Classification of elements and periodicity in properties practice · Mixed difficulty

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 D (All of these) is correct because the size of the atom, the charge on the nucleus, and the screening effect all influence the ionization energy.

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 (As and Sb) is correct because Arsenic (As) and Antimony (Sb) are both metalloids, exhibiting properties of both metals and nonmetals.

Option D (negative and positive) is correct because sulfide sols like CdS are typically negatively charged, while oxide sols like TiO2 are typically positively charged.

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 is correct because anions are larger than their neutral atoms, and among isoelectronic species, the size decreases with increasing nuclear charge.

Option C (1.1) is correct because it is the result of calculating the total ionization energy for all magnesium atoms.

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 B (Both (A) & (R) are true but the (R) is not the correct explanation of the (A)) is correct because while both the assertion and reason are true, the reason doesn't explain why Be and Al show a diagonal relationship.

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

Option B is correct because the period number in the periodic table corresponds to the highest principal quantum number (n) of the elements in that period.

Option A (5) is correct because the ionization energy is proportional to $Z^2$, and the difference between the ionization energies of $Li^{2+}$ and $He^+$ relative to H is 5.

Option A (${Na}$) is correct because, according to Newlands's law of octaves, elements with similar properties are arranged in a sequence where every eighth element resembles the first.

Option B is correct because elements with half-filled or fully-filled electronic configurations have higher ionization enthalpies due to their stability.

Option C (${\mathrm{C}_2}{\mathrm{H}_5}\mathrm{OH}$) is correct because shorter alkyl chains in alcohols lead to stronger hydrogen bonding due to reduced hydrophobic character.

The correct answer is B (Group 13) because there's a large jump in ionization energy between IE3 and IE4, indicating 3 valence electrons.

Option B (Be) is correct because Beryllium has a smaller atomic size and a higher effective nuclear charge compared to the other elements listed, leading to a higher ionization potential.

The correct answer is B (If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion) because the lanthanoid contraction is the primary reason for the similar sizes, not just belonging to the same group.

Option B is correct because He has a larger size than H due to the completely filled 1s² subshell.

Option D ($\mathrm{Br}, \mathrm{I}$) is correct because atomic size increases down the group in the periodic table.

Option C ($MB{r_3}$) is correct because there is a large jump in ionization energy between the 3rd and 4th ionization potentials, indicating that the element readily loses 3 electrons.

Option D (4) is correct because carbon has an electronic configuration of $1s^22s^22p^2$, meaning it has 4 electrons in its outermost (valence) shell.

Option D ($C{l^ - } < {K^ + } < C{a^{2 + }}$) is correct because it is the incorrect order of ionic radii; in isoelectronic species, size decreases with increasing nuclear charge.

Option B (Beryllium and Helium) is correct because they have fully filled electronic configurations, resulting in zero electron affinity.

Option B (${F^ - } < {O^{2 - }} < {N^{3 - }}$) is correct because for isoelectronic species, the ionic size increases with increasing negative charge (decreasing nuclear charge).