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p Block Elements ( XII CBSC TOPIC)

DEAR STUDENTS,<br> I have forund as a tutotr that preparing for topic p block elements is a very challenging job for students as there are so many things to remember . I have tried to make a ppt that can help students a lot for preoaring this topic and score good marks.<br>Dr Mona Srivastava <br>Founder , Masterchem classes.

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p Block Elements ( XII CBSC TOPIC)

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  1. Masterchemclasses CLASS-12 SUBJECT- CHEMISTRY CHAPTER - P BLOCK ELEMENTS Dr. Mona Srivastava M.Sc. Ph. D.

  2. Learning objective – GROUP 15 ELEMENTS – • Introduction • Occurrence • Electronic configuration PERIOD 1

  3. P block elements in periodic table-

  4. Group 15 includes • N, P, As, Sb , Bi • 2. As we go down the group, there is a shift from non-metallic to metallic through metalloid character. • N , P 2 non-metals, • As , Sb 2metalloids • Bi 1 metal. Introduction

  5. Molecular nitrogen (N2) - It comprises of 78% by volume of the atmosphere. • Ores of nitrogen- • 1. Sodium nitrate, NaNO3 (called Chile saltpetre and ) 2.Potassium nitrate (Indian saltpetre). • 3. Proteins - It is found in the form of proteins in plants and animals. • Phosphorus - occurs in minerals. • 1.Fluorapatite - Ca9 (PO4)6. CaF2 (the main components of phosphate rocks.) • Bones/ Living cells – P is an essential constituent of animal and plant matter. It is present in bones as well as in living cells. • Milk/eggs -Phosphoproteins are present in milk and eggs. • Arsenic, antimony and bismuth are found mainly as sulphide minerals. Occurrence

  6. Electronic configuration Their valence shell electronic configuration is ns2np3

  7. Covalent and ionic (in a particular state) radii increase in size down the group. • There is a considerable increase in covalent radius from N to P. • However, from As to Bi only a small increase in covalent radius is observed. • This is due to the presence of completely filled d and/ f orbitals in heavier members. Atomic and ionic radii

  8. Ionization enthalpy decreases down the group due to gradual increase in atomic size. • The ionization enthalpy of the group 15 elements is much greater than that of group 14 elements – ( In period) • Reason- Because of the extra stable half-filled p orbitals electronic configuration and smaller size, • The order of successive ionization enthalpies, as expected is  H1 <  H2 <  H3 . Ionization Enthalpies

  9. The electronegativity value, in general, decreases down the group with increasing atomic size. But amongst the heavier elements, the difference is not that much pronounced Electronegativities

  10. Learnin objectives - • Physical properties • Oxuidation states • Anomelous properties of Nitrogen Period 2

  11. All the elements of this group are polyatomic. • Dinitrogen is a diatomic gas while all others are solids. • Metallic character increases down the group. • Nitrogen and phosphorus are non-metals. • Arsenic and antimony metalloids . • Bismuth is a metal. • This is due to decrease in ionisation enthalpy and increase in atomic size. • The boiling points, in general, increase from top to bottom in the group . • The melting point increases upto arsenic and then decreases upto bismuth. • Except nitrogen, all the elements show allotropy. Physical Properties

  12. All the elements of group 15 have 5 electrons in their outermost orbit. They need only 3 electrons to complete their octet configuration. The octet can be achieved either by gaining 3 electrons or by sharing 3 electrons by means of covalent bonds. As a result, the common negative oxidation state of these elements is - 3.  • As we move down the group, the tendency to exhibit -3 oxidation state decreases. This is due to the increase in atomic size and metallic character. • Group 15 elements also show positive oxidation states of +3 and +5 by forming covalent bonds. Due to the inert pair affect the stability of +5 oxidation state decreases down the group, while that of +3 oxidation state increases. contd……. Oxidation states and trends in chemical reactivity

  13. Nitrogenhas only s- and p-orbitals, but no d-orbitals in its valance shell. Therefore, nitrogen can show a maximum covalency of 4. • A covalency of four is   obtained by sharing its lone pair of electron with another atom or ion. • Phosphorusand the remaining   elements can exhibit a covalency of five and a maximum covalency, also called expanded covalency of six. • This is possible because of the presence of vacant d-orbitals in the valence shell.  • All the compounds of group fifteen elements, which exhibit a +5 oxidation state, are covalent. Continued ……

  14. Nitrogen differs from the rest of the members of this group due to 1. Small size 2. non-availability of d orbitals. 3. high ionisation enthalpy 4. high electronegativity • Nitrogen has unique ability to form - • pp-pp multiple bonds with itself and with other elements (C, O). Thus, nitrogen exists as a diatomic molecule with a triple bond (one s and two p) between the two atoms. • Heavier elements of this group do not form pp-pp bonds because of their large atomic orbitals due to which they cannot have effective overlapping. Contd…… Anomalous properties of nitrogen

  15. Phosphorus, Arsenic And Antimony - Form single bonds as P–P, As–As and Sb–Sb smuth forms metallic bonds in elemental state. N–N bond is weaker than the single P–P bond . Reason- The N-N single bonds have high interelectronic repulsion of the non-bonding electrons, owing to the small bond length. As a result the catenation tendency is weaker in N2 , forming weaker covalent bonds. Contd……..

  16. Absence of d orbitals in valence shell in N2 -- • Due to absence of d orbitals N2 has unusual chemistry than rest of the members. • Nitrogen cannot form dp –pp bond as the heavier • elements can e.g., R3P = O or R3P = CH2 (R = alkyl group). • P and As can form dp –dp bond also with transition metals if their compounds like P(C2H5)3 and As(C6H5)3 act as ligands. Contd……

  17. Learning objectives- • Chemical reactivity of nitrogen towards- • Hydrogen • Oxygen • Halogen Period 3

  18. All the elements of Group 15 form hydrides of the type EH3 • E = N, P, As, Sb or Bi. • The hydrides show regular variation in their properties. • The stability of hydrides decreases from NH3 to BiH3 • The reducing character of the hydrides increases. • Ammonia is only a mild reducing agent while BiH3 is the strongest reducing agent of all the hydrides. • Basicity also decreases in the order NH3 > PH3 > AsH3 > SbH3 > BiH3. Reactivity towards hydrogen

  19. All elements of 15 group form two types of oxides: • E2O3 and E2O5. • The oxide in the higher oxidationstate of the element is more acidic than that of lower oxidation state. • Their acidic character decreases down the group. • E2O3 - Nitrogen and phosphorus - purely acidic, • E2O3 - Arsenic and antimony - amphoteric • E2o3 - Bismuth - Basic. Reactivity towards oxygen:

  20. These elements of 15 group form form types of halides: 1.EX3 , 2. EX5 • Nitrogen does not form pentahalide due to non-availability of the d orbitals in its valence shell. • Pentahalides are more covalent than trihalides. • All the trihalides of these elements except those of nitrogen are stable. • In N2 only NF3 is known to be stable. • Trihalides are covalent in nature except BiF3. Reactivity towards halogens:

  21. All the elements of group 15 react with metals to form binary compounds exhibiting –3 oxidation state. • Ex. Ca3N2 (calcium nitride) , Ca3P2 (calcium phosphide), • Na3As2 (sodium arsenide), Zn3Sb2 (zinc antimonide) • Mg3Bi2 (magnesium bismuthide). Reactivity towards metals

  22. Q.1. Though nitrogen exhibits +5 oxidation state, it does not form pentahalide. (Give reason) • Ans. - Nitrogen with n = 2, has s and p orbitals only. It does not have d orbitals to expand its covalence beyond four. That is why it does not form pentahalide. • Q.2.Give reason - PH3 has lower boiling point than NH3. Why? • Ans.-Unlike NH3, PH3 molecules are not associated through hydrogen bonding in liquid state. That is why the boiling point of PH3 is lower than NH3. Example- ( NCERT P -171 )

  23. Q.1. Why are pentahalides more covalent than trihalides ? Q.2. Why is BiH3 the strongest reducing agent amongst all Group 15 elements ? Assignment No. 3

  24. Learning objectuve – Dinitrogen – • A. commercial production • Physical properties • Chemical pproperties • Uses. Period 4

  25. Introduction- • Nitrogen was discovered in 1770 by Scheele and Priestley. • This nonmetallic element has no color, taste or odor and is present in nature as a noncombustible gas. • When compared with the rest of Group 15, nitrogen has the highest electronegativity which makes it the most nonmetallic of the group. • The common oxidation states of N2 are +5, +3, and -3. • N2 makes up about 0.002% of the earth's crust. • It constitutes 78% of the earth’s atmosphere by volume. DiNitrogen ( N2)

  26. N2 is produced commercially by – “The liquefaction and fractional distillation of air. Liquid dinitrogen (b.p. 77.2 K) distils out first leaving behind liquid oxygen (b.p. 90 K).” Commercial production-

  27. Dinitrogen in laboratory is prepared by treating - • Ammonium chloride (aq) + sodium nitrite. NH4CI(aq) + NaNO2(aq) N2(g) + 2H2O(l) + NaCl (aq) Small amounts of NO and HNO3 are also formed in this reaction; these impurities can be removed by passing the gas through aqueous sulphuric acid containing potassium dichromate. 2. It can also be obtained by the thermal decomposition of ammonium dichromate. Heat (NH4)2Cr2O7 N2 + 4H2O + Cr2O3 Laboratory method

  28. Physical properties 1. N2 is a colorless, odorless, tasteless and non-toxic gas. 2. Nitrogen atom has two stable isotopes: 14N and 15N. 3. It has a very low solubility in water (23.2 cm3 per litre of water at 273 K and 1 bar P ) 4.Low freezing and boiling points .

  29. N2 is inert at room temperature because of the high bond enthalpy of N N bond. • Reactivity, increases rapidly with rise in temperature. • 1. Reaction with metals and nonmetals ( at higher temp) • a. N2 reacts some metals to form ionic nitrides 6Li + N2 2Li3N 3Mg + N2 • b.Non-metals to form covalent nitrides. 3 Mg + N2 Mg3N2 • 2. Reaction with hydrogen- It combines with hydrogen at about 773 K in the presence of a catalyst (Haber’s Process) to form ammonia: N2(g) + 3H2(g) 2NH3(g) f H 0 = – 46.1 kJmol–1 Chemical properties -

  30. Chemical Properties - Continued……

  31. Learning objectives- Ammonia • Manufacture • Physical properties • Chemical properties • Uses Period -5

  32. Preparation – • 1. Formation in air- NH3 is present in small quantities in air and soil , it is formed by the decay of nitrogenous organic matter • e.g., urea. NH2CONH2 + 2H2O NH4 2CO3 • NH4 2CO3 ⇌ 2NH3 + H2O + CO2 • 2. On a small scale – NH3 in small scale obtained from ammonium salts by their decomposition followed by further treatment with caustic soda or calcium hydroxide. • 2NH4Cl + Ca(OH)2 2NH3 + 2H2O + CaCl2 (NH4)2 SO4 + 2NaOH 2NH3 + 2H2O + Na2SO4 Ammonia (NH3)

  33. Commercially NH3 is manufactured by Haber’s process. • N2(g) + 3H2(g) ⇌ 2NH3(g) f H0 = – 46.1 kJ mol–1 Commercial preparation of ammonia

  34. As per le Chatelier’s principle, following conditions woulf favour the formation of ammonia- • 1. High pressure would favour the formation of ammonia. • ( 200 × 105 Pa (about 200 atm),. • 2. A temperature of 400-450 0 C ( 700 K) helps in reaction to go in forward direction. • 3. The use of a catalyst such as iron oxide. • 4. Catalytic promotor - Small amounts of K2O and Al2O3 to increase the rate of attainment of equilibrium. • (The flow chart for the production of ammonia is shown in previous fig. , in which its shown that earlieriron was used as a catalyst with molybdenum as a promoter.) Conditions favoring formation of NH3

  35. 1.Ammonia is a colourless gas with a pungent odour. • 2. Its freezing and boiling points are 198.4 and 239.7 K . • 3. In the solid and liquid states, it is associated through hydrogen bonds and therefore it accounts for its higher melting and boiling points. • Geometry - • 1. The geometry of the molecule is • trigonal pyramidal with the • nitrogen atom at the apex. • 2. It has three bond pairs and • one lone pair of electrons as • shown in the structure. Physical properties of ammonia

  36. 1. Highly soluble in water- NH3 gas is highly soluble in water. Its aqueous solution is weakly basic due to the formation of OH– ions. • NH3(g) + H2O(l) ⇌ NH4+(aq) + OH– (aq) • 2. Formation of ammonium salts- It forms ammonium salts with acids, e.g., NH4Cl, (NH4)2 SO4, etc. • As a weak base, these salts sprecipitate the hydroxides (hydrated oxides in case of some metals) of many metals from their salt solutions. For example, • ZnSO4 aq + 2NH4OHaq Zn (OH)2 (s)+ (NH4) 2 SO4 aq (white )ppt Chemical Properties of ammonia

  37. The presence of a lone pair of electrons on the nitrogen atom of the ammonia molecule makes it 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 metal ions such as Cu2+, Ag+: • Ex. Cu2+ (aq) + 4 NH3(aq) ⇌ [Cu(NH3)4]2+ (aq) • (blue) (deep blue) • Ag +(aq) + Cl- (aq) AgCls (s) (colourless) (white ppt) • AgCls + 2NH3 (aq) [Ag (NH3)2] Cl (aq) • (white ppt) (colourless) Chemical properties continued…….

  38. 1. Ammonia is used to produce various nitrogenous fertilisers - Ex. 1. ammonium nitrate 2. urea, 3. ammonium phosphate 4. ammonium sulphate) • 2. In the manufacture of some inorganic nitrogen compounds, the most important one being nitric acid. • 3. Liquid ammonia is also used as a refrigerant. Uses of Ammonia

  39. Q.1. Why does nitrogen show catenation properties less than phosphorous ? • Q.2. How is ammonia manufactured industrially? • Q.3.Discuss the trends in the chemical reactivity of group 15 elements. • Q.3.Mention the conditions required to maximize the yield of ammonia. • Q.4. How does ammonia react with a solution of Cu2+? Assignment No. 5

  40. Learning objective Nitric acid • Preparation • Properties • uses Period 6

  41. Nitrogen forms 3 types of oxoacids • H2N2O2 (hyponitrous acid). • HNO2 (nitrous acid) • HNO3 (nitric acid). • Amongst them HNO3 is most important. Nitric acid (HNO3)

  42. Laboratory method - In the laboratory, nitric acid is prepared • NaNO3 + H2SO4 (conc) HeatingNaHSO4 + HNO3 Or KNO3 in a in a glass retort • 2. Large scale(commercial) production –By Ostwald’s process. • Method- By the catalytic oxidation of NH3 using atmospheric O2. • 4NH3 (g)+ 5O2 (g)  4NO(g)+ 6H2O(g) (from air) • Nitric oxide thus formed combines with oxygen giving NO2. • 2NOg + O2 g ⇌ 2NO2 g • Nitrogen dioxide so formed, dissolves in water to give HNO3. • 3NO2(g) + H2O (l)  2HNO3 (aq)+ NO(g) • NO thus formed is recycled and the aqueous HNO3 can be concentrated by distillation upto ~ 68% by mass. • Concentration upto98% can be achieved by dehydration with concentrated H2SO4. Preparation-

  43. Catalytic oxidation of ammonia by air- Ostwald's method of preparation of HNO3

  44. Physical properties- • 1. It is a colorless liquid (f.p. 231.4 K and b.p. 355.6 K). • Laboratory grade HNO3 has ~ 68% of the HNO3 by mass. • It has a specific gravity of 1.504. Structure- In the gaseous state, HNO3 exists as a planar molecule with the structure as shown- Chemical properties – 1. In aqueous solution, nitric acid behaves as a strong acid giving hydronium and nitrate ions. HNO3 (aq) + H2O (l)  H3O+(aq) + NO3- (aq) • 2. Concentrated nitric acid is a strong oxidizing agent . It attacks most metals except noble metals such as gold and platinum. Properties of Nitric acid

  45. 3. The products of oxidation depend upon the - • Concentration of acid 2. Temperature 3. The nature of the material undergoing oxidation. 3Cu + 8 HNO3 (dilute)  3Cu(NO3)2 + 2NO + 4H2O Cu + 4HNO3 (conc.)   Cu(NO3)2 + 2NO2 + 2H2O • Reaction of nitric acid on Zinc- • With dil. dilute nitric acid - 4Zn + 10HNO3(dilute)  4 Zn (NO3)2 + 5H2O + N2O b. With conc. nitric acid - Zn + 4HNO3(conc.)  Zn (NO3)2 + 2H2O + 2NO2 Some metals (e.g., Cr, Al) do not dissolve in concentrated nitric acid because of the formation of a passive film of oxide on the surface. Chemical Properties Continued-

  46. 5. Reaction concentrated nitric acid with non–metals and their comp. – ( oxidation takes place) I2 + 10HNO3 2HIO3 + 10NO2 + 4H2O C + 4HNO3 CO2 + 2H2O + 4NO2 S8 + 48HNO3  8H2SO4 + 48NO2 + 16H2O P4 + 20HNO34H3PO4 + 20NO2 + 4H2O 6.Brown Ring Test: Its performed for the detection of nitrates in solu. The test is usually carried out in two steps- Step.1-Adding dilute FeSO4 solu. to an aq. solu. containing nitrate ion. Step no.2. - Then carefully adding concentrated sulphuric acid along the sides of the test tube. A brown ring at the interface between the solution and H2SO4 acid layers indicates the presence of nitrate ion in solution. NO3- + 3Fe2+ + 4H+ NO + 3Fe3+ + 2H2O [Fe(H2O)6]2 + NO [Fe (H2O)5 (NO)]2+ + H2O (brown) Chemical properties ……..

  47. Uses: • 1. In the manufacture of ammonium nitrate for fertilizers and nitrates. • 2. In manufacture of explosives and pyrotechnics. • 3. Used for the preparation of nitroglycerin, trinitrotoluene and other organic nitro compounds. • 4. In the pickling of stainless steel, etching of metals and as an oxidizer in rocket fuels. Uses of nitric acid-

  48. Learning objective Group 16 elements – • Abundance • Electronic configuration • Atomic size and ionic radii • Ionization enthalpy • Electron gain enthalpy • Electronegativity Period – 7

  49. Oxygen ( O) Sulphur (S) Selenium(Se) Tellurium ( Te) Polonium (Po) Livermortium ( Lv) Group 16 elements (Chalcogens)

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