XI CHEMISTRY SAMPLE PAPER 5. Max. Time: Three Hours Max. Marks: 70

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1 XI CHEMISTRY SAMPLE PAPER 5 Max. Time: Three Hours Max. Marks: 70 General Instructions: 1. All questions are compulsory.. Question no. 1 to 8 are very short answer questions and carry 1 mark each.. Question no. 9 to 18 are short answer questions and carry marks each. 4. Question no. 19 to 7 are also short answer questions and carry marks each 5. Question no. 8 to 0 are long answer questions and carry 5 marks each 6. Use log tables if necessary, use of calculators is not allowed. Q1: What happens when sodium metal is heated in free supply of air? Q: Given: Al(s) + O (g) Al O (s), f H Ө = 1,670 kj/mol for Al O (s). Determine ΔH Ө for the reaction Al O (s) 4Al (s) + O (g). Q: Explain why BeH molecule has zero dipole moment although the Be- H bonds are polar? Q4: Predict the shape of the PH molecule according to VSEPR theory. (1mark) Q5: Which isotope of hydrogen is radioactive? Q6: Write the correct IUPAC name of the compound given below: Q7: How many mono substituted derivatives of naphthalene are possible?

2 Q8: Name any two gases responsible for greenhouse effect. Q9. Arrange the following ions in order of increasing ionic radius: K +, P, S, Cl. Give reason. ( marks) Q10: The successive ionization energies of a certain element are I 1 = kj/mol, I =1145 kj/mol, I = 4900 kj/mol, I 4 = 6500 kj/mol, and I 5 = 8100 kj/mol. This pattern of ionization energies suggests that the unknown element is: a) K b) Si c) Ca d) As Explain your answer. ( marks) Q11: A sample of gas occupies.00 L at 760 torr. Calculate the volume the gas will occupy if the pressure is changed to 1.45 atm and the temperature remains constant. ( marks) Q 1: A mixture of hydrogen and oxygen at 1 bar pressure contains 0 % by mass of hydrogen.calculate the partial pressure of hydrogen. ( marks) Q1: Complete the following reactions P O( s) H O(l) (i) (ii) S i C l( 4l) H O( l) ( marks) Q14: Explain: (i) Alkali metals are soft and can be cut with help of a knife. (ii) Potassium is more reactive than sodium. ( marks) Q 15: (i) How would you distinguish between BeSO4 and BaSO 4? (ii) Which is thermally most stable alkaline earth metal carbonate among MgCO, CaCO, SrCO, BaCO? Why? ( marks) Q16: Why SiCl 4 can be easily hydrolysed but CCl 4 cannot be hydrolysed easily? Explain with reaction. ( marks) Q17: Arrange benzene, hexane and ethyne in decreasing order of acidic behaviour.also give reasons for this behaviour. ( marks) Q 18. State the difference between classical smog and photochemical smog. ( marks) Q kg of N g & 10.0 kg of H g are mixed to produce NH g, identify the limiting reagent. Also, calculate the amount of NH formed. ( marks)

3 Q 0. (i) Calculate the wavelength in nanometers, of visible light having a 14-1 frequency of s. (ii) What are frequency and wavelength of a photon emitted during a transition from n = 6 to n = 1 state in the hydrogen atom. ( marks) Q 1. (i) Explain why the following electronic configuration is not possible: 1 n=1, l=0, m l =+1, m s=+ (ii) Write electric configurations of Cu & Cu. ( marks) Q. (i) Draw the resonating structures of carbon dioxide molecule. (ii) Why is NF trigonal pyramidal while BF is trigonal planar, though both are tetra atomic molecules? (iii) State the hybridization of carbon atoms numbered 1 & : O HC C CH CH C H Q. Consider the reaction: SO g + O g SO(g) kj 1 ( marks) Indicate the direction in which the equilibrium will shift when: (i) Concentration of SO is increased. (ii) Concentration of SO is increased. (iii) Temperature is increased. ( marks) Q 4. (i) Calculate the concentration of hydroxyl ion in 0.1 M solution of 5 NH 4 OH havingk (ii) b Ksp value of two sparingly soluble salts Ni OH and AgCN are 10 and respectively. Which salt is more soluble? ( marks) Q 5. Write the balanced equation by half-reaction method: H S + Cl S + Cl - (in acidic medium) ( marks) Q 6. (i) Draw the structural isomers of pentane. (ii) Give the IUPAC names of the following compounds.

4 a) CH l CH C C H CH CH CH b) l l CH CH c) CH CH CH CH CH l CH CH CH ( marks) Q 7. (i) g of an organic compound gave on combustion g of carbon dioxide and 0.05 g of water. Calculate the percentage of C and H in it. (ii) What will happen during Lassaigne s test for nitrogen if the compound also contains sulphur? ( marks) Q 8. (i) In a process, 701 J of heat is absorbed by a system and 94 J of work is done by the system. What is the change in internal energy for the process? (ii) The equilibrium constant for the reaction is 10. Calculate the value of G. Given R = 8.0 J mol -1 K -1 ; T = 00 K (5 marks) OR (i) Calculate lattice energy for the change + - Li g +Cl g LiCl(s) Given that: Δ H of Li = kj/mol sub diss Δ H of Cl =44.4 kj/mol Δ H ie eg f of Li g =50.07 kj/mol Δ H of Cl g = kj/mol Δ H of LiCl(s) = kj/ mol (ii) For a reaction; A g +B g D g U 10.5 kj & S 4.1 J

5 Calculate G for the reaction & predict whether the reaction is spontaneous or not at 98 K. (5 marks) Q 9. (i) When happens when borax solution is acidified. Write the chemical reactions for the reaction. (ii) Explain why BF exists whereas BH does not? (iii) SiO is solid but CO is a gas at room temperature. (5 marks) OR When a metal X is treated with NaOH a white precipitate (A) is obtained, which is soluble in excess of NaOH to give soluble complex (B). Compound (A) is soluble in dilute HCl to form compound (C).The compound (A) when heated strongly gives D which is used to extract metal. Identify (X), (A), (B), (C) & D. Write suitable equations to support their identities. (5 marks) Q 0. Complete the following reactions. CH COO Mn (i) CH CH + O Δ (ii) CH Br -CH Br +Zn (iii) l H CH C CH H O (iv) CH (v) OR (5 marks)

6 (i) Outline all the steps in the synthesis of the compound styrene from benzene. (ii) Give the products of ozonolysis of mesitylene. (5 marks) SOLUTIONS TO SAMPLE PAPER 5 Ans 1:On heating sodium in free supply of air, it forms sodium peroxide. (1/ mark) Na(s) + O (Air) Na O (Sodium peroxide) (1/ mark) Ans: ΔH Ө = x (+ (1,670)) kj/mol = +,40 kj/ mol Ans : BeH is a linear molecule with H-Be-H bond angle as 180. Although the Be-H bonds are polar, the bond polarities cancel each other and the net dipole moment is zero. Ans 4: Trigonal pyramidal Ans 5: Tritium Ans 6:,-Dimethylpentane Ans 7: Ans 8: Carbon dioxide and methane Ans9: K + < Cl < S < P Reason: All the ions are isoelectronic with 18 electrons. If the number of electrons is the same, as the number of protons increase, the nuclear charge increases and hence the outermost electrons will experience a greater force of attraction towards the nucleus. This results in the decrease in ionic radii. Since the nuclear charge decreases from K + to P -, the ionic radii increases from K + to P -. Ans 10: The unknown element is Ca. Here the third ionization energy is very high which suggest that the removal of the third electron is difficult. The electronic configuration of calcium is [Ar] 4s. First two electrons can be removed without much difficulty. But the removal of third electron from the stable electronic configuration of argon is difficult. Hence, the third ionization energy is high. Ans11: The given question is based on Boyle s law Therefore,

7 P V 1 1 P V(At cons tan t temperature)(1mark) 760 torr L 1.45 atm V L 760 torr / atm V.07L(1mark) Ans 1: Since 0 % by mass of hydrogen is present in the mixture, mass of hydrogen in the mixture is 0 g in 100 g of the mixture present. Remaining mass in the mixture is of oxygen = 80 g No. of moles of hydrogen =0 g / g mol -1 = 10 mol No. of moles of oxygen = 80 g / g mol -1 =.5 mol Mole fraction of hydrogen nh x(1 / mark) n n H H O (1 / mark) Partial pressure of hydrogen p x p(1 / mark) H H total 0.8 1bar 0.8bar(1 / mark) Therefore partial pressure of hydrogen is 0.8 bar. Ans 1: P O(s) 6 H O(l) 4H P O aq (i) (ii) S ic l(l) H O(l) S io s 4H C l(aq) 4 Ans 14: (i) Alkali metals have large atomic size with only one valence electron. Thus, they have weak metallic bonding between the atoms of the metal. Because of weak metallic bonding, alkali metals are soft and can be cut with a knife. (ii) Reactivity of metals depends on ionization enthalpy. Smaller is the ionization enthalpy, greater is the reactivity. Potassium has a larger atomic size than sodium. Thus, the ionization enthalpy of potassium is less than sodium. Hence, potassium is more reactive than sodium. Ans15: (i) BeSO 4 and BaSO 4 can be differentiated by solubility test. BeSO 4 is soluble in water and BaSO 4 is insoluble in water.

8 (ii) BaCO is thermally most stable alkaline earth metal carbonate because Ba + ion being larger in size is more stabilized by larger CO - ion through formation of stable lattice. Ans 16: In SiCl 4, Si atom has empty d-orbitals in its valence shell. These empty d orbitals of Si can accept lone pair of electrons from water molecule. Eventually this leads to hydrolysis of SiCl4 and Si(OH) 4 is formed. (1/ mark) (1/ mark) Carbon atom on the other hand does not have any vacant d-orbitals in its valence shell. Hence, it cannot accept the electron pair from water molecule. Thus, CCl 4 does not hydrolyse. Ans17: The decreasing order of acidic behaviour is: Ethyne > benzene > n-pentane The C-H bond in ethyne, benzene and n-pentane are formed by sp-s, sp s and sp -s overlap. Now, greater the percentage s character, greater is the electronegativity. Therefore, sp-hybridised carbon in ethyne is more electronegative then sp hybridised carbon of benzene which in turn is more electronegative than sp hybridised carbon of n-pentane. Therefore, the polarity of the C-H bond is in order of: Ethyne > benzene > pentane Hence the acidity order is: Ethyne > benzene > pentane Ans 18. Classical Smog 1. It occurs in cool humid climate. ( 1 mark). It is a mixture of smoke, fog & sulphur dioxide. ( 1 mark) Photochemical Smog 1. It occurs in warm, dry and sunny climate. ( 1 mark). Components of photochemical smog result from the action of sunlight on unsaturated hydrocarbons & oxides of nitrogen produced by automobiles &

9 factories. ( 1 mark) Ans 19. Moles of N Mass Molar mass g 8 g/ mol = mol ( 1 mark) Moles of H Mass = Molar Mass g = g/mol ( 1 mark) = mol N(g) + H g N H(g)(Eqn 1) According to equation (1), 1 mole of N (g) reacts with = moles of H (g) Therefore mol of N (g) will react with = moles of H (g) mol ( 1 mark) But we are having Hence, mol of H (g) only. H g is the limiting reagent. ( 1 mark) To calculate the amount of NH formed, moles of H (g) give = moles of NH (g) Therefore,

10 5.0 x 10 moles of H will give = x5 x10 moles of NH(1 / mark). x10 moles of NH Mass of NH produced. x10 x17 g of NH 56.1 kg(1 / mark) Ans 0. (i) c c 10 m / s s m 686nm(1 / mark) (1 / mark) (ii) Here n 1 = 6 & n = 1 The energy gap between two orbits for a hydrogen atom is given as 10 J -(1 / mark) ΔE =.18 n1 n J J 6 18 E J(1 / mark) Since E is negative energy is emitted, frequency of photon is given by ΔE =(1 / ) mark h J = Js = s = Hz(1 / ) mark Ans 1.

11 (i) For n =1, Value of l n For each value of l, Value of m l,...,0,...,+l (1/ mar k) Therefore, For n=1,l=0, m 0 l l Thus the value of m 1 is not possible. (1/ mark) l (ii) Electronic configuration of Cu Z 9 is 1s s p s p d 4s Electronic configuration of Cu Ans. (i) Resonating structures of CO molecule is 1s s p s p d O CO: O C O: :O C O: (ii) In NF, N atom involves sp hybridization and one position is occupied by a lone pair. Therefore the molecule is trigonal pyramidal. But in BF, B involves sp hybridization having trigonal planar geometry. Thus NF is trigonal pyramidal while BF is trigonal planar, even though both are tetra atomic molecules. (iii) C C 1 sp sp (1/ x = 1 mark) Ans. (i) If the concentration of SO is increased the equilibrium will shift in the forward direction to consume the reactant SO. (ii) If the concentration of SO is increased the equilibrium will shift in the backward direction to consume the product SO. (iii) If the temperature is increased, the equilibrium will shift in the backward direction as the increase in temperature will be compensated by absorbing heat. (1mark) Ans (i) NH OH aq NH(aq) + OH aq 4 4

12 + - NH 4 OH K b= NH 4 OH + - Now, NH 4 = OH NH4OH 0.1M K b - OH = NH 4 OH - -5 OH = = ( 1 Mark) ( 1 Mark) - - OH = mol/l ( 1 Mark) (ii) AgCN + - Ag + CN Let x mol/l be the solubility of AgCN + - Thus Ag =x, [CN ]=x + - K sp = Ag CN = x x= K sp ( 1 Mark) = = Ni(OH) Ni + OH Let y mol/l be the solubility of NiOH + - Thus Ni =y & OH =y + - K sp = Ni OH =y (y) =4 y K y = 4 sp 10 y= y = ( 1 Mark)

13 Since solubility of NiOH is more than AgCN, NiOH is more soluble than AgCN. ( 1 Mark) Ans 5. Step 1: Write the oxidation numbers and separate the reaction into oxidation half and reduction half reactions H S + Cl S + Cl - (1/ mark) Step : The half reactions are: Oxidation half reaction: H S S Reduction half reaction: Cl Cl - (1/ mark) Step : Balance oxidation number by adding electrons. Oxidation half reaction: H S S + e - Reduction half reaction: Cl + e - Cl - (1/ mark) Step 4: There are no oxygen atoms. So, balance the hydrogen atoms. Since the reaction takes place in acidic medium, the balancing of hydrogen atoms is done by adding the appropriate number of hydrogen ions to the deficient side. Oxidation half reaction: H S S + e - + H + Reduction half reaction: Cl + e - Cl - (1/ mark) Step 5: Add the two reactions to get a balanced redox reaction. Balanced reaction: H S + Cl S + H + + Cl - Ans 6. (i) There are structural isomers of pentane: CH CH CH CH CH n Pentane ( 1 Mark)

14 CH CH CH CH l CH Methylbu tan e CH l CH C CH l CH, Dimethylpropane (ii) a),, Trimethylhexane ( 1 Mark) ( 1 Mark) ( 1 Mark) b) Ethylcyclopentane ( 1 Mark) c) Ethylhexane ( 1 Mark) Ans 7. (i) Mass of organic compound = g Mass of CO produced = g Mass of H O produced = 0.05 g 1 Mass of CO % of C= Mass of compound taken = 100= Mass of HO % of H = Mass of compound taken 0.05 = 100= (ii) Blood red colouration due to Fe(CNS) will be produced. Ans 8. (i) Heat absorbed by the system (q) = J (1/ mark) Work done by the system (w) = -94 J (1/ mark) Change in internal energy (U) = q + w (1/ mark) = = +07 J (1/ mark) (ii)

15 G.0RT logk 1 1 R 8.0 JK mol T 00K K 10 G.0RT logk.0 x 8.0 x 00 xlog Jmol OR 1 Ans 8. (i) ΔfH =Δsub H +ΔieH Δ diss H + Δeg H + Δ lattice H (1/ mark) 1 ΔlatticeH = Δ lattice H =-89.1 kj/mol(1/ mark) (ii) Δ U =-10.5 kj, Δn g =-1mol, T =98 K R= kj K mol ΔH =ΔU + Δn g RT kj K mol 98K(1/ ma ΔH = mol = kj-.478 kj = kj(1/ mark) G H TS(1 / mark) 1978 J 98( 4.1J)(1 / mark) J(1 / mark) Since the value of (1/ mark) G is negative, the reaction is spontaneous. Ans 9. (i) Borax solution on acidification forms boric acid. N a B O + H C l + 5 H O N a C l + 4 H B O 4 7 rk) (ii) BF is trigonal planar molecule. Due to p p back bonding lone pair of electrons of F is back donated to B atom. This delocalization reduces the deficiency of electrons of boron thereby increasing the stability of BF molecule. ( 1 Mark) The mechanism is as follows:

16 ( 1 Mark) Due to absence of lone pair of electrons on H atom this compensation does not occur in BH. ( 1 Mark) In other words electron deficiency of B stays & hence it reduces its electron deficiency as BH dimerises to form B H 6. ( 1 Mark) (iii) Carbon is able to form p π - pπ bond with O atom and constitute a stable non - polar molecule O = C = O. Due to weak inter particle force its boiling point is low and it is gas at room temperature. Si on the other hand is not able to from p π - pπ bond with O atoms because of its relatively large size. In order to complete its octet Si is linked to four O atoms around it by sigma bond & these constitutes network structure, which is responsible for its solid state. OR Ans 9. marks for writing reactions 1 mark for identifying X ½ mark each for correctly identifying A, B, C and D. Ans 0.

17 CH COO Mn Δ (i) CH CH +O CH COOH+H O (ii) CH Br-CHBr+ Zn CH = CH + ZnBr (iii) (iv) OH l H l l CH C CH H O CH C CH CH CH (v) Ans 0. (i) OR

18 (ii) (1 mark for the correct product + 1 mark for the correct reagents)

Sectional Solutions Key

Sectional Solutions Key 1. For the equilibrium: 2SO 2 (g) + O 2 (g) 2SO 3 (g) + 188 kj, the number of moles of sulfur trioxide will increase if: a. the temperature of the system is increased (at constant