XI CHEMISTRY SAMPLE PAPER 4 Time: Three Hours
Transcription
XI CHEMISTRY SAMPLE PAPER 4 Time: Three Hours
XI CHEMISTRY SAMPLE PAPER 4 Time: Three Hours Max. Marks: 70 General Instructions 1. All questions are compulsory. 2. Question nos. 1 to 8 are very short answer questions and carry 1 mark each. 3. Question nos. 9 to 18 are short answer questions and carry 2 marks each. 4. Question nos. 19 to 27 are also short answer questions and carry 3 marks each 5. Question nos. 28 to 30 are long answer questions and carry 5 marks each 6. Use log tables if necessary, use of calculators is not allowed. Q 1: Write the formula of the compound Nickel (II) sulphate? Q 2: Temporary hardness in water is due to presence of which salts? Q 3: What is the mass of one atom of oxygen? Q 4: What are silicones? Q 5: Give the values for principal quantum number and magnetic quantum number for 19th electron of K (Potassium). Q 6: What shapes are associated with sp3d and sp3d2 hybrid orbitals ? Q 7: Why is an organic compound fused with sodium for testing halogens, nitrogen, sulphur and phosphorous?? Q 8: 2A(g) + B(g) 4C(g) + Heat What is the effect of adding He at a constant volume on above equilibrium? Q 9: Which of the following has largest size? Mg, Mg2+, Al3+, Al Q10: Give IUPAC name and symbol of element with atomic number 110 and 115. Q11: Which of the two is more concentrated and why? 1M or 1m aqueous solution of a solute + Q12: N2 N2 + O2 O2 Why is there an increase in bond order in going from O2 to O2+ while a decrease in going from N2 to N2+ ? Q13: Calculate the total pressure of a mixture of 8g of O2 (g) and 4g of H2 (g) in a vessel of 1dm3 at 270C. (Atomic mass of O=16gmol-1and H=1 gmol-1 .R=0.083 bar dm3/K mol). Q14: Give equations to prove amphoteric nature of water. Q15: Balance the following equation in an alkaline medium by half reaction method. 2Cr(OH)3 + IO3 I + CrO4 Q16: The wavelength of first spectral line in Balmer series is 6561 Å . Calculate the wavelength of second spectral line in Balmer series. Q17: On a ship sailing in Pacific Ocean where temperature is 23.4°C, a balloon is filled with 2 L air. What will be the volume of the balloon when the ship reaches Indian Ocean, where temperature is 26.1°C? Q18: All C-O bond lengths in CO32- are equal. Account for this observation. Q19: Give reasons: (i) Anhydrous AlCl3 is covalent but hydrated AlCl3 is electrovalent.Explain (ii)Boric acid behaves a Lewis acid? Explain (iii) Boron is unable to form BF63– ion. Explain Q20: At 60°C, dinitrogen tetroxide is fifty percent dissociated. Calculate the standard free energy change at this temperature and at one atmosphere. Q21: What type of isomerism is exhibited by following pair of compounds? (i) Ethanol and Methoxy methane (ii) o-cresol and m-cresol (iii) Pentan-3-one and pentane-2-one Q22: When a metal X is treated with sodium hydroxide, a white precipitate A is obtained, which is soluble in an excess of NaOH to give soluble complex B. Compound A is soluble in diluted HCl to form compound C. The compound A when heated strongly gives D, which is used to extract metal. Identify X, A, B, C and D. Write suitable equations to support their identities. Q23: Calculate the energy associated with the first orbit of He +. What is the radius of this orbit? Q 24: What are the necessary conditions for any system to be aromatic? Q 25: Calculate the enthalpy of combustion of glucose from the following data θ C (graphite) +O2 CO2 (g) ; rH = -395kJ H2 (g) + 1 O (g) 2 2 H2 O(l) θ rH = -269.4kJ ; 6C (graphite) + 6H2 (g) + 3 O2 (g) C6H12O6 (s) ; θ r H = -1169.9kJ Q 26: (a) Fish do not grow as well in warm water as in cold water. Why? (b) Why does rain water normally have a pH about 5.6? (c) Name two major greenhouse gases. Q27: 0.2325g of an organic compound was analysed for nitrogen by Duma’s method. 31.7mL of moist nitrogen was collected at 250C and 755.8mm Hg pressure. Calculate the percentage of N in the sample. (Aq. Tension of water at 25oC is 23.8mm). OR (a) Why cannot sulphuric acid be used to acidify sodium extract for testing S using lead acetate solution? (b)Which of the carbocations is most stable and why? + + + (CH3 ) 3C , CH3CH2CH2 , CH3CHCH2CH3 (c) Why does a liquid vaporize below its boiling point in steam distillation process? Q 28 (a) Convert: (i) Propene to propane-1,2-diol (ii) Isopropylbromide to n-propylbromide (b) An alkene on ozonolysis gives butan-2-one and 2-methylpropanal.Give the structure and IUPAC name of Alkene. What products will be obtained when it is treated with hot, concentrated KMnO4? OR Q 28: Complete the equations: NaNH2 Red hot iron tube A B (i) CH2 =CHBr 873K Anhdrous (ii) C6H6 +CH3COCl A+B AlCl 3 NaOH(aq) Sodalime B (iii) CH3COOH A No peroxide (iv) CH2 CH CH2 CH2 CH2 CH3 HBr Peroxide (v) CH2 CH CH2 CH2 CH2 CH3 HBr Q29: Calculate the pH of a 0.10M ammonia solution. Calculate the pH after 50.0 mL of this solution is treated with 25.0 mL of 0.10M HCl. The dissociation constant of ammonia, Kb = 1.77 x 10-5. OR Q29: Calculate the pH of the resultant mixtures: 10mL of 0.2M Ca(OH)2 + 25 mL of 0.1 M HCl Q30: Give reasons for the following (a) Unlike Na2CO3, K2CO3 cannot be prepared by Solvay process. Why? (b) Why are alkali metals not found in nature? (c) Sodium is less reactive than potassium why? (d) Alkali metals are good reducing agents. Why? (e) Alkali metals are paramagnetic but their salts are diamagnetic. Why? OR Q30: Complete the following reactions: (a) Why does the solubility of alkaline earth metal carbonates and sulphates in water decrease down the group? (b)Arrange the following alkali metal ions in decreasing order of their mobility: Li+, Na+ , K+, Rb+, Cs+ .Explain (c) NaOH is a stronger base than LiOH. Explain (d) Why are alkali metals kept in paraffin or kerosene ? (e) Why does lithium show properties uncommon to the rest of the alkali metals? SOLUTIONS TO SAMPLE PAPER 4 Ans 1: NiSO4 (1 mark) Ans 2: Temporary hardness in water is due to salts of magnesium and calcium in form of hydrogen carbonates. (1 mark) (1 mark) Ans 3: 23 NA or 6.023 × 10 atoms of oxygen have a mass of 16g where NA is Avogadro's number So, mass of one atom of oxygen = 16 23 6.023 × 10 = 2.65 10 23 g (1 mark) Ans 4: Silicones are a group of organosilicon polymers containing repeated (R2SiO) units. (1 mark) Ans 5: Electronic configuration of K = 1s22s22p63s23p64s1 This means that the 19th electron lies in the s- subshell, Thus it is 4s, so values for principal quantum number ,n=4 and magnetic quantum number, ml =0 (1 mark) Ans 6: 1 (a) sp3d hybrid orbitals -Trigonal bipyramidal mark 2 1 (b) sp3d2 hybrid orbitals-Octahedral 2 mark Ans 7: Organic compound is fused with sodium metal to convert N, S, P and halogens present in organic compound to their corresponding sodium salts. (1 mark) Ans 8: There will be no effect of adding He at constant volume. This is due to the fact that when an inert gas is added to equilibrium at constant volume, the total pressure will increase. But the concentration of reactants and products will not change. (1 mark) Ans 9: Atomic radii decrease across a period.Cations is smaller than their parent atoms. Among isoelectronic species, the one with the larger positive nuclear charge will have a smaller radius. Hence the largest species is Mg; the smallest one is Al3+. (2 marks) Ans 10: (a) Name and symbol of element with atomic number 110 1 Name: Ununnillium mark 2 1 Symbol: Uun 2 mark (b) Name and symbol of element with atomic number 115 1 Name: Ununpentium mark 2 1 Symbol: Uup 2 mark Ans 11: 1 molar solution contains 1mole of solute in 1 L of solution while 1 1 molal solution contains 1 mole of solute in 1000g of solvent. mark 2 Considering density of water as almost 1g/mL, then 1mole of solute is present in 1000mL of water in 1molal solution while 1mole of it is present in less than 1000mL of water in 1 molar solution (1000mL solution in molar solution = Volume of solute + Volume of solvent). (1 mark) 1 Thus 1M solution is more concentrated than 1m solution. mark 2 Ans 12: Electronic configuration of O2 is 2 2 2 2 2 2 2 1 1 (1s) ( * 1s) (2s) ( * 2s) (2pz ) (2px 2py )( * 2p x * 2p y ) 1 2 mark In going from O2 to O2+ , an electron is lost from an antibonding molecular Orbital . Bond order is calculated as one half of the difference in number of electrons in bonding and antibonding molecular orbital. Thus 1 bond order increases mark 2 Electronic configuration of N2 is 2 2 2 2 2 2 2 (1s) ( * 1s) (2s) ( * 2s) (2px 2py )(2pz ) In going from N2 to N2+ , the electron is lost from a bonding molecular Orbital. Bond order is calculated as one half of the difference in number of electrons in bonding and antibonding molecular orbital. Thus bond order decreases. 1 2 mark Ans 13: No. of moles of O2 = No. of moles of H2 = 8 = 0.25 32 4 =2 2 1 2 mark pV=nRT pO 2 pH = 2 0.25 0.083 300 6.23 bar 1 2 0.083 300 1 49.8 bar 1 2 mark 1 2 mark Total pressure = pO + pH 2 2 = 6.23 + 49.8 = 56.03 bar Ans 14: 1 2 mark Amphoteric nature means H2O can act as an acid as well as a base H2 O + NH3 + - NH4 + OH (1mark) (Acid) H2 O + HCl + - H3O + Cl (1mark) (Base) Ans 15: a) First, we will write down the oxidation number of each atom Cr(OH)3 3 + 2 IO3 5 - I 1 + 2 CrO4 6 b) Write separately oxidation & reduction half reactions Oxidation half reaction: Cr(OH)3 3 CrO4 6 2- Reduction half reaction: IO3 + 6e I 5 1 1 2 mark - 3e c) Balance O atoms by adding H2O molecules to the side deficient in 'O' atoms and then balancing H atoms Cr(OH)3 + H2 O - IO3 - + 6e CrO4 - I 2- - 3e + 3H2O 1 2 mark d) Balance H atoms. Since the medium is alkaline, therefore H2O molecules are added to the side deficient in H atoms and equal no. of OH ions to the other side : - 5OH Cr(OH)3 + 2- - 3e - 5OH ( Cr(OH)3 + H2 O - CrO4 - - IO3 + 6e 3H2 O I ( IO3 + 6e 6H2 O CrO4 +4H2O 2- - 3e +5H2O) - + 6OH I + 3H2O + 6OH ) 1 2 mark e) Equalise the electrons lost and gained by multiplying the oxidation half reaction with 2. 2Cr(OH)3 + - 10OH 2CrO4 2- - 6e +8H2O Adding the oxidation half reaction and reduction half reaction we get 2Cr(OH)3 + - - - 10OH 2CrO4 - 2- - 6e - +8H2O IO3 + 6e 3 H2 O I + 6OH _________________________________________________ 22Cr(OH)3 IO3 4OH 2CrO4 I +5 H2O 1 2 mark Ans 16 : According to Rydberg equation, R = 109,677 which is the Rydberg's constant 1 v= Wavelength For first line in Balmer series, n1 = 2, n2 = 3 Given wavelength of 1st spectral line = 6561 Å 1 1 1 5 1 Therefore, = R 2 2 =R (i) mark 6561 36 2 3 2 For second line in Balmer series, n1= 1 1 1 Therefore, = R 2 2 Wavelength 4 2 2, n2 =4 3 = R 16 1 (ii) mark 2 Dividing eq. (i) by (ii), we get: Wavelength 5 16 6561 36 3 1 2 mark 1 2 mark Wavelength=4860 Å Ans 17: V1 = 2 L T1 = (23.4 + 273) K = 296.4 K T2 = 26.1 + 273 = 299.1 K From Charles law V1 T1 V2 1 2 mark T2 V T V2 1 2 T1 V2 2 L 299.1 K 296.4 K 1 2 mark 2 L 1.009 2.018 L 1mark Ans 18: All C-O bond lengths in CO32- are equal because of resonance. All bonds have partial double bond character (1 mark) (1 mark) Ans 19 : (i)Anhydrous AlCl3 is covalent but hydrated AlCl3 is electrovalent because when it is dissolved in water the high heat of hydration is sufficient to break the covalent bond of AlCl3 into Al3+ and Cl- ions .(1 mark) (ii) Boric acid behaves as Lewis acid by accepting a pair of electron from 1 OH- ion (in water) . mark 2 - B(OH)3 + 2H-O-H [B(OH)4 ] + H3O 1 2 mark (iii)Due to non-availability of d orbitals, boron is unable to expand its octet. Therefore, the maximum covalence of boron cannot exceed 4 and it cannot form BF63– ion. (1 mark) Ans 20. N2O4 (g) 2NO2 (g) If N2O4 is 50% dissociated, the mole fraction of both the subs tan ces is given by 1 0.5 0.5 xN O ; 1 0.5 1.5 2 4 xNO 2 1 2 mark 1 2 mark 2 0.5 1 1 0.5 1.5 0.5 0.5 pN O 1 atm 1.5 1.5 2 4 pNO 2 1 2 mark 1 2 mark 1 1 1 atm 1.5 1.5 The equilibrium cons tan t KP is given by pNO2 2 KP pN2O4 1.5 2 (1.5) (0.5) 1 2 mark 1.33 atm Since r G RTln KP 1 1 1 1 r G (8.314 JK mol (8.314 JK mol 1 789.98 kJ mol ) (333 K) (ln(1.33)) ) (333 K) (2.303) (0.1239) 1 2 mark Ans 21: (i) Functional isomerism or functional group isomerism (ii) Position isomerism (iii) Metamerism (1 × 3) Ans 22: Since metal X reacts with NaOH to first give a white ppt. A which dissolves in excess of NaOH to give a soluble complex B , therefore metal X is aluminium ;ppt A is Al(OH)3 and complex B is sodium teterahydroxoaluminate (III) + 3NaOH Al X Al(OH)3 3Na Aluminium hydroxide Na [Al(OH)4 ] Al(OH)3 +NaOH A Sodium tetrahydroxoaluminate (III) B 1mark Since A is amphoteric in nature, it reacts with dilute HCl to form compound C which is AlCl3 Al(OH)3 3HCl AlCl3 3H2O A C 1mark Since A on heating gives D which is used to extract metal, therefore, D must be alumina (Al2O3) 2Al(OH)3 Al2O3 3H2O A D 1mark Ans 23: 2.18 10 J Z 18 En 2 1 atom 2 n 1 2 mark For He ,n 1, Z 2 2.18 10 J2 18 E1 2 2 1 8.72 10 18 J 1mark The radius of the orbit is given by equation rn 0.0529 nm n2 Z 1 2 mark S in ce n 1, and Z 2 rn 0.0529 nm 12 2 0.02645 nm . 1mark Ans 24: Necessary conditions for a compound to be aromatic is (i) The molecule should contain a cyclic cloud of delocalised electrons above and below plane of molecule 1mark (ii) For delocalisation of electrons the ring must be planar to allow cyclic overlap of p-orbitals 1mark (iii) The compound should contain (4n+2) -electrons where n=0,1,2 .This is known as Huckel’ rule 1mark Ans 25. Given C (graphite) +O2 H2 (g) + 1 O (g) 2 2 CO2 (g) H2 O(l) θ ; rH = -395kJ (Eq.1) θ ; rH = -269.4kJ C6H12 O6 (s) 6C (graphite) + 6H2 (g) + 3 O2 (g) ; (Eq.2) θ r H = 1169.9kJ (Eq.3) Combustion of graphite is given by the equation shown below C6H12 O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (g) (Eq.4) Thus enthalpy for combustion reaction can be obtained 1 2 mark by multiplying Eq.1 and Eq.2 by 6 and adding equation Eq.3 6C (graphite) + 6O2 6H2 (g) + 3O2 (g) 6CO2 (g) 6 H2 O(l) ; ; θ r H = -2370 kJ θ rH = -1614.4 kJ 1 2 mark (Eq.6) 1 2 mark (Eq.7) 1 2 mark 1 (Eq.8) mark 2 ______________________________________________________________________ C6H12 O6 (s) + 3O2 (g) 6CO2 (g) + 6H2O (g) C6H12 O6 (s) 6C (graphite) + 6H2 (g) + 3 O2 (g) Reaction enthalpy = (-2370 kJ-1614 kJ) + (1169.9 kJ+0) = -2814.1kJ ; θ r H = 1169.9kJ 1 2 mark Ans 26: (a) As the temperature increases solubility of gas in water decreases. Due to high temperature of water, amount of dissolved oxygen is less, which creates a problem for fish. So, fish do not grow well in warm water (1 mark) (b) Rain water is acidic due to dissolution of CO2 in it, leading to formation of H2CO3 which lowers the pH. Hence the pH is about 5.6 CO2 + H2O + H2CO3 (1 mark) 2- H2CO3 2H + CO3 (c) Carbon dioxide and methane (1 mark) Ans 27: Calculation of volume of nitrogen at S.T.P Experimental conditions Pressure of dry gas P1 = 755.8 - 23.8 = 732 mL V1 = 31.7 mL T1 = 25+273 = 298 K 1 2 mark S.T.P condition P2 = 760 mm V2 = ? T2 = 273 K Applying gas equation, P1 V1 P2 V2 = T1 T2 732×31.7 760×V2 = 298 273 1 2 mark 1 2 mark V2 = 27.97 mL % of Nitrogen = = 28× Volume of N2 at S.T.P×100 22400×Mass of compound 28×27.97×100 22400×0.2325 1 2 mark 1 2 mark 1 2 mark = 15.04 OR (a) This is because, in the presence of sulphuric acid, lead acetate will react with it forming white precipitate of PbSO4, thus interfering with the test. (1 mark) (b) (CH3) 3C+ is most stable because it has the maximum number of alkyl groups i.e. three. Greater the number of alkyl groups on the carbon carrying positive charge, greater would be the dispersal of charge and hence more will be the stability of carbocation. So, tertiary carbocation is the most stable due to maximum dispersal of charge. (1 mark) (c) In steam distillation, water and organic substance vaporize together and total pressure becomes equal to atmospheric pressure. Thus organic substance vaporizes and distils at a temperature lower than its boiling point. (1 mark) Ans 28: (a) (i) (ii) CH3 | CH3 -CHBr (1 mark) alc. KOH HBr CH3CH=CH2 Peroxide effect CH3CH2 CH2 Br (b) From the given products of Ozonolysis, CH3 -CH2 - C- CH3 || O and CH3 H | | CH3 CH-C=O The alkene would be CH3 | CH3 -CH2 - C=CH-CH-CH3 | CH3 1 mark IUPAC Name: 2, 4-dimethylhex-3-ene KMnO 4 CH3 -CH2 - C=CH-CH(CH3 )2 CH3 -CH2 -C=O | | CH3 CH3 Butane-2-one 1 2 mark OR (1 mark) + HOOC-CH(CH3 )2 2-Methylpropanoic acid 1 2 mark NaNH 2 CH2 =CHBr HC CH (i) A 1 2 mark C6H6 +CH3COCl Anhdrous AlCl 3 (ii) (iii) Red hot iron tube C6H6 873K NaOH (aq) CH3COOH (iv) C6H5COCH3 A 1 2 mark CH3COONa A 1 2 mark B 1 2 mark + HCl B 1 2 mark Sodalime B CH4 1 2 mark No peroxide CH2 CH CH2 CH2 CH2 CH3 HBr CH3 CH CH2 CH2 CH2 CH3 | Br Hex 1 ene 2 Bromohexane (v) peroxide CH2 CH CH2 CH2 CH2 CH3 HBr CH2 CH2 CH2 CH2 CH2 CH3 | Br Hex 1 ene 1 Bromohexane Ans 29: NH3 + + Kb = + H2O NH4 - [NH4 ][OH ] = 1.77 x 10-5 [NH3 ] Before neutralization, + [NH3 ] = 0.10 - x 0.10 Kb = - OH 1 mark 2 - [NH4 ] = [OH ] = x Thus + 1 mark 2 2 x -5 = 1.77 x 10 0.10 -5 x = 1.33 x 10 + - = [OH ] 1 mark 2 - Now Kw [H ][OH ] + [H ] = Kw / [OH ] = -14 10 -5 1.33 x 10 -12 = 7.51 x 10 -12 pH = -log(7.5 x 10 ) = 11.12 1 mark 2 On addition of 25 mL of 0.1M HCl solutions (i.e., 2.5 mmol of HCl) to 50 mL of 0.1M ammonia solution (i.e., 5 mmol of NH 3), 2.5 mmol of ammonia molecules are neutralized. The resulting 75 mL solution contains the remaining unneutralized 2.5 mmol of NH3 molecules and 2.5 mmol of NH4+. NH3 + 2.5 At equilibrium HCl 2.5 0 0 + NH4 + Cl 0 0 2.5 2.5 1 2 mark + The resulting 75 mL of solution contains 2.5 mmol of NH4 ions (i.e., 0.033 M) and 2.5 mmol (i.e., 0.033 M) of uneutralized NH3 molecules. This NH3 exists in the following equilibrium. + NH4OH NH4 + 0.033M - y y + Where, y = [OH ] = [NH4 ] - OH y The final 75 mL solution after neutralization already contains 2.5 m mol + + NH4 ions (i.e., 0.033M), thus total concentration of NH4 ions is given as: + [NH4 ] = 0.033 + y As y is small, [NH4OH] = 0.033 M and [NH4 +] = 0.033M. + Kb = - [NH4 ][OH ] 1 2 mark [NH4OH] y(0.033) = (0.033) -5 = 1.77 x 10 1 2 mark 1 2 mark M -5 Thus, y = 1.77 x 10 = [OH-] -14 10 + [H ] = -5 1.77 x 10 = 0.56 x 10-9 1 2 mark Hence, pH = 9.24 1 2 mark OR Ans29: 10 mL of 0.2 M Ca(OH)2 = 10 × 0.2 1000 -3 =2 ×10 25 mL of 0.1 M HCl = 25 × 1 2 mark moles 0.1 1000 =2.5 ×10 -3 1 2 mark moles Ca(OH)2 + 2 HCl CaCl2 + 2H2O 2 moles of HCl reacts with 1 mole of Ca(OH) 2.5 ×10 -3 2 -3 moles of HCl will react with 1.25 × 10 -3 Therefore moles of Ca(OH)2 left unreacted= 2 ×10 -3 =0.75 ×10 Total volume = 10+ 25 = 35 mL 1 2 mark moles of Ca(OH)2 -3 - 1.25 × 10 moles 1 2 mark -3 0.75 10 1000 35 Molarity of Ca(OH)2 = 1 2 mark = 0.0214 M - -2 [OH ] in Ca(OH)2 = 2 ×0.0214 =4.28 ×10 1 2 mark M -2 pOH = -log(4.28 × 10 ) = -(-2+0.6314) = 1.368 1 mark pH = 14-1.368 = 12.632 1 mark Ans 30: (a) Unlike NaHCO3, the intermediate KHCO3 formed during reaction, is highly soluble in water and thus cannot be taken out from solution to obtain K2CO3.Hence, K2CO3 cannot be prepared by Solvay process. (1 mark) (b)Alkali metals are highly reactive because of low ionization enthalpy value and therefore are not found in nature. They are present in combined state only in form of halides, oxides etc. (1 mark) (c) Ionization Energy of potassium is less than sodium because of large size or less effective nuclear charge. Thus, potassium is more reactive than sodium. (1 mark) (d)Alkali metals are strong reducing agents due to their greater ease to lose electrons. (1 mark) (e) Alkali metals have one unpaired electrons (ns1) and are paramagnetic. However, during the salt formation, the unpaired electron is lost by alkali metals to other atom forming anion. Their salts have all paired electrons and show diamagnetic nature. OR Ans 30: (a) The size of anions being much larger compared to cations, the lattice enthalpy will remain almost constant within a particular group. Since the hydration enthalpies decrease down the group, solubility will decrease as found for alkaline earth metal carbonates and sulphates. (1 marks) (b) Cs+ > Rb+> K+> Na+> Li+ Smaller the size of the ion greater is the degree of hydration. Lithium being small in size is hydrated to a large extent and cesium being large in size is hydrated to small extent. (1 mark) (c)The M-OH bond in hydroxides of alkali metal is very weak and can easily ionize to form M+ ions and OH- ions. This accounts for the basic character. Since ionization energy decreases down the group bond between metal and oxygen becomes weak. Therefore basic strength of hydroxides increases accordingly. Thus NaOH is a stronger base than LiOH. (1 marks) (d) Alkali metals are highly sensitive towards air and water and are kept therefore in kerosene or paraffin oil. (1 mark) (e) This is because of exceptionally small size of Lithium and high charge to radius ratio of Li+. (1 mark)