JEE Advanced 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.