Classification of elements and periodicity in properties practice · Mixed difficulty

Option B ($1.12\,\,\,\mathop {\text{A}}\limits^{\text{o}}$) is the most appropriate as the radius of Ne should fall between the radii of $F^-$ and $Na^+$, considering the trend of ionic radii.

Option D (+2) is correct because the large jump in ionization energy after the second ionization indicates a stable +2 oxidation state.

Option C (Cation) is correct because when an atom loses an electron, it becomes positively charged, forming a cation.

Option A (${Li}$) is correct because after losing one electron, it attains a stable noble gas configuration, requiring significantly more energy to remove another electron.

Option D ($N, Ne$) is correct because Neon (Ne) has Van der Waals radius, which is larger than the covalent radius of Nitrogen (N) in the same period.

Option D is correct because, according to Bohr's theory, the Coulombic force of attraction on the electron is proportional to $\frac{Z^2}{n^4}$.

Option A ($\mathrm{d^5, d^3, d^1, d^4}-$ increasing magnetic moment) is correct because the correct order for increasing magnetic moment is $\mathrm{d^1, d^3, d^4, d^5}$.

Option D (All of the above) is correct because the size of the cation is smaller, it's more difficult to remove an electron, and the effective nuclear charge is greater.

Option B (O) is correct because oxygen has the highest electronegativity among the given elements due to its small atomic size and high effective nuclear charge.

Option A is correct. As we go down a group, the atomic size increases, and the added electron experiences less attraction from the nucleus due to increased shielding, making the electron gain enthalpy less negative.

Formic acid (HCOOH) reduces Tollen's reagent due to the presence of an aldehyde group. Therefore, the statement that it does not reduce Tollen's reagent is incorrect, making option D the correct answer.

Option C (${F}^{-} < {O}^{2-} < {N}^{3-} < {C}^{4-}$) is correct because for isoelectronic species, ionic radii increase with increasing negative charge.

Option A (3) is correct because there is a significant jump in ionization enthalpy after the third ionization, indicating that the element has 3 valence electrons.

Option C (d-block) is correct because if the Aufbau principle is not followed, the 19th electron would enter the 3d subshell instead of the 4s subshell.

Option B (Cl) is correct because Chlorine has the highest electron affinity due to its electronic configuration and effective nuclear charge.

Option A ($s>p>d>f$) is correct because the screening effect decreases as the orbital shape becomes more diffused, leading to a weaker shielding of the nuclear charge.

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 C (b > c > d > a) is correct because ionization potential generally increases across a period and decreases down a group. Noble gases have the highest ionization potential, followed by elements with stable electronic configurations.

Option A is correct because the discovery of inert gases disrupted Newland's law of octaves, as they introduced elements that didn't fit the pattern.

Option B (Bond energies) is correct because Pauling's electronegativity scale is based on experimental bond energies.

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.

The correct answer is Lithium is amphoteric because Lithium is basic, not amphoteric. Amphoteric substances can act as both acid and base.

Option D is correct because both Al₂O₃ and As₂O₃ can act as both acids and bases, thus exhibiting amphoteric behavior.

Option C (Ti < Zr=Hf) is correct because atomic radii increase down the group, but Zr and Hf have similar radii due to lanthanide contraction.

Option A (Many elements has several isotopes) is correct because the existence of isotopes with different atomic masses challenged Mendeleev's periodic law, which was based on atomic weights.

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 B (Rb) is correct because Rubidium (Rb) is the first element in the fifth period of the periodic table.