p-Block Elements Exercises

p-BLOCK ELEMENTS IMPORTANT CONCEPTS 1. Back bonding 2. Dimerisation 3. Inert pair effect 4. Oxyacids 5. Hydrolysis 6. Allotropes 7. Silicates and Silicones 8. Drying agent and bleaching agents 9. Reactions and compounds (Group 13-18) 1.  BACK BONDING It is formed between two covalently bonded adjacent atoms. Conditions for back bonding : (i) Both bonded atoms must belong to 2nd period or one bonded atom must belong to 2nd period and the other must belong to 3rd period. (ii) One bonded atom must possess vacant orbital and the other bonded atom must possess lone pair. As a result of back bonding between the bonded atoms, bond length decreases and bond energy increases. Back bonding : It is a partial sideways overlap in which there is a slight electron transference from electron rich atom to electron deficient atom. As a result the electron deficiency of the deficient atom gets slightly compensated. Types of back bonding : Based on type of orbital : (2 types) (1) p-p back banding (2) p-d back bonding (1) p-p back bonding: It is used to explain following observations :(a) (b) Abnormal bond length and bond energy of B–F bond in BF3. Lewis acidic order of Boron and Beryllium halides. BeF2 < BeCl2 < BeBr2 < BeI2 (c)  Hybridisation : If a lone Pair participates in back bonding then it is not considered in hybridization. Ex. B3N3H6 (inorganic benzene or borazene or borazole) Hybridisation of B as well as N = sp2  Inorganic benzene is more reactive than organic benzene as in it the bonds are polar, although over all molecule is non polar. (d) If back bonding is present then tendency to form dimer or polymer decreases. Ex. BF3, BeF2 (2) p -d back bonding : It is used to explain following observations. (a) Hybridisation (b) 2. Acidic strength DIMERISATION / POLYMERISATION (A) By banana Bond or by 3C-2e bond or by e– deficient bond (a) B2H6  2C – 2e– bonds = 4 3C – 2e– bonds = 2 The structure of diborane, B2H6 3 Hybridisation state = sp Electron deficient molecule (act as Lewis acid) All four terminal hydrogen and two boron atoms are present in same plane both bridging H are present in perpendicular plane. If substitution reaction takes place than only four terminal hydrogen atom will be substituted. B2H6 ‡ˆ ˆˆ †ˆ BH 4 + BH 4 , B2H6 + 4CH3 – Cl B2H2(CH3)4 + 4HCl  (b) (BeH2)2 (dimer of BeH2 in vapour state) (c) (BeH2)n (polymer of BeH2 in solid state) (B) By-coordinate Bond I 3C - 4e bond (a) Al2Cl6 (dimer of AlCl3 in liquid or solid state) (b) (BeCl2)2 (dimer of BeCl2 in vapour state) (c) (BeCl2)n (polymer of BeCl2 in solid state) (d) I2Cl6 (dimer of ICl3) Note : (i) BCl3, BBr3 & BI3 do not form dimer due to smaller size of boron & large size of halogen (due to more steric repulsion) (ii) BF3 cannot form dimer due to its back bonding. (iii) AlF3 cannot form dimer due to its ionic nature. 1. BEGINNER’S BOX-1 In BF3 : (1) B–F bond has some double bond character and this bond is delocalised (2) All the B–F bonds are single covalent in nature (3) Bond energy and bond length of B–F bond indicate its single bond character (4) All the bonds are ionic 2. Which of the following statements is incorrect in the context of the B–F bond in BF3 :(1) All the three B–F bond lengths are equal and each of them is shorter than the sum of the covalent radii of boron and fluorine. (2) The bond energy of the. B–F bond is very high, higher than for any other single bond (3) The unusual shortness and strength of the B–F bond may be explained by a p-p interaction between boron and fluorine atoms. (4) The unusual shortness and strength of the bonds may be explained by a p-d interaction between the atoms of boron and fluorine. 3. In BF3, the B–F bond length is 1.30 Å, when BF3 is allowed to be treated with Me3N, it forms an adduct, [Me3N  BF3] The bond length of B–F in the adduct is : (1) Greater than 1.30 Å (2) Smaller than 1.30 Å (3) Equal to 1.30 Å (4) None of these 4. Which of the following structures correctly represents the boron trifluoride molecule : 5. (1) (2) (3) (4)  Trisilylamine [ N (SiH3)3] has a (1) Planar geometry (3) Pyramidal geometry (2) Tetrahedral geometry (4) None of these 6. In which of the following molecule, vacant orbitals do not participate in bonding :(1) B2H6 (2) Al2Cl6 (3) [H3N.BF3] (4) Si2H6 7. Which of the following is not possible due to back bonding. (1) State of hybridisation may change (2) Bond order increases (3) Bond angle always increases (4) Lewis acidic strength decreases 8. In which of the following compounds B–F bond length is shortest? (1) BF4 (2) BF3  NH3 (3) BF3 (4) BF3  N(CH3)3 9. In diborane(1) 2 bridged hydrogen and four terminal hydrogen are present (2) 3 bridged and three terminal hydrogen are present (3) 4 bridged hydrogen atoms are not in the same plane in dibroane (4) 1 bridged hydrogen and l terminal hydrogen are present 10. Which of the following statements is correct in the context of diborane (B2H6) ? (1) There are 12 valence electrons~ three from each of the two boron atoms and six from the six hydrogen atoms (2) Two of the six hydrogen atoms form two bridges between two boron atoms (3) The two bridging hydrogen atoms are in a plane perpendicular to the rest of the molecule and prevent rotation between the two boron atoms (4) All of these 3. INERT PAIR EFFECT Group-13 Group-14 Group-15 (Group oxidation state)   While moving down the group the stability of lower oxidation state (2 less than group oxidation state) progressively increases; for the last element of the group the stability of lower oxidation state becomes even greater than the group oxidation state. This is called inert pair effect. For the last element group oxidation state is highly oxidising in nature. Reason : As we move down the group there is presence of d & f-orbitals in inner shells which haw poor shielding effect hence Zeff increases. As a result the ns2 electron pair becomes more and more tightly held to the nucleus and becomes reluctant to participate in bonding. Examples: 1. PbCl2 is more stable than PbCl4 2. TlCl is more stable than TlCl3 3. GaCl3 is more stable than TlCl3 4. SnCl4 is more stable than PbCl4 5. Thalium (III) iodide does not exist. 6. PbI4 does not exist 7. Only BiF5 exists but BiCl5 does not exist? Ans. 4.   OXYACID General Formula of Oxy acid : HxZyOz (Z = non metal) Compounds which contain Z-OH bond are known as oxyacids. Acidic oxide Or + H2O ‡ˆ ˆˆ†ˆ Oxyacids Non metallicoxide E555555555555F Anhydrides of oxyacids  Oxidation state of central atom does not change. OXY ACIDS OF SULPHUR OXY ACID OF PHOSPHORUS OXY ACIDS OF HALOGENS OXY ACIDS CONTAINING S-S LINKAGE Important points : Trick for drawing structure of oxy compound Some important points (1) Oxyacid of phosphorus having +1 & +3 oxidation state (having P-H bond) can act as reducing agent or give disproportionation reaction on heating. 3 5 3  H3PO3   H3PO5  PH3 H3PO3  Ag2O   H3PO4  2Ag  Order of reducing nature. Note : Nitrogen compound having +1 & +3 oxidation state gives disproportion to reaction in acidic medium. 3  5 2 H HNO2   HNO3  NO (2) Heating effect of phosphoric acid (3) Heating effect of boric acid (4) SO3 exists as a cyclic trimer (S3O9) (5) Metaphosphoric acid exists as dirmer, cyclic trimer or polymer 100 C   2HBO2   B2O3 2H3BO3  2H 2 O  H2 O (6)  1. 2. Acidic nature of oxy acids. General concept : Oxidation number of   Acidic nature  For same elements   activeelement  of oxyacids  (Acidic nature of oxyacids  EN of active element) for different elements Exception: For oxy acids of P 1 Acidic nature  Oxidation number Example : H3PO2 > H3PO3 > H3PO4 (acidic order) BEGINNER’S BOX-2 Structures of metaboric acid and orthoboric acid respectively are : (1) (2) (3) Both the above (4) None Orthoboric acid(1) Donate proton to form H2BO03– (3) Donate OH– to form H2BO2+ (2) Accept proton of form H4BO3+ (4) Accept OH– to form [B(OH)4]– 3. Pick out the incorrect statement :(1) Orthophosphorus acid can be obtained by reacting P4O6 with H2O (2) Orthophosphoric acid can be obtained by reacting P4O10 with H2O (3) Pyrophosphoric acid can be obtained by heating orthophosphorus acid (4) Metaphosphoric acid is obtained by the dehydration of orthophosphoric acid at 316°C. 4. Hypophosphorus acid H3PO2 is- (1) Tribasic acid (3) Monobasic acid 5. (2) Dibasic acid (4) Not acidic at all The correct order of decreasing acid strength of oxy acids of group 15 elements is: (1) HNO3 > H3SbO4 > H3AsO4 > H3PO4 (2) H3PO4 > H3AsO4 > H3SbO4 > HNO3 (3) HNO3 > H3PO4 > H3AsO4 > H3SbO4 6. (4) HNO3 > H3AsO4 > H3PO4 > H3SbO4 Which one of the following is a mixed anhydride : (1) NO (2) NO2 (3) N2O3 (4) N2O5 7. Which of the following oxy acids of sulphur contains a sulphur-sulphur single bond : (1) H2S2O6 (2) H2S2P7 (3) H2S2O7 (4) H2S2O3 8. The structure of peroxodisufphuric acid is : 9. 10. (1) (2) (3) (4) Number of S-S bond in H2SnO6 (1) n (2) (n–1) (3) (n–2) Ga+ acts as a reducing agent because (1) Ga3+ state is less stable than Ga+1 (3) Ga3+ convert into Ga+1 reducing (2) Ga3+ state is more stable than Ga+1 (4) None of the above 5. HYDROLYSIS Chemical reaction of H2O with a covalent compound. Condition: One atom must possess vacant orbital and positive charge. Mechanism : (SN2 mechanism)  Weaker base goes out and a stronger base substitutes it. Example  For hybridisation of transition state Reactant  Transition state Ex.: sp  sp2 Sp2  sp3 sp3  sp3d Important examples of hydrolysis Hydrolysis of some important covalent molecules. 1. (4) (n+1)  Hydrolysis is not a redox reaction  Partial hydrolysis 2. Hydrolysis of interhalogen compounds 3. Hydrolysis of oxyacids 4. Hydrolysis of oxides (N, P, S, Cl) 5. Hydrolysis of Xe-fluoride (a) 2XeF2(s) + 2H2O()  2Xe (g) + 4 HF(aq) + O2(g) (It redox reaction not hydrolysis) Hydrolysis of XeF4 andXeF6 with water givesXeO3 (b) XeF6 + 3H2O  XeO3 + 6HF Partial hydrolysis of XeF6 gives oxyfluorides, XeOF4 and XeO2F2. XeF6 + 2H2O  XeOF4 + 2HF XeF6 + 2H2O XeO2 F2 + 4HF (c) 6XeF4 + 12H2O 4Xe + 2XeO3 + 24HF + 3O2 (Redox + hydrolysis both) 6. Hydrolysis of ionic compounds (a) Carbides (b) Nitrides (c) Phosphides 7. Hydrolysis of non-metals and halogen  'F2' is the only non-metal that can oxidise H2O as it is a powerful oxidising agent. Other halogens & non-metals undergo disproportionation with water in basic medium. (Redox reaction) 1. 2. 3. 4. BEGINNER’S BOX-3 Which of the following statements is correct : (1) Aluminium carbide as well as beryllium carbide produce methane gas on treatment with water (2) On reacting with water, calcium carbide (CaC2) produces acetylene white magnesium carbide (Mg2C3) gives propyne (3) Both of these (4) None of these N2 C  B (a) Al  (b) Al  A High temp. High temp. Product A and Bon hydrolysis yields respectively. (1) Ammonia and acetylene (2) Ammonia and methane (3) Nitric oxide and acetylene (4) None SbCl3 and BiCl3 on hydrolysis gives: (1) Sb+3 and Bi+3 (3) SbOCl and BiOCl (2) Sb(OH)3 and Bi(OH)3 (4) None The hydrolysis of PCl3, produces: (1) H3PO3 + HClO (3) H3PO4 + HCl (2) H3PO3 + HCl (4) PH3 + HClO 5. The number of molecule of water needed to convert one molecule of P2O5 into orthophosphoric acid is: (1) 2 (2) 3 (3) 4 (4) 5 6. XeF6 on partial hydrolysis produces:(1) XeF2 (2) XeOF4 7. 8. (3) XeF4 (4) XeO3 Which of the following halides does not hydrolysed? (1) PbCl4 (2) SiCl4 (3) CCl4 (4) SnCl4 Which of the following is an uncommon hydrolysis product of XeF2 and XeF4? (1) Xe (2) XeO3 (3) HF (4) O2 9. Consider the following reactions : (i) PCl3 + 3H2O  H3PO3 + 3HCl (ii) SF4 + 3H2O  H2SO3 + 4HF (iii) BCl3 + 3H2O  H3BO3 + 3HCl (iv) XeF6 + 3H2O  XeO3 + 6HF Then according to given information the incorrect statement is (1) During the (i) reaction the hybridisation of 15th group element does not change (2) During the (ii) reaction the hybridisation of 16th group element has been change· (3) During the (iii) reaction the hybridisation of 13th group element does not change (4) During the (iv) reaction the hybridisation of 18th group element does not change 10. Select correct statement about hydrolysis of BCl3 and NCl3. (1) NCl3 is hydrolysed and gives HOCl but BCl3 is not hydrofysed (2) Both NCl3 and BCl3 on hydrolysis gives HCl (3) NCl3 on hydrolysis gives HOCl but BCl3 gives HCl (4) Both NCl3 and BCl3 on hydrolysis gives HOCl 6. (A) ALLOTROPES Ex. Diamond Graphite Fullerene Ex. Coke Charcoal Coal, C-lamp black Special Point : (i) Thermodynamically graphite is more stable than diamond. (ii) Aqueous solution of graphite is known as aquadag. (iii) Graphite is also known as plumbago (used in lead pencil) (iv) Graphite use as a dry lubricants. (v) Hardest allotrope of carbon is diamond, softest allotrope is lampblack, purest allotrope is fullerene Fullerene  C-60 & C-70 are common fullerene,  C-60 is also known as Buckmeinster fullerene (Bucky ball) 12pentagonal  There are 32 ring 32 20hexagonal  Each carbon atom bonded with 3 other carbon by sigma & double bond (resonance)  sp2 hybridisation & aromatic in nature Dangling Bond In diamond or graphite some surface carbon atom have free valency. These carbon atoms form new bonds with impurities. These new bonds are known as dangling bond. (B) ALLOTROPES OF TIN (C) ALLOTROPES OF PHOSPHOROUS (a) white/yellow phosphorous (b) Red phosphorous Order of stability or MP or density white < red < black (D) ALLOTROPES OF SULPHUR (c) Black phosphorous II. (a) density of a S > S (b) Both are puckered crown shape having S8 units (c) S2 is paramagnetic sulphur which exist in vapour form at high temperature. (d) S6 is chair form of S III. Effect of temperature: 95.5 C 119 C 130  200 C  200 C  S8()   S(liq.)   Viscosity of liquid increase   S8()  viscosity decrease 7. (A) SILICATES Basic/ structural unit : Si O 4 4 Si O 4 4 have a tendency of polymerization (B) Sodium Zeolite [Na2Al2Si2O8] / [Na2O.Al2O3.2SiO2.xH2O) (i) It is a 3-D silicate (ii) It is used in (a) For softening of hard water (b) For cracking of hydrocarbon & isomerisation Ex. ZSM-5 (Zeolite) is used to convert ethyl alcohol into petrol. (C) SILICA : Generally silica is inert but it can react with NaOH & HF. (a) 2NaOH + SiO2 Na2SiO3 + H2O base acid [Na2SiO3 known as water glass or water soluble] (b) SiO2 + 4HF  2H2O + SiF4 SiF4 + 2HF  H2[SiF6] (hexa fluoro silicic acid) This process is also known as Etching of glass. SILICONES  (D) Silicones are organometallic polymer which contain, Ra2SiO as a basic unit. Properties : (i) Silicones are chemically inert due to presence of strong sigma bond. (ii) Silicones have water repelling nature due to presence of alkyl group. (iii) Silicones are insulator due to absence of free electrons. Preparation : hydrolysis condensation  product  Alkyl chloro silane  silicones R3SiCl2 : dimer (used to stop, chain formation) R2SiCl2 : linear polymer RSiCl 3 : cross link polymer (a) (b) Dimer silicone:  H2 O 2HCl  R3Si–OH + HO–SiR3  2R3Si–Cl + 2H–OH   R3Si–O–SiR3 Linear chain silicone (c) Cross linked silicone 8. BLEACHING AGENTS AND DEHYDRATING AGENTS (A) BLEACHING AGENTS Bleaching can be done by oxidation or reduction. [O] Coloured substance   Colourless (Permanent bleach) Oxidation    Examples By Oxidation (1) O3 (Dry bleach) (2) H2O2 (with moisture) (3) Cl2 (with moisture)  By Reduction SO2 (With moisture) Bleaching powder : Used in textile industries. Bleaches by oxidation. CaCl2 . Ca(OCl)2.Ca(OH)2.2H2O (average oxidation number of Cl = 0) or (B)  DEHYDRATING AGENTS Example: [P4O10, (Conc.) H2SO4, CaO (Quick line), CaCl2 (Anhydrous)] dehydrating  H2O + Oxide Acid  agent  Example: conc.  H2O + CO (Lab preparation of CO) HCOOH  H 2 SO 4      conc.   CO + CO2 + H2O (Lab preparation of CO & CO2 H 2 SO 4 conc.   2H2O + O=C=C=C=O (Carbon suboxide) H 2 SO 4 P4 O10 2HClO4   H2O + Cl2O7 P4 O10 2HNO3   H2O + N2O5 Charring of sugar (dehydration) : conc.  11H2O + C (Black mass) C12H22O11  H 2 SO 4 Some extra points : (i) NH3 can not be dried by conc. H2SO4, P4O10 and CaCl2 (ii) Alcohols, phenols can not be dried by metal hydrides. (iii) MgCl2.6H2O and AlCl3.6H20 can not be dried by heating alone 1. 2. BEGINNER’S BOX-4 Silicones have the general formula (1) SiO44– (2) Si2O76– (3) (R2SiO)n (4) (SiO3)n2– Glass or silica soluble in : (1) HClO4 (2) HF (4) H2SO4 (3) Aqua-regia 3. Si2O76– anion is obtained when (1) no oxygen of a SiO4–4 tetrahedron is shared with another SiO4–4 tetrahedron (2) one oxygen of a SiO4–4 tetrahedron is shared with another SiO4–4 tetrahedron (3) two oxygen of a SiO4–4 tetrahedron is shared with another SiO4–4 tetrahedron (4) three or all four oxygen of a SiO4–4 tetrahedron is shared with another SiO4–4 tetrahedron 4. Consider the following route of reactions · polymenisation  (B) R2SiCl2 + Water  (A)  Compound(B) in above reaction is (1) Dimer silicone (3) Cross linked silicone (2) Linear silicone (4) Polymerisation of (A) does not occur 5. (Si2O5)n2n– anion is obtained when (1) no oxygen of a SiO44– tetrahedron is shared with another SiO44– tetrahedron (2) one oxygen of a SiO44– tetrahedron is shared with another SiO44– tetrahedron (3) two oxygen of a SiO44– tetrahedron is shared with another SiO44– tetrahedron (4) three oxygen of a SiO44– tetrahedron is shared with another SiO44– tetrahedron 6. What is true about various allotropes of carbon? (1) Diamond is the hardest and graphite is the softest (2) Diamond is the hardest and coke is the softest (3) Diamond is the hardest and lamp black is the softest (4) Coke is the hardest and diamond is the softest 7. Different layers in graphite are held together by : (1) Ionic bonding (2) Metallic bonding (3) Covalent bonding (4) Vander Waals forces 8. Bucky ball or buck minster fullerene is : (1) An allotrope of carbon (2) It is referred as C-60 (3) It has sp2-hybridised nature and.· resembles with· soccer bali (4) All of these 9. Which of the following statement, is correct for allotropes of P :(1) White P ls soluble in CS2 but red P is insoluble in CS2 (2) P-P-P bond angle is 60° in red P (3) On heating in air, white P change into red P (4) White P change in black P' at ordinary temp. 10. Which of the following structural features of graphite best accounts for its use as a lubricant? (1) delocalized electrons (2) strong covalent bonds between carbon. atoms (3) van der Waal's forces between layers (4) limited three covalency of carbon BORON FAMILY (GROUP 13) INTRODUCTION The p-block elements are those elements of the periodic table in which the differentiating electron (or the last electron) enters the p-subshell of the last (outermost) shell. Since the maximum capacity of any atomic orbital is of two electrons with opposite spins (Pauli's principle), the maximum number of electrons that can be accommodated in any p-subshell is six. As a result, the p-block of the periodic table consists of six group: 13 (III A), 14(IVA), 15(VA), 16 (VIA), 17(VIIA) and 18 (zero group). On the basis of electronic configuration, the element He is to be excluded from the p-block as its last electron enters the s-subshell of the last shell. BORON FAMILY Boron is a typical non-metal, aluminium is a metal but shows many chemical similarities to boron, and gallium, indium and thallium are almost exclusively metallic in character, GENERAL CHARACTERISTICS Physical properties :  Boron is non-metallic in nature.  It is extremely hard and black coloured solid. It exists in many allotropic forms.  Due to very strong crystalline lattice, boron has unusually high melting point.  Rest of the member are soft metals with low melting point and high electrical conductivity. Gallium with low melting point (303 K), could exist in liquid state during summer. Its high boiling point (2676 K) makes it a useful material for measuring high temperatures. Chemical properties Reactivity towards air  Boron is unreactive in crystalline form.  Aluminium forms a very thin oxide layer on the surface which protects the metal from further attack.  Amorphous boron and aluminium metal on heating in air form B2O3 and Al2O3 respectively. With dinitrogen at high temperature they form nitrides.    2E2O3(s); 2E(s) + N2(g)   2EN (s). 2E(s) + 3O2 (g)  The nature of these oxides varies down the group. Boron trioxide is acidic and reacts with basic (metallic) oxides forming metal borates. Aluminium and gallium oxides are amphoteric and those of indium and thallium are basic in their properties. COMPOUNDS OF BORON (I) Borax (Na2B4O7.10H2O) 2Na+[B4O5(OH)4]–2 .8H2O Important points : B–O–B linkage = 5 Total B–O bonds = 14 Two Boron-sp2 hybridised Two Boron-sp3 hybridised (i) (ii) From Colemanite When colemanite powder is heated with Na2CO3 solution, the following reaction oecurs with the precipitation of CaCO3. Ca2B6O11 + 2Na2CO3  2CaCO3  + Na2B4O7 + 2NaBO2 The filtrate is cooled when white crystals of borax are precipitated. The mother liquor on treatment with CO2 converts NaBO2 to Na2B4O7 which precipitates out on crystallization. 4NaBO2 + CO2  Na2B4O7 + Na2CO3 From orthoboric acid Borax is obtained by the action of Na2CO3 on orthoboric acid. 4H3BO3 + Na2CO3 Na2B4O7 + 6H2O + CO2 Properties : (i) Borax is a white powder, less soluble in cold water, more soluble in hot water. (ii) Its aqueous solution is alkaline because of its hydrolysis to weak acid H3BO3 and strong alkali NaOH. Na2B4O7 + 7H2O  4H3BO3 + 2NaOH (iii) Action of heat When borax powder is heated, it first swells due to loss of water in the form of steam but at 740°C it becomes converted into colourless transparent borax bead.   Na2B4O7 + 10 H2O Na2B4O7. 10H2O  740 C  2NaBO2 + B2O3 (borax bead) Na2B4O7  Borax-bead test : Borax reacts with certain metal salts such as, Ni2+, Co2+, Cr3+, Cu2+, Mn2+ etc. to form coloured metaborates. The colour of the metaborates can be used to identify the metallic ions (cations) in salts. 740 C   2NaBO2  B2O3  CuO + B2O3    Na2B4O7  Na2B4O7·10H2O  10H 2 O 1 4 42 4 43 glassy mass Cu(BO2)2 (blue bead) Uses : Borax is used 1. in borax bead test 2. in purifying gold 3. as flux during welding of metals and 4. in production of glass. (II) ORTHO BORIC ACID [H3BO3/B(OH)3] Orthoboric acid is greasy to touch less soluble in cold water but more soluble in hot water. In the solid state, the B(OH)3 units are hydrogen bonded together into two dimensional sheets with almost hexagonal symmetry. (A) Preparation : (i) It is precipitated by treating a concentrated solution of borax with mineral add. Na2B4O7 + H2SO4 + 5H2O  Na2SO4 + 4H3BO3  Na2B4O7 + HCl + 5H20  2NaCl + 4H3BO3  (ii) From Colemanite : Powdered colemanite is suspended in water and excess SO2 is passed through it. On filtering and cooling the filtrate, white crystals of H3BO3 are obtained. Ca2B6O11 + 4SO2 + 11H2O  2Ca(HSO3)2 + 6H3BO3 (B) Properties: It is a weak monobasic acid soluble in water and in aqueous solution the boron atom completes its octet by accepting OH– from water molecules: B(OH)3(aq) + 2H2O() ‡ˆ ˆˆ †ˆ [B(OH)4]–(aq) + H3O+(aq), (Pka = 9.25) It, therefore, functions as a Lewis acid and not as a proton donor like most acids. (C) Test for Borate radical: When boric acid is heated with ethyl alcohol, the evolved gas is burned forming a green edged flame. · H3BO3 + 3C2H5OH  B(OC2H5)3 + 3H2O ethyl borate (volatile) Uses: 1. It is an antiseptic and its water solution is used as an eyewash. 2. It is also used in glass, enamel and pottery industry. (III) DIBORANE (B2H6] Binary compounds of boron wit}) hydrogen are called boron hydrides or boranes. Preparation : ether  2B2H6 + 3LiF + 3AlF3 (i) 4BF + 3LiAlH ether  ether  B2H6 + 2NaI + H2 (ii) 2NaBH4 + I2  453 K  B2H6 + 6NaF (Industrial method) (iii) 2BF3 + 6NaH  Properties : (i) B2H6 is colourless gas and highly reactive (boiling point 183 K). (ii) It catches fire spontaneously in air and explodes with O2. Reaction with oxygen is extremely exothermic. B2H6 + 3O2  B2O3 + 3H2O H = –2160 kJ mol–1 Mixtures of diborane with air or oxygen in flame spontaneously producing large amount of heat. Diborane has a higher heat of combustion per unit weight of fuel than most other fuels. It is therefore used as a rocket fuel. At red-heat the boranes decomposes to boron and hydrogen. (iii) Reaction with water is instantaneous. B2H6 + 6H2O  2B(OH)3 + 6H2 (iv) The electron deficient 3c-2e B–H–B bridges are sites of nucleophilic attack. Small amines such as NH3, CH3NH2 and (CH3)2NH give unsymmetrical Cleavage of diborane. B2H6 + 2NH3  [H2B (NH3)2]+ + [BH4]– Large amines such as (CH3)3N  pyridine give symmetrical cleavage of diborane . 2(CH3)3N + B2H6  2H3B  N(CH3)3 200 C,20 atm  2BH3CO (borane carbonyl) B2H6 + 2CO  (v) The reaction with ammonia depends on conditions. Excess NH3  B2H6.2NH3 or [H2B(NH3)2]+ [BH4]– (ionic compound). B2H6 + NH3  low temperature Excess NH3   (BN)x boron nitride/BORAZONE. (Inorganic graphite) higher temperature(  200 C) Ratio 2NH3 :1B2 H6   B3N3H6 borazole/ BORAZINE. (Inorganic benzene) higher temperature(200 C) Borazole is much more reactive than benzene. COMPOUNDS OF ALUMINIUM: ALUM (a) Alums are double sulphates with their general formula M2SO4.M'2(SO4)3.24H2O where M = monovalent radical like Na+, K+, NH4+ and M' =Trivalent radical like Al+3, Cr+3 Fe+3. (b) The different alums are (i) Potash alum K2SO4.Al2(SO4)3.24H2O (ii) Chrome alum K2SO4Cr2(SO4)3.24 H2O (iii) Iron alum (NH4)2SO4. Fe2(SO4)3. 24H2O (iv) Ammonium alum (NH4)2SO4.Al2(SO4)3.24H2O (c) In alums each metal ion is surrounded by six water molecules. (d) Lithium does not form alum because Li ion is too small to haw a coordination number of six. (e) Pseudo alums : Double sulphates of divalent ions and trivalent ions with 24 water molecules in their crystals are known as pseudo alums. Eg. MSO4.X2 (SO4)3.24H2O M = divalent or Bivalent ion X = Trivalent metal ion: Properties: (a) It is a white crystalline solid highly soluble in water. (b) On heating it undergoes dehydration and swells up. (c) It is highly acidic in aqueous solution due to the hyqrolysis of aluminium sulphate to sulphuric acid (d) It is a double salt and its aqueous solution gives reaction of an the constituents ions K+, Al+3, SO4–2 Uses: (i) Alum is used to stop bleeding. (ii) It is used for purification of water: (iii) It is used as· a mordant in dyeing industry. (iv) Alum is used for tanning of leather. 1. BEGINNER’S BOX-5 On the addition of mineral acid to an aqueous solution of borax, the compound formed is - " (1) Orthoboric acid (2) Boron hydride (3) Metaboric acid (4) Pyroboric acid 2. Which of the following can be detected by the borax-bead test? (1) Ni2+ (2) Co2+ +2 (3) Pb (4) Both (1) & (2) 3. mixture of boric acid with ethyl alcohol bums with green edged flame due to the formation of (1) Ethyl borax (2) Ethyl borate (3) Methyl borax (4) Methyl borate 4. The hydrolysis of borax produces (1) An acidic medium (3) A neutral medium (2) A basic medium (4) An acidic or an neutral medium 5. In alums, each metal ion is surrounded by(1) Two water molecules (2) Four water molecules (3) Six water molecules (4) Eight water molecules 6. Which of the following is false statements :(1) Boranes are easily hydrolysed (2) LiAlH4 reduces BCl3 to borane (3) BH3 is a Lewis acid (4) All the B–H distances in diborane (B2H6) are equal 7. Inorganic graphite is :(1) B3N3H6 (3) BN 8. (2) B2H6 (4) BF3 By which of the following reactions is borazine, prepared? low temperature  (1) B2H6 + NH3: (excess}  (ratio 2 NH3 :1B2 H6 )  (2) B2H6 + NH3  high temperature 3 2 6  (3) B2H6 +NH3  high temperature (ratio 2 NH :1B H ) (4) None of the above 9. When an inorganic compound(X) having 3c–2e as well as 2c–2 bonds reacts with ammonia gas at a certain temperature, gives a compound (Y), isostructural with benzene. Compound (X) with ammonia at a high temperature produces a substance (Z) ; (1) (X) is B2H6 (2) (Z) is known as inorganic graphite (3) (Y) is B3N3H6 (4) All of these 10. Which one of the following has the lowest m. p.? (1) B (2) Al (3) Ga (4) Tl CARBON FAMILY (GROUP l4) PHYSICAL PROPERTIES  All group 14 members are solids.  Carbon and silicon are non-metals, germanium is metalloid whereas tin and lead are soft metals with low melting points.  Melting points and boiling points of group 14 elements are much higher than those of corresponding elements of group 13 due to stronger metallic bonding. COMPOUNDS OF CARBON (A) Carbon Monoxide (CO) (I) Preparation (i) It is formed together with CO2, when carbon or carbonaceous matter is oxidized by air or oxygen. It is also produced when CO2 is reduced by red- hot carbon; this reaction is of importance in metal extractions. C(s) + CO2(g)  2CO(g) (ii) In the laboratory it can be prepared by dehydrating methanoic acid with concentrated sulphuric acid. 373K HCOOH (liq)  CO(g) + H2O conc. H 2 SO 4 (II) Physical Properties (i) Carbon monoxide is a colourless, odourless gas which burns in air with a blue flame, forming CO2. (ii) It is sparingly soluble in water and is a neutral oxide. (iii) CO is toxic, because it forms a complex with haemoglobin in the blood and this complex is more stable than oxy-haemoglobin. This prevents the haemoglobin in the red blood corpuscles from carrying oxygen round the body. This causes oxygen deficiency, leading to unconsciousness and then death. (iv) It acts as good reducing agents for all metal oxide. (B) CARBON DIOXIDE (CO2) Preparation : (i) Complete combustion of carbon containing compounds. ClI4 + O2 CO2 + H2O (ii) In the laboratory it can be conveniently made by the action of dilute hydrochloric add on marble chips : CO32–(aq) + 2H+(aq)  CO2(g) + H2O() (iii) Industrially It Is produced as a by-product during the manufacture of quicklime and in fermentation processes: CaCO3(s)  CaO(s) + CO2(g); C6H12O6(aq){glucose}  2C2H5OH(aq) + 2CO2(g) Properties : (i) It is a colourless, odourless and heavy gas. (ii) Carbon dioxide is the acid anhydride of carbonic acid which is a weak dibasic acid and ionises in two steps as follows.: H2CO3(aq) + H2O () ‡ˆ ˆˆ †ˆ HCO3– (aq) + H3O+ (aq) HCO3– (aq) + H2O () ‡ˆ ˆˆ †ˆ CO32– (aq) + H3O+ (aq) H2CO3 / HCO3– buffer system helps to maintain pH of blood between 7.26 to 7.42. A solution of carbonic acid in water will slowly turn blue. litmus red and when the solution is boiled, all 2 the CO2 is evolved. (iii) Carbon dioxide readily reacts with alkalies forming the carbonate and, if CO2 is in excess, the hydrogen carbonate. This is the basis of the lime-water test for CO2 gas. Ca(OH)2(aq) + CO2(g)  CaCO3(s) + H2O(liq) ; CaCO3(s) + H2O(liq) + CO2(g)  Ca(HCO3)2(aq) The above reaction accounts for the formation of temporarily hard water. (iv) Gaseous CO2 is extensively used to carbonate soft drinks. Being heavy and non-supporter of combustion it is used as fire extinguisher: A substantial amount of CO2 is used to manufacture urea. (C) Carbides : On the basis of type of bonding carbides are of three :(I) Salt like calbides (ionic carbides) (II) Covalent carbides (giant molecular carbides) (III) Interstitial carbides. (I) Salt like carbides (ionic carbides): (a) These are formed by strong electropositive elements of groups 1 : 2 and 13 (b) These are transparent crystalline substances and do not conduct electric current in the solid state. (c) These are easily hydrolysed by water or dilute acids to give aliphatic hydrolysis. (i) Methanides (Methides) – Methanides are the carbides which give methane on hydrolysis. They contain C4– ions. Be2C + 4H2O  2Be(OH)2 + CH4 Al4C3 + 12H2O  4Al(OH)3 + 3CH4 (ii) Acetylides - These are the carbides which yield acetylene on hydrolysis. They contain the ion C22–. e.g. CaC2 + 2H2O  Ca(OH)2 + C2H2 or C22– + 2H+  C2H2 (iii) Allylides- These carbides give rnethylacetylene or allylene.(CH3–CCH) on hydrolysis. They contain C34– ions. e.g. Mg2C3 + 4H2O  2Mg(OH)2 + CH3–CCH (II) Covalent carbides (giant molecular carbides) (a) Covalent carbides are the compounds of carbon with elements having similar electronegativity. (b) These are covalent, polymeric, thermally stable, extremely hard, high melting and technically inert solid. (c) B4C and SiC are two main covalent carbides. (III) Interstitial carbides Transition metals like Ti, Zr, Hf, V, Nb, Ta and Mo form these carbides. The metals retains their close packed lattice and carbons are incorporated in their in interstitial spaces. These carbides are extremely hard and have very high melting points. The carbides of V and Ta are very hard hence they are used for manufacture of high speed cutting tools. GLASS (a) Glass is a hard, brittle transparent atmorphous solid or supercooled liquid made up of silicates of Na, Ca and other metals like Pb, Zn etc. (b) being amorphous solid (not a true solid) it has no sharp melting and melts at high temperature (c) Glass has no chemical formula, however it may be represented as : xM2O. yMO.6SiO2 where M = monovalent metals (eg : Na, K) M = bivalent metals Ca, Pb, Zn etc. x and y are integers. 1700K (d) CaO + Na2CO3 + 6SiO2   Na2SiO3 + CaSiO3 + 4SiO2 + CO2 Glass Different Types of Glass (i) Soda Ol' Soft glass or Sodalime glass : It iS corm.non glass. Contain Na & Ca metal carbonate. Used for formation of tube, window. (ii) Potash or Hard glass : It is a mixture of potassium and calcium silicates. It fuses with difficulty. It is used for making hard glass apparatus. (iii) Flint glass : It is mixture of potassium and lead silicates. It has a very high refractive index and is used for making electric bulbs and optical instruments. (iv) Crooke's glass: It is a optical glass containing CeO2 which prevents the entry of UV rays. (v) Pyrex glass or borasil or borosilicate glass : It is used to make lab appliances as it is resistant to heat, shock and common reagents. It is a mixture of zinc and barium boro silicates and alumina silicate. Small part of SiO2 is replaced by boric oxide/borax. Low coefficient and thermal expansion, can with stand sudden change in temperature so used for making lab glass wares. SOME GASEOUS FUELS (a) Water gas or syn gas = CO + H2 (c) Semi water gas = CO + N2 + H2 (e) Coal gas = [CO2 + CO + H2 + CH4] (b) Producer gas = CO + N2 (d) Natural gas = CH4 BEGINNER’S BOX-6 1. Carbon has no tendency to form complex compounds because of : (1) its small size (2) The availability of vacant d-orbitals (3) Non availability of vacant d-orbitals (4) No tendency to form covalent bonds 2. Which of the following statement is false: (1) Dry ice is solid Co2 (2) CO2 is weakly acidic (3) CO2 and SiO2 are linear molecules (4) CO2 is a gas while SiO2 is solid 3. CO is : (1) Reductant (3) Neutral to litmus 4. (2) Poisonous gas (4) All Which is/are true statements: (1) Diamond is unaffected by conc. acids but graphite reacts with hot cone. HNO3 forming mellitic acid, C6(COOH)6 (2) CO is toxic because it forms a complex with haemoglobin in the blood (3) In C3O2, all carbons are sp hybridised (4) All are true statements NITROGEN FAMILY (GROUP 15) Group 15 includes nitrogen phosphorus, arsenic. Antimony and bismuth. PHYSICAL PROPERTIES :  Dinitrogen is a diatomic gas while all others are solids.  Except nitrogen, all the elements show allotropy.  As we go down the group, there is a shift from non-metallic to metallic through metalloidic character. Nitrogen and phosphorus are non-metal, arsenic and antimony are metalloid and bismuth is a typical metal. CHEMICAL PROPERTIES Oxidation States and trends in a chemical reactivity :  The common oxidation states of these elements are –3, +3 and +5. The tendency to exhibit –3 oxidation state decreases down the group , bismuth hardly forms any compound in –3 oxidation state.  Nitrogen also exhibits +1, +2, +4 oxidation states when it reacts with oxygen. Phosphorus also shows +1 and +4 oxidation states in some oxyacids.  In the case of nitrogen, all oxidation states from+ 1 to +4 tend to disproportionate in acid solution . For example, 3 HNO2  HNO3 + H2O + 2NO  Similarly, in case of phosphorus nearly all intermediate oxidation states disproportionate into +5 and –3 both in alkali and acid. However +3 oxidation state in case of arsenic, antimony and bismuth become increasingly stable with respect to disproportionation.      Reactivity towards hydrogen : All the elements of Group 15 form hydrides of the type EH3 where E=N, P, As, Sb or Bi The stability of hydrides decreases from NH3 to BiH3. Consequently the reducing character of the hydrides increases. Ammonia is only a mild reducing agent while BiH3 is the strongest reducing agent amongst all the hydrides. Due to weaker Bi–H bond. Reactivity towards oxygen : All these elements form two types of oxides : E2O3 and E2O5. The oxides of the type E2O3 of nitrogen and phosphorus are purely acidic, that of arsenic and antimony amphoteric and those of bismuth is predominantly basic. Reactivity towards halogens : These elements react to form two series of halides : EX3 and EX5. All the trihalides of these elements except those of nitrogen are stable. In case of nitrogen, only NF3 is known to be stable. Trihalides except BiF3 are predominantly covalent in nature. COMPOUND OF NITROGEN 1. NITROGEN GAS (N2) : Nitrogen is an important and essential constituent of proteins and amino acids. Nitrates and other nitrogen compounds are extensively used in fertilizers and explosive. (a) Preparation: (i) Laboratory method of preparation : NH4Cl(aq) + NaNO2(aq) N2(g) + H2O () + NaCl(aq) N2 is collected by the downward displacement of water. This reaction takes place in two steps as given below : (ii) (iii) (iv) (b) (i) (ii) (c) (i) (ii) (iii)   N2 + 2H2O NH4Cl +NaNO2 ® NH4NO2 + NaCl ; NH4NO2  By heating ammonium dichromate :   N2  + 4H2O + Cr2O3 (NH4)2Cr2O7  Very pure nitrogen can be obtained by heating sodium or barium azide.   Ba + 3N2 Ba(N3)2  300 C  3N2 + 2Na 2NaN3  Industrial methods of preparation: From liquified air by fractional distillation :The boiling point of N2 is –196°C and that of oxygen is – 183°C and hence they can be separated by fractional distillation of air. Properties: N2 is a colourless, odourless gas very less soluble in water. It is neither combustible nor a supporter of combustion. Reaction with oxygen: Dinitrogen combines with dioxygen only at very high temperature (at about 2000K) to form nitric oxide. This reaction is endothermic. N2 + O2  2NO Uses: For providing an inert atmosphere during many industrial processes where presence of air or O2 is to be avoided. For manufacture of NH3 by the Haber's process, Liquid dinitrogen is used as a refrigerant to preserve biological materials, food items & cryosurgery. 2. (a) AMMONIA (NH3) Preparation : (i) By the action of any base or alkali on any ammonium salt:   NH3  + NaNO3 + H2O NH4NO3 + NaOH    2NH3  + CaSO4 + 2H2O (NH4)2SO4 + Ca(OH)2  This is a general method and is used as a test for ammonium salts. Ammonia is present in small quantities in. air and soil where it is formed by the decay of nitrogenous organic matter e.g., urea. NH2CONH2 + 2H2O  (NH4)2CO3 ‡ˆ ˆˆ †ˆ 2NH3 +H2O +CO2 (ii) Industrial methods of preparation :  C,200 atm ˆ500 Haber’s process : N2 + 3H2 ‡ˆ ˆˆIronˆˆ oxide ˆ ˆˆ ˆˆK ˆˆO &ˆˆ Alˆˆ Oˆˆ †ˆ 2NH3 2 2 3 In accordance with Le Chatelier's principle, high pressure would favour the formation of ammonia. The optimum conditions for the production of ammonia are a pressure of 200×105 Pa (about 200 atm), a temperature of ~ 700 K. and the use of a catalyst such as iron oxide with small amounts of K2O and Al2O3 to increase the rate of attainment of equilibrium. (b) Physical properties : Ammonia fs a colourless gas with a pungent colour. (c) (i) Chemical.properties : Its aqueous solution is weakly basic due to the formation of OH– ions. NH3(g) + H2O() ‡ˆ ˆˆ †ˆ NH4+ (aq) + OH– (aq) (ii) It forms ammonium salts with acids, e.g., NH4Cl, (NH4)2 SO4 etc. As a weak base, it precipitates the hydroxides of many metals from their salt solutions. For example, FeCl3 (aq) + NH4OH (aq)  Fe2O3. xH2O (s) + NH4Cl (aq) (brown ppt) ZnSO4 (aq) + 2NH4OH (aq)  Zn(OH)2 (s) + (NH4)2 SO4 (aq) (white ppt) The presence of lone pair of electrons on the nitrogen atoms of the ammonia molecule makes it a Lewis base. It donates the electrons pair and forms linkage with metal ions and the formation of such complex compounds finds applications in detection of metal ions such as Cu2+, Ag+ (iii) 3. OXIDES OF NITROGEN: Nitrogen forms a number of oxides, N2O, NO, N2O3, NO2 or N2O4 and N2O5. All these oxides of nitrogen exhibit p-p: multiple bonding between nitrogen and oxygen. 4. NITRIC ACID (HNO3) (a) Preparation : In the laboratory, nitric acid is prepared by heating KNO3 or NaNO3 and concentrated H2SO4 in a glass retort. : · NaNO3 + H2SO4  NaHSO4 + HNO3 On a large scale it is prepared mainly by Ostwald’s process. This method is based upon catalytic oxidation of NH3 by atmospheric oxygen. Pt / Rh gauge cata;ust  4NO(g) + 6H2O(g) 4NH3(g) + 5O2(g)  500K, 9bar (from air) Nitric oxide thus formed combines with oxygen giving NO2. 2NO(g) + O2(g) ‡ˆ ˆˆ †ˆ 2NO2(g) Nitrogen dioxide so formed, dissolves in water to give HNO3 3NO2(g) + H2O()  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. (b) (i) Physical properties : It is a colourless liquid. Freezing point is 231.4 K and boiling point is 355.6 K. Laboratory grade nitric acid contains ~ 68% of the HNO3 by mass and has a specific gravity of 1.504. (ii) (iii) (iv) (v) In the gaseous state, HNO3 exists as a planar molecule. In aqueous solution, nitric acid behaves as a strong acid giving hydronium and nitrate ions. HNO3 (aq) + H2O ()  2H3O+ (aq) + NO3– (aq) Concentrate nitric acid is a strong oxidising agent and attacks most metals except Au & Pt. The product of oxidation depend upon the concentration of the acid, temperature and nature of metal. Concentrate HNO3 also oxidises non-metals and their compounds. conc.HNO3 dil.HNO3 Ex. Zn(NO3)2 + N2O   Zn(NO3)2 + NO2  Zn  conc.HNO3 dil.HNO3 Cu(NO3)2 + NO   Cu(NO3)2 + NO2  Cu  SOME EXCEPTIONAL POINTS  Mg and Mn form H2 gas on reaction with very dilute HNO3  Be, Al, Fe, Cr do not read with cone. HNO3 due to fromation of protective oxide layer.  Au and Pt (Noble metals) do not react with cone. HNO3  Sn reacts with cone. HNO3 and foms H2SnO3 (metastannic add)  Au and Pt react only with aqua regia (3-part of conc. HCl and 1 part of conc. HNO3) 3HCl + HNO3 NOCl + 2H2O + Cl– HCl Au + 3Cl– ® AuCl3  HAuCl4 Tetrachloro auric acid –  H2PtCl6 Pt + 4Cl  PtCl4  2HCl Hexachloro palatinic acid  Conc. HNO3 + skin  yellow spot (protein  x anthoprotein) The major use of nitric acid is in the manufacture of ammonium nitrate for fertilizers and other nitrates for use in explosives and pyrotechnics. It is also used for the preparation of nitroglycerin, trinitrotoluene and other organic nitro compounds. Other major uses are in the pickling of stainless steel, etching of metals and as an oxidiser in rocket fuels. COMPOUNDS OF PHOSPHORUS: 1. Phosphine (PH3) : (a) Preparation : (i) Phosphine is prepared by the reaction oh:akium phosphide with water. Ca3P2 + 6H2O  3Ca(OH)2 + 2PH3 (ii) In the laboratory, it is prepared byh heating white phosphorus with concentrated NaOH solution in an inert atmosphere of CO2, P4 + 3NaOH + 3H2O  PH3 + 3NaH2PO2 (sodium hypophosphite) (iii) When pure, it is non inflammable but becomes inflammable owing to ,the presence of P 2H4 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. PH4I + KOH  KI + H2O +PH3 (b) (i) Properties : 1 It is a colourless gas with rotten fish smell and is highly poisonous. It is slightly soluble in water. The solution of PH3 in water decomposes in presence of light giving red phosphorus and H2. When absorbed in copper sulphate or mercuric chloride, the corresponding phosphides are obtained. 3CuSO4 + 2PH3  Cu3P2 3H2SO4 3HgCl2 + 2PH3  Hg3P2 (brownish black)+ 6 HCl Phosphine is weakly basic and like ammonia, gives phosphonium compounds with adds e.g. PH3 + HBr  PH4Bt (ii) (iii) Uses : 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 bum arid serves as a signal. It is also used in the production of smoke screens. Calcium phosphide reacts with water producing phosphine which burns in air to give clouds of-phosphorus pentaoxide and that acts as smoke screens. 2. PHOSPHOROUS HALIDES Phosphorous forms two types of halides PX3 (X = F, Cl, Br, I) and PX5 (X = F, Cl, Br). Phosphorous Trichloride (PCl3) (a) Method of preparation (i) It is obtained by passing dry chlorine over heated white phosphorus. P4 + 6Cl2  4PCl3 (ii) It is also obtained by the action of thionyl chloride with white phosphorus. P4 + 8SOCl2  4PCl3 + 4SO2 + 2S2O2 (b) (i) (ii) Properties It is a colourless oily liquid and hydrolyses in the presence of moisture. PCl3 + 3H2O  H3PO3 + 3HCl It reacts with organic compounds containing –OH group such as CH3COOH, C2H5OH. 3CH3COOH + PCl3 CH3COCl + H3PO4 3C2O5OH + PCl3  3C2H5Cl + H3PO3 PHOSPHORUS PENTACHLORIDE (PCl5) (a) Method of 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 (b) (i) (ii) (iii) (iv) (v) Properties PCl5 is, a yellowish white powder and in moist air it hydrolyses to POCl3 and, finally gets converted to phosphoric acid. PCl5 + H2O  POCl3 + 2HCl POCl3 + 3H2O  H3PO4 + 3HCl When heated it sublimes but decomposes on stronger heating. Heat  PCl3 + Cl2 PCl5  It reads with organic compounds containing –OH group converting them to chloro derivatives. C2H5OH + PCl5  C2H5Cl + POCl3 + HCl Finally divided metals on heating with PCl5 give corresponding chlorides. 2Ag + PCl5  2AgCl + PCl3 Sn + 2PCl5  SnCl4 + 2PCl3 It is used in the synthesis of some organic compounds, e.g . C2H5CH3COCl. QUESTIONS (REASONING) Q.1 Ans. Why BiH3 is the strongest reducing agent amongst all the hydrides of Group 15 elements? In hydrides of nitrogen family on moving down the group M–H bond length increases so bond strength decreases hence tendency to release hydrogen increases and reducing nature increases. Increasing order of reducing nature is NH3< PH3 < AsH3 < SbH3 < BiH3 Q.2 Ans. Write the reaction of thermal decomposition of sodium azide. Thermal decomposition of sodium azide gives dinitrogen gas. 2NaN3  2Na + 3N2 Q.3 Ans. Why N2 is less reactive at room temperature? N2 is less reactive at room temperature because of the high bond enthalpy of NN bond. Q.4 Ans. Why does R3P = O exist but R3N = O does not (R = alkyl group)? Due to presence of vacant d orbital phosphorous can form five covalent bond, while nitrogen restricts its covalency to four due to absence of vacant d orbitals. Q.5 Ans. Why does nitrogen show catenation properties less than phosphorus? N–N bond is weaker than the single P–P bond due to high interelectronic respulsion of nonbonding electrons as a result the catenation tendency is weaker in nitrogen. Q.6 Ans. Mention the conditions required to maximise the yield of ammonia. In accordance with Le Chatelier's principle, high pressure would favour the formation of ammonia. The optimum conditions for the production of ammonia are a pressure of 200×105 Pa (about200 atm), a temperature of ~7 00 K and the use of a catalyst such as iron oxide with small amounts of K2O and Al2O3 to increase the rate of attainment of equilibrium. Q.7 How does ammonia react with a solution of Cu2+? Ans. The presence of a lone pair of electrons on the nitrogen atom of the ammonia molecule makes if a Lewis base. It donates the electron pair and forms linkage with metal ions and the formation of such complex compounds finds applications In detection of Cu2+. Cu2+ (aq) + 4NH3(aq) ‡ˆ ˆˆ †ˆ [Cu(NH3)4]2+(aq) (blue) (deep blue) Q.8 Ans. Illustrate how copper metal can give different products on react ion with HNO3. 3Cu + 8HNO3(dilute)  3Cu(NO3)2 + 2NO + 4H2O Cu + 4HNO3(conc.)  Cu(NO3)2 + 2NO2 + 2H2O Q.9 Why is nitrogen di-oxide paramagnetic in gaseous state but the solid obtained on cooling it is diamagnetic. NO2 contains odd number of electrons in its valence shell. On cooling it forms dimer and converted to stable N2O4 which is a colourless solid and diamagnetic in nature. Ans. Q.10 Why NH3 gas cannot be dried by passing over P2O5, CaCl2 and H2SO4? Ans. CaCl2 + 8NH3  CaCl2.8NH3 P2O5 + 6NH3 + 3H2O  2(NH4)3PO4 H2SO4 +2NH3  (NH4)2SO4 SO it is dried by passing over quick lime (CaO). CaO + H2O  Ca(OH)2 Q.11 Why inert atmosphere of CO2 is taken in the formation of PH3 by the reaction of white phosphorous with conc. NaOH solution. Ans. To decrease the partial pressure of O2 in atmosphere Q.12 Why does PCl3 fumes in moisture? Ans. PCl3 hydrolyses in the presence of moisture giving fumes of HCl PCl3 + 3H2O  H3PO3 +3HCl Q.13 What happens when PCl5 is heated? Ans. When heated, it sublimes but decomposes on stronger heating   PCl3 + Cl2 PCl5  Q.14 Can PCl5 act as an oxidising as well as reducing agent? Justify. Ans. It can act as an oxidising as well as reducing agent due to oxidising nature of P(V) and reducing nature of Cl–. 1. 2. 3. BEGINNER’S BOX-7 Which of the following halide of nitrogen is stable? (1) NF3 (2) NCl3 (3) NBr3 (4) NI3 The nitrogen oxide(s) that does not contain N–N bond(s) is (i) N2O (2) N2O3 (3) N2O4 (4) N2O5 What is false about N2O5? (1) It is anhydride of HNO3 (2) It Is a powerful oxidizing agent (3) Soli1d N2O5 is called nitronium nitrate (4) Structure of N2O5 contains no [NO] bond 4. Conc. HNO3 is yellow coloured liquid due to (1) Dissolution of NO in conc. HNO3 (2) Dissolution of NO2 in conc. HNO3 (3) Dissolution of N2O in conc. HNO3 (4) Dissolution of N2O3 in conc. HNO3 5. Which of the following process is not involved in Ostwald's process for the manufacture of HNO3? Pt  4NO(g) + 6H2O(g) (1) 4NH3 (g) + 5O2 (g)  500K, 9bar (2) 2NO(g)+ O2 ‡ˆ ˆˆ †ˆ 2NO2 (g) (3) 3NO2 (g) + H2O()  2HNO3(aq) + NO(g) (4) None 6. Which of the following salts give NH3 in alkaline medium? (l) (NH4)2CO3 (2) (NH4)2SO4 (3) NH4Cl (4) All of the above 7. Which of the following reaction is suitable for obtaining very pure nitrogen? (1) NH4Cl(aq) + NaNO2  N2(g) + 2H2O() + NaCl(aq)   N2+ 4H2O + Cr2O3 (2) (NH4)2Cr2O 7    Ba + 3N2 (3) Ba(N3)2  (4) All of the above 8. In warfare smoke screens are prepared from. (1) CaC2 (2) PH3 (3) P2O5 (4) COCl2 9. White phosphorouus on reaction with NaOH gives PH3as one of the products. This is a(1) dimerisation reaction (2) disproportionation reaction (3) condensation reaction (4) precipitation reaction 10. Which of the following is not correctly matched? (1) P4O10 + H2O  reactants involved in formation of H3P4 (2) CaC2 + H2O  C2H2 + Ca(OH)2; Ca3P2 + H2O  PH3 + Ca(OH)2 reactions involved in Holmes signal KOH  KI + H2O + PH3; purification of PH3 (3) PH3 + HI  PH4I  (4) PH3 + HI  PH4.1 ; shows Lewis acidic nature of PH3 OXYGEN FAMILY (GROUP 16 ) INTRODUCTION Oxygen, sulphur, selenium, tellurium and polonium constitute group 16 of the periodic table. This is sometimes known as group of chalcogens (ore forming elements) because a large number of metals ores are oxides or sulphides. Physical.Properties :  Oxygen and sulphur are non-metal, selenium and tellurium metalloids, whereas polonium is a metal. Polonium is radioactive and is short lived (Half-life 13.8 days).  The melting and boiling points increase with an increase in atomic number down the group. The larger difference between the melting and boiling points of oxygen and sulphur may be explained on the basis of their atomicity; oxygen exist as diatomic molecules (O2)whereas sulphur exists as polyatomic molecule (S8). CATENATION Tendency for catenation decreases down the group. This property is prominently displayed by sulphur (S8). · The S–S bond is important in biological system and is found in some proteins and enzymes such as cysteine. Chemical Properties Oxidation states and trends in chemical reactivity : The elements of group 16 exhibit a number of oxidation states. The stability of –2 oxidation state decreases down the group. Except oxygen other elements of the group exhibit +2, +4, +6 oxidation states but +4 and +6 are more common. Sulphur, selenium and tellurium usually show +4 oxidation in their compounds with oxygen and +6 oxidations state with fluorine. The stability of +6 oxidation state decreases down the group and stability of +4 oxidation state increases (inert pair effect). Bonding in + 4 and + 6 oxidation states are primarily covalent. (i) Reactivity with hydrogen : All the elements of group 16 formhydrides of the type H2E (E = O, S. Se, Te, Po). (ii) 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 {SOJ are gases while selenium dioxide (SeO2) is solid. Reducing property of dioxide decreases from SO2 toTeO2 ; SO2 is reducing while TeO2 is an oxidising agent. Besides EO2 type sulphur, selenium and tellurium also form EO3 type oxides (SO3, SeO3, TeO3}. Both types of oxides are acidic in nature. (iii) Reactivity toward the halogens: Elements of group 16 form a larger number of halides of the type EX6, EX4 and EX2 where E is an element of the group –16 and X is an halogen. The stabilities of the halides decrease in the order F > Cl > Br > l. Amongst hexahalides, hexafluorides are the only stable halides. All hexafluorides are gaseous in nature. They have octahedral structure. Sulphur hexafluoride SF6 is exceptionally stable for steric reasons. . Amongst tetrafluorides, SF4 is a gas, SeF4 liquid andTeF4 a solid. All elements except selenium form dichlorides and dibromides. The well known monohalides are dimeric in nature, Examples are S2F2, S2Cl2 S2Br2, Se2Cl2 and Se2Br2. These dimeric halides undergo disproportionation as given 2Se2Cl2 SeCl4 + 3Se COMPOUNDS OF OXYGEN 1. DIOXYGEN (O2) (a) Preparation of Di-Oxyges (O2) : By thermal decomposition of oxides of metals, 450 C 350 C  2Hg + O2; 2Ag2O   4Ag + O2 2HgO     Mn3O4 + O2 ; 2Pb3O4   6PbO + O2 3MnO2    2KCl + 3O2 (laboratory method) KClO3     4K2CrO4 + 2Cr2O3 + 3O2 ; 2 KMnO4   K2MnO4 + MnO2 + O2 4K2Cr2O7    2PbO(s) + O2(g) 2PbO2(s)  Properties :  Dioxygen is a colourless and odourless gas.  Oxygen atom has three stable isotopes: 16O, 17O and 18O. Molecular oxygen.  O2 is paramagnetic (by MOT)  Dioxygen directly reacts with nearly all metals and non-metals except some metals {e.g., Au, Pt) and –some noble gases. 2. OZONE (O3) : Preparation : It is prepared by passing silent electric discharge through pure and dry oxygen. Mixture obtained contains 5-10% ozone by volume and this mixture is called ozonised oxygen. 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. If concentrations of ozone greater than 10 per cent are required, a battery of ozonisers can be used, and pure ozone (b.p. 385 K) can be condensed in a vessel surrounded by liquid oxygen. Properties :  Pure ozone is a pale blue gas, dark blue liquid and violet-black solid.  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. Therefore, high concentrations of ozone can be dangerously explosive. Oxidising behaviour of ozone : Due to the ease with which it liberates atoms of nascent oxygen (O3  O2 + O), it acts as a powerful oxidizing agent For example, if oxidises tead sulphide to lead sulphate and Iodide ions to iodine. Tests for Ozone Tailing of mercury : Pure mercury is a mobile liquid but when brought in contact with O3 its mobility decreases and it starts sticking to glass surface forming a type of tail due to the dissolution of Hg2O (mercury sub-oxide) in Hg. 2Hg + O3  Hg2O + O2 Estimation of Ozone : When ozone reacts with an excess of potassium iodide solution buffered with a borate buffer (pH 9.2), iodine is titrated which can be titrated against a standard solution of sodium thiosulphate. This is a quantitative method for estimating O3 gas. 2KI + H2O + O3  O2 + I2 + 2KOH (Na 2S2 O3 .5H2 O) I2   2NaI + Na2S4O6 (sodium tetra thionate) Depletion of ozone layer : Experiments have shown that 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(g) + O3(g)  NO2(g) + O2(g) Another threat to this ozone layer is probably posed by the use of freons which are used in aerosol sprays and as refrigerants. COMPOUNDS OF SULPHUR : 1. Hydrogen Sulphide (H2S) Preparation : FeS + H2SO4  FeSO4 + H2S It is prepared in kipp's apparatus Properties : (i) Colourless gas with rotten egg smell (ii) Moderately soluble in water but solubility decreases with increasing temperature. (iii) It gives black ppt with lead acetate (CH3COO)2 Pb+ H2S  PbS + 2CH3COOH salt add black ppt Reducing behaviour : Acts as a strong reducing agent as it decomposes evolving hydrogen. moisture H2S + SO2  H2O + S; redox  2FeCl2 + S + 2HCl 2FeCl3 + H2S  green 2. SO2 (Sulphur Dioxide) Preparation : Burn  SO2 (i) S + O2 or air  (ii) By reaction of metal sulphites with dilute HCl (Laboratory method) Na2SO3 + 2HCl 2NaCl + SO2 + H2O Similarly bisulphites also give sp.~ with dilute HO NaHSO3 + HCl  NaCl + SO2 + H2O (iii) By heating sulphides (metal sulphide ores) in excess of air. 2ZnS + 3O2  2ZnO + 2SO2 By this method SO2 is obtained in large scale Properties : (i) Colourless gas with pungent smell. (ii) It is heavier than air and is highly soluble in water. (iii) Acidic Nature : Acidic oxide and thus dissolve in forming sulphurous acid. SO2 + H2O  H2SO3 Reducing nature : The above reactions are not given by CO2, so they are used to distinguish between SO2 and CO2 Oxidising nature : Acts as oxidizing agent with strong reducing agent 3. SULPHURIC ACID (H2SO4) : Manufacture by contacts process : Sulphurlc acid is one of the most important industrial chemicals worldwxide. Sulphuric acid is manufactured by the contact process which involves three steps : (i) burning of sulphur sulphide ores in air to generate SO2. (ii) Conversion of SO2 to SO3 by the reaction with oxygen in the presence of a catalyst (V2O3) (iii) Absorption of SO3 in H2SO4 to give Oleum (H2S2O7) The SO2 produced is purified by removing dust and other impurities such as arsenic compounds. The key step in the manufacture of H2SO4 is the catalytic oxidation of SO2 with O2 to give SO3 in the presence of V2O5 (catalyst). V2 O5 2SO2(g) + O2(g)   2SO3(g) FH– = –196.6 kJ mol–1 The reactiof1 is exothermic reversible and the forwardreactf6n leads to a decrease in volume. Therefore, row temperature and high pressure and the favourable conditions for maximum yield. But the temperature should not be very low other wise rate of reaction will become slow. Dilution of oleum with water gives H2SO4 of the desired concentration. In the industry two steps are carried out simultaneously to make the process a continuous one and slos to reuce the cost. SO3 + H2SO4  H2S2O7 ; H2S2O7 + H2O  2H2SO4 (Oleum) The sulphuric acid obtained by Contact process is 96-98% pure. Properties : Sulphuric acid is a colourless, dense, oily liquid with a specific gravity of 1.84 at 298 K. The acid freezes at 283 K and boils at 611 K It dissolves in water with the evolution of a larger quantity of heat. The chemical reaction 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 in aqueous solution, (i) Sulphuric acid ionises in two steps. H2SO4(aq) + H2O()  H3O+ (aq) + HSO4– (aq) ; Ka1 = very larger ( K a > 10) 1 HSO4– (aq) + H2O()  H3O+ (aq) + SO42- (aq) ; Ka2 = 1.2×10–2 The larger value of K a ( K a > 1 0) means that H2SO4 is largely dissociated into H+ and HSO4– Greater the value of dissociation constant (Ka) the stronger is the acid. The acid forms two series of salts : Normal sulphates (such as sodium sulphate and copper sulphate and acid sulphate (e.g., sodium hydrogen sulphate) Sulphuric acid, because of its low volatility can be used to manufacture more volatile acid from their corresponding salts. 2MX + H2SO4  2HX + M2SO4(X = F, Cl, NO3) NaCl + H2SO4  NaHSO4 + HCl 1 (ii) (iii) 1 (M =Metal) (iv) (v) KNO3 + H2SO4  KHSO4 + HNO3 Concentrated sulphuric acid is a strong dehydrating agent. Many wet gases can be dried by passing them through sulphuric acid, provided the gases do not react with the acid. Sulphuric acid removes water from organic compound; it is evident by its charring action on carbohydrates. H2 SO4 C12H22O11   12C + 11H2O (charring action of sugar) H2 SO4  CO + CO2 H2C2O4   H2 O Hot concentrated sulphuric acid is moderately strong oxidising agent In this respect it is intermediate between phosphoric and nitric acids. Both metals and non-metals are oxidised by concentrated sulphuric acid, which is reduced to SO2. Cu + conc. 2H2SO4  CuSO4 +2H2O 3S + conc. 2H2SO4  3SO2 + 2H2O C + conc. 2H2SO4  CO2 + 2SO2 + 2H2O KBr + conc.H2SO4  KHSO4 + 2H2O + Br2 + SO2 QUESTIONS (REASONING) Q.1 Ans. H2S is less acidic than H2Te. Why? Due to the decrease in bond (E–H) dissociation enthalpy down the group, acidic character increases. Q.2 Ans. Why is H2o a liquid and H2S a gas? Hydrogen bonds are present between Hp molecules while between H2S molecules, vander Waal's forces are present. Q.3 Ans. Why is dioxygen a gas but sulphur a solid? Oxygen exist as a O2 molecule while sulphur exist as a S8 molecule due to more molecular mass sulphur is solid. Q.4 Ans. Knowing the electron gain enthalpy values for O  O– and O  O2– as –141 and 702 kJ mol–1 respectively, how can you account for the formation of a large number of oxides having O2– species and not O–? Consider lattice energy factor in the formation of compounds. Q.5 Which of the following does not react with oxygen directly? Zn, Ti, Pt, Fe Ans. Pt is a noble metal which does not react directly with oxygen. Q.6 Complete the following reactions: (i) C2H4 + O2    (i) C2H4 + 3O2  2CO2 + 2H2O Ans.  (ii) 4Al + 3O2  (ii) 4Al + 3O2  2Al2O3 Q.7 Ans. How is O3 estimated quantitatively? 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. Q.8 Ans. What happens when sulphur dioxide is passed through an aqueous solution of Fe(III) salt? 2Fe+3 + SO2 + 2H2O  2Fe+2 + SO4–2 + 4H+ Q.9 Ans. How is the presence of SO2 detected? It has colourless gas with pungent smell and decolourise acidified KMnO4 solution. Q.10 Write the conditions to maximise the yield of H2SO4 by Contact process. Ans. The reaction is exothermic, reversible and the forward reaction leads to a decrease in volume. Therefore, low temperature and high pressure are the favourable conditions for maximum yield. But the temperature should not be very low otherwise rate of reaction will become slow. Q.11 Why is K a < < K a for H2SO4 in water? Ans. It is difficult to remove H+ ion from HSO4– ion. 2 1. 1 BEGINNER’S BOX-8 Which of the following is not oxidised by O3? (1) KI (2) FeSO4 (3) KMnO4 (4) K2MnO4 2. In which of the following reaction conc. H2SO4 is not used as on oxidising agent? (1) Cu + 2H2SO4 CuSO4 + SO2 + 2H2O (2) 2HI + H2SO4  I2 + SO2 + 2H2O (3) CaF2 + H2SO4  CaSO4 + 2HF (4) None 3. Hot conc. H2SO4 acts as strong oxidising agent which of the following element is oxidised by conc. H2SO4 into two gaseous products? (1) Cu (2) S (3) C (4) Zn 4. HCOOH reacts with conc. H2SO4 to produce (1) CO (2) CO2 (3) NO (4) NO2 HALOGEN FAMILY (GROUP-17) (A) (i) PHYSICAL PROPERTIES Fluorine and chlorine are gases, bromine is a liquid whereas iodine is a solid. (ii) (iii) (iv) (v) (B) (i)    Their melting and boiling points steadily increase with atomic number. 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. By absorbing different quanta of radiation, they display different colours. For example, F2, has yellow, Cl2, greenish yellow, Br2, red and I2, violet colour. Fluorine and chlorine react with water. Bromine and iodine are only sparingly soluble in water. But are soluble in organic solvents such as chloroform, carbon tetrachloride, carbon disulphide and hydrocarbons to give coloured solutions. Except the smaller enthalpy of dissociation of F2 compared to that of Cl2. The X–X bond disassociation enthalpies from chlorine onwards show the expected trend : Cl–Cl > Br– Br > F–F > I–I. The reason for the smaller enthalpy of dissociation of F2 is the relatively larger electrons - electrons repulsion among the lone pairs in F2 molecule where they are much closer to each other than in case of Cl2. CHEMICAL PROPERTIES Oxidation states and trends in chemical reactivity All the halogens exhibit –1 oxidation state. However, chlorine, bromine and iodine exhibit +1, +3, + 5. and +7 oxidation states also. The higher oxidation states of chlorine, bromine and iodine are realized mainly when the halogens are in combination with the small and highly electronegative fluorine and oxygen atoms e.g., in interhalogens, oxides and oxoacids. The fluorine atom has no d orbitals in its valence shell and therefore cannot expand its octet. Being the most electronegative, it exhibits only –1 oxidation state. All the halogens are highly reactive. They react with metals and non-metals to form halides. The reactivity of the halogens decreases down the group. Standard Reduction Potential (SRP) X2 + 2e–  2X– F2 + 2e– 2F– E– = +2.87V; Cl2 +2e–  2Cl– E– = +1.36 V – – – – – Br2 + 2e  2Br E = +1.09 V ; I2 + 2e  2I E– = +0.54 V More the value of the SRP; more powerful is the oxidising agent. Hence the order of oxidising power is F2 > Cl2 > Br2 > I2 Order of reducing behaviour is I– > Br– > Cl– > F– (ii)     Halogen oxides : Fluorine forms two oxides OF2 and O2F2. However, only OF2 is the thermally stable at 298 K These oxide are essentially oxygen fluorides because of the higher electronegativity of fluorine than oxygen. Both are strong fluorinating agents. O2F2 oxidises plutonium to PuF6 and the reaction is used in removing plutonium as PuF6 from spent nuclear fuel. Chlorine, bromine and iodine form oxides in which the oxidation states of these halogen vary from +1 to +7. A combination of kinetic and thermodynamic factors lead to the generally decreasing order of stability of oxides formed by halogens, I > Cl > Br. The higher oxides of halogens tend to be more stable than the lower ones. This is called middle row anamoly. Chlorine oxides, Cl2O, ClO2, Cl2O6 and Cl2O7 are highly reactive oxidising agents and tend to explode. ClO2 is used as a bleaching agent for paper pulp and textiles and in water treatment.   The bromine oxides, Br2O, BrO2, BrO3 are the least stable halogen oxides and exist only at low temperature. They are very powerful oxidising agents. The iodine oxides, I2O4, I2O5, I2O7 are insoluble solids and decompose on heating. I2O5 is very good oxidising agent and is used in the estimation of carbon monoxide. COMPOUNDS OF HALOGEN 1. CHLORINE GAS (Cl2) (a) Preparation : (i) By heating chloride with concentrated H2SO4 in presence of MnO2. 4H+ + MnO2 + 2X–  X2+Mn+2 + 2H2O Bromides and iodides also liberate Br2 and I2 respectively with concentrated H2SO4 and MnO2. (ii) • CaOCl2 + 2HCl  CaCl2 + Cl2 + H2O • 2KMnO4 + 16 HCl  2KCl + 2MnCl2 + 5Cl2 + 8H2O • PbO2 + 4HCl  PbCl2, + Cl2 + 2H2O (iii) Manufacture of chlorine :  Deacon's process : By oxidation of hydrogen chloride gas by atmospheric oxygen in the presence of CuCl2 (catalyst) at 723 K. CuCl2 4 HCl + O2   2Cl + 2H2O  Electrolytic process : Chlorine is obtained by the electrolysis of brine (concentrated NaCl solution). Chlorine is liberated at anode. It is obtained as a by-product in many chemical industries e.g.; in manufacturing of sodium hydroxide. NaX (aq)  Na+ (aq) + X– (aq) Anode : 2X–  X2 + 2e– (b) (i) (ii) Properties : It is a greenish-yellow gas with pungent and suffocating odour. It is aboul 2-5 times heavier than air. It can be liquefied into greenish-yellow liquid which boils at 239 K. It is soluble in water. Oxidising & bleaching properties : Chlorine dissolves in water giving HCl and HOCl. Hypochlorous acid (HOCl) so formed, gives nascent oxygen which is responsible for oxidising and bleaching properties of chlorine. It is a powerful bleaching agent; bleaching action is due to oxidation. 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. 2. (a) (i)    HALOGEN ACIDS (HCl) Preparation : By heating halide with concentrated add : 150 C  NaHSO4 + HCl NaCl + H2SO4  550 C  Na2SO4 + HCl NaHSO4 + NaCl  Above method is called as salt cake method as it involves the formation of NaHSO4 (salt cake). HCl cannot be dried over P2O5 (P4O10) or quick lime since they react with gas chemically. CaO + 2HCl  CaCl2 + H2O P4O10 + 3HCl  POCl3 + 3HPO3 HCl is, hence dried by passing through concentrated H2SO4. (b) (i) (ii) (iii) Properties: This is colourless, pungent smelling gas with acidic taste. This is neither combustible nor supporter of combustion. When perfectly dry, HX have no action on litmus, but in presence of moisture, they turn blue litmus red, showing acidic nature. Among HX, HI is the strongest and HF is the weakest acid. 3. Pseudo halogens and pseudo halides: Some inorganic compounds consisting of two or more atoms of which at least one is N have been found to behave like halogens & they are known as pseudo halogen solids. 4. INTERHALOGEN COMPOUNDS : We know that halogen atoms have different electronegativity. Due to this difference in electronegativity the halogen atoms combine with each other and give rise to the formation of binary covalent compounds, which are called interhalogen compounds. These are of four types. *Where A-halogen of smaller size and A is more electro positive than B. (a) (i) (ii) (iii) (iv) (v) Properties : hese compounds may be gases, liquids or solids. Gases: ClF, BrF, ClF3 , IF7 ; Liquids : BrF3, BrF5; Solids: ICI, IBr, IF3, ICI3. Interhalogens containing fluorine are generally colourless but inter halogens consisting of heavier halogens are coloured. The intensity of colour increases with increase in the molecular weight of the compounds. All interhalogens are covalent molecules and are diamagnetic in nature since all the valence electrons present as bonding or non-bonding electrons are paired. Interhalogen compounds are more reactive than the parent halogens but less reactive than F2. ICI + 2Na  NaI + NaCl The order of reactivity of some interhalogens is as follows : CIF3 > BrF3 > IF7 > BrF5 > BrF These compounds can be used as non aqueous solvents. Interhalogen compounds are very useful fluorinating agents. CIF3 and BrF3 are used for the production of UF6 in the enrichment of 235U. U(s) + 3 CIF3( )  UF6 (g) + 3 ClF (g) TEST OF HALOGENS 1. F– = Itching of glass 2. AgNO3 test (Cl– , Br, I–) Cl – Salt + AgNO3  AgCl (White ppt) Br salt+ AgNO3  AgBr (Pale yellow ppt) I – salt + AgNO3 AgI (Yellow ppt) 3. Test of I2 I2 + Starch  Navy blue I2 + CCl4  Violet 4. Layer test (Br–, I–) CCl4 Cl2 + 2Br  2Cl– + Br2   Reddish brown (Halogen displacement reaction) (Note : In case of I– violet colour is obtained.) 5. Chromyl chloride test (Test of Cl–): see d-block. QUESTIONS (REASONING ) Q.1 Covalent radius of fluorine is 64 pm but the bond length is not equal to 128 pm and that is 143 pm and bond energy is found to be comparable to I2. Ans. Q.2 Ans. This may be attributed to .p- .p repulsions due to small size of F atom. Electron affinity of chlorine is more than F. lnspite of this F 2 is the better oxidising agent. Why? SRP of F2 is much higher than that of Cl2 on account of smaller bond dissociation energy and high hydration energy of F– ion. Q.3 Ans. Layer test of Br– is successful with Cl2 but not with I2 Explain? Br– is oxidised by Cl2 but not by F2 Q.4 Ans. What is the difference between bleaching .action of SO2 and Cl2 The bleaching action of SO2 is temporary because it takes place through reduction. SO2 + 2H2O  H2SO4 + 2H SO32– + Coloured material  SO42– Reduced colourless material. O2 of air Reduced Colourless material    Coloured material The bleaching action of Cl2 is permanent because jt takes place through oxidation Q.5 (a) When HCl reacts with finely powdered iron, it forms ferrous chloride and not ferric chloride wily ? (b) Chlorine water turns blue litmus red but solution becomes colourless after sometime. After sometime. (a) It forms H2 gas Fe + 2 HCl  FeCl2 + H2. Liberation of hydrogen prevents the formation of ferric chloride. (b) Blue litmus change into red due to acidic nature (Cl2 + H2O  HOCl + HCl) but it is bleaching, agent, therefore, it decolourises the red litmus. Ans. ZERO GROUP (GROUP-18) INTRODUCTION Group 18 consists of six elements: helium, neon, argon, krypton, xenon and radon. All these are gases and chemically unreactive. They form very few compounds. Because of this they are termed noble gases. (A) (i) (ii) (iii) (iv) (B) PHYSICAL PROPERTIES All the noble gases are mono-atomic. They are colourless, and tasteless. They are sparingly soluble in water. They have very low melting and boiling points because the only type of interatomic interaction in these elements is weak dispersion forces. Helium has the lowest boding point (4.2K) of any known substance. It has a unusual property of diffusing through most commonly used laboratory materials such as rubber, glass or plastics. CHEMICAL PROPERTIES In general, noble gases are least reactive. Their inertness to chemical reactivity is attributed to the following reasons: (i) (ii) The noble gases except helium (1s2) have completely filled ns2 np6 electronic configuration in their valence shell. They have high ionisation enthalpy arid more positive electron gain enthalpy. Neil Bartlett, observed the reaction of a noble gas. First, he prepared a red compound which is formulated as O2+ [PtF6]–. He, then realised that the first ionisation enthalpy of molecular oxygen (1175 kJ mol–1) was almost identical with that xenon (1170 kJ mol–1). He made efforts to prepare same type of compound with Xe+ [PtF6]– by mixing PtF6 and Xenon. After this discovery, a number of xenon compounds mainly with most electronegative elements like fluorine and oxygen, have been synthesised . The compounds of krypton are fewer. Only the difluoride (KrF2) has been studied in detail. Compounds of radon have not been isolated but only identified (e.g., RnF2) by radiotracer technique. No true compounds of Ar, Ne or He are yet known. COMPOUNDS OF XENON Xenon-fluorine compounds Xenon forms three binary fluorides. XeF2, Xe4 and XeF6 by the direct reaction of elements uder appropriate experimental conditions. 673K1bar  XeF2(s) Xe (g) + F2(g)  (xenon in excess) 873K 7bar  XeF4 (s) Xe (g) + 2F2 (g)  (1:5 ratio) 873K 60  70bar  XeF6 (s) Xe (g) + 3F2 (g)  (1:20 ratio) XeF6 can also he prepared by the interaction of XeF4 and O2F2 at 143 K. XeF4 + O2F2 XeF6 + O2 XeF2, XeF4 and XeF6 are colourless crystalline solids and sublime readily at 298 K. They are powerful fluorinating agents. They are readily hydrolysed even by traces of water. For example, XeF2 is hydrolysed to give Xe, HF and O2. 2XeF2(s) + 2H2O()  2Xe (g) + 4HF(aq) + O2(g) Formation of addition compounds : XeF2 reacts with fluoride ion acceptors to form cationic species and fluoride ion donors to form fluoroanions. XeF2 + PF5  [XeF]+ [PF6]– IF5 + XeF2  [XeF]+ [IF6]– (lewis acid) 2SbF5 + XeF2  [XeF)]+[SbF6]– (lewis acid) CsF + XeF6 Cs[XeF7]– [With s-block flourides and alkyl flourides it is F– acceptor with others it is F– donor] USES OF INERT GASES: (1) He is non-inflammable and light gas, so it is used in filling balloons for meteorological observations. (2) He is used in gas cooled nuclear reactors. (3) Liquid He is used as cryogenic agent. (4) He is used to produce powerful superconducting magnets. (5) Ne is used in discharge tubes. (6) Ar is used as inert atmosphere in metallurgical process. (7) Xenon and Krypton are used in light bulbs designed for special purposes. IMPORTANT ORDER: (1) He < Ne < Ar < Kr < Xe Atomic radius (2) He > Ne > Ar > Kr > Xe Ionisation energy (3) He < Ne < Ar < Kr < Xe < Rn Density (4) He < Ne < Ar < Kr < Xe < Rn mpt. bpt [Atomic mass / size / polarisability / London force] (5) He < Ne < Ar < Xe < Rn [solublilty in water because of dipole-induced doipole attraction] Clatherate Compounds : Larger inert gas atoms are trapped into cavity or crystal structure of some organic compounds and form clatherate compounds. For eg xenondiquinol.  Force of attraction between noble gas and organic compound is dipole-induced dipole attraction. He, Ne do not form clatherate compounds due to smaller size. 1. 2. 3. BEGINNER’S BOX-9 Which is correct conc.H2 SO4 (1) NaCl  yellow green gas conc.H2 SO4 (2) NaBr  red brown vapour (3) NaF +Cl2 NaCl + F2 Middle row anamoly" is related with (1) stability of transition metal oxides (3) stability of halogen oxides True statement about I– will be (1) weak base (3) strong reducing agent (2) stability of alkali metal oxides (4)stability of alkaline earth metal oxides (2) strong nucleophile (4) All 4. Of the following statements : (a) Cl2 gas is dried by using conc.H2SO 4 (b) Fluorine have highest oxidising power (c) Oxidising power of halogens follow the order I2 > Br2 > Cl2 (d) HI is the strongest acid among HI, HBr, HCl (1) a, b and d are correct (2) a, c are correct (3) b, care corrects (4) c, d are correct 5. Chromyl chloride test is performed for the confirmation of the presence of the following in mixture: (1) SO42– (2) Cr+++ (3) Cl– (4) Cr+++ and Cl– 6. The ion that cannot undergo disproportionation is (1) ClO4– (2) ClO3– (3) ClO2– (4) ClO– 7. A greenish yellow gas reacts with an alkali metal hydroxide to form a halate which can be used in fire works and safety matches. The gas and halate respectively are : (1) Br2, KBrO3 (2) Cl2, KClO3 (3) I2, NaIO3 (4) Cl2, NaClO 8. In the clatherates of xenon with water, the nature of bonding between xenon and water molecule is : (1) covalent (2) hydrogen bonding (3) co-ordinate (4) dipole-induced dipole interaction 9. XeF2 reacts with PF5 to give : (1) XeF6 (2) [XeF]+[PF6] (4) [PF4]+[XeF3]– (3) XeF4 10. The first compound of noble gases prepared by Neil-Bartlett was:(1) Xe+ [PtF6]– (2) XeF4 (3) XeF6 (4)XeOF4 ANSWER KEY 1. 8. (1) (3) 2. 9. (4) (1) 3. 10. BEGINNER’S BOX-1 (1) 4. (1) 5. (4) (1) 6. (4) 7. (3) 1. 8. (2) (4) 2. 9. (4) (2) 3. 10. BEGINNER’S BOX-2 (3) 4. (3) 5. (2) (3) 6. (2) 7. (1) 1. 8. (3) (2) 2. 9. (2) (4) 3. 10. (2) 6. (2) 7. (3) 1. 8. (3) (4) 2. 9. (2) (1) 3. 10. BEGINNER’S BOX-3 (3) 4. (2) 5. (3) BEGINNER’S BOX-4 (2) 4. (2) 5. (3) (4) 6. (3) 7. (4) 1. 8. (1) (3) 2. 9. (4) (4) 3. 10. (2) (3) BEGINNER’S BOX-5 4. (2) 5. (3) 6. (4) 7. (3) 1. (3) 2. (3) 3. BEGINNER’S BOX-6 (4) 4. (4) 1. 8. (1) (2) 2. 9. (4) (2) 3. 10. BEGINNER’S BOX-7 (4) 4. (2) 5. (4) (4) 6. (4) 7. (3) 1. (3) 2. (3) 3. BEGINNER’S BOX-8 (3) 4. (1) 3. 10. BEGINNER’S BOX-9 (4) 4. (1) 5. (1) (3) 6. (1) 7. (2) 1. 8. (2) (4) 2. 9. (3) (2)