Model Question Paper I CHEMISTRY

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1 Model Question Paper I CEMISTRY I PUC Time:.15 hrs Max Marks: 70 General Instructions: 1) The question paper has four parts A, B, C and D. All Parts are compulsory ) Write balanced chemical equations and draw Labelled diagram wherever necessary PART A I Answer ALL the following questions in a sentence or a word: 10 x Equal volumes of all the gases at the same temperature and pressure should contain equal number of Molecules.. Pν nrt. CO 4. The ability of an atom in a chemical compound to attract the shared compound to attract the shared pair of electrons towards itself is called electronegativity. 5. Zn or Zinc Na CO 10 O Solid form of CO is called dry ice 8. sp hybridisation 9. 4-Methyl pent--en 10. Benzene COONa NaO CaO + Na CO Sodium benzoate Benzene PART -B II Answer Any FIVE questions. Each Question carries TWO marks: 5 x Mole is the amount of substance that contains as many particles or entities as there are atoms in exactly 1g of C 1 isotope Molar Mass of SO Mass No. of Moles of SO4 Molar Mass The volume of one mole of gas at critical temperature is called critical volume Unit of coefficient of viscosity is Poise or 1gcm s or 10 kgm s 1. Sigma bond Pi bond 1) The bond is formed by the axial overlap (head on) of the atomic orbitals 1) The bond is formed by the sideways overlapping of atomic orbitals ) Strong bond ) Weak bond

2 14. Sodium burns in air or oxygen to form sodium peroxide Na + O Na O Sodium peroxide 15. Carbon monoxide is poisonous because it combines with haemoglobin of the blood to form carboxy haemoglobin complex. This complex is 00 times More stable than haemoglobin this prevents haemoglobin from carrying oxygen round the body and resulting in death. 16. butane exhibits geometrical isomerism C C C C C C C Cis- But--ene C Trans But ene In Cis isomer the identical groups are on the same side of the double bond. In trans isomer the identical groups are on the opposite side of the double bond 17. The electron deficient species are called electrophiles Ex: Br or Cl or NO or C 18. i) The amount of oxygen required by bacteria to break down the organic Matter present in a certain volume of a sample of water is called Biochemical oxygen demand ii) Gaseous pollutants: CO or CO or NO or NO or SO PART -C III Answer any FIVE questions. Each Question carries TREE marks: 5 x S block element S- block elements are those in which the last electron enters into the S- orbital of the outer electronic configuration is ns 1- The S- block elements belong to group 1 and of the periodic table P Block elements P-block elements are those in which the last electron enters into the p-orbital of the outer most shell. The general outer electronic configuration is ns np 1-6 The P- block elements belong to group 1 to 18 of the periodic table d- block elements d- block elements are those in which added electron goes into inner d orbital of the penultimate shell. The general outer electronic configuration is (n-1)d 1-10 ns 1- The d- block elements belong to group to 1 of the periodic table d- block elements are also called as transition elements. 0. Electrons configuration of B in ground state is Electronic configuration of B in excited state is

3 Bron undergoes sp hybridisation by using one S and two p orbitals to give: three half filled sp hybrid orbitals which are oriented in trigonal planar arrangement and overlap with p z orbital of chlorine to form three B Cl bonds Cl B Cl 10 0 Cl The shape of a Molecule depends upon the number of valence shell electron pairs (bonding or non bonding) around the central atom. Pairs of electrons in the valence shell repel one another since their electron clouds are negatively charged. These pars of electrons tend to occupy such positions in space that Minimize repulsion and thus Maximize distance between them.. If the central atom is surrounded by lone pairs as well as bond pairs, then the Molecule will have distorted geometry.. C : ( σ1s) ( σ 1s) ( σs) ( σ s) ( π P π P ) x y Or KK( σs) ( σ s) π P π P ) x y Nb Na Bond order 8 4 Since there are no unpaired electrons it is diamagnetic in nature.. Oxidation Mno 4 + Br Mno +5 + BrO Reduction Oxidation Reduction MnO MnO units 4 6units Br BrO Reduction eq n x MnO 4 MnO Oxidation eq n x 1 Br BrO Adding MnO 4 + Br MnO + BrO As the reaction occur in basic medium to equalize ionic charges O are added on RS. To balance hydrogen atoms add O on LS MnO + Br + O MnO + BrO + O 4

4 4. i) C + O CO + (s) (g) (g) (g) 4 ii) iii) CO + O CO + (g) (g) (g) (g) Zn + Zn (s) (aq) (aq) (g) 5. N + O + CO (N 4) CO (N 4) CO + O + CO N4CO N4CO + Nacl N4Cl + NaCO NaCO NOCO + CO + O N4Cl + Ca(O) N + CaCl + O 6. a) In graphite each carbon atom has one outer shell electron that s not used to form covalent bonds. These electrons are delocalized over the whole structure. Electrons are mobile and therefore graphite conducts electricity b) B N 6 COO CO + O Conc. SO4 c) 7K PART -D IV Answer any FIVE questions. Each Question carries FIVE marks: 5 x a) Element C CL Percentage composition 4.6% 4.05% 71.67% Atomic Mass Percentage Atomic Mass Dividing the values by Smallest of them Empirical formula CCl Molecular formula (empirical formula )n Molecular Mass n Empiriacl formula Mass Molecular formula (C Cl) C 4 Cl b) Molecular formula of Glucose C 6 1 O 6 Molar Mass of Glucose (6 x 1) +(1 x 1) + (6 x 16) a) 1. Electrons revolve around the nucleus in certain selected circular paths called orbitals or energy levels. A long as electron revolve in a particular orbit it does not emit energy. If an electron jumps from higher energy level to a lower energy level, energy is emitted and when electron jumps from lower energy level to higher energy level. Energy is absorbed b) h h λ or λ mv p c) P

5 5 9. a) 1. Principal quantum number n Indicates the size or energy of the orbital. Azimuthal quantum number L Indicates the shape of the orbital ) Magnetic quantum number m Indicates orientation of orbitals in space 4) Spin quantum number S Indicates direction of spin of an electron b) It states that in the ground state of the atom the orbitals are filled in order of their increasing energies 0. a) 1. Gases consist of large number of minute discrete particles called Molecules. Molecules move randomly in straight lines in all directions and at various Speeds and the direction of Motion are changes when colliding with each other or with wall of the container.. There is no force of attraction between the particles of a gas at ordinary temperature and pressure. 4. The Collisions between the Molecules are perfectly elastic i.e., no energy loss on collision due to friction b) From Charge s law V1 V V1 L. T1 96.4K T T 1 V T V? T 99.1K 1 V T1 L 99.1K 96.4K.018L 1. a) ess s law states that the enthalpy change is same whether a reaction is carried out in one step or several steps b) 0 G.0 RTLogK p 0 9 G log Jmol KJ mol 1 c) Enthalpy of formulation of water Molecule is 85.8KJMol. a) Required equation + 0 6C(s) (g) C66 (l) f? Given 1) ) C + O 6CO + O 67 KJMol (l) (g) (g) (l) f (s) (g) (g) f C + O CO 9.5 KJMol ) + O O 85.8 KJMol 1 (g) (g) (l) f Reverse eqn 1, Equation x 6, Equation x and adding 6CO + O C + O 67KJ Mol 15 (g) (l) 6 6 (l) f (s) (g) (g) f 6C + 6O 6CO 61KJMol (g) (g) (l) f + O O 857.4KJMol Adding (s) (g) 6 6(l) f 6C + C KJMol b) Entropy increases

6 . a) Water is weak electrolyte, it undergoes partial ionsation. + O + O O + O (c) (c) (aq) (aq) + [O ][O ] K [ O] By Convention [O] is taken as constant + K [ ][O ] w Where Kw is called ionic product of water and its value at 5 0 C is b) A solution which resists the change in the p by the addition of little acid or base is called buffer solution example for acidic buffer CCOO + CCOONa or COO + COONa 4. a) Pka logka log b) p decreases, because N 4 Cl is a Salt of strong acid and weak base undergoes hydrolysis to give acidic solution PI Kp P P I PART D IV Answer any TWO questions. Each Question carries FIVE marks: x a) Principle: A known Mass of an organic compound is burnt in presence of excess of oxygen and cupric oxide carbon and hydrogen in the compound are converted into CO and O respectively. The Masses of CO and O produced is determined From which the percentage of carbon and hydrogen are calculated. 6 Calculation Percentage of carbon 1 Massof CO Mass of organic compound Percentage of ydrogen Mass of O Mass of organic compound b) 6. a) A known Mass of an organic compound is heated in a carius tube with Sodium peroxide or fuming NO, Sulphur present in the compound is oxidised: to Sulphuric acid. It is precipitated as barium Sulphate by adding Barium chloride solution. The precipitate is filtered, dried and weighed. The mass of Barium Sulphate is noted Calculation Mass of Organic Compound taken Wg Mass of BaSO 4 Formed Mg 1 mol of BaSO 4 g of BaSO 4 g of sulphur Wg of BaSO 4 contains W g Sulphur Percentage of Sulphur W 100 m b) Two or more compounds having. Same Molecular formula but differ in the position of functional group. Ex: C C C O C C C propanlol O propanol

7 7. a) Benzene on heating with nitrating mixture (Conc. NO +Conc. SO 4 ) gives nitrobenzene: 7 Benzene NO Conc Conc. SO 4 It involves th following steps i) Generation of an Electrophile: NO Nitrobenzene NO + SO NO + SO + O Nitronium ion ii) Attack of electrophile NO + NO + iii) Removal of proton + Carbocation NO O NO NO SO 4 SO 4 Nitrobenzene b) Alkyl halides on treatment with Sodium metal in dry ether give higher alkanes Ex: C Br + Na + C Br C C + NaBr dry ether Bromo methane Ethane

CHEMISTRY I PUC MODEL QUESTION PAPER -1

CHEMISTRY I PUC MODEL QUESTION PAPER - Time: 3 Hours 5 min Max Marks: 70 INSTRUCTIONS: i) The question paper has four parts A.B.C and D. All the parts are compulsory. ii) Write balanced chemical equations