Classification of elements and periodicity in properties practice · Mixed difficulty

Option C ($\begin{array}{llllllllllllllllllll}{1.71,\mathrm{ }1.40,\mathrm{ }1.36}\end{array}$) is correct because as the negative charge increases, the ionic radii increase due to greater electron repulsion, and all are isoelectronic.

Option D (${\mathrm{O}^{2 - }} > {\mathrm{F}^ - } > O > F$) is correct because anions are larger than their neutral atoms, and for isoelectronic species, the species with lower nuclear charge has a larger radius.

Option D is correct. Pitchblende is a uranium-rich ore that contains Actinium and Protoactinium, as well as Neptunium and Plutonium in trace amounts due to the radioactive decay of Uranium.

Option C ($\mathrm{O} > F > N > C$) is correct because after the first ionization, removing an electron from N+ gives it a stable half-filled configuration, making it easier than removing from F+.

Option C (-284 kJ/mol) is correct because it correctly sums the electron affinities for the two oxygen atoms.

Option C is correct because as you move down a group, the nuclear charge increases, but the shielding effect also increases, leading to a weaker effective nuclear charge and easier loss of electrons, hence increasing metallic nature.

Option D (None is incorrect) is correct because all the given orders are correct regarding the size of ions/atoms.

Option A ($1{s^2},2{s^2},2{p^5}$) is correct because this electronic configuration corresponds to a halogen, which are the most electronegative elements.

Option B ($\mathrm{Be} \to \mathrm{B}{\mathrm{e}^{2 + }}$) is correct because removing two electrons from Be requires significantly more energy due to its small size and relatively high nuclear charge compared to the other options.

Option C ([Ne] 3s² 3p⁵) is correct because it represents a halogen, while the others represent elements in the same group (group 12).

Option D (P) is correct because Phosphorus (P) exists as a solid at 25°C and 1 atm pressure.

Option A is incorrect because ionic radii decrease with increasing nuclear charge for isoelectronic species. The correct order should be Sc3+ > Mn3+ > Fe3+ > Cr3+.

Option A (Li) is correct because ionization energy decreases down a group, and Li is the first element in the group.

Option B is correct because noble gases have 8 outermost electrons (except for Helium which has 2), while their group number is 18 (or 8A), thus having a different number of outermost electrons than their group number.

Option A ($54\%$) is correct. The percentage of ionic character can be estimated using the Hannay-Smith equation: % ionic character = 16 * (electronegativity difference) + 3.5 * (electronegativity difference)².

Option C (Lanthanides) is correct because elements with atomic numbers 58 to 71 are defined as Lanthanides.

Option B ($51.40 \mathrm{~kJ}$) is correct because it involves calculating the total ionization energy for converting all Al atoms to Al+3 in the given mass.