Classification of elements and periodicity in properties practice · Medium

Option D (Be) is correct because after removing two electrons, Be has a stable $1s^2$ configuration, requiring a very high energy to remove the third electron.

Option B ($\mathrm{Cu(BO_2)_2}$) is correct because the blue-green color in the oxidizing flame is due to the formation of copper(II) metaborate.

Option C (O < Cl) is correct because chlorine has a more favorable electron gain enthalpy than oxygen due to its larger size and weaker electron-electron repulsion.

Option B ($3s^2 3p^3$) is correct because elements with half-filled or fully-filled electronic configurations have lower electron affinity values due to their stability.

Option D (He) is correct because inert gases have the highest ionization potential due to their stable electronic configurations.

Option B is correct because the first ionization enthalpy generally increases across a period, but there are exceptions due to electron configurations.

Option C is correct. Electron gain enthalpy becomes less negative down a group, but oxygen has a less negative value than sulfur due to its small size.

Option D is correct because Statement I is false; Be has a higher ionization potential than B due to its stable, filled s-orbital configuration, while Statement II is true.

The correct answer is C (${O^ - } \to {O^{2 - }}$) because adding an electron to an already negatively charged ion requires energy due to electron-electron repulsion.

Option D (Cl < P < Mg < Ca) is correct because atomic radius increases down a group and decreases across a period.

Option C (C < F < N < O) is correct because it reflects the increasing trend of second ionization potential considering electronic configurations and stability.

Option A (${N^{ - 3}}$) is correct because, among the given isoelectronic species, the one with the largest negative charge will have the largest size due to increased electron-electron repulsion and decreased effective nuclear charge.

Option B (p-block) is correct because after removing three electrons, the electronic configuration ends with 6s² 5d¹⁰ 4f¹⁴ 6p³, indicating that the element M belongs to the p-block.

Option D (All of these) is correct because all the given statements regarding ionization energy, ionization energies of Be and Mg, and atomic radii of Nb and Ta are factually true.

Option A (Both Assertion and Reason are correct; Reason is the correct explanation of Assertion) is correct because both ions are isoelectronic, and the higher nuclear charge of Al³⁺ results in a smaller ionic radius.

Option A (Both Statement I and Statement II are correct) is correct because $\mathrm{O}^{2-}$ and $\mathrm{F}^{-}$ are isoelectronic, and $\mathrm{O}^{2-}$ has a lower effective nuclear charge, resulting in a larger ionic radius.

Option B (Si) is correct because the large jump in IP5 indicates that the element has 4 valence electrons, which is characteristic of Silicon.

Option A (2.8) is incorrect. Option B (3) is correct because electronegativity is approximately the average of ionization energy and electron affinity, scaled appropriately.

The correct answer is C ($\mid$E.N. of $Y$ - E.N. of ${O}|>|$E.N. of $O$ -E.N. of ${H} \mid$ Acidic) because for a compound to be acidic, the electronegativity difference between Y and O should be less than that between O and H.

Option D ($6^{\text {th }}$) is correct because the 6th period contains 32 elements, which is the maximum number of elements in any period.

Option C (- 5.1 eV) is correct because the electron gain enthalpy of $Na^+$ is the negative of the ionization enthalpy of Na.

The correct answer is Si because there is a significant jump between the 4th and 5th ionization energies, indicating that the 4th electron is being removed from a stable, noble gas-like configuration.

Option C ($Al^{3+} < Mg^{2+} < Na^+ < F^- < O^{2-} < N^{3-}$) is correct because for isoelectronic species, ionic radii decrease with increasing nuclear charge.

Option D (All of the above) is correct because the long form of the periodic table reflects electron filling order, predicts valency, and shows property trends.

Option B is correct because both the assertion and the reason are correct, but the reason is not the correct explanation for the assertion.

Option D ($\mathrm{Na} > \mathrm{Mg} > \mathrm{Be} > \mathrm{Si} > \mathrm{P}$) is correct because metallic character decreases across a period and increases down a group.

Option D (${O_{(g)}} + {e^ - } \to {O_{(g)}^ - }$) is correct because the addition of an electron to an atom generally releases energy, which is electron affinity.

Option C is correct. Orders I and III are incorrect. The correct order for 2nd ionization potential is Mg < Si < Al < P < S, and the correct order for electron affinity is N < C < O < F.

Option D is correct because the standard electrode potential depends on the ionization of a gaseous metal atom, not a solid metal atom.

Option C ($\mathrm{C}{l}^{-} < \mathrm{S}^{2-} < \mathrm{P}^{3-}$) is correct because for isoelectronic species, ionic radii increase with increasing negative charge.