50 questions on this topic of Classification of Elements and Periodicity in Properties (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.
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Answer: (B) Decreases
Across a period the nuclear charge increases while electrons enter the same shell, pulling them closer, so atomic radius decreases.
NCERT Class 11
Answer: (A) Both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I
Phosphorus has a stable half-filled 3p³ subshell ([Ne] 3s² 3p³), whereas sulfur has configuration [Ne] 3s² 3p⁴. In sulfur, two electrons share the same 3p orbital, leading to increased inter-electronic repulsion, which makes it easier to remove one electron compared to phosphorus. Hence both statements are true and Statement II explains Statement I.
Answer: (B) B > Tl > Ga > Al > In
Due to the intervention of 3d and 4f electrons which provide poor shielding effect, the effective nuclear charge increases unexpectedly down the group. As a result, the correct decreasing order of first ionisation enthalpy in group 13 is B (801 kJ/mol) > Tl (589 kJ/mol) > Ga (579 kJ/mol) > Al (577 kJ/mol) > In (558 kJ/mol).
Answer: (B) A-III, B-II, C-IV, D-I
Ionisation of Na⁺(g) represents IE₂ of Na (A-III). The removal of an electron from Cl⁻(g) is the reverse of electron gain enthalpy, hence equal to -Δ(eg)H (B-II). Adding an electron to Cl(g) represents its electron gain enthalpy (C-IV). Converting neutral Na(g) to Na⁺(g) is its IE₁ (D-I).
Answer: (D) Ca²⁺ < K⁺ < Cl⁻ < S²⁻
All given species are isoelectronic having 18 electrons. For isoelectronic species, ionic radius decreases as the positive nuclear charge (atomic number Z) increases: Ca²⁺ (Z=20) < K⁺ (Z=19) < Cl⁻ (Z=17) < S²⁻ (Z=16).
Answer: (A) Te²⁻ > I⁻ > Se²⁻ > Br⁻
Te²⁻ and I⁻ have 54 electrons (isoelectronic with Xe); since Te has Z = 52 and I has Z = 53, Te²⁻ is larger than I⁻. Similarly, Se²⁻ and Br⁻ have 36 electrons (isoelectronic with Kr); Se²⁻ (Z = 34) is larger than Br⁻ (Z = 35). As principal quantum number increases down a group, the 5th shell species (Te²⁻, I⁻) are larger than the 4th shell species (Se²⁻, Br⁻). Thus, Te²⁻ > I⁻ > Se²⁻ > Br⁻.
Answer: (C) XCl₃
There is a very large jump in ionization enthalpy between IE₃ (2744 kJ/mol) and IE₄ (11577 kJ/mol). This demonstrates that the element has three valence electrons and attains a stable noble gas configuration after losing 3 electrons. Therefore, the element forms a stable trivalent ion X³⁺, resulting in the chloride XCl₃.
Answer: (A) Both Statement I and Statement II are correct
Mg²⁺ and Al³⁺ both possess 10 electrons and are isoelectronic. Al³⁺ has 13 protons while Mg²⁺ has only 12 protons, so the nuclear pull on the remaining electrons is greater in Al³⁺, making its ionic radius smaller than that of Mg²⁺. Thus, both statements are correct.
Answer: (B) Statement I is correct but Statement II is incorrect
Statement I is correct because ΔᵢH₁ of Ga (579 kJ/mol) is marginally higher than that of Al (577 kJ/mol). Statement II is incorrect because 3d-electrons have poor shielding power (diffuse shape), causing an increased effective nuclear charge on outer electrons.
Answer: (B) X = -Y
The first process represents the first ionisation enthalpy of sodium (endothermic, ΔH = X > 0). The second process is the exact reverse of the first process (gain of an electron by Na⁺ ion), so by the law of conservation of energy and Hess's law, Y = -X or X = -Y.
Answer: (A) Increases left to right
Nuclear charge increases → stronger pull on electrons → higher EN.
Quick tip: Period: EN left→right increases. Group: EN top→bottom decreases. Most EN: F.
NCERT Class 11, Chapter 3, page 86
Answer: (D) F(g) + e⁻ → F⁻(g)
The addition of an electron to a neutral halogen atom like fluorine releases energy, giving a negative electron gain enthalpy (ΔegH = -328 kJ/mol). For O⁻, adding a second electron faces strong electrostatic repulsion (ΔegH > 0). Be has a fully filled 2s² configuration and N has a half-filled 2p³ configuration, giving both slightly positive or unfavorable electron gain enthalpies.