JEE Advanced Chemistry questions with solutions

Q1. Which of the following contains the highest number of molecules?

1. 15 liters of hydrogen gas at STP
2. 5 liters of nitrogen gas at STP
3. 0.5 grams of hydrogen gas
4. 10 grams of oxygen gas

Answer: 15 liters of hydrogen gas at STP

Moles: 15 L H2 at STP = 0.67 mol; 5 L N2 = 0.22 mol; 0.5 g H2 = 0.25 mol; 10 g O2 = 0.31 mol. The 15 L H2 has the most molecules, so the correct option is index 0, not 2.

Q2. An organic compound with identical empirical and molecular formulas consists of 20% carbon, 6.7% hydrogen, 46.7% nitrogen, and the remainder as oxygen. When heated, it releases ammonia and leaves behind a solid residue. This residue produces a violet color when treated with an alkaline copper sulfate solution. Identify the compound.

1. Ammonium carbamate
2. Ammonium formate
3. Hydroxylamine
4. Urea

Answer: Urea

Mole ratios C:H:N:O = 20/12 : 6.7/1 : 46.7/14 : 26.6/16 = 1:4:2:1, giving CH4N2O. On heating it releases ammonia and the residue (biuret) gives a violet color with alkaline copper sulfate, which identifies the compound as urea.

Q3. Which of the following statements accurately describe a hydrogen peroxide solution with a concentration of 17 g/L?

1. The solution has a volume strength of 5.6 under conditions of 1 atm pressure and 273 K temperature.
2. The molarity of the hydrogen peroxide solution is 0.5 M.
3. 1 mL of this solution releases 2.8 mL of oxygen gas at 2 atm pressure and 273 K temperature.
4. The normality of this hydrogen peroxide solution is 2N.

Answer: The solution has a volume strength of 5.6 under conditions of 1 atm pressure and 273 K temperature.

The solution has a volume strength of 5.6 under conditions of 1 atm pressure and 273 K temperature, which is a measure of the amount of oxygen released by the hydrogen peroxide solution, making option A the correct answer.

Q4. When the kinetic energy of an electron is quadrupled, how does the wavelength of its associated de-Broglie wave change?

1. Reduced to a quarter
2. Reduced to half
3. Increased fourfold
4. Doubled

Answer: Reduced to half

The de Broglie wavelength lambda = h/p = h/sqrt(2 m KE) is inversely proportional to sqrt(KE). Quadrupling KE multiplies sqrt(KE) by 2, so lambda is reduced to half, not a quarter.

Q5. An electron labeled e₁ is in the fifth energy level, while another electron labeled e₂ is in the fourth energy level. The orbital radius of e₁ is five times that of e₂. What is the ratio of the speed of e₁ (v₁) to the speed of e₂ (v₂)?

1. 5: 1
2. 4: 1
3. 1: 5
4. 1: 4

Answer: 1: 4

From r∝n^2/Z, r1/r2=(25/Z1)(Z2/16)=5 gives Z2/Z1=16/5. From v∝Z/n, v1/v2=(Z1/5)/(Z2/4)=(4/5)(Z1/Z2)=(4/5)(5/16)=1/4. So v1:v2=1:4 (option index 3).

Q6. If the minimum wavelength of the spectral line in the Lyman series for a hydrogen atom is X, what is the maximum wavelength of the spectral line in the Balmer series for a Li²⁺ ion?

1. 9X
2. X/9
3. 5X/4
4. 4X/5

Answer: 4X/5

The maximum wavelength of the spectral line in the Balmer series for a Li²⁺ ion is 4X/5, where X is the minimum wavelength of the spectral line in the Lyman series for a hydrogen atom, due to the specific energy level transitions in hydrogen-like atoms.

Q7. For a sub-energy level with an azimuthal quantum number of 4, what are the highest and lowest possible spin multiplicities?

1. 4, -4
2. 9, 1
3. 10, 1
4. 10, 2

Answer: 10, 1

For l=4 (g subshell) there are 2l+1 = 9 orbitals. Maximum spin multiplicity occurs with 9 unpaired parallel electrons: S=9/2, multiplicity = 2S+1 = 10. Lowest multiplicity (all paired or empty) is 1. So highest and lowest are 10 and 1, not 9 and 1.

Q8. The count of d-electrons in Fe2+ differs from the number of:

1. d-electrons in an Iron atom
2. p-electrons in a Neon atom
3. p-electrons in a Chlorine atom
4. s-electrons in a Magnesium atom

Answer: p-electrons in a Chlorine atom

The number of d-electrons in Fe2+ is 6, which differs from the number of p-electrons in a Chlorine atom, making it the correct comparison.

Q9. Which statement accurately describes the trend in chemical reactivity of alkali metals and halogens as their atomic number increases within the group?

1. Reactivity rises with increasing atomic number for both alkali metals and halogens.
2. Reactivity grows for alkali metals but diminishes for halogens as atomic number increases.
3. Reactivity drops for alkali metals but rises for halogens with increasing atomic number.
4. Reactivity decreases for both alkali metals and halogens as atomic number increases.

Answer: Reactivity grows for alkali metals but diminishes for halogens as atomic number increases.

As atomic number increases within the group, alkali metals' reactivity grows due to lower ionization energy, while halogens' reactivity diminishes due to lower electronegativity.

Q10. The process of converting O(g) into O²⁻(g) requires 603 kJ mol⁻¹ of energy. If the first electron affinity of O(g) is -141 kJ mol⁻¹, what is the value of the second electron affinity of oxygen?

1. 603 kJ mol⁻¹
2. -603 kJ mol⁻¹
3. -744 kJ mol⁻¹
4. +744 kJ mol⁻¹

Answer: +744 kJ mol⁻¹

Forming O2- from O requires the sum of both electron affinities = 603 kJ/mol. With EA1 = -141 kJ/mol, EA2 = 603 - (-141) = +744 kJ/mol. The second electron gain is endothermic (positive) due to electron-electron repulsion adding to the negative O- ion, so the answer is +744 kJ/mol.

Q11. The sequence of increasing van der Waals radii for the elements O, N, Cl, F, and Ne is:

1. F, O, N, Ne, Cl
2. Ne, F, O, N, Cl
3. F, Cl, O, N, Ne
4. N, O, F, Ne, Cl

Answer: F, O, N, Ne, Cl

The van der Waals radii increase in the order F < O < N < Ne < Cl due to the increase in atomic size and electron cloud, making F, O, N, Ne, Cl the correct sequence.

Q12. Which of these sequences correctly represents the trend in electron affinity?

1. O > S > Se
2. Cl > F > Br
3. Cl > Br > I
4. O > C > N

Answer: Cl > F > Br

The trend in electron affinity is influenced by atomic size and nuclear charge, resulting in the correct sequence of Cl > F > Br.

Q13. Which of the following sequences of ionic sizes is accurate?

1. F⁻ is larger than Na⁺, which is larger than Mg²⁺
2. Al³⁺ is bigger than O²⁻, which is bigger than N³⁻
3. P³⁻ is greater in size than S²⁻, which is greater than Cl⁻
4. H⁻ has a larger radius than H⁺, which is larger than He

Answer: P³⁻ is greater in size than S²⁻, which is greater than Cl⁻

The ionic sizes follow the trend P³⁻ > S²⁻ > Cl⁻ due to the increase in nuclear charge and decrease in electron cloud size, making it the accurate sequence.

Q14. As you move across the elements carbon, nitrogen, oxygen, and fluorine, how does electronegativity change?

1. It reduces progressively from carbon to fluorine.
2. It grows steadily from carbon to fluorine.
3. It stays the same throughout the series.
4. It drops from carbon to oxygen and then rises again.

Answer: It grows steadily from carbon to fluorine.

Electronegativity increases steadily from carbon to fluorine due to the increase in nuclear charge and decrease in atomic radius.

Q15. Arrange the following elements in increasing order of their electronegativity.

1. Carbon, Nitrogen, Silicon, Phosphorus
2. Nitrogen, Silicon, Carbon, Phosphorus
3. Silicon, Phosphorus, Carbon, Nitrogen
4. Phosphorus, Silicon, Nitrogen, Carbon

Answer: Silicon, Phosphorus, Carbon, Nitrogen

The electronegativity order is Si < P < C < N based on the positions of the elements in the periodic table and the trends in electronegativity.

Q16. Identify the accurate statement regarding lattice energy:

1. Lattice energy is inversely proportional to the distance separating the ions (1/r₀).
2. When the size of the negative ion increases, the lattice energy of a positive ion also rises.
3. For a fixed negative ion, the lattice energy grows as the size of the positive ion becomes larger.
4. In the case of large positive ions, the lattice energy is primarily influenced by the size of the negative ion.

Answer: Lattice energy is inversely proportional to the distance separating the ions (1/r₀).

Lattice energy is inversely proportional to the distance separating the ions, as described by the Born-Lande equation, making it the accurate statement regarding lattice energy.

Q17. When N₂ is converted to N₂⁺, what happens to the bond dissociation energy of the N–N bond, and when O₂ is converted to O₂⁺, what happens to the bond dissociation energy of the O–O bond?

1. increases in the first case and decreases in the second
2. decreases in the first case and increases in the second
3. decreases in both cases
4. increases in both cases

Answer: decreases in the first case and increases in the second

When N₂ is converted to N₂⁺, the bond dissociation energy of the N–N bond decreases due to the loss of an electron, which reduces the bond order, whereas when O₂ is converted to O₂⁺, the bond dissociation energy of the O–O bond increases because the loss of an electron increases the bond order from 2 to 2.5.

Q18. In which of these ions is pπ–dπ bonding absent?

1. SO₃²⁻
2. PO₄³⁻
3. NO₃⁻
4. XeO₄⁴⁻

Answer: NO₃⁻

p-pi to d-pi back-bonding requires vacant d-orbitals on the central atom. S, P and Xe all have accessible d-orbitals, but nitrogen in NO3- is a second-period element with no d-orbitals, so p-pi-d-pi bonding is absent there. Correct option: NO3-.

Q19. What is the VSEPR formula for the chlorine trifluoride molecule?

1. AX5
2. AX3
3. AX3E2
4. AX5E2

Answer: AX3E2

Cl in ClF3 has 7 valence electrons; 3 are used in Cl-F bonds leaving 2 lone pairs. So the steric arrangement is 3 bonding + 2 lone = AX3E2 (T-shaped). Stored AX5 is wrong.

Q20. Consider the following molecules: CO2, SO2, and H2O. Arrange them in ascending order of their dipole moments.

1. CO2 < SO2 < H2O
2. SO2 < CO2 < H2O
3. H2O < SO2 < CO2
4. H2O < CO2 < SO2

Answer: CO2 < SO2 < H2O

The arrangement of CO₂, SO₂, and H₂O in ascending order of their dipole moments is based on the molecular shapes and the electronegativity differences between the atoms in each molecule, with CO₂ being linear and nonpolar, SO₂ having a bent shape with a dipole moment, and H₂O having a significant dipole moment due to its bent shape and electronegativity difference.

Q21. What form does the van der Waals equation take for a gas where intermolecular attractions are neglected, but the finite volume of molecules is considered?

1. P(V + nb) = nRT
2. P(V - nb) = nRT
3. P(V) = nRT
4. P(V) = nRT - an²/V

Answer: P(V - nb) = nRT

When intermolecular attractions are neglected but the finite volume of molecules is considered, the van der Waals equation reduces to P(V - nb) = nRT, where 'b' accounts for the volume occupied by the gas molecules.

Q22. In which of the following cases can the pH be determined using the equilibrium constant(s) provided?

1. A mixture containing 0.1 M NH4Cl and 0.1 M NH4OH; Kb value for NH4OH
2. A solution with 0.1 M NH4+; Kb of NH3 and Kw
3. A solution containing 0.1 M CH3COOH and 0.2 M CH3COONa; Ka value for CH3COOH
4. A solution with 0.1 M CH3COONa; Ka for CH3COOH and Kw

Answer: A solution containing 0.1 M CH3COOH and 0.2 M CH3COONa; Ka value for CH3COOH

The pH of a solution containing a weak acid and its conjugate base can be determined using the Ka value of the acid, as the solution is a buffer and its pH can be calculated using the Henderson-Hasselbalch equation.

Q23. In the equilibrium reaction NH2COONH4(s) ⇌ 2NH3(g) + CO2(g), which change will cause the partial pressure of NH3 to rise?

1. Adding more solid NH4COONH4 to the system at equilibrium
2. Introducing additional NH3 gas into the system
3. Increasing the concentration of CO2 gas in the system
4. Raising the temperature of the system

Answer: Raising the temperature of the system

Raising the temperature of an equilibrium reaction that is endothermic will cause the reaction to shift towards the products, increasing the partial pressure of NH3.

Q24. For the reaction XY2(g) ⇌ XY(g) + Y(g), if the value of α (degree of dissociation) is very small compared to 1, how is α related to the pressure (P)?

1. proportional to the square root of the volume (V)
2. inversely related to the volume (V)
3. inversely related to the pressure (P)
4. inversely proportional to the square root of the pressure (P)

Answer: inversely proportional to the square root of the pressure (P)

For a reaction with a small degree of dissociation, the relationship between the degree of dissociation and pressure can be derived from the equilibrium expression, showing that α is inversely proportional to the square root of the pressure.

Q25. At 460°C, 0.02 g of hydrogen reacts with 2.54 g of iodine until equilibrium is reached. Upon analysis, the equilibrium mixture contains 0.0021 moles of iodine. What is the equilibrium constant (Kc) for this reaction?

1. 40
2. 128
3. 566
4. 21

Answer: 566

To find the equilibrium constant Kc for the reaction involving hydrogen and iodine, the given masses are converted to moles, and then Kc is calculated based on the equilibrium concentrations, yielding a value of 566.

Q26. When carbonates from the s-block break down, a gas is released that is denser than air and extinguishes a burning splint. Identify the gas.

1. Oxygen (O2)
2. Carbon monoxide (CO)
3. Carbon dioxide (CO2)
4. Tricarbon monoxide (CO3)

Answer: Carbon dioxide (CO2)

When s-block carbonates decompose, they release carbon dioxide, which is denser than air and does not support combustion, thus extinguishing a burning splint.

Q27. Determine the pH of a 1.0 M solution of an unidentified hydroxide compound, assuming the metal ion has a +1 charge.

1. 11
2. 8.0
3. 7.5
4. 13.0

Answer: 13.0

A 1.0 M solution of a monovalent metal hydroxide, being a strong base, completely dissociates in water, resulting in a pH of 13.0.

Q28. What is the classification of Be2C and Al4C3 based on their chemical composition?

1. ethanides
2. methanides
3. carbonides
4. acetylides

Answer: methanides

Be2C and Al4C3 are classified as methanides because they react with water to produce methane, which is a characteristic property of methanides.

Q29. What is formed when an excess amount of sodium hydroxide is added to a solution of stannous chloride?

1. Sn(OH)2
2. SnO2 and H2O
3. Na2SnO2
4. No reaction occurs

Answer: Na2SnO2

SnCl2 first gives Sn(OH)2, which is amphoteric and dissolves in excess NaOH to form sodium stannite, Na2SnO2 (i.e. Na2[Sn(OH)4]). The product is Na2SnO2, not SnO2 + H2O, so option index 2 is correct.

Q30. Which of the following substituted silanes can form a cross-linked silicone polymer upon undergoing hydrolysis?

1. R2Si
2. R2SiCl2
3. RSiCl3
4. R3SiCl

Answer: RSiCl3

The correct option is RSiCl3 because it has three reactive chlorine atoms that can undergo hydrolysis to form a cross-linked silicone polymer, which is a characteristic of this type of silane.