Structure of Atom: JEE Advanced PYQ, 2007–2026

Answer: (C)

The azimuthal quantum number l = 1 identifies p subshells. Write the ground-state configuration of Br (Z = 35): 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁵. Check the total: 2+2+6+2+6+10+2+5 = 35. The p electrons are in 2p (6), 3p (6) and 4p (5), so the total is 6 + 6 + 5 = 17. The value of l depends only on the type of subshell, not on the shell number n, so all p electrons are counted.

Answer: (A)

Ground-state Cr is [Ar]3d⁵4s¹, not 3d⁴4s². By Hund's rule of maximum multiplicity, all six electrons in the 3d and 4s orbitals are unpaired with parallel spins, so they share one mₛ value. The filled subshells 1s², 2s², 2p⁶, 3s², 3p⁶ hold 18 electrons, with 9 of each spin. Electrons with the same mₛ as the 3d electrons number 9 + 5 + 1 = 15.

Answer: (A)

First find the total energy of a hydrogen-like ion: E = −13.6 Z²/n² eV = −13.6 × 4/9 = −6.04 eV. In a Bohr orbit, the Coulomb force supplies the centripetal force, so mv²/r = kZe²/r². This gives KE = ½mv² = kZe²/2r, while PE = −kZe²/r. So PE = −2 KE and E = KE + PE = −KE. Hence KE = +6.04 eV and PE = 2E = −12.09 eV.

Answer: (A)

A stable Bohr orbit holds a whole number of de Broglie wavelengths: 2πr = nλ. With λ = h/mv this gives mvr = nh/2π, which is Bohr's quantisation rule. A circumference of 3λ therefore means n = 3. For a hydrogen-like ion, Eₙ = −13.6 Z²/n² eV. With Z = 2 and n = 3, E = −13.6 × 4/9 ≈ −6.04 eV.

Answer: (B)

For n = 3 the allowed values are l = 0, 1, 2. The value mₗ = +1 requires l ≥ 1, so it occurs in one 3p orbital and one 3d orbital. Each orbital holds a maximum of 2 electrons (mₛ = ±½), so the total is 2 × 2 = 4.

Answer: (A)

Principle: momentum is conserved, and a photon carries momentum p = h/λ. The atom starts at rest, so after emission it recoils with equal and opposite momentum. Step 1: p = 6.626×10⁻³⁴ / 656×10⁻⁹ = 1.01×10⁻²⁷ kg m s⁻¹. Step 2: v = p/m = 1.01×10⁻²⁷ / 1.674×10⁻²⁷ ≈ 0.60 m s⁻¹. Do not equate the photon energy to the atom's kinetic energy, because that energy comes from the atom's internal transition.

Answer: (D)

The (n + l) rule says that the orbital with the lower n + l value has lower energy. The values are 4d: 4 + 2 = 6, 5p: 5 + 1 = 6, 5s: 5 + 0 = 5 and 4f: 4 + 3 = 7. So 5s is lowest. This is also why 5s is filled before 4d.

Answer: (A)

The ground-state ionisation energy of a hydrogen-like species is 13.6 Z² eV. Setting 13.6 Z² = 122.4 gives Z² = 9, so Z = 3 (Li²⁺). The orbit radius is rₙ = 0.529 × n²/Z Å. For n = 2 and Z = 3, r = 0.529 × 4/3 ≈ 0.705 Å.

Answer: (B)

By Pauli's exclusion principle, each orbital holds at most two electrons, and they must have opposite spins. So each orbital can contain at most one electron with m_s = −1/2. The number of orbitals in the n = 4 shell is n² = 16 (4s: 1, 4p: 3, 4d: 5, 4f: 7). The maximum number of electrons with n = 4 and m_s = −1/2 is therefore 16.

Answer: (D)

Each d orbital has l = 2, so two angular nodes, but their form differs. d_xy, d_yz and d_xz have lobes between the axes, and their nodal planes are the coordinate planes. d_xy vanishes on the xz and yz planes. d_(x²−y²) vanishes where x² − y² = 0, which gives the two planes x = ±y that bisect the axes. d_(z²) has two conical nodal surfaces, not planes. Only the statement on d_(x²−y²) is correct.