Classification of Elements and Periodicity: JEE Advanced PYQ, 2007–2026

Answer: (A)

In a multi-electron atom, orbitals of the same shell differ in energy because penetration decreases in the order s > p > d > f. Beryllium loses a 2s electron. Boron loses a 2p electron, which sits at higher energy and is partly shielded by the 2s electrons. This outweighs boron's extra proton, so the first ionization enthalpy of B is below that of Be.

Answer: (C)

Electron gain enthalpy is the enthalpy change when a gaseous atom gains an electron: Cl(g) + e⁻ → Cl⁻(g), ΔH = −349 kJ mol⁻¹. The reverse process, Cl⁻(g) → Cl(g) + e⁻, therefore has ΔH = +349 kJ mol⁻¹, so energy is absorbed. Moles of Cl⁻ = 3.55 g ÷ 35.5 g mol⁻¹ = 0.100 mol. Energy absorbed = 0.100 × 349 = 34.9 kJ.

Answer: (D)

The overall process is the sum of two steps. First, Na(g) → Na⁺(g) + e⁻ absorbs +496 kJ/mol. Second, Cl(g) + e⁻ → Cl⁻(g) releases 349 kJ/mol. Adding them gives +496 − 349 = +147 kJ/mol. Forming gaseous ions from gaseous atoms is therefore endothermic, and in solid NaCl the lattice enthalpy supplies the stabilization.

Answer: (A)

Successive ionization enthalpies rise steadily while valence electrons are removed. A sudden, very large jump shows that the next electron comes from a inner shell with a noble-gas-like configuration. Here the ratios are IE₂/IE₁ ≈ 3.1, IE₃/IE₂ ≈ 1.5 and IE₄/IE₃ ≈ 4.2. The largest jump is after IE₃, so X has three valence electrons (Group 13, like Al). It forms X³⁺ and the chloride is XCl₃.

Answer: (D)

In a one-electron species the energy is Eₙ = −(2.18 × 10⁻¹⁸ J) Z²/n². Ionization from the ground state (n = 1) needs the energy of the level with the sign reversed, so it is proportional to Z². For hydrogen Z = 1 gives 1312 kJ mol⁻¹. For Li²⁺, Z = 3, so the value is 3² × 1312 = 9 × 1312 = 11 808 kJ mol⁻¹.

Answer: (A)

The general trend is an increase across a period, but two irregularities occur. (1) Group 2 to group 13: Mg removes a 3s electron, which penetrates well and is held tightly. Al removes a 3p electron, which is shielded by 3s² and is higher in energy. So Al < Mg (578 vs 738 kJ mol⁻¹). (2) Group 15 to group 16: P has the stable half-filled 3p³ configuration. S has 3p⁴, where a paired electron experiences inter-electronic repulsion and is removed more easily. So S < P (about 1000 vs 1012 kJ mol⁻¹). Combining both anomalies gives Al < Mg < S < P. The other options each ignore one or both of these irregularities.

Answer: (C)

The carbon atom's electronegativity depends on its hybridisation. An s orbital lies closer to the nucleus than a p orbital, so the greater the s-character of the hybrid orbital, the more strongly the carbon holds the bonding electrons. Carbon is sp3 (25% s) in ethane, sp2 (33% s) in ethene and sp (50% s) in ethyne. The order of increasing electronegativity is therefore C2H6 < C2H4 < C2H2.

Answer: (A)

Successive ionisation enthalpies rise gradually while valence electrons are removed. They jump sharply when the next electron comes from a stable inner shell. Here the values go 578, 1817, 2745 and then leap to 11577, so the fourth electron is a core electron. X has 3 valence electrons and forms X³⁺ (this is aluminium). The energy to make 1 mol of gaseous X³⁺ is the cumulative sum: 578 + 1817 + 2745 = 5140 kJ. The option 2395 stops at X²⁺. The option 2745 uses only one stage. The option 16717 includes the core-electron removal.

Answer: (A)

Ionization enthalpy depends on the effective nuclear charge and on the distance of the outermost electron. Down a group the added shell normally dominates, so the value falls. From Al to Ga, however, the ten 3d electrons are added. d electrons shield the nucleus poorly, so the effective nuclear charge on the 4p electron of Ga rises more than expected and offsets the larger size. Hence IE₁(Ga) is slightly greater than IE₁(Al).

Answer: (D)

Principle: the ionisation enthalpy is very large when an electron must be taken from a stable noble-gas core. Application: removing one electron from Na gives Na⁺ (2s²2p⁶), so the second electron comes from an inner, stable octet. For the other elements the ion formed is Mg⁺ (3s¹), Al⁺ (3s²) and Si⁺ (3s²3p¹), each with electrons still in the valence shell. The approximate IE₂ values are Na 4562, Mg 1451, Al 1817 and Si 1577 kJ mol⁻¹. Na therefore has by far the highest IE₂. The IE₁ trend across the period cannot be applied to IE₂ directly.