DPP-21 to 22 Physical chemistry With Answer

FACULTY COPY DAILY PRACTICE PROBLEMS (DPP) Subject : Physical/Inorg.Chemistry Date : DPP No. 19 Class : XIII Course DPP No.1 Max. Time : 46 Total Time : 46 min. Single choice Objective ('–1' negative marking) Q.1 to Q.5 (3 marks 3 min.) [15, 15] Subjective Questions ('–1' negative marking) Q.6 to Q.8 (4 marks 4 min.) [12, 12] Match the Following (no negative marking) Q.9 (4 marks 4 min.) [4, 4] BooSt YoUr PreViouS ConCept Single choice Objective ('–1' negative marking) Q.10 to Q.14 (3 marks 3 min.) [15, 15] 1. The kinetic energy of an electron in a particular orbit in H atom is 3.4 eV. The angular momentum of electron in this orbit is : h (1) 2 h (2*)  13.6Z2 13.6 3h (3) 2 13.6 2h (4)  Sol. E = –KE = – 3.4 = – n n = 2 = n2 nh  K.E. = n2 2h h = 3.4 Angular momentum, J = 2 = 2 =  2. The ratio of the energy of a photon of wavelength 3000 Å to that of a photon of wavelength 6000Å is (1) 1/2 (2*) 2 (3) 3 (4) 1/3 Sol. E1  2 = 6000 = 2. E2 1 3000 3. Find out the value of (P) in the following cycle : NH (g) + HCl (g) H1 NH + Cl¯ (s) 3 D H(g) + Cl (g) 4 HL.E. .E. Heg NH (g) + H+(g) + Cl-(g) (P) NH +(g) + Cl- (g) Given H1 = – 400 kJ, I.E. = + 50 kJ , D = + 50 kJ, Heg = – 30 kJ, lattice energy = – 100 kg (1) – 400 (2) – 300 (3*) – 370 (4) – 310 4. Which of the following sequences represents the correct order of lattice energies ? (1) LiI > LiBr < LiCl < LiF (2) KBr < KCl < KF < KI (3) NaF < Nacl < Na < Br > NaI (4*) LiF > LiCl > LiBr > LiI 5. Condition for ionic bond formation between electropositive element M and electronegative element Z becomes favourable in all cases except : (1) with increase in electron affinity of Z (2) with decrease in ionization potential of M (3) with decrease in dissociation energy of Z2 (4*) with increase in sublimation energy of M. 6. How many photons of light of 7000 Å wavelength are equivalent to 1 J of energy ? Ans. 3.53 × 1018 photons nhc Sol. Use E =  . 7. How does bond polarity affect the % ionic character of the bond and the bond length? Explain determine the % ionic character of H–F from Henny Smith formula and calculate the bond lengths of N–O, and C–O bonds using EN values by the help of [d = rA + rB – 0.09 (XA – XB)]. (Given rN  0.75 Å; r0  0.74 Å; rc  0.77Å) Sol. As bond polarity  Bond Length  d = rA + rB – 0.09 (XA – XB) % ionic charac = 16(X ~ X ) + 3.5 (X ~ X )2 A B A B = 43% dN – O ~– 1.445 Å dC – O ~– 1.42 Å As polarity  B. Str.  B. Length  C – O < dN – O 8. Calculate the lattice energy of LiF given that enthalpy of (i) H for lithium is 155.2 kJ mol–1 (ii) H of 1 mole of F 2 2 is 75.3 kJ (iii) I.E of Lithium = 520 kJ mol–1 (iv) E.A. of Fluorine = – 333 kJ mol–1 (v) H for overall = – 594.1 Ans. – 1011.6 kJ mol–1 9. Match the following : (1) Li < Na < K < Rb (p) increase in radius (2) Li < B < C < N (q) decrease in ionisation energy (3) Mg++ < Na+ < F– < O– – (4) F < Cl < Br < I (r) increase in Zeff (s) decrease in Zeff Ans. [1 – p,q] ; [2 – r] ; [3 –p,q,s] ; [4 – p,q]. BooSt YoUr PreViouS ConCept 10. Which will be the proper alternative in place of A in the following equation- 2Fe3+(aq) + Sn2+(aq)  2Fe2+(aq) + A (1*) Sn4+ (2) Sn3+ (3) Sn2+ (4) Sn° 11. Which of the following reactions does not involve either oxidation or reduction ? (1) VO2+  V O (2) Na  Na+ (3) Zn+2  Zn (4*) CrO –2  Cr O –2 12. Which statement is wrong- (1*) Oxidation number of oxygen is +1 in peroxides (2) Oxidation number of oxygen is +2 in oxygen difluoride (3) Oxidation number of oxygen is  1 2 in superoxides (4) Oxidation number of oxygen is –2 in most of is compound 13. The oxidation number and covalency of sulphur in the sulphur molecule (S8) are respectively: (1*) 0 and 2 (2) +6 and 8 (3) 0 and 8 (4) +6 and 2 14. In the following change - 3Fe + 4H2O  Fe3O4 + 4H2. If the atomic weight of iron is 56, then its equivalent weight will be - (1) 42 (2*) 21 (3) 63 (4) 84 FACULTY COPY DAILY PRACTICE PROBLEMS (DPP) Subject : Physical/Inorg.Chemistry Date : DPP No. 20 Class : XIII Course : DPP No.2 Total Marks : 44 Max. Time : 44 min. Single choice Objective ('–1' negative marking) Q.1 to Q.5 (3 marks 3 min.) [15, 15] Mutiple choice Objective ('–1' negative marking) Q.6 to Q.8 (4 marks 4 min.) [12, 12] Subjective Questions ('–1' negative marking) Q.9 (4 marks 4 min.) [4, 4] BooSt YoUr PreViouS ConCept Single choice Objective ('–1' negative marking) Q.10 to Q.12 (3 marks 3 min.) [9, 9] Subjective Questions ('–1' negative marking) Q.13 (4 marks 4 min.) [4, 4] 1. Which is the correct order of solubility in water (1*) CsF > CsCl > CsBr > Cs (2) LiF > LiCl > LiBr > Li (3) RbC𝑙O4 > KC𝑙O4 > NaC𝑙O4 > LiC𝑙O4 (4) LiF > NaF > KF > RbF Sol. Larger is difference in size of cation & anion lorges is sohebility. 2. Which of the following is the right order of lattice energy (1) Na2 O < Al2O3 < MgO (2) MgO < Al2O3 < Na2O (3) Al2O3 < MgO < Na2O (4*) Na2O < MgO < Al2O3 1 Sol. As L.E.  r  r  |Z+| |Z–| hence Al2O3 will have highest value among these. 3. If it is known that on heating a ionic compound of a polyhalide with a cation it decomposese into more stable halide of that cation due to high lattice energy, for example Cs   Cs +  The complex compound Rb[IBrCl] after strong heating will : (1) RbI + BrCl (2*) RbCl + IBr (3) RbBr + ICl (4) None Sol. (L.E.)RbCl > (L.E.)RbBr > (L.E.)Rb and more stable compound is formed first. 4. LiCl is soluble in benzene. This property develops in LiCl because of : (1*) smaller size of lithium ion (2) Li+ is unable to cause distortion (3) formation of LiCl does not obey Fajan's rule (4) lithium is a s-block element 5. Which of the following cannot be explained on the basis of Fajan’s Rules. (1) Ag2S is much less soluble than Ag2O. (2) BeCO3 is thermally less stable than BaCO3 . (3*) BaCO3 is much less soluble than MgCO3. (4) Melting point of Al2O3 is much less than that of Na2O. Sol. The order of solubility of BeCO3 > MgCO3 > CaCO3 > SrCO3 > BaCO3 . can only be explained on the basis of lattice energy rest others can be explained by Fajan's rule. 6. Correct order of solubility of the compound is : (1) BeF2 < BeCl2 < BeBr2 < BeI (2*) SrSO4 < CaSO4 < MgSO4 (3) CsF < CsCl < CsBr (4*) CsBr < CsCl < CsF Sol. More is the size difference more is the solubility. 7. Identify Isomorphus structure : (1*) MgCO3 , NaNO3 (2) NaCO3 , Na2SO3 (3*) BaSO4 , KMnO4 (4) NaNO3, NaClO4 Sol. Mg CO3 NaNO3 CO 2– NO – 3 3 trigonal planar trigonal planar Cations & anians are isoshuctural & Both have similar formula type .so Isomorphus. BaSO4 K MnO4 These are I so morphus NO2CO3 Na2 SO3 CO 2– SO 2– 3 3 sp2 [Not I so morphus] NaNO3 NaCIO4 8. A d-block element which has magnetic moment about it. B.M. and 7 electrons has n + 𝑙 = 4. Which is correct (1*) it may be Cr (2) it may be Mo. (3*) 12 electrons are belongs to 𝑙 = 1. (4*) for the outermost electron n = 4 and 𝑙 = 0. Sol. m = = =  n = 6 e– s. Cr has 6 unpainred e– s  n  4,l  0   n  3 l  1  4s which have 1e–  3p which have 6e–s n + l = 4    Total e– s in n  l 4 is 7e– s For l = 1, 2P, & 3P which have 12e– s. 9. Arrange the solubility of NaCl in decreasing order in the following solvents : CH3COCH3 (dielectric constant = 12) ; CH3CH2OH (dielectric constant = 35) ; H2O (dielectric constant = 81) ; Ans. Solubility of NaCl lies in following sequence H2O > CH3CH2OH > CH3COCH3 Greater is the dielectric constant of solvent more is the solubility of an ionic compound in it. BooSt YoUr PreViouS ConCept 10. In the following reaction hydrazine is oxidized to N2. N2 H4 + OH–  N2 + H2O + e The equivalent weight of N2H4 (hydrazine) is (1*) 8 (2) 16 (3) 32 (4) 64 11. One mole of N2H4 loses ten moles of electrons to form a new compound Y. Assuming that there is no loss of nitrogen in the formation of the new compound, what is the oxidation state of nitrogen in Y ? There is no change in the oxidation state of hydrogen, (1) –1 (2) –3 (3*) + 3 (4) + 5 12. 2.00 g of a mixture of NaHCO3 and KClO3 requires 100 mL of 0.1 N HCl for complete neutralization. The weight of KClO3 present in the mixture is [Hint : KClO3 ] (1) 0.84 g (2) 1.84 g (3*) 1.16 g (4) 0.16 g 13. _ Calculate the valency factor of the reactant (oxidant or reductant) in following cases (i) 2 1 3 4 (ii) [Fe(CN) ]4–  Fe3+ + CO + NO – FeS2 Fe2O3  SO2 6 2 3 (iii) Cu2S  Cu2+ + SO (iv) Fe (SO )  Fe2+ + SO (v) Cr O 2–  Cr3+ (vi) MnO ¯  MnO 2– (highly basic) (vii) O ¯  ¯ (viii) ClO¯  Cl¯ (basic) (ix) H2O2  H2O (x) NO3¯  N2O (xi) NO3¯  NH + (xii) O  O + O2– Sol. (i) v.f. = 1(3 – 2) + 2(4 – (–1)) = 1 + 10 = 11 (ii) v.f. = 1(3 – 2) + 6(4 – 2) + 6(5–(–3)) = 61 (iii) v.f. = 2(2 – 1) + 1(4 – (–2)) = 2 + 6 = 8 (iv) v.f. = 2(3 – 2) + 3(6 – 4) = 8 (v) v.f. = 2(6 – 3) = 6 (vi) v.f. = 1(7 – 6) = 1 (vii) v.f. = 1(5 – (–1)) = 6 (viii) v.f. = 1(1 – (–1)) = 2 (ix) v.f. = 2(–1 – (–2)) = 2 (x) v.f. = 1(5 – 1) = 4 (xi) v.f. = 1(5 – (–3)) = 8 (xii) v.f. = 1(0 – (–2)) = 2