Classification of elements and periodicity in properties practice · Medium

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.

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 ($C{l^ - } < {K^ + } < C{a^{2 + }}$) 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).

Option D (0.736) is correct because the percentage of ionic character is estimated based on the electronegativity difference between the two atoms.

Option D ($B > Al < Ga < In < Tl$) is correct because ionization energy generally decreases down a group, but there are exceptions due to the filling of d and f orbitals.

Option B (Group 1) is correct because there is a large jump in ionization energy after the removal of one electron, indicating that the element has one valence electron.

Option A ($K < Cu < C{u^ + } < {K^ + }$) is correct because ionization energy increases as electrons are removed and noble gas configurations are more stable.

The correct answer is C because alkali metals have a significantly lower first ionization energy compared to the second due to the stable noble gas configuration after losing one electron.