SECTION – A

1. What are the number of oxygen molecules in 0.5 mol of CaCO₃?

(A) 1.0 x 6.022 x 10²³

(B) 1.5 x 6.022 x 10²³

(C) 0.5 x 6.022 x 10²³

(D) 3.0 x 6.022 x 10²³

2. Which of the following group of species are isoelectronic?

(A) O²⁻, F, Na, Mg²⁺

(B) O, F⁻, Na, Mg⁺

(C) O²⁻, F⁻, Na⁺, Mg²⁺

(D) O²⁻, F⁻, Na, Mg²⁺

3. Energy of an electron is given by Eₙ =

(A) 2.18 x 10⁻¹⁸ J [Z²/n²]

(B) -2.18 x 10⁻¹⁴ J [Z²/n²]

(C) 2.18 x 10⁻¹⁸ J [Z/n²]

(D) -2.18 x 10⁻¹⁸ J [Z²/n²]

4. The statement that is not correct for periodic classification of elements is

(A) The properties of elements are periodic functions of their atomic numbers

(B) Nonmetallic ¹ elements are less in number than metallic elements.

(C) For transition elements, the 3d-orbitals are filled with electrons after 3p-orbitals and before 4s-orbitals

(D) The first ionization ² enthalpies of elements generally increase with increase in atomic number as we go along a period

5. Which of the following statements are correct?

(A) Helium has the highest first ionization enthalpy in the periodic tables

(B) Chlorine has less negative electron gain enthalpy than fluorine

(C) Mercury and bromine are gases at room temperature

(D) In any period, atomic radius of alkali metal is the highest

6. The strongest conjugate base is

(A) Cl⁻

(B) Br⁻

(C) I⁻

(D) F⁻

7. Which one of the following pairs of solutions is not an acidic buffer?

(A) H₂CO₃ + Na₂CO₃

(B) H₃PO₄ + Na₃PO₄

8. Identify the correct statement in relation to the following reaction:

Zn + 2HCl → ZnCl₂ + H₂

(A) Zinc is acting as an oxidant

(B) Chlorine is acting as a reductant

(C) Hydrogen is acting as an reductant

(D) Zinc is acting as a reductant

9. Acetone (CH₃COCH₃) and Propanal (CH₃CH₂CHO) are

(A) Position isomers

(B) Functional group isomers

(C) Geometrical isomers

(D) Optical isomers

10. A mixture of o-nitrophenol and p-nitrophenol can be separated by

(A) Sublimation

(B) Steam distillation

(C) Fractional crystallization

(D) Simple distillation

12. Heating a mixture of sodium acetate and soda-lime gives

(A) Benzene

(B) Methane

(C) Ethane

(D) Ethyne

From Q.N. 13 to16, a statement of Assertion (A) is given followed by a corresponding statement of reason (R) just below it. Of the statements, mark the correct answer as:  

 

(A) Both A and R are true and R is correct explanation of A.

(B) Both A and R are true but R is not the correct explanation of A.

(C) A is true but R is false.

(D) A is false but R is true.  

13. Assertion: All isotopes of a given element show the same type of chemical behavior Reason: The chemical properties of an element are controlled by the number of electrons in the atom

(A) Both A and R are true and R is correct explanation of A.

14. Assertion: The number of radial nodes in 3s and 4p orbitals is not equal Reason: The number of radial nodes in any orbital depends upon the value of ‘n’ and ‘l’

(A) Both A and R are true and R is correct explanation of A.

15. Assertion: Electron gain enthalpy of elements become less negative as we go down a group Reason: Size of the atom increases on going down the group and the added electron would be farther from the nucleus  

(A) Both A and R are true and R is correct explanation of A.

16. Assertion: Lassaigne’s extract is boiled with dil. HNO₃ before testing for halogens by AgNO₃ Reason: CN⁻ and S²⁻ ions present in the extract interfere with the test of halide ions by AgNO₃

(A) Both A and R are true and R is correct explanation of A.

 

 SECTION – B

17. Chlorine is prepared in the laboratory by treating manganese dioxide (MnO2) with aqueous hydrochloric acid according to the reaction:

4HCl + MnO₂ (s) → 2H₂O (l) + MnCl₂ (aq) + Cl₂ (g)  

How many grams of HCl react with 5.0 g of manganese dioxide?

Atomic mass of Mn = 55 u

Answer:

1. Find the molar mass of MnO₂:

• Molar mass of MnO₂ = (55 g/mol Mn) + (2 * 16 g/mol O) = 87 g/mol

2. Calculate the moles of MnO₂:

• Moles of MnO₂ = mass / molar mass = 5.0 g / 87 g/mol = 0.0575 mol

3. Determine the moles of HCl required:

• From the balanced equation, 4 moles of HCl react with 1 mole of MnO₂.
• Moles of HCl = 0.0575 mol MnO₂ * (4 mol HCl / 1 mol MnO₂) = 0.23 mol HCl

4. Calculate the mass of HCl:

• Molar mass of HCl = 1 g/mol (H) + 35.5 g/mol (Cl) = 36.5 g/mol
• Mass of HCl = moles of HCl * molar mass of HCl = 0.23 mol * 36.5 g/mol = 8.395 g

Therefore, 8.395 grams of HCl react with 5.0 g of manganese dioxide.

18. Yellow light emitted from a sodium lamp has a wavelength (λ) of 580 nm. Calculate the frequency (ν) and the wave number (ν̅) of the yellow light.

Answer:

• Calculate frequency (ν):

  • Speed of light (c) = 3 x 10⁸ m/s
  • Wavelength (λ) = 580 nm = 580 x 10⁻⁹ m
  • ν = c / λ = (3 x 10⁸ m/s) / (580 x 10⁻⁹ m) = 5.17 x 10¹⁴ s⁻¹ or 5.17 x 10¹⁴ Hz

• Calculate wave number (ν̅):

  • ν̅ = 1 / λ = 1 / (580 x 10⁻⁹ m) = 1.72 x 10⁶ m⁻¹

Therefore, the frequency (ν) of the yellow light is 5.17 x 10¹⁴ Hz and the wave number (ν̅) is 1.72 x 10⁶ m⁻¹.

19. Describe hybridization in case of PCl5. Why are the axial bond longer as compared to the equatorial bond?

OR

What is meant by the term bond order? Calculate the bond order of O₂⁺.

Answer:

Hybridization in PCl5:

• Phosphorus (P) has five valence electrons.
• To form five bonds with five chlorine atoms, it undergoes sp³d hybridization.
• Five sp³d hybrid orbitals are formed, which overlap with the p orbitals of chlorine atoms to form five P-Cl bonds.
• The geometry of PCl5 is trigonal bipyramidal, with three equatorial bonds and two axial bonds.

Axial Bonds Longer Than Equatorial Bonds:

• The axial bonds are longer than the equatorial bonds due to greater repulsion from the equatorial lone pairs of electrons. The axial bonds experience more repulsion from the equatorial lone pairs than the equatorial bonds do from each other.

OR

Bond Order:

• Bond order is a measure of the strength of a bond between two atoms. It is defined as half the difference between the number of bonding electrons and the number of antibonding electrons.

• Bond Order of O₂⁺:  

  • Molecular orbital configuration of O₂⁺: σ1s², σ1s², σ2s², σ2s², σ2pz², π2px², π2py²
  • Number of bonding electrons: 10
  • Number of antibonding electrons: 5
  • Bond order = (Number of bonding electrons – Number of antibonding electrons) / 2 = (10 – 5) / 2 = 2.5

Therefore, the bond order of O₂⁺ is 2.5.

20. For the reaction, 2A(g) + B(g) → 2D(g), ΔU = -10.5 kJ and ΔS = -44.1 J/K. Calculate ΔG for the reaction and predict whether the reaction may occur spontaneously.

Answer:

• Calculate ΔG using the Gibbs-Helmholtz equation:

  • ΔG = ΔU + TΔS
  • Assuming a temperature of 298 K:
    • ΔG = -10.5 kJ + (298 K)(-44.1 J/K) = -10.5 kJ – 13.13 kJ = -23.63 kJ

• Prediction of spontaneity:

  • Since ΔG is negative, the reaction is spontaneous at 298 K.

21. What are electrophiles and nucleophiles? Explain with examples.

Answer:

• Electrophiles:

  • Electrophiles are electron-deficient species.
  • They are attracted to electron-rich centers and accept electron pairs.
  • Examples: H⁺, NO₂⁺, BF₃, carbocations

• Nucleophiles:

  • Nucleophiles are electron-rich species.
  • They are attracted to electron-deficient centers and donate electron pairs.
  • Examples: OH⁻, CN⁻, NH₃, H₂O

 SECTION – C

22. (a) State Pauli’s exclusion principle

• Pauli’s Exclusion Principle: No two electrons in an atom can have the same set of four quantum numbers (n, l, m_l, m_s). This means that each orbital can hold a maximum of two electrons with opposite spins.

(b) Write all the four quantum numbers of the last electron of chlorine

• Chlorine (Cl): Electronic configuration: 1s² 2s² 2p⁶ 3s² 3p⁵
• Last electron: 3p⁵
  • n (principal quantum number) = 3
  • l (azimuthal quantum number) = 1 (for p orbital)
  • m_l (magnetic quantum number) = -1, 0, or +1 (any of these values)
  • m_s (spin quantum number) = +1/2 or -1/2

 (c) Write electronic configuration of Cr (24)

• Cr (Chromium): 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d⁵
• Explanation: Chromium exhibits an exception to the Aufbau principle. One electron from the 4s orbital moves to the 3d orbital to achieve a more stable half-filled d-subshell.

23. Explain why?

a) Electron gain enthalpy of fluorine is less negative than that of chlorine

• Explanation:
  • Electron gain enthalpy is the energy released when an atom gains an electron.
  • Fluorine has a smaller atomic size compared to chlorine.
  • Due to the smaller size, electron-electron repulsion is higher in fluorine, making it slightly less favorable to add an extra electron compared to chlorine.

b) First ionization enthalpy of nitrogen is greater than that of oxygen

• Explanation:
  • Nitrogen has the electronic configuration 1s² 2s² 2p³.
  • The 2p subshell in nitrogen is half-filled, which is a stable electronic configuration.
  • Removing an electron from a half-filled stable configuration requires more energy.
  • Oxygen has the electronic configuration 1s² 2s² 2p⁴. Removing an electron from oxygen results in a more stable half-filled configuration, which is energetically favorable.

c) Write IUPAC name and symbol of element having atomic number 118.

• IUPAC Name: Oganesson
• Symbol: Og

24. Calculate the standard enthalpy of formation of CH₃OH(l) from the following data:

(i) CH₃OH(l) + 3/2O₂(g) → CO₂(g) + 2H₂O(l), ΔH = -726 kJ/mol (ii) C(s) + O₂(g) → CO₂(g), ΔH = -393 kJ/mol (iii) H₂(g) + 1/2O₂(g) → H₂O(l), ΔH = -286 kJ/mol

Answer:

Enthalpy of formation of CH₃OH (l):

1. Reverse equation (i): CO₂(g) + 2H₂O(l) → CH₃OH(l) + 3/2O₂(g), ΔH = +726 kJ/mol

2. Multiply equation (ii) by 1 and equation (iii) by 2: C(s) + O₂(g) → CO₂(g), ΔH = -393 kJ/mol 2H₂(g) + O₂(g) → 2H₂O(l), ΔH = -572 kJ/mol

3. Add the equations from steps 1, 2, and 3: C(s) + 2H₂(g) + O₂(g) → CH₃OH(l), ΔH = -239 kJ/mol

Therefore, the standard enthalpy of formation of CH₃OH(l) is -239 kJ/mol.

OR

State Hess’s Law of constant heat summation.

• Hess’s Law: The enthalpy change of a reaction is independent of the pathway taken and depends only on the initial and final states of the reactants and products.  
  

Using Hess’s Law, Calculate the enthalpy of formation of carbon monoxide (CO) from the following data:

(i) C(s) + O₂(g) → CO₂(g), ΔH = -393.5 kJ mol⁻¹ (ii) CO(g) + 1/2O₂(g) → CO₂(g), ΔH = -283.0 kJ mol⁻¹

Answer:

1. Reverse equation (ii): CO₂(g) → CO(g) + 1/2O₂(g), ΔH = +283.0 kJ mol⁻¹

2. Add equation (i) and the reversed equation (ii): C(s) + O₂(g) + CO₂(g) → CO₂(g) + CO(g) + 1/2O₂(g), ΔH = -393.5 kJ mol⁻¹ + 283.0 kJ mol⁻¹

3. Cancel out common terms (CO₂(g) and 1/2O₂(g)): C(s) + 1/2O₂(g) → CO(g), ΔH = -110.5 kJ mol⁻¹

Therefore, the enthalpy of formation of carbon monoxide (CO) is -110.5 kJ mol⁻¹.

25. Permanganate (VII) ion, in basic solution oxidizes iodide ion I⁻ to produce molecular iodine (I₂) and manganese (IV) oxide MnO₂. Write a balanced ionic equation to represent this redox reaction.

Balanced Ionic Equation:

2MnO₄⁻ (aq) + 10I⁻ (aq) + 16OH⁻ (aq) → 2MnO₂ (s) + 5I₂ (s) + 8H₂O (l)

OR

Balance the following equation by ion electron method

P₄ + OH⁻ (aq) → PH₃ + HPO₂⁻ [In basic medium]

Balanced Ionic Equation:

P₄ + 4OH⁻ + 4H₂O → 4PH₃ + 2HPO₂⁻

26. (a) Define Inductive effect

• Inductive Effect: The permanent displacement of sigma (σ) electrons along a saturated carbon chain due to the presence of an electronegative or electropositive group.

(b) Why tertiary carbocation is more stable than primary and secondary carbocation?

• Stability of Carbocations: The stability of carbocations increases with the number of alkyl groups attached to the positively charged carbon. This is due to the +I (inductive effect) of the alkyl groups, which release electron density towards the positively charged carbon, stabilizing it. 
  

(c) Write the name and structures of geometrical isomers of 2-Butene.

• 2-Butene has two geometrical isomers:
  • cis-2-Butene: The two methyl groups are on the same side of the double bond.
  • trans-2-Butene: The two methyl groups are on opposite sides of the double bond.

27. (a) Write IUPAC name of the following compound:

i) CH₃CH₂CH₂CH₂CHO

• IUPAC Name: Pentanal

ii) CH₃CH₂CH₂CH(CH₃)CH₂OH

• IUPAC Name: 3-Methylpentan-1-ol

(b) On complete combustion, 0.246 g of an organic compound gave 0.198 g of carbon dioxide and 0.1014 g of water. Determine the percentage composition of carbon and hydrogen in the compound.

 

Calculations:

• Molar mass of CO₂: 44 g/mol

• Moles of CO₂: 0.198 g / 44 g/mol = 0.0045 mol

• Moles of Carbon: 0.0045 mol (since 1 mole of CO₂ contains 1 mole of Carbon)

• Mass of Carbon: 0.0045 mol * 12 g/mol = 0.054 g

• Molar mass of H₂O: 18 g/mol

• Moles of H₂O: 0.1014 g / 18 g/mol = 0.00563 mol

• Moles of Hydrogen: 0.00563 mol * 2 = 0.01126 mol (since 1 mole of H₂O contains 2 moles of Hydrogen)

• Mass of Hydrogen: 0.01126 mol * 1 g/mol = 0.01126 g

• Percentage of Carbon: (Mass of Carbon / Mass of Compound) * 100 = (0.054 g / 0.246 g) * 100 = 21.95%

• Percentage of Hydrogen: (Mass of Hydrogen / Mass of Compound) * 100 = (0.01126 g / 0.246 g) * 100 = 4.57%

28. (a) An alkene “A” on ozonolysis gives a mixture of ethanal and pentan-3-one. Write the structure of “A”.

• Ozonolysis cleaves the double bond of an alkene and forms carbonyl compounds.

• Since ethanal (CH₃CHO) and pentan-3-one (CH₃CH₂COCH₂CH₃) are formed, the alkene “A” must be:

  CH₃-CH=CH-CH₂-CH₂-CH₃ 
  

  IUPAC Name: 3-Hexene

(b) Complete the following reaction:

i) CH₃-CH₂-Br + Na →

Answer: CH₃-CH₃ + NaBr (Wurtz Reaction: Formation of alkanes from alkyl halides)

ii) CH₃-CH₂-Cl + alc. KOH, heat →

Answer: CH₂=CH₂ + KCl + H₂O (Elimination reaction: Formation of alkene)

 SECTION – D

29. (a) State Le Chatelier’s principle.

Le Chatelier’s Principle: When a system at equilibrium is subjected to a stress, the system will shift in a direction that ¹ tends to minimize the effect of the stress. 

(b) Describe the effect of:

i) Addition of H₂

• Effect: The equilibrium will shift to the right (towards the product side) to consume the added H₂.

ii) Addition of CO

• Effect: The equilibrium will shift to the right (towards the product side) to consume the added CO.

iii) Removal of CO

• Effect: The equilibrium will shift to the left (towards the reactant side) to replenish the removed CO.

iv) Removal of CH₃OH

• Effect: The equilibrium will shift to the right (towards the product side) to replenish the removed CH₃OH.

(c) Classify the following as Lewis acids and bases: (a) OH⁻ (b) F⁻ (c) BF₃ (d) NH₄⁺

• Lewis Acid: Accepts electron pair.

  • BF₃ (electron-deficient due to incomplete octet)
  • NH₄⁺ (electron-deficient due to the positive charge)

• Lewis Base: Donates electron pair.

  • OH⁻
  • F⁻

OR

30. (a) What is meant by common ion effect?

• Common Ion Effect: The suppression of the ionization of a weak electrolyte by the addition of a strong electrolyte that has a common ion.

(b) Calculate the pH of a 1.0 x 10⁻³ M solution of HCl (given-log10 = 0.0414)

• HCl is a strong acid and completely ionizes in water.
• [H⁺] = 1.0 x 10⁻³ M
• pH = -log[H⁺] = -log(1.0 x 10⁻³) = -(-3) = 3

(c) Arrange the following in increasing order of acidic strength:

• (i) HF, HCl, HBr, HI

  • Order: HF < HCl < HBr < HI
  • Explanation: Acid strength increases down the group due to the increasing size of the halogen atom and decreasing bond strength of the H-X bond.

• (ii) CH₄, NH₃, H₂O, HF

  • Order: CH₄ < NH₃ < H₂O < HF
  • Explanation: Acid strength increases with the increasing electronegativity of the atom bonded to hydrogen.

31. (a) State Huckel’s rule of aromaticity. What are necessary and sufficient conditions for a molecule to be aromatic?

• Hückel’s Rule: A planar, cyclic, conjugated molecule with (4n+2) π-electrons (where n is an integer) is considered aromatic.

Necessary and Sufficient Conditions:

1. Planarity: The molecule must be planar.
2. Cyclic structure: The molecule must have a cyclic structure.
3. Conjugation: The molecule must have a continuous system of overlapping p-orbitals.
4. (4n+2) π-electrons: The molecule must have (4n+2) π-electrons, where n is an integer (0, 1, 2, etc.). This is also known as the Huckel’s rule.

(b) Give the following reactions:

i) Friedel-Crafts acylation of benzene with acetyl chloride

• Benzene + CH₃COCl (in presence of AlCl₃) → Acetophenone (C₆H₅COCH₃) + HCl

ii) Reduction of phenol with zinc dust.

• Phenol + Zn → Benzene + ZnO

iii) Sodium acetate react with soda lime

• CH₃COONa + NaOH (soda lime) → CH≡CH (acetylene) + Na₂CO₃

OR

32. (a) State Markovnikov’s rule

• Markovnikov’s Rule: In the addition of HX (where X is a halogen) to an unsymmetrical alkene, the hydrogen atom gets attached to the carbon atom with more hydrogen atoms, and the halogen atom gets attached to the carbon atom with fewer hydrogen atoms.

32. (b) Write the mechanism of addition of HBr to propene

• The mechanism involves the formation of a carbocation intermediate.
• The electrophilic H⁺ attacks the double bond of propene, forming a more stable secondary carbocation.
• The bromide ion (Br⁻) then attacks the carbocation to form 2-bromopropane.

32. (c) Perform the following conversion:

i) Ethyne to Ethanal

1. Hydroboration-Oxidation: Ethyne reacts with diborane (BH₃) followed by oxidation with hydrogen peroxide and sodium hydroxide to form ethanal.

ii) Benzene to Nitrobenzene

1. Nitration: Benzene reacts with a mixture of concentrated nitric acid and concentrated sulfuric acid (nitrating mixture) to form nitrobenzene.