Title 8A.HEAVY METALS SYNOPSIS_Final ( 205 - 223 ).pdf ✅

NISHITH Multimedia India (Pvt.) Ltd., 205 JEE ADVANCED VOL - VII HEAVY METALS PREPARATION AND PROPERTIES OF COMPOUNDS OF HEAVY METALS SYNOPSIS TIN & ITS COMPOUND 1. OXIDES OF TIN (i) Sn O2 1500 C     SnO2 [Burns with a bright flame] (ii) Sn + 2H2O (iii) Reactions with acid and base . Dil.HCl reaction is very slow due to non oxidising nature of the acid Hot conc.HCl Sn+2HCl SnCl2+H2 Dil.H2SO4 Dissolve Sn slowly forming SnSO4+H2 Hot .Conc.H2SO4 Sn+4H2SO4 Sn(SO4)2+2SO2+4H2O Cold veryDil.HNO3 4Sn+10HNO3 4Sn(NO3 )2+NH4NO3+3H2O Dil.HNO3 4Sn+10HNO3 4Sn(NO3 )2+N2O+5H2O Hot .Conc HNO3 5Sn+20HNO3 H2Sn5O11 .4H2O+20NO2+5H2O Metastanic acid SnO2 Sn+2NaOH+H2O Na2SnO3+2H2 Sn
Jr Chemistry E/M HEAVY METALS 206 NISHITH Multimedia India (Pvt.) Ltd., JEE ADVANCED VOL - VII KOH [In absence of air Na2 SnO2 forms and in contact with air it readily converts into Na2 SnO3 ] Oxides: grey white SnO &SnO2 heated strongly 1500 C    Sn + O2 SnC2O4 air contact of out of  SnO (grey) + CO + CO2 SnCl2  KOH Sn(OH)2 Both SnO and SnO2 are amphoteric in nature : SnO + H2 SO4  SnSO4 + H2O SnO + 2HCl  SnCl2 + H2O SnO + 2NaOH or KOH   cold Na2 SnO2 or K2 SnO2 +H2O But conc. hot alkali behaves differently. 2SnO + 2KOH or NaOH  K2 SnO3 or Na2 SnO3 + Sn + H2O Bi(OH)3 + [Sn(OH)4 ] 2–  Bi  + [Sn(OH)6 ] 2– (black) SnO2 + 2H2 SO4 Sn(SO4 ) 2 + 2H2O (Soluble only in hot conc. H2 SO4 ) SnO2 + 2NaOH  Na2 SnO3 + H2O SnO2 + 4S + 2Na2CO3  Na2 SnS3 + Na2 SO4 + 2CO2 (Insoluble in all acids even in aqaregia) 2. SnCl2 & SnCl4 : * Sn + 2HCl (hot conc.) SnCl2 + H2  SnCl2 .2H2O   Sn(OH)Cl + HCl  + H2O Hence anh. SnCl2 cannot be obtained.  SnO + HCl  {SnCl4 + 4H2O  Sn(OH)4 + 4HClfumes comes out} * A piece of Sn is always added to prepare a solution of SnCl2 . Explain. 6SnCl2 + 2H2O + O2  2SnCl4 + 4Sn(OH)Cl (white ppt) SnCl4 + Sn  2SnCl2 SnCl4 + 4H2O  Sn(OH)4  (white ppt.) + 4HCl * SnCl2 + HCl  HSnCl3   HCl H2 SnCl4 SnCl4 + 2HCl  H2 SnCl6 (Hexachloro stannic (IV) acid) SnCl4 + 2NH4Cl  (NH4 ) 2 SnCl6 (colourless crystalline compound known as " pink salt ")
NISHITH Multimedia India (Pvt.) Ltd., 207 JEE ADVANCED VOL - VII HEAVY METALS * Reducing prop. of SnCl2 : Sn+2 + 2Fe+3  2Fe+2 + Sn+4 2Cu+2 + Sn+2 2Cu+ + Sn+4 Hg+2 + Sn+2 Hg  +Sn+4 2 4 C H NO 3Sn 2H C H NH 3Sn 2H O 6 5 2 6 5 2 2         2 2 3 4 Cr O 3Sn 14H 2Cr 7H O 3Sn 2 7 2           2 4 Sn I Sn 2I 2       * This reaction is used to estimate tin. Formation of SnCl4 : * (molten) Sn + (dry) Cl (Excess) 2  SnCl4 (ii) 2HgCl2 + Sn  2Hg  + SnCl4 * 3Sn 2HNO 18HCl 3H SnCl 2NO 7H O 3 2 2 6              * SnCl4 . 5H2O is known as butter of tin  used as mordant. (NH4 ) 2 SnCl6 is known as 'pink salt'  used in calico printing. Mosaic gold : SnS2 yellow crystalline substance : Sn + 4NH4Cl  (NH4 ) 2 SnCl4 + 2NH3 + H2 2(NH4 ) 2 SnCl4 + 2S  SnS2 + 2NH4Cl + (NH4 ) 2 SnCl6 * Distinction of Sn+2 / Sn+4 : (i) H S2 : SnS: Chocolate Brown ppt. SnS2 : Yellow ppt. (ii) 2 Hg  : 2 Sn  reduces HgCl2 to Hg Cl 2 2 (white ppt.) and further addition leads to grey or black Hg ppt. 4 Sn  does not give directly white ppt. with HgCl2 . But on addition of 3 Fe  to 4 Sn  , 2 Sn  is obtained due to reduction. 3 4 2 2 2Fe Sn 2Fe Sn        . The solution gives white and grey ppt. with SnCl2 . COMPOUNDS OF LEAD 1. Oxides of lead : (i) PbO (ii) Pb3O4 (Red) (iii) Pb2O3 (reddish yellow) (Sesquioxide) (iv) PbO2 (dark brown) * PbO
Jr Chemistry E/M HEAVY METALS 208 NISHITH Multimedia India (Pvt.) Ltd., JEE ADVANCED VOL - VII Laboratory Prepn . : Pb(NO3 ) 2  2PbO + 4NO2 + O2 PbO2  above 600C Pb3O4  Pb2O3  H or C. easily reduced toPb by PbO, hot oxide  2     Preparation of Pb2O3 : 2PbO hotsolution . of it in NaOH + amount Limited NaOCl  Pb2O3 + NaCl Pb2O3 + 2HNO3  PbO2  + Pb(NO3 ) 2 + H2O This reaction suggests that Pb2O3 contains PbO2 . *PbO2 :Insoluble in water. HNO3 , But reacts with HCl + H2 SO4 (hot conc.) and in hot NaOH / KOH. (i) 3 4 3 2 3 2  2 Conc. Pb O 4HNO PbO 2Pb NO 2H O     (ii) Pb(OAc)2 + Ca(OCl)Cl + H2O  PbO2 + CaCl2 + 2CH3CO2H [Brown(dark)]  Excess bleaching powder is being removed by stirring with HNO3 Reaction : PbO2 + 4HCl  PbCl2 + Cl2 + 2H2O 2PbO2 + 2H2 SO4  2PbSO4 + 2H2O + O2 PbO2 + 2NaOH  Na2 PbO3 + H2O Sodium plumbate * Pb3O4 : 6PbO + O2 2Pb3O4 {In the same way, prove that its formula is 2PbO. PbO2 } Pb3O4 + 4HNO3 (cold.conc) or (hot dil.)  2Pb(NO3 ) 2 + PbO2 + 2H2O But 2Pb3O4 + 6H2 SO4  6PbSO4 + 6H2O + O2 Pb3O4 + 8HCl  3PbCl2 + 4H2O + Cl2 PbO2 : Powerful oxidising agent : (i) PbO2 + SO2  PbSO4 [spontaneously]    PbS + 4O3  PbSO4 + 4O2 PbS + 4H2O2  PbSO4 + 4H2O (ii) PbO2 + 2HNO3 + (COOH)2  Pb(NO3 ) 2 + 2CO2 + 2H2O (iii) 2MnSO4 + 5PbO2 + 6HNO3  3Pb(NO3 ) 2 + 2PbSO4  + 2HMnO4 + 2H2O Colorless Pink (Test for 2 Mn  ) 2. PbCl4 : Exists as H2 [PbCl6 ] PbO2 + 4HCl  PbCl4 + 2H2O {ice cold conc. saturated with Cl2 } PbCl4 + 2HCl  H2 PbCl6