

2), when

heated in excess of air.

Q. Caustic alkalies like NaOH are not stored in aluminium vessels. Explain.

A. Aluminium dissolves in NaOH to form soluble sodium meta aluminate.

2Al(s) + 2NaOH(aq) + 2H2O(l) 2NaAlO

2(aq) + 2H

2(g)

Q. Sodium metal can be used for drying ether but cannot be used for drying ethanol.

A. Sodium metal can be used for drying ether because it does not react with ether and removes water.In case of alcohol, sodium reacts with the alcohol to give sodium alkoxide. Thus, sodium cannot beused for drying alcohol.

Q. Lithium forms predominantly covalent compounds. Explain.

A. The polarising power of Li+ is greatest of all the alkali metals. This is why it has a tendency to formpredominantly covalent compounds.

Q. The basic strength of alkali metal hydroxides increases down the group. Explain.

A. The basic strength of alkali metal hydroxides increases down the group because the lattice energygoes on decreasing as we move down the group.

Q. Arrange the following in order of the increasing covalent character MCl, MBr, MF, MI(where M = alkali metal).

A. MF < MCl < MBr < MI .

Q. Why is KO2 paramagnetic ?

A. The superoxide O2– is paramagnetic, because of one unpaired electron in *2p molecular orbital

Q. Why metals like potassium and sodium cannot be extracted by reduction of their oxides by carbon;while lead, iron can be extracted ?

A. Potassium and sodium are strong electropositive metals and has great affinity for oxygen than thatof carbon. Hence, they cannot be extracted from their oxides by reduction with carbon. On the otherhand, lead and iron are less electropositive and possess lesser affinity for oxygen than carbon. Hence,they can be extracted by reduction with carbon.

Q. Group II hydroxides are weaker bases than Group I hydroxides.

A. Group II hydroxides are weaker bases than Group I hydroxides because the inter nuclear distancebetween the oxygen and the metal ion is less in Group II hydroxides. The increased charge density inGroup II metal ions also result in a stronger electrostatic attraction between the hydroxide ion andthe metal ion. This result in weaker dissociation of M(OH)

2.

Q. Discuss the various reactions that occur in the Solvay Process.

Q. Potassium carbonate cannot be prepared by Solvay process. Why ?

A. This is because potassium carbonate being highly soluble than sodium bicarbonate does not getprecipitated when CO

2 gas is passed through a concentrated solution of KCl saturated with

ammonia.

Q. Why is Li2CO

3 decomposed at a lower temperature whereas Na

2CO

3 at higher temperature ?

A. Li2CO

3 is formed with a weak acid CO

2 and a weak base LiOH. Both being weak cannot attract each

other strongly, therefore, Li2CO

3 decomposes at a lower temperature. However, NaOH being a stronger

base attract weak acid CO2 to form a stable Na

2CO

3 and hence decomposes at much higher tempera-

ture than Li2CO

3.

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Q. Starting with sodium chloride how would you proceed to prepare

(i) sodium metal (ii) sodium hydroxide (iii) sodium peroxide

(iv) sodium carbonate

A. (i) Sodium metal is prepared by Down’s method. In this 40%, NaCl mixed with 60% CaCl2 in the

molten state electrolysed at about 850 K using carbon (graphite) anode and an iron (steel) cathode.Chlorine is evolved at anode and sodium metal is obtained at cathode. (ii) Sodium hydroxide (NaOH)may be prepared by treating sodium metal with water. 2Na + 2H

2O 2NaOH + H

2(g). (iii) Sodium

peroxide (Na2O

2) is prepared by passing air over heated sodium metal. 2Na + O

2 (from air) Na

2O

2

(iv) Sodium carbonate (Na2CO

3) is prepared by passing CO

2 gas through NaOH solution. 2NaOH +

CO2 Na

2CO

3 + H

2O

Q. Describe two important uses of each of the following :

(i) caustic soda (ii) sodium carbonate (iii) quicklime

A. (i) Important uses of caustic soda : (a) It is used in the manufacture of soap, paper, artificial silk,dye-stuffs and a number of chemicals. (b) It is used for mercerizing cotton fabrices, in petroleumrefining and decolourisation of fats and oils and for absorption of acidic gases. (ii) Importance ofsodium carbonate : (a) It is used in glass, soap, paper, textile, paint and dye industry. (b) It is used inthe preparation of borax, caustic soda, sodium phosphate. (c) It is used in water softening, launder-ing and cleaning. (iii) Importance of quicklime : (a) Large quantities is used in theproduction ofslaked lime. (b) It is used as a constituent of mortar in building construction. (c) It is also used asdying agent as such or as sodalime.

Q. The hydroxides and carbonates of sodium and potassium are easily soluble in water; while thecorresponding salts of magnesium and calcium are sparingly soluble in water. Explain.

A. The hydroxides and carbonates of sodium and potassium are easily soluble in water, because theyhave low lattice energies owing to their smaller charge density and larger atomic size. Polar watermolecules can easily break the electrostatic force of attraction in their ionic crystal lattice. Moreover,these metals possess high heats of hydration. On the other hand, hydroxides of Mg and Ca aresparingly soluble in water, because they have higher lattice energies owing to their large chargedensity and smaller atomic size than sodium and potassium. Polar molecules can thus break onlypartially the electrostatic forces of attraction in their ionic crystal lattices. Moreover, these metalsposses low heats of hydration.

Q. State as to why

(a) a solution of Na2CO

3 is alkaline ?

(b) alkali metals are prepared by the electrolysis of their fused chlorides ?

(c) sodium is found to be useful than potassium ?

A. (a) A solution of sodium carbonate is alkaline in nature as it gives OH– and HCO3 ions in solution.

CO32– + H

2O HCO

3– + OH–. (b) Alkali metals are strong reducing agent and hence cannot be

isolated by reduction of their oxides or other compounds. Hence, they are prepared by theelectrolytic reduction of their fused chlorides. (c) Sodium is found more useful than potassiumbecause of its large availability and cheapness.

Q. Write balanced equation for reactions between

(a) Na2O

2 and water

(b) KO2 and water

(c) Na2O and CO

2

A. (i) Na2O

2 + 2H

2O 2NaOH + H

2O

2 (ii) 2KO

2 + 2H

2O 2KOH + H

2O

2 + O

2 (iii) Na

2O + CO

2

Na2CO

3

Q. How would you explain the following observations ?

(i) BeO is almost insoluble but BeSO4 in soluble in water.

(ii) BaO is soluble but BaSO4 is insoluble in water.

(iii) LiI is more soluble than KI in ethanol.

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A. (i) We know that a salt dissolves in water when H hydration > H lattice enthalpy. BeO is insolublebecause its H lattice enthalpy is greater than H hydration. BeSO

4 is soluble in water because its

H lattice enthalpy is less than H hydration (ii) In case of BaO, H hydration > H latticeenthalpy. In case of BaSO

4, H hydration < H lattice energy. (iii) LiI is covalent whereas KI is

electrovalent. Like dissolves like ethanol is a covalent compound.

Q. Which of the alkali metal is having least melting point ?

(a) Na (b) K (c) Rb (d) Cs

A. d, as the size of the alkali metal increases, the strength of metallic bonding decreases

Q. Which one of the following alkali metals gives hydrated salts ?

(a) Li (b) Na (c) K (d) Cs

A. a

Q. Which one of the alkaline earth metal carbonates is thermally the most stable ?

(a) MgCO3

(b) CaCO3

(c) SrCO3

(d) BaCO3

A. d

Q. Name a few important uses of sodium bicarbonate.

A. (i) in making baking powder, (ii) in fire extinguishers, (iii) in laboratory as a reagent, (iv) forpreparing pure Na

2CO

3, (v) in medicine to neutralise stomach acidity.

Q. Explain what happens when

(i) sodium hydrogen carbonate is heated ?

(ii) sodium amalgam reacts with water ?

(iii) fused sodium metal reacts with ammonia ?

A. (i) When sodium hydrogen carbonate is heated it decomposes to produce sodium carbonate, water

and carbon dioxide. 22323 COOHCONaNaHCO2

(ii) When sodium amalgam reacts with

water, it produces sodium hydroxide and dihydrogen gas is evolved.

)g(HHg2NaOH2OH2NaHg2 22 . (iii) When fused sodium metal reacts with ammonia,

the colour of the solution becomes blue due to the formation of solvated electron.

y3x33 )NH(e)NH(NaNH)yx(Na

Q. State as to why

(i) lithium on being heated in air mainly forms the monoxide and not the peroxide ?

(ii) an aqueous solution of sodium carbonate gives alkaline tests ?

(iii) sodium is prepared by electrolytic method and not by chemical method ?

A. (i) OLiO2

1Li2 22

(ii) OHHCOOHCO 3223 (iii) Sodium being a powerful reducing agent cannot be

isolated by ordinary reducing agent like C or CO. Hence, it is prepared by electrolytic reducingmethod.

Q. Discuss the general charcteristics and gradation in properties of alkaline earth metals.

Q. Beryllium and magnesium do not give colour to flame whereas other alkaline earth metals do so.Why ?

A. In case of Be and Mg, the energy required to excite the electron is very high because of their highionization energies and, therefore, their salts do not impart colour to the flame.

Q. Bones contain calcium ions. What do you think would be the anions associated with it ?

A. Bones and teeth are made of calcium phosphate, e.g., anion is PO43–

Q. Why are alkaline earth metals smaller in size than the alkali metals ?

A. This is because of greater effective nuclear charge in alkaline earth metals.

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Q. Why does beryllium forms mainly covalent compounds ?

A. The charge/radius ratio being very high, beryllium has a considerably polarising power. Thus mostof its compounds acquire covalent character.

Q. Name the soluble fluorides of Group I and Group II.

A. LiF (Group I) and BeF2 (Group II) are insoluble fluorides.

Q. Name the alkaline earth metal present in chlorophyl.

A. Magnesium

Q. Among the Group II hydroxides which is the most soluble in water ?

A. Ba(OH)2. The solubility of hydroxides follow the order : Be(OH)

2 < Mg(OH)

2 < Ca(OH)

2 < Sr(OH)

2

< Ba(OH)2

Q. The thermal stability of alkaline earth carbonates follow the order :

BeCO3 < MgCO

3 < CaCO

3 < SrCO

3 < BaCO

3. Explain giving reasons.

A. With increase in size the metals become more basic. The metal carbonates thus become more stableto heat with increase in size of the cation.

Q. Compare the solubility and thermal stability of the following compounds of the alkali metals withthose of the alkaline earth metals :

(a) Nitrates (b) Carbonates (c) Sulphates.

A. (a) Nitrates : (i) Solubility : Both Group 1 and Group 2 metal nitrates are highly soluble in water. (ii)Thermal stability : Group 1 nitrate decomposes on heating. 4LiNO

3 2Li

2O + 4NO

2 + O

2, 2NaNO

3

2NaNO2 + O

2. Other Group 1 nitrate also give nitrite and O

2 only. Whereas Group 2 nitrates on

heating give metal oxide, NO2 and O

2. 2M(NO

3)

2 2MO + 4NO

2 + O

2 (M = Be, Mg, Ca, Sr or Ba) (b)

Carbonates : (i) Solubility : Group 1 metal carbonates are highly soluble in water, whereas Group 2metal carbonates are sparingly soluble in water. (ii) Thermal stability : Group 1 metal carbonatesare highly stable to heat whereas Group 2 metal carbonates stability increases down the group. (c)Sulphates : (i) Solubility : All the sulphates of Group 1 are soluble in water; whereas among Group2 sulphates, BeSO

4 and MgSO

4 are readily soluble; and then the solubility decreases from CaSO

4 to

BaSO4. (ii) Thermal stability : The sulphates of Group 1 and Group 2 are quite stable.

Q. What happens when (i) magnesium is burnt in air (ii) quick lime is heated with silica (iii) chlorinereacts with slaked lime (iv) calcium nitrate is heated ?

A. (i) oxideMagnesium

2 )s(MgO2)g(O)s(Mg2

(ii) silicateCalcium

3Δ

Silica2

Quicklime

(s)CaSiO(s)SiOCaO(s) (iii) O2HCa(OCl)CaCl2Cl2Ca(OH) 2powderBleaching

22Δ

2limeSlaked

2

(iv) )g(O)g(NO4)s(CaO2)s()NO(Ca2 2223

Q. Draw the structure of (i) BeCl2 (vapour), (ii) BeCl

2 (solid).

A. (i) Structure of BeCl2 in vapour phase :

monomerClBeCl (ii) Structure of BeCl

2 in solid state :

Q. Describe the importance of the following : (i) limestone (ii) cement (iii) plaster of paris.

A. (i) Limestone : (a) It is widely used as building material and road aggregate. (b) It is used in themanufacture of quicklime. (c) It is used in the manufacture of iron. (d) It is a constituent oftooth-paste. (ii) Importance of cement : (a) It is very important next to iron and steel. (b) It is used inreinforced concrete for construction. (c) If is also used in plastering walls and in the construction ofbuilding, dams and bridges. (iii) Importance of Plaster of Paris : (a) It is used in surgical bandagesand in casting and moulding. (b) It is also used in density, in ornamental work and for taking casts ofstatues and busts.

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Q. Why does the solubility of alkaline earth metal hydroxides in water increase down the group ?

A. Among alkaline earth metal hydroxides, the anion is common e.g., OH–, the cationic radius wouldinfluence the lattice enthalpy. Since lattice enthalpy decreases much more than the hydrationenthalpy with increase in ionic size, the solubility of alkaline earth metal hydroxides in water in-creases as we go down the group.

Q. Why does the solubility of alkaline earth metal carbonates and sulphates in water decrease down thegroup ?

A. The size of anions e.g., carbonates and sulphates being much larger compared to cations (e.g., Be,Mg, Ca, Sr etc.), the lattice enthalpy will remain almost constant within a particular group. Since thehydration enthalpies decrease down the group, solubility of alkaline earth metal carbonates andsulphates will decrease.

Q. Why can zinc oxide be reduced to the metal by heating with carbon, but not chromium oxide ?

A. Carbon has greater affinity for oxygen than zinc because the decrease in free energy for the forwardreaction (ZnO + C Zn + CO) is more than the reverse reaction (Zn + CO ZnO + C). On theother hand, chromium has greater affinity for oxygen than carbon because the decrease in freeenergy for the backward reaction (Cr

2O

3 + 3C 2Cr + 3O) is more than that of forward reaction

(2Cr + 3CO Cr2O

3 + 3C) in the case of chromium oxide.

Q. Mention the main constituents of Portland cement.

A. (i) Limestone (ii) Clay (Al2O

3 + SiO

2 mixture) (iii) Powdered coal for heating (iv) Gypsum; CaSO

4.2H

2O

Q. What property of cement is utilised in the constructional activities ?

A. When cement is mixed with water and allowed to stand for sometimes, it sets to a hard rock-likemass.

Q. Write (i) which barium salt is insoluble in water ? (ii) which lead salt is soluble in water ? (iii) whichhydroxide is insoluble in water ? (iv) which iron salt are soluble in water ?

A. (i) BaSO4, (ii) Pb(NO

3)

2, (iii) Al(OH)

3, Fe(OH)

3, Cr(OH)

3 (iv) FeSO

4, Fe(NO

3)

2, FeCl

3

Q. A piece of burning magnesium ribbon continuous to burn in sulphur dioxide. Explain.

A. Magnesium is an active metal and it burns in sulphur dioxide by extracting oxygen of SO2 to form

MgO and sulphur. 2Mg + SO2 2MgO + S + heat

Q. Beryllium chloride gives acidic solution when dissolved in water. Explain.

A. When BeCl2 is added to water, it hydrolyses as follows : BeCl

2 + 2H

2O Be(OH)

2 + 2HCl. The

solution is acidic in nature due to the formation of HCl.

Q. How is Plaster of Paris prepared ? Describe its chief property due to which it is widely used.

A. Plaster of Paris is prepared by heating gypsum CaSO4.2H

2O at 393 K.

OH3OH.]CaSO[]OH2.CaSO[2 2ParisofPlaster

224K393

gypsum24 . Its chief property due to which it is widely used is

that when it is mixed with one-third of its weight of water it sets with expansion into a hard mass ofinterlocking crystals of gypsum within 5-15 minutes. The setting is due to hydration of Plaster ofParis into gypsum.

Q. Contrast the action of heat on the following and explain your answer :

(i) Na2CO

3 and CaCO

3

(ii) MgCl2.6H

2O and CaCl

2.6H

2O

(iii) Ca(NO3)

2 and NaNO

3

A. (i) Na2CO

3 is a stable compound whereas CaCO

3 is an unstale compound. Effect of heat (a)

OH9OH.CONaOH10.CONa 2emonohydrat232

K373

edecahydrat232 , OHCONaOH.CONa 232

K373232

.

Na2CO

3 is called soda ash and does not decompose on heating (b) 2

K12003 COCaOCaCO (ii)

(a) MgCl2.6H

2O on heating give MgO, water and HCl. HCl2OH5MgOOH6.MgCl 2

heat22 .

Anhydrous MgCl2 cannot be prepared by heating MgCl

2.6H

2O (b) CaCl

2.6H

2O on heating gives up

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its water of crystallization. OH6CaClOH6.CaCl 22heat

22 . Because of this CaCl2 is used as

dehydrating agent. (iii) (a) Ca(NO3)

2 on heating decomposes into CaO, NO

2 and O

2.

22heat

3 ONO4CaO2CaNO2 (b) NaNO3 on heating decomposes into NaNO

2 and O

2.

223 ONaNO2NaNO2

Q. Complete the following equations for the reaction between :

(i) Ca + H2O

(ii) Ca(OH)2 + Cl

2

(iii) BeO + NaOH

(iv) BaO2 + H

2SO

4

A. (i) 22heat

2 H)OH(CaOH2Ca (ii) OHCaOClCl)OH(Ca 2powderbleaching

222

(iii) OHBeONaNaOH2BeO 2berylatesodium

22 (iv)

peroxidehydrogen

22

sulphatebarium

4422 OHBaSOSOHBaO

P-Block :

Q. Boron does not usually form a cation.

A. B has (IE)1 = 801 kJ mol–1, (IE)

2 = 2419 kJ mol–1, (IE)

3 = 3646 kJ mol–1. Thus the total energy required

to give B3+ ions is far more than that which would be compensated by lattice energies of ioniccompounds or hydration of such ions in solution. Thus formation of cation (like B3+) is not possible

Q. The polarity of B — X bonds is in the order : B — F > B — Cl > B — Br but Lewis acidicity shows thesequence : BF

3 < BCl

3 < BBr

3.

A. With increase in polarity of B — X bond, acidity also increases and should be thus in the order :BF

3 > BCl

3 > BBr

3. But Lewis acidity is in reverse order : BF

3 < BCl

3 < BBr

3. There is lateral overlap

of the vacant 2p orbital of B with one completely filled orbital of F leading to p – p bonds betweenB and F. This B — F bond thus acquires double bond character. This also leads to compensateelectron deficiency of boron and thus Lewis acid character of BF

3 is reduced. This p - p bonding

decreases going from BF3 to BBr

3 and thus Lewis acidic nature increases in the order : BF

3 < BCl

3 <

BBr3

Q. Discuss the pattern of variation in the oxidation states of (i) B to Tl (ii) C to Pb.

A. (i) The first two elements e.g., B and Al exhibit an oxidation state of +3 because of the presence of twoelectrons in s- and one electron in the p-orbital of valence shell. On the other hand, all the otherelements from Ga and Tl contain only d-and f-electrons and hence, exhibit oxidation states of +1 and+3 due to inert pair effect. (ii) The first two elements e.g., C and Si exhibit an oxidation state of +4because of the presence of two electrons in s- and two electrons in the p-orbital of valence shell. Onthe other hand, all the other elements from Ge to Pb contain d- or d- and f-electrons and henceexhibit two oxidation states +2 and +4 due to inert pair effect.

Q. How can you explain higher stability of BCl3 as compared to TlCl

3 ?

A. Because of the absence of d- and f- electrons in B, all the three valence electrons e.g., two 2s and one2p, take part in the bond formation, showing an oxidation state of +3 in the formation of BCl

3. On

the other hand, because of the poor shielding of the s- electrons of the valence shell by the 3d-,4d- and 4f- electrons, inert pair effect is maximum in Tl. As a result, only 6p1 electron take part in thebond formation and hence, the most stable state of Tl is +1 rather than +3. Thus, BCl

3 is more stable

than TlCl3.

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Q. Why does boron trifluoride bahave as a Lewis acid ?

A. Boron trifluoride is an electron deficient species

Q. Consider the compounds, BCl3 and CCl

4. How will they behave with water ? Justify.

A. BCl3(s) + 3H

2O(l) H

3BO

3(s) + 3HCl(l). On the other hand, carbon tetrachloride is an electron rich

species, hence, it cannot accept any lone pair of electrons from H2O and hence, does not undergo

hydrolysis.

Q. Is boric acid a protic acid ? Explain.

A. No, boric acid is not a protic acid, as it does not ionise water to give a proton (H+). On the other hand,boric acid accepts a lone pair of electronto act as Lewis acid :

Q. Explain, what happens when boric acid is heated.

A. Boric acid, when heated, loses water at three different stages at different temperature to tiveborontrioxide.

Q. Describe the shapes of BF3 and [BF

4]–. Assign the hybridisation of boron in these species.

A. sp2 and sp3 respectively .

Q. Write reactions to justify amphoteric nature of aluminium.

A. It reacts, both with acids and alkalines to evolve dihydrogen, hence, it is amphoteric in nature.

)(OH3)aq()SO(Al)aq(SOH3)s(Al2 234242 l ,

(III)minatehydroxoalutetraSodium242 O(g)3H(aq)][Al(OH)2NaO(l)6H2NaOH(aq)2Al(s)

Q. What are electron deficient compounds ? Are BCl3 and SiCl

4 electron deficient species ? Explain.

A. Species in which the central atom either does not have eight electrons in the valency shell or thosewhich have eight electrons in the valency shell but can expand their covalency beyond four due tothe presence of d-orbitals are called electron deficient compound. Example : (i) In BCl

3, the central

boron atom has only six electrons, therefore, it is an electron deficient compound. (ii) In SiCl4, the

central silicon atom has 8 electrons it can expand its covalency beyond 4 due to the presence ofvacant d-orbitals. Therefore, according to the definition, SiCl

4 should also be taken as electron defi-

cient compound but, in fact, it does not accept two more Cl– ions to form [SiCl6]2–. Hence, it is not an

electron deficient molecule.

Q. Suggest a reason why the B–F bond length in BF3 (130 pm) and BF

4– (143 pm) differ.

A. In BF3, boron is sp2-hybridised and hence it is a polar molecule. It has an empty 2p- orbital. F-atom

also have lone pair of electrons in 2p-orbitals. Because of similar sizes, p-p back bonding takesplace in which free lone pair of fluorine is transferred to boron. As a result of this back bonding,B – F bond acquires some double bond character. On the other hand in [BF

4]– ion, B is sp3 hybridized

and does not have empty p-orbital to accept the electron donated by F atom. Consequently in[BF

4]–, B – F is purely single bond. We know that double bonds are shorter than single bonds, and

hence, bond length in BF3 is shorter (130 pm) than, B – F bond length (143 pm) in [BF

4]–.

Q. If B–Cl bond has a dipole moment, explain why BCl3 molecule has zero dipole moment.

Q. Aluminium trifluoride is insoluble in anhydrous HF but dissolves on addition of NaF. Aluminiumtrifluoride precipitates out of the resulting solution when gaseous BF

3 is bubbled through. Give

reasons.

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A. (a) Aluminium trifluoride is insoluble in anhydrous HF, because HF is strongly H-bonded covalentcompound and it does not give F– ions to dissolve AlF

3. On the other hand, KF is an ionic compound,

gives F– ions which combines with AlF3 to give soluble complex.

(III)aluminatehexafluorosodiumSoluble

633 ][AlFNaAlF3NaF

(b) Because of smaller size and higher electro-negativity of B, it has much higher tendency to formcomplexes than aluminium, therefore, when BF

3 is bubbled through soluble complex of aluminium,

AlF3 gets precipitate with the formation of soluble complex of boron.

(s)AlF]3Na[BF3BF][AlFNa 3

(III)oboratetetrafluorsodium.Soluble

4363 .

Q. What happens when

(a) Borax is heated strongly, (b) Boric acid is added to water,

(c) Aluminium is treated with dilute NaOH, (d) BF3 is reacted with ammonia ?

A. (a) When borax is heated strongly, a transparent bead which consists of sodium metaborate and

boric anhydride is formed.

beadglassyttransparen

rideBoricanhydmetaborate.Sod3222742

Borax2742 OBNaBO2OH10OBNaOH10.OBNa

(b) Boric acid acts as Lewis acid by accepting hydroxyl ion of water and releasing a proton into thesolution. HOH + B(OH)

3 [B(OH)

4]– + H+ (c) Sodium tetrahydroxoaluminate (III) is formed

alongwith the evolution of dihydrogen. 2Al(s) + 2NaOH(aq) + 6H2O(l) 2Na+[Al(OH)

4]–(aq) + 3H

2(g)

(d) BF3 being a Lewis acid accpets a pair of electrons from NH

3 to form a complex.

3Complexbase'LewisacidLewis

333 NHBFNH:BF .

Q. Explain the following reactions

(a) Silicon is heated with methyl chloride at high temperature in the presence of copper.

(b) Silicon dioxide is treated with hydrogen fluoride,

(c) CO is heated with zinc oxide,

(d) Hydrated alumina is treated with aqueous NaOH solution.

A. (a) When silicon is heated with methyl chloride at high temperature in the presence of copper,

dimethylchlorosilane is formed : lorosilaneDimethylch

223K570

powderCu3 SiCl)CH(SiClCH2 (b) When SiO

2 reacts with

HF, silicon tetrafluoride is formed. This dissolves in excess of HF to give hydrofluorosilicic acid.

acidosilicicHydrofluor624242 SiFHHF2SiF,OH2SiFHF4SiO (c) When C is heated with ZnO,

ZnO is reduced to zinc metal. ZnO + C Zn + CO. (d) When hydrated alumina is treated with

aqueous NaOH solution, soluble metaaluminate is formed.

O3H2NaAlO2NaOH(aq)O(s).2HOAl 2atemetaaluminSod.

2heat

232 or

(III)exoaluminattetrahydroSod.4

heat2232 ](aq)2Na[Al(OH)O(l)H2NaOH(aq)O(l).2HOAl

Q. Give reasons :

(i) Conc. HNO3 can be transported in aluminium container.

(ii) A mixture of dilute NaOH and aluminium pieces is used to open drain.

(iii) Graphite is used as a lubricant.

(iv) Diamond is used as an obrasive.

(v) Aluminium alloys are used to make aircraft body.

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(vi) Aluminium utensils should not be kept in water overnight.

(vii) Aluminium wire is used to make transmission cables.

A. (i) Conc. HNO3 reacts with aluminium to give a very thin film of aluminium oxide, which protects it

form further action. )O(3H(g)6NO(s)OAl(conc.)6HNO2Al(s) 22oxideAluminium

323 l . Thus, alu-

minium becomes passive and hence aluminium container can be used to transport conc. HNO3. (ii)

Dilute NaOH reacts with aluminium pieces to give dihydrogen. The dihydrogen has a highpressure which can be used to open clogged drains.

)g(H3)aq(NaAlO2)(OH22NaOH(aq)2Al(s) 222 l . (iii) Graphite has a layered structuree

in which the layers are held together by weak van der Waal’s forces and hence can made to slip oneover another. Therefore, graphite can be used as a lubricant. (iv) Diamond is the hardest materialknown, hence can be used as abrasive. (v) Aluminium alloys such as magnelium is light, tough andresitant to corrosion and hence are being used to make aircraft body. (vi) A thin film of Al

2O

3 is

produced by Al in presence of water and oxygen. A small part of this film dissolve in water to giveAl3+ ions, which are injurious to health. (vii) On weight to weight basis, aluminium conducts twice ascopper. Therefore, it is used in transmission cables.

Q. A certain salt X gives the following results.

(i) Its aqueous solution is alkaline to litmus.

(ii) It swells up to a glassy material Y on strong heating.

(iii) When conc. H2SO

4 is added to a hot solution X of white crystal of an acid Z separates out.

Write equations for all the above reactions and identify X, Y and Z.

A. (ii)

materialglassy)y(322

2742heat

)x(2742

OBNaBO2

OH10OBNaOH10.OBNa

heat

(iii) OH5SONaBOH4SOHOH10.OBNa 242)Z(

3342)X(

2742

Q. Write balanced equations for :

(i) LiHBF3 (ii) OHHB 262

(iii) 62HBNaH (iv)

33BOH

(v) NaOHAl (vi) 362 NHHB

A. (i) Diborane

62 LiF6HB (ii) acidOrthoboric

233 H6BOH2 (iii) eborohydridSodium

4 ]BH[Na2 (iv) acidMetaboric

22K370

33 OHHBOBOH

(v)

(III)minatehydroxoalutetraSodium

24 H3])OH(Al[Na2 (vi) 2BH3NH

3

Q. Boric acid is polymeric due to

(a) its acidic nature (b) the presence of hydrogen bonds

(c) its monobasic nature (d) its geometry

A. (b). Hint : Boric acid is polymeric due to the presence of hydrogen bonds

Q. The type of hybridisation of boron in diborane is

(a) sp (b) sp2 (c) sp3 (d) dsp2

A. c

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Q. If the starting material for the manufacture of silicones is RSiCl3, write the structure of product

formed.

A.

Q. Write the resonance structures of CO32– and HCO

3–.

Q. What is the state of hybridisation of carbon in (a) CO32– (b) diamond (c) graphite ?

A. (a)CO32– is sp2 hybridised. (b) diamond is sp3 hybridised. (c) graphite is sp2 hybridized

Q. Explain the difference in properties of diamond and graphite on the basis of their structures.

A. (i) Diamond is an extremely hard substance, whereas graphite is soft. Explanation : The compactthree dimensional structure in which carbon atoms are tetrahedrally bonded (sp3 hybridisation) tothe other carbon atoms by strong single covalent bond accounts for the hardness of diamond.

. Whereas graphite has a sheet of layered structure (sp2 hybridisation)

in which each layer a carbon atom is bonded to only three neighbouring carbon atom by singlecovalent bonds. This result in a hexagonal network of carbon and accounts for its softness.

(ii) Diamond is a non-conductor of electricity while graphite is a good conductor..

Explanation : All the four electrons of each carbon atom in diamond are utilised in the formation offour single covalent bonds and thus does not have free electrons and hence act as non-conductor,whereas in graphite each carbon atom is bonded to only three neighbouring carbon atoms by singlecovalent bonds and one electron being free makes it good conductor of electricity. (iii) Graphite isless dense than diamond. Explanation : In graphite, the distance between two alternate layers isquite large ( 3.40 Å), whereas in diamond the distance between the two adjacent carbon atoms issmall ( 1.54 Å). (iv) Diamond has high melting point, while melting point of graphite iscomparatively much lower. Explanation : Diamond has sp3 hybridisation having compact structure,whereas graphite has sp2 hybridisation having layered structure. Hence, the explanation.

Q. Rationalise the given statements and give chemical reactions :

(i) Lead (II) chloride reacts with Cl2 to give PbCl

4

(ii) Lead (IV) chloride is highly unstable towards heat.

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(iii) Lead is known not to form an iodide, PbI4.

A. (i) Because of inert pair effect, Lead (II) is more stable in +2 than in +4 oxidation state. Hence, Lead(II) chloride does not react with Cl

2 to give Lead (IV) chloride. (ii) This is also because of inert pair

effect. (iii) PbI4 does not exist. This happens because of oxidising power of Pb4+ ion and reducing

power of I– ion.

Q. Suggest a reason as to why CO is poisonous ?

A. Oxygen combines loosely and reversibly with the haemoglobin present in red blood cells to produceoxyhaemoglobin. This oxyhaemoglobin formed in lungs then travel to the different parts of the bodythrough blood steam and supply O

2 to them. However, CO if present, combines with haemoglobin to

give stable carboxyhaemoglobin, which destroys the oxygen carrying capacity of haemoglobin andhence acts as poisonous gas.

Q. How is excessive content of CO2 responsible for global warming ?

A. CO2 is produced through combustion and other industrial process. It is utilised in photosynthesis

process by plant and CO2 is released in the atmosphere. Through CO

2 cycle, a constant percentage of

oxygen (about 21%) is maintained in the atmosphere. However, if the concentration of CO2

increased beyond a limit, some of it will remain unutilised. This excess CO2 absorbs heat radiated by

the earth. Some of it is dissipated into the atmosphere while the remaining part is radiated back tothe earth and other bodies present on the earth. As a result, temperature of the earth and otherbodies on the earth increases. This is called greenhouse effect. As a result of greenhouse effect, globalwarming occurs which has serious consequences.

Q. Explain the structures of diborane and boric acid.

A. (a) Structure of diborane is shown below (a) and (b)

Nature of bonding in B2H

6 : Each B atom uses sp3 hybrids for bonding. Out of the four sp3 hybrids

on each B atom, one is without an electron shown in broken lines. The terminal B-H bonds arenormal 2c-2e bonds but the two bridge bonds are 3c-2e bonds. The 3c-2e bridge bonds are alsoreferred to an banana bonds. (b) The structure of boric acid is shown in figure :

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Structure of boric acid; the dotted lines represent hydrogen bonds.

Q. Explain, why is there a phenomental decrease in ionization enthalpy from carbon to silicon ?

A. This effect is due to increase in atomic size and screening effect. Because of this the force ofattraction of the nucleus for the valence electron decreases in silicon as compared to carbon. As aresult, there is a phenomenal decrease in ionization enthalpy from carbon to silicon.

Q. How would you explain the lower atomic radius of Ga as compared to Al ?

A. Due to poor shielding effect of the valence electrons of Ga by the inner 3d-electrons, the effectivenuclear charge of Ga is higher in magnitude than that of Al. As a result, the electrons in Gaexperience a larger force of attraction by the nucleus than Al and hence atomic size of Ga (135 pm)is slightly less than that of Al (143 pm).

Q. What are allotropes ? Sketch the structure of two allotropes of carbon, namely, diamond andgraphite. What is the impact of structure of physical properties of two allotropes ?

A. The phenomenon of existence of an element in two or more forms which have different physicalproperties but similar chemical properties is called allotropy and the different forms are calledallotropes. (i) Diamond (ii) Graphite.

Q. Classify following oxides as neutral, acidic, basic or amphoteric : CO, B2O

3, SiO

2, CO

2, Al

2O

3, PbO

2,

Tl2O

3.

A. Neutral oxides : CO, Acidic oxides : B2O

3, SiO

2, CO

2, Amphoteric oxide : Al

2O

3, PbO

2, Basic oxide :

Tl2O

3.

Q. When metal (X) is treated with sodium hydroxide, a white precipitate (A) is obtained, which is solublein excess of NaOH to give soluble complex (B). Compound (A) is soluble in dilute HCl to formcompound (C). The compound (A) when heated strongly gives (D), which is used to extract metal.Identify (X), (A), (B), (C) and (D). Write suitable equations to support their identities.

A. Since metal (X) when treated with sodium hydroxide, gives a white ppt. (A), which dissolves inexcess of NaOH to give soluble complex (B), therefore (X) seems to be metal aluminium, ppt. (A)seems to be Al(OH)

3 and complex (B) seems to be sodium tetra hydroxoaluminate (III).

)B(4

)A(33

)X(])OH(Al[NaNaOH)OH(Al,Na3)OH(AlNaOH3Al2 . Since, (A) is

amphoteric in nature, it seems to react with dil. HCl to give (C) which may be AlCl3.

OH3AlClHCl3)OH(Al 2)C(

3)A(

3 . Since (A) on heating gives (D) which is used for metal

extraction, therefore, (D) is alumina (Al2O

3) : OH3OAl)OH(Al2 2

(D)alumina

32)A(

3

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Q. What do you inderstand by :

(a) inert pair effect

(b) allotropy

(c) catenation ?

A. Catenation : It may be defined as the ability of lime atoms to link with one another through covalentbonds. This is due to smaller size and higher electronegativity of carbon atom and unique strength ofC – C bonds.

Q. Give one method for industrial preparation and one for laboratory preparation of CO and CO2

each.

Q. An aqueous solution of borax is

(a) neutral (b) amphoteric (c) basic (d) acidic

A. (c). Hint : Borax is salt made up of weak acid. (H3BO

3) and strong base (NaOH). Therefore, it is basic

in nature

Hydrogen :

Q. Justify the position of hydrogen in the periodic table on the basis of its electronic configuration.

A. There is always confusion because of position of hydrogen in the periodic table. It have theconfiguration 1s1, which is similar to alkali metal, which have the electronic configuration ns1. But itis also similar to halogen family because they also need a one electron to gain noble gasconfiguration.

Q. Write the names of isotopes of hydrogen. What is the mass ratio of these isotopes ?

A. Hydrogen has 3 isotopes : protium (11H), deuterium (2

1H) or D and tritium (3

1H) or T. The mass ratio

of H : D : T is 1 : 2 : 3.

Q. Why does hydrogen occur in diatomic form rather than in monoatomic form under normalconditions ?

Q. How can the production of dihydrogen, obtained from ‘coal gasification’, be increased ?

A. The process of producing synthesis gas or syn gas from coal is called coal glasification.

)g(H)g(CO)s(OH)s(C 2Ni

K1270

Steam2

Coal

Q. Describe the bulk preparation of dihydrogen by electrolytic method. What is the role of anelectrolyte in this process ?

A. This is the best method of manufacturing dihydrogen where electricity is cheep. A small quantity(15-25%) of acid or alkali is added to water to make it a good conductor and electrolysed in a cell.

In this cell, iron sheet is used as a cathode while nickel plated iron sheet acts as anode.

The two electrodes are separated from each other by an asbestors diaphragm which prevents mixingof dihydrogen and dioxygen. On passing electric current, dihydrogen gas is collected at cathodewhile dioxygen at anode. When 20% NaOH solution is used for electrolysis, the decomposition ofwater takes place as follows :

H2O H+ + OH–

At cathode, H+ + e– H

H + H H2

At anode, 4OH– 4OH + 4e– ; 4OH– 2H2O + O

2

Q. Complete the following reactions :

(i)

)s(OM)g(H om2 (ii)Catalyst

2 )g(H)g(CO

(iii)Catalyst

283 )g(OH3)g(HC (iv)

heat)aq(NaOH)s(Zn

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A. (i) mM(s) + oH2O(l) (ii) CH

3OH (l) (iii) 3CO(g) + 7H

2(g) (iv) )g(H)aq(ZnONa 2

ZincateSodium22

Q. Discuss the consequences of high enthalpy of H–H bond in terms of chemical reactivity of dihydrogen.

A. Due to high enthalpy (436 kJ mol–1) of H–H bond, it is quite unreactive at room temperature.However, it combines with many metals and non-metals, at high temperatures and in presence ofcatalyst to form hydrides.

Q. What do you understand by the term ‘auto-protolysis’ of water ? What is its significance ?

A. Auto-protolysis means self ionization of water. It may be represented as

21 Base2

Acid2 )(OH)(OH ll

12 BaseAcid3 )aq(OH)aq(OH

Due to auto-protolysis, water is amphoteric in nature i.e., it reacts with both acids and bases. It actsas a base towards acids stronger than itself and as an acid towards bases stronger than itself. Forexample.

1221 BaseAcid4

Base3

Acid2 )aq(OH)aq(NH)aq(NH)(OH l

2121 BaseAcid3

Acid2

Base2 )aq(HS)aq(OH)aq(SH)(OH l

Q. Consider the reaction of water with F2 and suggest, in terms of oxidation and reduction, which

species are oxidised/reduced ?

A. (aq)4F(aq)4H(g)O)O(2H(g)2F 2Reductant

2Oxidant

2 l (11) (a) AlCl

3 when treated with (i) normal water,,

it undergoes hydrolysis to give Al(OH)3 and HCl (ii) acidic water, the H+ ions react with Al(OH)

3 to

give Al3+(aq) and H2O(l). Thus, in acidic water, AlCl

3 exists as Al3+(aq) and Cl(aq) ions.

)aq(Cl3)aq(Al)s(AlCl 3

waterAcidic3 . (iii) alkaline water, Al(OH)

3 reacts with (OH–) ions to

give tetrahydroxoaluminate ion. exoaluminatTetrahydro

43 (aq)][Al(OH)(aq)OH(s)Al(OH) (b) KCl when treated

with (i) normal water, it does not hydrolyse.

Q. Write chemical reactions to show the amphoteric nature of water.

A. Water is amphoteric in character, i.e., it behaves both as an acid as well as a base. With acids (e.g.H

2S stronger than itself, it behaves as a base and with bases (e.g., NH

3) stronger than itself, it acts as

an acid.

2121 BaseAcid3

Acid2

Base2 )aq(HS)aq(OH)aq(SH)(OH l

1221 BaseAcid4

Base3

Acid2 )aq(OH)aq(NH)aq(NH)(OH l

Q. What properties of water make it useful as a covalent ? What types of compound can it (i) dissolve,and (ii) hydrolysis ?

A. (i) Water has a high dielectric constant (79.39) and high dipole moment (1.84 D). Because of theseproperties, water dissolves most of the inorganic (ionic) compounds and many covalent compounds.That is why water is called a universal solvent. Whereas ionic compounds dissolve in water due toion-dipole interaction or solvation of ions, covalent compounds such as alcohols, amines, urea,glucose, sugar, etc. dissolve in water due to H-bonding.

(ii) Water can hydrolyse many oxides (metallic or non-metallic), hydrides, carbides, nitrides,phosphides and other salts. In these reactions, H+ and OH– ions of water interact with anions andcations respectively leading to the formation of an acid or a base or both as shown below :

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Acetylene222

2222

3222

22

)g(CHHC)aq()OH(Ca)(OH2)s(CaC

)g(H2)aq()OH(Ca)(OH2)s(CaH

)aq(SOH)(OH)g(SO

)aq()OH(Ca)(OH)s(CaO

l

l

l

l

Q. Do you expect different products in solution when aluminium (III) chloride and potassium chloridetreated separately with (i) normal water (ii) acidified water and (iii) alkaline water ? Write equationswherever necessary.

A. KCl is the salt of a strong acid has a strong base. It does not undergo hydrolysis in normal water. Itjust dissociates to give K+ (aq) and Cl– (aq) ion.

)aq(Cl)aq(K)s(KClWater

Since the aqueous solution of KCl is neutral, therefore, in acidified water or in alkaline water, theions do not react further and stay as such.

AlCl3, on the other hand, is a salt of a weak base Al(OH)

3 and a strong acid HCl. Therefore, in

normal water, it undergoes hydrolysis to form Al(OH)3, H+ and Cl– ions.

)aq(Cl3)aq(H3)s()OH(AlOH3)s(AlCl 323

In acidic water, the H+ ions react with Al(OH)3 to form Al3+ (aq) ions and H

2O. Thus, in acidic water,

AlCl3 exists as Al3+(aq) and Cl–(aq) ions.

)aq(Cl3)aq(Al)s(AlCl 3waterAcidified3

In alkaline water, Al(OH)3 reacts to form soluble tetrahydroxoaluminate complex or meta-alumi-

nate ion,

i.e., )O(2H(aq)AlOor(aq)][Al(OH)(aq)OH(s)Al(OH) 2ionaluminatemeta

2exoaluminatTetrahydro

43 l

The complete equation may be written as :

Q. What do you understant by the terms :

(i) hydrogen economy (ii) hydrogenation

(iii) ‘syngas’ (iv) water-gas shift reaction

(v) fuel-cell.

A. (i) Use of liquid hydrogen as an alternative source of energy is called hydrogen economy. This tech-nology involves the production, transportation and storage of energy in the form of liquid hydrogenin bulk quantities. (ii) Hydrogenation is used for the conversion of unsaturated oils into saturatedfats. (iii) Syngas is produced from sewage, saw dust, scrap wood, newspapers etc. The process of

producing syngas from coal is called coal gasification. )g(H)g(CO)g(OH)s(C 2Ni

K12702 . (iv)

By ‘water-gas shift reaction’ we mean that more hydrogen is prepared by water-gas. (v) Fuel-cell isa devide of converting combustion energy directly to electrical energy.

Q. What do you understand by (i) electron-deficient, (ii) electron-precise, and (iii) electron-richcompounds of hydrogen ? Provide justification with suitable examples.

A. (i) Hydrides produced from the elements of group 13 e.g., BH3, AlH

3, etc. do not have sufficient

number of electrons to form normal covalent bonds. Hence, they are called electron deficienthydrides. To make up this deficiency, they exist in polymeric forms i.e., B

2H

6, (AlH

3)

n, etc.

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(ii) Hydrides produced from the element of group 14 e.g., CH4, SiH

4, SnH

4 have sufficient number of

electrons to form normal covalent bonds and hence are called electron-precise hydrides.(iii) Hydrides of group 15, 16 and 17 e.g., NH

3, H

2O, H

2S, HF etc. have more electrons than required

to form normal covalent bonds and hence are called electron-rich hydrides.

Q. What characteristics do you expect from an electron-deficient hydride with respect to its structureand chemical reactions ?

A. Electron deficient hydrides do not have sufficient number of electrons to form normal covalentbonds. Therefore, to make up this deficiency, they react with many metals and non-metals and theircompounds. Because of this, electron deficient hydrides are very reactive i.e., B

2H

6(g) + 3O

2(g)

B2O

3(s) + 3H

2O(g). Being electron deficient, they act as Lewis acid.

Q. Do you expect the carbon hydrides of the type (CnH

2n + 2) to act as Lewis’ acid or base ? Justify your

answer.

A. they do not have a tendency to gain or lose electrons and hence they do not act as Lewis acid or Lewisbases.

Q. Whay do you understand by the term “non-stoichiometric hydrides” ? Do you expect this type of thehydrides to be formed by alkali metals ? Justify your answer.

A. Hydrides in which the ratio of the metal to hydrogen is fractional are called non-stoichiometrichydrides. These hydrides are not formed by alkali metals. This type of hydrides are formed byd- and f-block elements. In these hydrides, the hydrogen atoms occupy holes in the metal lattice.

Q. How do you expect the metallic hydrides to be useful for hydrogen storage ? Explain.

A. n the formation of metallic hydrides, hydrogen atom is absorbed on the metal surface. On heatingmetallic hydrides, hydrogen leaves the metal surface. The hydrogen thus evolved can be used as afuel. Thus, metals can be used to store and transport hydrogen, which can be used as a fuel.

Q. How does the atomic hydrogen or oxy-hydrogen torch function for cutting and welding purpose ?Explain.

A. Atomic hydrogen is produced by passing dihydrogen through an electric arc stuck between tungsten

rods at about 4000 K. 1

K4000

arcElectric2 molkJ436H;H2H . Atomic hydrogen exists for a few

seconds, as its life is about 0.3 seconds, and hence immediately it gets converted into the molecularhydrogen liberating a large amount of heat which is utilised for cutting and welding purpose.Oxy-hydrogen torch produced by passing dioxygen and dihydrogen in presence of active catalystlike spongy platinum. In this situation, explosion is caused, and produces some atomic hydrogen,which start the reaction and then it goes on its own. The temperature produced by oxy-hydrogentorch is very high and hence used for cutting and welding.

Q. Among NH3, H

2O and HF, which would you expect to have highest magnitude of hydrogen bonding

and why ?

A. Nitrogen, oxygen and fluorine have higher electronegativity than hydrogen, hence all these undergohydrogen bonding. As the electronegativity of fluorine is highest amongst N, O and F, therefore,magnitude of positive charge on hydrogen and negative charge on fluorine is the highest and hencethe electrostatic attraction is strongest in H–F.

Q. Saline hydrides are known to react with water violently producing fire. Can CO2, a well-known fire

extinguisher, be used in this case ? Explain.

A. Saline hydrides like NaH, CaH2 etc. react with water violently to give corresponding metal hydrox-

ide, and dihydrogen is liberated. e.g., NaH(s) + H2O(l) NaOH (aq) + H

2(g), CaH

2(s) + 2H

2O(l)

Ca(OH)2(aq) + H

2(g). The reaction being vigorous H

2 evolved catches fire. The fire produces cannot

be extinguished by CO2 as it gets reduced by the hot metal hydrides to give sodium formate.

Q. Arrange the following :

(i) CaH2, BeH

2 and TiH

2 in order of increasing electrical conductance.

(ii) LiH, NaH and CsH in order of increasing ionic character.

(iii) H–H, D–D and F–F in order of increasing bond dissociation enthalpy.

(iv) NaH, MgH2 and H

2O in order of increasing reducing property.

A. (i) BeH2 < CaH

2 < TiH

2 (ii) LiH < NaH < CsH (iii) F–F < H–H < D–D (iv) H

2O < MgH

2 < NaH

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Q. How can saline hydrides remove traces of water from organic compounds ?

A. Saline hydrides like CaH2 and NaH react with water forming corresponding metal hydroxides with

liberation of hydrogen gas. Thus, traces of water present in organic solvents can be easily removedby distilling them over saline hydrides when H

2 escapes into the atmosphere, metal hydroxide is left

in the flask while dry organic solvent distils over.

Q. What do you expect the nature of hydrides is, if formed by elements of atomic numbers 15, 19, 23and 44 with dihydrogen ? Compare their behaviour towards water.

A. (i) covalent hydride PH3 (ii) saline hydride K+H– (iii) metallic hydride VH

3 (iv) it does not form any

hydride.

Q. Knowing the properties of H2O and D

2O, do you think that D

2O can be used for drinking purpose ?

A. Heavy water is injurious to human beings, plants and animals since it slows down the rates ofreactions occurring in them. Thus, heavy water does not support life so well as does ordinary water.

Q. What is the difference between the terms ‘hydrolysis’ and ‘hydration’ ?

A. Interaction of H+ and OH– ions of H2O with the anion and the cation of a salt respectively to give the

original acid and the original base is called hydrolysis. For example,

Acid32

Base2

Salt32 COHNaOH2OH2CONa

Hydration, on the other hand, means addition of H2O to ions or molecules to form hydrated ions or

hydrated salts. For example,

)aq(Cl)aq(Na)(OH)s(ClNa 2Salt

l

)Blue(242

)Colourless(4 )s(OH5.CuSO)(OH5)s(CuSO l

Q. What causes the temperory and permanent hardness of water ?

A. Presence of bicarbonates of calcium and magnesium, i.e., Ca(HCO3)

2 and Mg(HCO

3)

2 in water causes

temporary hardness and presence of soluble chlorides and sulphates of calcium and magnesium, i.e,CaCl

2, CaSO

4, MgCl

2 and MgSO

4 in water causes permanent hardness.

Q. Discuss the principle and method of softening of hard water by snythetic ion-exchange resins.

A. Synthetic ion exchange resins are of two types :

Cation exchange resins and anion exchange resins.

Cation exchange resins are either carboxylic acids or sulphonic acids having the general formula,R-COOH or R-SO

2OH where R represents the giant hydrocarbon framework. These resins exchange

their H+ ions with Ca2+ and Mg2+ ions present in hard water.

24

resin)(Exhausted22

water)hard(From4

resin)exchange(Cation2

resin)(Exhausted2

water)hard(From2

resin)exchange(Cation

SO2HgMO)(RSOgSOMHOSO2R

2Cl2HCa(RCOO)CaClHCOO2R

Anion exchange resins, on the other hand, are substituted ammonium hydroxides having the

general formula,

OHHNR 3 where R denotes the giant hydrocarbon framework. These resins

exchange their OH– ions with Cl– and SO42– ions present in hard water.

OHClHNRClOHHNRresin)(Exhausted3

water)hard(Fromresin)exchange(Anion3

OH2SO)HN(RSOOHHN2Rresin)(Exhausted

423water)hard(From

24

resin)exchange(Anion3

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Simultaneously, the H+ ions produced from cation excharge resins and OH– ions produced fromanion exchange resins combine to form H

2O.

Process. The hard water is first passed through cation exchange resin and then through anionexchange resin.

The resulting water is freed from both cations and anions and hence is called demineralised water ordeionised water and is as good as distilled water.

Q. What is meany by ‘demineralised water’ and how can it be obtained ?

A. A ‘demineralised water’ is the one which is free from all cations and anions. It is obtained by passinghard water through cation and anion exchange resin respectively.

Q. Is demineralised or distilled water useful for drinking purpose ? If not, how can it be made useful ?

A. No, demineralised or distilled water is not useful for drinking purpose, as it does not contain usefulminerals required for the development of the body. Therefore, for making it useful for drinkingpurpose, useful minerals are added in required amounts.

Q. Describe the usefulness of water in biosphere and biological systems.

A. Water is essential for all forms of life i.e., for animals and plants. As compared to other liquids, waterhas high specific heat, surface tension, dipole moment, thermal conductivity and dielectric constant.These properties of water play an important role in biosphere and biological systems. The high heatcapacity and heat of vaporisation are responsible for maintaing body temperature of living beingsand regularisation of the climate. As an excellent solvent it helps in transportation of minerals andother neutrients for plant and animal metabolism. Water also help for photosynthesis in plants.

Q. Compare the structures of H2O and H

2O

2.

Q. Complete the following chemical reactions :

(i) PbS (s) + H2O

2 (aq) (ii) MnO

4– (aq) + H

2O

2 (aq)

(iii) CaO (s) + H2O (g) (iv) AlCl

3(g) + H

2O (l)

(v) Ca3N

2(s) + H

2O (l)

A. (i) PbSO4(s) + 4H

2O(l) (ii) 2Mn2+(aq) + 8H

2O(l) + 5O

2(g) (iii) Ca(OH)

2(aq) (iv) Al(OH)

3(s) + 3HCl(aq)

(v) 3Ca(OH)2(aq) + 2NH

3(aq)

Q. Describe the structure of the common form of ice.

Q. Write chemical reactions to justify that hydrogen peroxide can function as an oxidising as well asreducing agent.

A. H2O

2 can act as an oxidising as well as a reducing agent both in acidic and basic media as illustrated

below :

(i) Oxidising agent in acidic medium

2Fe2+(aq) + 2H+(aq) + H2O

2(aq) 2Fe3+(aq) + 2H

2O(l)

(ii) Oxidising agent in basic medium

Mn2+(aq) + H2O

2(aq) + 2OH–(aq) )(OH2)s(MnO 2

dioxideManganese2 l

(iii) Reducing agent in acidic medium

2MnO4–(aq) + 6H+(aq) + 5H

2O

2(aq) 2Mn2+(aq) + 8H

2O(l) + 5O

2(g)

(iv) Reducing agent in basic medium

I2(s) + H

2O

2(aq) + 2OH–(aq) 2I–(aq) + 2H

2O(l) + O

2(g)

Q. What do you understand by the terms :

(i) hydrogen economy (ii) hydrogenation (iii) ‘syngas’ (iv) water-gas shift reaction and (v) fuel cell ?

A. (i) Hydrogen economy. The proposal to use hydrogen as a fuel in industry, power plants and possiblyalso in homes and motor vehicles is called hydrogen economy. The basic principle of hydrogen economyis production, storage and transportation of energy in the form of liquid or gaseous dihydrogen.

(ii) Hydrogenation means addition of hydrogen across double and triple bonds to form saturatedcompounds. The vegetable oil such as soyabean oil, cotton seed oil, groundnut oil, etc. are calledpolyunsaturated oil since they contain many C = C bonds. When these oils are exposed to air for

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prolonged periods, the double bonds present in them undergo oxidation, i.e., develop unpleasanttaste. To avoid this, double bonds are hydrogenated. For this purpose, dihydrogen is bubbled throughedible oils in presence of finely divided nickel at 473 K when the oils are converted into solid fats.

GheeVegetableHoilVegetableK473,Ni

2

This process is called hydrogenation or hardening of oils and is used in the manufacture of vegetableghee like Dalda, Gagan, Rath, etc. from vegatable oils. It may, however, be noted that hydrogenationreduces the number of double bonds but does not completely eliminate them.

(iii) Syngas. The mixtures of CO and H2 are called synthesis gas or syngas. It can be produced by the

reaction of steam on hydrocarbon or coke at high temperature in the presence of nickel as catalyst.

2Ni

K127022n2n H)1n2(nCO)g(OnHHC

e.g. )g(H3)g(CO)g(OH)g(CH 2Ni

K127024

These days ‘syngas’ is produced from sewage, saw dust, scrap wood, newspapers etc. The process ofproducing syngas form coal is called ‘coal gasification’.

)g(H)g(CO)g(OH)s(C 2Ni

K12702

(iv) Water-gas shift reaction. The amount of hydrogen in water gas can be further increased byoxidising CO to CO

2 by mixing it with steam and passing the mixture over FeCrO

4 as catalyst at

673 K.

Syngas22

FeCrO

K673

Steam2

gasWater2 )g(H2)g(CO)g(OH)g(OH)g(CO

4

This conversion of water gas to syngas is called water-gas shift reaction.

(v) Fuel cell. Fuel cell is a device which converts the energy produced during the combustion of a fueldirectly into electrical energy. Dihydrogen is used in hydrogen-oxygen fuel cells for generatingelectrical energy. It has many advantages over the conventional fossil fuels. It does not cause anypollution and releases more amount of energy per unit mass of fuel as compared to gasoline and othefuels.

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