Tiêu đề DPP-9 SOLUTION.pdf ✅

CLASS : XIth SUBJECT : CHEMISTRY DATE : DPP No. : 9 1 (d) HF is weaker acid due to H-bonding. 2 (c) Fe(II) has four unpaired electrons (3d 6 ) where Fe(III) has five unpaired electrons (3d 5 ). This can be obtained by measuring magnetic moment of molecule in solid state. 3 (b) NH3 + 3Cl2⟶NCl3 + 3HCl 4 (b) Yhe structure of H3PO4 is It can loose three H+ ions so its basicity is three. 6 (d) Chlorine, being only a slightly stronger oxidizing agent than bromine can not oxidise it to +7 oxidation state as is required for the formation of the compound BrCl7 7 (c) The true peroxide contains O 2― 2 (O ― O) 2― ion. ∵ Out of given choices only BaO2 has O 2― 2 in its structure. ∴ BaO2 is true peroxide. 8 (d) SO2 + 2H2O + Br2 ⟶2HBr + H2SO4 9 (c) Nitrogen does not have d-orbitals 10 (d) Pernitric acid is HNO4. 11 (a) Platinum acts as catalyst in the oxidation of ammonia to form nitric oxide .This reaction is P HO OH OH O Topic :-THE P-BLOCK ELEMENTS-2 Solutions
used in the ostwald ‘s method of nitric acid preparation. 4NH3+5O2⟶4NO+6H2O 2NO+O2⟶2NO2 4NO2+O2+2H2O⟶4HNO3 12 (d) Frankland and Lockyer pointed out the new D3 line observed in the yellow region of the sun’s spectrum observed by Jonsen in 1868 was due to a new element which they named Helium. It was the first noble gas to be discovered. The two known line D1 and D2 were of sodium 13 (b) 3Cl2(g) + 6KOHaq. ∆ KClO3 + 5KCl + 3H2O (Green yellow (Used in fire- gas) works and safety match box) 14 (d) It is a fact. 15 (b) NH3 + 3Cl2⟶NCl3 + 3HCl 16 (d) He, because of its small size can diffuse through rubber, glass PVC etc. easily 18 (a) Orthophosphate + Amm. Molybdate HNO3 ∆ yellow ppt ↓ + AgNO3 Red ppt 19 (a) 2HNO2 + H2SO4⟶2NO2 + SO2 + 2H2O 20 (c) CN ― acts as complexing agent and reducing agent. CuSO4 + 2KCN⟶Cu2(CN)2 + K2SO4 + (CN)2 (Reducing agent) Cu2(CN)2 +6KCN⟶2K3Cu(CN)4(Complexing agent)
ANSWER-KEY Q. 1 2 3 4 5 6 7 8 9 10 A. D C B B A D C D C D Q. 11 12 13 14 15 16 17 18 19 20 A. A D B D B D B A A C