Tiêu đề DPP-7 SOLUTION.pdf ✅

CLASS : XIth SUBJECT : CHEMISTRY DATE : DPP No. : 7 1 (c) Only He and Ne are remained unadsorbed on the coconut charcoal at -100°c (173K) as their boiling points are less than -100°c .(He=4K, Ne=27K). 2 (c) ClO ― 2 has sp 3 hybridisation and two lone pairs on halogen which produces V-shape bent structure 4 (d) S8 has puckered ring structure. 5 (d) Ti has configuration 1s 2 ,2s 2 2p 6 ,3s 2 3p 6 3d 2 ,4s 2 . Thus, Ti 4+ has configuration 1s 2 ,2s 2 2p 6 ,3s 2 3p 6 ,i.e., of Ar. 6 (b) 3CuSO4 + 2PH3⟶Cu3P2 + 3H2SO4 Black 7 (a) Anhydrous Ba(ClO4 )2 is an effective drying agent. It is used under the trade name desicchlora 8 (a) Neil Bartlett prepared first noble gas compound, xenon hexafluoride (IV) 10 (a) Cl O O - Topic :-THE P-BLOCK ELEMENTS-2 Solutions
The structure of H3PO3 is given as In this structure two —OH group are present, so it is dibasic acid. In it one P—H bond is present, so it provides hydrogen and due to such hydrogen it acts as reducting agent. 11 (d) When chlorine reacts with dilute and cold NaOH sodium chlorine and sodium hypochlorite are formed. 2NaOH(cold) +Cl2⟶NaCl +NaClO +H2O Sodium hypochloride Let oxidation state of Cl in NaCl is x + 1 + x =0 X = -1 Let oxidation state of Cl in NaClO is x. NaClO +1 + x - 2=0 x - 1=0 x = + 1 ∴ oxidation states of chlorine changes from 0 to -1 and +1. 12 (b) It is a fact. 13 (d) These are uses of H2SO4. 14 (c) Hydrolysis of NCl3 gives NH3or NH4OH and HCIO as NCl3+4H2O⟶NH4OH+3HOCl 15 (c) Xe in XeF2, XeF4, XeF6 has sp3d, sp3d 2 and sp3d 3 hybridisation with 3, 2, 1 lone pair of electrons respectively. 16 (c) N2O is itself non-combustible but supports combustion S + 2N2O→SO2 + 2N2 18 (b) (NH4 )2Cr2O7 ∆ N2↑ + Cr2O3 + 4H2O NH4NO2→N2↑ + 2H2O 19 (c) Fluorapatite is CaF2 ∙ 3Ca3 (PO4 )2 . 20 (d) It is a fact. P H O O O H H
ANSWER-KEY Q. 1 2 3 4 5 6 7 8 9 10 A. C C A D D B A A A A Q. 11 12 13 14 15 16 17 18 19 20 A. D B D C C C B B C D