Chemistry Past Paper Notes 2024-25 PDF

SIMPLE NOTES & HSE PREVIOUS QUESTIONS AND ANSWERS PREPARED BY : YOOSAFALI T K , GHSS DESAMANGALAM (8196) , THRISSUR (DT) , 9947444175 Name : …………………………………………......... School : …………………………………………………. Class : …………...

SIMPLE NOTES & HSE PREVIOUS QUESTIONS AND ANSWERS PREPARED BY : YOOSAFALI T K , GHSS DESAMANGALAM (8196) , THRISSUR (DT) , 9947444175 Name : …………………………………………......... School : …………………………………………………. Class : …………………………………………………. Roll No. : …………………………………………………  XI CHEMISTRY CONTENTS UNIT 1 SOME BASIC CONCEPTS OF CHEMISTRY UNIT 2 STRUCTURE OF ATOM UNIT 3 CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES UNIT 4 CHEMICAL BONDING AND MOLECULAR STRUCTURE UNIT 5 THERMODYNAMICS UNIT 6 EQUILIBRIUM UNIT 7 REDOX REACTIONS UNIT 8 ORGANIC CHEMISTRY- SOME BASIC PRINCIPLES AND TECHNIQUES UNIT 9 HYDROCARBONS PREPARED BY: YOOSAFALI T K ,HSST CHEMISTRY , GVHSS DESAMANGALAM (08196) , THRISSUR (DT)  UNIT 1 : SOME BASIC CONCEPTS OF CHEMISTRY PREPARED BY: YOOSAFALI T K , GVHSS DESAMANGALAM (08196), THRISSUR (DT) ====================================================== 1. What is Matter? How will you classify it? Matter is anything that has mass and occupies space. Chemically matter is classified as two. (i) Pure substances and (ii) Mixtures. Matter Pure substances Mixtures Elements Compounds Homogeneous Heterogeneous 2. What are Pure substances? How will you classify it? Explain each. Pure substances contain only one substance. It cannot be physically separated. Pure substances are classified as (i) Elements and (ii) Compounds. Elements Compounds Elements contain only one type of atoms Compounds contain different types of atoms Examples: Hydrogen , Oxygen , Carbon , Iron , Gold etc Examples: Marble (CaCO3), ice (H2O) etc. 3. What are Mixtures? How will you classify it? Differentiate them. Mixtures contain more type of substances. It can be physically separated. These are classified as (i) homogeneous mixture and (ii) heterogeneous mixture. Homogeneous mixture Heterogeneous mixture Mixtures having uniform composition through out Mixtures having different composition in different parts Examples:- Air, Sugar solution, Kerosene, Petrol, Examples:- Iron and Sulphur, Muddy water, Sand, Alloys (Brass), 916 gold Smoke, Gun powder, Soil 4. Classify the following matter as homogeneous mixture, heterogeneous mixture, element and compounds. ( Silver, Air, Muddy water, Water ) [March 2024] Ans: Homogeneous mixture: Air , Heterogeneous mixture: Muddy water Element: Silver Compound: Water 5. What are Significant figures? Significant figures are the number of digits in a measurement about which we are certain plus one additional digit which is uncertain. 6. Which are the rules for determining the number of significant figures? (I) All non-zero digits are significant. Example: 285 → 3 significant figures (II) Trapped zeros are significant. Example: 2.005 → 4 significant figures (III) Before zeros are never significant. Example: 0.0052 → 2 significant figures (IV) End zeros are significant if it is after the decimal point. Example: 0.200 g → 3 significant figures (V) End zeros are non significant if there is no decimal point. Example: 100→ 1 significant figure (VI) In scientific notation, powers of ten are non- significant. Example: 6.022 X 10 23→4 significant figures (VII) Counting numbers have infinite number of significant figures. Example: 20 eggs → Infinite significant figures 7. Which are Laws of chemical combination? (I) Law of conservation of mass (II) Law of definite proportion (III) Law of multiple proportion (IV) Gay Lussac’s law of gaseous volume (V) Avogadro law 8. State and explain Law of conservation of mass [December 2020] [March 2020] Matter can neither be created nor destroyed. OR In a chemical reaction, total mass of reactants is equal to total mass of products. This law is proposed by Lavoisier. Example: C + O2 → CO2 12g 32 g 44 g Total mass of reactants = 12+32=44 g Total mass of products =44 g 9. State and explain Law of definite proportion [March 2023] A given compound always contains exactly the same proportion of elements by weight. This law is proposed by Joseph Proust. Example:- Carbon dioxide can be obtained by many methods. Its formula → CO2 Mass ratio→ 12:32 Simple ratio →3:8 10. State and explain Law of multiple proportion [August 2018] [SAY 2021] [March 2021] [March 2024] If two elements can combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element, are in the ratio of small whole numbers. This law is proposed by John Dalton. Example:- Hydrogen and oxygen combine to form two compounds, water and hydrogen peroxide. 𝟏 𝑯𝟐 + 𝑶𝟐 → 𝑯𝟐 𝑶 , 𝑯𝟐 + 𝑶𝟐 → 𝑯𝟐 𝑶𝟐 𝟐 2g 16 g 18 g 2g 32 g 34 g Here hydrogen has fixed mass ( 2 g). Oxygen has different masses ( 16 and 32) Its ratio → 16:32 = 1:2 It is simple whole number ratio More examples:- (i) CO , CO2 (ii) NO , NO2 11. State and explain Gay Lussac’s law of gaseous volume When gaseous reactants combine to form gaseous products, there exist a simple whole number ratio between their volumes. This law is proposed by Gay Lussac Example :- H2 + Cl2 → 2 HCl Volume ratio of reactants and products → 1: 1: 2 12. State and explain Avogadro law Equal volumes of all gases under similar conditions of temperature and pressure contain equal number of molecules. This law is proposed by Avogadro. For example, if we take hydrogen, nitrogen and oxygen in different flasks of same capacity, we will find that all flasks contain the same number of molecules. 13. Hydrogen combines with oxygen to form two different compounds, namely water (H 2O) and hydrogen peroxide (H2O2). Which law is obeyed by this combination? [March 2024] [March 2020] Ans : Law of multiple proportions 14. NO and NO2 are two oxides of nitrogen. Which law of chemical combination is illustrated by these compounds? [July 2017] [SAY 2023]  Ans: Law of multiple proportions 15. Two elements, carbon and hydrogen combine to form C 2H6, C2H4 and C2H2. Identify the law [March 2017] Ans : Law of multiple proportions 16. When nitrogen and hydrogen combines to form ammonia, the ratio between the volumes of gaseous reactants and products is 1: 3: 2. Name the law of chemical combination illustrated here. [March 2016] Ans: Gay – Lussac’s law of Gaseous volumes 17. What is atom? An atom is the smallest particle of an element which may or may not have free existence. Atoms of Helium and Neon can exist freely. Atoms of hydrogen, oxygen, nitrogen etc. cannot exist freely 18. What are the postulates of Dalton’s atomic theory? (I) Matter is made up of extremely small, indivisible particles called atoms. (II) Atoms of the same element are identical. (III) Atoms of different elements are different. (IV) Compounds are formed when atoms of different elements combine in a fixed ratio (V) Atoms can neither be created nor destroyed. 19. Define atomic mass. The atomic mass of an element means how many times an atom of an element is heavier than one-twelfth of a carbon-12 atom. Relative atomic mass of Hydrogen , H =1 Relative atomic mass of Carbon , C = 12 Relative atomic mass of Oxygen , O = 16 Relative atomic mass of Chlorine , Cl = 35.5 20. Define Atomic mass unit (amu) or unified mass (u). What is its value? [March 2013]. One atomic mass unit is equal to one-twelfth ( 1/12 ) the mass of an atom of carbon-12. 1 amu (u) = 1.66056 x 10 -24 g 21. Define molecule Molecule is the simplest particle of an element or a compound which has free existence. 22. Define molecular mass The molecular mass of a substance (element or compound) means how many times the mass of a molecule is heavier than one-twelfth ( 1/12) of a carbon-12 atom. It is obtained by multiplying the atomic mass of each element by the number of its atoms and adding them together. Molecular mass of water ( H2O) = (1 X 2 ) + (16 X 1) = 2 + 16 = 18 23. Define Mole. [March 2013] Mole is the amount of a substance which contain 6.022 x 10 23 particles (Avogadro’s number) 1 mol of hydrogen atoms = 6.022 x 10 23 atoms 1 mol of water molecules = 6.022 x 10 23 molecules 24. Define Molar mass? The mass of one mole substance is called molar mass. 25. Give the equation of percentage composition (mass percentage)? [March 2010] 𝑴𝒂𝒔𝒔 𝒐𝒇 𝒕𝒉𝒂𝒕 𝒆𝒍𝒆𝒎𝒆𝒏𝒕 𝒊𝒏 𝒕𝒉𝒆 𝒄𝒐𝒎𝒑𝒐𝒖𝒏𝒅 𝑿 𝟏𝟎𝟎 𝑴𝒂𝒔𝒔 𝒑𝒆𝒓𝒄𝒆𝒏𝒕𝒂𝒈𝒆 𝒐𝒇 𝒕𝒉𝒂𝒕 𝒆𝒍𝒆𝒎𝒆𝒏𝒕 = 𝑴𝒐𝒍𝒂𝒓 𝒎𝒂𝒔𝒔 𝒐𝒇 𝒕𝒉𝒆 𝒄𝒐𝒎𝒑𝒐𝒖𝒏𝒅 26. What are empirical formula and molecular formula? [SAY 2012] Molecular formula is the actual formula of a compound. Empirical formula is the simplest formula of a compound. 27. Write the Molecular formula and Empirical formula of the following species. Species Molecular formula Empirical formula Glucose C6H12O6 CH2O Ethane C2H6 CH3 Benzene C6H6 CH Hydrogen peroxide H2O2 HO Butene C4H8 CH2 Carbon dioxide CO2 CO2 Acetic acid CH3COOH ( C2H4O2) CH2O Benzene hexachloride C6H6Cl6 CHCl Water H2O H2O 28. Write the relation between empirical formula and molecular formula. [September 2016] [March 2023] Ans : Molecular formula = n x Empirical formula , Where n = 1,2,3,……… 𝑴𝒐𝒍𝒆𝒄𝒖𝒍𝒂𝒓 𝒎𝒂𝒔𝒔 𝒏= 𝑬𝒎𝒑𝒊𝒓𝒊𝒄𝒂𝒍 𝒇𝒐𝒓𝒎𝒖𝒍𝒂 𝒎𝒂𝒔𝒔 29. What are stoichiometric calculations? Calculations of masses or volumes of reactants and products in a chemical reaction are called stoichiometric calculations. Consider a reaction, CH4 + 2 O2 → CO2 + 2 H2O 1 mol CH4 + 2 mol O2 → 1 mol CO2 + 2 mol H2O 16 g of CH4 + 2 X 32 g of O2 → 44 g of CO2 + 2 X 18 g of H2O 6.022 X 1023 molecules of CH4 + 2 X 6.022 X 1023 molecules of O2 → 6.022 X 1023 molecules of CO2 + 2 X 6.022 X 10 23 molecules of H2O 22.7 L of CH4 + 2 X 22.7 L of O2 → 22.7 L of CO2 + 2 X 22.7 L of H2O 30. What is Limiting reagent (limiting reactant) ? [SAY 2023] [SAY 2022] [March 2024] The reactant that is consumed completely in a reaction is called Limiting reagent (limiting reactant ). Example: 2 H2(g) + O2(g) → 2H2O(g) 2 mol 1 mol 2 mol In this reaction , 2 mole hydrogen react with 1 mole oxygen to produce 2 mole water. If this reaction is carried out using 2mol H2 and 2mol O2 , 2mol H2 is used up completely and so H2 is the limiting reagent. Here O2 is excess reagent 31. What is a solution? Solution is a homogeneous mixture of two or more substances. Solutions having only two components are called binary solutions. Example: Salt water The component which is in larger amount is called solvent. The component which is in smaller amount is called solute. Solute + Solvent → Solution Salt + Water → Salt water 32. What is concentration of a solution? The amount of solute present in a given amount of solvent or solution is called concentration. 33. Which are the ways of expressing concentration of a solution? (I) Mass percent (II) Mole fraction (III) Molarity (IV) Molality 34. Give equation for mass percentage of solute. 𝑴𝒂𝒔𝒔 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒆 𝑿 𝟏𝟎𝟎 𝑴𝒂𝒔𝒔 𝒑𝒆𝒓𝒄𝒆𝒏𝒕𝒂𝒈𝒆 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒆 = 𝑴𝒂𝒔𝒔 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 35. Define mole fraction. [SAY 2016] Mole fraction is the ratio of number of moles of one component to the total number of moles of the solution.  𝐍𝐮𝐦𝐛𝐞𝐫 𝐨𝐟 𝐦𝐨𝐥𝐞𝐬 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐞 𝐧𝐁 𝑴𝒐𝒍𝒆 𝒇𝒓𝒂𝒄𝒕𝒊𝒐𝒏 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒆 = = 𝐓𝐨𝐭𝐚𝐥 𝐧𝐮𝐦𝐛𝐞𝐫 𝐨𝐟 𝐦𝐨𝐥𝐞𝐬 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧 𝐧𝐀 + 𝐧𝐁 𝐍𝐮𝐦𝐛𝐞𝐫 𝐨𝐟 𝐦𝐨𝐥𝐞𝐬 𝐨𝐟 𝐬𝐨𝐥𝐯𝐞𝐧𝐭 𝐧𝐀 𝑴𝒐𝒍𝒆 𝒇𝒓𝒂𝒄𝒕𝒊𝒐𝒏 𝒐𝒇 𝒔𝒐𝒍𝒗𝒆𝒏𝒕 = = 𝐓𝐨𝐭𝐚𝐥 𝐧𝐮𝐦𝐛𝐞𝐫 𝐨𝐟 𝐦𝐨𝐥𝐞𝐬 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧 𝐧𝐀 𝐧𝐁 36. Define molarity [MARCH 2023] Molarity is defined as the number of moles of solute in one litre of the solution. 𝑵𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒎𝒐𝒍𝒆𝒔 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒆 𝑴𝒐𝒍𝒂𝒓𝒊𝒕𝒚 (𝑴) = 𝑽𝒐𝒍𝒖𝒎𝒆 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝒊𝒏 𝒍𝒊𝒕𝒓𝒆 Molarity depends on temperature because it is related to volume, which changes with temperature. 37. Define molality Molality is defined as the number of moles of solute present in one kilogram of solvent. 𝑵𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒎𝒐𝒍𝒆𝒔 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒆 𝑴𝒐𝒍𝒂𝒍𝒊𝒕𝒚 (𝒎) = 𝑴𝒂𝒔𝒔 𝒐𝒇 𝒔𝒐𝒍𝒗𝒆𝒏𝒕 𝒊𝒏 𝒌𝒊𝒍𝒐𝒈𝒓𝒂𝒎 Molality does not depend on temperature 38. The number of moles of solute in one litre of the solution is ……………………. [March 2024] Ans: Molarity SOME HSE PREVIOUS QUESTIONS AND ANSWERS 1. Which among the following measurements contains the highest number of significant figures? [SAY 2018] a) 1.123 x 10-3 kg b) 1.2 x 10-3 kg c) 0.123 x 103 kg d) 2 x 105 kg Ans: a) 1.123 x 10-3 kg 2. Round off 0.0525 to a number with two significant figures. [March 2019] Ans: 0.052 3. What is the number of hydrogen atoms in 1 mole of methane (CH 4)? Ans: 4 x 6.022 x 1023 atoms (Hint : CH4 contains 4 hydrogen atoms) 4. The number of moles of CO2 present in 220g of CO2is…………… [March 2023] Ans : 5 mol ( Hint : 220/44 = 5 ) 5. The number of oxygen atoms present in 5 moles of glucose (C 6H12O6) is.............. [March 2018] Ans: 30 x 6.022 x 1023 atoms (Hint : Glucose contains 6 oxygen atoms) 6. The mass of 2 moles of ammonia gas is......... [October 2015] Ans : 34g (Hint : 2 X 17= 34) 7. Calculate the mass of a magnesium atom in grams. [July 2017] Mass of 1 mol Magnesium = 24 g Mass of 6.02 x 10 23 Magnesium atoms = 24 g 𝟐𝟒 𝐌𝐚𝐬𝐬 𝐨𝐟 𝐚 𝐌𝐚𝐠𝐧𝐞𝐬𝐢𝐮𝐦 𝐚𝐭𝐨𝐦 = = 𝟑. 𝟗𝟗 𝑿 𝟏𝟎 𝟐𝟑 𝟔. 𝟎𝟐𝟐 𝒙 𝟏𝟎𝟐𝟑 8. Determine the number of moles present in 0.55 mg of electrons.1 mole ii) 2 moles iii) 1.5 moles iv) 0.5 mole [March 2017] Ans: 1 mol 9. Give the empirical formula of the following. C 6H12O6, C6H6, CH3COOH, C6H6Cl6 [March 2017] Ans : Empirical formulae are: CH2O, CH, CH2O, CHCl. 10. Benzene has empirical formula CH. Its molecular mass is 78. What is its molecular formula [March2018 ] Ans : Empirical formula mass= 12 + 1 = 13 𝑴𝒐𝒍𝒆𝒄𝒖𝒍𝒂𝒓 𝒎𝒂𝒔𝒔 𝟕𝟖 𝒏= = =𝟔 𝑬𝒎𝒑𝒊𝒓𝒊𝒄𝒂𝒍 𝒇𝒐𝒓𝒎𝒖𝒍𝒂 𝒎𝒂𝒔𝒔 𝟏𝟑 Molecular formula = n x Empirical formula =6 (CH) = C 6H6 11. Determine the empirical formula of an oxide of iron which has 69.9% iron (Fe) and 30.1% oxygen (O) by mass. [Hint: Atomic mass of Fe = 56]. ([December 2020] Elements Atomic Percentage (%) 𝑹𝒆𝒍𝒂𝒕𝒊𝒗𝒆 𝒏𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒎𝒐𝒍𝒆𝒔 = Simple ratio Whole mass 𝑷𝒆𝒓𝒄𝒆𝒏𝒕𝒂𝒈𝒆 number ratio 𝑨𝒕𝒐𝒎𝒊𝒄 𝒎𝒂𝒔𝒔 Fe 56 69.9 69.9 /56 =1.25 1.25/1.25 =1 2 O 16 30.1 30.1/16 = 1.88 1.88/1.25 =1.5 3 Empirical formula = Fe2O3 12. An organic compound has the following percentage composition C = 12.36%, H = 2.13%, Br = 85%. Its vapour density is 94. Find its molecular formula. (Hint: Molecular mass = 2 x vapour density) [September 2016] Elements Atomic Percentage 𝑹𝒆𝒍𝒂𝒕𝒊𝒗𝒆 𝒏𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒎𝒐𝒍𝒆𝒔 Simple ratio Whole number mass (%) 𝑷𝒆𝒓𝒄𝒆𝒏𝒕𝒂𝒈𝒆 ratio = 𝑨𝒕𝒐𝒎𝒊𝒄 𝒎𝒂𝒔𝒔 Hydrogen 1 2.13% 2.13/1 = 2.13 2.13/1.03 =2.06 2 Carbon 12 12.36% 12.36/12 =1.03 1.03/1.03 =1 1 Bromine 80 85%. 85/80 = 1.06 1.06/1.03 = 1.02 1 Empirical formula = CH2Br Empirical formula mass = (12 x1) + (1 x 2) + (80 x 1) = = 12 + 2 + 80 = 94 Molecular mass = 2 x vapour density = 2 X 94 = 188 𝑴𝒐𝒍𝒆𝒄𝒖𝒍𝒂𝒓 𝒎𝒂𝒔𝒔 𝟏𝟖𝟖 𝒏= = =𝟐 𝑬𝒎𝒑𝒊𝒓𝒊𝒄𝒂𝒍 𝒇𝒐𝒓𝒎𝒖𝒍𝒂 𝒎𝒂𝒔𝒔 𝟗𝟒 Molecular formula = n x Empirical formula =2 (CH2Br) = C2H4Br2 13. An organic compound on analysis gave the following composition. Carbon = 40%, Hydrogen = 6.66% and oxygen = 53.34%. Calculate its molecular formula if its molecular mass is 90. [MARCH 2022] Elements Atomic Percentage 𝑹𝒆𝒍𝒂𝒕𝒊𝒗𝒆 𝒏𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒎𝒐𝒍𝒆𝒔 Simple ratio Whole mass (%) 𝑷𝒆𝒓𝒄𝒆𝒏𝒕𝒂𝒈𝒆 number = 𝑨𝒕𝒐𝒎𝒊𝒄 𝒎𝒂𝒔𝒔 ratio Hydrogen 1 6.66 % 6.66 /1 = 6.66 6.66/3.33 = 2 2 Carbon 12 40 % 40 /12 = 3.33 3.33/3.33 = 1 1 Oxygen 16 53.34 % 53.34 /16=3.33 3.33/3.33 = 1 1 Empirical formula = CH2O Empirical formula mass =( 12 x1) + (1 x 2) + (16 x 1 ) = 12 + 2 + 16 = 30 Molecular mass = 90 𝑴𝒐𝒍𝒆𝒄𝒖𝒍𝒂𝒓 𝒎𝒂𝒔𝒔 𝟗𝟎 𝒏= = =𝟑 𝑬𝒎𝒑𝒊𝒓𝒊𝒄𝒂𝒍 𝒇𝒐𝒓𝒎𝒖𝒍𝒂 𝒎𝒂𝒔𝒔 𝟑𝟎 Molecular formula = n x Empirical formula =3(CH2O) = C3H6O3 14. A compound contain A 70% , B 30 %. The relative number of moles of A ad B in the compound are 1.25 and 1.88 respectively. Its molecular mass is 160. What are its EF and MF? [March 2016] Solution: 𝑷𝒆𝒓𝒄𝒆𝒏𝒕𝒂𝒈𝒆 𝑹𝒆𝒍𝒂𝒕𝒊𝒗𝒆 𝒏𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒎𝒐𝒍𝒆𝒔 𝒐𝒇 𝑨 = = 𝟏. 𝟐𝟓 𝑨𝒕𝒐𝒎𝒊𝒄 𝒎𝒂𝒔𝒔  𝑷𝒆𝒓𝒄𝒆𝒏𝒕𝒂𝒈𝒆 𝟕𝟎 𝑨𝒕𝒐𝒎𝒊𝒄 𝒎𝒂𝒔𝒔 𝒐𝒇 𝑨 = = = 𝟓𝟔 𝑹𝒆𝒍𝒂𝒕𝒊𝒗𝒆 𝒏𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒎𝒐𝒍𝒆𝒔 𝒐𝒇 𝑨 𝟏. 𝟐𝟓 𝑷𝒆𝒓𝒄𝒆𝒏𝒕𝒂𝒈𝒆 𝑹𝒆𝒍𝒂𝒕𝒊𝒗𝒆 𝒏𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒎𝒐𝒍𝒆𝒔 𝒐𝒇 𝑩 = = 𝟏. 𝟖𝟖 𝑨𝒕𝒐𝒎𝒊𝒄 𝒎𝒂𝒔𝒔 𝑷𝒆𝒓𝒄𝒆𝒏𝒕𝒂𝒈𝒆 𝟑𝟎 𝑨𝒕𝒐𝒎𝒊𝒄 𝒎𝒂𝒔𝒔 𝒐𝒇 𝑩 = = = 𝟏𝟔 𝑹𝒆𝒍𝒂𝒕𝒊𝒗𝒆 𝒏𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒎𝒐𝒍𝒆𝒔 𝒐𝒇 𝑩 𝟏. 𝟖𝟖 𝟏. 𝟐𝟓 𝐒𝐢𝐦𝐩𝐥𝐞 𝐫𝐚𝐭𝐢𝐨 𝐨𝐟 𝐀 = =𝟏 𝟏. 𝟐𝟓 𝟏. 𝟖𝟖 𝐒𝐢𝐦𝐩𝐥𝐞 𝐫𝐚𝐭𝐢𝐨 𝐨𝐟 𝐁 = = 𝟏. 𝟓 𝟏. 𝟐𝟓 Whole number ratio of A = 1 x 2 = 2 Whole number ratio of B = 1.5 x 2 = 3 Empirical formula = A2B3 Empirical formula mass = (2 x 56) + (3 x 16) = 112 + 48 = 160 Molecular mass = 160 𝑴𝒐𝒍𝒆𝒄𝒖𝒍𝒂𝒓 𝒎𝒂𝒔𝒔 𝟏𝟔𝟎 𝒏= = =𝟏 𝑬𝒎𝒑𝒊𝒓𝒊𝒄𝒂𝒍 𝒇𝒐𝒓𝒎𝒖𝒍𝒂 𝒎𝒂𝒔𝒔 𝟏𝟔𝟎 Molecular formula = 1 ( A2B3 ) = A2B3 15. Calculate the amount of oxygen required for the complete combustion of 48 g of methane [March 2023] CH4 + 2 O2 → CO2 + 2 H2O 16g 64g 16g CH4 require 64 g oxygen 48 g (3X16 ) CH4 require 192 g ( 3X 64 ) oxygen 16. The balanced chemical equation for combustion of CH4 is CH4 + 2 O2 → CO2 + 2 H2O Calculate the amount of water formed by the combustion of 32g of CH4. [March 2020] Ans: CH4 + 2 O2 → CO2 + 2 H2O 16g 64g 44g 36g 16g CH4 produces 36g water. So the amount of water formed by the combustion of 32g CH 4 = 72 g. 17. Calculate the mass of SO3 produced, if 500 g SO2 reacts with 200 g of O2 according to the equation 2 SO2 + O2→2 SO3 , identify the limiting reactant. [July 2019] Ans : 2 SO2 + O2 → 2 SO3 2 mol SO2 + 1 mol O2→2 mol SO3 𝟓𝟎𝟎 𝟓𝟎𝟎𝐠 𝐨𝐟 𝐒𝐎𝟐 = = 𝟕. 𝟖 𝒎𝒐𝒍 𝐒𝐎𝟐 𝟔𝟒 𝟐𝟎𝟎 𝟐𝟎𝟎 𝐠 𝐨𝐟 𝐎𝟐 = = 𝟔. 𝟐𝟓 𝒎𝒐𝒍 𝐎𝟐 𝟑𝟐 According to equation 2 mol SO2 require 1 mol O2 𝟏 𝟏 𝐦𝐨𝐥 𝐒𝐎𝟐 𝐫𝐞𝐪𝐮𝐢𝐫𝐞𝐬 𝐦𝐨𝐥 𝐎𝟐 𝟐 𝟏 ∴ 𝟕. 𝟖 𝒎𝒐𝒍 𝐒𝐎𝟐 𝐫𝐞𝐪𝐮𝐢𝐫𝐞𝐬 𝐗 𝟕. 𝟖 𝐦𝐨𝐥 𝐎𝟐 = 𝟑. 𝟗 𝐦𝐨𝐥 𝐎𝟐 𝟐 Here SO2 is used up completely and so SO2 is limiting reagent. 2 mol SO2 → 2 mol SO3 Therefore 1 mol SO2 → 1 mol SO3  Therefore 7.8 mol SO2 → 7.8 mol SO3 = 7.8 ( 32 + 48) = 7.8 X 80 = 624 g SO3 18. A reaction mixture for the production of NH3 gas contains 250 g of N2 gas and 50 g of H2 gas under suitable conditions. Identify the limiting reactant if any and calculate the mass of NH 3 gas produced. [March 2019] Solution: N2(g) + 3H2(g) → 2NH3(g) 1 mol N2 3 mol H2 2 mol NH3 28g N2 6g H2 34 g NH3 𝟐𝟓𝟎 𝟐𝟓𝟎 𝐠 𝐨𝐟 𝐍𝟐 = = 𝟖. 𝟗 𝒎𝒐𝒍 𝐍𝟐 𝟐𝟖 𝟓𝟎 𝟓𝟎 𝐠𝐨𝐟 𝐇𝟐 = = 𝟐𝟓 𝒎𝒐𝒍 𝐇𝟐 𝟐 According to equation 1 mol N2 require 3 mol H2. Hence for 8.9 mol nitrogen , the moles of hydrogen required = 8.9X 3 = 26.7 mol H 2 Here only 25 mol hydrogen is present. So hydrogen is limiting reagent. According to equation 3 mol H2 gives 2 mol NH3 1mol H2 gives 2/3 mol NH3 𝟐 𝟐𝟓 𝒎𝒐𝒍 𝐇𝟐 𝐠𝐢𝐯𝐞𝐬 𝟐𝟓 𝐗 𝐦𝐨𝐥 𝐍𝐇𝟑 = 𝟏𝟔. 𝟔𝟔 𝐦𝐨𝐥 𝐍𝐇𝟑 = 𝟐𝟖𝟑𝐠 𝐚𝐦𝐦𝐨𝐧𝐢𝐚 𝟑 19. Calculate the mass of NaOH required to make 500 ml of 0.5M aqueous solution. (Molar mass of NaOH = 40) [March 2013] 𝑾𝑩 𝑿 𝟏𝟎𝟎𝟎 𝑾𝑩 𝑿 𝟏𝟎𝟎𝟎 𝑴𝒐𝒍𝒂𝒓𝒊𝒕𝒚 = = 𝑴𝑩 𝑿 𝑽(𝒊𝒏 𝒎𝒍) 𝟒𝟎 𝑿 𝟓𝟎𝟎 𝑾𝑩 𝑿 𝟏𝟎𝟎𝟎 𝟎. 𝟓 = 𝟒𝟎 𝑿 𝟓𝟎𝟎 𝟎. 𝟓 𝑿𝟒𝟎 𝑿 𝟓𝟎𝟎 𝑾𝑩 = = 𝟏𝟎 𝒈 𝟏𝟎𝟎𝟎 The mass of NaOH required = 10 g 20. Calculate the mass of oxalic acid dihydrate (H2C2O4.2H2O) required to prepare 0.1M, 250 ml of its aqueous solution. [March 2018]. Molar mass of Oxalic acid dihydrate = 126 , Molarity = 0.1M , Volume of the solution = 250 mL 𝑾𝑩 𝑿 𝟏𝟎𝟎𝟎 𝑴𝒐𝒍𝒂𝒓𝒊𝒕𝒚 = 𝑴𝑩 𝑿 𝑽(𝒊𝒏 𝒎𝒊) 𝑾𝑩 𝑿 𝟏𝟎𝟎𝟎 𝟎. 𝟏 = 𝟏𝟐𝟔 𝑿 𝟐𝟓𝟎 𝟎. 𝟏 𝑿 𝟏𝟐𝟔 𝑿 𝟐𝟓𝟎 𝑾𝑩 = = 𝟑. 𝟏𝟓 𝒈 𝟏𝟎𝟎𝟎 𝟏 𝟏𝟐 −𝟐𝟒 𝟏 𝒂𝒎𝒖 = 𝑿 𝟐𝟑 = 𝟏. 𝟔𝟔𝟎𝟓𝟔 𝐱 𝟏𝟎 𝒈 𝟏𝟐 𝟔. 𝟎𝟐𝟐 𝑿 𝟏𝟎 =============================================================================================  UNIT 2 : STRUCTURE OF ATOM PREPARED BY: YOOSAFALI T K , GVHSS DESAMANGALAM (08196), THRISSUR (DT) ======================================================= 1. Electron was discovered by J J Thomson by cathode ray discharge tube experiment. 2. What are the properties of cathode rays? [ SAY 2017] (I) They start from cathode, more rays are produced from the space between cathode and anode and move towards anode (II) They travel in straight lines. (III) They are deflected by both electric and magnetic field. Deflection in the electric field is towards positive plate shows that they are negatively charged particles called electrons (IV) They does not depend on the nature of the gas inside discharge tube (V) The charge to mass ratio (e/m) is same for all gases 3. What is the mass of electron? 𝑴𝒂𝒔𝒔 𝒐𝒇 𝒆𝒍𝒆𝒄𝒕𝒓𝒐𝒏 = 𝟗. 𝟏 𝑿 𝟏𝟎 𝟑𝟏kg 4. What are the properties of anode rays (Canal rays OR Positive rays)? [MARCH 2020] (I) They are produced from the space between cathode and anode and move towards cathode (II) They travel in straight lines. (III) They are deflected by both electric and magnetic field. Deflection in the electric field is towards negative plate shows that they are positively charged particles (IV) They depend on the nature of the gas inside discharge tube (V) The charge to mass ratio (e/m) is different for different gases 5. What is proton? What is the mass of proton? The lightest positively charged particles are called protons. Mass of proton =1.672 x 10-27 kg. It is equal to the mass of hydrogen atom 6. Who discovered proton? Ans: Rutherford OR Goldstein 7. How neutron is discovered? What is its mass? Neutron is discovered by James Chadwick by bombarding beryllium sheet by alpha (α) particles. They are neutral particles. Mass of neutron =1.674 x 10-27 kg. Mass of neutron is slightly greater than proton. 8. Give the equation of Atomic number and Mass number Atomic number = Number of protons = No. of electrons Mass number = Number of protons + Number of neutrons = No. of nucleons 9. How will you find number of neutrons Number of neutrons= Mass number - Atomic number 10. Define isotopes and isobars. Give examples for each. (I) Isotopes are atoms of same element having the same atomic number but different mass number. They contain different number of neutrons. Example : 126C , 136C , 14 6C (II) Isobars are atoms of different elements which have the same mass number. Example : 146C , 147N 11. Which are different atom models? (I) Thomson’s atom model (Watermelon model , Plum-pudding model) (II) Rutherford’s atom model (III) Bohr atom model (IV) Quantum mechanical model 12. Explain Rutherford’s alpha ray scattering experiment. Give its important observations and conclusions. [MARCH 2023] Rutherford bombarded thin gold foil by alpha particles (positive charge) and the movements of rays are detected by circular zinc sulphide screen. Observations Conclusions 1 Most of the alpha particles passed through the Most of the space in an atom is empty. gold foil without any deflections. 2 A few alpha particles were deflected through A heavy positive centre is present at the centre small angles. of the atom called nucleus. 0 3 Very few alpha particles are deflected back (180 ). Nucleus is very small in size. 13. What are the postulates of Rutherford atom model? [SAY 2022] [SAY 2023] [MARCH 2022] (I) Most of the mass and all positive charge is concentrated at the centre of the atom called nucleus. (II) Electrons are revolving around the nucleus with very high speeds. (III) Most of the space inside the atom is empty. (IV) Electrons and nucleus are held by electrostatic forces of attraction. 14. What are the draw backs (failure) of Rutherford atom model? [MARCH 2024] (I) Failed to explain the stability of the atom. (II) Failed to explain hydrogen spectrum. (III) It does not say anything about the electronic structure of atom. 15. Arrange electromagnetic radiations in order of the wavelength. Gamma-rays < X-rays Na+ > Mg2+ > Al3+ Among the isoelectronic species, greater the nuclear charge , smaller the size. 25. What is Ionization energy (ionization enthalpy).? [SAY 2023] The amount of energy required to remove the most loosely bound electron from an isolated gaseous atom is called ionization energy (ionization enthalpy). 26. Explain the variation of Ionization energy along a period and in a group. [March 2020] In a period, from left to right ionization enthalpy increases due to increase in nuclear charge and so valence electrons are more tightly held by the nucleus. But some irregularities are observed. Beryllium has higher ionization energy than Boron due to stable electronic configuration ( 1s 22s2). Nitrogen has higher ionization energy than oxygen due to half filled stable electronic configuration ( 1s22s22p3) In a group, Ionization enthalpy decreases from top to bottom due to increase in atomic size. 27. Beryllium has higher ionization energy than Boron. Give reason. [SAY 2023] 2 2 Beryllium has stable electronic configuration( 1s 2s ). 28. Account for the following: First ionisation enthalpy of Boron is less than that of carbon. [SAY 2016] 2 2 1 5B – 1s 2s 2p. After the removal of one electron, B gets the stable fully filled electronic configuration. So its first ionisation enthalpy is low. 29. Nitrogen has higher ionization energy than oxygen. Why? [MARCH 2023]  Nitrogen has half filled stable electronic configuration ( 1s 22s22p3). 30. Give reasons for the following : 'O' has lower ionization enthalpy than N [SAY 2018] 2 2 4 Ans: The electronic configuration of O is 1s 2s 2p. After the removal of one electron, oxygen gets the stable half filled electronic configuration. So it has lower ionisation enthalpy. 31. What are the factors affecting ionization energy? (i) Atomic size (ii) Nuclear charge (iii) Shielding effect or Screening effect of inner electrons (iv) Electronic configuration. 32. First ionization enthalpy of sodium is lower than that of magnesium but its second ionization energy is higher than that of magnesium. Explain. For sodium (1S22S22P63S1) , first electron is removed from 3s orbital , that is easy , because removal of electron can form stable electronic configuration. So first ionization energy is low (Na +→1S22S22P6 ) For Magnesium (1S22S22P63S2) , first electron is removed from completely filled 3s orbital , that is difficult , because removal of electron is from stable electronic configuration. So first ionization energy is high. But for Na+, its second ionization energy is high because of stable electronic configuration. 33. What is electron gain enthalpy? [SAY 2023] The amount of energy released when an electron is added to isolated gaseous atom is called electron gain enthalpy. X (g) +e− → X(g) − ∆H = ∆egH 34. Explain the variation of electron gain enthalpy along a period and in a group. In a period, from left to right electron gain enthalpy becomes more and more negative. It is due to increase in nuclear charge. In a group, Electron gain enthalpy decreases from top to bottom due to increase in atomic size. 35. Electron gain enthalpies of noble gases (Example: Neon) are zero or positive.Why? [MARCH 2018] 2 6 Due to completely filled electronic configuration ( ns np ). 36. Electron gain enthalpy of fluorine is less than that of chlorine.Why? [SAY 2023] [MARCH 2023] This is due to the very small size of fluorine atom. As a result, inter electronic repulsion in the 2p sub shell of F is more than that in the relatively larger 3p sub shell in chlorine atom. 37. 'Chlorine has the most negative electron gain enthalpy'. Justify the statement. [MARCH 2024] Ans: For Cl, the incoming electron goes to the larger 3rd shell. So the electronic repulsion is low and hence Cl adds electron more easily than F. 38. What are Factors affecting electron gain enthalpy ? [MARCH 2017] (i)Atomic size (ii) Nuclear charge, (iii) electronic configuration 39. What is Electronegativity? [MARCH 2024] Electro negativity is the ability of an atom in a molecule to attract the shared pair of electrons towards it. Halogens have highest electro negativity in their periods. Fluorine is the most electro negative element. 40. Name electronegativity scales [MARCH 2022] Ans: Pauling scale , Mulliken-Jaffe scale 41. Explain the variation of Electronegativity along a period and in a group. Electro negativity increases from left to right in a period due to increasing nuclear charge. Electro negativity decreases from top to bottom in a group due to increase in atomic size. 42. What are Anomalous properties? The first element of each group in s and p blocks differs from the rest of the elements in many properties are called anomalous properties. 43. Aluminium forms [AlF6] 3- whereas boron cannot form [BF6] 3- but forms [BF4] - even though both belong to the same group. Explain. [IMP 2015] Ans: Due to the presence of vacant d orbitals in Al, Al can extend its covalency beyond 4. So it can form [AlF6] 3- But in B, there is no vacant d-orbitals. So its maximum covalency is 4. 44. What are the reasons for Anomalous properties? [MARCH 2016] (i) Small size (ii)High charge/radius ratio.(iii)High electro negativity (iv) Non availability of d orbitals. 45. What is Diagonal relationship? First element in any group shows similarities with second element in ne next xt group. This is called diagonal relationship. Examples: (i) Lithium resembles Magnesium due to diagonal relationship (ii) Beryllium resembles aluminium due to diagonal relationship. 46. What are the reasons for Diagonal relationship? (i) Similar size, (ii)Similar imilar ionization energy , (iii) Similar electronegativity, (iv)Similar imilar charge/radius ratio 47. What is Valency? Valency is the combining capacity of an element. It is determined by valence electrons. Valency = Numberr of valence electrons OR 8 − No. of valence electrons In a group,, valency is same, since valence electrons are same. In a period , valency increases up to 4 and then decreases with respect to hydrogen and with respect to oxygen, valency increases up to seven. 48. Explain the behavior of different type Oxides? The normal oxide formed by the element on extreme left is the most basic (e.g. Na 2O, CaO). The normal oxide formed by the element on extreme right is acidic ( e.g. Cl 2O7). Oxides of the elements in the centre are amphoteric (e.g. Al 2O3 , As2O3 ) or neutral (e.g. CO, NO, N2O) 49. Show by chemical reaction with water that Na2O is basic and Cl2O7 is acidic. Na2O with water forms a strong base whereas Cl 2O7 forms a strong acid. Na2O + H2O → 2 NaOH , Cl2O7 + H2O → 2 HClO4 50. Notation for IUPAC nomenclature of elements digit name abbreviation digit name abbreviation 0 nil n 5 pent p 1 un u 6 hex h 2 bi b 7 sept s 3 tri t 8 oct o 4 quad q 9 enn e 51. A B Most electronegative element Fluorine (F) Most electron gain enthalpy element Chlorine (Cl) Most electropositive element Francium (Fr) Most abundant element in the universe Hydrogen Most abundant element in the earth crust Oxygen Most abundant element in the atmosphere Nitrogen  SOME HSE PREVIOUS QUESTIONS AND ANSWERS 1. Identify the positions of Al (Z=13) and S (Z=16) in the periodic table with the help of their electronic configurations. Predict the formula of the compound formed between them. [MARCH 2019] 2 1 Ans: 13Al – [Ne] 3s 3p , Period – 3, Group - 13 , 2 4 16S – [Ne] 3s 3p , Period – 3, Group - 16 Formula : Al2S3. 2. Identify the group and period of an element having atomic number (Z) 25 in the periodic table. Ans: Electronic configuration : [Ar] 3d5 4s1 Period = 4 , Group = 6 3. The element that has outer electronic configuration 3d5 4s1 belongs to: (a) s-block (b) p-block (c) d-block (d) f-block [MARCH 2020] Ans: d-block 4. Predict the formula of the stable binary compound that would be formed by the combination of the following pairs of elements: (i) Lithium and oxygen (ii) Aluminium and iodine. [MARCH 2020] Ans: (i) Li2O and (ii) AlI3 5. Account for the following : The ionic radius of fluoride ion (F- ) is 136 pm, while the atomic radius of fluorine (F) is only 64 pm. [IMP 2019] Ans: This is due to lesser effective nuclear charge in F –. 6. Among N3-, O2-, F- , Na+ and AI3+, which one has the smallest size? [IMP 2018] 3+ Ans: AI 7. Select isoelectronic species from the following: O – , F– , Na+ , Mg+ [IMP 2017] – + Ans: F and Na 8. A group of ions are given below. Find one pair which is not Isoelectronic. Na + , Al3+, Ca2+, Br – ,F – [MARCH 16] Ans: Ca2+and Br – 9. Justify the following : The size of Al3+ is lower than that of F−. [MARCH 2018] 3+ Ans: Due to greater effective nuclear charge in AI. 10. Write the oxidation state and covalency of Al in [AlF6] 3- [MARCH 2017] Ans: Oxidation state = +3, Covalency = 6. 11. Account for the following : The second ionization enthalpy of an element is always greater than that of the first ionization enthalpy. Ans: This is because it is more difficult to remove an electron from a positively charged ion than from a neutral atom. 12. Which one of the following has the highest ionisation enthalpy? (a) P (b) S (c) Cl (d) F [IMP 2020] Ans: F 13. (a) Complete the reactions: (i) Na2O + H2O →............. (ii) Cl2O7 + H2O →............ (b) Identify the nature of the above oxides by examining the products of the above reactions. [Sept 2020] Ans: (a) (i) NaOH (ii) HClO4 (b) Na2O is basic and Cl2O7 is acidic. 14. Which is the acidic oxide among the following? a) Cl2O7 b) Na2O c) AI2O3 d) CO [MARCH 2018] Ans: Cl2O7 15. Give the IUPAC name of the element with Atomic number 117. [IMP 2019] Ans: Ununseptium (Uus) 16. The atomic number of element with IUPAC name ‘Ununbium’ is ……… [MARCH 2015] Ans: 112  UNIT 4 :CHEMICAL BONDING AND MOLECULAR STRUCTURE PREPARED BY: YOOSAFALI T K , GVHSS DESAMANGALAM (08196), THRISSUR (DT) ======================================================= 1. What is Chemical bond? The attractive force between the atoms in a molecule is called chemical bond. Chemical bonds are (I) Covalent bond :- Formed byy the sharing of electrons (II) Ionic bond :- Formed byy the transfer of electrons 2. What is Covalent bond? The bond formed by mutual sharing of electrons between combining atoms. (I) Cl2 molecule :- Cl-Cl Cl SSingle ingle bond is formed by sharing of one electron each. each (II) O2 molecule :- O=O Double bond is formed by sharing of two electrons each. (III) N2 molecule :- N≡N ≡N Triple bond is formed by the sharing of three electrons each. 3. State Octet rule [IMP 2022] Atoms of various elements enter into chemical combination to attain eight electrons (octet of electrons) in their valence shell ( outer most shell). 4. What are the limitations of octet rule rule? [IMP 2022] (I) It cannot explain the formation of molecules with incomplete octet (eg: BeF2 , BF3) (II) It cannot explain the formation of molecules with expanded (super) octet (eg: PCl5, SF6) (III) It cannot explain the formation of compounds by Xe and Kr( eg : XeF2 , XeF6) (IV) It cannot explain the formation of odd electron molecules (eg: NO, NO 2) 5. Give two examples of compounds having expanded octet. [March 2020] Ans: PCl5 and SF6 6. Define ionic bond. How is it formed? Give examples.  The attractive force between etween opposite charged ions is called ionic bond. Ionic bond or electrovalent bond is formed by the complete transfer of one or more electrons from one atom to another atom. Positive ion called cation and is formed by the loss of electrons. Negative ion on called anion and is formed by the gain of electrons. Examples: NaCl , CaF2 , CaO 7. What are the factors favouring ionic bond? [MARCH 2016 2016] (i) Low ionization energy of the electropositive atom (i.e., metal atom) (ii) High negative electron gain enthalpy of the electronegative atom. (iii) High lattice enthalpy of the ionic compound formed. 8. What is lattice energy and its importance? The lattice enthalpy of an ionic solid is defined as the energy required to completely separate one mole of the ionic compound in to its gaseous seous ions. The higher the lattice energy, higher the stability of the ionic compound formed. 9. What is Formal Charge? The formal charge is the charge assigned to some atoms in the lewis structure of certain compounds. FC = V-N- B/2 FC = Formal Charge , V= Number of valence electrons in free atom, N= Number of non bonding electrons , B= Number of bonding electrons 10. What is resonance? The properties of some compounds cannot be explained by single lewis structure. Such compounds exist as a combination of two or more structures. This phenomenon is called resonance. Its Characteristics are (I) Resonance stabilizes the mo molecule. (II) Resonance averages the bond characteristics as a whole. 11. Draw the resonance structures of ozone. 12. Polar molecules and non polar molecules It is a polar molecule Polar molecules Non polar molecules Covalent bonded molecules having partial Covalent bonded molecules having nocharge positive charge and negative charge are called are called nonpolar molecules. polar molecules. Hetero nuclear diatomic molecules( HCl, HBr ) Homo nuclear diatomic molecules (O2 ,H2 N2) Irregular geometry molecules (H2O , NH3) Regular geometry molecules (BeF2, BF3 ,CH4) 13. What is Dipole moment?? Give its unit. [IMP 2023] Dipole moment is defined as the product of the magnitude of charge and the distance between the centre of charges. Its unit is Debye (D).. Dipole moment ( µ) = charge (q) x distance (r) 14. The dipole moment of BeF2 is zero. Why? [March 2019]  BeF2 is linear molecule. The two equal bond dipoles are in opposite directions and cancel each other. So the dipole moment of BeF2 is zero. 15. The dipole moment of H2O is not zero. Why? [March 2019] Water molecule has bent structure.. The bond dipoles of two O-HH bonds do not cancel each other. So water molecule has net dipole moment. 16. The dipole moment of BF3 zero. Why? [IMP 2023] BF3 has trigonal planar structure. Here the resultant of any two bond dipole is equal and opposite to third and the dipole moments of these bonds cancel one another giving net dipole moment equal to zero. 17. Ammonia (NH3) has higher dipole moment than NF3, even though F is more electronegative than hydrogen. Why? [MARCH 2024] Both have pyramidal structure. The individual dipole moments do not cancel each other. So they have net dipole moment. But ammonia has higher dipole moment. It is due to the orbital dipole due to the lone pair is in the same direction of three N-H bonds. But in nitrogen tri fluoride, the resultant dipole of three N N-F bonds is in n opposite direction to the orbital dipole of lone pair. So partially cancelled and dipole moment is low. 18. State Fajan’s rule regarding the partial covalent character of an ionic bond. [IMP 2021] The partial covalent ovalent character of ionic bond increases with (i) Small size of cation and large size of anion (ii) Large charge on both the cation and anion. (1) LiCl > NaCl > KCl (Here LiCl is more covalent. Reason: Reason:- Lithium is small size cation) (2) NaF < NaCl < NaBr < NaI (Here NaI is more covalent. Reason:- Iodine odine is large size anion) 19. Among NaCl, BeCl2 and AlCl3, which one is more covalent? Justify the answer. [MARCH 2019]2019 Ans: AlCl3. According to Fajans rule, smaller the size and greater the charge of the cation, greater the covalent character. So AlCl3 has the most covalent character. NaCl < MgCl2 < AlCl3 20. What are the main postulates of valence shell electron pair repulsion theory (VSEPR) [IMP 2023] (I) The shape of the molecule depends on the number of valence electron pairs of the central atom. (II) The electron pairs repel each other other.. As a result, the electron pairs try to stay as far apart to acquire a state of minimum energy or maximum stability. (III) A multiple bond is treated as if it is a ssingle electron pair. (IV) The repulsive interaction decreases in the order. Lone pair-Lone Lone pair > Lone pair pair-Bond pair > Bond pair-Bond pair  21. Explain the shape of following molecules on the basis of VSEPR theory. BeCl2 BF3 Two bond pairs around Beryllium. Three bond pairs around Linear geometry. Boron. Bond angle 180 0 Trigonal planar geometry. Bond angle 120 0 CH4 PCl5 Four bond pairs around Carbon. Tetrahedral geometry. Five bond pairs around P. P Bond angle 109.5 0 Trigonal bipyramid geometry. Bond angle 120 0 and 90 0 SF6 NH3 Nitrogen has three bond pairs and one lone pair. Six bond pairs around Sulphur. Bond pair-lone lone pair repulsion is greater. Octahedral geometry. Bond angle is 107 0. Bond angle 90 0 Geometry is trigonal pyramidal. H 2O XeF4 Oxygen has two bond pairs and two lone pairs around it. Bond pair-bond bond pair repulsion < bond pair pair- lone pair repulsion < lone pair – lone pair repulsion. Square planar geometry Bond angle is 104.5 0. Four bond pairs and two lone pairs Geometry is bent shape or inverted V shape.  Type No. of electron pairs Shape of the molecule Examples Bond angle AB2 2 (bp) Linear BeF2 , BeCl2 180 0 AB3 3(bp) Trigonal planar BF3 120 0 AB4 4(bp) Tetrahedral CH4 109.5 0 AB5 5(bp) Trigonal bi pyramid PCl5 120 0, 90 0 AB6 6(bp) octahedral SF6 90 0 AB3E 3(bp), 1 (lp) Trigonal pyramidal NH3 107 0 AB2 E2 2(bp), 2 (lp) Bent or inverted V shape H2O 104.5 0 AB4E 4(bp), 1 (lp) See- saw SF4 AB3 E2 3(bp), 2 (lp) T shape ClF3 AB2 E3 2 (bp), 3(lp) Linear XeF2 AB5E 5 (bp), 1 (lp) Square Pyramid BrF5 AB4 E2 4(bp), 2 (lp) Square planar XeF4 22. Define Bond angle. [IMP 2020] Ans: It is defined as the angle between the orbitals containing bonding electron pairs around the central atom in a molecule. 23. The bond angle in water is lower than the tetrahedral angle. Why? [IMP 2020] Ans: Because of the presence of lone pairs of electrons in water 24. Draw the potential energy curve for the formation of a hydrogen molecule on the basis of inter nuclear distance of the hydrogen atoms. [IMP 2021] 25. The orbital overlap concept of covalent bond formation.(VALENCE BOND THEORY) (I) Covalent bonds are formed by the overlapping of half filled atomic orbitals present in the valence shell of the combining atoms. (II) The orbitals undergoing overlapping must have electrons with opposite spins. (III) The greater the overlapping , the stronger the bond formed. 26. What are the difference between sigma bond and pi bond? Covalent bond is classified in to two types (i) Sigma bond (σ bond) (ii) Pi bond (π bond) Sigma bond (σ bond) Pi bond (π bond) Sigma bond is formed by the end to end (or axial ) Pi bond is formed by the side wise overlap of atomic overlap of atomic orbitals orbitals This can be formed by overlap of s-s ,s-p ,p-p orbitals This can be formed mainly by overlap of p-p orbitals Sigma bond is strong bond Pi bond is weak bond Pz + Pz → sigma bond (σ) , Px + Px →pi bond (π) , Py + Py →pi bond (π) 27. What is Hybridisation? [MARCH 2015 2015] Inter mixing of atomic orbitals of same element with slightly different energies and different shape to get orbitals of same energy and shape is called hybridisation. 28. Write Characteristics of Hybridisation ation [IMP 2021] (I) The number of hybridised orbitals formed is equal to the number of atomic orbitals that get hybridized. (II) Hybridized orbitals have same energy and shape and so more effective in forming stable bonds. (III) The hybrid orbitals are directed in some directions, and give geometry to the molecules. 29. Explain sp3 hybridisation using CH4 as example. [MARCH 2015] One s orbital + Three p orbitals → Four sp3 hybridized orbitals Tetrahedral geometry. Bond angle is 109.50. 30. Explain sp2 hybridisation using BF3 as example [IMP 2022] One s orbital + Two p orbitals → Three sp2 hybridized orbitals Trigonal planar geometry. Bond angle is 1200 31. Explain sp hybridisation using BeCl2 as example One s orbital + One p orbital → Two sp hybridized orbitals Linear geometry. Bond angle is 1800 32. Explain sp3d hybridisation using PCl5 as example. [MARCH 2017]  One s orbital + Three p orbitals + One d orbital → Five sp 3d hybridized orbitals Trigonal bipyramid geometry. Bond angle is 120 0 and 90 0 33. Explain the geometry of PCl5 molecule and account for its high reactivity. [MARCH 2020] Hybridisation is sp3d. Trigonal bipyramid geometry. There are three equatorial bonds and two axial bonds. The axial bonds are slightly longer than equatorial bonds due to greater repulsion from equatorial bonds. Due to different bond lengths, it unsymmetric and highly reactive. 34. Explain sp3d2 hybridisation using SF6 as example. [IMP 2016] One s orbital + Three p orbitals + Two d orbitals → Six sp3d2 hybridized orbitals Octahedral geometry. Bond angle is 90 0. 35. Hybridisations in hydrocarbons In alkanes, all carbon atoms are in sp3 hybridisation. In alkenes , double bonded carbons are in sp2 hybridisation In alkynes , triple bonded carbons are in sp hybridization Ethane → sp3 hybridisation →7 sigma bonds Ethene (Ethylene) → sp2 hybridisation → 5 sigma bonds and 1 pi bond Ethyne ( Acetylene) → sp hybridisation → 3 sigma bonds and 2 pi bond 36. Hybridization and shape of the molecules Hybridization No. of electron pairs Shape of the molecule Examples Bond angle sp 2 (bp) Linear BeF2, BeCl2 180 0 sp 2 3(bp) Trigonal planar BF3 120 0 sp 3 4(bp) Tetrahedral CH4 109.5 0 sp 3d 5(bp) Triogonal bi pyramid PCl5 120 0, 90 0 sp 3d 2 6(bp) octahedral SF6 90 0 sp 3 3(bp), 1 (lp) Trigonal pyramidal NH3 107 0 sp 3 2(bp), 2 (lp) Bent or inverted V shape H2O 104.5 0  37. By using the concept of hybridization, explain the structure of H 2O molecule. [IMP 2017] Ans: sp3 hybridisation. Four pair of electrons. Two bond pairs and two lone pairs of electrons Due to the greater repulsion between lone pairs, the shape is distorted to angular shape or bent structure or inverted ‘v’ shape and the bond angle becomes 104.5 0. 38. The hybridization of C in ethene is............ Ans: sp2 [MARCH 2019] 39. The hybridization of C in ethyne molecule is............ Ans: sp [MARCH 2024] 40. Give 1 example of a molecule in which the central atom is in sp hybridisation. Predict its geometry. [IMP 2020] Ans: BeCl2. Its geometry is linear 41. Predict the hybridisation of phosphorous atom in PCl5 molecule. [March 2020] Ans: sp3d 42. What are the postulates of Molecular orbital theory (MOT)? [MARCH 2024] (I) In molecules, electrons are present in molecular orbitals. (II) Molecular orbitals are formed by the combina combination tion of atomic orbitals of same energy and proper geometry. (III) The number of molecular orbitals formed is equal to the number of combining atomic orbitals. (IV) Molecular orbitals are associated with the nuclei of all the bonded atoms in a molecule. (V) In molecular orbitals rbitals electrons are filled according to Aufbau principle, Pauli’s exclusion principle and Hund’s rule. 43. What are the differences between bonding molecular orbital and anti bonding molecular orbital? BMO ψA + ψ B ABMO ψA - ψ B BMO is formed by the addition (attraction) of atomic ABMO is formed by the substraction (repulsion) of atomic orbitals orbitals It has greater electron density between the nuclei of It has less electron density between the nuclei of bonded bonded atoms atoms Its energy is less than the energy of atomic orbitals Its energy is more than the energy of atomic orbitals 44. Define bond order. [MARCH 2024] Bond order is defined as half of the difference between the number of electron electronss in the bonding molecular orbitals and the number of electrons in the anti bonding molecular orbitals. Bond order = ½ [Nb - Na ] If the bond order is positive , molecule is stable. If the bond order is zero, molecule is unstable. Such molecule will not exist. Bond order = 1 , single bond ,Bond Bond order = 2 , double bond , Bond order = 3 , triple bond 45. How is bond order related to bond strength? Bond order is directly proportional to bond strength and bond dissociation energy. 46. How is bond order related to bond length? Bond order is inversely proportional to bond length. 47. Explain the stability and magnetic property of H 2 molecule (2 electrons) Molecular orbital electronic configuration of H2 = σ1s2 Bond order = ½ [Nb - Na ] = ½ [2 - 0 ] = 1 Here the bond order is positive , molecule is stable. Bond order = 1 , single bond, No unpaired electrons, diamagnetic. 48. Why He2 molecule will not exist? (4 electrons) [MARCH 2021] Molecular orbital electronic configuration of He2 = σ 1s2 σ*1s2 Bond order = ½ [Nb - Na ] = ½ [2 - 2 ] = 0 Here the bond order is zero , molecule is unstable. So it will not exist. 49. Calculate the bond order of Lithium molecule (6 electrons) [IMP 2017] Molecular orbital electronic configuration of Li 2 = σ1s2 σ*1s2 σ2s2 Bond order = ½ [Nb - Na ] = ½ [4 - 2] = 1 Stable molecule, Diamagnetic 50. Be2 molecule will not exist. Give reason. ( 8 electrons) Molecular orbital electronic configuration of Be2 = σ1s2 σ*1s2 σ2s2 σ*2s2 Bond order = ½ [Nb - Na ] = ½ [4 - 4 ] = 0 Here the bond order is zero , molecule ecule is unstable. So it will not exist. 51. Find the bond order of B2 molecule (10 electrons) Molecular orbital electronic configuration of B2 = σ1s2 σ*1s2 σ2s2 σ*2s2 π2px1 = π2py1 Bond order = ½ [Nb - Na ] = ½ [6 - 4] = 1 Bond order positive, molecule is Stable Unpaired electrons present →Paramagnetic Paramagnetic 52. Find the bond order of C2 molecule (12 electrons) Molecular orbital electronic configuration of C 2 = σ1s2 σ*1s2 σ2s2 σ*2s2 π2px2 = π2py2 Bond order = ½ [Nb - Na ] = ½ [8 - 4] = 2 Bond order positive, molecule is Stable Bond order=2 , Double bond Unpaired electrons present →Paramagnetic Paramagnetic 53. With the help of molecular orbitals, find bond order of N 2 and explain its stability [MARCH 2022]  Molecular orbital electronic configuration of N2 = σ1s2 σ*1s2 σ2s2 σ*2s2 π2px2 = π2py2 σ2pz2 Bond order = ½ [Nb - Na ] = ½ [10 - 4 ] = 3 Here the bond order is positive , molecule is stable. Bond order = 3 , triple bond. Here all electrons are paired , the molecule is diamagnetic. 54. Calculate the bond order and predict the magnetic property of O 2 molecule [MARCH 2022] [IMP 2021] Molecular orbital electronic configuration of O2 = σ1s2 σ*1s2 σ2s2 σ*2s2 σ2pz2 π2px2 = π2py2 π*2px1 = π*2py1 Bond order = ½ [Nb - Na ] = ½ [10 - 6 ] = 2 Here the bond order is positive , molecule is stable. Bond order = 2 , double bond. Here unpaired electrons are present in π*2px and π*2py , so oxygen molecule is paramagnetic. 55. Write the molecular orbital electronic configuration of F2 molecule find bond order [IMP 2023] Molecular orbital electronic configuration of F2 = σ1s2 σ*1s2 σ2s2 σ*2s2 σ2pz2 π2px2 = π2py2 π*2px2 = π*2py2 Bond order = ½ [Nb - Na ] = ½ [10 - 8 ] = 1. Here the bond order is positive , molecule is stable. Bond order = 1 , single bond , Here all electrons are paired , the molecule is diamagnetic. 56. The diatomic species Ne2, does not exist, but Ne2 – can exist. Explain on the basis of molecular orbital theory. [IMP 2018] M.O. configuration of Ne2 is σ1s2 σ*1s2 σ2s2 σ*2s2 σ2pz2 π2px2 = π2py2 π*2px2 = π*2py2 σ*2pz2 Bond order (B.O) = ½ [Nb – Na] For Ne2, B.O = ½ [10 – 10] = ½ x 0 = 0 For Ne2 – , σ1s2 σ*1s2 σ2s2 σ*2s2 σ2pz2 π2px2 = π2py2 π*2px2 = π*2py2 σ*2pz2 σ3s1 B.O = ½ [11 – 10] = ½ x 1 = 0.5 Since B.O of Ne2 is zero, it does not exist. But Ne2 – has a +ve bond order, so it exists. 57. Calculate ulate bond order. Compare the stability ( Bond dissociation enthalpy ). Arrange in the increasing order of bond length O2, O2+ and O2- , O222- O2 (16 electrons) → σ1s2 σ*1s2 σ2s2 σ*2s2 σ2pz2 π2px2 = π2py2 π*2px1 = π*2py1 Bond order = ½ [Nb - Na ] = ½ [10 - 6 ] = 2 O2+ (16-1 = 15 electrons) → σ1s2 σ*1s2 σ2s2 σ*2s2 σ2pz2 π2px2 = π2py2 π*2px1 = π*2py0 Bond order = ½ [Nb - Na ] = ½ [10 - 5 ] = 2.5 O2- (16 + 1 = 17 electrons) → σ1s2 σ*1s2 σ2s2 σ*2s2 σ2pz2 π2px2 = π2py2 π*2px2 = π*2py1 Bond order = ½ [Nb - Na ] = ½ [10 - 7 ] = 1.5 O22- (16 + 2 = 18 electrons) → σ1s2 σ*1s2 σ2s2 σ*2s2 σ2pz2 π2px2 = π2py2 π*2px2 = π*2py2 Bond order = ½ [Nb - Na ] = ½ [10 - 8 ] = 1 Bond order order : O22- < O2- < O2 < O2+ Bond strength (Bond stability , Bond dissociation enthalpy ) order : O22- < O2- < O2 < O2+ Bond length order O22- > O2- > O2 > O2+ 58. What is hydrogen bond? Hydrogen bond is defined as the attractive force between hydrogen atom bonded to fluorine, oxygen or nitrogen and an electronegative atom of the same or adjacent molecule. 59. Which are different type hydrogen bonds? Explain each [MARCH 2024] (I) Inter molecular hydrogen bond ::- Hydrogen bond between different molecules of same type or different type. It increasess the boiling point. Example : H bonding in HF, …….H-F……H-F…….H-F……..H-F….. (II) Intra molecular hydrogen bond: bond:- Hydrogen bond within the same molecule. It decreases the boiling point. Example : Hydrogen bonding in Ortho nitro phenol 60. Ortho nitro phenol and para nitro phenol can be separated by steam distillation. Explain Ortho nitro phenol →Intra molecular hydrogen bond → Boiling oiling point is low and steam volatile. Para nitro phenol → Inter nter molecular hydrogen bonding →Boiling point is high and so it is not steam volatile. So these can be separated by steam distillation. 61. H2O is liquid ,H2S is gas at room temperature. Give reason [IMP 2008] In water → Inter nter molecular hydrogen bonds → Liquid at room temperature. In hydrogen sulphide → Noo hydrogen bond → Gas. 62. HF is liquid while HCl is gas at room temperature. Give reason [IMP 2019] In HF → Inter molecular hydrogen bonds → Liquid …….H-F……H-F…….H-F……..H-F….. In HCl → No hydrogen bond → Gas. SOME HSE PREVIOUS QUESTIONS AND ANSWERS 1. Draw the Lewis dot symbols of (i) Cl2 (ii) NF3 [March 2020] 2. Represent the Lewis structure of Ozone (O3) molecule and assign the formal charge on each atom. [MARCH 2019] FC = V-N- B/2 Formal charge = Total number of valence electrons on the free atom – Total no. of lone pairs of electron – ½ [Total no. of bonding electrons] Formal charge on first O atom = 6 – 2 – ½ (6) = +1 Formal charge on second O atom = 6 – 4 – ½ (4) = 0 Formal charge on third O atom = 6 – 6 – ½ (2) = -1 3. A molecule of the type AB4E has 4 bond pairs of electrons and 1 lone pair of electron. Predict the most stable structure of this compound. [IMP 2021] Ans: See-saw shape 4. The geometry of SF6 molecule is …….. Ans : Octahedral 5. Give the shape of the following species. i) NH4+ ii) HgCl2 Ans: i)NH4+ Tetrahedral ii) HgCl2 Linear 6. A molecule of the type AB3E2 has three bond pairs and two lone pairs of electrons. Predict the most stable arrangement of electron pairs in this molecule. [IMP 2012] Ans : T-shape 7. Predict the shape of XeF4 molecule, according to VSEPR theory. [IMP 2017] Ans: XeF4 →Square planar geometry 8. If Z-axis is the internuclear axis, name the type of covalent bond formed by the overlapping of two p y orbitals. [IMARCH 2018] Ans: π bond 9. Which one of the following correctly represents the formation of bonding molecular orbital from the atomic orbitals having wave functions ψA and ψB? i)ψA x ψB ii) ψA / ψB iii) ψA + ψBiv) ψA – ψB [IMP 2015] Ans : ψA + ψB 10. Isoelectronic species have the same bond order. Among the following choose the pair having same bond order. CN –, O2–, NO+, CN+ [MARCH 2017] Ans: For CN –, number of electrons =6 + 7 +1 = 14 For O2 –, number of electrons = 8 + 2 = 10 For NO+, number of electrons =7 + 8 - 1 = 14 For CN+ , number of electrons =6 + 7 - 1 = 12 Pairs having same bond order → CN – , NO+ ( Reason: No. of electrons are same) 11. Match the molecules given in column A with their hybridization given in column B [March 2023] A B BeCl2 sp PCl5 sp3d CH4 sp3 BCl3 Sp2 12. Match the molecules in column I with their shapes in column II [March 2024] A B PCl5 Trigonal bipyramidal SF6 Octahedral CH4 Tetrahedral NH3 Trigonal pyramidal  UNIT 5 : THERMODYNAMICS PREPARED BY: YOOSAFALI T K , GVHSS DESAMANGALAM (08196), THRISSUR (DT) ====================================================== 1. Define system and surrounding. System :- The part of universe under study Surroundings :- The remaining part of the universe that interact with system Universe = System + Surrounding 2. Explain open system, closed system and isolated system with example. (I) Open system :- A system which can exchange both energy and matter with the surroundings. e.g. Hot water in a cup (II) Closed system :- A system can exchange only energy but not matter with the surroundings. e.g. Hot water in a closed steel tumbler. (III) Isolated system :- A system which can neither exchange matter nor energy with the surroundings. e.g. Hot water in a perfectly insulated thermos flask. 3. Differentiate state functions and path functions. Give examples for each. [MARCH 2023] [IMP 2018] State functions Path functions A function or property that depends only on the A function or property that depends on the initial initial state and final state of the system and not on state and final state of the system and on the path the path followed is called state function followed also is called path function Examples :- Temperature(T) , Pressure (P), Volume Examples:- Heat(q) , work (w) (V) , Internal energy (U) , Enthalpy (H) , Entropy (S) , Gibbs free energy (G) 4. Differentiate extensive and intensive properties. Give examples for each. [IMP 2019] [IMP 2023] Extensive properties Intensive properties These are properties which depend on the amount These are properties which are independent on the of matter present in the system. amount of matter present in the system. Examples: - Mass (m) , Volume (V) , Length (l), Examples : - Temperature ,Pressure, Density , Internal energy (U) , Enthalpy (H) , Entropy (S) , Refractive index , Viscosity , Surface tension , Gibbs free energy (G), Heat capacity etc Specific heat , Molar heat capacity 5. Intensive properties:- An extensive property is......... i) density ii) pressure iii) temperature iv) mass [SAY 2017] Ans : Mass 6. Some macroscopic properties are given below. Help Reena to classify them into two groups under suitable titles. [Heat capacity, Entropy, Refractive index, Surface tension] [MARCH 2017] Ans: Extensive properties: Heat capacity, Entropy Intensive properties: Refractive index, Surface tension 7. Classify the following into intensive and extensive properties. i) Internal energy ii) Density iii) Heat capacity iv) Temperature [MARCH 2015] Ans: Intensive properties: Density, Temperature Extensive properties: Internal energy, Heat capacity 8. What is isothermal process? A process which takes place at constant temperature. ∆T= 0 9. What is Adiabatic process? A process which takes place at constant heat. d q= 0 10. What are reversible and irreversible processes Reversible process:- A process is thermodynamically reversible if it can be reversed at any stage by a very small change in some conditions such as temperature, pressure or concentration. At every stage the process is in thermodynamic equilibrium. Irreversible process:- A process which cannot be reversed by a small change in any one of the controlling properties is called an irreversible process. An irreversible process proceeds to one direction and it does not remain in equilibrium 11. What is Internal energy (U) ? How internal energy can be changed? Internal energy is the total energy present within a substance. Internal energy is the sum of all types of molecular energies like translational energy, rotational energy, vibrational energy, electronic energy , nuclear energy etc. It is a state function and extensive property. It can be changed by the following ways: (i) By allowing heat to flow in to the system or out of the system (ii) By doing work on the system or by the system. 12. State First law of thermo dynamics and give its mathematical form. [MARCH 2024] It is law of conservation of energy. It states that energy can neither be created nor destroyed. Mathematical form is ∆U = q + w ∆U = change in internal energy, q = heat, w = work For expansion work ( w = - P∆V) , ∆U = q − P∆V Work done on the system, w = +ve, Work done by the system, w = −ve, Heat absorbed by the system, q = +ve , Heat liberated by the system , q = −ve 13. Write the mathematical expression of First Law of thermodynamics. [March 2020] Ans: ∆U = q + w 14. Write the expression for (i) Work done by the gas in irreversible expansion or compression W = −Pext∆V = −Pext(Vf-Vi) (ii) Work done in isothermal reversible expansion or compression of an ideal gas. Wrev = −2.303 nRT log Vf/Vi OR Wrev = − 2.303 nRT log Pi/Pf 15. What is the significance of ∆U ? Change in internal energy (∆U) is the heat absorbed or evolved at constant volume. ∆U = qv 16. Define Enthalpy (H) Enthalpy is the heat content of the system. Enthalpy is the sum of the internal energy and pressure volume energy. H = U + PV It is a state function and extensive property. 17. Give the relation connecting ∆H and ∆U. ∆H = ∆U + P ∆V OR ∆H = ∆U + ∆n RT Where ∆n= nP − nR 18. What is the significance of ∆H? Change in enthaly (∆H) is the heat absorbed or evolved at constant pressure. ∆H = qp. 19. Write the relation connecting q p and q v q p = q v + ∆n RT q p = Heat absorbed or evolved at constant pressure. q v = heat absorbed or evolved at constant volume. 20. What are exothermic and endothermic reactions? Give its sign of ∆H. Exothermic reactions Endothermic reactions The reactions which takes by the liberation of The reactions which takes by the absorption of heat is called exothermic reactions heat is called endothermic reactions Eg. C+O2→CO2 ∆H = −393.5 kJ Eg. N2+O2→ 2 NO ∆H = 180.5 kJ For exothermic reactions, ∆H = −ve. For endothermic reactions, ∆H = +ve. 21. Define Heat capacity? Heat capacity of a substance is defined as the amount of heat required to raise its temperature through 1 0C It is an extensive property. 22. Define Specific heat capacity (specific heat)? It is the amount of heat required to raise the temperature of 1 gram of the substance through 1 0C. Specific heat capacity is intensive property. 23. Define Molar heat capacity? It is the amount of heat required to raise the temperature of 1 mol substance through 1 0C. Molar heat capacity is intensive property. 24. Give the relation between Cp and Cv for an ideal gas. Cp − C v = R Cp → Molar heat capacity at constant pressure. C v →Molar heat capacity at constant volume. 25. What is Thermo chemical equation? A chemical equation which indicates the enthalpy change occurring during the reaction is called thermo chemical equation. Eg. C(s) +O2(g)→CO2 (g) ∆H = −393.5 kJ (I) For exothermic reactions, ∆H = −ve, For endothermic reactions, ∆H = +ve. (II) Physical states of reactants and products should be specified. (III) When the coefficients in the chemical equations are multiplied or divided, the value of ∆H must be multiplied or divided. (IV) When a chemical equation is reversed, the sign of ∆H is reversed. (magnitude remain same) 26. Define Enthalpy of reaction The enthalpy change during a chemical reaction. ∆rH = Sum of enthalpies of products − Sum of enthalpies of reactants The enthalpy change during a chemical reaction when all participating substances are in standard state is called standard enthalpy of reaction. (Standard states: Pure forms, 1 bar pressure, 298 K temperature) 27. Define Standard enthalpy of combustion It is defined as the enthalpy change when one mole of a substance is completely burned in the presence of air or oxygen when all their reactants and products are in their standard state. Enthalpy combustion is always negative. 28. Define Standard enthalpy of formation [MARCH 2023] The enthalpy change when one mole of a compound is formed from its elements in their stable states. The standard enthalpy of formation of all elements in their standard state is taken as zero. O2→ 0, H2→ 0, N2→ 0, Cgraphite→ 0, Srhombic→ 0 29. What is the importance of standard enthalpy of formation It is useful for calculating standard enthalpies of a reaction Standard enthalpy change of a reaction = Standard enthalpies of formation of products - Standard enthalpies of formation of reactants. 30. Define Enthalpy of solution The enthalpy change when one mole of a substance is dissolved in a specified amount of solvent. 31. Define Standard enthalpy of fusion The enthalpy change when one mole of a solid is converted to its liquid state at its melting point under standard pressure 1 bar. 32. Define Standard enthalpy of vapourisation The enthalpy change when one mole of a liquid is converted to its gaseous state at its boiling point under standard pressure 1 bar. 33. Define Standard enthalpy of sublimation The enthalpy change when one mole of a solid is directly converted to its gaseous state at a constant temperature and 1 bar pressure. 34. State and illustrate Hess’s Law of Constant Heat of Summation. [IMP 2022] [MARCH 2023] [MARCH 2021] It states that the enthalpy change in a chemical reaction is the same whether the reaction takes place in one step or several steps. ∆Hr = ∆H1 +∆H2 + ∆H3 35. Give some applications of Hess’s law. (I) It is used to determine the enthalpies of reaction. (II) It is used to determine the enthalpy of formation (III) It is used to determine the enthalpy of transition of allotropic forms. (IV) It is used to determine the bond enthalpy and lattice enthalpy. 36. Define Bond enthalpy The amount of heat required to break one mole of covalent bonds to gaseous products is called bond dissociation enthalpy. For diatomic molecules, bond dissociation enthalpy and bond enthalpy are same. In the case of poly atomic molecules, bond dissociation enthalpy is different for different bonds within the same molecule. In such cases, bond enthalpy is the average of bond dissociation enthalpies of various similar bonds. 37. How will you calculate enthalpy change of a reaction from bond enthalpies? ∆𝑯𝟎𝒓 = 𝑺𝒖𝒎 𝒐𝒇 𝒃𝒐𝒏𝒅 𝒆𝒏𝒆𝒓𝒈𝒊𝒆𝒔 𝒐𝒇 𝒓𝒆𝒂𝒄𝒕𝒂𝒏𝒕𝒔 − 𝑺𝒖𝒎 𝒐𝒇 𝒃𝒐𝒏𝒅 𝒆𝒏𝒆𝒓𝒈𝒊𝒆𝒔 𝒐𝒇 𝒑𝒓𝒐𝒅𝒖𝒄𝒕𝒔 38. What is Lattice enthalpy? [MARCH 2024] The lattice enthalpy of an ionic compound is the enthalpy change when one mole of the ionic compound dissociates in to its gaseous ions. We cannot calculate lattice enthalpy directly by experiment. So Born Haber cycle is used to calculate it. 39. Draw Born Haber cycle for the determination of lattice enthalpy of sodium chloride (NaCl). [MARCH 2024] ∆𝑯𝟎𝒇 = ∆𝑯𝒔𝒖𝒃 + 𝑰𝑬 + ∆𝑯𝒅𝒊𝒔𝒔 + 𝑬𝑨 + 𝑼 𝑳𝒂𝒕𝒕𝒊𝒄𝒆 𝒆𝒏𝒆𝒓𝒈𝒚 (𝑼) = ∆𝑯𝟎𝒇 − (∆𝑯𝒔𝒖𝒃 + 𝑰𝑬 + ∆𝑯𝒅𝒊𝒔𝒔 + 𝑬𝑨) 40. What is spontaneous process and non spontaneous process? [IMP 2023] Spontaneous process is a process that takes place without the help of any external agency. e.g. Flow of water from high level to low level, flow of heat from hot body to cold body Non spontaneous process is a process that takes place with the help of an external agency. E.g. flow of water from low level to high level. 41. Which are driving forces for spontaneous process? (a) Decrease in energy (b) increase in disorder or randomness of the system 42. Define Entropy (S) ? [MARCH 2023] Entropy (S) is a measure of degree of disorder or randomness of the system. Its unit is J/K/mol 𝐪𝐫𝐞𝐯𝐞𝐫𝐬𝐢𝐛𝐥𝐞 𝐓 43. Predict in which of the following entropy increases(∆S = +ve) entropy decreases(∆S = -ve) (i) Melting of ice → Entropy increases (∆S = +ve) (ii) 2 N2O5(g) → 4 NO2(g) + O2(g) → Entropy increases (∆S = +ve) (iii) Condensation of steam in to water→ Entropy decreases (∆S = −ve) [MARCH 2024] (iv) Freezing of water in to ice→ Entropy decreases (∆S = −ve) [MARCH 2024] (v) Liquid crystallizes in to solid → Entropy decreases (∆S = −ve) [IMP 2022] [MARCH 2022] (vi) Evaporation of water → Entropy increases (∆S = +ve) [MARCH 2024] (vii) H2(g) → 2 H (g) → Entropy increases (∆S = +ve) (viii) 2 NaHCO3 (s) → Na2CO3 (s) + CO2 (g) + H2O (g) → Entropy increases (∆S = +ve) (ix) H2 (273 K)→ H2 (300 K) → Entropy increases (∆S = +ve) (x) Temperature of a crystalline solid raised from 0K to 115K → Entropy increases (∆S = +ve) [MARCH 2022] (xi) I 2 (s)→ I 2 (g) → Entropy increases (∆S = +ve) 44. Which of the following is a process taking place with increase in entropy? [SAY 2016] i) Freezing of water ii) Condensation of steam iii) Cooling of a liquid iv) Dissolution of a solute Ans: Dissolution of a solute 45. What happens to the entropy during the following changes? a) A gas condenses into liquid. b) CaCO3(s)→ CaO(s) + CO2 (g) [MARCH 2018] Ans: a) Entropy decreases , b) Entropy increases. 46. State Second law of thermo dynamics and gives equation [MARCH 2023] It states that the entropy of the universe increases in the course of every spontaneous (natural) change. ∆Stotal = ∆Ssystem + ∆Ssurrounding > 0 ∆Suniverse > 0 47. State third law of thermo dynamics The entropy of any pure crystalline substance approaches zero as the temperature approaches absolute zero. This is called third law of thermodynamics 48. Define Gibb’s free energy (G) [MARCH 2024] Gibbs energy is defined as the maximum amount of available energy that can be converted to useful work. 49. Give the relation connecting Gibb’s free energy (G) , Enthalpy (H) and Entropy (S) [MARCH 2022] G =H – TS 50. Write Gibb’s equation OR Give the relation connecting ∆G, ∆H, and ∆S. [MARCH 2024] [IMP 2020] ∆G= ∆H −T ∆S. 51. Explain Gibb’s energy and spontaneity. [MARCH 2023] (i) If ∆G is negative, the process will be spontaneous. (ii) If ∆G is zero, the process is in equilibrium. (iii) If ∆G is positive, the process will be non spontaneous. 52. What are the conditions for ∆G to be negative (spontaneous process) (I) Exothermic process (∆H = −ve) and ∆S = +ve , ∆G will be always negative and the reaction is spontaneous. (II) Exothermic process (∆H = −ve) and ∆S = −ve , ∆G will be negative only when ∆H > T ∆S and the reaction is spontaneous only at low temperature. (III) Endothermic process (∆H = +ve) and ∆S = +ve , ∆G will be negative only when ∆H < T ∆S and the reaction is spontaneous only at high temperature.  (IV) Endothermic process (∆H = +ve) and ∆S = − ve , ∆G will be always positive and the reaction is always non spontaneous. 53. Write the condition of temperature for a process to be spontaneous whose ∆H and ∆S values are positive. [Hint : ∆G = ∆H – T∆S] [March 2020] Ans: When ∆H and ∆S values are positive, the process will be spontaneous at higher temperatures. 54. Give the criterion of spontaneity in terms of ∆G for a process taking place at constant temperature and pressure. [MARCH 2019] Ans: For a spontaneous process ∆G should be –ve. (Or, ∆G < 0) 55. Give the relation connecting standard Gibb’s energy (∆G 0 ) and equilibrium constant(K). 0 ∆G = − 2.303 RT logK SOME HSE PREVIOUS QUESTIONS AND ANSWERS 1. In a process 701 J of heat is absorbed by a system and 394 J of work is done by the system. The change in internal energy for the process is ……………………. [SAY 2019] Ans: 307 J [Here q = 701 J and w = -394 J. ∆U = q + w = 701 + -394 = 307 J] 2. According to the first law of thermodynamics, for an isolated system, ∆U = …………… [MARCH 2019] Ans: Zero 3. Expansion of a gas in vacuum is called free expansion. Which one of the following represents free expansion of an ideal gas under adiabatic conditions? i) q = 0, ∆T ≠ 0, w = 0 ii) q ≠ 0, ∆T = 0, w = 0 iii) q = 0, ∆T = 0, w = 0 iv) q = 0, ∆T < 0, w ≠ 0 [SAY 2015] Ans: iii)q = 0, ∆T = 0, w = 0 [For adiabatic process q = 0. For free expansion no work is done. So w = 0. On applying these values to the mathematical form 1st law, ΔU = q+w = 0. Also since ΔU = q. ΔT, ΔT = 0] 4. First law of thermodynamics can be stated as ∆U = q + w. How can this equation be expressed for : a) An isothermal reversible change? b) A process carried out at constant volume? [SAY 2018] Ans: (a) For an isothermal reversible change, ΔU = 0. So q = -w (b) For a process taking place at constant volume, ΔV = 0. So ΔU = q v 5. The reaction of cyanamide, NH2CN (s), with dioxygen was carried out in a bomb calorimeter, and ∆U was found to be –742.7 kJ mol–1 at 298 K. Calculate enthalpy change for the reaction at 298 K. NH2CN(s) + 3/ 2 O2(g) → N2(g) + CO2(g) + H2O(l) [MARCH 2018] Solution : ∆U =–742.7 kJ mol = –742700 J mol , ∆n= nP - nR = 2 –3/2 =1/2 , R = 8.314 J K–1mol–1 –1 –1 ∆H = ∆U + ∆n RT = –742700 + ½ X 8.314 X 298 =–741500 J 6. Calculate the enthalpy change for the reaction 𝑪𝑯𝟒 (𝒈) + 𝟐 𝑶𝟐 (𝒈) → 𝑪𝑶𝟐 (𝒈) + 𝟐 𝑯𝟐 𝑶(𝒍) Given that the standard enthalpies of formation of 𝑪𝑯𝟒 (𝒈), 𝑪𝑶𝟐 (𝒈) 𝒂𝒏𝒅 𝑯𝟐 𝑶(𝒍) are −74.8 k Jmol−1, −393.5 kJ mol−1 and −285.8 k Jmol−1 respectively. [SAY 2019] [SAY 2016] Solution: 𝟎( ∆𝑯 𝑶𝟐 ) = 𝟎 Standard enthalpy change of a reaction = Standard enthalpies of formation of products – Standard enthalpies of reactants ∆𝑯𝟎 = ∆𝑯𝟎 (𝑪𝑶𝟐 ) + 𝟐 ∆𝑯𝟎 (𝑯𝟐 𝑶) − ∆𝑯𝟎 (𝑪𝑯𝟒 ) + 𝟐 ∆𝑯𝟎 ( 𝑶𝟐 ) = [−393.5 + 2 X (−285.8)] – [ −74.8 + (2 X0 ) ]= −965.1 + 74.8 = −890.3 kJ Jmol −1 7. Calculate the enthalpy of formation of methanol (CH3OH) from the following data. [MARCH 2023] [MARCH 2019] CH3OH + 3/2 O2 → CO2 + 2 H2O ∆H0 = -726 kJ C + O2 → CO2 ∆H0 = -393kJ H2 + ½ O2 → H2O ∆H0 = - 286 kJ Solution: C + 2 H2 + ½ O2 → CH3OH , ∆H0 = ? CH3OH + 3/2 O2 → CO2 + 2 H2O ∆H0 = -726 kJ ……………….(1) C + O2 → CO2 ∆H0 = -393kJ ……………… (2) H2 + ½ O2 → H2O ∆H0 = - 286 kJ …………………..(3) Equation (2) + 2 X Equation (3) – Equation (1) C + O2 + 2 (H2 + ½ O2 ) – (CH3OH + 3/2 O2 ) → CO2 + 2 H2O – (CO2 + 2 H2O) , ∆H0 = -393 + 2 X (- 286) – (−726) = –239 kJ 0 C + 2 H2 + ½ O2 → CH3OH , ∆H = –239 kJ 8. Calculate the enthalpy of formation of ethane from the following data. [OCTOBER 2022] (a) C + O2 → CO2 ∆H0 = -393.5kJ (b) H2 + ½ O2 → H2O ∆H0 = - 285.8 kJ (c) C2H6 + 7/2 O2 →2 CO2 + 3 H2O ∆H0 = -1560 kJ Solution: 2C + 3 H2 → C2H6 , ∆H0 = ? 2 X Equation (a) + 3 X Equation (b) – Equation (c