Study package for p block-12

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  1. STUDY PACKAGE FOR P-BLOCK ELEMENTS FOR CLASS-12 SUPRATIM DAS DURGAPUR WEST BENGAL│9434008713 1. Why are pentahalides more covalent than trihalides? Ans- In…
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  • 1. STUDY PACKAGE FOR P-BLOCK ELEMENTS FOR CLASS-12 SUPRATIM DAS DURGAPUR WEST BENGAL│9434008713 1. Why are pentahalides more covalent than trihalides? Ans- In pentahalides, the oxidation state is +5 and in trihalides, the oxidation state is +3. Since the metal ion with a high charge has more polarizing power, pentahalides are more covalent than trihalides. 2. Why is BiH3 the strongest reducing agent amongst all the hydrides ofGroup 15 elements? Ans- As we move down a group, the atomic size increases and the stability of the hydrides of group 15 elements decreases. Since the stability of hydrides decreases on moving from NH3 to BiH3, the reducing character of the hydrides increases on moving from NH3 to BiH3. 3. Why is N2 less reactive at room temperature? Ans- The two N atoms in N2 are bonded to each other by very strong triple covalent bonds. The bond dissociation energy of this bond is very high. 4. Mention the conditions required to maximise the yield ofammonia in Haber process. Ans. Ammonia is prepared using the Haber’s process. The yield of ammonia can be maximized under the following conditions: (i) High pressure (∼ 200 atm) (ii) A temperature of ∼700 K (iii) Use of a catalyst such as iron oxide mixed with small amounts of K2O and Al2O3 5.Howdoes ammonia react with a solution ofCu2+ ? Ans. NH3 acts as a Lewis base. It donates its electron pair and forms a linkage with metal ion. 6. Bond angle in PH4 + is higher than that in PH3. Why? Ans. In PH3, P is sp3 hybridized. Three orbitals are involved in bonding with three hydrogen atoms and the fourth one contains a lone pair. As lone pair-bond pair repulsion is stronger than bond pair-bond pair repulsion, the tetrahedral shape associated with sp3 bonding is changed to pyramidal. PH3 combines with a proton to form PH4 + in which the lone pair is absent. Due to the absence of lone pair, there is no lone pair-bond pair repulsion. Hence, the bond angle in PH4 + higher than the bond angle in PH3. 7. What happens when PCl5 is heated? Ans. All the bonds that are present in PCl5 are not similar. It has three equatorial and two axial bonds. The equatorial bonds are stronger than the axial ones. Therefore,when PCl5 is heated strongly, it decomposes to form PCl3. 8. What is the basicity ofH3PO4? Ans. In H3PO4,since there are three OH groups present in H3PO4, its basicity is three i.e., it is a tribasic acid. 9. What happens when H3PO3 is heated? Ans. H3PO3,on heating, undergoes disproportionation reaction to form PH3 and H3PO4. The oxidation numbers of P in H3PO3,PH3,and H3PO4 are +3,−3, and +5 respectively. As the oxidation number of the same element is decreasing and increasing during a particular reaction, the reaction is a disproportionation reaction.
  • 2. 10. Write the important sources of sulphur. Ans. Sulphur mainly exists in combined form in the earth’s crust primarily as sulphates [gypsum (CaSO4.2H2O), Epsom salt (MgSO4.7H2O),baryte (BaSO4)] and sulphides [(galena (PbS), zinc blends (ZnS), copper pyrites (CuFeS2)]. 11. Write the order ofthermal stability of the hydrides ofGroup 16 elements. Ans. The thermal stability of hydrides decreases on moving down the group. This is due to a decrease in the bond dissociation enthalpy of hydrides on moving down the group. 12. Why is H2O a liquid and H2S a gas? Ans. H2O has oxygen as the central atom. Oxygen has smaller size and higher electronegativity as compared to sulphur. Therefore, there is extensive hydrogen bonding in H2O,which is absent in H2S. Molecules of H2S are held together only by weak Vander Waal’s forces of attraction. Hence,H2O exists as a liquid while H2S as a gas. 13. Which of the following does not react with oxygen directly? Zn, Ti, Pt, Fe Ans. Pt is a noble metal and does not react very easily. All other elements, Zn, Ti, Fe, are quite reactive. Hence, oxygen does not react with platinum (Pt) directly. 14. Why does O3 act as a powerful oxidising agent? Ans. Ozone is not a very stable compound under normal conditions and decomposes readily on heating to give a molecule of oxygen and nascent oxygen. Nascent oxygen, being a free radical, is very reactive.Therefore,ozone acts as a powerful oxidising agent. 15. Howis O3 estimated quantitatively? Ans. Quantitatively, ozone can be estimated with the help of potassium iodide. When ozone is made to react with potassium iodide solution buffered with a borate buffer (pH 9.2), iodine is liberated. This liberated iodine can be titrated against a standard solution of sodium thiosulphate using starch as an indicator. 16. What happens when sulphur dioxide is passed through an aqueous solution ofFe(III)salt? Ans. SO2 acts as a reducing agent when passed through an aqueous solution containing Fe(III) salt. It reduces Fe(III) to Fe(II) i.e., ferric ions to ferrous ions. 18. Comment on the nature oftwo S−O bonds formed in SO2 molecule. Are the two S−O bonds in this molecule equal? Ans. The electronic configuration of S is 1s2 2s2 2p6 3s2 3p4 . During the formation of SO2, one electron from 3p orbital goes to the 3d orbital and S undergoes sp2 hybridization. Two of these orbitals form sigma bonds with two oxygen atoms and the third contains a lone pair. p-orbital and d-orbital contain an unpaired electron each. One of these electrons forms pπ-pπ bond with one oxygen atom and the other forms pπ- dπ bond with the other oxygen. This is the reason SO2 has a bent structure. Also, it is a resonance hybrid of structures I and II. Both S−O bonds are equal in length (143 pm) and have a multiple bond character. 19. Howis the presence ofSO2 detected? Ans. SO2 is a colourless and pungent smelling gas. It can be detected with the help of potassium permanganate solution. When SO2 is passed through an acidified potassium permanganate solution, it decolonizes the solution.
  • 3. 20. Mention three areas in which H2SO4 plays an important role. Ans. Sulphuric acid is an important industrial chemical and is used for a lot of purposes. Some important uses of sulphuric acid are given below. (i) It is used in fertilizer industry. It is used to make various fertilizers such as ammonium sulphate and calcium super phosphate. (ii) It is used in the manufacture of pigments, paints, and detergents. (iii) It is used in the manufacture of storage batteries. 21. Considering the parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy, compare the oxidising power ofF2 and Cl2. Ans. Fluorine is a much stronger oxidizing agent than chlorine. The oxidizing power depends on three factors. 1. Bond dissociation energy 2. Electron gain enthalpy 3. Hydration enthalpy The electron gain enthalpy of chlorine is more negative than that of fluorine. However,the bond dissociation energy of fluorine is much lesser than that of chlorine. Also, because of its small size, the hydration energy of fluorine is much higher than that of chlorine. Therefore, the latter two factors more than compensate for the less negative electron gain enthalpy of fluorine. Thus, fluorine is a much stronger oxidizing agent than chlorine. Oxides of nitrogen Oxy -Acids of Nitrogen: Oxy Acids Name of oxy – acid
  • 4. 1. H2N2O2 Hyponitrous acid 2. H2 NO2 Hydronitrous acid 3. HNO2 Nitrous acid 4. HNO3 Nitric acid 5. HNO4 Per nitric acid Oxides of Phosphorus: Oxy – Acids of Phosphorus: Oxo acid Name H3PO2 Hypophosphorus acid H3PO3 Phosphorus acid H4P2O6 Hypophosphoric acid H3PO4 Orthophosphoric acid H4P2O7 Pyrophosphoric acid HPO3 Metaphosphoric acid Oxygen Family (Group 16 Elements): Sr. No. Property Oxygen Sulfur Selenium Tellurium Polonium 1. Configuration [He]2s2 2p4 [Ne]3s2 3p4 [Ar]4s2 4p4 [Kr]5s2 5p4 [Xe]6s2 6p4 2. Common oxidation state -2 -2, +4, +6 +4, +6 +4, +6 3. Atomic radius (pm) 66 104 116 143 167 4. First ionization energy (KJ/mol) 1314 1000 941 869 812 5. Electronegativity 3.5 2.5 2.4 2.1 2.0 Oxy-acids of Group-16 elements Sulphur Selenium Tellurium Sulphurous acid H2SO3. Sulphuric acid H2SO4 Peroxomonosulphuric acid Selenious acid H2SeO3 Selnenic acid H2SeO4 Tellurous acid H2TeO3. Telluric acid H2TeO4.
  • 5. H2SO5(Caro’s acid) Peroxodisulphuric acid H2S2O8 (Marshell’s acid) Thio sulphuric acid H2S2O3 Dithiconic acid H2S2O6 Pyrosulphuric acid H2S2O7 Interhalogens Type XX’1 (n = 1) (with linear shape) Type XX’3 (n = 3) (with T- shape) XX’5 (n = 5) (with square pyramidal shape) XX’7 (n = 7) with pentagonal bipyramidal shape) CIF ClF3 ClF5 BrF BrCl BrF3 BrF5 ICl, IBr, IF ICl3, IF3 IF5 IF7 Hydrogen Halides: i) Dipole moment : HI < HBr < HCl < HF ii) Bond length: HF < HCl < HBr < HI iii) Bond strength: HI < HBr < HCl < HF iv) Thermal stability: HI < HBr < HCl < HF v) Acid strength: HF < HCl < HBr < HI vi) Reducing power: HF < HCl < HBr < HI Pseudohalide ions Pseudohalide ions Name CN– Cyanide ion OCN– Cyanate ion SCN– Thiocyante ion SeCN– Selenocyanate ion NCN2– Cyanamide ion N3 – Azide ion Oxy-acids of Chlorine Formula Name Corresponding Salt HOCl Hypochlorous acid Hypochlorites HClO2 Chlorous acid Chlorites HClO3 Chloric acid Chlorates HClO4 Perchloric acid Perchlorates Acidic Character: Acidic character ofthe same halogen increaseswith the increase in oxidation number ofthe halogen: HClO4 > HClO3 > HClO2 > HOCl Compounds of Xenon Molecule Total electron pairs (BP + LP) Hybridisation Shape XeF2 5 Sp3 d Linear XeF4 6 Sp3 d2 Square planar
  • 6. XeF6 7 sp3 d3 Distorted octahedral
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