CHEMISTRY-25-02- 11th (PQRS & J) Partial Marking

CHEMISTRY FINALTEST PART -A Q.1 If % ionic character of O–H bond is 10% & O–H bond distance is 1 4.8 Å, then calculate dipole moment of H2O molecule (in Debye) given H–O–H bond angle is 105°. [Use sin 52.5° = 0.8] [5] [Ans. 0.12 D] [Sol.  q × d = 4.8 × 10–10 × 0.1 × 1 × 10–8 4.8 = 0.1 × 10–18 ese cm = 0.1 D  {2}  net = 2 cos 52.5 = 2 × 0.1 × (1 – 0.64) D = 2 × 0.1 × 0.6  net = 0.12 D  {3} 2 Marks if only bond dipole is calculated ] Q.2 Arrange in ascending order (reasons not required) (a) | Hhydrogenation| of : 1, 2 Pentadiene, 1, 3 Pentadiene & 1, 4 Pentadiene (b) Ionic conductance of : F–, Na+ & Cs+. (c) Dipole Moment of : , , , (d) Boiling point : n-pentane, isopentane, neo-pentane [1.5+1+1.5+1] [Sol.(a) Reasons not required for marks. No marks for partially correct order. (1) (2) (3) C = C = C – C – C, C = C – C = C – C, C = C – C – C = C Most stable (2) > (3) > (1) |HHE|  (2) < (3) < (1)  {1.5} (b) I.C  1  aq.size 1 Hydration  Na+ < F– < Cs+  {1} (c) < < <  {1.5} (d) neo-pentane < iso-pentane < n-pentane  {1} ] Q.3 Balance the reaction (a) Oxidation of cyclobutylmethanol to cyclobutane carboxylic acid by MnO– in acidic medium. (b) Ca(OCl2) + KI  I2 + CaCl2 + KCl (Acidic medium) [3+2] [Sol. No partial marks for incorrectly balance equation + MnO–  + Mn2+ Mn+7 + 5e–1  Mn+2 ] × 4 C–1  C+3 + 4e– ] × 5  + 4 MnO–  + 4Mn+2 + 12H+ + 11H2O  5 + 4 MnO– + 12H+  5 (5, 4, 12, 5, 4, 11) + 4Mn+2 + 11H2O  {3} (b) 0 –1 0 Ca(OCl2) + KI  CaCl2 + KCl + I2 0 Cl2 + 2e–  2Cl– 2I–  I2 + 2e–  Ca(OCl2) + 2KI  CaCl2 + I2 + 2KCl +2Cl– + 2H+ +H2O  Ca(OCl2) + 2KI + 2HCl  CaCl2 + I2 + 2KCl + H2O  {2} (1, 2, 2, 1, 1, 2, 1) ] Q.4 (a) Write the IUPAC name of [2] (i) (ii) (b) Predict the final product [2] O || (i) CH3  CH2  C  NH2 + NaOH H2O ? (ii) + NaOH H2O ? (c) Which transition in Be3+ emits same wavelength as the 2nd line of Balmer series of Hydrogen. [1] [Sol.(a) Deduct marks for wrong/missing numbering and incorrect basic name (neglect small errors) (i) 3[2-chloro,4-formyl cyclobutyl]cyclobutane carbonyl chloride  {1} (ii) (2 chloro cyclopropyl)spiro(2,2) pentane carboxylate  {1} (b) (i) CH3 – CH2 O || – C  N–H O– |  CH3  CH2  C = NH  CH3 – CH2 OH | – C = NH  {0.5} 0.5 Marks (ii) + NaOH H2O   H2O  {0.5} 0.5 Marks (c) Marks for final answer, procedure not essential 1  1 1  = R × = R  1 ×16 1   1 1  = R ×16  H  22 42  H  4 16 16 16  H 82 162   Transition from 16 to 8  {1} ] Q.5(a) Draw an Andrew's isotherm of a real gas for three temperatures (T > Tcritical, T = Tcritical, T < Tcritical) on a same graph. Highlight the liquifaction curve in it. (b) Argon at a temperature of 273 K & pressure 1 atm has a density of (48/22.4) gm/lit. Predict whether it behaves ideally or acts as real. If real, then which Vander Waal constant is more influential. Why? [2+3] [Sol.(a) Deduct some marks for non mentioning of liquifaction curve  {2} (b) P = 1 atm, M = 40, T = 273  PM = ZdRT  1 × 40 = Z × 48  22.4 22.4  Z = 5/6  {2} □ Z < 1  Vander Waal constant a is influential  {1} ] Q.6(a) If internuclear distance between Cl atoms in Cl2 is 10 Å & between H atoms in H2 is 2 Å, then calculate internuclear distance between H & Cl (Electronegativity of H = 2.1 & Cl = 3.0). [2] (b) Compare the following giving reasons (i) Acidic nature of oxides: CaO, CO, CO2, N2O5, SO3 (ii) Bond length of P—O, S—O & Cl—O in PO3– , SO2– , ClO– respectively. [3] 4 4 4 [Sol.(a) rA–B = rA + rB – 0.09(XA – XB) rA = 10/2; rB = 2/2 = 5 + 1 – 0.09 × 0.9 = 6 – 0.081 = 5.919 Å  {2} Award only 0.5 marks if EN considerations not done. (b)(i) CaO < CO < CO2 < N2O5 < SO3  {1} Metal oxide Neutral Acidic Acidic Max. Acidic oxide basic oxide oxide Along a period acid nature increases on moving right  {0.5} (ii) Bond order 1.25 1.5 1.75  {1}  B.L. of Cl–O < S–O < P–O  {0.5} as B. O  B.L.  ] PART -B Q.7 A solution of Ba3(PO4)2 having molarity = 2M initially dissociates to give Ba2+ & PO43–. If extent of dissociation of salt is 75% & density of solution is 3.202 gm/ml, then calculate molality (m) of each of the specie. [Ba3(PO4)2, Ba2+ & PO43–] [6] [Sol. Ba (PO ) l 3Ba+2 + 2 PO3– 3 4 2 4 m of salt = 2 100 = 1 m  {3} 1000 3.202  2  601  m of Ba3(PO4)2 = 0.25 m  {1} m of Ba+2 = 2.25 m  {1} m of PO3– = 1.5 m  {1} Check for alternate method also ] Q.8 A 117.6 mg sample of impure KSCN on reaction with KMnO4 give Mn2+, SO42– & CN–. All the SO2– produced was reacted with KI to give SO2 gas & I2 gas. I2 gas liberated required 10 ml of 0.1 M K Cr O solution to get converted to IO– alongwith Cr3+. Write all the unbalanced skeleton 2 2 7 3 reactions. Calculate percent purity of KSCN sample & moles of KMnO4 required in first step. [6] [Sol. KSCN + MnO–  Mn+2 + SO2– + CN–  {0.5} SO2– + KI  SO2 + I2  {0.5} I + K Cr O  IO– + Cr+3  {0.5} 2 2 2 7 3 meq. of K2Cr2O7 = meq. of I2 (3rd reaction) = 10 × 0.1 × 6 m moles of I2 = 10 × 0.1 × 6 × 1/10  {1} meq. of I2 (2nd reaction) = meq of SO2– = 10  0.1 6  1  2 10 = m moles of SO2– = m. m of KSCN = 10  0.1 6  1  2 × 10 1  {1} 2  wt. of KSCN = 0.6 × (98) = 58.8 mg  % purity = 58.8 100 = 50%  {1} 117.6 n factor of KSCN = 8 – 2 = 6  meq. = 0.6 × 6  moles of KMnO4  {1} 0.36 m. m = 5  moles = 0.36 5 × 10–3  {0.5} ] Q.9 A metal was subjected to electromagnetic radiations having varied frequency & a graph of maximum KE of ejected photon vs frequency was plotted. From the graph calculate (a) Work function of the metal in eV/atom (b) Possible range of KE (in eV) of ejected photoelectron if they are accelerated by a potential difference of 2V if  of electromagnetic radiation is 20 nm. [6] [Sol. From the graph (a) 10 eV = ( 6.4 6.62 1016 × 6.62 × 10–34) / (1.6 × 10–19) – W  {1.5}  10 eV = 40 eV – W  W = 30 eV  {1.5} 1240 (b) Fpho = 20 nm  {0.5}  Max. KE = 62 – 30 = 32 eV  {1.5}  Range of KE = 30 eV – 34 eV {1} ] Q.10 Suppose in the Bohr's model of H–atom, instead of electrostatic force, some other force is acting between nucleus & electron which cause the potential energy to vary as k . What should be the sign of k so that 5r5 electron can move in uniform circular motion about nucleus.Assuming quantization of angular momentum derive expression for radius of nth orbit. [6] k [Sol. PE(r) = 5r5  F(r) = – dE k dr = + r6 □ For the above system to be possible PE  as r   k < 0  {2} Check for other justifications mv2 Also, r  v = k = r6  {1} nh & mvr = 2  {0.5}  m × nh = 2 42  k  m  r3 = n2  h2  r =  {1.5} & KE = 1 mv2 2  1 mv2 = 1 k  {1} ] 2 2 r5 Q.11 Three flasks of identical volume are connected through tubes of negligible volume with two valves both initially closed. Container I is at temperature 2T & contains 1 mole H2, 2 mole O2 & 3 mole SO2 & container II and III are maintained at temperature T. Valve I is opened for a very long time (till movement of gases stop). Valve I is then closed & valve II is opened for a very short time (so that rate of effusion remains constant) & then closed again.Assuming that container II & III contained nothing initially answer the following questions. (a) What is the number of moles in the two container I & II just before opening valve II? (b) What is the ratio of moles of each specie in container III after opening value II? [6] [Sol. After opening valve I for a long time gases will move till partial pressure of each is same.  {1}  Vsame Psame  nT should be same for container I & II  moles get divided in the ratio 1:2 in the three containers  Container I = 1 H , 3 2 3 O2 , 3 SO 3 2  {1} Container II = 2 3 H2 , 4 3 O2 , 6 SO 3 2  {1} Check for alternate solution After opening valve II for a short time, effusion will occur  {1} rH 2 r = = 2 2  {0.5} 1 r 2 & r = 2 = 4 2 3  {0.5}  r : r : r = 4 2 : 2 2 : 1  {1} ] H2 O2 SO2 3 3 PART -C Q.12 An organic compound contains 1 – OH group, 1  bond & 1 ketone group other than carbon and hydrogen. It is known that '  ' bond is non terminal & alcohol is a primary alcohol. 224 ml of the gaseous molecule at STP is exploded with unknown but excess amount of O2 and the resultant reaction caused a contraction of V ml at STP. The acidic gas obtained in the reaction was treated with excess of NaOH & the resultant solution was divided into two equal halves. One half required 35 ml of 1 N HCl using phenolphthalein indicator & other half required 60 ml of 1 N HCl using MeOH indicator. Using the above information answer the questions (a) Predict the general formula of compound (b) Predict the molecular formula of compound (c) What will be the value of V ml (contraction due to combustion & cooling) (d) Predict the possible structural formula(e) [1.5+3+1.5+1.5=7.5] [Sol. Gen. formula = CnH2n–4O2  {1.5} O. C + O2  CO2 + H2O (g) (g) (g) (l) CO2 + NaOH  Na2CO3 x n 0 n – 2x x x  n  2x   × 1 +  2   × 1 = 35  x  n  2x  & × 2 +  2   × 1 = 6   x = 50 n – 2x = 20  n = 120  m.m of NaOH = 120 m. m of CO2 = 50  {1} Also, since all CO2 formed O.C whose I D m.m reacted. (b)  n = 5  C5H6O2  {2} (c) C H O + 13 O  5CO + 3H O  {0.5} 5 6 2 2 2 2 2  Contraction in moles = 15 - 5 = 5 2 2  Contraction in volume = 25 ml  {1} OH O (d) | ||  {0.75} C – C – C  C – C O OH || |  {0.75} ] C – C  C  C – C Q.13 An atomic orbital has radial wave function represented as (r) = ker / k' r2[(r2  5.5a r + 7a 2 )(r2  8.5a r +18a 2 )] where k, k' & a0 are constants. (a) How many radial nodes are present at what distance from the nucleus? (answer in terms of a0) (b) Draw a radial probability distribution curve for the above orbital clearly highlighting the parameters which are being plotted & nodes & antinodes. (c) In which period of periodic table does the first element with last electron in this orbital is expected to go. What will be the total number of elements which can be present in that period? [4+1.5+2=7.5] [Sol.(a)  Polynomial is of 4th degree  4 roots  4 radial nodes  {1} (r2 – 5.5a r + 7 a 2 ) = 0 0 0  (r – 3.5a0)(r – 2a0) = 0  r = 2a0 & 3.5a0  {1.5} Also, (r2 – 8.5a r + 18 a 2 ) = 0 0 0  (r – 4.5a0)(r – 4a0) = 0  r = 4a0, 4.5a0  {1.5}  4r nodes at 2a0, 3.5a0, 4a0, 4.5a0 (b) n–l–1 = 4 l = 2  n = 7d  {0.5}  {1} (c) 7d goes in 8th period  {0.5} no. of subshells (8s, 8p, 7d, 6f, 5g) no. of orbitals (1 + 3 + 5 + 7 + 9) no. of elements = 48  {1.5} ] Q.14 Complete the missing data in the following graph. (a) moles, n = 2, ideal gas subjected to a process in which P(atm) × 0.0821 = V (lit) Calculate tan  & C. (b) Calculate slope & intercept of the graph if wt. of silver salt of succinic acid is plotted with wt. of silver residue obtained due to heating the silver salt. [4+3.5=7.5] [Sol.(a) P × 0.0821 = V log P = log V – log 0.0821  {1}  y = mx + c  c = – log 0.0821  {1.5} & slope = 1  {1.5} O (b) || O O || || Ag AgO  O ||  Ag HO  C  CH2  CH2  C OH Succinic Acid  C  CH2  CH2  C  OAg Silver Salt  Wss × 2 × 108 = W 332 Ag  {1.5}  W = 332  W ss 2 108 Ag  y = mx + c c = 0 Slope(m) = 332 2 108 = 1.537  {2} ] Q.15(a) Calculate H of the following reaction from the data (all in kJ/mole) NH3 (g) + HCl (g)  NH4Cl (s) (i) H proton gain enthalpy of NH3 (g) = – 720 (ii) H bond dissociation of HCl (g) = 100 (iii) H ionization energy of H = 1300 (iv) H Electron gain enthalpy of Cl = – 350 (v) H lattice energy of NH4Cl (s) = – 700 (b) If 107 gm of a mixture containing NH3 (g) and HCl (g) is to be divided so as to obtain maximum amount of energy. What should be the wt. of each specie taken for this and what is the amount of energy. [5+2.5] [Sol.(a) NH3(g) + HCl (g)  NH4Cl(g) H = ?  {5} For correct Born Haber cycle award 2 marks For calculation error deduct only 1 mark (b) For maximum energy equal moles so that no. L. R.  {1}  a × 53.5 = 107  a = 2  ER = 2 × 370 & wt = 34 & 73  {1.5} ]