DPP-23 to 24 With Answer

Max. Time : 28 Total Time : 28 min. Single choice Objective ('–1' negative marking) Q.1 to Q.6 (3 marks 3 min.) [18, 18] Subjective Questions ('–1' negative marking) Q.7 (4 marks 4 min.) [4, 4] BooSt YoUr PreViouS ConCept Single choice Objective ('–1' negative marking) Q.8 to Q.9 (3 marks 3 min.) [6, 6] 1. The five successive ionisation energies of an element ‘X’ are 800, 1427, 2658, 25024 and 32824 KJ mole–1 respectively. The valency of ‘X’ is- (1*) 3 (2) 5 (3) 1 (4) 2 2. A, B and C have the oxidation number of +6, –2 and –1 repectively. The possible molecular formula when these atoms combine. (1) A2BC (2*) AB2C2 (3) ABC2 (4) AB2C 3. Which is the correct order of E.A. (1) Li > Be > Na > Mg (2) Li > Na > Be > Mg (3) Li < Be =Na < Mg (4*) Li > Na > Mg  Be Be ~– Mg  Na  Li Sol. –– Hhighly  ve  Regular order 4. If the ionic radii of K+ and F– are about 1.34 Å each, then the expected values of atomic radii of K and F should be respectively : (1*) 2.31 and 0.64 Å (2) 2.31 and 1.34 Å (3) 0.64 and 2.31 Å (4) 1.34 and 1.34 Å 5. Amongst the following elements (whose electronic configurations are given below) the one having the highest ionization enthalpy is : (1) [Ne] 3s2 3p1 (2*) [Ne] 3s2 3p3 (3) [Ne]3s2 3p2 (4) [Ar] 3d10 4s2 4p3 6. Tick the correct order of second ionisation enthalpy in the following : (1) F > O > N > C (2*) O > F > N > C (3) O > N > F > C (4) C > N > O > F 7. Among the elements with atomic no. 9,12, 36. Identify by atomic number of an element which is : (a) highly electronegative (b) an inert gas (c) highly electropositive Sol. (a) 9F (b) 36Kr (c) 12Mg BooSt YoUr PreViouS ConCept 8. For the redox reaction MnO – + C O 2– + H+  Mn2+ + CO + H O 4 2 4 2 2 the correct stoichiometric coefficients of MnO –, C O 2– and H+ are respectively 4 2 4 (1*) 2, 5, 16 (2) 16, 5, 2 (3) 5, 16, 2 (4) 2, 16, 5 9. Two flask of equal volumes are evacuated. At the same temperature and pressure one is filled with gas A and other with gas B. The weight of gas B was found to be 0.8 g while the weight of gas A is found to be 1.4 g. The weight of one molecule of B is : (1) 1.4 times as that of A (2) 0.4 times as that of A (3*) 0.57 times as that of A (4) 0.8 times as that of A Sol. For two gases, volume, pressure and temperature are equal, hence moles of two gases must be equal. so nA = nB 1.4  0.8 MA MB  M = 0.8  MA 1.4 = 0.57 MA. Total Marks : 30 Max. Time : 30 min. Single choice Objective ('–1' negative marking) Q.1 to Q.7 (3 marks 3 min.) [21, 21] BooSt YoUr PreViouS ConCept Single choice Objective ('–1' negative marking) Q.8 to Q.10 (3 marks 3 min.) [9, 9] 4 3 2 1 1. The hybridization of carbon atoms in C2 – C3 single bond of HC  C CH  CH2 is : (1) sp3 – sp3 (2*) sp2 – sp (3) sp – sp2 (4) sp3 – sp 2. In C3O2, the hybridization state of carbon is : (1*) sp (2) sp2 (3) sp3 (4) dsp3 3. In which of the following ‘N’ atom is sp2 hybridised : (1) NH3 (2) NH + (3) (4*) B N H Sol. ‘N’ atom is sp3 4. Which of the following species has the highest electron affinity ? (1) F– (2) O– (3) Na+ (4*) O 5. Which of the following processes is endoergic in nature ? (1*) O– + e–  O2– (2) Cl + e–  Cl– (3) S + e– 6. Select the correct answer : Na cannot exhibit +2 oxidation state, because its  S– (4) F + e–  F– (1) IE2 < IE1 (2*) IE2>> IE1 (3) IE2 = IE1 (4) IE2 = 0 7. Which of the following isoelectronic ions has the lowest first ionization enthalpy. (1) K+ (2) Ca2+ (3) Cl¯ (4*) S2– BooSt YoUr PreViouS ConCept 8. Number of electrons in 1.8 mL of H2O are : (1*) 6.02 × 1023 (2) 6.02 × 1024 (3) 6.02 × 1022 (4) 6.02 × 1025 Sol. V = 1.8 ml (d = 1 g/ml) mass = 1.8 g 1.8 So moles of H2O = 18 = 0.1. Now, no. of electrons in 1 molecule of H2O is 10 So in 0.1 mole of H2O = 10 × 0.1 × NA. 9. A mixture of N2 and H2 is caused to react in a closed container to form NH3 . The reaction ceases before either reactant has been totally consumed. At this stage, 2.0 moles each of N2, H2 and NH3 are present. The moles of N2 and H2 present originally were respectively : (1) 4 and 4 moles (2*) 3 and 5 moles (3) 3 and 4 moles (4) 4 and 5 moles Sol. N2 + 3H2  2NH . a b 0 (a–x) (b–3x) 2x Given, 2x = 2, So x = 1. a – x = 2, So a = 2 + 1 = 3. b – 3x = 2, So b = 2 + 3 × 1 = 5. 10. Which d -orbital does not have four lobes? (1) dx2 – y2 (2) dxy (3) dyz (4*) dz2

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