CHEMISTRY-19-11- 11th (PQRS) SOLUTION

Q.1 (a) The first ionisation energy of Mg is higher than that of Na, but on the other hand the second ionisation energy of Na is very much higher than that of Mg. Explain [2] (b) Select the one having lower first ionisation energy in each given pair with explanation (i) I and I– (ii) Ba and Sr (iii) Be and B [3] [Sol:(a) Mg  1s2, 2s2 2p6, 3s2 , Na  1s2, 2s22p6, 3s1 Mg+  1s2, 2s22p6, 3s1 , Na+  1s2, 2s22p6 Electronic configuration of Mg (3s2) is more stable than that of Na(3s1) is less stable than Na+ (2, 8) so IE1 of Mg is higher than IE1 of Na but IE2 if Na is very much higher than IE2 of Mg. (b) (i) I–, since Zeff decreases due to increase in size. (ii) Ba, since in the same group IE decreases down the group. (iii) B, since B has EC 1s2, 2s22p1 & B has EC 1s2, 2s2 which is more stable . ] Q.2 Select those species from the following having X–X linkage (X-represent central atom) and draw their Lewis structure. [5] (i) H N O (ii) N O (iii) H P O (iv) S O2 2 2 2 2 5 4 2 5 2 7 (v) S O2 2 6 [Sol: (i) H–O–N=N–O–H (ii) (iii) (iv) (v) Species having X–X linkage are (i) H2N2O2 (v) S2O62– ] Q.3 Arrange the following in (a) order of increasing IE1 Mg, F, Na, Al (b) order of decreasing EA1 Na, Be, B, K (c) order of increasing ionic mobility in aq. solution Mg2+, Ba2+, Cs+, Al+3 [1.5+1.5+2] [Sol: (a) Na < Al < Mg < F (b) B > Na > K > Be (c) Al3+ < Mg2+ < Ba2+ < Cs+ ] r+ + r– = r– |H K K |  & |H |  LE r  HE r + Q.8  |HLE| < |HHE| So the above order is correct OR Since small cation stabilizes with small anion so as size of anion increases lattice energyof halide decreases more than hydration energy. ] (a) Draw the structure of the following compounds : [2] (i) Hydrogen phosphite ion (H2PO2–) (ii) Dihydrogen phosphate ion (H2PO4–) (b) Arrange in order of increasing magnitude of energy released in the formation of adduct (addition product) between BF3 & MH3 where M may be N, P, As. [3] [Sol:(a) (i) H2PO2– (ii) H2PO – ] (b) BF3  NH3 BF3  PH3 BF3  AsH3 As we move from N  P  As, electron cloud charge density decreases so extent of overlapping decreases and hence bond become weaker. So order of increasing magnitude of energy released is BF3  AsH3 < BF3  PH3 < BF3  NH3 ] Q.9 With the help of following information compare the reducing power of Na or Cs in aqueous solution. [5] Type of enthalpy change (kJ ) Na Cs  H Sub 108 78  H I . E . 496 376  H H .E . – 406 – 276 [Sol: For Na (4) HBDE[Cl2] = 2 [110 – 30] = 160 kJ/mol HEA[Cl(g)] = 60 – 110 = –50 kJ/mol  HBDE [Cl2] + HEA[Cl(g)] = 160 – 50 = 110 kJ/mol ] Q.11 The maximum amount of energy which can be obtained bya mixing 30 gm of carbon with 48 gm oxygen will be? Given:– C(s) + O2  CO2 H = – 80 kJ/mol 1 C(s) + 2 O2  CO H = – 50 kJ/mol [5] [Sol: nC = 2.5, n = 2 48 32 = 1.5 1 C + 2 t = 0 1.5 1.5 O2  CO on completion LR 0.25 2.5 energy released = 2.5 × 50 = 125 kJ 1 CO + 2 t = 0 2.5 0.25 O2  CO2 on completion 2 LR 0.5 energy released = 0.5 × 30 = 15 kJ so total energy released = 125 + 15 = 140 kJ Ans. ] Q.12 Electron present in H atom jumps from energy level 3 to 1. Emitted photons when passed through a sample containing excited He+ ion causes further excitation to some higher energy level. Determine Z2 principal quantum number of initial excited level & higher energy level of He+. (Given En = –13.6 2 ) [5] [Sol: For H E = E – E = 13.61  1  eV 31 3 1 1 9  For He+ = 13.6 × 4 1   4 1  36  eV (1) E = 13.6 × 4  1  n2 1  – n2  eV (2) On comparison  1 2  n1 = 2, n2 = 6 ] Q.13 Find the number of diffusion steps required to separate the isotopic mixture initially containing some amount of H2 gas and 1 mol of D2 gas in a container of 3 lit capacity maintained at 24.6 atm & 27 °C to  w D   2  1 the final mass ratio   H  equal to . (R = 0.082 Lt-atm/mol-K) [5] 2  (i) Molarity of MgCl2 = 0.3 1000 = 0.6 M 500 Molarity of Cl– = 2 × 0.6 = 1.2 M (ii) Moles of Mg2+ = 0.3 moles of Cl– = 0.6 mass of solution = 500 × 1.2 = 600 gm mass of MgCl2 = 28.5 gm  mass of water = 600 – 28.5 = 571.5 gm 571.5 moles of H2O = 18 = 31.75  mole fraction of Mg2+ = 0.3 0.3 + 0.6 + 31.75 = 0.0092 (iii) mass of Mg2+ = 0.3 × 24 = 7.2 gm 7.2 106 ppm of Mg2+ = 600 = 12000 ppm ] Q.16 NO2 partially converts into NO, O2 and N2O4 according to following reaction. NO2(g) NO(g) + 1 O 2 2 (g) 1 2 N2O4 (g) Pressure increases from 5 atm to 5.25 atm due to above reactions at constant temperature and the ratio of number of moles of O2 & N2O4 becomes 3/2. Calculate the final partial pressure of each gas in the final state. [5] [Sol: NO2 (g)  NO(g) + ½ O2(g) t = 0 5 – p – t 5–p–p2 p2/2 n 2 n N O p = pN O p1/ 2 = p2 / 2 p1 3 = p2 = 2 .............(1) Total pressure pT = p – p1 + p1 + p1 + 5 – p – p + 2 2 p2 = 5.25 2  p1 – 2 p2 = 5.25 – 5 = 0.25 (2) 2 On solving eq. (1) and (2) 3 p1 = 2 atm = 1.5 atm p2 = 1 atm Hence pNO = 1.5 atm (Z )   2   (Vavg )   N * 2  11 A =  A   A   A  (Z11 )   2   (Vavg )   N *  B (Z11 )A  B   1 B   B   1 2 1 (Z11 ) = 4 × 2 ×  2  = 2 ] B   Q.18 A balloon containing 1 mole air at 1 atm initially is filled further with air till pressure increases to 3 atm. The initial diameter of the balloon is 1 m and the pressure at each state is proportion to diameter of the balloon. Calculate (a) No. of moles of air added to change the pressure from 1 atm to 3 atm. (b) balloon will burst if pressure increases to 7 atm. Calculate the number of moles of air that must be added after initial condition to burst the balloon. [2+3] [Sol: P  diameter d = Kp Initially 1 = K × 1 K = 1 m atm (a) Now initially PV = nRT 4  1 3 1 ×    3 2 = niR T (1)   Finally d = K × P = 1 × 3 = 3m 4  3 3 Volume =    m3 3  2  4  3 3  3 ×    3 2 = nf RT (2)   n f Dividing (2) by (1)  nf = 81 moles added = 80 (b) For Pf = 7 atm d = 7m ni = 3 × 27 = 81 4  7 3 V =    m3 3  2  4  7 3 7 ×    3 2 = nf RT (3)   n f Dividing (3) by (1) ni = 7 × 7 × 7 × 7 nf = 2401  moles added = 2400 ]