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Grade XIChemistry

Exam Preparation Booklet

Chapter-wise Important Questions

Some Basic Concepts of Chemistry

Q1. Differentiate between diffusion and osmosis. (2 marks)

Q2. An organic compound contains 56.8 % carbon, 7.32 % hydrogen and the remaining oxygen. The vapour density of the compound is 68. Calculate the molecular formula of the compound. (2 marks)

Q3. Calculate the number of moles produced when 400 g of hydrogen reacts with nitrogen to form ammonia. (2 marks)

Q4. What is the limiting reactant when 0.45 mole of Zn reacts with 0.7 mole of HCl to form ZnCl2 and H2? (2 marks)

Structure of Atom

Q1. Explain the following: (i) Dual behaviour of matter (ii) Heisenberg’s Uncertainty Principle and its significance (2+3 marks)

Q2. If the number of photons emitted by a 100 watt bulb is 4.01 × 1020, then what is the wavelength of the electromagnetic radiation to be emitted? (2 marks)

Q3. Calculate the energy associated with the second orbit of Li+ and the radius of this orbit. (2 marks)

Q4. What is the ratio of the first spectral line of Balmer and Paschen series of hydrogen spectrum? (2 marks)

Classification of Elements and Periodicity in Properties

Q1. (i) Arrange the following in the order of increasing ionic size: Al3+, F-, O2-, Na+ (ii) How electronegativity affects the nonmetallic character of elements? (iii) Why does the ionic radius decreases on going from left to right in a periodic table?

(3 marks)

Q2. The variation of ionisation enthalpy with atomic numbers for the elements of 2nd period of the Modern Periodic table is shown below. Which elements are likely to represent A and F? (2 marks)

Q3. Use the following information to answer the question.

Element Ionisation energy (kJ/mol)

P 2080

Q 899

R 520

S 1681

(i) Which of the following elements has the highest metallic character? (ii) Which of the following elements belongs to the noble gas group? (iii) Which of the following elements is most likely to be a halogen? (iv) Which of the following elements acts as the strongest reducing agent? (v) Out of element P and R, which one has high hydration energy? (5 marks)

Q4. Account for the following statement: Li shows anomalous behaviour compared to other group members. (3 marks)

Chemical Bonding and Molecular Structure

Q1. Draw the molecular orbital diagram of O2 and calculate its bond order. (3 marks)

Q2. Consider the information given in column I and column II of the following table.

Column I Column II

A. Trigonal planar P. NF3 and PH3

B. Tetrahedral Q. NO3− and CO32−

C. Pyramidal R. SO42− and ClO4-

Match the given columns and draw the structures of the corresponding species.

(3 marks)

Q3. What gives rise to the polarity in covalent bonds? What do you understand by dipole moment and how will you explain that the resultant dipole moment in NF3 is equal to 0.80 × 10−30C m, whereas the resultant dipole moment in NH3 is 4.90 × 10−30C m. (3 marks)

Q4. Giving reasons, arrange the given species in order of their decreasing stabilities. (5 marks)

States of Matter

Q1. A gas occupies a volume of 300 cm3 at a temperature of 27°C and pressure of 760 mm. What will be the volume of the gas at – 4°C and 760 mm pressure? (2 marks)

Q2. (i) Define Boyle’s Law. (ii) A gas occupies a volume of 900 cm3 at a pressure P. If the pressure is altered to 3.5 atms, the volume of the gas is found to be 1000cm3. Calculate the value of initial pressure (P). (2 marks)

Q3. (i) State Dalton’s law of partial pressure. (ii) A mixture of gases A (molecular mass = 23) and B (molecular mass = 30) contains 90 g of A and 119 g of B. If this mixture is present in a container at 10 bar, then what will be the partial pressure exerted by the two gases? (1+2 marks)

Q4. What gives rise to surface tension in liquids? What do you understand by viscosity of a liquid? What is the effect of increase in temperature on surface tension and viscosity? (3 marks)

Q5. What is the formula of the compound if a solid has a cubic structure in which X atoms are located at the corners of the cube, Y atoms at the cube centres and O atoms are at the edge centres?

(2 marks)

Thermodynamics

Q1. Two moles of a gas is allowed to expand isothermally and reversibly at 30°C from a volume of 10 m3 to 80 m3. Calculate w, U and q. (3 marks)

Q2. The complete combustion of 5.6 g of benzene evolved 424 kJ of heat. This heat was measured at constant volume and at a temperature of 30°C. What will be the heat of combustion of benzene at constant pressure? (2 marks)

Q3. What do you understand by the heat capacity of a substance? Derive the relation between Cp and Cv. (3 marks)

Q4. Calculate the heat of combustion of glucose (C6H12O6) from the given data. C (graphite) + O2 (g) → CO2 (g) (ΔH = − 465.0 kJ) H2 (g) +(½)O2 (g) → H2O (l) (ΔH = − 342.8 kJ) 6C (graphite) + 6H2 (g) + 3O2 (g) → C6H12O6 (s) (ΔH = −1368.9 kJ) (3 marks)

Q5. Calculate for the reaction if the bond enthalpy of C−H bond = 413 kJ/mol, Cl−Cl bond = 243 kJ/mol, C−Cl bond = 328 kJ/mol and H−Cl bond = 432 kJ/mol. (2 marks)

Equilibrium

Q1. Derive a relation between Kp and Kc for the given reaction.

(3 marks)

Q2. (i) Write the conjugate bases for each of the following acids: HS−, H2SO4 (ii) Write the conjugate acids for each of the following bases:

(2 marks)

Q3. Calculate the percent dissociation of PCl5, if 0.10 mol of PCl5 is kept in 16.0 L flask at 600 K. The value of Kc for the reaction is 4.2 × 10−2. (3 marks)

Q4. Calculate the value of ΔG° at 50°C for the given reaction, for which the equilibrium constant is 4.0 × 10−7. What will be the effect on equilibrium constant if the concentration of SO2 is increased?

(2 marks)

Q5. Total pressure for the reaction at equilibrium was found to be 8.97 bar. If 15.8 g of N2O4 was placed in a 1 L reaction vessel at 400 K, then calculate Kc, Kp and partial pressures at equilibrium. (3 marks)

Redox Reactions

Q1. What are disproportionation reactions? Explain giving suitable examples. (2 marks)

Q2. What products are obtained when ClO3- and SF6 undergo disproportionation reaction? (2 marks)

Q3. Determine the oxidation number of each of the bromine atom in the given structure of Br3O7.

(2 marks)

Q4. Balance the following reaction in acidic medium using ion electron method.

(3 marks)

Hydrogen

Q1. What is the strength of a solution of H2O2 marked as 15 V? Why is H2O2 kept away from dust? (2+1 marks)

The S-Block Elements

Q1. Why is an ammoniated solution of sodium metal paramagnetic and blue in colour? (1 mark)

Q2. Despite of having more negative reduction potential than sodium, Li does not react readily with water. Explain. (2 marks)

Q3. Among BaCO3, CaCO3, SrCO3 and BeCO3, which will require the (i) highest energy to undergo decomposition? (ii) least energy to undergo decomposition? (2 marks)

Q4. Answer the following: (i) How do the properties of beryllium differ from rest of the alkali earth metals? (ii) What are the similarities in the properties of beryllium and aluminium? (iii) Draw the structure of BeCl2 in solid state

(5 marks)

The p-Block Elements

Q1. Explain the inert pair effect. Tl(I) compounds are more common than Tl(III) compounds. Comment. (3 marks)

Q2. Justify the given statement: (CH3)3 N is pyramidal while (SiH3)3 N is tetrahedral. (3 marks)

Q3. Answer the following: (i) How is boric acid prepared and what are its physical properties? (ii) Why boric acid acts like a weak acid? (2+1 marks)

Q4. Phosphorus forms compounds containing many phosphorus atoms, while nitrogen compounds do not have more than three nitrogen atoms. Why? (1 mark)

Organic Chemistry - Some Basic Principles and Techniques

Q1. Answer the following: (i) Explain inductive effect. (ii) Comment on the stability of carbocations, carbanions and free radicals on the bases of the inductive effect. (2+3 marks)

Q2. Write the IUPAC names of the following compounds. (i) (ii) (iii)

(3 marks)

Q3. 0.4950 g of an organic compound gave on combustion 0.9900 g of carbon dioxide and 0.4050 g of water. Calculate the percentage of carbon and hydrogen in it.

(2 marks)

Q4. How does the hyperconjugation effect contribute towards the stability of ethyl carbocation?

(2 marks)

Hydrocarbons

Q1. What do you understand by conformations? Explain Sawhorse and Newman projections using the example of ethane. (3 marks)

Q2. Which of the given compounds is not aromatic?

(3 marks)

Q3. The ozonolysis of an alkene forms methanol and propanone. What is the most likely structure of the alkene? (2 marks)

Environmental Chemistry

Q1. What are the harmful effects of photochemical smog? (2 marks)

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Grade XIChemistry

Exam Preparation Booklet

Chapter-wise Solutions of Important Questions

Some Basic Concepts of Chemistry

Ans 1.

Diffusion Osmosis

(i) It is the movement of molecules from a region of higher concentration to a region of lower concentration.

(i) It is a special case of diffusion wherein the medium is water.

(ii) It does not require a membrane. (ii) It requires a semi permeable membrane.

Ans 2.

Element Percentage Atomic mass

Moles of atom

Atomic ratio

Simplest ratio

C 56.8 12 56.8/12 = 4.73

4.73/2.24=2.1

2

H 7.32 1 7.32/1 = 7.32 7.32/2.24=3.3

3

O 35.88 16 35.88/16 = 2.24

2.24/2.24=1

1

∴ Empirical formula = C2H3O Empirical formula mass = 12 × 2 + 1 × 3 + 16 × 1= 43 g Molecular mass = 2 × Vapour density

= 2 × 68 = 136 g Hence, the molecular formula = (E.F.)n = (C2H3O)3 = C6H9O3

Ans 3. The given chemical reaction can be represented as: The number of moles of H2 in 400 g

Now, 3 moles of hydrogen react with 1 mole of hydrogen to produce 2 moles of ammonia. 200 moles of hydrogen react with (200/3) moles of hydrogen to produce (2x200)/3 moles of NH3. Therefore, moles of NH3 produced = 133.33 moles

Ans 4. The balanced equation for the reaction is Zn + 2HCl → ZnCl2 + H2 1 mole of Zn reacts with 2 moles of HCl. 0.45 mole Zn will react with (0.45 x 2)/1 = 0.9 moles of HCl. The HCl used in the reaction is only 0.7 mole (given). Thus, Zn will not completely react with HCl. ∴ 0.7 mole of HCl will react with (0.7 x 1)/2 = 0.35 moles of Zn. HCl will completely react with Zn. Hence, HCl is the limiting reactant in the given reaction.

Structure of Atom

Ans 1.

(i) Dual behaviour of matter: de-Broglie proposed that like radiation, matter also exhibits dual nature, i.e., matter also exhibits properties of both a wave and a particle. Therefore, like a photon, an electron should also possess wave as well as particle nature. With regard to this, he derived an equation, known as de-Broglie equation. Where, λ= Wavelength of the particle h = Planck’s constant p = Momentum of the particle m = Mass of the particle v = Velocity of the particle

(ii) Heisenberg’s Uncertainty Principle: According to this rule, it is impossible to determine the exact position and the exact momentum (velocity) of an electron. To locate the position of an electron, it needs to be illuminated with high intensity photons. When these high energy photons collide with electrons, they change their velocity by transferring energy to them. Hence, the position and the momentum of an electron can never be determined accurately at the same point of time. This can be shown mathematically as follows: Where, x = Uncertainty in position px = Uncertainty in momentum vx = Uncertainty in velocity m = Mass of electron Significance of Heisenberg’s Uncertainty Principle: It rules out that electrons and other similar particles have specific paths or trajectories. This is because the trajectory of a substance is determined by its location and velocity at various moments/instances.

Ans 2.

Number of photons emitted by the bulb = 4.01 × 1020 Energy of 1 photon = 100/ (4.01 × 1020) = 24.938 × 1020 J

Ans 3.

We know that, Therefore, energy for Li+, n = 2, Z = 3

Ans 4.

According to Rydberg equation, For first line of Balmer series, For first line of Paschen series, n1 = 3; n2 = 4 Hence, the ratio of Balmer and Paschen series is 7:20.

Classification of Elements and Periodicity in Properties

Ans 1. (i) Al3+ < Na+ < F- < O2- (ii) Nonmetallic character is directly related to electronegativity. As we move from left to right of a period, the outermost electrons feel greater pull of the nucleus. This is because the nuclear charge increases and atomic size decreases from left to right. Thus, the elements on the right of the periodic table do not lose electrons easily, so these are nonmetallic in nature. In this way, electronegativity affects the nonmetallic character of elements. (iii) On going from left to right, the effective nuclear charge increases, as number of electrons keep on increasing in the outermost shell. Due to this, electrostatic attraction between the nucleus and the electrons increases and ionic size decreases.

Ans 2. A represents the element lithium (Li) and F represents the element neon (Ne). This is because the ionisation energy for A is very less. So, A must lie at the extreme left of the 2nd period. F on the other hand, has the maximum ionisation energy, so it must have a stable electronic configuration.

Ans 3. (i) R (ii) P (iii) S (iv) R (v) R

Ans 4. Li shows anomalous behaviour compared to other group members due to the following reasons:

● Li ion (Li+) is very small. ● Li has high electronegativity. ● Li ion (Li+) has large charge/radius ratio. ● Li does not have any delectron in its valence shell.

The positive values of the second electron gain enthalpy can be explained on the basis of the following reason: After the addition of one electron, the atom acquires a negative charge. Now, the second electron has to be added to a negatively charged ion and this addition is opposed by coulombic repulsions. Thus, the energy has to be supplied to force the second electron into an ion.

Chemical Bonding and Molecular Structure

Ans 1. Number of electrons in O2 = 2 × 8 = 16 Electronic configuration of the molecule = σ1s2, σ*1s2, σ2s2, σ*2s2, σ2pz2, π2px2 = π2py2, π*2px1 = π*2py1 Bond order = (10 - 6)/2 = 4/2 = 2 Thus, oxygen molecule has two bonds (one sigma and one pi bond). There are two unpaired electrons in the molecule. Thus, it is paramagnetic.

Ans 2. The structures of the given species are as under:

Ans 3. In the case of homoatomic molecules like H2 and O2, the shared pairs of electrons are attracted by the two atoms in a similar manner. Hence, they are present at the centre of the bond. However, in the case of heteronuclear molecules, the shared pairs of electrons are attracted more strongly by one atom depending on the electronegativity of the atoms. Thus, the shared pairs of electrons remain closer to the more electronegative atom. This gives rise to the polarity in covalent bonds. Dipole moment is defined as the product of the magnitude of charge and the distance between the centres of the positive and the negative charges. Dipole moment = Charge × Distance of separation We know that N is more electronegative than H, but less electronegative than F. Therefore, as shown in the figure, the dipole moment in NF3 due to NF bonds opposes the dipole moment due to the lone pair present on N. However, in NH3, the dipole moment due to NH bonds enhances the dipole moment due to the lone pair of electrons, making the value of the resultant dipole moment extremely high.

Ans 4. The respective electronic configurations of the given molecules are as follows: In O22- there are 10 electrons in the bonding molecular orbitals, while there are 8 electrons in the antibonding molecular orbitals.

Hence, bond order of = In , there are 10 electrons in the bonding molecular orbitals, while there are 7 electrons in the antibonding molecular orbitals.

Hence, bond order of = In O2, there are 10 electrons in the bonding molecular orbitals, while there are 6 electrons in the antibonding molecular orbitals.

Hence, bond order of O2 = In , there are 10 electrons in the bonding molecular orbitals, while there are 5 electrons in the antibonding molecular orbitals.

Hence, bond order of = Greater the bond order, more stable is the molecule. Therefore, the decreasing order of the stability of given species are as follows:

States of Matter

Ans 1. Given that, Initial volume of gas (V1) = 300 cm3 Initial temperature of gas (T1) = 27°C = (273 + 27) K = 300 K Final temperature of gas (T2) = – 4°C = (273 – 4) = 269 K Final volume of gas (V2) = ?

According to Charles’ Law : ∴ ⇒ V2 = 269 cm3

Ans 2. (i) Boyle’s Law states that the absolute pressure and volume of a given mass of gas within the closed system, are inversely proportional to each other, if the temperature remains unchanged. (ii) Given that, P2 = 3.5 atm, V1= 900 cm3 , V2= 1000 cm3 According to Boyle’s Law, P1V1= P2V2 P × 900 = 3.5 × 1000

⇒ P = Therefore, P = 3.88 atm

Ans 3. (i) Dalton’s law of partial pressure: It states that the total pressure exerted by the mixture of non reactive gases is equal to the sum of the partial pressures of the individual gases. ptotal = p1 + p2 + p3 +……. (ii)

Ans 4. In liquids, the molecules present in the bulk experience equal force from all directions. However, the molecules present on the surface experience force only from below, as shown in the figure. Therefore, a molecule present on the surface experiences a net inward force. Due to these net forces, the surface of a liquid behaves like a stretched membrane. This is known as surface tension. It is defined as the force acting per unit length perpendicular to the line drawn on the surface of the liquid. The viscosity of a liquid is defined as its measure of resistance to flow. It arises due to the internal friction between the layers of a fluid. Stronger the forces of attraction of a liquid, more viscous it will be. We know that the kinetic energy of particles increases on increasing temperature. Therefore, intermolecular forces of attraction become less effective when temperature is increased. As intermolecular forces of attraction decreases, both the surface tension and viscosity decrease.

Ans 5.

Each atom of X at the corners makes contribution towards a particular unit cell.

∴ The number of atoms of X per unit cell = Y atom present at the body centre belongs to a particular unit cell. ∴The number of atoms of Y per unit cell = 1 × 1 = 1

There are 12 edges of a cube. Each atom of O present at the edge centre makes contribution towards a particular unit cell.

∴ The number of O atoms per unit cell The formula of the compound is XYO3.

Thermodynamics

Ans 1. It is given that, V1 = 10 m3 , V2 = 80 m3, n = 2 moles R = 8.314 JK−1 mol−1 T = 30 + 273 = 303 K (i) Work done, w can be calculated as follows: (ii) ΔU = 0 for an isothermal expansion. (iii) Heat, q can be calculated as follows:

ΔU = q + w ∴q = −w Or, q = 5239.35 J mol−1

Ans 2. The combustion of benzene is represented as follows: C6H6 (l) + (15/2)O2 (g) → 6CO2 (g) + 3H2O (l) Heat evolved in the complete combustion of 5.6 g of benzene = 424 kJ ∴Heat evolved in the complete combustion of 78 g (1 mole) of benzene =

{(78/5.6) × 424} = 5 905.71 kJ Thus, heat liberated at constant volume = −5 905.71 kJ Here, n = (6 − 7.5) = −1.5 T = 273 + 30 = 303 K ΔH = ΔE + ΔnRT = −5 905.71 + {(−1.5) × 8.3 × 10−3 × 303} = −5 909.48 kJ mol−1 Hence, the heat of combustion of benzene at constant pressure is −5 909.48 kJ mol−1.

Ans 3. The increase in the temperature of a substance is directly proportional to the heat supplied to it. q ∝ T q = C ΔT Here, C is known as the heat capacity of the substance and depends upon its size, nature and composition. At constant volume, heat capacity is denoted as Cv and at constant pressure, heat capacity is denoted as Cp. Therefore, at constant volume qv = Cv ΔT = ΔU At constant pressure, qp = Cp ΔT = ΔH Now, we know that for 1 mole of an ideal gas, ΔH = ΔU + Δ(pV) =Δ U + Δ(RT) = ΔU + RΔT Now, substituting the values of ΔH and ΔU in the equation, we get Cp ΔT = Cv ΔT + RΔT Cp = Cv + R Cp − Cv = R

Ans 4. The required equation is represented as follows: C6H12O6 (s) + 6O2 (g) → 6CO2 (g) + 6H2O (l) ...(i) C (graphite) + O2 (g) → CO2 (g) (ΔH = − 465.0 kJ) ...(ii) H2 (g) + (½)O2 (g) → H2O (l) (ΔH = − 342.8 kJ) ...(iii) On multiplying equations (ii) and (iii) by 6 and adding them up, the reaction that is obtained is represented as follows: 6C (graphite)+ 6H2 (g) + 9O2 (g) → 6CO2 (g) + 6H2O (l) ...(iv) {ΔH = −4846.8 kJ} On subtracting equation (i) from equation (iv), the reaction obtained is as follows: C6H12O6 (s) + 6O2 (g) → 6CO2 (g) + 6H2O (l) {ΔH = −3477.9 kJ}

Hence, the heat of combustion of glucose is −3 477.9 kJ.

Ans 5.

= Bond energy of reactants - bond energy of products Bond energy of reactants = (4 × C − H) + (4 × Cl − Cl) = (4 × 413) + (4 × 243) = 1652 + 972 = 2624 kJ Bond energy of products = 4 × C − Cl + 4 × H–Cl = (4 × 328) + (4 × 432) = 1312 + 1728 = 3040 kJ

= 2624 − 3040 = −416 kJ mol−1

Equilibrium

Ans 1. Using the relation, PV = nRT

Or, P = CRT PNO = [NO] RT ...(i) PCl2 = [Cl2] RT ...(ii) PNOCl = [NOCl] RT ...(iii)

Substituting the values from equations (i), (ii) and (iii) in the expression of Kp,, we obtain Or, Kp = KcRT

Ans 2. (i) Following are the conjugate bases: HS−: S2−

H2SO4 : HSO4 (ii) Following are the conjugate acids:

Ans 3.

Hence, the percent dissociation is 2.6 × 10−1× 100 = 26%

Ans 4. Using the relation, ΔG° = 2.303 RT log K It is given that, K = 4.0 × 10−7, R = 8.314 J K−1mol−1, T = 323 K ΔG° = 2.303 × 8.314 × 323 log (4.0 × 10−7) = − 39.56 kJ Hence, the value of ΔG° at 50°C for the given reaction is − 39.56 kJ. If the concentration of SO2 is increased, then according to Le-Chatelier’s principle, more and more of SO2 will react with O2 to form SO3. Hence, the reaction moves in the forward direction.

Ans 5. Total volume (V) = 1 L Molecular mass of N2O4 is 92 g Number of moles of N2O4 = 15.8/92 = 0.172 Now, Or, Initial = 5.6 bar 0 pressure Equilibrium = (5.64 - x) bar 2x bar pressure P total at equilibrium = PN2O4 + PNO2 Partial pressures of individual components at equilibrium is

Redox Reactions

Ans 1. The reaction in which the same element is oxidised and reduced simultaneously is called disproportionation reaction. For this reaction to take place, the element should exist in at least 3 oxidation states. The reaction takes place when the element is in the intermediate oxidation state and both the lower and the higher oxidation states of the element are obtained as products. For example, oxygen can display three oxidation states, 0, −1, −2. The oxidation state of oxygen in H2O2 is −1(intermediate oxidation state). When H2O2 undergoes disproportionation reaction, water and molecular oxygen are produced. The oxidation number of oxygen in water is −2 and in molecular oxygen is 0. It can be observed from the reaction that the oxidation number of hydrogen remains the same in the reaction.

Ans 2. Cl−and ClO4 are produced as products when ClO3 undergoes disproportionation reaction. SF6, on the other hand, cannot undergo disproportionation reaction as the oxidation state of S in SF6 is +6, which is the highest oxidation state displayed by sulphur.

Ans 3. The oxidation numbers of bromine atoms (1) and (3) are +6 each and that of (2) is +4. The oxidation number of Br (1) and Br (2) can be calculated as follows: x + 3 (−2) = 0 x − 6 = 0 x = 6 The oxidation number of Br (2) is calculated as follows:

x + 2 (−2) = 0 x − 4 = 0 x = 4

Ans 4. Step 1: Indicating the oxidation numbers of each atom

Thus, Cr in Cr2O72− and Fe change their oxidation numbers. Step 2: Writing the oxidation and reduction half-reactions

Step 3: Addition of electrons to make up the difference in the oxidation numbers

Step 4: Balancing O by adding equal numbers of H2O molecules to the side that is deficient in O atom

Step 5: Balancing H by adding H+ ions to the side that is deficient in H atoms

Step 6: Multiplying oxidation half-reaction with 6 so that the electrons lost and gained are equal. This is followed by the addition of the two equations.

Hence, the balanced chemical equation is as follows:

Hydrogen

Ans 1. (i) 15 V H2O2 solution means that 1 L of this H2O2 solution will give 15 L of oxygen at STP. H2O2 gives O2 according to the following reaction. That is, at STP, 22.4 L of O2 is produced from 68 g of H2O2. Therefore, at STP, 15 L of O2 is produced from of H2O2. = 45.54 g of H2O2 (ii) H2O2 is kept away from dust because dust can induce explosive decomposition of H2O2.

The S-Block Elements

Ans 1. Sodium metal dissolves in liquid ammonia forming sodium ammoniated cation and ammoniated electron. The presence of ammoniated electron is responsible for the blue colour and the paramagnetic nature of ammoniated solution.

Ans 2. When Na reacts with water, it releases hydrogen gas with a lot of heat. This released heat is sufficient to burn the produced hydrogen. This makes Na more reactive for water.

In case of reaction of Li with water, Li+ ion is totally surrounded by the water due to its smaller size and high effective nuclear charge. This provides high hydration enthalpy, which makes reduction electrode potential highly negative. However, this is not sufficient to burn Li. Hence, Lithium having more E0 reacts less vigorously with water than sodium.

Ans 3. (i) Among the given carbonates, BaCO3 needs the highest energy to undergo decomposition. This is because Ba ion is the largest in size and is stabilised by the large anion . Hence, highest amount of energy is required to break the crystal lattice to undergo decomposition. (ii) Among the given carbonates, BeCO3 needs the least energy to undergo decomposition. This is because Be ion being smallest in size, is not much stabilised by the ion. Hence, it requires the least energy to undergo decomposition.

Ans 4. (i) Anomalous Properties of Beryllium ● Exceptionally small size of the atom and the cation ● Forms covalent compounds (due to small size and high ionisation enthalpy) ● Highest coordination number is four while other members exhibit six. ● Reason: Be has four orbitals in its valence shell, while other members can

use d-orbitals. ● Unlike other elements, the oxides and hydroxides of Be are amphoteric in

nature. (ii) Similarities Between Be and Al

● Like Al, Be is not readily attacked by acids (Reason − presence of an oxide film on the surface of the metal)

● Both dissolve in excess of alkali, giving beryllate ion [Be(OH)4]2−and aluminate ion [Al(OH)4]−

● Like Al2Cl6, BeCl2 has a Cl−bridged polymeric structure. Both these compounds are Lewis acids, and are soluble in organic solvents.

● Both form fluoro complex ions, [BeF4]2−, [AlF6]3− (iii) Structure of BeCl2 in solid state:

The p-Block Elements

Ans 1. Group 13 elements have ns2np1 electronic configuration. Hence, they would be expected to be trivalent. In most of their compounds, this is the case; however, for

the heavy elements, lower oxidation states are more stable. This is explained by the s electrons that remain paired and not participating in bond formation. This inertness of s-subshell electrons towards the bond formation is called inert pair effect. This happens because the s orbitals are held closer to the nucleus. The electrons present in s orbitals are held strongly by the nucleus because of large electrostatic forces. Hence, the energy required to unpair the selectrons is high because of which they remain paired. For example: (1) In 13th group, thallium can exhibit +1 and+3 oxidation states, but it is stable in +1 oxidation state only due to inert pair effect (2) In 14th group, lead shows both +2 and +4 oxidation states, but it is stable in +2 oxidation state due to inert pair effect. Tl(I) compounds are more common due to the inert pair effect. The valence shell electronic configuration of Tl is 6s2 6p1. After the removal of one p-electron, the 6s-electrons do not take part in compound formation.

Ans 2. In (CH3)3N, lone pair of electrons are present in nitrogen atom resulting in sp3 hybridisation. Hence, it is pyramidal. In case of (SiH3)3 N, the lone pair of electrons in nitrogen atom overlaps with the empty d-orbital of silicon atom. This results in dπpπ backbonding. Hence, the lone pairs are no longer available in nitrogen atom. Thus, it involves sp2 hybridisation and has trigonal planar geometry.

Ans 3. (i) Boric acid is prepared by acidifying an aqueous solution of borax. The chemical equation for the process is: Boric acid exists as a white crystalline solid and is soapy to touch. It is sparingly soluble in cold water and highly soluble in hot water. (ii) Boric acid is a weak acid as it does not release H+ ions on its own. When added to water, it accepts OH ions from it thereby furnishing H+ ions.

Ans 4. Nitrogen does not form compounds containing more than three N-atoms because of repulsion between its lone pair of electrons. The N−N single bond is thus very weak. On the other hand, no such repulsions are present on phosphorus atom because of its large atomic size.

Organic Chemistry - Some Basic Principles and Techniques

Ans 1. (i) Inductive effect is an electronic effect due to the polarisation of σ bonds within a molecule or ion. In a covalent bond, formed by the atoms of different electronegativity, the electron density moves towards the more electronegative atom of the bond. There are two types of inductive effect: +I effect: The groups that push the electron density away from themselves exhibit positive inductive effect (+I). Example: The groups exhibiting + I effect are −CH3,−NH2. −I effect: Atoms or groups that pull the electron density towards themselves exhibit negative inductive effect (−I). Example: The groups exhibiting − I effect are −COOH,−NO2. (ii) Stability of carbonium ions: The stability of carbonium ions increases with increase in number of alkyl groups, due to their +I effect. The alkyl groups release electrons to carbon, bearing positive charge and hence stabilize the ion. The order of stability of carbonium ions is: Stability of free radicals: In the same way, the stability of free radicals increases with increase in the number of alkyl groups. Thus, the stability of different free radicals is: Stability of carbanions: Stability of carbanions decreases with increase in the number of alkyl groups, since the electron donating alkyl groups destabilise the carbanions by increasing the electron density. Thus, the order of stability of carbanions is:

Ans 2. (i)

1 − Ethoxypropan − 2 − ol (ii) Ethyl 2 − bromo − 2 − (3 −nitrophenyl) propanoate (iii) 2 − (2− bromophenyl) ethanal

Ans 3. It is given that, Weight of organic compound = 0.4950 g Weight of carbon dioxide = 0.9900 g Weight of water =0.4050 g Percentage of carbon Percentage of water ∴The percentages of carbon and hydrogen in the organic compound are 54.54% and 9.09% respectively.

Ans 4. The hyperconjugation in ethyl carbocation can be represented as follows:

The electron density from the adjacent sigma bond helps in dispersing the positive charge, hence stabilising the cation.

Hydrocarbons

Ans 1. Carbon atoms joined by only a sigma bond can rotate freely about the bond. This rotation gives rise to different spatial arrangements of atoms. These spatial arrangements are interchangeable, and are known as conformations. For example, in a molecule of ethane, the two carbon bonds are connected by a single bond, and are thus, free to rotate. This gives rise to infinite number of conformations. However, there are two extreme cases − staggered and eclipsed. In the former, the hydrogen atoms joined to the carbon atoms are at the maximum distance, and in the latter, the hydrogen atoms are at the minimum distance. There are two ways in which these conformations can be represented. These representations are called Sawhorse and Newman projections. Sawhorse projection: In this projection, the molecule is drawn in such a way that its molecular axis is clearly visible. The staggered and eclipsed conformations for ethane are as follows: Newman projection: In this projection, the molecule is viewed from the side in such a way that only one carbon atom is visible. Therefore, in this representation, the carbon atom present nearer to the eye is represented by a dot and the atom away from the eye is represented by a circle. The staggered and eclipsed conformations for ethane are as follows:

Ans 2. For Aromaticity, a compound must obey Huckel’s rule. According to this rule, there should be (4n + 2) π electrons in the ring, where n is an integer (n = 0, 1, 2…). In structure (I), there are (4×1 + 2) i.e. 6π electrons in the ring. This means that it obeys Huckel’s rule. Hence, it is aromatic. In structure (II), there are (4×1 + 2) i.e., 6π electrons in the ring. This means that it obeys Huckel’s rule. Hence, it is aromatic. Structure (III) has 8π electrons in its ring. Hence, it does not obey Huckel’s rule. Hence, it is not aromatic.

Ans 3. The products methanol and propanone are obtained by the ozonolysis of 2methyl propene. The reaction takes place as follows:

Environmental Chemistry

Ans 1. Harmful effects of photochemical smog: ● Ozone and PAN (Peroxyacetyl nitrate) − the components of photochemical

smog cause irritation to the eyes. Ozone and nitric acid cause irritation to the nose and throat. When these chemicals are present in higher concentrations, they can cause headaches, chest pain and breathing problems.

● Photochemical smog causes cracking of rubber. It also corrodes metals, painted surfaces and building materials.

● It causes extensive damage to plants.